Explorations of edge-weighted Cayley graphs and p-ary bent functions
Charles Celerier, David Joyner, Caroline Melles, David Phillips, Steven Walsh
EExplorations of edge-weighted Cayley graphsand p -ary bent functions Charles Celerier, David Joyner ∗ , Caroline Melles,David Phillips, Steven WalshJune 18, 2018 Abstract
Let f : GF ( p ) n → GF ( p ). When p = 2, Bernasconi et al haveshown that there is a correspondence between certain properties of f (eg, if it is bent) and properties of its associated Cayley graph. Anal-ogously, but much earlier, Dillon showed that f is bent if and onlyif the “level curves” of f had certain combinatorial properties (again,only when p = 2). The attempt is to investigate an analogous the-ory when p > f can be characterizedin terms of function-theoretic properties of f , and which function-theoretic properties of f correspond to combinatorial properties of theset of “level curves” f − ( a ) ( a ∈ GF ( p )). While the natural general-izations of the Bernasconi correspondence and Dillon correspondenceare not true in general, using extensive computations, we are able todetermine a classification in small cases: ( p, n ) ∈ { (3 , , (3 , , (5 , } .Our main conjecture is Conjecture 67. Contents ∗ USNA, Mathematics Department; email: [email protected] a r X i v : . [ m a t h . C O ] J un Partial difference sets 7 GF (3) → GF (3) . . . . . . . . . . . . . . . . . . . . . . . . . 485.2 GF (3) → GF (3) . . . . . . . . . . . . . . . . . . . . . . . . . 555.3 GF (5) → GF (5) . . . . . . . . . . . . . . . . . . . . . . . . . 63 GF (3) → GF (3) . . . . . . . . . . . . . . . . 766.3 Bent functions GF (3) → GF (3) . . . . . . . . . . . . . . . . 836.4 Bent functions GF (5) → GF (5) . . . . . . . . . . . . . . . . 85 Fix n ≥ V = GF ( p ) n , where p is a prime. Definition 1. ( Walsh(-Hadamard) transform ) The
Walsh(-Hadamard)transform of a function f : GF ( p ) n → GF ( p ) is a complex-valued functionon V that can be defined as W f ( u ) = (cid:88) x ∈ V ζ f ( x ) −(cid:104) u,x (cid:105) , (1)where ζ = e πi/p .We call f bent if | W f ( u ) | = p n/ , u ∈ V . The class of p -ary bent functions are “maximally non-linear”in some sense, and can be used to generate pseudo-random sequences rathereasily.Some properties of the Walsh transform:1. The Walsh coefficients satisfy Parseval’s equation (cid:88) u ∈ V | W f ( u ) | = p n .
2. If σ = σ k : Q ( ζ ) → Q ( ζ ) is defined by sending ζ (cid:55)→ ζ k then W f ( u ) σ = W kf ( ku ).If f : V → GF ( p ) then we let f C : V → C be the function whose valuesare those of f but regarded as integers (i.e., we select the congruence classresidue representative in the interval { , . . . , p − } ). Definition 2. ( Fourier transform ) When f is complex-valued, we definethe analogous Fourier transform of the function f asˆ f ( y ) = f ∧ ( y ) = (cid:88) x ∈ V f C ( x ) ζ −(cid:104) x,y (cid:105) . (2)Note ˆ f (0) = (cid:88) x ∈ V f C ( x ) , and note W f ( y ) = ( ζ f ) ∧ ( y ) . We say f is even , if f ( − x ) = f ( x ) for all x ∈ GF ( p ) n . It is not hard to seethat if f is even then the Fourier transform of f is real-valued. (However,this is not necessarily true of the Walsh transform.) Example 3.
It turns out that there are a total of 3 = 81 even functions f : GF (3) → GF (3) with f (0) = 0, of which exactly 18 are bent. Section6.2 discusses this in more detail. 3 xample 4. It turns out that there are a total of 3 = 1594323 evenfunctions f : GF (3) → GF (3) with f (0) = 0, of which exactly 2340 arebent. Section 6.3 discusses this in more detail. Example 5.
It turns out that there are a total of 5 = 244140625 evenfunctions f : GF (5) → GF (5) with f (0) = 0, of which exactly 1420 arebent. Section 6.4 discusses this in more detail. Definition 6. ( Hadamard matrix ) We call an N × N { , } -matrix M a Hadamard matrix if M · M t = N I N , where I N is the N × N identity matrix. Remark 7.
There is a concept similar to the notion of bent, called “CAZAC.”We shall clarify their connection in this remark.Constant amplitude zero autocorrelation (CAZAC) functions have beenstudied intensively since the 1990’s [BD]. We quote the following character-ization, which is due to J. Benedetto and S. Datta [BD]:Theorem: Given a sequence x : Z /N Z → C , and let C x be a circulant matrixwith first row x = ( x [0] , x [1] , . . . , x [ N − . Then x is a CAZAC sequence ifand only if C x is a Hadamard matrix.This is due to the fact that the definition of CAZAC functions uses theFourier transform on Z /N Z . The corresponding definition of bent functions f uses the Fourier transform (of ζ f ) on GF ( p ) n , N = p n . The analogousHadamard matrix for a bent function f is not circulant but “block circulant.” In the Boolean case, there is a nice simple relationship between the Fouriertransform and the Walsh-Hadamard transform. In equation (24) below, weshall try to connect these two transforms, (1) and (2), in the GF ( p ) case aswell. In this context, it is worth noting that it is possible (see Proposition73) to characterize a bent function in terms of the Fourier transform of itsderivative.Suppose f : GF ( p ) n → GF ( p ) is bent. Definition 8. ( regular ) Suppose f is bent. We say f is regular if and onlyif W f ( u ) /p n/ is a p th root of unity for all u ∈ V .4f f is regular then there is a function f ∗ : GF ( p ) n → GF ( p ), called the dual (or regular dual ) of f , such that W f ( u ) = ζ f ∗ ( u ) p n/ , for all u ∈ V . Wecall f weakly regular , if there is a function f ∗ : GF ( p ) n → GF ( p ), calledthe dual (or µ -regular dual ) of f , such that W f ( u ) = µζ f ∗ ( u ) p n/ , for someconstant µ ∈ C with absolute value 1. Proposition 9. (Kumar, Scholtz, Welch) If f is bent then there are func-tions f ∗ : GF ( p ) n → Z and f ∗ : GF ( p ) n → GF ( p ) such that W f ( u ) p − n/ = (cid:26) ( − f ∗ ( u ) ζ f ∗ ( u ) , if n is even , or n is odd and p ≡ ,i f ∗ ( u ) ζ f ∗ ( u ) , if n is odd and p ≡ . The above result is known (thanks to Kumar, Scholtz, Welch [KSW])but the form above is due to Helleseth and Kholosha [HK3] (although wemade a minor correction to their statement). Also, note [KSW] Property 8established a more general fact than the statement above.
Corollary 10. If f is bent and W f (0) is rational (i.e., belongs to Q ) then n must be even. The condition W f (0) ∈ Q arises in Lemma 56 below, so this corollaryshall be useful later.Suppose f : V = GF ( p ) n → GF ( p ) is bent. In this case, for each u ∈ V ,the quotient W f ( u ) /p n/ is an element of the cyclotomic field Q ( ζ ) havingabsolute value 1.Below we give a simple necessary and sufficient conditions to determineif f is regular. The next three lemmas are well-known but included for thereader’s convenience. Lemma 11.
Suppose f : V → GF ( p ) is bent. The following are equivalent. • f is weakly regular. • W f ( u ) /W f (0) is a p -th root of unity for all u ∈ V .Proof. If f is weakly regular with µ -regular dual f ∗ , then W f ( u ) /W f (0) = ζ f ∗ ( u ) − f ∗ (0) , for each u ∈ V .Conversely, if W f ( u ) /W f (0) is of the form ζ i u , for some integer i u (for u ∈ V ), then let f ∗ ( u ) be i u (mod p ) and let µ = W f (0) / ( p n/ ). Then f ∗ ( u )is a µ -regular dual of f . (cid:3) If µ is fixed and we want to be more precise, we call this µ -regular . emma 12. Suppose f : V → GF ( p ) is bent and weakly regular. Thefollowing are equivalent. • f is regular. • W f (0) /p n/ is a p -th root of unity.Proof. One direction is clear. Suppose that f is a weakly regular bent func-tion with µ -regular dual f ∗ and suppose that W f (0) / ( p n/ ) = ζ i . Note that W f (0) = µζ f ∗ (0) p n/ = ζ i p n/ so that µ = ζ i − f ∗ (0) . Let g ( u ) = f ∗ ( u ) − f ∗ (0)+ i (where we are treating i as an element of GF ( p )). Then W f ( u ) = µζ f ∗ ( u ) p n/ = ζ i − f ∗ (0) ζ g ( u )+ f ∗ (0) − i p n / ζ g ( u ) p n/ , so f is regular. (cid:3) Lemma 13.
Suppose that f is bent and weakly regular, with µ -regular dual f ∗ . Then f ∗ is bent and weakly regular, with µ − -regular dual f ∗∗ given by f ∗∗ ( x ) = f ( − x ) . If f is also even, then f ∗ is even and f ∗∗ = f .Proof. Suppose that f is bent and weakly regular with µ -regular dual f ∗ .Then W f ( u ) = µζ f ∗ ( u ) p n for all u in V . The Walsh transform of f ∗ is given by W f ∗ ( u ) = (cid:88) y ∈ V ζ f ∗ ( y ) ζ −(cid:104) u,y (cid:105) = (cid:88) y ∈ V µ − p − n W f ( y ) ζ −(cid:104) u,y (cid:105) = µ − p − n (cid:88) y ∈ V (cid:88) x ∈ V ζ f ( x ) ζ −(cid:104) y,x (cid:105) ζ −(cid:104) u,y (cid:105) = µ − p − n (cid:88) y ∈ V (cid:88) x ∈ V ζ f ( x ) ζ −(cid:104) y,x + u (cid:105) = µ − p − n (cid:88) w ∈ V ζ f ( w − u ) (cid:88) y ∈ V ζ −(cid:104) y,w (cid:105) . (3)Next we note that (cid:88) y ∈ V ζ −(cid:104) y,w (cid:105) = (cid:40) p n if w = 00 if w (cid:54) = 06ince, if y = ( y , y , . . . , y n ) and w = ( w , w , . . . , w n ), we have (cid:88) y ∈ V ζ −(cid:104) y,w (cid:105) = (cid:88) y ∈ GF ( p ) (cid:88) y ∈ GF ( p ) . . . (cid:88) y n ∈ GF ( p ) ζ − y w ζ − y w . . . ζ − y n w n = n (cid:89) i =1 (cid:88) y ∈ GF ( p ) ζ − yw i and, if w i (cid:54) = 0, (cid:88) y ∈ GF ( p ) ζ − yw i = ζ + ζ + . . . + ζ p − = 0 . Therefore equation 3 reduces to W f ∗ ( u ) = µ − p − n ζ f ( − u ) p n = µ − ζ f ( − u ) p n . It follows that f ∗ is bent with µ − -regular dual f ∗∗ given by f ∗∗ ( x ) = f ( − x )and that if f is even, f ∗∗ = f .Furthermore, if f is even, ζ f ∗ ( − u ) = µ − p − n W f ( − u )= µ − p − n (cid:88) x ∈ V ζ f ( x ) ζ −(cid:104)− u,x (cid:105) = µ − p − n (cid:88) w ∈ V ζ f ( − w ) ζ −(cid:104) u,w (cid:105) = µ − p − n (cid:88) w ∈ V ζ f ( w ) ζ −(cid:104) u,w (cid:105) since f is even= µ − p − n W f ( u )= ζ f ∗ ( u ) . Since f ∗ takes values in GF ( p ), it follows that f ∗ ( − u ) = f ∗ ( u ) for all u in V , so f ∗ is even. (cid:3) Dillon’s thesis [D] was one of the first publications to discuss the relationshipbetween bent functions and combinatorial structures, such as difference sets.His work concentrated on the Boolean case. Consider functions7 : GF ( p ) n → GF ( p ) , where p is a prime, n > f − (1) gives rise to a difference set in GF (2) n . Definition 14. ( difference set ) Let G be a finite abelian multiplicativegroup of order v , and let D be a subset of G with order k . D is a ( v, k, λ ) -difference set (DS) if the list of differences d d − , d , d ∈ D , represents everynon-identity element in G exactly λ times.A Hadamard difference set is one whose parameters are of the form(4 m , m − m, m − m ), for some m >
1. It is, in addition, elementary if G is an elementary abelian 2-group (i.e., isomorphic to ( Z / Z ) n ).Let D − = { d − | d ∈ D } . Lemma 15.
Let G be a finite abelian multiplicative group of order v andlet D be a subset of G with order k , such that D = D − . If ( G, D ) is an ( v, k, λ, λ ) -partial difference set then it is also a ( v, k, λ ) -difference set.Proof. This follows from character theory: combine Theorem 1 and Theorem2, and (the proof of) Proposition 1 in Polhill [Po]. (cid:3)
Theorem 16. (Dillon Correspondence, [D], Theorem 6.2.10, page 78) Thefunction f : GF (2) n → GF (2) is bent if and only if f − (1) is an elementaryHadamard difference set of GF (2) n . Two (naive) analogs of this are formalized below in Analog 34 and Analog35.
In this paper, we consider the “level curves” f − ( a ) ⊂ GF ( p ) n ( a ∈ GF ( p ), a (cid:54) = 0) and investigate the combinatorial structure of these sets, especiallywhen f is bent. Definition 17. ( PDS ) Let G be a finite abelian multiplicative group of order v , and let D be a subset of G with order k . D is a ( v, k, λ, µ ) -partial differenceset (PDS) if the list of differences d d − , d , d ∈ D , represents every non-identity element in D exactly λ times and every non-identity element in G \ D exactly µ times. 8his notion can be characterized algebraically in terms of the group ring C [ G ]. Lemma 18.
With the notation as in the defintion above, ( G, D ) forms a ( v, k, λ, µ ) -PDS if and only if (6) holds. The well-known proof is omitted.
Example 19.
Consider the finite field GF (9) = GF (3)[ x ] / ( x + 1) = { , , , x, x + 1 , x + 2 , x, x + 1 , x + 2 } , written additively. The set of non-zero quadratic residues is given by D = { , , x, x } . One can show that D is a PDS with parameters v = 9 , k = 4 , λ = 1 , µ = 2 . We shall return to this example (with more details) below, in Example26.
Definition 20. ( Latin square type PDS ) Let (
G, D ) be a PDS. We say itis of
Latin square type (resp., negative Latin square type ) if there exist
N >
R >
N <
R <
0) such that( v, k, λ, µ ) = ( N , R ( N − , N + R − R, R − R ) . The example above is of Latin square type ( N = 3 and R = 2) and ofnegative Latin square type ( N = − R = − G be a finite abelian multiplicative group and let D be a subset of G . Decompose D into a union of disjoint subsets D = D ∪ · · · ∪ D r , (4)and assume 1 / ∈ D . Let k i = | D i | . Definition 21. ( weighted PDS ) Let W be a weight set of size r , and v ∈ Z , k ∈ Z | W | , λ ∈ Z | W | , and µ ∈ Z | W | . We say D is a weighted ( v, k, λ, µ ) -PDS ,if the following properties hold 9 The list of “differences” D i D − j = { d d − | d ∈ D i , d ∈ D j } , represents every non-identity element of D (cid:96) exactly λ i,j,(cid:96) times andevery non-identity element of G \ D exactly µ i,j times (1 ≤ i, j, (cid:96) ≤ r ). • For each i there is a j such that D − i = D j (and if D − i = D i for all i then we say the weighted PDS is symmetric ).This notion can be characterized algebraically in terms of the group ring C [ G ]. Lemma 22.
With the notation as in the definition above, ( G, D , . . . , D s ) forms a symmetric weighted ( v, k, λ, µ ) -PDS if and only if D = D − and (21)holds. The straightforward proof is omitted.
Remark 23. If D = D ∪ · · · ∪ D r is a symmetric weighted PDS then µ i,j = µ j,i and λ i,j,(cid:96) = λ j,i,(cid:96) . How does the above notion of a weighted PDS relate to the usual notionof a PDS?
Lemma 24.
Let ( G, D ) , where D = D ∪ · · · ∪ D r (disjoint union) is as in(4), be a symmetric weighted PDS, with parameters ( v, ( k i ) , ( λ i,j,(cid:96) ) , ( µ i,j )) . If (cid:88) i,j λ i,j,(cid:96) does not depend on (cid:96) , ≤ (cid:96) ≤ r , then D is also an unweighted PDS withparameters ( v, k, λ, µ ) where k = (cid:88) i k i , λ = (cid:88) i,j λ i,j,(cid:96) , µ = (cid:88) i,j µ i,j . Remark 25.
Case 1 in Proposition 91 does not satisfy this hypothesis. roof. The claim is that (
G, D ) is a PDS with parameters ( v, k, λ, µ ). Since v = | G | and k = | D | = | D | + · · · + | D r | = k + · · · + k r , we need only verify the claim regarding λ and µ .Does each element d of D occur the same number of times in the list DD − ? Suppose d ∈ D (cid:96) , where 1 ≤ (cid:96) ≤ r . By hypothesis, d occurs in D i − D j exactly λ i,j,(cid:96) times. Since DD − is the concatenation of the D i D − j ,for 1 ≤ i, j ≤ r , d occurs in D · D − exactly (cid:88) i,j λ i,j,(cid:96) times. By hypothesis, this does not depend on (cid:96) , so the claim regarding λ has been verified.Does each non-zero element d of G \ D occur the same number of timesin the list D · D − ? By hypothesis, d occurs in D i D − j exactly µ i,j times.Since D · D − is the concatenation of the D i D − j , for 1 ≤ i, j ≤ r , d occursin D · D − exactly (cid:88) i,j µ i,j times. This verifies the claim regarding µ and completes the proof of thelemma. (cid:3) Example 26.
Consider the finite field GF (9) = GF (3)[ x ] / ( x + 1) = { , , , x, x + 1 , x + 2 , x, x + 1 , x + 2 } , written multiplicatively The set of non-zero quadratic residues is given by D = { , , x, x } . Let D = { , } and D = { x, x } .Translating the multiplicative notation to the additive notation, we find D D − = [ d − d | d ∈ D , d ∈ D ] = [0 , , , , D − = [ d − d | d ∈ D , d ∈ D ] = [ x + 1 , x + 2 , x + 1 , x + 2] ,D D − = [ d − d | d ∈ D , d ∈ D ] = [ x + 1 , x + 2 , x + 1 , x + 2] ,D D − = [ d − d | d ∈ D , d ∈ D ] = [0 , , x, x ] . Therefore, this describes a weighted PDS with parameters k , = 2 , k , = 2 , k , = k , = 0 ,λ , , = 1 , λ , , = 0 , λ , , = 0 , λ , , = 0 ,λ , , = 0 , λ , , = 0 , λ , , = 0 , λ , , = 1 , and µ , = 0 , µ , = 1 , µ , = 1 , µ , = 0 . As we will see, there a weighted analog of the correspondence betweenPDSs and SRGs.We have the following generalization of Theorem 42.
Theorem 27.
Let G be an abelian multiplicative group and let D ⊂ G be asubset such that / ∈ D and with disjoint decomposition D = D ∪ D ∪· · ·∪ D r .The following are equivalent:(a) ( G, D ) is a symmetric weighted partial difference set having parameters ( v, k, λ, µ ) , where v = | G | , k = { k i } with k i = | D i | , λ = { λ i,j,(cid:96) } , and µ = { µ i,j } .(b) Γ( G, D ) is a strongly regular edge-weighted (undirected) graph with pa-rameters ( v, k, λ, µ ) as in (a).Proof. Let D (cid:48) = G \ D − { } .((a) = ⇒ (b)) Suppose ( G, D ) is a weighted partial difference set sat-isfying D i = D − i , for all i . The graph Γ = Γ( G, D ) has v = | G | vertices,by definition. Each vertex g of Γ has k i neighbors of weight i , namely, dg where d ∈ D i . (We say two vertices are “neighbors having edge-weight 0”if they are not connected by an edge in the unweighted graph.) Let g , g be distinct vertices in Γ. Let x be a vertex which is a neighbor of each: x ∈ N ( g , i ) ∩ N ( g , j ). By definition, x = d g = d g , for some d ∈ D i , d ∈ D j . Therefore, d − d = g g − . If g g − ∈ D (cid:96) , for some (cid:96) (cid:54) = 0, then thereare λ i,j,(cid:96) solutions, by definition of a weighted PDS. If g g − ∈ D (cid:48) then thereare µ i,j solutions, by definition of a weighted PDS.12(b) = ⇒ (a)) Note f even implies the symmetric condition of a weightedPDS. For the remainder of the proof, note the reasoning above is reversible.Details are left to the reader. (cid:3) Let f be a GF ( p )-valued function on V . The Cayley graph of f is definedto be the edge-weighted digraphΓ f = ( GF ( p ) n , E f ) , (5)whose vertex set is V = V (Γ f ) = GF ( p ) n and the set of edges is defined by E f = { ( u, v ) ∈ GF ( p ) n | f ( u − v ) (cid:54) = 0 } , where the edge ( u, v ) ∈ E f has weight f ( u − v ). However, if f is even thenwe can (and do) regard Γ f as a weighted (undirected) graph. Theorem 28.
Let f : GF ( p ) n → GF ( p ) be an even function such that f (0) = 0 . Let D i = f − ( i ) , for i = 1 , , . . . , p − , and D = { } . Let D p = GF ( p ) n \ D ∪ D ∪ · · · ∪ D p − . If ( GF ( p ) n , D , . . . , D p ) is a weighted partialdifference set, where G = GF ( p ) n , then the associated (strongly regular)graph is the (edge-weighted) Cayley graph of f . Remark 29.
Roughly speaking, this theorem says that “if the level curves of f form a weighted PDS then the (edge-weighted) Cayley graph correspondingto f agrees with the (edge-weighted) strongly regular graph associated to theweighted PDS.”Proof. The adjacency matrix A for the Cayley graph of f is defined by A ij = f ( j − i ) for i, j ∈ GF ( p ) n . So the top row of A is defined by A j = f ( j ).The adjacency matrix for any Cayley graph can be determined from its toprow, since it is a circulant matrix. Therefore, it is enough to show that thatthe adjacency matrix of the weighted partial difference set has the same toprow as A . Let B be the adjacency matrix of the weighted partial differenceset. Then B is defined by B ij = k if j − i ∈ D k (for all i, j ∈ GF ( p ) n and k ∈ GF ( p )). The top row of B is defined by B j = k if j ∈ D k . But if j ∈ D k ,then f ( j ) = k , so B j = f ( j ). The top rows of A and B are equivalent, so A = B . Therefore, the strongly regular graph associated with the weightedpartial difference set ( G, D ) is the Cayley graph of f . (cid:3) onjecture 30. (Walsh) If f : GF ( p ) n → GF ( p ) , p > is weakly regularand bent and corresponds to a weighted SRG (via Analog 61) then µ ii = 0 , ≤ i ≤ s , for the associated weighted PDS. Remark 31.
If you drop the hypothesis that f be weakly regular then theconjecture is false. The following definition is standard, but we give [PTFL] as a reference.
Definition 32. ( association scheme ) Let S be a finite set and let R , R , . . . , R s denote binary relations on S (subsets of S × S ). The dual of a relation R isthe set R ∗ = { ( x, y ) ∈ S × S | ( y, x ) ∈ R } . Assume R = ∆ S = { ( x, x ) ∈ S × S | x ∈ S } . We say ( S, R , R , . . . , R s ) isa s -class association scheme on S if the following properties hold. • We have a disjoint union S × S = R ∪ R ∪ · · · ∪ R s , with R i ∩ R j = ∅ for all i (cid:54) = j . • For each i there is a j such that R ∗ i = R j (and if R ∗ i = R i for all i thenwe say the association scheme is symmetric ). • For all i, j and all ( x, y ) ∈ S × S , define p ij ( x, y ) = |{ z ∈ S | ( x, z ) ∈ R i , ( z, y ) ∈ R j }| . For eack k , and for all x, y ∈ R k , the integer p ij ( x, y ) is a constant,denoted p kij .These constants p kij are called the intersection numbers or parameters or structure constants of the association scheme.Next, we recall (see Herman [He]) the matrix-theoretic version of thisdefinition. 14 efinition 33. ( adjacency ring ) Let S be a finite abelian multiplicativegroup of order m . Let ( S, R , . . . , R s ) denote a tuple consisting of S withrelations R i for which we have a disjoint union S × S = R ∪ R ∪ · · · ∪ R s , with R i ∩ R j = ∅ for all i (cid:54) = j . Let A i ∈ M at m × m ( Z ) denote the adjacencymatrix of R i , i = 0 , , . . . , s .We say that the subring of Z [ M at m × m ( Z )] is an adjacency ring (also calledthe Bose-Mesner algebra) provided the set of adjacency matrices satisfyingthe following five properties: • for each integer i ∈ [0 , . . . , s ], A i is a (0 , • (cid:80) si =0 A i = J (the all 1’s matrix), • for each integer i ∈ [0 . . . , s ], t A i = A j , for some integer j ∈ [0 , s ], • there is a subset J ⊂ G such that (cid:80) j ∈ J A j = I , and • there is a set of non-negative integers { p kij | i, j, k ∈ [0 , . . . , s ] } such thatequation (20) holds for all such i, j .Regarding the Dillon correspondence, we have the following combinatorialanalogs (which may or may not be true in general). Analog 34. If f is an even bent function then the tuple ( GF ( p ) n , D , D , D , · · · , D p − , D p ) defines a weighted partial difference set. We reformulate this in an essentially equivalent way using the languageof association schemes.
Analog 35.
Let f be as above and let R , R , . . . , R p denote binary relationson GF ( p ) n given by R i = { ( x, y ) ∈ GF ( p ) n × GF ( p ) n | f ( x, y ) = i } , ≤ i ≤ p. If f is an even bent function then ( GF ( p ) n , R , R , . . . , R p ) is a p -class as-sociation scheme.
15t is well-known that a PDS (
G, D ) is naturally associated to a 2-classassociation scheme, namely (
G, R , R , R ) where R = ∆ G ,R = { ( g, h ) | gh − ∈ D } ,R = { ( g, h ) | gh − / ∈ D, g (cid:54) = h } . To verify this, let consider the “Schur ring.”For the following definition, we identify any subset S of G with the formalsum of its elements in C [ G ]. Definition 36. ( Schur ring ) Let G be a finite abelian group and let C , C , . . . , C s denote finite subsets with the following properties. • C = { } is the singleton containing the identity. • We have a disjoint union G = C ∪ C ∪ · · · ∪ C s , with C i ∩ C j = ∅ for all i (cid:54) = j . • for each i there is a j such that C − i = C j (and if C − i = C i for all i then we say the Schur ring is symmetric ). • for all i, j , we have C i · C j = s (cid:88) k =0 ρ kij C k , for some integers ρ kij .The subalgebra of C [ G ] generated by C , C , . . . , C s is called a Schur ring over G .In the cases we are dealing with, the Schur ring is commutative, so ρ kij = ρ kji , for all i, j, k .If ( G, C , . . . , C s ) is a Schur ring then R i = { ( g, h ) ∈ G × G | gh − ∈ C i } , for 0 ≤ i, j ≤ s , gives rise to its corresponding association scheme.16 emark 37. Weighted PDSs in the notation of Definition 21 naturally cor-respond to association schemes of class r + 1 . For a precise version of this,see Proposition 70 below. Example 38.
For an example of a Schur ring, we return to the PDS, (
G, D ).Let D (cid:48) = G \ ( D ∪ { } ) . We have the well-known intersection D · D = ( k − µ ) · I + ( λ − µ ) · D + µ · G = k · I + λ · D + µ · D (cid:48) , (6)and D · D (cid:48) = ( − k + µ ) · − − λ + µ ) · D + ( k − µ ) · G == 0 · I + ( k − − λ ) · D + ( k − µ ) · D (cid:48) . (7)Provided k ≥ max( µ, λ + 1), | G | ≥ max( k + 1 , k − µ + 2), with these, onecan verify that a PDS naturally yields an associated Schur ring, generatedby D , D (cid:48) , and D = { } in C [ G ], and a 2-class association scheme.Using (7), one can verify that D (cid:48) is ( v, k (cid:48) , λ (cid:48) , µ (cid:48) )-PDS with ( D (cid:48) ) − = D (cid:48) and 1 / ∈ D (cid:48) , where k (cid:48) = v − k − λ (cid:48) = v − k − µ, and µ (cid:48) = v − k + λ. (8). We include a proof here for convenience. Proof.
We will show that D (cid:48) is a ( v, k (cid:48) , λ (cid:48) , µ (cid:48) )-partial difference set. The firstof these three equations is immediate, from the definition of D (cid:48) . The factthat D (cid:48) = ( D (cid:48) ) − also follows immediately the hypotheses.By the definition of D , and because D − = D , we have D · D = k λD + µD (cid:48) . (9)To find D · D (cid:48) , we note that kG = D · G = D · ( { } + D + D (cid:48) )= D + D · D + D · D (cid:48)
17o that D · D (cid:48) = ( k − λ − D + ( k − µ ) D (cid:48) . (10)Similarly, we note that k (cid:48) G = G · D (cid:48) = (( { } + D + D (cid:48) ) · D (cid:48) = D (cid:48) + D · D (cid:48) + D (cid:48) · D (cid:48) so that D (cid:48) · D (cid:48) = k (cid:48) { } + ( k (cid:48) − k + λ + 1) D + ( k (cid:48) − k − − + µ ) D (cid:48) = k (cid:48) { } + ( v − k + λ ) D + ( v − k − µ ) D (cid:48) . (11)Equation 11 shows that D (cid:48) is a ( v, k (cid:48) , λ (cid:48) , µ (cid:48) )-partial difference set, with λ (cid:48) and µ (cid:48) as in Equation 8. (cid:3) It can be shown that µ (cid:48) = k (cid:48) (cid:0) − µk (cid:1) . With the identities in the above example, one can verify that a PDSnaturally yields an associated Schur ring and a 2-class association scheme.We will now state a more general proposition concerning weighted partialdifference sets.
Proposition 39.
Let G be a finite abelian group. Let D , · · · , D r ⊆ G suchthat D i ∩ D j = ∅ if i (cid:54) = j , and • G is the disjoint union D ∪ · · · ∪ D r • for each i there is a j such that D − i = D j , and • D i · D j = r (cid:80) k =0 p kij D k for some positive integer p kij .Then the matrices P k = ( p kij ) ≤ i,j ≤ l satisfy the following properties: • P is a diagonal matrix with entries | D | , · · · , | D r |• For each k , the j th column of P k has sum | D j | ( j = 0 , · · · , l ). Likewise,the i th row of P k has sum | D i | ( i = 0 , · · · , l ). Proof.
We begin by taking the sum 18 i · D j = r (cid:80) k =0 p kij D k over all i, ≤ i ≤ l . G · D j = r (cid:80) k =0 ( r (cid:80) i =0 p kij ) D k We know that G · D j = | D j | · G , and all the D k are disjoint. As an identityin the Schur ring, each element of G must occur | D j | times on each side ofthis equation. Therefore, | D j | = r (cid:80) i =0 p kij .So the sum of the elements in the j th row of P k is | D j | for each j and k . Theanalogous claim for the row sums is proven similarly. (cid:3) Let (
G, D ) be a PDS.
Definition 40. ( Cayley graph ) The
Cayley graph
Γ = Γ(
G, D ) associatedto the PDS ( G, D ) is a graph constructed as follows: from a subset D of G ,let the vertices of the graph be the elements of the group G . Two vertices g and g are connected by a directed edge if g = dg for some d ∈ D .If D is a partial difference set such that λ (cid:54) = µ , then D = D − (Proposition1 in [Po]). Thus, if g = dg , then g = d − g , so the Cayley graph Γ( G, D )is an undirected graph.
Definition 41. ( SRG ) A connected graph Γ = (
V, E ) is a ( v, k, λ, µ ) -strongly regular graph if: • Γ has v vertices such that each vertex is connected to k other vertices • Distinct vertices g and g share edges with either λ or µ commonvertices , depending on whether they are neighbors or not.The neighborhood of a vertex g in a graph Γ is the set N ( g ) = { g (cid:48) ∈ V | ( g, g (cid:48) ) is an edge in Γ } . The following result is well-known, but the proof is included for conve-nience. 19 heorem 42.
Let G be an abelian multiplicative group and let D ⊆ G be asubset such that (cid:54)∈ D . D is a ( v, k, λ, µ ) -PDS such that D = D − if and onlyif the associated (undirected) Cayley graph Γ( G, D ) is a ( v, k, λ, µ ) -stronglyregular graph.Proof. Suppose D is a ( v, k, λ, µ )-PDS such that D = D − . Then Γ( G, D )has v vertices. D has k elements, and each vertex g of Γ( G, D ) has neighbors dg , d ∈ D . Therefore, Γ( G, D ) is regular, degree k . Let g and g be distinctvertices in Γ( G, D ). Let x be a vertex that is a common neighbor of g and g , i.e. x ∈ N ( g ) ∩ N ( g ). Then x = d g = d g for some d , d ∈ D ,which implies that d d − = g − g . If g − g ∈ D , then there are exactly λ ordered pairs ( d , d ) that satisfy the previous equation (by Definition 17). If g − g / ∈ D , then g − g ∈ G \ D , so there are exactly µ ordered pairs ( d , d )that satisfy the equation. If g − g ∈ D , then g = dg for some d ∈ D , so g and g are adjacent. By a similar argument, if g − g ∈ G \ D , then g and g are not adjacent. So Γ( G, D ) is a ( v, k, λ, µ )-strongly regular graph.Conversely, suppose Γ(
G, D ) is a ( v, k, λ, µ )-strongly regular graph. IfΓ(
G, D ) is undirected, then for vertices g and g , there is an edge from g to g if and only if there is an edge from g to g . By definition, g and g are connected by an edge if and only if g = dg , d ∈ D . This means that g = d g if and only if g = d g , for some d , d ∈ D . This implies that d = d − , so D = D − . Since Γ( G, D ) is ( v, k, λ, µ )-strongly regular, it is k -regular, so the order of D is k . Let x be a vertex in Γ( G, D ) such that x ∈ N ( g ) ∩ N ( g ). Then x = d g = d g for some d , d ∈ D , which impliesthat d d − = g − g . If g and g are adjacent, then g − g ∈ D , so thereare exactly λ ordered pairs ( d , d ) that satisfy the previous equation. If g and g are not adjacent, then g − g ∈ G \ D , so there are exactly µ orderedpairs ( d , d ) that satisfy the equation. Therefore, D is a ( v, k, λ, µ )-PDS and D = D − . (cid:3) For any graph Γ = (
V, E ), let dist: V × V → Z ∪ {∞} denote the distancefunction. In other words, for any v , v ∈ V , dist( v , v ) is the length of theshortest path from v to v (if it exists) and ∞ (if it does not). The diameterof Γ, denoted diam(Γ), is the maximum value (possibly ∞ ) of this distancefunction. Definition 43.
Let Γ = (
V, E ) be a graph, let dist: V × V → Z denote thedistance function, and let G = Aut(Γ) denote the automorphism group. Forany v ∈ V , and any k ≥
0, let 20 k ( v ) = { u ∈ V | dist( u, v ) = k } . For any subset S ⊂ V and any u ∈ V , let N u ( S ) denote the subset of s ∈ S which are a neighbor of u , i.e., let N u ( S ) = S ∩ Γ ( u ) . We say a graph is distance transitive if, for any k ≥
0, and any ( u , v ) ∈ V × V , ( u , v ) ∈ V × V with dist( u , v ) = k , there is a g ∈ G such that g ( u ) = v and g ( u ) = v .We say a graph is distance regular if for for any k ≥ v , v ) ∈ V × V with dist( v , v ) = k , the numbers a k = | N v (Γ k ( v ) | ,b k = | N v (Γ k +1 ( v ) | ,c k = | N v (Γ k − ( v ) | , are independent of v , v . Remark 44.
The following “conjecture” is false: If f : GF ( p ) n → GF ( p ) isany even bent function then the (unweighted) Cayley graph of f is distancetransitive. In fact, this fails when p = 2 for any bent function of variableshaving support of size . Indeed, in this case the Cayley graph of f is isomor-phic to the Shrikhande graph (with strongly regular parameters (16 , , , ),which is not a distance-transitive (see [BrCN], pp 104-105, 136). Proposition 45. If f : GF ( p ) n → GF ( p ) is any even function then eachconnected component of the (unweighted) Cayley graph of f is distance reg-ular. Proof.
First, we prove the following Claim:Γ k (0) = { v ∈ GF ( p ) n | v is the sum of k support vectors , and no fewer } . We prove this by indiction. The statement for k = 1 is obvious, since Γ (0) =Supp( f ). Assume the statement is true for k . We prove it for k + 1. Let v (cid:48) ∈ Γ k +1 (0), so dist(0 , v (cid:48) ) = k + 1. There is a v (cid:48)(cid:48) ∈ Γ k (0) such that v (cid:48) = v (cid:48)(cid:48) + v (cid:48)(cid:48)(cid:48) ,21or some v (cid:48)(cid:48)(cid:48) ∈ Supp( f ). By the induction hypothesis, v (cid:48)(cid:48) can be written asthe sum of k support vectors, so v (cid:48) is the sum of k + 1 vectors (thus, provingthe claim).Claim: If v ∈ GF ( p ) n is arbitrary thenΓ k ( v ) = v + Γ k (0) , for 1 ≤ k ≤ diam(Γ). This follows from the definitions.Claim: For all i, j with 1 ≤ i, j ≤ diam(Γ), and for all u, v with u, v ∈ GF ( p ) n , the cardinalities | ( u + Γ i (0)) ∩ ( v + Γ j (0)) | , are independent of u, v . This follows from the definitions.From these claims, the Proposition follows. (cid:3) If Γ is any weighted graph (without loops or multiple edges), we fix alabeling of its set of vertices V (Γ), which we often identify with the set { , , . . . , N = | V (Γ) |} . Moreover, we assume that the edge weights of Γ arepositive integers. If u, v are vertices of Γ, then we say a walk P from u to v has weight sequence ( w , w , . . . , w k ) if is there is a sequence of edges in Γconnecting u to v , ( v = u, v ), ( v , v ), . . . , ( v k − , v k = v ), say, where edge( v i − , v i ) has weight w i . If A = ( a ij ) denotes the N × N weighted adjacencymatrix of Γ, so a ij = (cid:26) w, if ( i, j ) is an edge of weight w, , if ( i, j ) is not an edge of Γ . From this adjacency matrix A , we can derive weight-specific adjacency ma-trices as follows. For each weight w of Γ, let A ( w ) = ( a ( w ) ij ) denote the N × N (1 , a ( w ) ij = (cid:26) , if ( i, j ) is an edge of weight w, , if ( i, j ) is not an edge of weight w. Let us impose the following conventions. • If u, v are distinct vertices of Γ but ( u, v ) is not an edge of Γ then wesay the weight of ( u, v ) is w = 0. • If u = v is a vertex of Γ (so ( u, v ) is not an edge, since Γ has no loops)then we say the weight of ( u, v ) is w = − A ( − , A (0)as well, and we can (and do) extend the weight set of Γ by appending 0 , − a ( w ) ij (cid:54) = 0 then a ( w (cid:48) ) ij = 0 for all weights w (cid:48) (cid:54) = w .The well-known matrix-walk theorem can be formulated as follows. Proposition 46.
For any vertices u, v of Γ and any sequence of non-zeroedge weights w , w , . . . , w k , the ( u, v ) of A ( w ) A ( w ) . . . A ( w k ) is equal to thenumber of walks of weight sequence ( w , w , . . . , w k ) from u to v . Moreover,tr A ( w ) A ( w ) . . . A ( w k ) is equal to the total number of closed walks of Γ ofweight sequence ( w , w , . . . , w k ).Let us return to describing the Cayley graph in (5) above. We identify Z /p n Z with { , , . . . , p n − } , and let η : Z /p n Z → GF ( p ) n (12)be the p -ary representation map. In other words, if we regard x ∈ Z /p n Z as a polynomial in p of degree ≤ n −
1, then η ( x ) is the list of coefficients,arranged in order of decreasing degree. This is a bijection. (Actually, forour purposes, any bijection will do, but the p -ary representation is the mostnatural one.) Lemma 47.
A graph Γ having vertices V is a (edge-weighted) Cayley graphfor some even GF ( p ) -valued function f on V with f (0) = 0 if and only if Γ isregular and the adjacency matrix A = ( A ij ) of Γ has the following properties:(a) A ,i = 1 if and only if f ( η ( i )) (cid:54) = 0 , (b) A i,j = 1 if and only if A ,k = 1 ,where η ( k ) = η ( i ) − η ( j ) .Proof. Let w ∈ GF ( p ). We know that A i,j = w if and only if there is an edgeof weight w from η ( i ) to η ( j ) if and only if f ( η ( i ) − η ( j )) = w . The resultfollows. (cid:3) We assume, unless stated otherwise, that f is even. For each u ∈ V ,define • N ( u ) = N Γ f ( u ) to be the set of all neighbors of u in Γ f , • N ( u, a ) = N Γ f ( u, a ) to be the set of all neighbors v of u in Γ f for whichthe edge ( u, v ) ∈ E f has weight a (for each a ∈ GF ( p ) × = GF ( p ) −{ } ),23 N ( u,
0) = N Γ f ( u,
0) to be the set of all non-neighbors v of u in Γ f (i.e.,we have ( u, v ) / ∈ E f ), • supp( f ) = { v ∈ V | f ( v ) (cid:54) = 0 } to be the support of f .It is clear that supp( f ) = N (0) is the set of neighbors of the zero vector.More generally, for any u ∈ V , N ( u ) = u + supp( f ) , (13)where the last set is the collection of all vectors u + v , for some v ∈ supp( f ).We call a map g : GF ( p ) n → GF ( p ) balanced if the cardinalities | g − ( x ) | ( x ∈ GF ( p )) do not depend on x . We call the signature of f : GF ( p ) n → GF ( p ) the list | S | , | S | , | S | , . . . , | S p − | , where, for each i in GF ( p ), S i = { x | f ( x ) = i } . (14)We can extend equation (13) to the more precise statement N ( u, a ) = u + S a , (15)for all a ∈ GF ( p ). We call N ( u, a ) the a -neighborhood of u .A connected simple graph Γ (without edge weights) is called stronglyregular if it consists of v vertices such that | N ( u ) ∩ N ( u ) | = k, u = u ,λ, u ∈ N ( u ) ,µ, u / ∈ N ( u ) . In the usual terminology/notation, such a graph is said to have parameters srg ( ν, k, λ, µ ). Remark 48.
Let V = GF ( p ) n , D = supp( f ) , D (cid:48) = S − { } . These sets, because f is even, have the property that D − = D , ( D (cid:48) ) − = D (cid:48) .For each d ∈ D , let d = |{ ( g, h ) ∈ D × D | g − h = d }| , and, for each d (cid:48) ∈ D (cid:48) , let µ d (cid:48) = |{ ( g, h ) ∈ D × D | g − h = d (cid:48) }| . It is known that if D is a partial difference set (PDS) on the additive groupof V then (a) λ d does not depend on d ∈ D (the common value is denoted λ ), and (b) µ d (cid:48) does not depend on d (cid:48) ∈ D (cid:48) (the common value is denoted µ ).See Theorem 42 for an equivalence between Cayley graphs of PDSs andstrongly regular graphs.Let k = | D | and ν = | V | . Since g − h ∈ D if and only if f ( g − h ) (cid:54) = 0 (for distinct g, h ∈ V ), there are kλ non-neighbors in V . Likewise, since g − h ∈ D (cid:48) if and only if f ( g − h ) (cid:54) = 0 (for distinct g, h ∈ V ), there are ( ν − k − µ neighbors in V . Therefore, since vertex pairs must be neighborsor non-neighbors, k − k = kλ + ( ν − k − µ. (16)The concept of strongly regular simple graphs generalizes to edge-weightedgraphs. Definition 49. ( edge-weighted SRG ) Let Γ be a connected edge-weightedgraph which is regular as a simple (unweighted) graph. The graph Γ iscalled strongly regular with parameters v , k = ( k a ) a ∈ W , λ = ( λ a ) a ∈ W , µ =( µ a ) a ∈ W , denoted SRG W ( v, k, λ, µ ), if it consists of v vertices such that, foreach a = ( a , a ) ∈ W | N ( u , a ) ∩ N ( u , a ) | = k a , u = u ,λ a ,a ,a , u ∈ N ( u , a ) , u (cid:54) = u ,µ a , u / ∈ N ( u ) , u (cid:54) = u , (17)where k = ( k a | a ∈ W ) ∈ Z | W | , λ = ( λ a | a ∈ W ) ∈ Z | W | , µ = ( µ a | a ∈ W ) ∈ Z | W | , and W = GF ( p ) is the set of weights, including 0 (recall an“edge” has weight 0 if the vertices are not neighbors).How does the above notion of an edge-weighted strongly regular graphrelate to the usual notion of a strongly regular graph?25 emma 50. Let Γ be an edge-weighted strongly regular graph as in (49),with edge-weights W and parameters ( v, ( k a ) , ( λ a ,a ,a ) , ( µ a ,a )) . If (cid:88) ( a ,a ) ∈ W λ a ,a ,a does not depend on a , for a ∈ W , then Γ is strongly regular (as an un-weighted graph) with parameters ( v, k, λ, µ ) where k = (cid:88) a ∈ W k a , λ = (cid:88) ( a ,a ) ∈ W λ a ,a ,a , µ = (cid:88) ( a ,a ) ∈ W µ a ,a . The proof follows directly from the definitions.Let (
G, D ) be a symmetric weighted PDS.
Definition 51.
The edge-weighted Cayley graph
Γ = Γ(
G, D ) associated tothe symmetric weighted PDS ( G, D ) is the edge-weighted graph constructedas follows. Let the vertices of the graph be the elements of the group G . Twovertices g and g are connected by an edge of weight i if g = dg for some d ∈ D i . Since ( G, D ) is symmetric, the graph Γ is undirected.
Remark 52.
This notion of an edge-weighted strongly regular graph differsslightly from the notion of a strongly regular graph decomposition in [vD],in which the individual graphs of the decomposition must each be stronglyregular.
Definition 53.
We say that an edge-weighted strongly regular graph is amor-phic if it’s corresponding association scheme is amorphic in the sense of [CP].
The following result is due to van Dam [vD] (see [CP]).
Proposition 54. (van Dam) Let f : GF ( p ) n → GF ( p ) be an even bentfunction with f ( x ) = 0 if and only if x = 0 . If the weighted Cayley graphof f , Γ f , is an edge-weighted strongly regular amorphic graph then Γ f hasa strongly regular decomposition into subgraphs Γ i all of whose edges haveweight i (where i ∈ GF ( p ) , i (cid:54) = 0 ), and each Γ i is, as an unweighted graph, astrongly regular graph of either Latin square type or of negative Latin squaretype. A of the Cayley graph of f is the matrixwhose entries are A i,j = f ( η ( i ) − η ( j )) , where η ( k ) is the p -ary representation as in (12). Note Γ f is a regular di-graph (each vertex has the same in-degree and the same out-degree as eachother vertex). The in-degree and the out-degree both equal wt ( f ), where wt denotes the Hamming weight of f , when regarded as a vector of integervalues (of length p n ). Let ω = ω f = wt ( f )denote the cardinality of supp( f ) = { v ∈ V | f ( v ) (cid:54) = 0 } . Note that ˆ f (0) = ω ≥ | supp( f ) | . If f is even then Γ f is an ω -regular graph.If A is the adjacency matrix of a (simple, unweighted) strongly regulargraph having parameters ( v, k, λ, µ ) then A = kI + λA + µ ( J − I − A ) , (18)where J is the all 1s matrix and I is the identity matrix. This is relativelyeasy to verify, by simply computing ( A ) ij in the three separate cases (a) i = j , (b) i (cid:54) = j and i, j adjacent, (c) i (cid:54) = j and i, j non-adjacent .If A is the adjacency matrix of an edge-weighted strongly regular graphhaving parameters ( v, k a , λ a ,a ,a , µ a ,a ) and positive weights W ∈ Z one cancompute ( A ) ij explicitly, again by looking at the three separate cases (a) i = j , (b) i (cid:54) = j and i, j adjacent, (c) i (cid:54) = j and i, j non-adjacent. We obtain( A ) i,j = (cid:80) a ∈ W a k a , i = j, (cid:80) ( a,b ) ∈ W abλ ( a,b,c ) , i (cid:54) = j, i ∈ N ( j, c ) , (cid:80) ( a,b ) ∈ W abµ ( a,b ) , i (cid:54) = j, i / ∈ N ( j ) . (19)As the following lemma illustrates, it is very easy to characterize Cayleygraphs in terms of its adjacency matrix. Lemma 55.
A graph Γ having vertices V is a Cayley graph for some even GF ( p ) -valued function f on V with f (0) = 0 if and only if Γ is regular andthe adjacency matrix A = ( a ij ) of Γ has the following properties: for each It can also be proven by character-theoretic methods, but this method seems harderto generalize to the edge-weighted case. ∈ GF ( p ) , (a) a ,i = w if and only if f ( η ( i )) = w , (b) a i,j = w if and onlyif a ,k = w , where η ( k ) = η ( i ) − η ( j ) . This statement follows from the definitions and its proof is omitted.Note that W f (0) = | S | + | S | ζ + · · · + | S p − | ζ p − , which we can regard as an identity in the ( p − Q -vector space Q ( ζ ). The relation 1 + ζ + ζ + · · · + ζ p − = 0 , gives W f (0) − | S | + | S | = ( | S | − | S | ) ζ + · · · + ( | S p − | − | S | ) ζ p − . We have proven the following result.
Lemma 56. If f : GF ( p ) n → GF ( p ) has the property that W f (0) is arational number then | S | = | S | = · · · = | S p − | , and W f (0) = | S | − | S | . In particular, | supp( f ) | = | S | + | S | + · · · + | S p − | = ( p − | S | = ( p − | S | − W f (0)) . Remark 57.
It is also known that if n is even and f is bent then | S | = | S | = · · · = | S p − | . We have more to say about these sets later.28
12 3 4 5 6 7811 1 2 2 12212 1 2 2122 2 12 211 1 12 21 2 12 2 21 11 1
Figure 1: The undirected Cayley graph of an even GF (3)-valued bent func-tion of two variables from Example 105. (The vertices are ordered as in theExample.)) For example, the Cayley graph of the even bent function in Example 105 isgiven in Figure 1.
Remark 58.
In Chee et al [CTZ], it is shown that if n is even then theunweighted Cayley graph of certain weakly regular even bent functions f : GF ( p ) n → GF ( p ) , with f (0) = 0 , is strongly regular. Problem 59.
Some natural problems arise. For f even,1. find necessary and sufficient conditions for Γ f to be strongly regular,2. find necessary and sufficient conditions for Γ f to be connected (andmore generally find a formula for the number of connected componentsof Γ f ), By “certain” we mean that f , regarded as a function GF ( p n ) → GF ( p ), is homoge-neous of some degree. . classify the spectrum of Γ f in terms of the values of the Fourier trans-form of f ,4. in general, which graph-theoretic properties of Γ f can be tied to function-theoretic properties of f ? Theorem 60. (Bernasconi Correspondence, [B], [BC], [BCV]) Let f : GF (2) n → GF (2) . The function f is bent if and only if the Cayley graph of f isa strongly regular graph having parameters (2 n , k, λ, λ ) for some λ , where k = | supp( f ) | . The (naive) analog of this for p >
Analog 61.
Assume n is even. If f : GF ( p ) n → GF ( p ) is even bent then,for each a ∈ GF ( p ) × , we have • if u , u ∈ V are a -neighbors in the Cayley graph of f then | N ( u , a ) ∩ N ( u , a ) | does not depend on u , u (with a given edge-weight), for each a , a , a ∈ GF ( p ) × ; • if u , u ∈ V are distinct and not neighbors in the Cayley graph of f then | N ( u , a ) ∩ N ( u , a ) | does not depend on u , u , for each a , a ∈ GF ( p ) × .In other words, the associated Cayley graphs is edge-weighted strongly regularas in Definition 49. Unfortunately, it is not true in general.
Remark 62.
1. This analog is false when p = 5 .2. This analog remains false if you replace “ f : GF ( p ) n → GF ( p ) is evenbent” in the hypothesis by “ f : GF ( p ) n → GF ( p ) is even bent andregular.” However, when p = 3 and n = 2 , see Lemma 107(a).3. In general, this analog remains false if you replace “ f : GF ( p ) n → GF ( p ) is even bent” in the hypothesis by “ f : GF ( p ) n → GF ( p ) iseven bent and weakly regular.” However, when p = 3 and n = 2 , seeLemma 107(b). . This analog is false if n is odd.5. The converse of this analog, as stated, is false if p > . Parts 2 and 3 in Problem 59 are addressed below (see Lemmas 63 and 64,resp., and § A = A f is the matrix whose entries are A i,j = f C ( η ( i ) − η ( j )) , where η ( k ) is the p -ary representation as in (12). Ignoring edge weights, welet A ∗ i,j = (cid:26) , f C ( η ( i ) − η ( j )) (cid:54) = 0 , , otherwise . Note Γ f is a regular edge-weighted digraph (each vertex has the same in-degree and the same out-degree as each other vertex). The in-degree and theout-degree both equal wt ( f ), where wt denotes the Hamming weight of f ,when regarded as a vector (of length p n ) of integers. Let ω = ω f = wt ( f )denote the cardinality of supp( f ) = { v ∈ V | f ( v ) (cid:54) = 0 } and let σ f = (cid:88) v ∈ V f C ( v ) . Note that ˆ f (0) = σ f ≥ | supp( f ) | . If f is even then Γ f is an σ f -regular (edge-weighted) graph. If we ignore weights, then it is an ω f -weighted graph.Recall that, given a graph Γ and its adjacency matrix A , the spectrum σ (Γ) = { λ , λ , . . . , λ N } , where N = p n , is the multi-set of eigenvalues of A .Following a standard convention, we index the elements λ i = λ i ( A ) of thespectrum in such a way that they are monotonically increasing (using thelexicographical ordering of C ). Because Γ f is regular, the row sums of A areall ω whence the all-ones vector is an eigenvector of A with eigenvalue ω . Wewill see later (Corollary 77) that λ N ( A ) = σ f .Let D denote the identity matrix multiplied by σ f . The Laplacian of Γ f can be defined as the matrix L = D − A .31 emma 63. Assume f is even. As an edge-weighted graph, Γ f is connectedif and only if λ N − ( A ) < λ N ( A ) = σ f . If we ignore edge weights, then Γ f isconnected if and only if λ N − ( A ∗ ) < λ N ( A ∗ ) = ω f .Proof. We only prove the statement for the edge-weighted case.Note that for i = 1 , . . . , N , λ i ( L ) = ω − λ N − i +1 ( A ), since det( L − λI ) =det( σ f I − A − λI ) = ( − n det( A − ( σ f − λ ) I ). Thus, λ i ( L ) ≥
0, for all i .By a theorem of Fiedler [F], λ ( L ) > f is connected. But λ ( L ) > σ f − λ N − ( A ) > (cid:3) Clearly, the vertices in Γ f connected to 0 ∈ V is in natural bijection withsupp( f ). Let W j denote the subset of V consisting of those vectors whichcan be written as the sum of j elements in supp( f ) but not j −
1. Clearly, W = supp( f ) ⊂ W ⊂ · · · ⊂ Span (supp( f )) . For each v ∈ W = supp( f ), the vertices connected to v are the vectorsin supp( f v ) = { v ∈ V | f ( v − v ) (cid:54) = 0 } , where f v ( v ) = f ( v − v ) denotes the translation of f by − v . Therefore,supp( f v ) = v + supp( f ) . In particular, all the vectors in W are connected to 0 ∈ V . For each v ∈ W ,the vertices connected to v are the vectors in supp( f v ) = v + supp( f ) , soall the vectors in W are connected to 0 ∈ V . Inductively, we see that Span (supp( f )) is the connected component of 0 in Γ f . Pick any u ∈ V rep-resenting a non-trivial coset in V /Span (supp( f )). Clearly, 0 is not connectedwith u in Γ f . However, the above reasoning implies u is connected to v ifand only if they represent the same coset in V /Span (supp( f )). This provesthe following result. Lemma 64.
The connected components of Γ f are in one-to-one correspon-dence with the elements of the quotient space V /Span (supp( f )) . We note here some useful facts about the action of nondegenerate lineartransforms on p -ary functions. Suppose that f : V = GF ( p ) n → GF ( p ) and32 : V → V is a nondegenerate linear transformation (isomorphism of V ),and g ( x ) = f ( φ ( x )). The functions f and g both have the same signature,( | f − ( i ) | | i = 1 , . . . , p − W f g ( u ) = W f (( φ − ) T u )(where T denotes transpose).It follows that if f is bent, so is g = f ◦ φ , and if f is bent and regular, sois g . If f is bent and weakly regular, with µ -regular dual f ∗ , then g is bentand weakly regular, with µ -regular dual g ∗ , where g ∗ ( u ) = f ∗ (( φ − ) T u ).Next, we examine the effect of the group action on bent functions andthe corresponding weighted PDSs. Proposition 65.
Let f : GF ( p ) n → GF ( p ) be an even, bent function suchthat f (0) = 0 and define D i = f − ( i ) for i ∈ GF ( p ) − { } . Suppose φ : GF ( p ) n → GF ( p ) n is a linear map that is invertible (i.e., det φ (cid:54) = 0 (mod p )).Define the function g = f ◦ φ ; g is the composition of a bent function and anaffine function, so it is also bent. If the collection of sets { D , D , · · · , D p − } forms a weighted partial difference set for GF ( p ) n then so does its imageunder the function φ . Proof.
We can explore this question by utilizing the Schur ring generated bythe sets D i . Define D = { } , where 0 denotes the zero vector in GF ( p ) n ,and define D p = GF ( p ) n − ∪ ≤ i ≤ p − D i .( D , D , D , · · · , D p − , D p ) forms a weighted partial difference set for GF ( p ) n if and only if ( C , C , C , · · · , C p ) forms a Schur ring in C [ GF ( p ) n ], where C = { } (where 0 denotes the zero element of C [ GF ( p ) n ]), C = D , · · · , C p − = D p − C p = GF ( p ) n − ( C ∪ · · · ∪ C p − ) C i · C j = p (cid:80) k =0 ρ kij C k ,for some intersection numbers ρ kij ∈ Z . Note that f is even, so C i = C − i for all i , where C − i = {− x | x ∈ C i } . Define S i = g − ( i ) = { v ∈ GF ( p ) n : g ( v ) = i } . D i = f − ( i ) = ( g ◦ φ − ) − ( i ) = ( φ · g − )( i ) = φ ( S i ). So the map φ sends S i to D i . φ can be extended to a map from C [ GF ( p ) n ] → C [ GF ( p ) n ]33uch that φ ( g + g ) = φ ( g )+ φ ( g ) and φ ( S i ) = D i . So φ is a homomorphismfrom the Schur ring of g to the Schur ring of f . Therefore, the level curves of g give rise to a Schur ring, and the weighted partial difference set generatedby f is sent to a weighted partial difference set generated by g under themap φ − . We conclude that the Schur ring of g corresponds to a weightedpartial difference set for GF ( p ) n , which is the image of that for f . (cid:3) Remark 66.
It is known that for “homogeneous” weakly regular bent func-tions , the level curves give rise to a weighted PDS. In fact, the weightedPDS corresponds to an association scheme and the dual association schemecorresponds to the dual bent function (see [PTFL], Corollary 3, and [CTZ]).We know that any bent function equivalent to such a bent function also hasthis property, thanks to the proposition above. Our data seems to support the following statement.
Conjecture 67.
Let f : GF ( p ) n → GF ( p ) be an even bent function, with p > and f (0) = 0 . If the level curves of f give rise to a weighted partialdifference set then f is homogeneous and weakly regular. This section is devoted to stating some results on the p kij ’s. Theorem 68.
Let f : GF ( p ) n → GF ( p ) be a function and let Γ be itsCayley graph. Assume Γ is a weighted strongly regular graph. Let A = ( a k,l )be the adjacency matrix of Γ. Let A i = ( a ik,l ) be the (0 , a ik,l = (cid:40) a k,l = i i = 1 , , . . . , p −
1. Let A be the p n × p n identity matrix. Let A p be the (0 , A + A + · · · + A p − + A p = J , the p n × p n matrix with all entries 1. Let R denote the matrix ring generated by { A , A , · · · , A p } . The intersection numbers p kij defined by Here, “homogeneous” is meant in the sense of [PTFL], not in the sense we use in thispaper. In the sense of Remark 29. i A j = p (cid:88) k =0 p kij A k (20)satisfy the formula p kij = (cid:18) p n | D k | (cid:19) T r ( A i A j A k ) , for all i, j, k = 1 , , . . . , p .This is (17.13) in [CvL]. We provide a different proof for the reader’sconvenience. Proof.
By the Matrix-Walk Theorem, A i A j can be considered as countingwalks along the Cayley graph of specific edge weights. Supposed ( u, v ) is anedge of Γ with weight k . If k = 0, then u = v and the edge is a loop. If k = p , then ( u, v ) is technically not an edge in Γ, but we will label it as anedge of weight p .The ( u, v )-th entry of A i A j is the number of walks of length 2 from u to v where the first edge has weight i and the second edge has weight j ; theentry is 0 if no such walk exists. If we consider the ( u, v )-th entry on eachside of the equation (20) we can deduce that p kij is the number of walks oflength 2 from u to v where the first edge has weight i and the second edge hasweight j (it equals 0 if no such walk exists) for any edge ( u, v ) with weight k in Γ.Similarly, the Matrix-Walk Theorem implies that T r ( A i A j A k ) is the totalnumber of walks of length 3 having edge weights i, j, k . We claim that if (cid:52) is any triangle with edge weights i, j, k , then by subtracting an element v ∈ GF ( p ) n , we will obtain a triangle in Γ containing the zero vector as avortex with the same edge weights. Suppose (cid:52) = ( u , u , u ), where ( u , u )has edge weight i , ( u , u ) has edge weight j , and ( u , u ) has edge weight k .Let (cid:52) (cid:48) = (0 , u − u , u − u ). We compute the edge weights of (cid:52) (cid:48) :edge weight of (0 , u − u ) = f (( u − u ) −
0) = f ( u − u ) = i edge weight of ( u − u , u − u ) = f (( u − u ) − ( u − u ) = f ( u − u ) = j edge weight of ( u − u ,
0) = f (0 − ( u − u )) = f ( u − u ) = k (cid:18) | GF ( p ) n | (cid:19) T r ( A i A j A k ) = (cid:18) p n (cid:19) T r ( A i A j A k )is the number of closed walks of length 3 having edge weights i, j, k and con-taining the zero vector as a vertex, incident to the edge of weight i and theedge of weight k .There are | D k | edges incident to the zero vector, so (cid:18) p n (cid:19) (cid:18) | D k | (cid:19) T r ( A i A j A k )is the number of walks of length 2 from the zero vector to any neighbor of italong an edge of weight k . This is equivalent to the definition of the number p kij in the Matrix-Walk Theorem. (cid:3) The following corollary is well-known (see [CvL], page 202).
Corollary 69.
Let G = GF ( p ) n . Let D , · · · , D r ⊆ G such that D i ∩ D j = ∅ if i (cid:54) = j , and • G is the disjoint union of D ∪ · · · ∪ D r • for each i there is a j such that D − i = D j , and • D i · D j = r (cid:80) k =0 p kij D k for some positive integer p kij .Then, for all i, j, k , | D k | p kij = | D i | p ikj . Proof.
For all i, j, k, we have the following identity of adjacency matrices:Tr( A i A j A k ) = p n | D k | p kij where p n is the order of G and p kij is an intersection number. Since Tr( AB )= Tr( BA ) for all matrices A and B , Tr( A i A j A k ) = Tr( A k A j A i ), and theproposition follows. (cid:3)
36e can apply this concept to a weighted partial difference set and achievesimilar results. If G is a set and D = D ∪ D ∪ · · · ∪ D r (all D i distinct) isa weighted partial difference set of G , then we can construct an associationscheme as follows: • Define R = ∆ G = { ( x, x ) ∈ G × G | x ∈ G } . • For 1 ≤ i ≤ r , define R i = { ( x, y ) ∈ G × G | xy − ∈ D i , x (cid:54) = y }• Define R r +1 = { ( x, y ) ∈ G × G | xy − / ∈ D, x (cid:54) = y } Proposition 70. If R r +1 is non-empty then the collection ( G, R , R , . . . , R r , R r +1 )as defined above produces an association scheme of class r + 1. Proof.
Consider the subring S of C [ G ] generated by D , · · · , D r +1 , where D = { } and D r +1 = G \ ( D ∪ { } ). First, we show that S is a Schur ring.We know that for 0 ≤ i ≤ r , D − i = D j for some j . D − r +1 = D r +1 because( G, D ) is a partial difference set if and only if (
G, G \ D ) is a partial differenceset.We can then compute D i · D j in C [ G ]; by the definition of a weightedpartial difference set, D i · D j = α ij · r (cid:88) l =1 λ i,j,l D l + µ i,j D r +1 , (21)for some integer α ij . So the Schur ring decomposition formula D i · D j = r +1 (cid:80) p kij D k holds for some integer p kij .Furthermore, p ij = δ ij k i for 0 ≤ i, j ≤ r + 1 and p l j = δ jl for 0 ≤ j, l ≤ r + 1.By expanding out expressions for D i · G and D r +1 · G , it can be shownthat p li,r +1 = k i δ il − r (cid:88) j =1 λ ijl for 1 ≤ i, l ≤ r , 37 r +1 i,r +1 = k i − r (cid:88) j =1 µ ij , for 1 ≤ i ≤ r , and p r +1 r +1 ,r +1 = k r +1 − − r (cid:88) i =1 k i + r (cid:88) i =1 r (cid:88) j =1 µ ij . Also, p lij = p lji for all i and j , by symmetry.Next, we will show that for all i, j, k ∈ { , · · · , r + 1 } and for ( x, y ) ∈ R k , |{ z ∈ G | ( x, z ) ∈ R i , ( z, y ) ∈ R j }| is a constant that depends only on k (and i, j ).Choose ( x, z ) ∈ R i , ( z, y ) ∈ R j ; then xz − ∈ D i , zy − ∈ D j . Consider( xz − )( zy − ) = xy − ∈ D i · D j . This is independent of z . There are exactly p kij such elements z by the Schur ring structure identity, since every elementin D k (e.g. xy − ) is repeated p kij times. (cid:3) In the case p = 2, the spectrum of Γ f is determined by the set of values ofthe Walsh-Hadamard transform of f when regarded as a vector of (integer)0 , n ). Does this result have an analog for p > Definition 71. ( Butson matrix ) We call an N × N complex matrix M a Butson matrix if M · M t = N I N , where I N is the N × N identity matrix. Lemma 72.
Consider a map g : GF ( p ) n → GF ( p ) , where we identify GF ( p ) with { , , , . . . , p − } . The following are equivalent.(a) g is balanced.(b) | g − ( x ) | = p n − , for each x ∈ GF ( p ) .(c) The Fourier transform of ζ g satisfies ˆ ζ g (0) = 0 . roof. It is easy to show that (a) and (b) are equivalent. Also, it is not hardto establish (a) implies (c).We show (c) implies (a). This is proven by an argument similar to thatused for Lemma 56.Note thatˆ ζ g (0) = | supp( g ) | + | supp( g ) | ζ + · · · + | supp( g ) p − | ζ p − , which we can regard as an identity in the ( p − Q -vector space Q ( ζ ). If ˆ ζ g (0) is rational then relation1 + ζ + ζ + · · · + ζ p − = 0 , implies all the | supp( g ) j | are equal, for j (cid:54) = 0. It also implies ˆ ζ g (0) = | supp( g ) | − | supp( g ) | . Therefore, ˆ ζ g (0) = 0 implies g is balanced. (cid:3) The following equivalences are known (see for example [T] and [CD]), butproofs are included for the convenience of the reader.
Proposition 73.
Let f : GF ( p ) n → GF ( p ) be any function. The followingare equivalent.(a) f is bent.(b) The matrix ζ F = ( ζ f ( η ( i ) − η ( j )) ) ≤ i,j ≤ p n − is Butson, where η is as in(12).(c) The derivative D b f ( x ) = f ( x + b ) − f ( x ) , is balanced, for each b (cid:54) = 0 . Remark 74.
From Proposition 73, we know D b f ( x ) is balanced, for each b (cid:54) = 0 , if and only if f is bent. Therefore, we know that |{ u ∈ V | f ( u − u ) = f ( u − u ) }| = p n − (which is obviously independent of u , u ). On the other hand, if f is bent,it is not true in general that, for each a ∈ GF ( p ) , |{ u ∈ V | f ( u − u ) = f ( u − u ) = a }| , is independent of u , u ∈ V . See Example 105 for a counterexample. h ( b ) = ( ζ D b f ) ∧ (0) = (cid:88) x ∈ V ζ f ( x + b ) − f ( x ) . Proof. (a) = ⇒ (c) : Note thatˆ h ( y ) = (cid:80) b ∈ V (cid:80) x ∈ V ζ f ( x + b ) − f ( x ) ζ −(cid:104) y,b (cid:105) = (cid:80) b ∈ V (cid:80) x ∈ V ζ f ( x + b ) − f ( x ) −(cid:104) y,b (cid:105)−(cid:104) y,x (cid:105) + (cid:104) y,x (cid:105) = (cid:80) x ∈ V ζ − f ( x )+ (cid:104) y,x (cid:105) (cid:80) b ∈ V ζ f ( x + b ) −(cid:104) y,x + b (cid:105) = ˆ ζ f ( y ) ˆ ζ f ( y ) = | ˆ ζ f ( y ) | = | W f ( y ) | . Therefore, if f is bent then ˆ h is a constant, which means that h is supportedat 0. By Lemma 72, D b f ( x ) is balanced. (c) = ⇒ (a) : We reverse the above argument. Suppose D b f ( x ) is bal-anced. By Lemma 72, h is supported at 0, so ˆ h is a constant. Plug in y = 0 and using the fact D b f ( x ) is balanced, we see that the constant must | V | = p n , Thus = | W f ( y ) | = p n/ . (c) = ⇒ (b) : Note that p n − (cid:88) j =0 ζ f ( η ( i ) − η ( j )) − f ( η ( j ) − η ( k )) = (cid:88) x ∈ V ζ f ( x ) − f ( x + η ( i ) − η ( k )) = (cid:88) x ∈ V ζ f ( x + b ) − f ( x ) , where b = η ( i ) − η ( k ). If D b f ( x ) is balanced then by Lemma 72, this sumis zero for all b (cid:54) = 0. These are the off-diagonal terms in the product ζ F ζ F t .Those terms when i = k are the diagonal terms. They are obviously | V | = p n .This implies ζ F is Butson. (b) = ⇒ (c) : This follows by reversing the above argument. The detailsare omitted. (cid:3) Recall a circulant matrix is a square matrix where each row vector is acyclic shift one element to the right relative to the preceding row vector.Our Fourier transform matrix F is not circulant, but is “block circulant.”Like circulant matrices, it has the property that (cid:126)v a = ( ζ −(cid:104) a,x (cid:105) | x ∈ V ) isan eigenvector with eigenvalue λ a = ˆ f ( − a ) (something related to a value ofthe Hadamard transform of f ). Thus, the proposition below shows that it“morally” behaves like a circulant matrix in some ways.40 roposition 75. The eigenvalues λ a = ˆ f ( − a ) of this matrix F are valuesof the Fourier transform of the function f C , ˆ f ( y ) = (cid:88) x ∈ V f C ( x ) ζ −(cid:104) x,y (cid:105) , and the eigenvectors are the vectors of p -th roots of unity, (cid:126)v a = ( ζ −(cid:104) a,x (cid:105) | x ∈ V ) . Proof. In F = ( F i,j ), we have F i,j = f C ( η ( i ) − η ( j )) for i, j ∈ { , , . . . , p n − } .For each a ∈ GF ( p ) n , let (cid:126)v a = ( ζ −(cid:104) a,η ( i ) (cid:105) | i ∈ { , , . . . , p n − } )Then F (cid:126)v a = ( (cid:88) y ∈ V f C ( x − y ) ζ −(cid:104) a,y (cid:105) | x ∈ V ) . The entry in the i th coordinate, where x = η ( i ) is given by (cid:80) y ∈ V f C ( x − y ) ζ −(cid:104) a,y (cid:105) = (cid:80) y ∈ V f C ( − y ) ζ −(cid:104) a,y + x (cid:105) = ζ −(cid:104) a,x (cid:105) (cid:80) y ∈ V f C ( − y ) ζ −(cid:104) a,y (cid:105) = ζ −(cid:104) a,x (cid:105) (cid:80) y ∈ V f C ( y ) ζ (cid:104) a,y (cid:105) = ζ −(cid:104) a,x (cid:105) ˆ f ( − a ) . Therefore, the coordinates of the vector
F (cid:126)v a are the same as those of (cid:126)v a , up toa scalar factor. Thus λ a = ˆ f ( − a ) is an eigenvalue and (cid:126)v a = ( ζ −(cid:104) a,x (cid:105) | x ∈ V )is an eigenvector. (cid:3) Corollary 76.
The matrix F is invertible if and only if none of the valuesof the Fourier transform of f C vanish. Corollary 77.
The spectrum of the graph Γ f is precisely the set of values ofthe Fourier transform of f C . Examples of Cayley graphs
Let V = GF ( p ) n and let f : V → GF ( p ). If we fix an ordering on GF ( p ) n ,then the p n × p n matrix F = ( f C ( x − y ) | x, y ∈ V ) , (22)is a Z -valued matrix. Here x indexes the rows and y indexes the columns. Example 78.
I can be shown that Example 26 (or an isomorphic copy)arises via the bent function b (see also Example 108). For this example of b , we compute the adjacency matrix associated to the members R and R of the association scheme ( G, R , R , R , R ), where G = GF (3) , R i = { ( g, h ) ∈ G × G | gh − ∈ D i } , i = 1 , , and D i = f − ( i ).Consider the following Sage computation: Sage sage: attach "/home/wdj/sagefiles/hadamard_transform.sage"sage: FF = GF(3)sage: V = FFˆ2sage: Vlist = V.list()sage: flist = [0,2,2,0,0,1,0,1,0]sage: f = lambda x: GF(3)(flist[Vlist.index(x)])sage: F = matrix(ZZ, [[f(x-y) for x in V] for y in V])sage: F sage: F1.eigenmatrix_right()% (% [ 2 0 0 0 0 0 0 0 0] [ 1 0 0 1 0 0 0 0 0]% [ 0 2 0 0 0 0 0 0 0] [ 0 1 0 0 1 0 0 0 0]% [ 0 0 2 0 0 0 0 0 0] [ 0 0 1 0 0 1 0 0 0]% [ 0 0 0 -1 0 0 0 0 0] [ 0 1 0 0 0 0 1 0 0]% [ 0 0 0 0 -1 0 0 0 0] [ 0 0 1 0 0 0 0 1 0]% [ 0 0 0 0 0 -1 0 0 0] [ 1 0 0 0 0 0 0 0 1]% [ 0 0 0 0 0 0 -1 0 0] [ 0 0 1 0 0 -1 0 -1 0]% [ 0 0 0 0 0 0 0 -1 0] [ 1 0 0 -1 0 0 0 0 -1]% [ 0 0 0 0 0 0 0 0 -1], [ 0 1 0 0 -1 0 -1 0 0]sage: F2 = matrix(ZZ, [[eval2(f(x-y)) for x in V] for y in V])sage: F2[0 1 1 0 0 0 0 0 0][1 0 1 0 0 0 0 0 0][1 1 0 0 0 0 0 0 0][0 0 0 0 1 1 0 0 0][0 0 0 1 0 1 0 0 0][0 0 0 1 1 0 0 0 0][0 0 0 0 0 0 0 1 1][0 0 0 0 0 0 1 0 1][0 0 0 0 0 0 1 1 0]sage: F1*F2-F2*F1 == 0Truesage: delta = lambda x: int((x[0]==x[1]))sage: F3 = matrix(ZZ, [[(eval0(f(x-y))+delta([x,y]))%2 for x in V] for y in V])sage: F3[0 0 0 1 1 0 1 0 1][0 0 0 0 1 1 1 1 0][0 0 0 1 0 1 0 1 1][1 0 1 0 0 0 1 1 0][1 1 0 0 0 0 0 1 1][0 1 1 0 0 0 1 0 1][1 1 0 1 0 1 0 0 0][0 1 1 1 1 0 0 0 0][1 0 1 0 1 1 0 0 0]sage: F3*F2-F2*F3==0Truesage: F3*F1-F1*F3==0Truesage: F0 = matrix(ZZ, [[delta([x,y]) for x in V] for y in V])sage: F0[1 0 0 0 0 0 0 0 0][0 1 0 0 0 0 0 0 0][0 0 1 0 0 0 0 0 0][0 0 0 1 0 0 0 0 0][0 0 0 0 1 0 0 0 0][0 0 0 0 0 1 0 0 0][0 0 0 0 0 0 1 0 0][0 0 0 0 0 0 0 1 0][0 0 0 0 0 0 0 0 1]sage: F1*F3 == 2*F2 + F3True The Sage computation above tells us that the adjacency matrix of R is43 = , the adjacency matrix of R is A = , and the adjacency matrix of R is A = Of course, the adjacency matrix of R is the identity matrix. In the abovecomputation, Sage has also verified that they commute and satisfy A A = 2 A + A in the Schur ring. 44 xample 79. We take V = GF (3) and consider an even function. Sage sage: flist = [0,1,1,2,0,1,2,1,0]sage: f = lambda x: GF(3)(flist[Vlist.index(x)])sage: x = V.random_element()sage: f(x) == f(-x)Truesage: Gamma = boolean_cayley_graph(f, V)sage: A = Gamma.adjacency_matrix(); A[0 1 1 2 0 1 2 1 0][1 0 1 1 2 0 0 2 1][1 1 0 0 1 2 1 0 2][2 1 0 0 1 1 2 0 1][0 2 1 1 0 1 1 2 0][1 0 2 1 1 0 0 1 2][2 0 1 2 1 0 0 1 1][1 2 0 0 2 1 1 0 1][0 1 2 1 0 2 1 1 0]sage: Gamma.connected_components_number()1
The plot returned by
Graph(A).show(layout="circular", edge_labels=True, graph_border=True,dpi=150) is shown in Figure 2.This example shall be continued below.
Example 80.
We take V = GF (3) and consider an even function whoseCayley graph has three connected components. Sage sage: flist = [0,0,0,1,0,0,1,0,0]sage: f = lambda x: GF(3)(flist[Vlist.index(x)])sage: x = V.random_element()sage: f(x) == f(-x)Truesage: Gamma = boolean_cayley_graph(f, V)sage: A = Gamma.adjacency_matrix(); A[0 0 0 1 0 0 1 0 0][0 0 0 0 1 0 0 1 0][0 0 0 0 0 1 0 0 1][1 0 0 0 0 0 1 0 0][0 1 0 0 0 0 0 1 0][0 0 1 0 0 0 0 0 1][1 0 0 1 0 0 0 0 0][0 1 0 0 1 0 0 0 0][0 0 1 0 0 1 0 0 0]sage: Gamma.connected_components_number()3
12 3 4 5 6 7811 2 1 2111 2 211 21 21 1 211 1 2 12 11 1
Figure 2: The undirected unweighted Cayley graph of an even GF (3)-valuedfunction of two variables from Example 79. (The vertices are ordered as inthe Example.)
06 3 14 7 2 581 1 1 1 111 1 1
Figure 3: The undirected Cayley graph of an even GF (3)-valued function oftwo variables from Example 80. (The vertices are ordered as in the Example.)The plot returned by Graph(A).show() is shown in Figure 3.
Example 81.
We return to the ternary function from Example 79.
Sage sage: V = GF(3)ˆ2sage: Vlist = V.list()sage: Vlist[(0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2)]sage: flist = [0,1,1,2,0,1,2,1,0]sage: f = lambda x: GF(3)(flist[Vlist.index(x)])sage: Gamma = boolean_cayley_graph(f, V)sage: Gamma.adjacency_matrix() This shows that, in this case, the spectrum of the Cayley graph of f agreeswith the values of the Fourier transform of f C .Suppose we want to write the function ζ f ( x ) as a linear combination oftranslates of the function f : ζ f ( x ) = (cid:88) a ∈ V c a f ( x − a ) , (23)for some c a ∈ C . This may be regarded as the convolution of f C with afunction, c . One way to solve for the c a ’s is to write this as a matrix equation, ζ (cid:126)f = F · (cid:126)c, where (cid:126)c = (cid:126)c f = ( c a | a ∈ V ) and ζ (cid:126)f = ( ζ f ( x ) | x ∈ V ). If F is invertible, thatis if the Fourier transform of f is always non-zero, then (cid:126)c = F − ζ (cid:126)f . If (23) holds then we can write the Walsh transform f , W f ( u ) = (cid:88) x ∈ GF ( p ) n ζ f ( x ) −(cid:104) u,x (cid:105) , as a linear combination of values of the Fourier transform,47 f ( y ) = (cid:88) x ∈ V f ( x ) ζ −(cid:104) x,y (cid:105) . In other words, W f ( u ) = (cid:80) a ∈ V c a (cid:80) x ∈ GF ( p ) n ζ −(cid:104) u,x (cid:105) f ( x − a )= (cid:80) a ∈ V c a (cid:80) x ∈ GF ( p ) n ζ −(cid:104) u,x + a (cid:105) f ( x )= (cid:80) a ∈ V c a ζ −(cid:104) u,a (cid:105) (cid:80) x ∈ GF ( p ) n ζ −(cid:104) u,x (cid:105) f ( x )= ˆ f ( u ) (cid:80) a ∈ V c a ζ −(cid:104) u,a (cid:105) . (24)This may be regarded as the product of Fourier transforms (that of the func-tion f C and that of the function c , which depends on f ). In other words, thereis a relationship between the Fourier transform of a GF ( p )-valued functionand its Walsh-Hadamard transform. However, it is not explicit unless oneknows the function c (which depends on f in a complicated way). GF (3) → GF (3) Using Sage, we verified the following fact (originally discovered by the last-named author, Walsh).
Proposition 82.
There are even bent functions f : GF (3) → GF (3) such that f (0) = 0 . The group G = GL (2 , GF (3)) acts on the set B of allsuch bent functions and there are two orbits in B /G : B /G = B ∪ B , where | B | = 12 and | B | = 6 .The bent functions b , b , . . . , b are given here in table form and al-gebraic normal form. The orbit B consists of the functions b , b , b , b , b , b , b , b , b , b , and b . These functions are all regular. The orbit B consists of the functions b , b , b , b , b , and b . These functions areweakly regular (but not regular).Each of the bent functions give rise to a weighted PDS.
12 3 4 5 6 7811 2 1 2111 2 211 21 21 1 211 1 2 12 11 1
Figure 4: The weighted Cayley graph of a non-bent even GF (3)-valued func-tion. Example 83.
Consider the even function f : GF (3) → GF (3) with thefollowing values: GF (3) (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1) (0, 2) (1, 2) (2, 2) f The Cayley graph Γ of f is given in Figure 4.The values of the Hadamard transform of f are listed below (showingthat f is not bent). Sage sage: V = GF(3)ˆ2sage: flist = [0,1,1,2,0,1,2,1,0]sage: Vlist = V.list()sage: f = lambda x: GF(3)(flist[Vlist.index(x)])sage: [hadamard_transform(f,a) for a in V][eˆ(2/3*I*pi) + eˆ(4/3*I*pi) + eˆ(2/3*I*pi) + eˆ(4/3*I*pi) + 2*eˆ(2/3*I*pi) + 3,eˆ(2/3*I*pi) + 5*eˆ(4/3*I*pi) + 3,3*eˆ(4/3*I*pi) + eˆ(2/3*I*pi) + 2*eˆ(4/3*I*pi) + 3,eˆ(2/3*I*pi) + 2*eˆ(4/3*I*pi) + 3*eˆ(2/3*I*pi) + 3,eˆ(4/3*I*pi) + 4*eˆ(2/3*I*pi) + eˆ(4/3*I*pi) + 3,eˆ(4/3*I*pi) + eˆ(2/3*I*pi) + eˆ(4/3*I*pi) + 6,eˆ(4/3*I*pi) + eˆ(2/3*I*pi) + eˆ(4/3*I*pi) + 3*eˆ(2/3*I*pi) + 3,eˆ(4/3*I*pi) + eˆ(2/3*I*pi) + eˆ(4/3*I*pi) + 6,4*eˆ(2/3*I*pi) + 2*eˆ(4/3*I*pi) + 3] age: [CC(hadamard_transform(f,a)) for a in V][-2.22044604925031e-16 + 1.73205080756888*I,-2.22044604925031e-15 - 3.46410161513775*I,-1.99840144432528e-15 - 3.46410161513775*I,-2.22044604925031e-16 + 1.73205080756888*I,1.73205080756888*I,4.50000000000000 - 0.866025403784438*I,1.73205080756888*I,4.50000000000000 - 0.866025403784438*I,1.73205080756888*I] This f is not bent and has algebraic normal form2 x x + x + x x + 2 x . In particular, it is non-homogeneous. The weighted adjacency matrix of itsCayley graph is . The matrix N whose u, v -entry is | N ( u, ∩ N ( v, | , u, v vertices of Γ, is: . • Are all the values of N [ u, v ] the same if u, v are distinct vertices whichare not neighbors? Yes: N [ u, v ] = 0 for all such u, v . Therefore, µ (2 , = 0 in the notation of (17).50 Are all the values of N [ v, v ] the same? Yes: N [ v, v ] = 2 for all v .Therefore, k (2 , = 2 in the notation of (17). • Are all the values of N [ u, v ] the same if u, v are neighbors with edge-weight 2? Yes: N [ u, v ] = 1 for all such u, v . Therefore, λ (2 , , = 1 inthe notation of (17). • Are all the values of N [ u, v ] the same if u, v are neighbors with edge-weight 1? Yes: N [ u, v ] = 0 for all such u, v . Therefore, λ (2 , , = 0 inthe notation of (17).The matrix N , whose u, v -entry is | N ( u, ∩ N ( v, | = | N ( u, ∩ N ( v, | , is: . Define N similarly. Sage verifies that N = N . • Are all the values of N [ u, v ] the same if u, v are distinct vertices whichare not neighbors? Yes: N [ u, v ] = 2 for all such u, v . Therefore, µ (2 , = 2 in the notation of (17). • Are all the values of N [ v, v ] the same? Yes: N [ v, v ] = 0 for all v .Therefore, k (2 , = 0 in the notation of (17). • Are all the values of N [ u, v ] the same if u, v are neighbors with edge-weight 2? Yes: N [ u, v ] = 0 for all such u, v . Therefore, λ (2 , , = 0 inthe notation of (17). • Are all the values of N [ u, v ] the same if u, v are neighbors with edge-weight 1? Yes: N [ u, v ] = 1 for all such u, v . Therefore, λ (2 , , = 1 inthe notation of (17). 51he matrix N , whose u, v -entry is | N ( u, ∩ N ( v, | , is: . • Are all the values of N [ u, v ] the same if u, v are distinct vertices whichare not neighbors? Yes: N [ u, v ] = 2 for all such u, v . Therefore, µ (1 , = 2 in the notation of (17). • Are all the values of N [ v, v ] the same? Yes: N [ v, v ] = 4 for all v .Therefore, k (1 , = 4 in the notation of (17). • Are all the values of N [ u, v ] the same if u, v are neighbors with edge-weight 2? Yes: N [ u, v ] = 2 for all such u, v . Therefore, λ (1 , , = 2 inthe notation of (17). • Are all the values of N [ u, v ] the same if u, v are neighbors with edge-weight 1? Yes: N [ u, v ] = 1 for all such u, v . Therefore, λ (1 , , = 1 inthe notation of (17).In summary, we have µ (1 , = 2 , k (1 , = 4 , λ (1 , , = 1 , λ (1 , , = 2 ,µ (1 , = 2 , k (1 , = 0 , λ (1 , , = 1 , λ (1 , , = 0 ,µ (2 , = 0 , k (2 , = 2 , λ (2 , , = 0 , λ (2 , , = 1 . This verifies the statements in the conclusion of Analog 61 for this func-tion. In other words, the associated edge-weighted Cayley graph is stronglyregular. (However, f is not bent.) Example 84.
Consider the even function f : GF (3) → GF (3) with thefollowing values: 52 F (3) (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1) (0, 2) (1, 2) (2, 2) f This f has algebraic normal form x x + 2 x + x x + 2 x , and is not bent and non-homogeneous. The weighted adjacency matrix of itsCayley graph is . We have µ (1 , = 0 , k (1 , = 2 , λ (1 , , = 1 , λ (1 , , = 0 ,µ (1 , = 2 , k (1 , = 0 , λ (1 , , = 0 , λ (1 , , = 1 ,µ (2 , = 2 , k (2 , = 4 , λ (2 , , = 2 , λ (2 , , = 1 . This verifies the statements in the conclusion of Conjecture 61 for this func-tion. (Again, f is not bent.) Example 85.
Consider the even function f : GF (3) → GF (3) with thefollowing values: GF (3) (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1) (0, 2) (1, 2) (2, 2) f This f has algebraic normal form x x + 2 x , and is bent and homogeneous. The weighted adjacency matrix of its Cayleygraph is 53 . We have µ (1 , = 0 , k (1 , = 2 , λ (1 , , = 1 , λ (1 , , = 0 ,µ (1 , = 1 , k (1 , = 0 , λ (1 , , = 0 , λ (1 , , = 0 ,µ (2 , = 0 , k (2 , = 2 , λ (2 , , = 0 , λ (2 , , = 1 . This verifies the statements in the conclusion of Conjecture 61 for this func-tion, f = b .The last-named author (SW) has made the following observation. Proposition 86.
Let f : GF (3) → GF (3) be an even bent function with f (0) = 0 . If the level curves of f , D i = { v ∈ GF (3) | f ( v ) = i } , yield a weighted PDS with intersection numbers p kij then one of the followingoccurs.1. We have | D | = | D | = 2 , and the intersection numbers p kij are givenas follows: ij p ij p ij p ij Furthermore, D = D ∪ D is a (9 , , , -PDS of Latin square type( N = 3 and R = 2 ) and negative Latin square type ( N = − and R = − ).2. We have | D | = | D | = 4 , D = ∅ , and the intersection numbers p kij are given as follows: p ij p ij p ij p ij This is verified using a case-by-case analysis. GF (3) → GF (3) We can classify some bent functions on GF (3) in terms of the correspondingcombinatorial structure of their level curves. Unlike the GF (3) case, not allsuch bent functions have “combinatorial” level curves.55 f ( x , x , x ) = x + x + x B f ( x , x , x ) = x x + 2 x + 2 x x B f ( x , x , x ) = − x − x − x B f ( x , x , x ) = − x x − x − x x Table 1: Representatives of orbits in B /G Proposition 87.
There are even bent functions f : GF (3) → GF (3) such that f (0) = 0 . The group G = GL (3 , GF (3)) acts on the set B of allsuch bent functions and there are orbits in B /G : B /G = B ∪ B ∪ B ∪ B , where | B | = 234 , | B | = 936 , | B | = 234 , and | B | = 936 .The bent functions which give rise to a weighted PDS are those in orbits B and B . The other bent functions do not.The functions in orbits B and B are weakly regular but not regular. Thefunctions in orbits B and B are not weakly regular. Remark 88.
The result above agrees with the results of Pott et al [PTFL],where they overlap.
Example 89.
Consider the example of the even function f : GF (3) → GF (3) given in § V, E ) is given below.
Sage sage: FF = GF(3)sage: V = FFˆ3sage: Vlist = V.list()sage: Vlist[(0, 0, 0), (1, 0, 0), (2, 0, 0), (0, 1, 0), (1, 1, 0), (2, 1, 0), (0, 2, 0),(1, 2, 0), (2, 2, 0), (0, 0, 1), (1, 0, 1), (2, 0, 1), (0, 1, 1), (1, 1, 1),(2, 1, 1), (0, 2, 1), (1, 2, 1), (2, 2, 1), (0, 0, 2), (1, 0, 2), (2, 0, 2),(0, 1, 2), (1, 1, 2), (2, 1, 2), (0, 2, 2), (1, 2, 2), (2, 2, 2)]sage: flist = [0,2,2,1,1,1,1,1,1,2,0,1,1,2,0,1,0,0,2,1,0,1,0,0,1,0,2]sage: f = lambda x: GF(3)(flist[Vlist.index(x)])sage: [CC(hadamard_transform(f,a)).abs() for a in V][5.19615242270663, 5.19615242270663, 5.19615242270664,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663, Note, the weighted PDSs are given in the examples below. .19615242270663, 5.19615242270664, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270664, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270664]sage: Gamma = boolean_cayley_graph(f, V)sage: Gamma.spectrum()[18, 3, 3, 3, 3, 3, 3, 0, 0, 0, 0, 0, 0, 0, 0, -3, -3, -3, -3, -3, -3, -3,-3, -3, -3, -3, -3] The algebraic normal form of f is x x + 2 x + 2 x x , which is non-homogeneous but bent. It is not regular, nor merely weaklyregular. The weighted adjacency matrix is We have µ (1 , = { , } , k (1 , = 12 , λ (1 , , = 5 , λ (1 , , = 6 , (1 , = 3 , k (1 , = 0 , λ (1 , , = { , } , λ (1 , , = 2 ,µ (2 , = { , } , k (2 , = 6 , λ (2 , , = { , } , λ (2 , , = 1 . In this example, Analog 61 is false.
Example 90.
Consider the example of the bent even function f : GF (3) → GF (3) given by f ( x , x ) = x x + x , which is homogeneous but bent. It is not weakly regular. The adjacencymatrix of the associated edge-weighted Cayley graph is . Sage sage: p = 3; n = 3sage: FF = GF(p)sage: V = GF(p)**nsage: f = lambda x: FF(x[0]*x[1]+x[2]ˆ2)sage: flist = [f(v) for v in V]sage: flist[0, 0, 0, 0, 1, 2, 0, 2, 1, 1, 1, 1, 1, 2, 0, 1, 0, 2, 1, 1, 1, 1, 2, 0, 1, 0, 2]sage: age:sage: [CC(hadamard_transform(f,a)).abs() for a in V][5.19615242270663, 5.19615242270663, 5.19615242270664,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270664, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270664, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270664]sage: Gamma = boolean_cayley_graph(f, V)sage: Gamma.spectrum()[18, 3, 3, 3, 3, 3, 3, 0, 0, 0, 0, 0, 0, 0, 0, -3, -3, -3, -3, -3, -3, -3,-3, -3, -3, -3, -3]sage: Gamma.is_strongly_regular()False The unweighed Cayley graph of f is regular but has four distinct eigen-values, so is not strongly regular (as Sage indicates above). However, a Sagecomputation shows | W f ( a ) | = 3 / for all a ∈ GF (3) , f is bent. Since W f (0) / / is a 6-th root of unity but not a cube root, f is not weaklyregular.By a Sage computation, we have µ (1 , = 6 , k (1 , = 12 , λ (1 , , = 5 , λ (1 , , = 4 ,µ (1 , = 3 , k (1 , = 0 , λ (1 , , = 2 , λ (1 , , = 4 ,µ (2 , = 0 , k (2 , = 6 , λ (2 , , = 2 , λ (2 , , = 1 . In this example, Analog 61 is true.Let f : GF (3) → GF (3) be an even bent function with f (0) = 0, let D i = { v ∈ GF (3) | f ( v ) = i } , i = 1 , , let D = { } and D = GF (3) \ ( D ∪ D ∪ D ).The next result extends Proposition 86. Proposition 91.
Let f : GF (3) → GF (3) be an even bent function with f (0) = 0 . If the level curves of f , D i , yield a weighted PDS with intersectionnumbers p kij then one of the following occurs. . We have | D | = 6 , | D | = 12 , and the intersection numbers p kij aregiven as follows: p ij p ij p ij p ij
2. We have | D | = 12 , | D | = 6 , and the intersection numbers p kij aregiven as follows: p ij p ij p ij p ij f : GF (3) → GF (3) satisfies the hypothesis of the aboveproposition then f is necessarily quadratic .One way to investigate this question is to partition the set of even func-tions into equivalence classes with respect to the group action of GL (3 , GF (3)),then pick a representative from each class and test for bentness. Once we However, this may very likely have more to do with the fact that p and n are so small. GL (3 , GF (3)) are bent, we can check the conjectureand the question for a representative from each orbit.What are these orbits?Consider the set E of all functions f : GF (3) → GF (3) such that • f is even, • f (0) = 0, and • the degree of the algebraic normal form of f is at most 4.The algebraic normal form of such a function must be of the form f ( x , x , x ) = a x + a x x + a x x + a x + a x x + a x + b x x + b x x x + b x x + b x x x + b x x x + b x x where a , ..., a , b , ..., b are in GF (3) . Thus there are 3 = 531 ,
441 suchfunctions. Recall the signature of f is the sequence of cardinalities of thelevel curves D i = { x ∈ GF (3) | f ( x ) = i } . Let G = GL (3 , GF (3)) be the set of nondegenerate linear transformations φ : GF (3) → GF (3) . This group acts on E in a natural way and we say f ∈ E is equivalent to g ∈ E if and only if f is sent to g under some elementof G . An equivalence class is simply an orbit in E under this action of G . Mathematica was used to calculate that | G | = 11232. However, since f ( φ ( x )) = f ( − φ ( x )) for all φ in G and x in GF (3) , there are at most 5616functions in the equivalence class of any nonzero element of E .If f is bent, then so is f ◦ φ , for φ in G . Therefore, one way to find all bentfunctions in E is to partition B into equivalence classes under the action of G and test an element of each equivalence class to see if it is bent. However,the computational time for attacking this problem directly was prohibitive.We next note that the size of the level curves f − (1) and f − (2) is pre-served under the action of elements of G , i.e., the signature of f is the samefor all functions in each equivalence class. Mathematica was used to parti-tion E into sets with the same signature. There are 120120 elements of E or signature (6 ,
12) or (12 , | D | = | D | = 8. 61athematica was then used to find all equivalence classes of functionsin E under transformations in G for each of the 35 signature equivalenceclasses. There are a total of 281 equivalence classes of functions in E underthe action of GL (3 , GF (3)). Of these, 4 classes consist of bent functions. Inother words, if B denotes the subset of E consisting of bent functions then G acts on B and the number of orbits is 4.There were two equivalence classes of bent functions of type | D | = 6 and | D | = 12. The other two bent classes were of type | D | = 12 and | D | = 6and consisted of the negatives of the functions in the first two classes. Wewill call the classes B , B , B , and B : B /G = B ∪ B ∪ B ∪ B . Note the (6 ,
12) classes are negatives of the (12 ,
6) classes, so after a possiblereindexing, we have B = − B and B = − B .A representative of B is x + x + x . There are 234 bent functions in its equivalence class under nondegeneratelinear transformations. Note that the algebraic normal form of each functionin this class is quadratic.A representative of B is x x + 2 x + 2 x x . There are 936 bent functions in its equivalence class under nondegeneratelinear transformations.Thus there are a total of 2340 bent functions in B .We know that if W f (0) is rational then the level curves f − ( i ), i (cid:54) = 0,have the same cardinality . The following Sage computation shows that, inthis case, W f (0) is not a rational number for the representatives of B , B displayed above. Sage sage: PR.
There are even bent functions f : GF (5) → GF (5) such that f (0) = 0 . The group G = GL (2 , GF (5)) acts on the set B of allsuch bent functions and there are orbits in B /G : B /G = B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B , where | B | = 40 , | B | = 60 , | B | = · · · = | B | = 120 , and | B | = | B | =240 . The bent functions which give rise to a weighted PDS are f , f , f , f , f in Table 2. The other f i ’s do not. Remark 93.
The result above agrees with the results of Pott et al [PTFL],where they overlap.
Example 94.
Consider the example of the even function f : GF (5) → GF (5) given in § Sage sage: FF = GF(5)sage: V = FFˆ2sage: Vlist = V.list()sage: R.
Using Sage, we have µ (1 , = 0 , k (1 , = 4 , λ (1 , , = 0 , λ (1 , , = 2 , λ (1 , , = 0 , λ (1 , , = 1 ,µ (1 , = 1 , k (1 , = 0 , λ (1 , , = 2 , λ (1 , , = 0 , λ (1 , , = 0 , λ (1 , , = 0 ,µ (1 , = 1 , k (1 , = 0 , λ (1 , , = 0 , λ (1 , , = 0 , λ (1 , , = 2 , λ (1 , , = 0 ,µ (1 , = 1 , k (1 , = 0 , λ (1 , , = 1 , λ (1 , , = 0 , λ (1 , , = 0 , λ (1 , , = 1 ,µ (2 , = 0 , k (2 , = 4 , λ (2 , , = 0 , λ (2 , , = 0 , λ (2 , , = 1 , λ (2 , , = 2 ,µ (2 , = 1 , k (2 , = 0 , λ (2 , , = 0 , λ (2 , , = 1 , λ (2 , , = 1 , λ (2 , , = 0 ,µ (2 , = 1 , k (2 , = 0 , λ (2 , , = 0 , λ (2 , , = 2 , λ (2 , , = 0 , λ (2 , , = 0 ,µ (3 , = 0 , k (3 , = 4 , λ (3 , , = 2 , λ (3 , , = 1 , λ (3 , , = 0 , λ (3 , , = 0 ,µ (3 , = 1 , k (3 , = 0 , λ (3 , , = 0 , λ (3 , , = 0 , λ (3 , , = 0 , λ (3 , , = 2 ,µ (4 , = 0 , k (4 , = 4 , λ (4 , , = 1 , λ (4 , , = 0 , λ (4 , , = 2 , λ (4 , , = 0 . In this example, Analog 61 is true.The parameters as an unweighted strongly regular graph are (25 , , , xample 95. Consider the example of the even function f : GF (5) → GF (5) given by f ( x , x ) = x + 2 x x . This is non-homogeneous, but bent and regular.
Sage sage: p = 5; n = 2sage: FF = GF(p)sage: V = GF(p)**nsage: Vlist = V.list()sage: f = lambda x: FF(x[0]ˆ4+2*x[0]*x[1])sage: [CC(hadamard_transform(f,a)).abs() for a in V][5.00000000000000, 5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000]sage: Gamma = boolean_cayley_graph(f, V)sage: Gamma.spectrum()[16, 3.236067977499790?, 3.236067977499790?, 3.236067977499790?, 3.236067977499790?,1, 1, 1, 1, -0.3819660112501051?, -0.3819660112501051?, -0.3819660112501051?,-0.3819660112501051?, -1.236067977499790?, -1.236067977499790?, -1.236067977499790?,-1.236067977499790?, -2.618033988749895?, -2.618033988749895?, -2.618033988749895?,-2.618033988749895?, -4, -4, -4, -4]sage: Gamma.is_strongly_regular()False
Its edge-weighted Cayley graph Γ = (
V, E ) has weighted adjacency matrixgiven by 66
Using Sage, we have µ (1 , = 0 , k (1 , = 4 , λ (1 , , = 3 , λ (1 , , = 0 , λ (1 , , = 0 , λ (1 , , = 0 ,µ (1 , = 1 , k (1 , = 0 , λ (1 , , = 0 , λ (1 , , = 0 , λ (1 , , = 1 , λ (1 , , = 1 ,µ (1 , = 1 , k (1 , = 0 , λ (1 , , = 0 , λ (1 , , = 1 , λ (1 , , = 0 , λ (1 , , = 1 ,µ (1 , = 1 , k (1 , = 0 , λ (1 , , = 0 , λ (1 , , = 1 , λ (1 , , = 1 , λ (1 , , = 0 ,µ (2 , = { , } , k (2 , = 4 , λ (2 , , = 0 , λ (2 , , = 0 , λ (2 , , = 2 , λ (2 , , = 0 ,µ (2 , = { , } , k (2 , = 0 , λ (2 , , = 1 , λ (2 , , = 2 , λ (2 , , = 0 , λ (2 , , = 0 ,µ (2 , = { , } , k (2 , = 0 , λ (2 , , = 1 , λ (2 , , = 0 , λ (2 , , = 0 , λ (2 , , = 2 ,µ (3 , = { , } , k (3 , = 4 , λ (3 , , = 0 , λ (3 , , = 0 , λ (3 , , = 0 , λ (3 , , = 1 ,µ (3 , = { , } , k (3 , = 0 , λ (3 , , = 1 , λ (3 , , = 0 , λ (3 , , = 1 , λ (3 , , = 1 ,µ (4 , = 0 , k (4 , = 4 , λ (4 , , = 0 , λ (4 , , = 2 , λ (4 , , = 1 , λ (4 , , = 0 . In this example, Analog 61 is false.67he number of even (polynomial) functions f of degree less than or equalto 4 is 5 = 390625. The number of such functions having signature (4 , , , , , ,
6) is 2920.Using Sage, we discovered there are 1420 even bent functions f : GF (5) → GF (5) such that f (0) = 0. The group G = GL (2 , GF (5)) acts on the set B of all such bent functions and there are 11 orbits in B /G : B /G = B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B ∪ B , where | B | = 40, | B | = 60, | B | = · · · = | B | = 120, and | B | = | B | = 240.A representative of B is f ( x , x ) = − x + 2 x . A representative of B is f ( x , x ) = − x x + x . A representative of B is f ( x , x ) = − x + 2 x + 2 x x . A representative of B is f ( x , x ) = − x + x x − x . A representative of B is f ( x , x ) = x x + 2 x . A representative of B is f ( x , x ) = − x x + x A representative of B is f ( x , x ) = x − x x A representative of B is f ( x , x ) = 2 x − x x + 2 x .
68 representative of B is f ( x , x ) = − x x + x A representative of B is f ( x , x ) = 2 x x + x − x . A representative of B is f ( x , x ) = x x − x − x . These 11 bent functions form a complete set of representatives of the G -equivalence classes of B . We write f ∼ g if and only if f = g ◦ φ , for some φ ∈ G . The group GF (5) × also acts on B . • for i ∈ { , , } , the functions af i , for a ∈ GF (5) × , are all G -equivalent, • f ∼ f ∼ f ∼ f , • f ∼ f ∼ f ∼ f , • f ∼ f ∼ f ∼ f , • f ∼ f ∼ f ∼ f , • f ∼ f ∼ f ∼ f , • f ∼ f ∼ f ∼ f , • f ∼ f ∼ f ∼ f , • f ∼ f ∼ f ∼ f .It follows that f , f , f and f all must have the same signature. Similarly, f and f must have the same signature, and f and f must have the samesignature.Note f and f are not GL (2 , GF (5))-equivalent but they both corre-sponding to weighted PDSs with the same intersection numbers. In partic-ular, the adjacency ring corresponding to f is isomorphic to the adjacencyring corresponding to f . 69 xample 96. The example of f above can be used to construct an edge-weighted strongly regular Cayley graph, hence also a weighted PDS attachedto its level curves.Define the level curve D i ( i = 1 , , ,
4) as above, the let D = { } and D = GF (5) \ ∪ i =0 D i . We can interprete p kij to be the number of times eachelement of D k occurs in D j D − i . By computing these numbers directly usingSage, we obtain the intersection numbers p kij : p ij p ij p ij p ij p ij p ij Example 97.
The example of f above can be used to construct an edge-weighted strongly regular Cayley graph, hence also a weighted PDS attachedto its level curves.Define the level curve D i ( i = 1 , , ,
4) as above, the let D = { } and D = GF (5) \ ∪ i =0 D i . We can interprete p kij to be the number of times eachelement of D k occurs in D j D − i . By computing these numbers directly using70age, we obtain the intersection numbers p kij : p ij p ij p ij p ij p ij p ij Example 98.
The level curves of f above do not give rise to a weightedPDS.On the other hand, we can define the adjacency matrix A i attached tothe level curve D i ( i = 1 , , ,
4) as the 25 ×
25 matrix obtained by taking theweighted adjacency matrix A of the corresponding Cayley graph and puttinga 1 in every entry where the corresponding entry of A is equal to i , and a 0otherwise. In this case, D = { (2 , , (3 , , (4 , , (1 , } ,D = { (1 , , (3 , , (2 , , (4 , } ,D = { (4 , , (3 , , (2 , , (1 , } , = { (1 , , (2 , , (3 , , (4 , } . The “adjacency matrix” A is the 25 ×
25 identity matrix and the “adjacencymatrix” A is the 25 ×
25 matrix which has the property that A + A + A + A + A + A is the all 1’s matrix.If the Cayley graph were a strongly regular edge-weighted graph then,according to [CvL], equation (17.13) (proven in Theorem 68 above), theintersection numbers p kij could be computed usingtrace ( A i A j A k ) = | GF (5) || D k | p kij . Using
Sage , we computetrace ( A ) = 0 , trace ( A A ) = 0 , trace ( A A ) = 200 , trace ( A ) = 0 , trace ( A A ) = 0 , trace ( A A ) = 0 , trace ( A ) = 300 , trace ( A A ) = 200 , trace ( A A ) = 0 , trace ( A ) = 0 , trace ( A A ) = 100 , trace ( A A ) = 300 , trace ( A ) = 300 , trace ( A A ) = 0 , trace ( A A ) = 100 , trace ( A A ) = 0 , trace ( A A ) = 100 , trace ( A A ) = 0 , trace ( A A ) = 0 , trace ( A A ) = 0 , trace ( A A ) = 400 , trace ( A A ) = 0 , trace ( A A ) = 200 , trace ( A A ) = 200 , trace ( A A ) = 200 , trace ( A A A ) = 100 , trace ( A A A ) = 0 , trace ( A A A ) = 100 , trace ( A A A ) = 200 , trace ( A A A ) = 100 , trace ( A A A ) = 100 , trace ( A A A ) = 200 , trace ( A A A ) = 200 , trace ( A A A ) = 100 , trace ( A A A ) = 100 . Using these, we can compute the p kij ’s.We have | D | = | D | = | D | = | D | = 4, and the intersection numbers p kij are given as follows: 72 ij p ij p ij p ij p ij p ij / / / /
22 0 1 / / / / / / Example 99.
Consider the bent function f ( x , x ) = − x − x + x x . This function represents a GL (2 , GF (5)) orbit of size 120. The level curvesof this function do not give rise to a weighted PDS. By the way, if we try acomputation of all the p kij ’s as in the above example, we do not get integers.Similarly, the level curves of f , f , f , and f do not give rise to a weightedPDS, since the p kij ’s are not always integers. Example 100.
The example of f above can be used to construct an edge-weighted strongly regular Cayley graph, hence also a weighted PDS attachedto its level curves. 73he intersection numbers p kij are given by: p ij p ij p ij p ij p ij p ij f and f above have the same p kij ’s.Note f and f are not multiples. Therefore, the p kij ’s do not determinethe equivalence class of the bent function nor even the (larger) equivalenceclass “up to a scalar factor” . Similar to how Carlet [C] shows that every Boolean function can be writtenin algebraic normal form, we can show that each GF ( p )-valued function over GF ( p ) n can be written in algebraic normal form as well.74n atomic p -ary function is a function GF ( p ) n → GF ( p ) supported ata single point. For v ∈ GF ( p ) n , the atomic function supported at v is thefunction f v : GF ( p ) n → GF ( p ) such that f v ( v ) = 1 and for every w ∈ GF ( p ) n such that w (cid:54) = v f v ( w ) = 0. We begin by showing how to write the algebraicnormal form of the atomic p -ary functions, where Theorem 101.
Let f v be an atomic p -ary function. Then f v ( x ) = n − (cid:89) i =0 (cid:32) p − p − (cid:89) j =1 ( j + v i − x i ) (cid:33) . (25) Proof.
First, we start by showing that f v ( v ) = 1. We can do this by plugging v directly into (25). f v ( v ) = n − (cid:89) i =0 (cid:32) p − p − (cid:89) j =1 ( j + v i − v i ) (cid:33) = n − (cid:89) i =0 (cid:32) p − p − (cid:89) j =1 j (cid:33) = n − (cid:89) i =0 (cid:18) ( p − p − (cid:19) = 1 . Second, we show that f v ( w ) = 0 for every w (cid:54) = v . Let w (cid:54) = v . Then,pick k such that w k (cid:54) = v k . So there exists j ∈ { , . . . , n − } such that j + v k − w k = 0 (mod p ). Thus, the inside product of (25) is 0 for i = k andthe whole equation is 0. So f v ( w ) = 0. (cid:3) It easily follows that every GF ( p )-valued function over GF ( p ) n can bewritten in algebraic normal form. Corollary 102.
Let g : GF ( p ) n → GF ( p ) . Then g ( x ) = (cid:88) v ∈ GF ( p ) n g ( v ) f v ( x ) . (26) Example 103.
Sage can easily list all the atomic functions over GF (3)having 2 variables: 75 age sage: V = GF(3)ˆ2sage: x0,x1 = var("x0,x1")sage: xx = [x0,x1]sage: [expand(prod([2*prod([GF(3)(j)+v[i]-xx[i] for j in range(1,3)])for i in range(2)])) for v in V][x0ˆ2*x1ˆ2 + 2*x0ˆ2 + 2*x1ˆ2 + 1,x0ˆ2*x1ˆ2 + x0*x1ˆ2 + 2*x0ˆ2 + 2*x0,x0ˆ2*x1ˆ2 + 2*x0*x1ˆ2 + 2*x0ˆ2 + x0,x0ˆ2*x1ˆ2 + x0ˆ2*x1 + 2*x1ˆ2 + 2*x1,x0ˆ2*x1ˆ2 + x0ˆ2*x1 + x0*x1ˆ2 + x0*x1,x0ˆ2*x1ˆ2 + x0ˆ2*x1 + 2*x0*x1ˆ2 + 2*x0*x1,x0ˆ2*x1ˆ2 + 2*x0ˆ2*x1 + 2*x1ˆ2 + x1,x0ˆ2*x1ˆ2 + 2*x0ˆ2*x1 + x0*x1ˆ2 + 2*x0*x1,x0ˆ2*x1ˆ2 + 2*x0ˆ2*x1 + 2*x0*x1ˆ2 + x0*x1]sage: f = x0ˆ2*x1ˆ2 + x0ˆ2*x1 + x0*x1ˆ2 + x0*x1sage: [f(x0=v[0],x1=v[1]) for v in V][0, 0, 0, 0, 1, 0, 0, 0, 0] Proposition 104. (Hou)
The degree of any bent function f : GF ( p ) n → GF ( p ), when represented in ANF, satisfiesdeg( f ) ≤ n ( p − . The degree of any weakly regular bent function f : GF ( p ) n → GF ( p ), whenrepresented in ANF, satisfiesdeg( f ) ≤ n ( p − . For a proof of these results, see Hou [H] (and see also [CM] for furtherdetails). GF (3) → GF (3) We focus on examples of even functions GF (3) → GF (3) sending 0 to 0.There are exactly 3 = 81 such functions. Example 105.
Here is an example of a bent function f of two variables over GF (3). This function f is defined by the following table of values: GF (3) (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1) (0, 2) (1, 2) (2, 2) f age sage: V = GF(3)ˆ2sage: Vlist = V.list()sage: Vlist[(0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2)]sage: f00 = 0; f10 = 1; f01 = 1; f11 = 2; f12 = 2sage: flist = [f00,f10,f10,f01,f11,f12,f01,f12,f11]sage: f = lambda x: GF(3)(flist[Vlist.index(x)])sage: [CC(hadamard_transform(f,a)).abs() for a in V][3.00000000000000, 3.00000000000000, 3.00000000000000,3.00000000000000, 3.00000000000000, 3.00000000000000,3.00000000000000, 3.00000000000000, 3.00000000000000]sage: pts = [CC(hadamard_transform(f, a)) for a in V]sage: t = var(’t’)sage: P1 = points([(x.real(), x.imag()) for x in pts],pointsize=40, xmin=-12,xmax=12,ymin=-12,ymax=12)sage: P2 = parametric_plot([(3)*cos(t),(3)*sin(t)], (t,0,2*pi), linestyle = "--")sage: (P1+P2).show() The plot of the values of the Hadamard transform (2) of f is in Figure 5.The set of such functions has some amusing combinatorial properties weshall discuss below.There are exactly 18 such bent functions. GF (3) (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1) (0, 2) (1, 2) (2, 2) b b b b b b b b b b b b b b b b b b The unweighted
Cayley graph of b (as well as b , b , b , b , b , b , b , b , b , b and b ) is a strongly regular graph having parameters SRG ( ν, k, λ, µ )where ν = 9, k = 4, λ = 1, µ = 2. We say that these bent functionsare of type (9 , , , type (9 , , , f in thecomplex plane of the even GF (3)-valued bent function of two variables fromExample 105. (The vertices are ordered as in Example 105.)Up to isomorphism, there is only one (unweighted) strongly regular graphhaving parameters SRG (9 , , ,
2) [Br], [Sp]. We shall see later that the edge-weighted Cayley graphs arising from these 12 bent functions of type (9 , , , as weighted (strongly regular) graphs. Likewise, these6 bent functions of type (9 , , ,
0) are also isomorphic as weighted (stronglyregular) graphs.
Example 106.
Let b , . . . , b denote the bent functions defined in § b is b and the dual of b is b ,but in one case we must pre-multiply by − Sage sage: FF = GF(3)sage: V = FFˆ2 We say edge-weighted graphs are isomorphic if there is a bijection of the verticeswhich preserves the weight of each edge. age: Vlist = V.list()sage: flist = [0, 1, 1, 1, 2, 2, 1, 2, 2] Note the pre-multiplication by −
1. This bent function f = b : GF (3) → GF (3) is weakly regular, and the weakly regular dual of b is b . Sage sage: flist = [0, 2, 2, 1, 0, 0, 1, 0, 0]
This bent function f = b : GF (3) → GF (3) is regular, and the regulardual of b is b . 79e similar computations verify the following: • b and b are both weakly regular and − • b and b are regular and dual to each other, • b and b are regular and dual to each other, • b and b are regular and dual to each other, • b and b are regular and dual to each other, • b and b are regular and dual to each other, • b and b are regular and dual to each other, • b , b , b and b are all weakly regular and are each − b = − b , b = − b , b = − b , b = − b , b = − b ,b = − b , b = − b , b = − b , b = − b ,b = b + b = b + b , b = b + b = b + b , b = b + b = b + b ,b = b + b = b + b , b = b + b = b + b , b = b + b = b + b . Recall from Example 106, the following are regular b ∗ = b , b ∗ = b , b ∗ = b , b ∗ = b , b ∗ = b , b ∗ = b , whereas b ∗ = − b are weakly regular and ( − − b ∗ = − b , b ∗ = − b , b ∗ = − b , b ∗ = − b . Supports: { , , , } = supp( b ) = supp( b ) , { , , , } = supp( b ) = supp( b ) , , , , } = supp( b ) = supp( b ) , { , , , } = supp( b ) = supp( b ) , { , , , } = supp( b ) = supp( b ) , { , , , } = supp( b ) = supp( b ) , andsupp( b ) = supp( b ) = supp( b ) = supp( b ) = supp( b ) = supp( b )are all equal to { , , , , , , , } . Note that • All these functions are weight 4 or weight 8. • If you pick any two support sets of weight 4, S and S say, then theysatisfy either S ∩ S = ∅ or | S ∩ S | = 2 . • The 12 which are regular, but not µ -regular for some µ (cid:54) = 1) can allbe obtained from f ( x , x ) = x + x x by linear transformations ofthe coordinates, i.e. ( x , x ) (cid:55)→ ( ax + bx , cx + dx ) where ad − bc (cid:54) =0. Each such isomorphism of GF (3) induces an isomorphism of theassociated edge-weighted Cayley graphs. • Similarly, the 6 which are weakly regular can all be obtained from x + x by linear transformations of the coordinates.In fact, if you consider the set S = {∅} ∪ { supp( f ) | f : GF (3) → GF (3) , f (0) = 0 } , then S forms a group under the symmetric difference operator ∆. In fact, S ∼ = GF (2) . Question 1.
To what extent is it true that if f , f are bent functions on GF ( p ) n with f (0) = f (0) = 0 , then supp( f ) ∆ supp( f ) = supp( f ) for some bent function f satisfying f (0) = 0 ? More general version: 81 uestion 2.
Over GF ( p ) , p (cid:54) = 2 , does the set of supports {∅} ∪ { supp( f ) | f is bent , f (0) = 0 , f is even } form a group (under ∆ )? This does not seem to hold in the binary case .The following was verified with direct (computer-aided) computations. Lemma 107.
Assume p = 3 , n = 2 .(a) The edge-weighted Cayley graph of b i is strongly regular and not com-plete as a simple (unweighted) graph if and only if b i is regular if andonly if i ∈ { , , , , , , , , , , , } . (b) The edge-weighted Cayley graph of b i is strongly regular and completeas a simple (unweighted) graph if and only if b i is weakly regular if andonly if i / ∈ { , , , , , , , , , , , } . Example 108.
This example is intended to illustrate the bent function b listed in the table above, and to provide more detail on Example 26.Consider the finite field GF (9) = GF (3)[ x ] / ( x + 1) = { , , , x, x + 1 , x + 2 , x, x + 1 , x + 2 } . The set of non-zero quadratic residues is given by D = { , , x, x } . Let Γ be the graph whose vertices are GF (9) and whose edges e = ( a, b ) arethose pairs for which a − b ∈ D .The graph looks like the Cayley graph for b in Figure 6, except8 → , → x + 2 , → x + 1 , → x, → x + 2 , → x + 1 , → x, → , → , → . This is a strongly regular graph with parameters (9 , , , What is true in the binary case is an oddly similar result: the vectors in the supportof a bent function form a Hadamard difference set in the additive group GF (2) n . b v N ( v,
0) 3,4,6,8 4,5,6,7 3,5,7,8 0,2,6,7 0,1,7,8 1,2,6,8 0,1,3,5 1,2,3,4 0,2,4,5 N ( v,
1) 5,7 3,8 4,6 1,8 2,6 0,7 2,4 0,5 1,3 N ( v,
2) 1,2 0,2 0,1 4,5 3,5 3,4 7,8 6,8 6,7
The axioms of an edge-weighted strongly regular graph can be directly veri-fied using this table. GF (3) → GF (3) Let f : GF (3) → GF (3) be an even bent function with f (0) = 0, let D i = { v ∈ GF (3) | f ( v ) = i } , i = 1 , , let D = { } and D = GF (3) − ( D ∪ D ∪ D ).There are a total of 2340 even bent functions on GF (3) .We know that if W f (0) is rational then the level curves f − ( i ), i (cid:54) = 0,have the same cardinality . The following Sage computation shows that,in this case, W f (0) is not a rational number for therepresentatives of B , B displayed above. Sage sage: PR.
Since W f (0) / / is a 6-th root of unity, but not a cube root of unity, itfollows from Lemma 11 that f is not weakly regular. Sage sage: a = 2; b = 4; c = 4sage: flist = [0,a,a,c,b,1,c,1,b,2,0,1,1,2,0,1,0,0,2,1,0,1,0,0,1,0,2]sage: f = lambda x: GF(3)(flist[Vlist.index(x)])sage: [f(a)- f(-a) for a in V][0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]sage: [CC(hadamard_transform(f,a)).abs() for a in V][5.19615242270663, 5.19615242270663, 5.19615242270664, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270664, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270664, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270663, 5.19615242270663,5.19615242270663, 5.19615242270663, 5.19615242270664] age: supp_f = [Vlist.index(x) for x in V if f(x)<>0]; supp_f[1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 18, 19, 21, 24, 26] Here are some algebraic examples:
Sage sage: R.
Note the first example is even but the second one is not. GF (5) → GF (5) Using Sage, we give examples of bent functions of 2 variables over GF (5)and study their signatures (14).Do the “level curves” of a bent function GF (5) → GF (5) give rise to aPDS? An association scheme (see below for a definition)? Example 109.
Let G = GF (25) = GF (5)[ x ] / ( x + 2), D = { , , x + 2 , x + 3 } , D = { x + 1 , x + 3 , x + 2 , x + 4 } , = { x + 1 , x + 2 , x + 3 , x + 4 } , D = { , , x + 4 , x + 1 } , and D = D ∪ D ∪ D ∪ D . If f ( x , x ) = x + x x then each subset D i ( i = 1 , , ,
4) is the image of the level curve f − ( i ) under the GF (5)-vectorspace isomorphism φ : GF (5) → GF (25) , ( a, b ) (cid:55)−→ bx + a ,D i = φ ( f − ( i )), i = 1 , , ,
4. As in the previous example, we can com-pute the k i,j ’s, µ i,j ’s, and λ ki,j ’s (see Example 94 for the details). This f ishomogeneous, bent and regular (hence also weakly regular).The weighted adjacency matrix A of the edge-weighted Cayley graphassociated to f , Γ f , is given below: 86 This is verified using the following Sage commands:
Sage sage: attach "/home/wdj/sagefiles/hadamard_transform.sage"sage: FF = GF(5)sage: V = FFˆ2sage: R.
Next, we consider f ( x , x ) = x + x , having signature [9 , , , , Sage sage: ff = x0ˆ2+x1ˆ2sage: flist = [FF(x) for x in flist]sage: f = lambda x: FF(flist[Vlist.index(x)])sage: ff = x0ˆ2+x1ˆ2sage: flist = [ff(x0=v[0],x1=v[1]) for v in V]sage: f = lambda x: FF(flist[Vlist.index(x)])sage: flist = [FF(x) for x in flist]sage: [CC(hadamard_transform(f,a)).abs() for a in V][5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000]sage: flist[0, 1, 4, 4, 1, 1, 2, 0, 0, 2, 4, 0, 3, 3, 0, 4, 0, 3, 3, 0, 1, 2, 0, 0, 2]sage: fcount = [flist.count(x) for x in FF]; fcount[9, 4, 4, 4, 4] f ( x , x ) = x +2 x + x , having signature [4 , , , , Sage sage: ff = x0ˆ2-x1ˆ2+x0sage: flist = [ff(x0=v[0],x1=v[1]) for v in V]sage: f = lambda x: FF(flist[Vlist.index(x)])sage: [CC(hadamard_transform(f,a)).abs() for a in V][5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000, 5.00000000000000, 5.00000000000000,5.00000000000000]sage: flist = [FF(x) for x in flist]; flist[0, 2, 1, 2, 0, 4, 1, 0, 1, 4, 1, 3, 2, 3, 1, 1, 3, 2, 3, 1, 4, 1, 0, 1, 4]sage: fcount = [flist.count(x) for x in FF]; fcount[4, 9, 4, 4, 4]
Example 110.
Consider the bent function f ( x , x ) = − x − x + x x . This function represents a GL (2 , GF (5)) orbit of size 120. The level curvesof this function do not give rise to a weighted PDS. By the way, if we try acomputation of all the p kij ’s as in the above example, we do not get integers.Similarly, the level curves of f , f , f , and f do not give rise to aweighted PDS. Example 111.
The example of f above can be used to construct an edge-weighted strongly regular Cayley graph, hence also a weighted PDS attachedto its level curves. The examples of f and f above have an isomorphicweighted PDS attached to their (respective) level curves. Acknowledgements : We are grateful to our colleague T. S. Michael for manystimulating conversations and suggestions on this paper.89
Appendix: A new search algorithm for bentfunctions
The following algorithm and code is due to the first author C. Celerier.
Pythonfrom collections import defaultdictfrom copy import deepcopyfrom random import shufflefrom sage.crypto.boolean_function import BooleanFunctionclass NoBentFunction(Exception):passclass BentFinder(object):def __init__(self, n):self.V = GF(2)**nself.n = ndef searchForBent(self):self.walshTrace = []A = defaultdict(int)W = defaultdict(int)B = list(self.V.list())shuffle(B)R = self.__searchForBent(A,B,W,0)self.walshTrace.reverse()return BooleanFunction([R[tuple(x)] for x in self.V]), self.walshTracedef __searchForBent(self, A, B, W, wgt):n = self.nif len(A) == 2**n:return Aif wgt > 2**(n-1)-2**(n/2-1):raise NoBentFunctionA,B,W,wgt = self.__deepcopy(A,B,W,wgt)v = B.pop()cf = self.__coinFlip()values = (cf, (1+cf) % 2)for a in values:try:A[tuple(v)] = aupdate = self.__getUpdate(v, a)W_a = self.__applyUpdate(W, update, 2**n-len(A))R = self.__searchForBent(A,B,W_a,wgt+a)self.walshTrace.append((v, A[tuple(v)], [W_a[x] for x in W_a]))return Rexcept NoBentFunction:passraise NoBentFunctiondef __applyUpdate(self, W, update, leftToFill):n = self.nW = deepcopy(W)for u in self.V:W[tuple(u)] += update[tuple(u)]Wmin = W[tuple(u)] - leftToFillWmax = W[tuple(u)] + leftToFillif not ((Wmin <= 2**(n/2) and 2**(n/2) <= Wmax) or (Wmin <= -2**(n/2) and -2**(n/2) <= Wmax)):raise NoBentFunctionreturn Wdef __getUpdate(self,v,a):update = {}for u in self.V:update[tuple(u)] = Integer(-1)**(a+u.dot_product(v))return updatedef __deepcopy(self, *args):R = []for a in args: .append(deepcopy(a))return Rdef __coinFlip(self):return 1 if random() > .5 else 0 An example:
Sage sage: %attach bentFunctions.sagesage: B=BentFinder(4)sage: B.searchForBent()(Boolean function with 4 variables,[((0, 0, 0, 1), 1, [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]),((0, 1, 0, 0), 0, [2, 2, 0, 0, 0, 0, 2, 2, 2, 0, 2, 0, 0, 2, 0, 2]),((1, 1, 0, 1), 1, [1, 3, 1, 1, -1, -1, 3, 3, 3, 1, 1, 1, -1, 1, -1, 1]),((0, 1, 1, 1), 0, [2, 2, 2, 2, 0, -2, 4, 4, 2, 0, 2, 0, 0, 0, -2, 0]),((0, 0, 1, 0), 1, [3, 3, 1, 3, 1, -3, 3, 5, 1, -1, 1, 1, -1, 1, -1, -1]),((0, 0, 1, 1), 0, [2, 4, 2, 2, 0, -4, 2, 6, 0, 0, 2, 2, -2, 0, 0, 0]),((0, 1, 0, 1), 0, [1, 5, -1, 3, 1, -3, 1, 7, -1, -1, 3, 1, -1, 1, -1, 1]),((0, 1, 1, 0), 0, [2, 4, 4, 2, 2, -4, 0, 6, -2, 0, 4, 0, -2, 0, -2, 2]),((1, 1, 0, 0), 1, [1, 3, -1, 5, 1, -5, 1, 5, -1, -1, 5, 3, -3, 1, -1, 3]),((1, 0, 1, 1), 0, [0, 4, 6, 2, 2, -6, 0, 4, 0, -2, 4, 2, -2, 0, -2, 4]),((1, 0, 0, 0), 1, [1, 3, -1, 5, 1, -7, -1, 5, 1, -3, 3, 3, -1, 3, -1, 5]),((1, 1, 1, 0), 0, [2, 4, 6, 4, 2, -6, -2, 4, 2, -4, 4, 0, -2, -2, 0, 4]),((1, 0, 0, 1), 0, [3, 5, -3, 5, 1, -5, -1, 3, 1, -5, 3, 3, -3, 5, -1, 5]),((1, 1, 1, 1), 0, [4, 6, 4, 4, 4, -6, -2, 2, 2, -4, 4, 2, -2, -4, -2, 4]),((1, 0, 1, 0), 1, [3, 5, -3, 3, 3, -5, -3, 3, 3, -5, 5, 5, -3, 5, -3, 3]),((1, 0, 1, 0), 1, [4, 4, 4, 4, 4, -4, -4, 4, 4, -4, 4, 4, -4, -4, -4, 4])])
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