Anticoncentration versus the number of subset sums
aa r X i v : . [ m a t h . C O ] J a n ANTICONCENTRATION VERSUS THE NUMBER OF SUBSET SUMS
VISHESH JAIN, ASHWIN SAH, AND MEHTAAB SAWHNEY
Abstract.
Let ~w = ( w , . . . , w n ) ∈ R n . We show that for any n − ≤ ǫ ≤ , if { ~ξ ∈ { , } n : h ~ξ, ~w i = r } ≥ − ǫn · n for some r ∈ R , then {h ~ξ, ~w i : ~ξ ∈ { , } n } ≤ O ( √ ǫn ) . This exponentially improves a recent result of Nederlof, Pawlewicz, Swennenhuis, and Węgrzyckiand leads to a similar improvement in the parameterized (by the number of bins) runtime of binpacking. Introduction
For ~w := ( w , . . . , w n ) ∈ R n and a real random variable ξ , recall that the Lévy concentrationfunction of ~w with respect to ξ is defined for all r ≥ by L ξ ( ~w, r ) = sup τ ∈ R P [ | w ξ + · · · + w n ξ n − τ | ≤ r ] , where ξ , . . . , ξ n are i.i.d. copies of ξ . In combinatorial settings (where ~w ∈ Z n ) a particularlynatural and interesting case is when r = 0 and ξ is a Bernoulli random variable, i.e., ξ = 0 withprobability / and ξ = 1 with probability / . For lightness of notation, we will denote this specialcase by ρ ( ~w ) = L Rad ( ~w,
0) = sup τ ∈ R P [ h ~w, ~ξ i = τ ] . In this note, we study the following question.
Question 1.1.
For a vector ~w = ( w , . . . , w n ) ∈ R n with ρ ( ~w ) ≥ ρ , how large can the range R ( ~w ) = { w ξ + · · · + w n ξ n : ξ i ∈ { , }} be?The two extremal examples here are ~w = (0 , , . . . , , which corresponds to ρ ( ~w ) = 1 , |R ( ~w ) | = 1 and ~w = (1 , , . . . , n − ) , which corresponds to ρ ( ~w ) = 2 − n , |R ( ~w ) | = 2 n . Motivated by theseexamples, one may ask if there is a smooth trade-off between ρ ( ~w ) and |R ( ~w ) | . This turns outnot to be the case. Indeed, for any ǫ > , Wiman [6] gives an example of a ~w ∈ Z n for which |R ( ~w ) | ≥ (1 − ǫ ) n and ρ ( ~w ) ≥ − . n . At the other end of the spectrum, when ρ ( ~w ) ≥ − ǫn , theso-called inverse Littlewood–Offord theory [4,5] heuristically suggests that ~w is essentially containedin a low-rank generalized arithmetic progression of ‘small’ volume so that |R ( ~w ) | is also ‘small’.However, the number of ‘exceptional elements’ in the inverse Littlewood–Offord theorems (cf. [3])is unfortunately too large to be able to rigorously establish such a statement.Nevertheless, in a recent work on the parameterized complexity of the bin packing problem (seeSection 1.1), Nederlof, Pawlewics, Swennenhuis and Węgrzycki [2] showed that for any ǫ > , ρ ( ~w ) ≥ − ǫn = ⇒ | R ( ~w ) | ≤ δ ( ǫ ) n , here δ ( ǫ ) = O log log( ǫ − ) p log( ǫ − ) ! . (1.1)In particular, δ ( ǫ ) → as ǫ → . Moreover, we must have δ ( ǫ ) ≥ (2 − o (1)) ǫ , as can be seen byconsidering ~w = ( C , . . . , C , C , . . . , C , . . . , C n/k , . . . , C n/k ) ∈ R n , where each C i is repeated k times, and C i is sufficiently small compared to C i +1 for all i . Indeed,for such ~w , we have ρ ( ~w ) = 2 − ( + o k (1)) nk log k while |R ( ~w ) | ≤ (1+ o k (1)) nk log k .We conjecture that this example is essentially the worst possible, so that δ ( ǫ ) ≤ ǫ . We are ableto show that δ ( ǫ ) = O ( √ ǫ ) , (1.2)thereby obtaining an exponential improvement over (1.1). More precisely, Theorem 1.2.
Let ǫ > . For any n ≥ ǫ − / and any ~w ∈ R n satisfying ρ ( ~w ) ≥ exp( − ǫn ) , we have |R ( ~w ) | ≤ exp( C . ǫ / n ) , where C . is an absolute constant. We prove this theorem in Section 2.1.1.
Application to bin packing.
The bin packing problem is a classic NP-complete problemwhose decision version may be stated as follows: given n items with weights w , . . . , w n ∈ [0 , and m bins, each of capacity , is there a way to assign the items to the bins without violating thecapacity constraints? Formally, is there a map f : [ n ] → [ m ] such that P i ∈ f − ( j ) w i ≤ for all j ∈ [ m ] ?Björklund, Husfeldt, and Koivisto [1] provided an algorithm for solving bin packing in time ˜ O (2 n ) where the tilde hides polynomial factors in n . It is natural to ask whether the base of the exponentmay be improved at all i.e. is there a (possibly randomized) algorithm to solve bin packing in time ˜ O (2 (1 − ǫ ) n ) for some absolute constant ǫ > ?In recent work, Nederlof, Pawlewics, Swennenhuis and Węgrzycki [2] showed that this is trueprovided that the number of bins m is fixed. More precisely, they showed that there exists afunction σ : N → R > and an algorithm for solving bin packing which, on instances with m bins,runs in time ˜ O (2 (1 − σ ( m )) n ) , where ˜ O hides polynomials in n as well as exponential factors in m .Their analysis, which crucially relies on (1.1), gives a very small value of σ ( m ) satisfying σ ( m ) ≤ − /m . (1.3)Using Theorem 1.2 instead of (1.1) in a black-box manner in the analysis of [2], we exponentiallyimprove the bound on σ ( m ) . Corollary 1.3.
With notation as above, the algorithm of [2] solves bin packing instances with m bins in time ˜ O (2 (1 − σ ( m )) n ) for σ : N → R > satisfying σ ( m ) = ˜Ω( m − ) , (1.4) where ˜Ω hides logarithmic factors in m . We remark that the conjecturally optimal bound δ = O ( ǫ ) , plugged into the analysis of [2], wouldlead to σ ( m ) = ˜Ω( m − ) . .2. Notation.
We use
Ber(1 / to denote the balanced { , } Bernoulli distribution and
Bin( k ) todenote the binomial distribution on k trials with parameter / . Recall that Bin( k ) is the sum of k independent Ber(1 / random variables. We also use the following standard additive combinatoricsnotation: C + D = { c + d : c ∈ C, d ∈ D } is the sumset (if C, D are subsets of the same abeliangroup), and for a positive integer k , k · C = C + · · · + C ( k times) is the iterated sumset.2. Proof of Theorem 1.2
We begin by recording the following key comparison bound, which will be proved at the end ofthis section.
Lemma 2.1.
Let n ≥ k ≥ C . , where C . is an absolute constant and let δ > . For any A ⊆ { , } n with | A | ≤ exp( δn ) , the following holds. Let ~x,~b ∼ Bin( k ) ⊗ n be independent n -dimensional random vectors. Then, E ~x (cid:20) sup ~a ∈ A P ~b [ ~b = ~x − ~a ] P ~b [ ~b = ~x ] (cid:21) ≤ exp C . k + r δk ! n ! . Let ~w be as in Theorem 1.2. Let τ be such that P [ h ~w, ~ξ i = τ ] = ρ ( ~w ) , where ~ξ is a random vectorwith i.i.d. Ber(1 / components. Let B = { ~ξ ∈ { , } n : h ~w, ~ξ i = τ } . In particular, | B | ≥ exp( − ǫn ) · n . Let |R ( ~w ) | = exp( δn ) . For each r ∈ R ( ~w ) , let ~ξ ( r ) be a fixed(but otherwise arbitrary) element of { , } n such that h ~w, ~ξ ( r ) i = r . Let A = { ~ξ ( r ) ∈ { , } n : r ∈ R ( ~w ) } . Note that, by definition, for any distinct ~a , ~a ∈ A , we have that h ~w, ~a i 6 = h ~w, ~a i and that | A | = |R ( ~w ) | = exp( δn ) .We will make use of the simple, but crucial, observation from [2] that A and k · B have a fullsumset for all k ≥ . Lemma 2.2 ([2, Lemma 4.2]) . The map ( ~a, ~c ) ~a + ~c from A × ( k · B ) to A + k · B is injective.Proof. Indeed, if ~a + ( ~b (1)1 + · · · + ~b (1) k ) = ~a + ( ~b (2)1 + · · · + ~b (2) k ), where ~a i ∈ A and ~b ( i ) j ∈ B , thentaking the inner product of both sides with ~w and using h ~w,~b i = τ for all b ∈ B , we see that h ~w, ~a i = h ~w, ~a i , which implies that ~a = ~a by the definition of A . (cid:3) We are now ready to prove Theorem 1.2.
Proof of Theorem 1.2.
Let k ≥ be a parameter which will be chosen later depending on ǫ . ByLemma 2.2, for each ~x ∈ { , . . . , k + 1 } n for which there exist ~a ∈ A and ~c ∈ k · B with ~a + ~c = ~x ,there exists a unique such choice ~a = ~a ( ~x ) ∈ A .Now, let ~a be uniform on A , let ~b , . . . ,~b k be uniform on B , and let ~v , . . . , ~v k be uniform on { , } n . Let C i ⊆ { , . . . , k + 1 } n be the set of vectors with i coordinates equal to k + 1 . For ~x ∈ { , . . . , k + 1 } n , we let ~x ∗ ∈ { , . . . , k } n denote the vector obtained by setting every occurrenceof k + 1 in ~x to k . We have P [ ~a + ~b + · · · + ~b k ∈ { , . . . , k + 1 } n ]= n X i =0 X ~x ∈ C i P [ ~a + ~b + · · · + ~b k = ~x ] n X i =0 X ~x ∈ C i P [ ~a = ~a ( ~x )] P [ ~b + · · · + ~b k = ~x − ~a ( ~x )] ≤ | A | n X i =0 X ~x ∈ C i (cid:18) n | B | (cid:19) k P [ ~v + · · · + ~v k = ~x − ~a ( ~x )] ≤ e kǫn | A | n X i =0 X ~x ∈ C i P [ ~v + · · · + ~v k = ~x ∗ ] sup ~a ∈ A P [ ~v + · · · + ~v k = ~x − ~a ] P [ ~v + · · · + ~v k = ~x ∗ ]= e kǫn | A | n X i =0 (1 / k ) i X S ∈ ( [ n ] i ) E ~x ∼ Bin( k ) ⊗ ([ n ] \ S ) ×{ k +1 } S (cid:20) sup ~a ∈ A P [ ~v + · · · + ~v k = ~x − ~a ] P [ ~v + · · · + ~v k = ~x ∗ ] (cid:21) . Let A S be the set of elements in A ⊆ { , } n whose support contains S . Let A ′ S = { ~a ′ ∈ { , } [ n ] \ S : ∃ ~a ∈ A S with ~a | [ n ] \ S = ~a ′ } . Recall that | A | = exp( δn ) . Abusing notation so that the supremum of an empty set is , cancontinue the above chain of inequalities to get that ≤ e kǫn | A | n X i =0 (1 / k ) i X S ∈ ( [ n ] i ) E ~x ∼ Bin( k ) ⊗ ([ n ] \ S ) ×{ k +1 } S (cid:20) sup ~a ∈ A P [ ~v + · · · + ~v k = ~x − ~a ] P [ ~v + · · · + ~v k = ~x ∗ ] (cid:21) = e kǫn | A | n X i =0 (1 / k ) i X S ∈ ( [ n ] i ) E ~x ∼ Bin( k ) ⊗ ([ n ] \ S ) (cid:20) sup ~a ∈ A S P [ ~v + · · · + ~v k = ~x − ~a × { } S ] P [ ~v + · · · + ~v k = ~x × { k } S ] (cid:21) = e kǫn | A | n X i =0 (1 / k ) i X S ∈ ( [ n ] i ) E ~x ∼ Bin( k ) ⊗ ([ n ] \ S ) (cid:20) sup ~a ∈ A S P [ ~v + · · · + ~v k = ~x − ~a ] P [ ~v + · · · + ~v k = ~x ] (cid:21) ≤ e kǫn | A | n/ X i =0 · + n X i = n/ − ki · n · max ℓ (cid:0) kℓ ± (cid:1)(cid:0) kℓ (cid:1) ! i ≤ e kǫn | A | n/ X i =0 · + n · − kn/ · n · k n ≤ e kǫn | A | (cid:18) n/ X i =0 − ki exp( C . k − + δ / k − / ) n ) + 2 − kn/ (cid:19) ≤ exp( − δn ) exp (cid:16) O ( kǫ + k − + δ / k − / ) n (cid:17) by Lemma 2.1 applied to A S , as long as n/ ≥ k ≥ C . ≥ . Hence, δ ≤ C ( kǫ + k − + δ / k − / ) for some absolute constant C > . Now letting k = ǫ − / / (note that this satisfies k = ǫ − / ≤ n ),we find that δ = O ( ǫ / ) , as desired. (cid:3) The proof of Lemma 2.1 relies on the following preliminary estimate. emma 2.3. If ≤ s ≤ k/ (16 π ) , then E x ∼ Bin( k ) (cid:18) xk + 1 − x (cid:19) s ≤ exp(10 πs /k ) + 2 k s (4 / k . Proof.
We let x ∼ Bin( k ) and y = x − k/ ∼ Bin( k ) − k/ throughout. We let z ∼ N (0 , kπ/ . Wehave E x ∼ Bin( k ) (cid:18) xk + 1 − x (cid:19) s = E y (cid:18) y − k/ − y (cid:19) s ≤ E y (cid:18) yk/ − y (cid:19) s | y |≤ k/ + k s P [ | y | ≥ k/ ≤ E y (cid:18) yk/ − y (cid:19) s | y |≤ k/ + 2 k s (4 / k . Since for | y | ≤ k/ , y ( k/ − y ) ≤ yk/ y ( k/ , and using (1 + x ) ≤ exp( x ) . we can continue the previous inequality as ≤ E y (cid:18) yk/ y ( k/ (cid:19) s | y |≤ k/ + 2 k s (4 / k ≤ E y exp (cid:18) syk + 2 + 32 sy k (cid:19) + 2 k s (4 / k . Now, let z , . . . , z k be i.i.d. N (0 , random variables. Then, y ∼
12 (sgn z + · · · + sgn z k ) . Moreover, for any − k ≤ ℓ ≤ k , E [ z + · · · + z k | sgn( z ) + · · · + sgn( z k ) = ℓ ] = r π ℓ. In particular, under this coupling of y, z , . . . , z k , we have E [ z + · · · + z k | y ] = r π y. Let z = z + · · · + z k , so that z ∼ N (0 , k ) . Then, by the convexity of f ( r ) = exp (cid:18) syk + 2 + 32 sy k (cid:19) , and using Jensen’s inequality, we have E y f ( y ) = E y,z ,...,z k f ( y )= E y,z ,...,z k f (cid:18)r π E [ z | y ] (cid:19) ≤ E z f ( √ πz/ √ E w ∼N (0 , exp (cid:18) s √ kπk + 2 w + 4 sπk w (cid:19) = (cid:18) − πsk (cid:19) − / exp (cid:18) πs k ( k + 2) ( k − πs ) (cid:19) exp (cid:18) πsk + 2 πs k (cid:19) ≤ exp(10 πs /k ) . (cid:3) Finally, we can prove Lemma 2.1
Proof of Lemma 2.1.
We may assume that δ ≥ /k since the statement for δ < /k followsfrom the statement for δ = 2000 /k . Also, note that δ ≤ log 2 . For any t ∈ R , we have P ~x (cid:20) sup ~a ∈ A P ~b [ ~b = ~x − ~a ] P ~b [ ~b = ~x ] ≥ e tn (cid:21) ≤ | A | sup ~a ∈ A P ~x (cid:20) P ~b [ ~b = ~x − ~a ] P ~b [ ~b = ~x ] ≥ e tn (cid:21) ≤ | A | sup ~a ∈ A inf s ≥ exp( − stn ) E ~x (cid:20)(cid:18) P ~b [ ~b = ~x − ~a ] P ~b [ ~b = ~x ] (cid:19) s (cid:21) = | A | sup ~a ∈ A inf s ≥ exp( − stn ) n Y i =1 E x ∼ Bin( k ) (cid:20)(cid:18) P [Bin( k ) = x − a i ] P [Bin( k ) = x ] (cid:19) s (cid:21) ≤ | A | inf s ≥ exp( − stn ) (cid:18) E x ∼ Bin( k ) (cid:18) xk + 1 − x (cid:19) s (cid:19) n . In the last line, we have used that E x ∼ Bin( k ) (cid:20)(cid:18) xk + 1 − x (cid:19) s (cid:21) ≥ (cid:18) E x ∼ Bin( k ) (cid:20) x ( k + 1 − x ) (cid:21)(cid:19) s/ = (cid:18) k − X ℓ =0 ℓ + 1 k − ℓ (cid:18) kℓ (cid:19) − k (cid:19) s/ = (cid:18) k − X ℓ =0 (cid:18) k + 2 k + 4( k + 1)( ℓ − k/ k + ( k + 1)( k − ℓ ) k ( k − ℓ ) (cid:19)(cid:18) kℓ (cid:19) − k (cid:19) s/ ≥ (cid:18) k − X ℓ =0 (cid:18) k + 2 k + 4( k + 1)( ℓ − k/ k (cid:19)(cid:18) kℓ (cid:19) − k (cid:19) s/ = (cid:18) k + 2 k − k + 4 k − k (cid:19) s/ ≥ if k ≥ . Therefore, by Lemma 2.3, we have P ~x (cid:20) sup ~a ∈ A P ~b [ ~b = ~x − ~a ] P ~b [ ~b = ~x ] ≥ e tn (cid:21) ≤ | A | inf s ≥ exp( − stn ) (cid:18) E x ∼ Bin( k ) (cid:18) xk + 1 − x (cid:19) s (cid:19) n ≤ | A | inf ≤ s ≤ k/ (16 π ) exp( − stn ) (cid:18) exp(10 πs /k ) + 2 k s (4 / k (cid:19) n ≤ | A | inf ≤ s ≤ k/ (10 log k ) exp( − stn ) (cid:18) exp(12 πs /k ) (cid:19) n ≤ | A | exp (cid:16) − kt n π (cid:17) if q πδk ≤ t ≤ (log k ) − | A | exp (cid:16) − kn π (log k ) (cid:17) if (log k ) − ≤ t ≤ log k. ere, the second case follows by plugging in s = k/ (10 log k ) and simplifying, and the first casefollows from plugging in s = kt/ π which satisfies ≤ s ≤ k/
10 log k by the restriction on t and δ .Finally, since ≤ sup ~a ∈ A P ~b [ ~b = ~x − ~a ] P ~b [ ~b = ~x ] ≤ max ℓ (cid:0) kℓ ± (cid:1)(cid:0) kℓ (cid:1) ! n ≤ k n , we have E ~x (cid:20) sup ~a ∈ A P ~b [ ~b = ~x − ~a ] P ~b [ ~b = ~x ] (cid:21) = Z log k −∞ P (cid:20) sup ~a ∈ A P ~b [ ~b = ~x − ~a ] P ~b [ ~b = ~x ] ≥ e tn (cid:21) ne tn dt ≤ Z log k / log k · + Z / log k √ πδ/k · + Z √ πδ/k −∞ ne tn dt ≤ e √ πδ/kn + Z / log k √ πδ/k | A | exp (cid:18) − kt n π (cid:19) ne tn dt + Z log k / log k | A | exp (cid:18) − kn π (log k ) (cid:19) ne tn dt ≤ exp (cid:16) O ( p δ/k ) n (cid:17) + Z / log k √ πδ/k ne − tn dt + 1 ≤ exp (cid:16) O ( p δ/k ) n (cid:17) . (cid:3) References [1] Andreas Björklund, Thore Husfeldt, and Mikko Koivisto,
Set partitioning via inclusion-exclusion , SIAM Journalon Computing (2009), 546–563.[2] Jesper Nederlof, Jakub Pawlewicz, Céline M. F. Swennenhuis, and Karol Węgrzycki, A Faster Exponential TimeAlgorithm for Bin Packing with a Constant Number of Bins via Additive Combinatorics , arXiv:2007.08204.[3] Hoi H. Nguyen and Van H. Vu,
Small ball probability, inverse theorems, and applications , Erdős Centennial,Springer, 2013, pp. 409–463.[4] Mark Rudelson and Roman Vershynin,
The Littlewood–Offord problem and invertibility of random matrices , Ad-vances in Mathematics (2008), 600–633.[5] Terence Tao and Van H Vu,
Inverse Littlewood–Offord theorems and the condition number of random discretematrices , Annals of Mathematics (2009), 595–632.[6] Mårten Wiman,
Improved constructions of unbalanced uniquely decodable code pairs , 2017.
Department of Statistics, Stanford University, Stanford CA 94305, USA
Email address : [email protected] Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139, USA
Email address : {asah,msawhney}@mit.edu{asah,msawhney}@mit.edu