aa r X i v : . [ m a t h . C O ] S e p The achromatic number of K (cid:3) K is Mirko Horňák ∗ Institute of Mathematics, P.J. Šafárik UniversityJesenná 5, 040 01 Košice, SlovakiaSeptember 18, 2020
Abstract
A vertex colouring f : V ( G ) → C of a graph G is complete if for anytwo distinct colours c , c ∈ C there is an edge { v , v } ∈ E ( G ) suchthat f ( v i ) = c i , i = 1 , . The achromatic number of G is the maximumnumber achr( G ) of colours in a proper complete vertex colouring of G .In the paper it is proved that achr( K (cid:3) K ) = 18 . This result finalisesthe determination of achr( K (cid:3) K q ) . Keywords: complete vertex colouring, achromatic number, Cartesian prod-uct
Consider a finite simple graph G and a finite colour set C . A vertex colouring f : V ( G ) → C is complete if for any two distinct colours c , c ∈ C one canfind an edge { v , v } ∈ E ( G ) ( { v , v } is usually shortened to v v ). The achromatic number of G , in symbols achr( G ) , is the maximum cardinalityof C admitting a proper complete vertex colouring of G . The invariant wasintroduced by Harary, Hedetniemi, and Prins in [4], where the followinginterpolation theorem was proved: ∗ E-mail address: mirko.hornak @ upjs.sk heorem 1. If G is a graph, and χ ( G ) ≤ k ≤ achr( G ) for an integer k ,there exists a k -element colour set C and a proper complete vertex colouring f : V ( G ) → C . Let G (cid:3) H denote the Cartesian product of graphs G and H (the nota-tion follows the monograph [10] by Imrich and Klavžar). So, V ( K p (cid:3) K q ) = V ( K p ) × V ( K q ) , and E ( K p (cid:3) K q ) consists of all edges ( x, y )( x, y ) with y = y and all edges ( x , y )( x , y ) with x = x . The problem of determining achr( K p (cid:3) K q ) is motivated by the fact that, according to Chiang and Fu [2], achr( G (cid:3) H ) ≥ achr( K p (cid:3) K q ) for arbitrary graphs G, H with achr( G ) = p and achr( H ) = q . Clearly, achr( K q (cid:3) K p ) = achr( K p (cid:3) K q ) , so it is natural tosuppose p ≤ q . The case p ∈ { , , } was solved by Horňák and Puntigán in[9], and that of p = 5 by Horňák and Pčola in [7, 8]. Proposition 2 (see [1]) . achr( K (cid:3) K ) = 18 . More generally, in [3] Chiang and Fu proved that if r is an odd projectiveplane order, then achr( K ( r + r ) / (cid:3) K ( r + r ) / ) = ( r + r ) / .The aim of the present paper is to finalise the determination of K (cid:3) K q (for q ≥ see Horňák [5, 6]) by proving Theorem 3. achr( K (cid:3) K ) = 18 . For k, l ∈ Z we denote integer intervals by [ k, l ] = { z ∈ Z : k ≤ z ≤ l } , [ k, ∞ ) = { z ∈ Z : k ≤ z } . Further, for a set A and m ∈ [0 , ∞ ) let (cid:0) Am (cid:1) be the set of m -element subsetsof A .If we suppose that V ( K r ) = [1 , r ] for r ∈ [1 , ∞ ) , then a vertex colouring f : V ( K p (cid:3) K q ) → C can be conveniently desribed using the p × q matrix M = M ( f ) , in which the entry in the i th row and the j th column is ( M ) i,j = f ( i, j ) .If f is proper, then each line (row or column) of M consists of pairwisedistinct entries.If f is complete, then each pair { c , c } ∈ (cid:0) C (cid:1) (of colours in C ) is good in M in the sense that either there are i ∈ [1 , p ] and j , j ∈ [1 , q ] suchthat { c , c } = { ( M ) i,j , ( M ) i,j } , which is expressed by saying that the pair { c , c } is row-based (in M ) or there are i , i ∈ [1 , p ] and j ∈ [1 , q ] suchthat { c , c } = { ( M ) i ,j , ( M ) i ,j } , i.e. , the pair { c , c } is column-based (in2 ). For ( i, j ) ∈ [1 , p ] × [1 , q ] and C ′ ⊆ C \ { ( M ) i,j } let g ( i, j, C ′ ) denote thenumber of pairs { ( M ) i,j , γ } with γ ∈ C ′ that are good in M , while this factis documented by the ( i, j ) -copy of the colour ( M ) i,j .Let M ( p, q, C ) be the set of p × q matrices M with entries from C suchthat entries of any line of M are pairwise distinct, and all pairs in (cid:0) C (cid:1) aregood in M . Thus if f : [1 , p ] × [1 , q ] → C is a proper complete vertex colouringof K p (cid:3) K q , then M ( f ) ∈ M ( p, q, C ) .Conversely, if M ∈ M ( p, q, C ) , it is immediate to see that the mapping f M : [1 , p ] × [1 , q ] → C determined by f M ( i, j ) = ( M ) i,j is a proper completevertex colouring of K p (cid:3) K q .So, we have just proved Proposition 4. If p, q ∈ [1 , ∞ ) and C is a finite set, then the followingstatements are equivalent: (1) There is a proper complete vertex colouring of K p (cid:3) K q using as colourselements of C . (2) M ( p, q, C ) = ∅ . The following straightforward proposition comes from [5].
Proposition 5. If p, q ∈ [1 , ∞ ) , C, D are finite sets, M ∈ M ( p, q, C ) ,mappings ρ : [1 , p ] → [1 , p ] , σ : [1 , q ] → [1 , q ] , π : C → D are bijections, and M ρ,σ , M π are p × q matrices defined by ( M ρ,σ ) i,j = ( M ) ρ ( i ) ,σ ( j ) and ( M π ) i,j = π (( M ) i,j ) , then M ρ,σ ∈ M ( p, q, C ) and M π ∈ M ( p, q, D ) . Let M ∈ M ( p, q, C ) . The frequency of a colour γ ∈ C is the numberof entries of M equal to γ . An l-colour ( l + colour ) is a colour of frequency l (at least l ), and C l ( C l + ) is the set of l -colours ( l + colours). Further, for k ∈ { l, l + } let c k = | C k | , R ( i ) = { ( M ) i,j : j ∈ [1 , q ] } , R k ( i ) = C k ∩ R ( i ) , r k ( i ) = | R k ( i ) | , i ∈ [1 , p ] , C ( j ) = { ( M ) i,j : i ∈ [1 , p ] } , C k ( j ) = C k ∩ C ( j ) , c k ( j ) = | C k ( j ) , j ∈ [1 , q ] . Finally, denote R ( i , i ) = C ∩ R ( i ) ∩ R ( i ) , r ( i , i ) = | R ( i , i ) | , i , i ∈ [1 , p ] , i = i , C ( j , j ) = C ∩ C ( j ) ∩ C ( j ) , c ( j , j ) = | C ( j , j ) | , j , j ∈ [1 , q ] , j = j . Considering a nonempty set S ⊆ [1 , p ] × [1 , q ] we say that a colour γ ∈ C occupies a position in S ( appears in S , has a copy in S or simply is in S ) ifthere is ( i, j ) ∈ S such that ( M ) i,j = γ .3 Properties of a counterexample to Theorem 3
We prove Theorem 3 by the way of contradiction. It is well known that achr( G ) ≥ achr( H ) if H is an induced subgraph of a graph G . So, by Propo-sition 2, achr( K (cid:3) K ) ≥ achr( K (cid:3) K ) = 18 . Provided that Theorem 3 isfalse, by Theorem 1 and Proposition 4 there is a set C with | C | = 19 and amatrix M ∈ M (6 , , C ) ; henceforth the whole notation corresponds to this(hypothetical) matrix M .For a colour γ ∈ C denote V γ = f − M ( γ ) ⊆ [1 , × [1 , , and let N ( V γ ) be the neighbourhood of V γ (the union of neighborhoods of vertices in V γ ).The excess of γ is defined to be the maximum number exc( γ ) of vertices ina set S ⊆ N ( V γ ) such that the partial vertex colouring of K (cid:3) K , obtainedby removing colours of S , is complete concerning the colour class γ . Claim 1. If γ ∈ C l , then exc( γ ) = − l + 12 l − .Proof. The vertex colouring f M of K (cid:3) K is proper, hence | N ( V γ ) | = 7 l + l (6 − l ) − l = l (12 − l ) . Further, f M is complete, and so each colour of C \ { γ } occupies a position in N ( V γ ) . As a consequence, exc( γ ) = l (12 − l ) − (19 −
1) = − l + 12 l − . (cid:4) Claim 2.
The following statements are true: . c = 0 ; . if l ∈ [7 , ∞ ) , then c l = 0 ; . c ∈ [15 , ; . c ∈ [1 , ; . c ≤ c − ; . if c = 0 , then c = c = 4 ; . if c ≥ , then c ≤ ; . c + c ≤ ; . if c ≥ , then c + c ≤ ; . Σ = P i =3 ic i ∈ [6 , ; . if i , i ∈ [1 , , i = i , then r ( i , i ) ≤ .Proof.
1. If γ ∈ C , then, by Claim 1, exc( γ ) = − < , a contradiction.2. If γ ∈ C l for some l ∈ [7 , ∞ ) , by the pigeonhole principle the colouring f M is not proper, a contradiction.3. By Claims 2.1, 2.2 we have c + c = P i =2 c i = | C | = 19 and P i =2 ic i = | V ( K (cid:3) K ) | = 42 , hence c +6(19 − c ) = 2 c +6 c ≥ P i =2 ic i ≥ c + 3 c = 2 c + 3(19 − c ) , which yields − c ≥ ≥ − c and ≤ c ≤ .4. A consequence of Claim 2.3.5. We have · − c + c = 3( c + c + c ) − c + c ≤ P i =2 ic i = 42 and c ≤ c − .6. If c = 0 , then c + c = 19 , c + 3 c = 42 and c = c = 42 − ·
19 =4 . 7. The assumption c ≥ and c = 4 would mean Σ ≥ · > , a contradiction.8. If c = 0 , then c + c = c ≤ . With c = 1 we have, byClaim 2.5, c ≤ c − , c ≥ , c = 19 − c − c ≤ , c ≤ and c + c ≤ . Finally, from c ≥ it follows that ≤ c ≤ c − , c ≥ ,
19 = c + c + c ≥
17 + c + 2 = 19 + c ≥ , c = 17 , c = 0 , c = 2 = c and c + c = 4 .9. We have − c )+5 c = 2( c + c + c )+5( c + c ) ≤ P i =2 ic i = 42 , c ≤ and c + 2 c = 42 − − c − c ) − c − c ≤ − c , hence c ≥ yields c = 1 , c + 2 c ≤ and c + c ≤ .10. The assertion of Claim 2.3 leads to P i =3 ic i = P i =2 ic i − c =42 − c ∈ [6 , .11. The inequality is trivial if r ( i , i ) = 0 . If γ ∈ R ( i , i ) , then eachcolour of R ( i , i ) \ { γ } contributes one to the excess of γ , hence, by Claim 1, r ( i , i ) − ≤ exc( γ ) = 2 and r ( i , i ) ≤ . (cid:4) A set D ⊆ C is of the type ( m a . . . m a k k , n b . . . n b l l ) if ( m , . . . , m k ) , ( n , . . . , n l ) are decreasing sequences of integers from the interval [1 , | D | ] , |{ i ∈ [1 ,
6] : | D ∩ R ( i ) | = m i }| = a i ≥ for each i ∈ [1 , k ] , and |{ j ∈ [1 ,
7] : | D ∩ C ( j ) | = n j }| = b j ≥ for each j ∈ [1 , l ] ; clearly, P ki =1 m i a i = 2 | D | = P lj =1 n j b j .Forthcoming Claims 3, 6, 4, 7, and 9 state that certain types of 2- and 3-element subsets of C are impossible in a matrix M contradicting Theorem 3.The mentioned claims are proved by contradiction. When proving that M avoids a type T , we suppose that there is D ⊆ C , which is of the type T (without explicitly mentioning it), and we reach a contradiction with someof the properties following from the fact that M ∈ M (6 , , C ) . Claim 3.
No set { α, β } ⊆ C is of the type (1 , ) .Proof. Since we have at our disposal Proposition 5, we may suppose withoutloss of generality that ( M ) , = α = ( M ) , and ( M ) , = β = ( M ) , . We use5w) to express briefly that it is Proposition 5, which enables us to simplify ourreasoning by restricting our attention to matrices with a special structure.With A = C (1) ∪ C (2) we have | A | ≤ . If γ ∈ C \ A , the fact that both pairs { γ, α } and { γ, β } are good forces γ to occupy a position in S α = { , }× [3 , and in S β = { , }× [3 , as well. So, | C \ A | ≤ and | C \ A | = | C |−| A | ≥ .By Claim 2.4, | ( C \ A ) ∩ C | = | C \ A | − | ( C \ A ) ∩ C | ≥ − c ≥ ,hence there is ( i, k ) ∈ { , } × { , } such that δ ∈ R ( i, k ) ∩ ( C \ A ) . If ( i, k ) = (1 , , (w) ( M ) , = δ = ( M ) , .If | C \ A | = 10 , then all ten positions in both S α , S β are occupied bycolours of C \ A , and all twelve bullet positions in Fig. 1 are occupied bycolours of ( C \ A ) \ { δ } , which means that exc( δ ) ≥ − | ( C \ A ) \ { δ }| =12 − (10 −
1) = 3 in contradiction to Claim 1.If | C \ A | = 9 , then C (1) ∩ C (2) = { α, β } . Let B be the set of four coloursoccupying a position in [5 , × [1 , . Using exc( α ) = exc( β ) = 2 we see thatexactly two positions in [1 , × [3 , are occupied by a colour of B . Thus, if ε ∈ B is not in [1 , × [3 , , it must be in [5 , × [3 , , and so ε ∈ C . Then,however, the number of pairs { ζ , ε } with ζ ∈ ( C \ A ) ∩ C that are good is atmost four, while | ( C \ A ) ∩ C | ≥ | C \ A | − | C | ≥ − , a contradiction.Provided that ( i, k ) = (1 , , a contradiction can be reached in a similarmanner. (cid:4) α . δ • • • • β . • δ • • • . α • • . . .. β • • . . .. . . . . . .. . . . . . . α β . . . . .. . α β . . .γ • • • . . . • γ • • . . . • • . . . . . • • . . . . . α β . . ◦ . .ζ /. ./ζ α β ◦ ◦ ◦ γ . . . . • • . γ . . . • • . . . . ζ . .. . . . . . . Fig . . . Claim 4.
No set { α, β } ⊆ C is of the type (2 , ) .Proof. Now (w) ( M ) , = α = ( M ) , and ( M ) , = β = ( M ) , . With A = C (1) ∪ C (2) ∪ R (2) each colour γ ∈ C \ A has a copy in { i } × [3 , , i = 1 , ( { γ, α } and { γ, β } are good). From | A | ≤ it follows that | C \ A | ≥ −
15 = 4 , and then C \ A ⊆ C : indeed, if δ ∈ ( C \ A ) ∩ C , then exc( δ ) ≥ | ( C \ A ) \ { δ }| ≥ , a contradiction. Thus C ⊆ A , c ≤ , hence,by Claims 2.3, 2.4, c = 15 , c = c = 4 , C \ A = C = C , each colour6f C \ { α, β } has exactly one copy in ([1 , × [1 , ∪ ( { } × [3 , , and (w) ε = ( M ) , , ζ = ( M ) , are (distinct) 2-colours.First note that ε, ζ / ∈ R (1 , , for otherwise max(exc( ε ) , exc( ζ )) ≥ c = 4 .So, the second copies of ε, ζ are in [4 , × [3 , , and the pair { ε, ζ } is goodin the corresponding × submatrix of M .If the pair { ε, ζ } is column-based, (w) ε = ( M ) , and ζ = ( M ) , , then,with a (2-) colour η appearing in { } × [4 , , both pairs { η, ε } and { η, ζ } aregood only if η occupies a position in { , , } × { } , a contradiction.If the pair { ε, ζ } is row-based, (w) ε = ( M ) , and ζ = ( M ) , , consider sixpositions in [5 , × [5 , . Since r (1) = r (3) = 4 = c , at most four of thosepositions are occupied by 3-colours and at least two of them by 2-colours.Let B be the set of 2-colours having a copy in ([5 , × [5 , ∪ ( { } × [5 , .If ϑ ∈ B , then, having in mind that both pairs { ϑ, ε } and { ϑ, ζ } are good, ϑ must have a copy in { (1 , , (3 , , (4 , , (4 , } ; this yields a contradiction,because | B | ≥ . (cid:4) Claim 5. If j, l ∈ [1 , , j = l , then c ( j, l ) ≤ .Proof. The assumption c ( j, l ) ≥ would contradict Claim 3 or Claim 4. (cid:4) Claim 6.
No set { α, β } ⊆ C is of the type (2 , ) .Proof. Here (w) ( M ) , = α = ( M ) , and ( M ) , = β = ( M ) , . With A = R (1) ∪ R (2) we have | A | ≤ , each colour of C \ A is in both sets S α =[3 , × { , } , S β = [3 , × { , } , and ≤ | C \ A | ≤ . As | ( C \ A ) ∩ C | ≥ ,there is ( j, l ) ∈ { , } × { , } such that γ ∈ ( C \ A ) ∩ C ( j, l ) .If ( j, l ) = (1 , , (w) ( M ) , = γ = ( M ) , .If | C \ A | = 8 , all eight positions in both sets S α , S β are occupied by coloursof C \ A . Further, all ten bullet positions in Fig. 2, which are positions ofvertices in ( N ( V α ) ∪ N ( V β )) ∩ N ( V γ ) , are occupied by colours of ( C \ A ) \{ γ } ) ,and so exc( γ ) ≥ − (8 −
1) = 3 , a contradiction.Suppose that | C \ A | = 7 (and A | = 12 ). For m ∈ { , , } and n ∈ [0 , let C nm be the set of colours in C m having n copies in [5 , × [5 , ,and let c nm = | C nm | . If δ ∈ C ∪ C , then, since the pairs { δ, α } , { δ, β } and { δ, γ } are good, δ must appear in { } × [1 , , and so c + c ≤ ; further, c = 0 . Using Claim 2.8 we obtain c + c + c + 2 c + 2 c ≤ c + c + X n =0 ( c n + c n ) + c ≤ c + c + c + c ≤ , (1)7hich implies c = c = c = 0 , (2) c = c , (3) c + c = 2 , (4) c + c = 4 , and so, by Claim 2.9, c = 0 .For δ ∈ { α, β } choose a set S δ ′ ⊆ S δ with | S δ ′ | = 7 occupied by sevendistinct colours of C \ A , and let P = ([3 , × [1 , \ ( S α ′ ∪ S β ′ ) ; then | P | = 2 . Since | N ( V γ ) ∩ ([3 , × [1 , | = 10 , we have γ ) ≥ − ( | P | + | ( C \ A ) \ { γ }| = 4 − | P | = 2 , hence both positions in P are necessarilyoccupied by a colour of A , and all sets S α ′ , S β ′ , P are unique. We express thisproperty of γ by saying that γ is A -exact . Besides that, the two positions in P are occupied by two distinct colours of A , say λ and µ ; indeed, otherwise thecolour of A , which occupies both positions in P , by (3), would be a 5+colour,a contradiction. Let P = { ( i λ , j λ ) , ( i µ , j µ ) } , where λ = ( M ) i λ ,j λ and µ =( M ) i µ ,j µ . The excess of both α, β is 2, therefore ( j λ , j µ ) ∈ { , } × { , } (acolour occupying a position in P contributes to the excess of either α or β ,and α, β are contributing to the excess of each other).The above reasoning concerning γ can be repeated to prove that anycolour in ( C \ A ) ∩ C is A -exact.Suppose that ε ∈ ( C \ A ) ∩ C ; then ε is A -exact and ε ∈ C ( j ′ , l ′ ) , where ( j ′ , l ′ ) ∈ { , } × { , } . Let { l λ , l µ } = [1 , \ { j λ , j µ } .Assume first that i λ = i µ . By Claim 4, | ( C \ A ) ∩ C ( j λ , j µ ) | ≤ . If ( j ′ , l ′ ) = ( j λ , j µ ) , then either ε = ( M ) i λ ,l λ or ε = ( M ) i λ ,l µ .The second possibility is i λ = i µ . By Claim 4, | ( C \ A ) ∩ C ( l λ , l µ ) | ≤ . Onthe other hand, if ( j ′ , l ′ ) = ( l λ , l µ ) , then either ε = ( M ) i λ ,j µ or ε = ( M ) i µ ,j λ .In both cases | ( C \ A ) ∩ C | ≤ and | ( C \ A ) ∩ C | ≥ − . (5)From (1) and (3) we obtain ( C \ A ) ∩ C ⊆ C ∪ C , hence c + c ≥ .Let us show the following:(*) No 3+colour occupies a position in [3 , × [5 , , and c ≥ .Because of (2) and (3) we know that colours of ( C \ A ) ∩ C appearonly in ([3 , × [1 , ∪ ([5 , × [5 , . If c ≥ , then, by Claim 2.7, ≥ c ≥ c + c ≥ , c = c + c = 3 , C = C ∪ C = ( C \ A ) ∩ C , c = 0 , c = 2 (see (4)), and so (*) is true.8f c = 0 , from (1), (4) and Claim 2.4 it follows that c + c = 4 = c , C = C ∪ C = (( C \ A ) ∩ C ) ∪ C and, by (5), c ≥ ; since a colourof C is only in ( { } × [1 , ∪ ([5 , × [5 , , c ≤ and c ≥ , (*) is trueagain.Now, by (*), six positions in [3 , × [5 , are occupied by six distinct 2-colours belonging to A \ { λ, µ } , and there is a colour ζ ∈ C , (w) ζ = ( M ) , ,see Fig. 3. If a 2-colour η appears in a bullet position, then, since the pair { η, ζ } is good, the second copy of η must occupy a circle position. In sucha case, however, it is easy to check that there is a set { ϑ, ι } ⊆ A ∩ C of 2-colours occupying two bullet positions and two circle positions, whichcontradicts either Claim 3 or Claim 4.The case ( j, l ) = (1 , can be treated similarly. (cid:4) Claim 7.
No set { α, β } ⊆ C is of the type (2 , ) .Proof. Let (w) ( M ) , = α = ( M ) , and ( M ) , = β = ( M ) , . First ofall, we have R (1 ,
2) = { α, β } . Indeed, if (w) γ ∈ R (1 , \ { α, β } , then, byClaim 6, necessarily ( M ) , = γ = ( M ) , . Each colour δ ∈ C \ R (1 , occupies at least two positions in [3 , × [1 , (all pairs { δ, α } , { δ, β } , { δ, γ } are good), hence | C | ≤
11 + ⌊ · ⌋ = 17 < , a contradiction.With A = R (1) ∪ R (2) ∪ C (2) any colour γ ∈ C \ A occupies a position in [3 , × { } as well as in [3 , × { } , hence | C \ A | ≤ , | A | ≤ , | C \ A | =19 − | A | ≥ and | A | ≥ .Assume first that | A | = 15 and | C \ A | = 4 , which yields C \ A ⊆ C (a 2-colour γ ∈ C \ A would satisfy exc( γ ) ≥ ), c = c = 4 and A = C . For colours γ = ( M ) , and δ = ( M ) , their second copies appear in [3 , × ( { } ∪ [4 , , and the pair { γ, δ } is good in the corresponding × submatrix of M . However, at most one of γ, δ is in [3 , × { } , hence { γ, δ } is good in the submatrix of M corresponding to [3 , × [4 , .If the pair { γ, δ } is column-based, (w) γ = ( M ) , and δ = ( M ) , , atmost one of colours in [5 , × { } belongs to R (1) ∪ R (2) , hence there is a2-colour ε and i ∈ [5 , such that ε = ( M ) i, = ( M ) − i, (so that both pairs { ε, γ } , { ε, δ } are good). For every colour ζ / ∈ ( C (2) ∪ { ( M ) i, } occupying aposition in [1 , × [5 , (the number of such colours is at least 4) there is η ∈ { γ, δ, ε } such that the pair { η, ζ } is not good, a contradiction.If the pair { γ, δ } is row-based, (w) γ = ( M ) , and δ = ( M ) , . If a colour ε occupies a position in [4 , × { } and does not belong to R (1) ∪ R (2) (thereare at least two such colours), it must appear in { } × [6 , (pairs { ε, γ } and9 ε, δ } are good), (w) ( M ) , = ε = ( M ) , and ( M ) , = ζ = ( M ) , . If a 2-colour η is in { }× [4 , , then η = ( M ) , (all pairs { η, ϑ } with ϑ ∈ { γ, δ, ε, ζ } are good), r (6) ≥ > c , and so the colouring f M is not proper, acontradiction.From now on | A | = 16 and | C \ A | = 3 . Suppose first that C \ A ⊆ C .From c ≤ we obtain | A ∩ C | ≤ .If ( R (1) ∪ R (2)) ∩ C = ∅ , (w) γ = ( M ) , , δ = ( M ) , , ε = ( M ) , are2-colours, and their second copies appear in [3 , × [4 , . Let J = { j ∈ [4 ,
7] : C ( j ) ∩ { γ, δ, ε } 6 = ∅} ; by Claim 5 we know that ≤ | J | ≤ . If ( i, j ) ∈ S = { (1 , , (2 , } ∪ ([1 , × ([4 , \ J )) , then g ( i, j, { γ, δ, ε } ) = 0 ;note that | S | = 10 − | J | . On the other hand, it is easy to see that thenumber of pairs ( i, j ) ∈ [3 , × [4 , , satisfying g ( i, j, { γ, δ, ε } ) = 3 , is lessthan | S | (at most 3 if | J | = 3 and at most 4 if | J | = 2 ). Thus, there is a2-colour ζ in S and η ∈ { γ, δ, ε } such that the pair { ζ , η } is not good.If | ( R (1) ∪ R (2)) ∩ C | = 1 , then c = c = 4 and c (2) = 6 .Suppose first that both γ = ( M ) , and δ = ( M ) , are 2-colours. Thesecond copies of γ, δ are then in [3 , × [4 , , for if not, max(exc( γ ) , exc( δ )) ≥ | C \ A | = 4 .If the pair { γ, δ } is column-based, (w) γ = ( M ) , and δ = ( M ) , , then ( M ) , = ε = ( M ) , and ( M ) , = ζ = ( M ) , (all pairs { ε, γ } , { ε, δ } , { ζ , γ } , { ζ , δ } are good). For ( i, j ) ∈ [1 , × [5 , then g ( i, j, { γ, δ, ε, ζ } ) = 1 , and atleast three positions in [1 , × [5 , are occupied by a 2-colour that is in [3 , × [5 , ; on the other hand, for ( i, j ) ∈ [3 , × [5 , we have g ( i, j, { γ, δ, ε, ζ } ) ≤ , a contradiction.If the pair { γ, δ } is row-based, (w) γ = ( M ) , and δ = ( M ) , . Then g ( i, j, { γ, δ } ) = 0 for ( i, j ) ∈ [4 , × { } and g ( i, j, { γ, δ } ) ≤ for ( i, j ) ∈ [4 , × [4 , ; this leads to a contradiction, since at least one of colours in [4 , × { } has its second copy in [4 , × [4 , .So, one of γ, δ is a 2-colour and the other a 3-colour, (w) γ ∈ C and δ ∈ C . As above, the second copy of γ appears in [3 , × [4 , , (w) γ =( M ) , . All colours of the set B = { ε, ζ , η, ϑ } , where ε = ( M ) , , ζ = ( M ) , , η = ( M ) , and ϑ = ( M ) , , are 2-colours. By Claim 3, the second copy ofa colour ι ∈ B does not appear in C (1) ∪ C (3) , hence is in [3 , × [4 , and,additionally, in R (3) ∪ C (4) , provided that ι = ε (the pair { ι, γ } is good).Clearly, | B ∩ R (3) | ≤ , since otherwise exc( ε ) ≥ . So, by Claim 5, with B ′ = { ζ , η, ϑ } we have ≤ | B ′ ∩ C (4) | ≤ .If | B ′ ∩ C (4) | = 2 , (w) η = ( M ) , , ζ = ( M ) , (here we use Claim 4)and ϑ = ( M ) , . For both l ∈ [6 , then g (2 , l, B ′ ∪ { γ } ) = 0 . This leads to10 contradiction, since ( M ) , , ( M ) , are 2-colours, and g ( i, j, B ′ ∪ { γ } ) = 4 only if ( i, j ) = (6 , .If | B ′ ∩ C (4) | = 1 , (w) ζ = ( M ) , , η = ( M ) , and ϑ = ( M ) , , then g (2 , , B ′ ∪ { γ } ) = 0 . A contradiction follows from the fact that g ( i, j, B ′ ∪{ γ } ) ≤ for each ( i, j ) .Now suppose that ( C \ A ) ∩ C = ∅ , (w) γ = ( M ) , = ( M ) , ∈ ( C \ A ) ∩ C .For m ∈ { , , } , n ∈ [1 , let C nm be the set of colours in C m having n copies in [5 , × [4 , and c nm = | C nm | ; then c + c + 2 c = 8 . (6)Since g ( i, j, { α, β, γ } ) = 0 for ( i, j ) ∈ [5 , × [4 , and g ( i, j, { α, β, γ } ) = 3 ifand only if ( i, j ) ∈ S = { (1 , , (2 , , (3 , , (4 , } , we have c + c ≤ . (7)Let us first show that c ≤ . Indeed, if c = 4 , it is easy to see that allpairs { δ, ε } ∈ (cid:0) C (cid:1) are good only if there is i ∈ [5 , such that C ⊆ R ( i ) . Thisimmediately implies c = 0 and, by Claim 2.4 and (6), ≥ c ≥ c = 4 , c = 4 and C ⊆ R (11 − i ) . Then δ = ( M ) − i, ∈ C , the second copy of δ is in [3 , × [4 , (by Claim 3), hence at least one of pairs { δ, ε } with ε ∈ C is not good, a contradiction.Further, we prove that c + c = c , (8)which is equivalent to C ∪ C = C . (9)If c ≥ , Claim 2.7 yields c ≤ . Because of (6) we obtain c + c ) = 8 + c − c , c + c = (8 + c − c ) ≥ (8 −
3) = , ≥ c ≥ c + c ≥ , and so (24) is true under the assumption c ≥ (implying c = 3 ).On the other hand, c = 0 implies c = c = 4 (Claim 2.6). In thiscase, using (6) and (7), we see that c + c ) + ( c + c ) ≤ c = 8 ,hence c = 4 = c + c = c + c , and (24) holds again.Note that now necessarily | ( C \ A ) ∩ C | = 1 . Indeed, | ( C \ A ) ∩ C | = 3 is impossible by Claim 5, since in such a case c (1 , ≥ | ( C \ A ) ∩ C | = 3 .Moreover, by Claims 3 and 4, the assumption | ( C \ A ) ∩ C | = 2 would meanthat for the unique colour δ ∈ ( C \ A ) ∩ C there is i ∈ [5 , such that11 M ) i, = δ = ( M ) − i, ; however, according to (9), δ has an exemplar in [5 , × [4 , , and so the colouring f M is not proper, a contradiction.Thus | ( C \ A ) ∩ C | = 2 . (10)Because of a reasoning analogous to that above we see that each colour of ( C \ A ) ∩ C occupies exactly one position in [5 , × { , } , ( C \ A ) ∩ C = C , (11)and then, using (9), C ⊆ R (5) ∩ R (6) . (12)Now if c ≥ (and, consequently, c = 3 , which we have seen already),then, by (24), (10) and (11), c = 2 and c = 1 ; since c ≤ , in such acase c + c + 2 c ≤ in contradiction with (6).Therefore, in the rest of the proof of Claim 7 we work with c = 0 , c = 4 , c = 2 , c = c = 2 and c = 2 , see (6), (24), (10), (11).Moreover, all positions in S are occupied by colours of C ∪ C . If δ =( M ) i,j ∈ C with ( i, j ) ∈ { (1 , , (2 , } , then, because of (10), (11) and (12), exc( δ ) ≥ | ( C \ A ) ∩ C | = 1 + c = 3 , a contradiction.Thus { ( M ) , , ( M ) , } ⊆ C , and for a 2-colour ε occupying a position in [5 , × { , } (there are two such colours), by (12) we have exc( ε ) ≥ c − , a contradiction again. (cid:4) Claim 8. If i, k ∈ [1 , , i = k , then r ( i, k ) ≤ .Proof. The assumption r ( i, k ) ≥ would be in contradiction with Claim 6or Claim 7. (cid:4) Claim 9.
No set { α, β, γ } ⊆ C is of the type (3 , ) .Proof. Having in mind Claim 4 (or else Claim 7), assume (w) ( M ) , = α =( M ) , , ( M ) , = β = ( M ) , and ( M ) , = γ = ( M ) , . Let A be the set ofcolours occupying a position in ( { }× [4 , ∪ ([4 , ×{ } ) ∪{ (2 , , (3 , } , A n the set of colours in ( { n } × [4 , ∪ ([4 , × { n } ) for n = 2 , , A nm = C m ∩ A n and a nm = | A nm | for m ∈ { , } , n ∈ [1 , . Then | A | = a + a = 9 , (13) | A | = a + a = 7 , (14) A ∩ A = ∅ , (15)12ince otherwise exc( α ) ≥ . Moreover, | A | ≤ and A ⊆ A (each pair { γ, η } with η ∈ A is good), hence, by (14), A = A . (16)Let us show that distinct colours δ = ( M ) , , ε = ( M ) , (a consequenceof (13)) satisfy { δ, ε } ⊆ C . (17)Indeed, if η = ( M ) i, − i ∈ { δ, ε } ∩ C for some i ∈ [2 , and (w) η = ( M ) , ,then all colours appearing in ( { i } × [4 , ∪ ([4 , × { − i } ) ∪ { (5 − i, , (4 , i ) } belong to A \ { η } , hence, by (14) and (16), exc( η ) ≥ − (7 −
1) = 3 , acontradiction.Further, with ζ = ( M ) , we have ζ ∈ C ∩ A . (18)To see it realise first that, since the pair { ζ , α } is good and f M is proper, weget ζ / ∈ A ∪ { δ, ε } and ζ ∈ A . Moreover, ζ ∈ C , because the assumption ζ ∈ R (1) ( ζ ∈ C (1) ) contradicts Claim 7 (Claim 4, respectively).By (13), (14) and Claim 2.4, we have a + a ≥ (9 + 7) − c ≥ . (19)Further, by (13), (14) and (17)–(19), ≤ a ≤ , (20) ≤ a ≤ . (21)Consider a colour η ∈ A ∩ R (1) (from(20) we see that there are at least twosuch colours), (w) η = ( M ) ,j = ( M ) ,l . Then from g (1 , j, A ) ≤ g (1 , j, A ) =2 and g (4 , l, A ) ≤ g (4 , l, A ) ≤ it follows that a ≤ , hence, by(14), (20), (21), a = a = 6 and a = 1 .Suppose first ζ ∈ C (1) so that all positions in { } × [4 , are occupied bycolours of A . If η = ( M ) ,j , j ∈ [4 , , then proceeding similarly as aboveit is easy to see that both positions in [2 , × { j } are occupied by coloursof A . Thus R (2 , consists of at least two colours of A . By Claim 8 weobtain r (2 ,
3) = 2 , (w) ( M ) , = ϑ = ( M ) , and ( M ) , = ι = ( M ) , .Now κ = ( M ) , satisfies κ ∈ C (4) ∪ C (5) (the pair { κ, ϑ } is good) and,analogously, λ = ( M ) , ∈ C (4) ∪ C (5) . By Claim 6, the copies of κ, λ that13re in [4 , × [4 , do not share a row, (w) one of them is in R (4) and the otherin R (5) . Then, however, reasoning similarly as above again, all positions in [4 , × [2 , are occupied by colours of A . Consequently, the unique colourof A is ( M ) , = ( M ) , , and f M is not proper.If ζ ∈ R (1) , (w) ζ = ( M ) , , then all positions in ( { }× [4 , ∪ ([4 , ×{ } ) are occupied by colours of A , which implies that all positions in ([2 , × [4 , ∪ ([4 , × [2 , are occupied by colours of A . So, the unique colourof A is ( M ) , = ( M ) , , a contradiction. (cid:4) We are now ready to do the final analysis. Suppose (w) that r (1) ≥ r ( i ) for i ∈ [2 , , which, by Claim 2.3, implies ≥ r (1) ≥ (cid:24) c (cid:25) ≥ (cid:24) (cid:25) = 5 . (22)Given r (1) we assume (w) that the sequence S = ( r (1 , i )) i =2 is nonincreas-ing. Since r (1) ∈ [5 , , we have r (1 , ≥ ⌈ r (1)5 ⌉ ≥ , r (1 , ≤ ⌊ r (1)5 ⌋ = 1 ,and Corollary 8 yields r (1 , ≤ . We suppose (w) that j ∈ [1 , r (1)] ⇒ ( M ) ,j ∈ C , and, more precisely, ( M ) , = α, ( M ) , = β, ( M ) , = γ, ( M ) , = δ, ( M ) , = ε,r (1) ≥ ⇒ ( M ) , = ζ , r (1) = 7 ⇒ ( M ) , = η. Let p be the smallest integer in [2 , such that r (1 , i ) ≤ for every i ∈ [ p, ; p is correctly defined since r (1 , ≤ . Then r (1 , i ) = 1 ⇔ i ∈ [ p, r (1) + 3 − p ] , and, counting the number of positions in [2 , × [1 , occupied by colours of R (1) , we obtain p − ≤ r (1) ≤ p −
2) + (7 − p ) , which yields r (1) − ≤ p ≤ (cid:22) r (1) + 42 (cid:23) ≤ . (23)14oreover, because of Claims 6 and 7 we have p ≥ ⇒ (( M ) , = β ∧ ( M ) , = α ) ,p ≥ ⇒ (( M ) , = δ ∧ ( M ) , = γ ) ,p = 5 ⇒ (( M ) , = ζ ∧ ( M ) , = ε ) . Let q j = | R (1) ∩ C ( j ) | for j ∈ [1 , . By Claim 9 we know that a 2-colour µ , which occupies a position in { } × [2 p − , r (1)] , satisfies µ / ∈ C ( j ) forevery j ∈ [1 , p − , hence q j = 2 for any j ∈ [1 , p − , and X j =2 p − q j = 2[ r (1) − (2 p − r (1) + 8 − p ; (24)clearly, j ∈ [2 p − , ⇒ q j ≤ , (25)since with q j ≥ and µ ∈ R (1) ∩ C ( j ) for some j ∈ [2 p − , we have exc( µ ) ≥ q j − ≥ . Notice also that j ∈ [2 p − , r (1)] ⇒ ≤ q j ≤ min(3 , r (1) + 4 − p ) , (26)because 2-colours occupying a position in [1 , × { j } are distinct from p − (2-)colours appearing in { } × [1 , p − . Moreover, we assume (w) thatthe sequence ( q j ) r (1) j =2 p − is nonincreasing, and that, if ( r (1) , p ) = (5 , , thesequence ( q , q , q , q , q ) is nonincreasing.For every pair ( r (1) , p ) obeying (22) and (23), we analyse the set Q ( r (1) , p ) of sequences ( q j ) j =2 p − satisfying all restrictions (24)–(26). More precisely,we show that the assumption that M is characterised by an arbitrary se-quence Q ∈ Q ( r (1) , p ) leads to a contradiction, mostly because of Σ ≥ (a contradiction to Claim 2.10) or the existence of a line of M containing atleast five 3+colours (by Claim 2.4 then the colouring f M is not proper).15he structure of the sets Q ( r (1) , p ) with ( r (1) , p ) = (5 , follows: Q (7 ,
5) = ∅ , Q (7 ,
4) = { (3 , , , (2 , , } , Q (6 ,
5) = { (0) } , Q (6 ,
4) = { (2 , , , (2 , , , (1 , , } , Q (6 ,
3) = { (3 , , , , , (3 , , , , , (3 , , , , , (3 , , , , , (2 , , , , , (2 , , , , , (2 , , , , , (2 , , , , } , Q (5 ,
4) = { (1 , , } , Q (5 ,
3) = { (3 , , , , , (3 , , , , , (2 , , , , , (2 , , , , , (2 , , , , , (2 , , , , , (1 , , , , , (1 , , , , } . As we shall see later, it is not necessary to know the structure of Q (5 , explicitly.Our analysis is organised according to the following rules: All visible colours in M (those represented by Greek alphabet letters) are 2-colours,and both copies of a visible colour are present in M . Asterisk entries in M represent colours. Some asterisk entries appear in M by definition, e.g. , each asterisk entry in the first row of M occupies a position in { } × [ r (1) + 1 , . Another reason why an asterisk entry appears in M is that, ifthe corresponding position is occupied by a 2-colour λ , then putting anothercopy of λ to a free position ( i.e. , one that is not occupied by a visible colour)in any proper way (so that the resulting partial vertex colouring f ′ of K (cid:3) K is proper) leads to a situation, in which no continuation of f ′ to a propercomplete vertex colouring of K (cid:3) K is possible, because at least one pair { λ, µ } , where µ is a visible colour, is not good.If Q = (3 , , , then (w) ( M ) , = ζ , ( M ) , = η and ( M ) , = ε , hencethe set { ε, ζ } is of the type (2 , ) , which contradicts Claim 4.In the case Q = (2 , , we are (w) in the situation of Fig. 4. If a 2-colour µ occupies a position in { k } × [2 l − , l ] for some k ∈ [4 , and l ∈ [1 , ,then µ = ( M ) − l,h ( k ) , where h ( k ) = (3 k − k + 90) , and ν ∈ C for eachcolour ν occupying a position in ([4 , \ { k } ) × [5 − l, − l ] (with ν ∈ C the pair { µ, ν } is not good). As a consequence, it is easy to see that at leastnine positions in [4 , × [1 , are occupied by 3+colours. Besides that, if µ = ( M ) i,j ∈ C with ( i, j ) ∈ { (2 , , (2 , , (3 , , (3 , } , the second copy of16 must occupy one of the positions (4 , , (5 , , (6 , . Altogether we have Σ ≥ . α β γ δ ε ζ ηβ α . . . . .. . δ γ . . .. . . . ζ ∗ .. . . . . η ∗ . . . . ∗ . ε α β γ δ ε ζ ∗ β α . . . . .. . δ γ . . .. . . . ζ ε .. . . . . . ∗ . . . . . . ∗ α β γ δ ε ζ ∗ β α . . . . .. . δ γ . . .. . . . ζ ∗ .. . . . . . ε. . . . . ∗ ∗ Fig . . . Q = (0) , Fig. 5: Because of c (7) ≥ − c ≥ we suppose (w) η =( M ) , ∈ C so that η is in { (3 , , (3 , , (4 , , (4 , } .Under the assumption η ∈ R (3) we have (w) η = (3 , . Let C ′ be theset of 2-colours occupying a position in [5 , × [1 , ; clearly, c ≤ implies | C | ≥ . If µ ∈ C ′ , then from the fact that each pair { µ, ν } with ν ∈ C ′′ = { α, β, γ, δ, η } is good one easily gets that the second copy of µ occupies aposition in [2 , × [1 , . As g (4 , , C ′′ ) = 0 and g ( i, j, C ′′ ) ≤ provided that ( i, j ) ∈ [2 , × [1 , is a dot position, we obtain ω = ( M ) , ∈ C (noticethat ω / ∈ C ′ ), hence ( M ) , ∈ C . Then exc( η ) ≥ , since we can uncolourvertices ( i, j ) with i ∈ [2 , and ( M ) i,j ∈ C (here we use r ( i ) ≥ and C ⊆ C (7) ), as well as the vertices (5 , , (6 , (independently from thefrequencies of ( M ) , and ( M ) , ) without affecting the completeness of thecolour class η in the resulting partial colouring.In the case η ∈ R (4) we obtain a contradiction similarly as above.The assumption Q = (2 , , means that (w) ( M ) , = ζ and ( M ) , = ε ,hence the type of the set { ε, ζ } is (2 , ) in contradiction to Claim 4.For Q = (2 , , the situation is (w) depicted on Fig. 6. If λ = ( M ) ,j ∈ C , where j ∈ [2 k − , k ] with k ∈ [1 , , then λ = ( M ) − k, . As a con-sequence, r (6) = 4 , η = ( M ) , ∈ C , and with µ = ( M ) , , ν = ( M ) , we have { µ, ν } ⊆ R (6) . Then, however, exc( η ) ≥ ( µ, ν and at least one3+colour contribute to the excess of η ). Q = (1 , , , Fig. 7: Similarly as above we see that r (6) = 4 , η =( M ) , ∈ C , { ( M ) , , ( M ) , } ⊆ R (6) , and so exc( η ) ≥ .If Q = (3 , , , , , then (w) ( M ) , = δ, ( M ) , = ε and γ ∈ { ( M ) , , ( M ) , } so that the type of the set { γ, δ } contradicts Claim 4.17 α β γ δ ε ζ ∗ β α . . . . .. . δ γ . . .. . . . ∗ . ε. . . . . ∗ ζ. . . . ∗ ∗ . α β γ δ ε ζ ∗ β α . . . . .. . δ . . . ∗ . . ζ . . . ∗ . . . ε . . .. . . . γ . ∗ α β γ δ ε ζ ∗ β α . . • • •• • δ . ∗ • ∗• • ε . ∗ • •• • . ζ • • •• • . . • • γ Fig . . . If Q = (3 , , , , , then, having in mind Claim 4, we are (w) in thesituation of Fig. 8. Further, η = ( M ) , ∈ C and ϑ = ( M ) , ∈ C , whichimplies η = ( M ) , and ϑ = ( M ) , . Consequently, it is easy to see that bothpositions in { (3 , , (4 , } are occupied by 3+colours, and, provided that µ = ( M ) i,j ∈ C for some ( i, j ) ∈ [3 , × [1 , , then ( i, j ) ∈ { (5 , , (5 , } and µ = ( M ) , . Therefore, Σ ≥ r (2) ≥ . Q = (3 , , , , , (w) Fig. 9 (using Claim 4 again): If a bullet position isoccupied by a 2-colour µ , then the second copy of µ occupies a dot position.Therefore, Σ ≥ −
6) = 17 . Q = (3 , , , , , (w) Fig. 10: Analogously as in the case of Fig. 9 weobtain Σ ≥ −
6) = 18 .Under the assumption Q = (2 , , , , we have g ( i, , { α, β, γ, δ, ε, ζ } ) =1 for any i ∈ [3 , and g ( k, l, { α, β, γ, δ, ε, ζ } ) ≤ for any position ( k, l ) ∈ [2 , × [1 , occupied by a colour of C \ { α, β, γ, δ, ε, ζ } , hence c (7) ≥ , acontradiction. α β γ δ ε ζ ∗ β α . • • • . • • δ ∗ • • . • • ε • ∗ • . • • . ∗ ∗ • γ • • . • • • ζ α β γ δ ε ζ ∗ β α . . . . .. . δ • . • • . . . ε • • • . . • . γ • • . . . . . ∗ ζ α β γ δ ε ζ ∗ β α . . . . .. . δ ∗ . ∗ .. . . ε . . ∗ . . . . ζ ∗ .. . . . . . γ Fig .
10 Fig .
11 Fig . If Q = (2 , , , , , because of Claim 4 (w) there are two possibilities forthe structure of M , see Figs. 11 and 12.In the case of Fig. 11 a bullet position can be occupied by a 2-colouronly if the second copy of that colour appears in { } × [3 , . A position in18 (2 , , (2 , , (6 , , (6 , } is occupied by a 2-colour only if the second copyof that colour is in { (3 , , (4 , , (5 , } . Further, it is easy to see that atmost one of two colours in { i } × [1 , with i ∈ [3 , is a 2-colour (which isin { } × [3 , ). Therefore Σ ≥ −
3) + (4 −
3) + 3 · r (2) ≥ .In the situation of Fig. 12 let k = max( i ∈ { , , } : ( M ) i, ∈ C ) and η = ( M ) k, . The assumption k = 2 implies η ∈ { ( M ) , , ( M ) , } . α β γ δ ε ζ ∗ β α . . . . η. . δ ∗ η ∗ ∗∗ ∗ . ε . . ∗∗ ∗ . . ζ ∗ ∗∗ ∗ . . . . γ α β γ δ ε ζ ∗ β α . . . . η ∗ ∗ δ ∗ . ∗ ∗ . . . ε . . ∗ . . . η ζ ∗ ∗∗ ∗ . . . . γ α β γ δ ε ζ ∗ β α . . η . .. . δ ∗ . ∗ η ∗ ∗ . ε . . ∗∗ ∗ . . ζ ∗ ∗ . . ∗ . . . γ Fig .
13 Fig .
14 Fig . If η = ( M ) , , see Fig. 13, then Σ ≥
13 + r (2) ≥ .In the case η = ( M ) , depicted in Fig. 14 we have r (3) ≥ .If k = 3 (Fig. 15), then η = ( M ) , , and from r (2) ≥ it follows that exc( α ) ≥ , a contradiction.Fig. 16 corresponds to k = 5 , requiring η = ( M ) , . If ϑ ∈ C is in { } × [1 , , then ϑ = ( M ) , , and, if ι ∈ C is in { } × [1 , , then ι = ( M ) , .So, c (1) + c (2) ≥ , which implies exc( α ) ≥ . α β γ δ ε ζ ∗ β α . η . . . ∗ ∗ δ ∗ . ∗ .. . . ε . . ∗ . . . . ζ ∗ η. . . . . ∗ γ α β γ δ ε ζ ∗ β α . . . . .. . δ . . ∗ .. . . ε ∗ . .. . ∗ . ∗ . γ. . . . . ∗ ζ α β γ δ ε ζ ∗ β α . • • • . • • δ • • • . • • . • ∗ ∗ γ • • . • ∗ • ε • • . • • ∗ ζ Fig .
16 Fig .
17 Fig . In the case Q = (2 , , , , , using Claim 4, (w) the description by Fig. 17applies. Claim 9 implies that a 2-colour occupying a position in [3 , × [1 , does not appear in { } × [3 , . Therefore, it is easy to see that for any i ∈ [3 , at most one of the positions in { i } × [1 , is occupied by a 2-colour;as a consequence of c ≤ and r (2) ≥ then exc( α ) ≥ .19f Q = (2 , , , , , then (w) we have the situation of Fig. 18 with Σ ≥ −
6) = 18 (reasoning as in Fig. 9). α β γ δ ε ∗ ∗ β α . . . . .. . δ γ . . .. . . . . ε .. . . . . . ∗ . . . . . . ∗ α β γ δ ε ∗ ∗ β α . . . . .. . δ • . . • . . ε . • . • . . . . . γ • . . . • • • • α β γ δ ε ∗ ∗ β α . . . • • . . δ . . . .. . . ε . . .. . . . γ . . • • . . . ∗ ∗ Fig .
19 Fig .
20 Fig . Clearly, if r (1) = 5 , then r ( i ) = 5 and r ( i ) = 2 for each i ∈ [1 , ,hence c = P i =1 r ( i ) = 15 , and, by Claims 2.5, 2.6, c = c = 4 .If Q = (1 , , , we are (w) in the situation of Fig. 19. Each colourof C (7) has its second copy in [2 , × [1 , , hence at least c (7) + P i =2 r ( i ) = 11 + 2 c (7) ≥ positions in [2 , × [1 , are occupied bycolours of { α, β, γ, δ, ε } ∪ C (7) ∪ C . Since a colour in R (5) ∪ R (6) has itssecond copy in [2 , × [1 , , we have r (5) + r (6) ≤ − (15 −
3) = 6 and r (5) + r (6) = 14 − [ r (5) + r (6)] ≥ , a contradiction.If Q = (3 , , , , , then the set { γ, δ } ⊆ C is of the type (2 , ) ,which contradicts Claim 4. Q = (3 , , , , , (w) Fig. 20: A bullet position can be occupied by acolour µ ∈ C only if µ = ( M ) , . That is why r (6) ≥ , a contradiction. Q = (2 , , , , , (w) Fig. 21: If a bullet position is occupied by a colour µ ∈ C , then µ ∈ { ( M ) , , ( M ) , , ( M ) , } . One can easily see that if i ∈ [3 , , then at most one of colours in { i } × [6 , is a 2-colour. Therefore, if ( M ) ,j ∈ C for both j = 6 , , then c (6) + c (7) ≥ · · , andthere is j ∈ [6 , with c ( j ) ≥ , a contradiction. Thus, there is j ∈ [6 , with ( M ) ,j ∈ C . Then, however, since ( M ) , , ( M ) , ∈ C (a consequenceof r (6) = 2 ), the pair { ( M ) ,j , ( M ) ,l } is not good for l = 1 , .If Q = (2 , , , , , then (w), by Claim 4, the situation is depicted inFig. 22. If a 2-colour µ is in { (2 , , (6 , , (6 , } , then µ ∈ { ( M ) , , ( M ) , } ,and if a 2-colour ν is in { (5 , , (6 , } , then ν = ( M ) , . From r (6) =2 it follows that there is a 2-colour ζ in { } × [1 , ; as a consequencethen ω = ( M ) , ∈ C (with ω ∈ C the pair { ω, ζ } is not good), η =( M ) , = ( M ) , ∈ C , ( M ) , ∈ C , and each colour, occupying a positionin { } × [1 , , is a 2-colour. In such a case, however, with ϑ = ( M ) , ∈{ ( M ) , , ( M ) , } the pair { ϑ, η } is not good.20 = (2 , , , , , (w) Fig. 23: If ζ ∈ { ( M ) , , ( M ) , } ∩ C , then ζ =( M ) , , and if η ∈ { ( M ) , , ( M ) , } ∩ C , then η = ( M ) , . Therefore, at leasttwo positions in { (3 , , (5 , , (6 , , (6 , } are occupied by 3+colours. Since c (7) ≤ , at most one position in { (3 , , (5 , } and at least one position in { (6 , , (6 , } is occupied by a 3+colour. Further, from r (6) = 2 it followsthat exactly one position in { (6 , , (6 , } and in { (3 , , (5 , } as well isoccupied by a 3+colour. Consequently, by Claim 6, ( M ) , , ( M ) , and ( M ) , are three distinct 2-colours; this, however, leads to a contradiction, becauseif ϑ ∈ { ( M ) , , ( M ) , , ( M ) , } ∩ C , then necessarily ϑ ∈ { ( M ) , , ( M ) , } .If Q = (1 , , , , , we have (w) { γ, δ, ε } ∩ R (6) = ∅ . If a position in { } × ([1 , ∪ { } ) is occupied by a 2-colour ζ , then ζ = ( M ) , , which yields r (6) ≥ , a contradiction. α β γ δ ε ∗ ∗ β α . . . . .. . δ . . . ∗ . . . ε . . .. . . . . γ .. . . . . . ∗ α β γ δ ε ∗ ∗ β α . . . . .. . δ . . . .. . . . . γ ∗ . . . . . ε .. . . . . . ∗ α β γ δ ε ∗ ∗ β α . . . . .. . ∗ . . γ .. . . ∗ . δ .. . . . . . ε. . ∗ ∗ . . . Fig .
22 Fig .
23 Fig . If Q = (1 , , , , , the situation is (w) described by Fig. 24. If a 2-colour ζ is in { } × [1 , , then ζ = ( M ) , , hence r (6) ≥ .If Q ∈ Q (5 , , we have P j =1 q j = 10 . Let J = { j ∈ [1 ,
7] : q j ≥ } . If | J | ≤ , realise that any colour ζ ∈ C \ { α, β, γ, δ, ε } requires existence of a sufficient pair ( i, j ) ∈ [2 , × [1 , , i.e. , one satisfying g ( i, j, { α, β, γ, δ, ε } ) ≥ . If ( i, j ) is a sufficient pair, then necessarily j ∈ J . Moreover, given j ∈ J ,the number of sufficient pairs ( i, j ) is at most three. This is certainly true if q j = 3 . On the other hand, if q j = 2 and ( M ) k,l = ( M ) ,j with k = 1 , then,by Claim 7 and the fact that p = 2 , ( M ) k,j / ∈ { α, β, γ, δ, ε } , which means that g ( k, j, { α, β, γ, δ, ε } ) = 2 , and there are at most three i ’s such that the pair ( i, j ) is sufficient. Therefore, c ≤ · , which contradicts Claim 2.3.So, we have | J | ≥ . If q = 3 , then
10 = P j =1 q j ≥ · · ,hence q = q = q = 2 , q = 1 and q = q = 0 . If ζ = ( M ) i,j ∈ C with ( i, j ) ∈ [2 , × [6 , , then, since g ( i, j, { α, β, γ, δ, ε } ) = 1 , we have ζ ∈ C (1) \ { α, β, γ, δ, ε } . Thus c (6) + c (7) ≥ −
3) = 9 , and thereis j ∈ [6 , with c ( j ) ≥ , a contradiction.21f q ≤ , then g ( i, j, { α, β, γ, δ, ε } ) ≤ q j + 1 ≤ q + 1 ≤ for every ( i, j ) ∈ [2 , × [1 , , hence g ( k, l, { α, β, γ, δ, ε } ) ≥ whenever ( k, l ) ∈ [2 , × [6 , and ( M ) k,l ∈ C , which implies q l ≥ , l = 6 , . As a consequence then
10 = P j =1 q j ≥ | J | + (7 − | J | ) = | J | + 7 ≥ , a final contradiction provingTheorem 3. Acknowledgements.
This work was supported by the Slovak Researchand Development Agency under the contract APVV-19-0153.
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