Monochromatic solutions of linear equations

MONOCHROMATIC SOLUTIONSOF LINEAR EQUATIONS.

CHRISTIAN ELSHOLTZ, FRANK WEBER

Abstract. Landman and Robertson [4] define n(a, b) as the least positive integersuch that every 2-colouring of [1, n(a, b)] admits a monochrome solution to ax +by = (a + b)z for distinct x, y, z (where a, b ≥ 1 and gcd(a, b) = 1). Referring tounpublished results by Burr and Loo [2] they show that n(a, b) ≤ 4(a + b) + 1,n(1, b) = 4b+ 5 if b 6≡ 0 mod 4 and n(1, b) ≥ 4b+ 3 if b ≡ 0 mod 4 and ask for theremaining cases. We show that for 0 < a < b, n(a, b) = 4(a + b) + 1 if a > 1 andn(1, b) = 4b+ 4 if b ≡ 0 mod 4 and thereby settle all open cases.

1. Introduction

A partition of a set I into r blocks, conventionally represented as a mappingχ : I → {1, . . . , r}, is frequently referred to as an r-colouring of I. The van derWaerden number W (k, r) is the minimal length of intervals that guarantees a mono-chrome arithmetic progression of length k in the interval regardless of the r-colouring.The numbers studied here can be considered generalisations of the van der Waerdennumber W (3, r) arising in the context of linear equations a1x1 + · · · + akxk = t, oralternatively, as a special case of Rado numbers [5]. Following Ruzsa [6, 7] (alsoHegarty [3]) we classify these as homogeneous if t = 0 (when solutions are invariantunder rescaling); and as invariant if s = 0, where s = a0 + · · · + ak (when solutionsare invariant under additive shifts). We consider the case of homogeneous invariantlinear equations in three variables:

(1) ax− by + cz = 0 where a− b+ c = 0 and 0 < a ≤ c ≤ b.

Definition 1. Let W (a, b, c; r) be the smallest integer w such that for all r-colouringsof a w-interval I there is a (non-trivial) monochrome solution x, y, z ∈ I to equation(1).

(Trivial solutions x = y = z obviously do not count.) The equation x− 2y+ z = 0generating arithmetic progressions of length 3 is the archetype. The choice of signsensures that solutions x, y, z occur as progressions x < y < z or z < y < x. By aclassical result of Rado [5] such equations are in fact regular (i.e., admit monochromesolutions for every r-colouring of N) and the minimal lengths are all finite. Ournumbers relate to Landman and Robertson’s via n(a, b) = W (a, a + b, b; 2) if a ≤ band W (a, b, c; 2) = n(a, c) if a ≤ c. There are many results about Rado numbersfor special types of linear equations (and inequalities). In this paper we shall limitourselves completely to the homogeneous invariant three variable case.

Date: December 1, 2014.1

2 C. ELSHOLTZ, F. WEBER

By shift-invariance, only the length of the interval sourcing the solutions matters.We use the term w-interval to refer to an arbitrary interval in the integers of lengthw. (“[1, w]” can be substituted for our “w-interval”, but “[0, w)” will be found moreconvenient.) We shall assume gcd(b, c) = 1 (and hence gcd(a, b, c) = 1). Landmanand Robertson, referring to unpublished results by Burr and Loo [2], present thefollowing result ([4], Theorems 9.12 and 9.15):1

Theorem 1. (i) W (a, b, c; 2) ≤ 4b+ 1, (ii) W (a, b, c; 2) ≥ 4b+ 1 if a = 1 and c 6≡ 0mod 4, (iii) W (a, b, c; 2) ≥ 4b− 1 if a = 1 and c ≡ 0 mod 4, whenever 0 < a ≤ c < b.

They also ask for the exact value in the remaining cases (Problem 9.3). We answertheir call by showing that:

Theorem 2. (i) W (a, b, c; 2) = 4b if a = 1 and c ≡ 0 mod 4 (and b, c coprime), (ii)W (a, b, c; 2) = 4b+ 1 for all other coprime b, c. whenever 0 < a ≤ c < b.

The classical van der Waerden result W (3, 2) = 9 is immediate (case a, b, c =1, 2, 1). We shall indicate how, dispensing with the assumption of coprimality, onecan obtain results for arbitrary b, c by substituting b/ gcd(b, c) for b in the bounds ofthe theorem.

2. Upper bound

In preparation, we present in somewhat different language the derivation of theupper bound found in [4].The essential observation is that solutions parameterise in 2 dimensions: a = b− c,

so

(2) 0 = ax− by + cz = b(x− y) + c(z − x),

so the set of solutions (x, y, z) translates into a set (u, v) of solutions to bu+ cv = 0.Letting d = gcd(b, c) > 0 and b = db′, c = dc′, that set is {(−kc′, kb′) : k ∈ Z}.Such solutions translate back into distinct solutions (x, y, z) = (x0, x0 + kc′, x0 + kb′)for arbitrary x0, which also comprises of course the solutions (x, y, z) = (x0, x0 +kc, x0 + kb). We thus replace b and c by b′ and c′ (and a by a′ = b′ − c′) wheneverconsidering solution sets of this form. Henceforth assume gcd(b, c) = 1. Each x0

is then represented as x0 = ic + jb for some i, j, and each solution (x, y, z) as(x, y, z) = (ic + jb, (i + k)c + jb, ic + (j + k)b) for some i, j, k. The coordinates(i, j), (i + k, j), (i, j + k) of such solutions form (lower left) “hooks” in the planewhich we call corners, and we also then call (i, j) the joint of this corner:

(i, j + k)...

(i, j) . . . (i+ k, j)

corner in the plane

←→

ic+ (j + k)b...

ic+ jb . . . (i+ k)c+ jb

solution triple

1Part (i) is also implicit in [1]. We rediscovered it late in 2008, and our main result in 2010, butmissed their significance until [4] came to our attention.

MONOCHROMATIC SOLUTIONS OF LINEAR EQUATIONS. 3

For i, j ≥ 0, ic+ jb ∈ [0, (i+ j)b]. Restricting the coordinates (i, j) to a triangulararray ∆n = {(i, j) : 0 ≤ i, j and i + j < n} restricts the solution triples (x, y, z) =(ic+ jb, (i+ k)c+ jb, ic+ (j+ k)b) to the interval [0, (n− 1)b], of length (n− 1)b+1.E.g., for n = 4, b = 7, c = 5 the triangular array ∆ = ∆5 is:

4b3b c+ 3b2b c+ 2b 2c+ 2bb c+ b 2c+ b 3c+ b0 c 2c 3c 4c

7→

j

4 283 21 262 14 19 241 7 12 17 220 0 5 10 15 20

0 1 2 3 4 i

Here, each point (i, j) is represented by its value e(i, j) = ic + jb.2 Any 2-colouringχ of an interval [0, (n− 1)b] gives rise to a 2-colouring φ of ∆n via

(3) φ(i, j) = χ(ic+ jb)

for (i, j) ∈ ∆n. A φ-monochrome corner in ∆n yields in turn a χ-monochrome solutionof (1) in [0, (n−1)b]. Theorem 1(i) now follows from the fact that any 2-colouring φ of∆5 creates a φ−monochrome corner. (We use the abbreviation m.c. for ‘monochromecorner’ from here on.)The 32 possible 2 colourings on the diagonal i+ j = 4 reduce to just three (modulo

colour-swaps and order-reversal) by the following observation:(†) Any three or more monochrome points in ‘symmetrical’ position on any slope

-1 diagonal (or any horizontal or vertical) in ∆5 immediately force a m.c. there.Such are, specifically:(A) Monochrome triples in symmetric positions, either on a diagonal (i + j = h

for 3 ≤ h ≤ 4), or on a horizontal (j = constant), or on a vertical (i = constant).Symmetric triples on a diagonal are of form:

•. . .•

. . .•

(i− k, h− i+ k)

(i, h− i)

(i+ k, h− i− k)

Here, 0 < i < h and 0 < k ≤ min{i, h − i}. Those on a horizontal or a vertical arelike

(i, j − k) . . . (i, j) . . . (i, j + k)• • •

or

(i− k, j) •...

(i, j) •...

(i+ k, j) •

2Our argument does not require uniqueness of the representation, only that corners representsolutions; the case b = 2, c = 1 is instructive.


Each of the three coordinates in a diagonal triple gives rise to a corner with a jointoff the diagonal, and the three joints of these corners also form a corner. Either oneof the three joints shares a colour with the triple, or the three joints have the samecolour.

•∗ •∗ ∗ •

→•• •∗ ∗ ∗

or∗∗ •∗ • •

or•∗ ∗• ∗ •

or∗◦ ∗◦ ◦ ∗

(Here, • stands for one of the two colours, ◦ for the other and ∗ for the ‘indeterminate’colour.)Configurations on horizontals or verticals are complemented by triples in symmetric

positions on diagonals to form the kind of corner cluster just encountered. Thus, thederivation of a m.c. is in these cases exactly as for the diagonal case, starting fromeither

∗∗ ∗• • •

or•• ∗• ∗ ∗

(B) Pairs of equidistant coordinates on the diagonal i+ j = h, i.e., quadruples ofform

••

••

α = (i− k, h− i+ k)β = (i, h− i)

γ = (j, h− j)δ = (j + k, h− j − k)

for 0 < i < j < h and 0 < k ≤ min{i, h − i, j, h − j}. These can be thought of asresulting from a splitting of the middle element (i, h − i) in a symmetric triple – orconversely, the latter as resulting from a fusion of the inner antipodes of a quadruple.The joints of the two corners spanned, respectively, by the equidistant coordinatepairs 〈α, β〉 and 〈γ, δ〉 continue to form a corner together with the joint of the cornerspanned by the outer antipodes 〈α, δ〉 – there is in fact always a second pair of cornersfor which that is true:

•∗ •

•∗ ∗ •

or

••

∗ •∗ ∗ •

αβ

γδ

and the derivation of m.c.s is, in either case, again exactly as before.

The remaining (asymmetrical) 2-colourings on the diagonal i + j = 4 are handledby observing that successive adoption of the colour choices forced by m.c.-avoidance(based on †) in the remainder i+ j < 4 of ∆5 eventually results in a clash: If four (ormore) coordinates on the diagonal i + j = 4 share the same colour, there will be aquadruple in split symmetric position (i.e., a pattern of type B), or some triple willbe in symmetric position (a pattern of type A), so we can assume that exactly threecoordinates on the diagonal share one colour (which we label 1) and two coordinates


share the other (labelled 0). Ignoring, among the resulting 10 combinations, the fourcases of monochrome triples in symmetric position leaves, up to swapping order, thefollowing three distributions of colours on the diagonal:

S =

0∗ 1∗ ∗ 0∗ ∗ ∗ 1∗ ∗ ∗ ∗ 1

, T =

1∗ 0∗ ∗ 0∗ ∗ ∗ 1∗ ∗ ∗ ∗ 1

, U =

0∗ 1∗ ∗ 1∗ ∗ ∗ 0∗ ∗ ∗ ∗ 1

The following sequences for S and T are defined by adding at each step at leastone colour forced by the avoidance of m.c.s or triples in symmetric positions ondiagonals, verticals or horizontals. Here, i denotes a newly forced 1, o a newly forced0. Since each sequence eventually does engender a m.c., highlighted with Is and Os,the desired conclusion follows:

S →

0∗ 1i ∗ 0∗ o ∗ 1∗ o ∗ o 1

→

0∗ 11 i 0∗ 0 ∗ 1∗ 0 i 0 1

→

0o 11 1 0∗ 0 ∗ 1o 0 1 0 1

→

0O 11 1 0∗ 0 ∗ 1O 0 1 O 1

Similarly:

T →

1∗ 0∗ i 0o ∗ ∗ 1o ∗ ∗ o 1

→

1i 0i 1 00 ∗ ∗ 10 i ∗ 0 1

→

1I 0I I 00 ∗ ∗ 10 1 ∗ 0 1

U →

0∗ 1∗ o 1i ∗ ∗ 0∗ o o ∗ 1

→

0∗ 1∗ 0 11 i ∗ 0i 0 0 i 1

→

0∗ 1o 0 11 1 o 01 0 0 1 1

→

0I 10 0 11 1 0 0I 0 0 I 1

If follows that any 2-colouring of the interval [0, 4b] results in some monochromesolution of (1).

3. Avoiding monochrome solutions in 4b-intervals

It fairly easy to establish exactness of the 4b+ 1 bound for some special classes ofcoefficients:

Proposition 3. W (a, b, c; 2) = 4b+ 1 for coprime b, c such that b+ c ≡ 0 mod 4.

Proof. It suffices to exhibit a 2-colouring χ of a 4b-interval that admits nomonochrome solution to Equation (1). The latter two conditions imply that b, c areboth odd, and that neither ±b, nor ±c, nor in fact ±(b− c) are congruent 0 mod 4.


We shall in fact define a 2-colouring χ on the integers that admits no monochromesolution of (1) in any 4b-interval, by setting

χ(x) = ⌊x mod 4

2⌋,

e.g.,

x . . . −1 0 1 2 3 4 5 6 . . .χ(x) . . . 1 0 0 1 1 0 0 1 . . .

Suppose χ(x) = χ(y) = χ(z) for some non-trivial solution of (1) in the integers. Notethat x = z would imply x = y = z, so x and z must differ. Since by (2)

0 = b(x− y) + c(z − x) ≡ b(x− y)− b(z − x) ≡ −b(z − 2x+ y) ≡ 0 mod 4,

and b is odd, we get z − 2x + y ≡ 0 mod 4. From χ(x) = χ(y) = χ(z), either{x, y, z} mod 4 ⊆ {0, 1} or {x, y, z} mod 4 ⊆ {2, 3}, so

(x− y) mod 4 = (z − x) mod 4 ∈ {0,±1}

and hence x ≡ y ≡ z mod 4. Thus, 0 = b(x − y) + c(z − x), and x − y and z − xare non-zero multiples of 4, so again from gcd(b, c) = gcd(b, 4) = gcd(c, 4) = 1, theymust also be non-zero multiples of c and b, respectively. Thus |x − z| ≥ 4b, i.e., xand z cannot lie in the same 4b-interval. ✷

4. Complete arrays and periodic colourings

Since the values e(∆n) omit the tail portion ((n−1)b, nb) of the full interval [0, nb),colourings of the triangular array do not support conclusions about lower bounds. Weproceed to consider an extension that does.

Definition 4. Call an array A ⊆ Z × Z full for b, c, N , if each x ∈ [0, N) isparametrised in A (each x is ic + jb for some (i, j) ∈ A). A full array consisting ofa set of adjacent columns in the complete array is a standard array. The completearray for b, c, n is {(i, j) : 0 ≤ ic + jb < nb}. Finally, say that a planar colouringφ : A → {0, 1} for A ⊆ Z× Z and a colouring on the line χ : L → {0, 1} for L ⊆ Z

correspond if equation (3) holds whenever both (i, j) ∈ A and ic+ jb ∈ L.

Example A full array for b, c, 4b = 7, 5, 28 (each point (i, j) being represented by itsvalue ic+ jb, with the triangular subset e(∆5) highlighted):

j

4 3 8 13 18 233 1 6 11 16 21 262 4 9 14 19 241 2 7 12 17 22 270 0 5 10 15 20 25

-5 -4 -3 -2 -1 0 1 2 3 4 5 i


In this example each element of [0, 28) occurs eaxctly once and is parametrised withthe smallest possible non-negative j-coordinate. A standard array is highlighted here:

j

4 . 233 . 16 21 262 . 9 14 19 246 2 7 12 17 22 275 0 5 10 15 20 254 3 8 13 18 233 1 6 11 16 21 262 4 9 14 19 241 2 7 12 17 .0 0 5 10 .

. .-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 i

Call array colourings that avoid m.c.s evasive. The objective is to find evasivecolourings of the complete strip for b, c, n and pull these back to colourings avoidingmonochrome solution to Equation (1) of [0, nb), for n = 3, 4. The set of corners inminimal full arrays generally does not represent all solution triples to (1) — someare only represented by corners straddling boundaries between neighbouring copiesof such arrays. (In the example b, c, N = 7, 5, 28, positions labelled 13, 18 and 20do not form a corner in the depicted array, but they do when allowed to lie acrossthe border between neighbouring arrays.) Since it is technically simpler, we requireevasion in the complete array, rather than in a sufficiently large neighbourhood ofsome minimal full array.Two intuitive observations help us find evasive colourings: The complete array for

n has cross-sectional height n − 1 and it is periodic with period (−b, c), or (b,−c).Technically, setting qj := ⌊jb/c⌋:

Lemma 1. The complete array for b, c, n covers a (−b, c)-periodic strip S(b, c, n) inthe plane Z × Z of cross-sectional height n − 1 composed of rectangular segmentsS(b, c, n; j) in the plane of form

S(b, c, n; j) =

−qj, j + n− 1 . . . −qj−1 + 1, j + n− 1...

...−qj, j . . . −qj−1 + 1, j

S(b, c, n) =⋃

j∈Z S(b, c, n; j), contains every full array for b, c, N = nb.


Schematically,

. −qj+1, j + nc

. −qj+1, j + n− 1 −qj, j + n− 1 . −qj−1 + 1, j + n− 1

. . −qj, j + n− 2 . −qj−1 + 1, j + n− 2 −qj−1, j + n− 2 .

. . . . . −qj−1, j + n− 3 .

. −qj+1, j + 2 . . . . .

. −qj+1, j + 1 −qj, j + 1 . −qj−1 + 1, j + 1 .−qj, j . −qj−1 + 1, j −qj−1, j .

−qj−1, j − 1 .

Note that S(b, c, n; j−1) is to the right of S(b, c, n; j) with a vertical offset of 1, andS(b, c, n; j − c) is S(b, c, n; j) shifted b units right and c units down: S(b, c, n; j − c)= S(b, c, n; j) + (b,−c). Calling any sequence {S(b, c, n; j) : k ≥ j > k − c} a cyclein S(b, c, n) and denoting it by S(b, c, n; [k, k − c)>), each such cycle contains a fullarray for b, c, N = nb. The proof of Lemma 1 and of the preceding observations isstraightforward; we omit the details.Say that a corner 〈(i, j), (i+k, j), (i, j+k)〉 has size k and call a corner in a coloured

array a size k m.c. if it is monochrome and of size k. An immediate consequence ofLemma 1 is that the complete array for b, c, n can only contain corners of size lessthan n (since it cannot contain a corner of height k for k ≥ n). Thus a colouringof [0, nb) can only engender size k m.c.s for k = 1, . . . , n − 1 in the correspondingcolouring on the strip.Lemma 1 also suggests when array colourings can be pulled back to intervals. Since

the evaluation map e : Z × Z → Z, (i, j) 7→ ic + jb, is periodic on Z × Z with thesame period (−b, c), e(i, j) = ic + jb = (i − kb)c + (j + kc)b = e((i, j) + k(−b, c)),it follows that, as matrices, e(S(b, c, n; j)) = e(S(b, c, n; j + c)), where e(S) denotes{e(s) : s ∈ S}. From the equivalence: ic + jb = uc + vb iff (u − i)c + (v − j)b = 0iff (u − i, v − j) = ±(−kb, kc) for some k ∈ Z, it is clear that any mapping f onZ × Z is consistent with evaluation (in the sense that f(i, j) = f(u, v) whenevere(i, j) = e(u, v)) precisely when it is periodic with period (−b, c). To summarise:

Lemma 5. A colouring φ of S(b, c, n) represents a colouring χ of [0, nb) via χ(ic +jb) = φ(i, j) (i.e., such a colouring χ is well-defined), precisely when φ is periodicwith period (−b, c).

We also say that a global colouring φ of Z × Z is consistent with S(b, c, n) if itsrestriction onto S(b, c, n) is consistent with evaluation. Taking n = 2, we immediatelyobtain that the (−b, c) periodic 2-colouring which colours the bottom row of eachS(b, c, 2; j) with one colour and the top row with another will avoid m.c.s in S(b, c, 2)altogether. Thus generally W (a, b, c; 2) ≥ 2b + 1. But there are some rather simplecolourings of the entire plane that avoid size k m.c.s for k ≤ 3, and that are (−b, c)-periodic depending on b, c.

Lemma 6. The 2-colourings of the plane Z× Z defined mod 2 by

ξ(i, j) = ⌊(j − i)/2⌋, η(i, j) = (i+ ⌊j/2⌋), ζ(i, j) = η(j, i),


. . . . . .

. − − + + .

. − + + − .

. + + − − .

. + − − + .

. . . . . .

. . . . . .

. − + − + .

. − + − + .

. + − + − .

. + − + − .

. . . . . .

. . . . . .

. − − + + .

. + + − − .

. − − + + .

. + + − − .

. . . . . .

Figure 1. Patches of ξ, η and ζ

are (4, 4)-periodic and admit no size k m.c. for k = 1, 2, 3, hence no m.c. in anyarray of height 3.

Proof. Clearly, ξ is a (4, 4)-periodic tiling of the plane, while η and ζ are (2, 4)and (4, 2)-periodic tilings, respectively. We deal with ξ first: If ξ(i, j) = ξ(i+ k, j) =ξ(i, j + k), then ⌊(j − i)/2⌋ ≡ ⌊(j − i− k)/2⌋ ≡ ⌊(j + k − i)/2⌋ mod 2, whence

{j − i, j − i− k, j − i+ k} mod 4 ⊆ {0, 1} or ⊆ {2, 3}

which is only possible if k is a multiple of 4. Thus any m.c. in this colouring has sizeat least 4.Next consider η. (By symmetry this suffices, because the property size k corner

is invariant under the reflection i ↔ j.) If η(i, j) = η(i + k, j) = η(i, j + k), theni + ⌊j/2⌋ ≡ i + k + ⌊j/2⌋ ≡ i + k + ⌊j/2⌋ mod 2 (where / represents arithmeticdivision), hence ⌊j/2⌋ ≡ k + ⌊j/2⌋ ≡ ⌊(j + k)/2⌋ mod 2, which requires that k andk/2 are both even, i.e., k must be a multiple of 4. Thus any corner in this colouringhas height at least 4 and hence requires a strip of cross-sectional height 4. ✷

The colouring ξ is clearly (−b, c)-periodic whenever b + c ≡ 0 mod 4: With, say,b+ c = 4h, we have mod 2:

ξ(i− b, j + c) ≡ ⌊(j + c− i+ b)/2⌋ ≡ ⌊(j − i+ 4h)/2⌋

≡ ⌊(j − i)/2 + 2h⌋ ≡ ⌊(j − i)/2⌋ ≡ ξ(i, j).

Presently this gives just another proof of Proposition 3, which we now proceed toextend.3

5. Proof of Theorem 2.

We begin by proving:

Proposition 7. W (a, b, c; 2) = 4b + 1 for coprime b, c such that any one of the fol-lowing holds: (i) b is odd and c ≡ 2 mod 4; (ii) b is even; (iii) b > c+1. Furthermore,W (a, b, c; 2) ≥ 4b for all coprime b, c.

Proof.(i) Observe that consistency of η with S(b, c, 4), i.e.,

η((i, j) + (−b, c)) = η(i− b, j + c) = η(i, j)

3This is how we discovered this and similar facts originally.


− − + + −− + + − −+ + − − ++ − − + +0 1 2 3 0

− − + +− + + + + − − −+ + + − − − − ++ − − − − + + +− + + −

0 2 1 3 2 0 3 1

Figure 2. Colour columns and (intra/inter segment) transitions

for all i, j, is equivalent to

i− b+ ⌊(j + c)/2⌋ ≡ i+ ⌊j/2⌋ mod 2,

⌊(j + c)/2⌋ − ⌊j/2⌋ ≡ b mod 2

hence to

b ≡ 0 and ⌊(j + c)/2⌋ ≡ ⌊j/2⌋

or b ≡ 1 and ⌊(j + c)/2⌋ 6≡ ⌊j/2⌋

mod 2. Now,

⌊j/2⌋ ≡ 0 mod 2 iff j ≡ 0, 1 mod 4,

⌊j/2⌋ ≡ 1 mod 2 iff j ≡ 2, 3 mod 4,

and likewise for j + c in place of j, the condition becomes

b ≡ 0 mod 2 and ({j, j + c} mod 4 ⊆ {0, 1} or ⊆ {2, 3})

or b ≡ 1 mod 2 and ({j} mod 4 ⊆ {0, 1} iff {j + c} mod 4 ⊆ {2, 3})

These conditions must hold for arbitrary (i and) j. Now {j, j + c} mod 4 = {0, 1} or{j, j + c} mod 4 = {2, 3} for arbitrary j is only possible if c ≡ 0 mod 4, so from thefirst condition we get the condition b even and c ≡ 0 mod 4 which tells us nothingnew. On the other hand, {j} mod 4 ⊆ {0, 1} iff {j + c} mod 4 ⊆ {2, 3} for arbitraryj results in the condition b odd and c ≡ 2 mod 4. This proves (i).A completely symmetrical argument applied to ζ (swapping the roles of i and j –

admissible because this reflection preserves corners and m.c.s) leads to the conditionc odd and b ≡ 2 mod 4, but this case is subsumed by part (ii).

Applying ξ (or any shifted copy of it) to S(b, c, 4) gives rise to four colour columnslisted below left in the order of their succession within a segment S(b, c, 4; j), aswell as four inter-segment transition rules (the possible successions from one segmentS(b, c, 4; j) to the next adjacent, S(b, c, 4; j−1)), listed below right: The same colourcolumns, but different transition rules, derive from η.

We can identify a colouring of S(b, c, 4) by a sequence whose members alternatebetween colour columns and the transition type identifiers →, for an intra segmenttransition, and ↓, for a segment border transition. We shall call this the colouring’ssignature (on an array). For instance, ξ applied to a segment of width 3 followed


by a segment of width 4 has signature 0 → 1 → 2 ↓ 0 → 1 → 2 → 3 ↓ 1. Inthe following we frequently drop the → identifier altogether and replace the ↓ by a| and call this the reduced signature. The reduced signature fragment for the aboveexample reads 012|0123|1. In the signature of S(b, c, 4) including either an initial offinal inter-segment transition identifier |, the number of colour columns equals b andthe number of transition identifiers equals c, and the length of the signature is b+ c.Array colourings based on the four ξ (or η) colour columns above automatically

avoid size 2 m.c.s. Moreover, the intra- and inter-segment transition rules for ξ (aswell as those for η) prevent size 1 m.c.s. We shall call such transitions legal. For a(finite) signature fragment σ that either starts or ends on an arrow we denote by [σ]n

its n-fold repetition, e.g., (|012)2 = |012|012.

(ii) Specify a colouring of the cycle S(b, c, 4; [c, 0)>) by stipulating that its signaturecontain only the transitions 0 → 2, 2 → 0, 0 ↓ 2 and 2 ↓ 0. Thus colour columnsalternate between types 0 and 2. Assume we start with colour column 0 on the left.This sequence is therefore consistent with S(b, c, 4) (to mean: results in the samecolouring for S(b, c, 4; 0) as for S(b, c, 4; c)) precisely when b is even. Observe thatall these transitions avoid size 1 m.c.s (intra-segment transitions being η-like, inter-segment transitions being ξ-like). By the remark above this colouring also avoidssize 2 m.c.s. Finally, since it uses exclusively colour columns 0 and 2, it avoids size3 m.c.s. Periodic extension thus results in an evasive colouring of S(b, c, 4). Thisproves (ii).

(iii) Begin with the alternating 0-2 types colouring of the cycle S(b, c, 4, [c, 0)>)specified in (ii) and assume without loss that b is odd. The signature of this colouringon the cycle begins and ends with colour column 0. Periodic extension results in anillegal transition 0 ↓ 0 crossing from S(b, c, 4; 1) into S(b, c, 4; 0) (hence a transitionengendering two size 1 m.c.s on the border). By the remarks concluding the proof of(ii) no m.c.s of larger sizes can arise.Write wj for the width of each segment S(b, c, 4; j), wj = qj − qj−1. Note that

w0 ≥ wj for all j and wj = wj+c by periodicity. If w0 ≥ 3, then there is an immediateresolution of the problem: Change the colouring on S(b, c, 4; 1) by substituting colourcolumn ℓ+1 for ℓ (i.e., 1 for 0 and 3 for 2) and extend c-periodically. In the segmentblock S(b, c, 4; [2, 0]>) the resulting colouring is a shifted version of η and hence evasivethere. Any new m.c.s would have to staddle the boundaries between S(b, c, 4; 3) andS(b, c, 4; 2) or between S(b, c, 4; 0) and S(b, c, 4;−1). No size 3 m.c. can arise with joinin S(b, c, 4; [c, 2)>), since columns are coloured with types 0 or 2 there. No size 3 m.c.can arise with right limb in S(b, c, 4; (0,−c]), since (owing to w0 ≥ 3) its join wouldhave to lie in S(b, c, 4; 0) where again column colours are ot types 0 or 2. (Figure 3.The signature fragment in the top row represents the initial colouring, the one belowthe modified one.)By cyclic shifts of the colour signature on the cycle S(b, c, 4; [c, 0]>), this argument

applies to any segment of width at least 3. (Move the illegal transition to the leftborder of this segment.)


j

2 0 2 0 0 2 0 0 2 0 2 0 0 2 0 2 0 2 0 2 0 0 2 0

2 0 3 1 0 2 0 0 2 1 3 1 0 2 0 2 0 3 1 3 1 0 2 0

+ - - + +

+ - + - - + - + - + - + - + -

- + - + - + - + - + - + - + - - + - + - + - + -

2 - + - + - + - + - + - + - + - - + - + - + - + -

1 + - + - + - + - + - + + - + - + - +

0 + - + + - + + - +

Figure 3. Examples for w0 ≥ 3 with w1 = 2, 3, 4

Since b > 2c implies w0 > 2, we can assume c < b < 2c (equality being ruled outby the hypotheses). This means that S(b, c, 4) contains only segments of width atmost 2 and at least one segment of width 1.Assume there are two adjacent width 2 columns. Fix such a pair that is followed by

a width 1 column: wj+2 = wj+1 = 2, wj = 1. Rotate the initial colouring so that theillegal 0 ↓ 0 transition occurs in S(b, c, n; [j + 1, j]>). The following local operationimmediately removes any m.c.s: Substitute signature |23| for |20| in segment j + 2.(I.e., replace the colour type 0 by 3 in the rightmost column of S(b, c, n; j + 2).)Substitute signature |12| for |20| in S(b, c, n; j + 1). (Figure 4.)

0 2 0 2 0 0 2 .0 2 3 1 2 0 2 .−− + ++ + − − ++ − − + + −

j + 2 − + + − − + .j + 1 − − + + .j + − .

− ..

Figure 4. Case wj+2 = wj+1 = wj + 1 = 2

So we can assume that no two width 2 segments are adjacent. On the other handwe have at least two width 2 segments, since we get from the assumptions that b ≡ cmod 4, and hence b ≥ c + 4: By the preceding proposition we can assume b + c 6≡ 0mod 4. By (i) and (ii), we can assume b ≡ c ≡ 1 mod 2. So either b ≡ c ≡ 1 orb ≡ c ≡ 3 mod 4.Fix c > k > 1 such that wk = w0 = 2 and wk−1 = . . . = w1 = 1. Shift the

initial alternating 0-2 types colouring so that the illegal 0 ↓ 0 transition occurs inS(b, c, 4; [k, k− 1]>). Change colour type in the rightmost column of S(b, c, 4; 0) from


0 to 3. Exchange colour-column ℓ + 1 for ℓ (i.e., 1 for 0 and 3 for 2) in the width 1segments S(b, c, 4; [k− 1, 1]>) and do the same in the leftmost column of S(b, c, 4; 0).(Figure 5.)

j

0 2 0 0 2 0 2 0 0 2 0 0 2 0 2 0 2 0 2 0 0 2 0 2 0 2 0

0 2 3 1 3 0 2 0 0 2 3 1 3 1 2 0 2 0 2 3 1 3 1 3 0 2 0

k=2 k=3 - k=4

- - + -

- - + + + + - -

- + + + + - - + - + + +

4 + + - - + - - + + - + + - -

3 + - - + + - - + + - - + - - + + -

2 - + + - - + - - + + - + + - - +

1 - - + + - + + - - + - - + + -

0 + + - - - - + + + + - -

- + + - - +

+ - +

Figure 5. Examples for w0 ≥ 3 with w1 = 2, 3, 4

All transitions are now legal, so no size 1 m.c.s can arise. Since only colour types0 through 3 are involved there are also no size 2 such. On the interval of alternating1-3 types the colouring is equivalent to ξ, so no m.c.s can arise in that interval. Nosize 3 hook can have joint in the interval of alternating 0-2 types to the left, i.e.,straddling S(b, c, 4; j + k). Since the signature in the vicinity of S(b, c, 4; j) is alwayseither 3|1|30|2|0 or 1|3|12|0|2, inspection shows that no right limb of such a hookcan come to lie in the alternating 0-2 types interval to the right. Thus the resultingcolouring is evasive.This proves part (iii) of the proposition.

For the final statement we can by the preceding assume b = c + 1, b odd andc ≡ 0 mod 4. We use colouring ζ which on S(b, c, 4) gives rise to two colour columns

− ++ −− ++ −4 5

Observe that colourings consisting only of these columns automatically avoid m.c.sof sizes 1 and 3 (and so in particular all transitions are legal within segments andacross segment borders).First define a colouring on S(b, c, 4; (0, c]) for the ‘minimal’ case b = 5, c = 4 by

the following signature: 4 → 5 ↓ 4 ↓ 5 ↓ 5. Periodic extension results in a 5 ↓ 4transition from S(b, c, 4; 1) into S(b, c, 4; 0). Two size 2 m.c.s arise with joints in thesecond column of S(b, c, 4; c). We eliminate these by changing this second column to


have column colour 0. This introduces two m.c.s with common upper limb in the topright element of S(b, c, 4; c) ∼= S(b, c, 4; 0). We remove this without creating furtherm.c.s elsewhere by replacing colour-column 5 by 0∗ (that is colour-column 0 withthe top element omitted). This top element has coordinates (i, j) = (1, 3) and value3b + c = 4b − 1. Omitting this element results in a colouring of the 4b − 1-interval[0, 4b− 1) free of monochrome solutions.Under the current assumption about b and c, S(b + 4, c + 4, 4; (0, c]) extends

S(b, c, 4; (0, c]) by exactly four width 1 segments. We can thus extend the abovecolouring to one for b = 9, c = 8 by adding the signature ↓ 4 ↓ 5 ↓ 4 ↓ 5. Formally,assume by induction that we have a periodic colouring on S(b, c, 4; (0, c])∗ (the arrayfor b, c, 4 omitting the element(s) of value 4b − 1), of form 4 → 0∗ ↓ 4 ↓ 5 ↓ 5[↓ 4 ↓5 ↓ 4 ↓ 5]n. We can extend this to a peridic colouring on S(b + 4, c + 4, 4; (0, c])∗, ofform 4→ 0∗ ↓ 4 ↓ 5 ↓ 5[↓ 4 ↓ 5 ↓ 4 ↓ 5]n+1. ✷

Proof. (Theorem 2) By Propositions 3 and 7, W (a, b, c; 2) = 4b+1 if b > c+1 orb + c ≡ 0 mod 4. If b = c + 1 and b + c 6≡ 0 mod 4, then 2c + 1 = b + c = 4k + ℓ forsome integer k and some ℓ ∈ {1, 3}. If ℓ = 1 then c = 2k is even and b odd. If alsoc ≡ 2 mod 4 we get W (a, b, c; 2) ≥ 4b + 1 from part (i) of the proposition. If ℓ = 3then c = 2k + 1 is odd and b even and we get W (a, b, c; 2) ≥ 4b+ 1 from part (ii) ofthe proposition. This leaves only the case b = c + 1 and c ≡ 0 mod 4, and by part(i) of Theorem 1 it follows that part (ii) of the theorem holds. The final case (andpart (i) of the theorem) will follow from Proposition 7 if any periodic colouring ofS(b, c, 4) for such b, c can be shown to generate a m.c..

This argument is tedious, so we provide a sketch first. By an easy observationnone of the three standard colourings (ξ, η, ζ) are (−b, c)-periodic and that anyperiodic extension on the entire array produces m.c.s on the cycle boundaries. Thisis straightforward by looking at the length 4 colour-column sequences resulting foreach of these colourings covering the cycle boundary, which only depend on the valueof b+ c mod 4.Next observe that S(b, c, 4; (0, c]) now consists of exactly one width 2 segment,

S(b, c, 4; c) ∼= S(b, c, 4; 0) and b − 2 = c − 1 ≡ 3 mod 4 width 1 segments. Thestrategy is to consider evasive colourings on the trunk of width 1 segments, startingwith ‘colouring generators’ on a slope -1 diagonal. Here, we use principle (†) in thefollowing form:

Definition 8. Call a set of points A in the plane (corner-) closed if it contains thethird member of a corner whenever it contains some two. The (corner-) closure ofa subset A of the plane is the smallest (by containment) closed super-set A. Pointsalong a slope -1 diagonal (respectively horizontal, vertical) are in symmetrical positonif they are preserved set-wise by a reflection about some slope 1 diagonal (vertical,horizontal, respectively).

Lemma 9. Any three or more monochrome points in symmetrical position on anyslope -1 diagonal (or any horizontal or vertical) force a m.c.in the closure of theirset.


Thus any such monochrome symmetrical arrangement in the strip for b, c, 4 whoseclosure also lies in the strip results in a m.c. in the strip.Proof. (Lemma 9) Consider the complement of the original set in its closure.

Each point in that complement completes a pair of points on the original set to acorner, so they must all be of the opposite colour. But the complement itself containsa corner. ✷

Essentially then, any ‘colour generator’ on the width 1 segments, with a chance ofnot immediately being resolved by (†), propagates upwards or downwards throughthe entire cycle to produce η-, ξ- or ζ-like colourings – which then force a m.c. giventhe conditions on b and c.Manual computations (or computer searches) show that W (1, b, c; 2) = 4b for

(b, c) = (5, 4), (9, 8) and (13, 12). We can assume that we have at least four ad-jacent width 1 segments.We consider the slope -1 diagonal consisting of the top elements in this set.If there is a preponderance of one colour over the other (or a local imbalance in

the distribution), we are either done by (†), or we generate a left cascading ζ- anda right cascading ξ-like colouring or a left cascading ξ- and a right cascading η-likecolouring, each propagating until they enter the width 2 segment from opposite sides.If the colours are equally distributed then we generate, in the case of the period 2

distribution: two kinds of ξ-like colourings propagating in opposite directions; in thecase of the period 4 distribution: ζ-like or η-like colourings propagating in oppositedirections, each depending on the choice of one additional ‘balance tipping’ colourpoint. Propagation, either upwards or downwards through the cycle, again continuesinto the width 2 segment.To be precise, we sometimes get a sequence of choices between either (permanently)

staying with a colouring of one kind, or admitting a ‘pattern flipper’ that generatesa colouring of another kind; i.e., we avoid cascading into one pattern by adhering toanother.Resolution occurs when the propagating colourings filter through the width 2 seg-

ment from opposite sides. This segment itself will continue to adhere partially tothe patterns obtained for the initial partial colourings — just enough to ensure some‘clash’ that forces a m.c.. The condition on b and c is crucial in producing m.c.sacross the frontiers of the generated partial colourings.In the next section we implement this plan.

6. The case b = c+ 1, c ≡ 0 mod 4

We refer to the portion of the array consisting of the width 1 segments as the trunkand to the unique width 2 segment (modulo period) as the root. Since the cycle is oflength b, the trunk has width (or length) b− 2 ≡ 4 mod 3. We can assume that thetrunk has width (or length) b ≥ 13 (since the first few cases of b = c + 1 ≡ 1 mod 4easily resolve by direct calculation).Consider the colour distribution on the column top elements in the trunk which we

refer to as top colours on the trunk (TCTs). By (†), we can assume that the there are


no three contiguous TCTs of the same colour (e.g., Figure 6). Thus the only periodic

+. +. . +. . .

. ..

or

−. −. . −. . .

. ..

Figure 6. TCTs immediately resulting in a m.c..

(or alternating) colour sequences on contiguous TCT sequences have period 2 or 4(i.e., are of the form · · ·+−+− . . . or · · ·++−−++−− . . . .) The only exceptionto these are colour sequences that contain somewhere on the trunk an unbalancedlength 4 stretch of the form · · ·++−+ . . . or · · ·+−++ . . . as in Figure 7. In this

.

. +0

. . +0

. . . −0. . . +0. . . . .

.

. +0

. . −0

. . . +0. . . +0

. . . .

Figure 7. Unbalanced TCTs.

figure and those to follow, alpha-numeric characters 0, 1, . . . , 9, A,B,C, . . . placednext to colours ‘+’ or ‘−’ indicate the order in (or stages at) which these coloursare generated in the array, either by initial choice (stage 0) or by forcing (i.e., m.c.avoidance via (†): stages 1, 2, . . .). Numbers on the bottom row identify the colourcolumn generated.

Case 1. Suppose the first of the unbalanced cases, TCT sequence + + −+, occurssomewhere in the trunk. (Figure 8.) This set of TCTs generates a left-cascading ζpattern and a right-cascading ξ one. Each of these continues until it enters the root.The ζ pattern propagating leftwards will enter the second column of this segment

(i.e., the second column of the root), from the right. The ξ pattern propagating tothe left will straddle both columns of the root on a size 3 triangular region enteringfrom the left and exiting on the right; see Figure 9. In this figure, the (cyclicallyrecurring) root is indicated by the two columns between vertical bars.Note that these root overlap shapes arise in all cases, specifically in the cases where

the balance breaking + +−+ TCT sequence is adjacent to the left to the root or toits right, or in fact where the right-most column of this sequence lies on this segment.(Figure 10.)Both propagations will eventually come to a halt, but enough colours are deter-

mined to produce clashes on the overlap.


.

. −B

. +E +A

. −D −B +6+C +A −9 −5

−9 +8 +6 −1−7 −5 +4 +0

+4 −3 −1 +0+2 +2 −3 −0

−1 −1 +2 +0+4 +6 −7 −5

−7 −9 +A +8+A +C −D −B

−D −F +G +E+G . .

. ..

4 5 5 4 4 5 3 1 3 1 3

Figure 8. Case 1.

| . ∗ || . ∗ | ∗| . ∗ | ∗ ∗| . ∗ | ∗ ∗ .

∗ ∗ .∗ . .

. . ∗. ∗ ∗. ∗ ∗ | ∗ . |∗ ∗ | ∗ ∗ |∗ | ∗ ∗ | ∗| . . |

Figure 9. Root overlap shapes.

The following subcases arise, depending on the location of the balance tipping+ + −+ TCTs relative to the root. Let r and ℓ denote the lengths of the ζ andξ patterns (starting from the left-most TCT in the unbalanced + + −+ sequence).Clearly from b = c+ 1 ≡ 1 mod 4 we have

r + ℓ = b− 2 ≡ 3 mod 4.


| . −1 || . −4 | +0| . +3 | −1 +0| . +2 | +2 −3 −0

−1 −1 +2 +0+4

+0−1 +0+2 −3 −0−1 −1 +2 | +0 . |

+4 +6 | −7 −5 |−7 | −9 +A | +8

| . . |

Figure 10. Root overlap with unbalanced TCTs on the border.

Subcase 1.1. r ≡ 1 mod 4, ℓ ≡ 2 mod 4 ≡ 0 mod 2. The following signatures resultnear root-entry:

✷ ↓ 5 ↓ 5 ↓ 4 ↓ . . . ↓ 4 ↓ 5 ↓ 3 ↓ 1 ↓ . . . ↓ 1 ↓ 3 ↓ 1 ↓ ✷

This immediately yields a size 3 m.c. straddling the root (Figure 11, with limbs ofcorner circled).

−+ ++ − ⊖− − + | . . |

+ + | . . | +⊖ | . . | ⊖ +| . . | + − −

− + +− −

+

1 3 1 5 5 4

Figure 11. Subcase 1.1.

Subcase 1.2. r ≡ 0 mod 4, ℓ ≡ 3 mod 4 ≡ 3 mod 2. Signatures at root-entry:

✷ ↓ 5 ↓ 4 ↓ 4 ↓ . . . ↓ 4 ↓ 5 ↓ 3 ↓ 1 ↓ . . . ↓ 3 ↓ 1 ↓ 3 ↓ ✷

Figure 12 shows the colouring engendered by m.c. avoidance (†) from the initialcolouring fragment. Either colour on the bottom left point in the root (marked ‘?’)then yields a m.c. (with limbs as circled).


+− −− + ++ + −0 | . . |− −0 | ⊕1 . | +

+ | ⊖2 +1 | −0 −0| ±? ⊖1 | ⊕0 +0 −

− − ++ −

+

3 1 3 5 4 4


Subcase 1.3. Signatures at root-entry: r ≡ 3 mod 4, ℓ ≡ 0 mod 4 ≡ 0 mod 2.

✷ ↓ 4 ↓ 4 ↓ 5 ↓ . . . ↓ 4 ↓ 5 ↓ 3 ↓ 1 ↓ . . . ↓ 1 ↓ 3 ↓ 1 ↓ ✷

The result is the same as for the preceding subcase with colours reversed (Figure 13).

−+ ++ − −− − +0 | . . |

+ +0 | ⊖1 . | −− | ⊕2 −1 | +0 +0| ±? ⊕2 | ⊖0 −0 +

+ + −− +−

1 3 1 4 4 5


Subcase 1.4. r ≡ 2 mod 4, ℓ ≡ 1 mod 4 ≡ 1 mod 2. Signatures at root-entry:

✷ ↓ 4 ↓ 5 ↓ 5 ↓ . . . ↓ 4 ↓ 5 ↓ 3 ↓ 1 ↓ . . . ↓ 3 ↓ 1 ↓ 3 ↓ ✷

This case resolves in the given columns on a size 3 corner exactly like the first subcase(1.1, because 3 = 1 and 4 = 5; see Figure 14).This completes the unbalanced ++−+ case. We can extract the following general

observation:


+− −− + ⊕+ + − | . . |− − | . . | −⊕ | . . | ⊕ +| . . | − − +

+ + −− +−

3 1 3 4 5 5


Corollary 10. (1) Column colour patterns ↓ 1 ↓ 3 ↓ and ↓ 3 ↓ 1 ↓ right propagateby ‘self-replication’. (2) Column colour patterns ↓ 4 ↓ 4 ↓ 5 ↓ and ↓ 5 ↓ 5 ↓ 4 ↓ leftpropagate by ‘mutation’.

Proof. (Corollary 10) Figure 15 shows how new generations (n) arise out of old(o) in the first pattern. The statement for the second pattern follows because columns1 and 3 are dual. ✷

−0+0 +0+0 −0 −2−0 −0 +1 +1

+0 +3 −4 −1−4 −6 +7 +5

−6

1 3 1 3o o n n

−6+9 +5−8 −6 +1+7 +5 −4 −0

−4 +3 +0 −0−2 −0 +0 +0

+0 −0 −0+0 +0

−0

5 5 4 4 5o o o

n n n

Figure 15. Corollary 10

Case 2. Supose, otherwise, that the unbalanced TCT sequence + − ++ occurssomewhere on the trunk. We may assume that the left adjacent TCT has colour ‘−’(else we are back in the preceding case, possibly exploiting the remark in passingabout admissible locations of its right-most column), but this choice is in fact forced.


(I.e., if this TCT is right adjacent to the root we still get the patterns observed inCase 1.) Aided by the left bordering TCT of colour ‘−’ there result a left-cascadingξ pattern and a right-cascading η one, each lasting again until it penetrates the root.(See Figure 16.) The overlap shapes of the left and right cascading colour patternson the root are as in case 1.

.

. +F

. +H −B

. −G −E +8−E +D +A −5

+A −9 −7 +0−7 +6 +2 −0

+4 −1 −3 +0−1 +2 −1 +0

+1 −3 +4 −1+4 −5 +6 −5

−7 +8 −9 +6−9 +A −B +A

+C −D +E −B+E −F .

. ..

2 0 2 0 2 0 3 2 1 0 3

Figure 16. Case 2.

Subcases arise depending on r and ℓ as defined before.

Subcase 2.1. ℓ ≡ 1 mod 4, r ≡ 2 mod 4 ≡ 0 mod 2. Signatures at root-entry:

✷ ↓ 2 ↓ 0 ↓ 2 ↓ . . . ↓ 2 ↓ 0 ↓ 3 ↓ 2 ↓ . . . ↓ 1 ↓ 0 ↓ 3 ↓ ✷

Outcome is immediately as for Subbcase 1.4 (Figure 17).


✷ ↓ 0 ↓ 2 ↓ 0 ↓ . . . ↓ 2 ↓ 0 ↓ 3 ↓ 2 ↓ . . . ↓ 2 ↓ 1 ↓ 0 ↓ ✷

Outcome is as for Subcase 1.3 (Figure 18).


✷ ↓ 2 ↓ 0 ↓ 2 ↓ . . . ↓ 2 ↓ 0 ↓ 3 ↓ 2 ↓ . . . ↓ 3 ↓ 2 ↓ 1 ↓ ✷

Colours engendered by (†) from the initial choices are as in Figure 19. Again, eithercolour on the top left point in the root (marked ‘?’) results in a m.c. (potential limbsas circled).


−+ −+ − ⊕− + − | . . |

+ − | . . | +⊕ | . . | ⊕ −| . . | − − +

− + ++ −−

1 0 3 2 0 2


++ −− + −− + − | ⊕1 . |− +0 | . . | −

+0 | ⊖1 +2 | −0 +| ±? ⊖3 | +0 ⊕0 −

+ − −− +

+

2 1 0 0 2 0



✷ ↓ 0 ↓ 2 ↓ 2 ↓ . . . ↓ 2 ↓ 0 ↓ 3 ↓ 2 ↓ . . . ↓ 0 ↓ 3 ↓ 2 ↓ ✷

Colours forced as in Figure 20. There result m.c.s from either colour on the rootpoint marked ‘?’ (indicated by circles).This settles the second unbalanced TCT case. As before we can extract a general

observation:

Corollary 11. (1) Column colour patterns ↓ 0 ↓ 2 ↓ 0 ↓ and ↓ 2 ↓ 0 ↓ 2 ↓ leftpropagate by ‘mutation’. (2) Column colour patterns ↓ 3 ↓ 2 ↓ 1 ↓ and ↓ 1 ↓ 0 ↓ 3 ↓left propagate by ‘mutation’.

Proof. (Corollary 11) New generations arise out of the old as in Figure 21. ✷


⊕0− +− + −⊕0 − +0 | ±? −2 |

− +0 | ⊖1 ⊖4 | +− | −3 +3 | + −0| +2 +1 | −0 −0 +0

− + ++ −

−

3 2 1 2 0 2


−− +0+ − ++ − + | ⊖1 . |

+0 −0 | ±? ⊖1 | −−0 | ⊕1 ⊕2 | −0 +

| . −1 | + + −+ − −− +

+

0 3 2 0 2 0


Case 3. Assuming that no unbalanced TCT sequences exist there can then onlybe alternating periodic patterns on the TCTs. We can assume these start on ‘+’ onthe leftmost TCT.

Subcase 3.1. First assume there is a period 2 TCT sequence, e.g., + − +−, onthe trunk. The pattern of Figure 22 arises immediately via (†). The colour pat-terns compatible with this distribution are completely determined by the followingalternatives:Scan the trunk starting from the top left element proceeding down through each

column before moving into the column next right. (This defines an ordering of thetrunk elements.) Either the colours observed are consistent with a 2 ↓ 0 ↓ 2 ↓ 0 . . .signature (i.e., one kind of ξ pattern), or there is a first break from that pattern inthat order, in which case we obtain a . . . ↓ 3 ↓ 1 ↓ 3 ↓ 1 signature or its dual (i.e., asecond kind of η pattern) on a tail of the trunk.


.

. −5

. +7 +2

. −6 +4 −0+4 −3 −0 +0

−1 +0 +0 −0+0 −0 −0

−0 +0+0

2 0 2 0o o o

n n n

+0−0 +0−0 +0 −0+0 −0 +0 −1

−0 +0 −1 +2−0 +2 −2 +1

+4 −5 +6+6 −9

−9

3 2 1 0 3o o o

n n n

Figure 21. Corollary 11

+0−0

−1 +0+1 −0

−1 +0+1 −0

. . .

Figure 22. Case 3.1.

By the preceding diagramme, a 02 pattern breaker can only occur in one of twopositions in a column (1st down or 3rd down). Since columns to the right of thefirst 02 pattern breaker play no role in the resulting right propagation, there arejust four cases, determined by the colour on the top and the position of the patternbreaker in the column. By symmetry on the column top colour, there are just twoessentially distinct cases. See Figure 23, where pattern breakers are indicated withboxes around colours. Note that colours further down the top diagonal are actuallyforced as indicated by the stage identifiers, so this pattern continues into the rootleading to the same overlap as before.If the pattern breaker resides at distance 3 or greater from the root, the colour

distribution in its vicinity is as in Figure 24, yielding a m.c.. (No clash would ariseif the trunk were of even length!) Since the clash only involves the two columnsadjacent left and right to the root, the same clash arises if the pattern breaker is atdistance 1 or greater from the root. Given the width of the trunk, it cannot resideat distance less than 3 of both sides of the root.Suppose it lies in the column immediately to the right of the root. If it’s located

in 3rd position down from the top, the resulting clash is as before. Otherwise we


+0⊟2 −0−1 +3 +0+3 +1 −4 −0

−4 −6 +7 +5+7 +9 −A −8

−A −C +D +B

3 1 3 1(2) (0) (2)

+0+0 −0−1 . +0⊞2 +1 −3 −0

. −6 +7 +5+7 +9 −A −8

−A −C +D +B

∗ ∗ 3 1(2) (0) (2) (0)

Figure 23. Pattern breakers.

+0−0 −0−0 +0 ⊕0+0 +0 −0 | . . |

−0 −0 | . . | +0⊕0 | . . | ⊕0 −0

| . . | −0 −0 +0−0 +0 +0

+0 −0−0

3 1 3 2 0 2(2) (0) (2)

Figure 24. Pattern breaker at distance 3

essentially get a complete 13 ξ pattern (which we know cannot avoid a m.c. given theconditions on b and c); see Figure 25.Suppose, finally it lies immediately to the left; in that case we get nearly complete

02 ξ-patterns, with much the same outcome (see Figure 26).


⊕0−0 −0−0 +0 +0⊕0 +0 −0 | ±? . |

−0 −0 | +1 . | +0+0 | ⊖2 +1 | ⊖0 −0

| . . | −0 +0 +0+0 +0 −0

−0 −0+0

3 1 3 3 1 3(2) (0) (2)

Figure 25. Pattern breaker at distance 1 to the right

This completes the period 2 alternating case.

Subcase 3.2. In the period 4 alternating case we immediately get the distributionof Figure 27.The colour distribution is now completely determined by a balance tipping choice

in the second column of a monochrome TCT pair (as indicated by the ‘?’ above).Scan the balance tipping candidates (BTCs) from left to right. Either they are all

chosen to result in a ζ-like pattern 5 ↓ 5 ↓ 4 ↓ 4 ↓ . . . . Or the first such choicebreaking that pattern results in an η-like pattern . . . ↓ 3 ↓ 2 ↓ 1 ↓ 0 ↓ 3 . . . on a tailof the trunk.By symmetry, it suffices to consider BTCs in the + topped columns, as indicated

in Figure 28. (Where colours on TCTs from the fourth column left are determinedan alpha-numeric on the colour indicates the order of generation.)We conclude this case by showing that either balance tipping scenario generates

a desired m.c. in the root, using the condition b ≡ 1 mod 4 to determine the colourcolumn signatures near root entry. See Figure 29.This exhausts all cases and completes the verification of Part (i) and hence all of

Theorem 2. ✷


+0+0 −0−0 −0 ⊕0−0 +0 +0 | . . |

+0 −0 | . . | +0⊕0 | . . | ⊕0 −0

| . . | −0 −0 +0−0 +0 +0

+0 −0−0

2 0 3 2 0 2

+0+0 −0−0 −0 ⊕0−0 +0 +0 | . . |

+0 −0 | . . | +0⊕0 | . . | ⊕0 −0

| . . | −0 −0 +0−0 +0 +0

+0 −0−0

2 0 ∗ 2 0 2

Figure 26. Pattern breaker at distance 1 to the left

+0−1 +0

? −0+1 −0

−1 +0−1 +0

. . .

Figure 27. Case 3.2.

.


+0−1 +0. ⊞2 −0. −3 +1 −0−2 +4 −5 +1

−5 +6 −7 +6+8 −9 +A −7

+A −C +D −B−D +E −F +C

−F

2 1 0 3 2 1

+0−1 +0+3 ⊟2 −0−4 +3 +1 −0

−2 −4 +3 +1+5 −4 −2 +0

+2 +5 −4 −1−6 +5 +2 −0

−2 +1

5 5 4 4 5 5

Figure 28. Balance tippers.

.


+0−0 +0−0 +0 ⊖0+0 −0 +0 | . . |

−0 −0 | . . | +0⊖0 | . . | ⊖0 +0

| . . | . +0 −0. −0 +0−0 +0

−0

3 2 1 ∗ 2 1

+0−0 +0+0 −0 −0−0 +0 +0 | ⊖3 −1 |

−0 −0 | +2 +2 | +0+0 | ⊕4 −1 | −0 +0

| ±? ⊕2 | +0 ⊖0 −0−0 +0 +0

−0 −0+0

5 5 4 5 5 4

Figure 29. Balance tipping scenarios.

References

[1] Axenovich, M. and Manske, J.: On monochromatic subsets of a rectangular grid, Integers: Elec-tronic Journal of Combinatorial Number Theory 8 (2008), #A21

[2] Burr, S. and Loo, S.: On Rado Numbers I / II, unpublished[3] Hegarty, P.: Extremal subsets of {1, . . . , N} avoiding solutions to linear equations in three vari-

ables, Electronic Journal of Combinatorics, 14 (2007), #R74[4] Landman, B. and Robertson, A.: Ramsey Theory on the Intergers, AMS Student Mathematical

Library, Vol. 24, Providence RI, 2003[5] Rado, R.: Studien zur Kombinatorik, Mathematische Zeitschrift, (1933), 424-480[6] Ruzsa, I. Z.: Solving a linear equation in a set of integers I, Acta Arith. 65 (1993), 259-283[7] Ruzsa, I. Z.: Solving a linear equation in a set of integers II, Acta Arith. 72 (1995), 385-397

C. Elsholtz, Institut fur Mathematik A, Technische Universitat Graz, Steyr-

ergasse 30, A-8010 Graz, Austria, [email protected]

F. P. Weber, 94 Spencer Place, Leeds LS7 4DT, UK, [email protected]

Monochromatic solutions of linear equations

Documents

Transcript of Monochromatic solutions of linear equations