Copy-squaring of a two-colour triple system loses a factor of two

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This page records obstructions proved by Plasma AI for the squaring interface of two-colour stepping-up for triples, whose target is Problem 564. The proofs are author-recorded and have not been independently reviewed. They concern only the rules named here.

Setting

Let $\chi$ be a red/blue colouring of the triples of a set $V$ with no monochromatic $n$-set. A copy-squaring colours the triples of $V\times V$, whose points $(t,\alpha)$ have a block $t$ and a position $\alpha$, as follows:

• three distinct blocks: $\chi$ of the three blocks;
• one block: $\chi$ of the three positions;
• two blocks, a pair $(t,\alpha),(t,\alpha')$ in one block and an outsider $(s,\beta)$ in the other: a mixed rule $\mu$.

The squaring interface of the route asks for a class of such systems closed under one operation that squares the vertex count and adds a constant $C$ to the forcing size $n$. The first two items alone already bound a monochromatic set $S$ by $(n-1)^2$: its blocks form a monochromatic set, and so do the positions in each block. Everything rests on $\mu$.

Two natural mixed rules:

• the positions rule: $\mu=\chi(\alpha,\alpha',\beta)$, the colour of the three positions when they are distinct;
• the outsider-block rule: $\mu=\chi(\alpha,\alpha',s)$, the colour of the pair's positions and the outsider's block when they are distinct.

Proposition

Proposition. Let $U\subseteq V$ be a red set of $k\ge3$ points of the seed, $t_0=\min U$ and $t_1=\max U$. In the copy-squaring with the positions rule,

$$S_1=\{(t_0,\alpha):\alpha\in U\setminus\{t_1\}\} \cup\{(r,t_1):r\in U\setminus\{t_0\}\}$$

is a red set of $2(k-1)$ points. In the copy-squaring with the outsider-block rule,

$$S_2=\{(t_0,\alpha):\alpha\in U\setminus\{t_1\}\} \cup\{(t_1,\beta):\beta\in U\setminus\{t_0\}\}$$

is a red set of $2(k-1)$ points.

Proof. In $S_1$, block $t_0$ holds $k-1$ points and every other block one point; the positions in block $t_0$ avoid $t_1$. A triple inside block $t_0$ is coloured by three positions in $U$; a triple with two points in block $t_0$ is coloured by $\chi(\alpha,\alpha',t_1)$ with all three in $U$; a triple meeting three blocks is coloured by three blocks in $U$. All are red. In $S_2$ the blocks are $t_0<t_1$. Triples inside a block are coloured by positions in $U$. A pair in block $t_0$ with an outsider in block $t_1$ is coloured by $\chi(\alpha,\alpha',t_1)$ with $\alpha,\alpha'\ne t_1$; a pair in block $t_1$ with an outsider in block $t_0$ by $\chi(t_0,\beta,\beta')$ with $\beta,\beta'\ne t_0$. All are triples of $U$, so red. $\square$

With $k=n-1$ both rules give monochromatic sets of $2n-4$ points, so an increment $C$ fails as soon as $n\ge C+4$, and iterating such a squaring multiplies rather than adds. This also rules out restricting to a specially chosen class closed under either operation. Write $a_j$ for the largest monochromatic size after $j$ iterations. Once the order is at least three, $a_j\ge3$, and the proposition gives $a_{j+1}\ge2a_j-2$. Therefore $a_{j+s}\ge2+(a_j-2)2^s$, which eventually exceeds any proposed bound $n_0+C(j+s)-1$. A seed of order at least two reaches order at least three after at most one squaring. The witnesses never invoke a mixed triple with repeated seed arguments, so arbitrary choices in those exceptional cases cannot repair the recurrence. The finite checks in evidence/main.py build both sets for a seed on 14 points with $k=7$ and recompute every product triple's colour.

Where the mixed rule must act

The two sets show the two configurations every mixed rule must control.

• Fat and thin blocks ($S_1$). One block holds a monochromatic set $A$ of positions and each other block a single point $(r,\beta_r)$. The blocks $r$ together with the fat block form a monochromatic set, and the only further condition is that $\mu$ be red on each pair of $A$ with the outsider $(r,\beta_r)$. The position $\beta_r$ is chosen separately for each thin block, so an estimate that completing points are rare does not by itself exclude the configuration.
• Two fat blocks ($S_2$). Blocks $t_0<t_1$ hold position sets $A$ and $B$. If $\mu$ does not read the outsider's position, the condition splits: $A$ must be red under $\chi$ and a red clique of the pair colouring $\mu(t_0,t_1;\cdot,\cdot)$ on lower-block pairs, and $B$ likewise for upper-block pairs, independently. An increment $C$ then needs, for every $t_0<t_1$, the two largest such sets to have total size at most $n+C-1$, which neither natural rule provides.

No mixed rule controlling both configurations is supplied here. A rule depending only on blocks and on the pair's positions, as in the second item, reduces the question to a property of pair colourings attached to the seed, which the inductive hypothesis (no monochromatic $n$-set) does not supply; the class of admitted systems would have to carry it as an extra invariant and reproduce it after squaring.