lvii. Chameleons

Problem. An island has $a$ red, $b$ green and $c$ blue chameleons. When two of different colours meet, both change to the third colour. Can all become blue?

Answer: yes iff $a\equiv b(\mathrm{mod}3)$ and at least two colours are present (or $a=b=0$ already).

bool allBecome(long long a, long long b, long long c) {   // a, b: counts of the other two colours, c: target colour
    int nonzero = (a > 0) + (b > 0) + (c > 0);
    if (a == 0 && b == 0) return true;
    return (a - b) % 3 == 0 && nonzero >= 2;
}

Proof: the mod-3 invariant and a construction

Invariant. A meeting changes the three counts by $(-1,-1,+2)$ in some order, so every pairwise difference changes by $0$ or $\pm 3$: the differences mod 3 are invariant. All blue means $a=b=0$, so $a\equiv b(\mathrm{mod}3)$ is necessary; and if only one colour is present, no meeting can happen. Construction. Assume $a\ge b$, $a\equiv b$. While $b>0$, a red-green meeting decreases both by one; this ends at $(a-b,0,c+2b)$ with $a-b\equiv 0(\mathrm{mod}3)$. If $a-b=3t>0$, a red-blue meeting gives $(3t-1,2,\cdot )$, then two red-green meetings give $(3t-3,0,\cdot )$; repeat $t$ times. A blue chameleon is needed for that step; it exists because at least two colours were present (if $c=0$ initially, then $b>0$, which creates blue ones). Checked against BFS for all counts up to 7.