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What's our data?

$ A_{m,n} $ pieces and they have 3^4 possible shapes. We will assign them randomly at the beginning, satisfying boundary conditions.

Let $A_{m,n,s}$ represent original "pokiness" in each direction. $0 =|$, $-1 = \langle$, and $+1= \rangle$.

Let m represents row,n represents column as a helpful reminder to TOP/BOT, LEFT/RIGHT relations.

Which satisfies formula 
\begin{align*}
A_{0, n, \text{TOP}} = 0 \\
A_{k, N-1, \text{RIGHT}} = 0 \\
A_{k, 0, \text{LEFT}} = 0 \\
A_{M-1,0, \text{BOT}} = 0 \\
A_{m, n, \text{BOT}} + A_{m + 1, n, \text{TOP}} = 0 \\ 
A_{m, n, \text{RIGHT}} + A_{m, n + 1, \text{LEFT}} = 0
\end{align*}

Define $\text{TOP}=0$, $\text{RIGHT}=1$, $\text{BOT}=2$, $\text{LEFT}=3$ all $(\text{mod} 4)$, and $\oplus$ to indicate sum modulo 4.

Once scrambled, we have $ B_{k} $ pieces $K= M\times N $ each with $3^4$ possible shapes. Let their post rotation pokiness be also be handled through $B_{k,s}$ notation. We will assume pieces don't get flipped, or easy to figure out which side is up.

We have to determine mapping $ k \rightarrow (m, n) $ and the orientation $ k \rightarrow s $ . So one natural way to do is one hot encode
$X_{k, m, n}$ and $Y_{k, s}$ :

\begin{align*}
\sum_{m,n} X_{k,m,n} = 1 \\
\sum_{s} Y_{k,s} = 1 \\
\end{align*}

With jigsaw constraints (All 6 above have analogues). The binary ones are more complex which would look like:
\begin{align*}
X_{k,m,n} \land Y_{k, s} \land 
X_{l,m+1,n} \land Y_{l, t} 
\implies (B_{k,BOT \oplus s} + B_{l,TOP \oplus t} == 0)
\end{align*}

Since rhs is constant, this would simply reduce to a constraint only when it is false. Meaning if rhs is false, then lhs must be false, otherise lhs can be anything.

Tue, 19 Dec 2023 05:02 GMT