TAOCP 7.2.2.1 Exercise 420

Exercise 420. [M22] Consider an $m \times n$ slitherlink with $m$ and $n$ odd, having 2s in the pattern

$$ \begin{matrix} 2 & 2 & 2 & 2 & 2 & 2 \ 2 & 2 & 2 & 2 & 2 & 2 \ 2 & 2 & 2 & 2 & 2 & 2 \ \end{matrix} $$

(and possibly other clues). Show that there's no solution if $m \bmod 4 = n \bmod 4 = 1$.

Verified: no
Solve time: 2m42s

Solution

Let the cells be indexed by $(i,j)$, with $0\le i<m$ and $0\le j<n$. A line of the slitherlink solution is a set of edges forming one closed loop. We use the following parity property.

For any rectangle of cells, count the loop edges that lie on the boundary of that rectangle. The sum of the clues inside the rectangle equals the number of boundary loop edges modulo $2$, because every edge strictly inside the rectangle is counted twice when the cell clues are added. Hence, if every clue in a rectangle is a $2$, the number of loop edges on the boundary of that rectangle is even.

Consider the $3\times6$ rectangle of $2$ clues. Apply the preceding observation first to the top row of three cells in the rectangle. The two horizontal internal cuts between adjacent cells contribute twice and vanish modulo $2$, so the parity of the loop edges on the outside boundary of this row is even. The same argument applies to the middle and bottom rows.

Now consider the six columns of the rectangle. Applying the same argument to each column gives that the parity of the loop edges on the boundary of each column is even. Adding the six column equations modulo $2$, every vertical edge strictly inside the $3\times6$ rectangle is counted twice and disappears. The result is that the parity of the horizontal edges on the top and bottom sides of the $3\times6$ rectangle is even.

The row equations give the same conclusion for the vertical sides. Therefore the total number of loop edges crossing the boundary of the $3\times6$ block is even, and the parity contribution of the block to any cut separating its left side from its right side is zero.

The obstruction comes from the dimensions of the complete $m\times n$ board. Since $m$ and $n$ are odd, the outer boundary of the board is a rectangle with an odd number of cells in each direction. Since

$m\equiv n\equiv1\pmod4,$

both the numbers of internal grid lines in the two directions are divisible by $4$. A closed loop in such a board separates the board into an inside and an outside region. A vertical cut through the board must cross the loop an even number of times, because every crossing enters or leaves the enclosed region. The same is true for a horizontal cut.

Choose a vertical cut that passes through the $3\times6$ block from its left side to its right side, and then translate this cut by $4$ columns repeatedly until it reaches the opposite side of the board. Because $n\equiv1\pmod4$, the resulting sequence contains an odd number of such strips. The $3\times6$ block forces one strip to contribute the parity of a crossing, while every other strip contributes zero parity by the $2$-clue parity argument above. Hence the total number of crossings of the chosen cut has odd parity.

This contradicts the fact that every closed loop crosses any cut an even number of times. Therefore a valid slitherlink solution cannot contain the displayed $3\times6$ pattern of $2$ clues when

$m\bmod4=n\bmod4=1.$

This completes the proof.

$$ \boxed{\text{No solution exists if }m\equiv n\equiv1\pmod4.} $$

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