TAOCP 7.2.2.1 Exercise 411

Edit The statement is false.

Section 7.2.2.1: Dancing Links

Exercise 411. [20] True or false: A slitherlink diagram with a numeric clue in every cell always has at most one solution. Hint: Consider the $2 \times 2$ case.

Verified: no
Solve time: 2m17s


Edit

Solution

The statement is false. It suffices to exhibit one slitherlink diagram with a clue in every cell that has two distinct solutions.

Consider the (2\times2) diagram

[ \begin{matrix} 2&3
3&2 \end{matrix}. ]

Label the vertices of the grid by ((r,c)), with (0\le r,c\le2).

First consider the loop consisting of the eight edges

[ (0,0)(0,1),\quad (0,1)(0,2),\quad (0,2)(1,2),\quad (1,2)(1,1), ]

[ (1,1)(2,1),\quad (2,1)(2,0),\quad (2,0)(1,0),\quad (1,0)(0,0). ]

These edges form the single cycle

[ (0,0)\to(0,1)\to(0,2)\to(1,2)\to(1,1)\to(2,1)\to(2,0)\to(1,0)\to(0,0). ]

Now check the four clues. The upper-left cell has boundary edges

[ (0,0)(0,1),\quad (0,0)(1,0),\quad (0,1)(1,1),\quad (1,0)(1,1), ]

and the loop contains exactly the first two, so its clue is (2).

The upper-right cell has boundary edges

[ (0,1)(0,2),\quad (0,2)(1,2),\quad (0,1)(1,1),\quad (1,1)(1,2), ]

and the loop contains the first, second, and fourth edges, so its clue is (3).

The lower-left cell has boundary edges

[ (1,0)(1,1),\quad (1,0)(2,0),\quad (2,0)(2,1),\quad (1,1)(2,1), ]

and the loop contains the second, third, and fourth edges, so its clue is (3).

The lower-right cell has boundary edges

[ (1,1)(1,2),\quad (1,2)(2,2),\quad (2,1)(2,2),\quad (1,1)(2,1), ]

and the loop contains exactly the first and fourth edges, so its clue is (2).

Therefore this is a valid solution.

A second solution is given by the loop consisting of

[ (0,1)(0,2),\quad (0,1)(1,1),\quad (0,2)(1,2),\quad (1,0)(1,1), ]

[ (1,0)(2,0),\quad (2,0)(2,1),\quad (2,1)(2,2),\quad (1,2)(2,2). ]

These edges form the single cycle

[ (0,1)\to(0,2)\to(1,2)\to(2,2)\to(2,1)\to(2,0)\to(1,0)\to(1,1)\to(0,1). ]

For the upper-left cell, the loop contains the two boundary edges

[ (0,1)(0,1+1)=(0,1)(0,2) ]

and

[ (0,1)(1,1), ]

so the clue is (2).

For the upper-right cell, the loop contains

[ (0,1)(0,2),\quad (0,2)(1,2),\quad (0,1)(1,1), ]

and no other boundary edge, so the clue is (3).

For the lower-left cell, the loop contains

[ (1,0)(1,1),\quad (1,0)(2,0),\quad (2,0)(2,1), ]

and no other boundary edge, so the clue is (3).

For the lower-right cell, the loop contains

[ (1,2)(2,2),\quad (2,1)(2,2), ]

and no other boundary edge, so the clue is (2).

Thus the same (2\times2) diagram has two valid slitherlink solutions. They are distinct because the first solution contains the edge

[ (0,0)(0,1), ]

whereas the second solution does not.

Hence a slitherlink diagram with a numeric clue in every cell need not have a unique solution.

[ \boxed{\text{False}} ]