There are 16 cards: 8 black and 8 white. On each card is a number between 1 and 4 inclusive. There are:

4 white cards with a 1     6 black cards with a 1
2 white cards with a 2     2 black cards with a 2
1 white card with a 3 
1 white card with a 4 

Arrange the cards into a 4x4 grid such that the number on each card represents the number of black cards among its 2, 3 or 4 orthogonally adjacent cards, and so that the top-rightmost card is a black card with a 1 on it.

Consider the sum of all sixteen cards' numbers, and how each black card contributes to that sum.

Each corner black card has two neighbors, and so contributes 2 to the sum. Each edge black card contributes 3 and each center black card contributes 4. The total for these 16 cards in the problem is 25 (4 + 3 + 4*2 + 10*1).
Suppose there are k corner black cards, e edge, and c center. Then we can write:

k + e + c = 8 (there are 8 black cards total)
2k + 3e + 4c = 25 (using the logic above)

Now, since 25 is odd, then the lhs is also odd and so e is odd.

Since there's a white card with 4 black neighbors, that card is a center card. There are at most three black center cards, therefore. But that white card has two center-card neighbors so there are also at least 2 black center cards. So c is either 2 or 3.

Suppose c is 3. Then the center cards contribue 12 to the total leaving 13 -- e is then either 1 or 3, for if it's 5 there are too many neighbors and if it's 7 there are too many black cards. But if e is 1 then k is 5 to make 25 and that's too many black cards (e+k+c = 9 > 8) so if c is 3 then e is also 3 and k is 2.

Suppose c is 2. Then the center cards contribute 8 and the rest 17. e can be 1,3, or 5. But if e is 1 then k is 7 which is too many black cards. Ditto for e = 3 which makes k 4. So if c is two, e is 5 and k is 1.

It's important to note that there are no cards with zeros on them.

For convenience, number the spots like so:

1  2  3  4
5  6  7  8
9  10 11 12
13 14 15 16

Consider the case of c = 3. Then one of the black center cards already has two black neighbors. No black card has a number greater than 2 on it, so the remaining two neighbors must be white. But these two neighbors are also the *only* neighbors of the corner diagonally adjacent to the black center card in question. If both (all) of that corner's neighbors are white, it would have a zero on it. Since no card has a zero, there is no solution when c = 3.

Accordingly, the only solutions are those with two center black cards (which must be on a diagonal to share a neighbor--the W4), one black corner card, and five black edge cards. Without loss of generality, we can put the W4 at #6 and draw a partial diagram:

?   B   ?   *
B   W4  B   ?
?   B   W   -
*   ?   -   ?

Suppose the lone black corner is either of 4 or 13 (marked * in the diagram.) Then the W3 spot is taken by 3 or 9 respectively and so the edges marked - must be W or else the lower right center white will also have 3 or 4 written on it and both of those cards are used up. But if both - cards are W then #16 gets a zero, so neither * can be black; both are white. Also, since no edge spot can be surrounded entirely by black at this point, the W3 must be at #11. Since the diagram is thus far symmetric around the upper left - lower right diagonal, we lose no generality in assuming #15 is the W3's third black neighbor:

?   B   ?   W
B   W4  B   ?
?   B   W3  W
W   ?   B   ?

CASE ONE

Assume the last B is at #1. Then it's B2 and the #16 is W1. Two of the remaining 4 are B, but they can't both be neighbors of the same corner or the *other* corner would be W0. Whichever neighbor of #13 is B, it's B2 and so the #3 must be W or there'd be three B2s. That forces #8 to be B. Now, #9 can't be B because if it were, then #1, #5 and #9 would all be B2 and that's too many. So #9 is W and that leaves #14 as B for this solution:



B2  B1  W2  W1
B1  W4  B1  B1
W2  B1  W3  W1
W1  B2  B1  W1

CASE TWO

Assume the last B is at #16. Then #1 is W2. As before, both neighbors of either #4 or #13 can't be black or the opposite corner would be W0. Now, though, we have no B2s so far, and only one can come from the neighbor of #13, and #8 can't ever be a B2, so the 2nd B2 is #3 and #8 is W. But #9 can't be W or #5 would be B0, so #9 is B and #14 is W giving the second solution:


W2  B1  B2  W1
B1  W4  B1  W1
B2  B1  W3  W1
W1  W2  B1  B1

Since there is only one decision point that yields exactly one solution for each of its two possible answers, these are necessarily (minus reflections and rotations) the only solutions.

Posted by Paul on 2010-11-30 22:50:57