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Black Mirror II Hints
Elimination-Style Logic Process
The elimination board will start out as shown below. Notice that for each space except the one we know must end up empty, there's a B for a blue lion head and a G for a gold lion head. This is because we don't yet know which color of lion head any of the spaces needs to have.
BG BG BG BG BG BG BG BG BG BG BG BG BG BG BG
Looking at the marks to the left of the rows, we can see that the top row and the third row are easy to figure out. The top row has no gold marks, so every space in it must contain a blue lion head. The third row has four gold marks, so it has to contain all gold lion heads. That gives us this adjusted board:
B B B B BG BG BG BG G G G G BG BG BG
Now let's look at the columns. Columns two and four are easy to figure out now that we've made a few eliminations. Column two has three blue marks above it, and our board shows only three possible spaces left for blue lion heads in column two, so we can eliminate gold lion heads from all three of those spaces. With column four, there can only be one blue lion head, and it has to be in the top space, so we can eliminate blue lion heads from all other spaces in the column. That gives:
B B B B BG B BG G G G G G BG B BG
That's all the easy, straightforward eliminations we can make. The best way to proceed now is with a guess-and-check strategy. Let's guess that the second row's other gold lion head goes in the first slot, then see if that causes any problems. Using straightforward eliminations, we find that our guess yields the following viable solution board:
B B B B G B B G G G G G B B G
Actually, the other way works, too. Guessing that the second row's other gold lion head goes in the third space, we get this viable solution board:
