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- Clue A1 immediately tells us that A3, A5, E1, E3 and E5 are safe.
- Clue E5 tells us that A4 was rigged by the black spy.
- We know that there’s a pair of consecutive safe spots in column 2, thanks to A3’s clue. Let’s lay out all the possibilities: the two safe spots can be A2 and B2, B2 and C2, C2 and D2, D2 and E2, or E2 and F2.
- But neither A2 nor E2 can be safe, because that would give us three consecutive safe spots in a row, which violates A4’s clue. So we can eliminate every possibility that includes either A2 or E2.
- That leaves us with two possibilities: B2 and C2, or C2 and D2. However, consider the clue you got from E1. There’s one corner that’s adjacent to three rigged spaces (and no safe spaces). That corner has to be A1, since A6, F1 and F6 are each adjacent to at least one safe space.
- That means that B1, A2 and B2 are all rigged. Since we know that B2 isn’t safe, the two consecutive safe spots in column 2 have to be C2 and D2.
- Now let’s look at Column 4. Using the same logical process, we can immediately rule out any pairs that include A4 or E4. That leaves us with two possibilities: B4 and C4, or C4 and D4.
- We don’t have the information necessary to narrow this down any further, but we do know that either way, C4 is safe.

- There are three consecutive black traps in Row F, according to C4’s clue. Once again, let’s narrow down the possibilities.
- A5’s clue told us that two corners contain white traps and two corners are safe. In other words, none of the corners contain black traps, which means that the three consecutive black spaces in Row F do not include F1 or F6.
- That leaves us with two possibilities. The three consecutive black traps occupy either F2, F3 and F4 or F3, F4 and F5. Either way, F3 and F4 contain black traps.

- We know that E4 cannot be safe, or else it would contradict A4’s clue. F4’s clue tells us that it can’t have been rigged by the white spy, either. Thus, E4 contains a black trap.
- We’ve already identified two safe spaces adjacent to E4 (E3 and E5), so none of the other spaces surrounding it are safe. This includes F5, which we know cannot contain a white trap either (F4’s clue). Thus, F5 was rigged by the black spy.
- Thus far, you’ve identified nine safe spaces (A1, A3, A5, E1, E3, E5, C2, D2 and C4). F5’s clue tells us that there are 12 safe spaces in total, so there are three more you need to find. From A3’s clue, we can conclude that two of those are in Column 6, and the last remaining one is either B4 or D4. From this point forward, we can assume that every remaining space outside Column 4 and 6 is rigged.
- Therefore, we know that F2 was rigged, but there are exactly three consecutive black traps in Row F (and we found them all), so F2 cannot contain a black trap. F2 was rigged by the white spy.
- We also know that F1 was rigged, but it couldn’t have been rigged by the black spy either, since none of the corners contain black traps (A5’s clue). F1 was rigged by the white spy.

- 8. Since Row F must contain at least one safe space (F1’s clue), we can conclude that F6 is safe.
- That leaves us with two remaining safe spots; one in Column 6, and one in Column 4. But the two safe spots in Column 6 have to be consecutive (A3’s clue), so E6 is safe.
- And now that we’ve identified two safe corners (A1 and F6), we can conclude that the other two corners contain white traps (A5’s clue). A6 contains a white trap.

- A6’s clue helps us figure out D4’s status. D4 cannot contain a black trap, since that would give us three consecutive black traps in Column 4. However, we know that it cannot be safe either, since E4 is only adjacent to two safe spaces (E4’s clue). D4 contains a white trap.
- Now that we know that D4 isn’t safe, we can conclude that B4 is (A3’s clue).

- C2’s clue told us that one column contains four white traps. Let’s figure out which one it is.
- In order for Column 1 to have four white traps, B1, C1 and D1 would all have to contain white traps, which would contradict A6’s clue.
- Column 3 only contains one white trap (B4’s clue).
- We’ve already identified every space in Column 4, and we only found one white trap.
- We haven’t found any white traps in Column 5 so far, and there are only 3 spaces remaining.
- In order for column 6 to have four white traps, F2, F3 and F4 would all have to contain white traps, which, again, would contradict A6’s clue.
- The only remaining column is Column 2. We’ve identified one white trap in that column so far, and there are only three more spaces remaining, so A2, B2 and E2 were rigged by the white spy.

- B2’s clue tells us that one corner is adjacent to three spaces that all have the same status. We know that this can’t apply to F1 (adjacent to two safe spaces and one white trap) or F6 (adjacent to two safe spaces and one black trap) so that leaves us with A1 and A6.
- A6 is adjacent to one safe space. In order for it to fit B2’s clue, both B5 and B6 would have to be safe as well. However, we have already identified all 12 safe spaces, so this cannot be true.
- Therefore, A1 must fit B2’s clue, which means that it must be surrounded by white traps. B1 contains a white trap.

- B4’s clue tells us that there is only one white trap in Column 3, and as we have established, none of the remaining spaces are safe. So if B3 contains a white trap, both C3 and D3 would have to contain black traps.
- But this can’t be true. Both C3 and D3 lie on main diagonals, so according to B1’s clue, they can’t both contain black traps. B3 contains a black trap.
- Let’s look at B4’s clue some more. Of the three remaining spaces that lie on main diagonals (C3, D3 and B5), exactly one contains a black trap.
- This means that if B5 contains a black trap, then neither C3 nor D3 can. They’d both have to contain white traps, but that would contradict B4’s clue.
- B5 contains a white trap.

- E4 is adjacent to eight spaces in total, so B5’s clue leaves us with two possibilities. E4 is either adjacent to one white trap and two black traps, or two white traps and four black traps. But we’ve already identified three black traps adjacent to E4 (F3, F4 and F5), so it must be the latter. That means that of the two remaining spaces surrounding E4 (D3 and D5), one contains a white trap and one contains a black trap.
- Look back at the clues in F3 and F5. There are 12 safe spaces in total, and the remaining 24 spaces are divided equally between the black spy and the white spy. This means that each spy rigged 12 spaces.
- So far we’ve identified 9 spaces that were rigged by the white spy (B1, F1, A2, B2, E2, F2, D4, B5 and A6). This means that there are 3 remaining; two are in Row C (B3’s clue), and the last remaining one is either D3 or D5. Thus, we can safely assume that all other spaces contain black traps.
- B6, D1 and D6 contain black traps.
- Now that we know the status of B6 and D6, we can tell that C6 contains a white trap (A6’s clue).

- C6’s clue tells us that there are two columns that contain an odd number of black traps. We already know that one of them is Column 4 (A4, E4 and F4), but what about the other one?
- Let’s look at Column 3. There are two remaining spaces in that column that we haven’t identified, and they both lie on main diagonals. Thanks to B1’s clue, we know that exactly one of them contains a black trap.
- We’ve already identified two other black traps in the column (B3 and F3), so that makes three in total. Columns 3 and 4 each contain an odd number of black traps.
- This means that each of the other columns either contains an even number of black traps, or none at all. C1 contains a black trap.

- C1’s clue tells us that C5 contains a white trap, meaning that D5 has to contain a black trap (A6’s clue). This, in turn, means that D3 has to contain a white trap (B5’s clue), which means that C3 contains a black trap (B1’s clue).

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