Puzzle Pack #2 Puzzle 36 Answer

Tina’s clue says that Olof is one of Gary’s 7 innocent neighbors. Olof at C3 is indeed a neighbor of Gary at B2, so this clue is directly telling us that Olof is innocent. Therefore, we can determine that C3 is INNOCENT.

Gary’s neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Tina says Olof is one of Gary’s 7 innocent neighbors, so among those eight neighbors exactly seven are innocent, and since Olof at C3 is already known innocent, Gary himself must be criminal. Eve’s neighbors are A1, B1, B2, A3, and B3. Olof says Eve has exactly 3 innocent neighbors, and A3 and B3 are the same people as two of Gary’s neighbors while B2 is Gary. Since Gary’s eight surrounding spaces contain exactly one non-innocent and we now know that non-innocent is Gary at B2, the other shared neighbors force A3 and B3 to be innocent, so Eve’s third innocent neighbor must be Eve herself rather than Gary. That makes Eve innocent, and it also means the two remaining neighbors around Gary that fit Tina’s count, C1 and C2, are innocent as well. Therefore, we can determine that A2 is INNOCENT, C1 is INNOCENT, C2 is INNOCENT, and B2 is CRIMINAL.

Olof is at C3, so his neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Tina says 7 of those 8 neighbors are innocent, and Gary at B2 is already known to be criminal, so all seven of Olof’s other neighbors must be innocent, including D2, B3, D3, B4, and D4. Eve is at A2 and has three innocent neighbors: B1, A1, and A3 can contribute, while B2 is criminal and the others are already innocent, so Celia’s clue means Xena must have more than Eve’s innocent-neighbor total. Xena at C5 can only have up to five neighbors, and with C4 already innocent, that forces all of Xena’s other neighbors B4, D4, B5, and D5 to be innocent as well. This also matches Eve’s clue, because among Olof’s five innocent neighbors that also neighbor Xena, the only ones strictly between both areas are B4 and D4. Therefore, we can determine that B5 is INNOCENT and D5 is INNOCENT.

Olof is at C3, so his five neighbors are B2, C2, D2, B3, and D3. Since B2 is Gary the criminal, the innocent neighbors among those five are Isaac at C2 and Olof’s other innocent neighbors must come from D2, B3, and D3; Eve’s clue says exactly two of Olof’s innocent neighbors also neighbor Xena at C5. Among Olof’s neighbors, the ones that also neighbor Xena are only D2, B3, and D3, because Isaac at C2 does not touch Xena, so the total works only if D2, B3, and D3 are all innocent. Now use Zoe’s clue about row 3: among Mark at A3, Nicole at B3, Olof at C3, and Phil at D3, only one person has exactly five innocent neighbors. With B3 and D3 now innocent, Sue at B4 is one of Nicole’s neighbors and also one of Olof’s neighbors; this fixes the row 3 neighbor counts so that Sue must be innocent for the clue to hold with exactly one such person in row 3. Therefore, we can determine that B4 is INNOCENT.