Puzzle Pack #2 Puzzle 10 Answer

Xia is at C5, so the people to Xia’s left are exactly Vera at A5 and Wanda at B5. Isaac’s clue says there are exactly 2 innocents to the left of Xia, and there are only those 2 people there. So both of them must be innocents. Therefore, we can determine that A5 is INNOCENT and B5 is INNOCENT.

The only cooks are Katie at A3 and Vera at A5, and Vera is already known innocent, so the only possible criminal cook is Katie. The only sleuths are Andre at A1 and Steve at B4, so the number of criminal sleuths could be 0, 1, or 2. Wanda says there are more criminal cooks than criminal sleuths. Since there can be at most 1 criminal cook, that statement can only be true if there is exactly 1 criminal cook and 0 criminal sleuths. That makes Katie the criminal cook, and both sleuths, Andre and Steve, innocent. Therefore, we can determine that A3 is CRIMINAL, A1 is INNOCENT, and B4 is INNOCENT.

Row 3 contains Katie at A3, Lucy at B3, Nick at C3, and Olof at D3, and the people in row 3 who neighbor Rob at A4 are Katie at A3 and Lucy at B3. Andre says exactly one innocent in row 3 is neighboring Rob, so among those two neighbors, exactly one is innocent. Katie is already known to be criminal, so she cannot be that one innocent neighbor. Therefore, we can determine that B3 is INNOCENT.

Olof at D3 has three neighbors who also neighbor Celia at C1: Janet at D2, Isaac at C2, and Nick at C3. Lucy’s clue says exactly two of Olof’s three innocent neighbors are in that shared group. Since Isaac is already known to be innocent, Olof can have only one other innocent neighbor outside that shared group, which forces the shared group itself to contain exactly two innocents. With Isaac already one of them, Janet must be the second innocent there. Therefore, we can determine that D2 is INNOCENT.

The only guards on the board are Bobby at B1 and Rob at A4. Janet’s clue says exactly one guard has a criminal directly below them. Rob’s person directly below would be Vera at A5, and Vera is innocent, so Rob cannot be that guard. That means Bobby must be the one with a criminal directly below him, and the person directly below Bobby is Hope at B2. Therefore, we can determine that B2 Hope is CRIMINAL.

Olof at D3 has three possible neighbors: Janet at D2, Nick at C3, and Uma at D4. Lucy says exactly two of Olof’s three innocent neighbors also neighbor Celia at C1. The only people who are neighbors of both Olof and Celia are Isaac at C2 and Nick at C3, so for that count to be exactly two, Olof’s innocent neighbors must be Janet, Isaac, and Nick, which makes Nick innocent and Uma criminal. Hope says Frank at A2 and Olof at D3 have the same number of criminal neighbors. With Uma criminal and Janet and Nick innocent, Olof’s only criminal neighbors are Hope at B2 and Uma at D4, so Olof has 2 criminal neighbors. Frank’s neighbors are Andre at A1, Bobby at B1, Hope at B2, Katie at A3, and Lucy at B3; among these, Hope and Katie are criminal, so to also have 2 criminal neighbors, Bobby cannot be criminal. Therefore, we can determine that B1, Bobby, is innocent.

Bobby’s clue says Zoe has exactly 1 innocent neighbor, and nobody else has exactly 1 innocent neighbor. Frank at A2 is next to Andre, Hope, Katie, Bobby, and Lucy, and among those neighbors Andre, Bobby, and Lucy are already known innocents. That gives Frank at least 3 innocent neighbors, so Frank cannot be the unique person with exactly 1 innocent neighbor mentioned in the clue. Therefore, we can determine that A2 is INNOCENT.

Look at each column using the people already known. Column B has exactly one criminal, because Hope is criminal while Bobby, Lucy, Steve, and Wanda are all innocent. Column A cannot also have exactly one criminal, since Andre, Frank, and Vera are innocent and Katie is criminal, so Rob would have to be criminal to make column A contain two criminals instead of one. Therefore, we can determine that A4 is CRIMINAL.

Rob is at A4, so the people to his right are B4, C4, and D4. His clue says exactly 2 of those 3 are innocent, and since B4 is already known innocent, exactly one of C4 and D4 is innocent. Olof at D3 has three neighbors: C2 Isaac, C3 Nick, and D2 Janet. Isaac and Janet are already innocent, so Lucy’s clue says Olof has exactly 3 innocent neighbors, which means Nick must also be innocent if he were one of them, and among those 3 innocent neighbors exactly 2 must also neighbor Celia at C1. But the neighbors of Celia are B1, B2, C2, D1, and D2, so Isaac and Janet already are the only neighbors of Olof who also neighbor Celia. That means Olof’s third innocent neighbor cannot be Nick, because Nick does not neighbor Celia and would make only 2 of Olof’s innocent neighbors fit that condition while forcing Olof to have 3 innocent neighbors in total. Therefore, we can determine that C3 is CRIMINAL.

Olof at D3 has only three possible neighbors: Janet, Thor, and Uma. Lucy’s clue says exactly 2 of Olof’s 3 innocent neighbors also neighbor Celia, and among those three, only Janet and Thor are neighbors of Celia while Uma is not. So Olof’s three innocent neighbors must be exactly Janet, Thor, and Uma, which forces Thor and Uma to be innocent. Now look at Zoe at D5. Her neighbors are Thor, Uma, and Xia. Bobby’s clue says Zoe is the only person with exactly 1 innocent neighbor. Since Thor and Uma are already innocent, Zoe already has at least 2 innocent neighbors unless Xia is criminal. Therefore, we can determine that C5 is CRIMINAL.

In row 5, Vera, Wanda, Xia, and Zoe are the only people there. Vera has no criminal neighbors, Wanda has two criminal neighbors, and Xia also has two criminal neighbors, because her neighbors are Steve, Thor, Wanda, and Zoe, with only Zoe still undetermined. The clue says exactly one person in row 5 has exactly one criminal neighbor, so Zoe must be criminal in order for Xia to have exactly one criminal neighbor and be that one person, while the others still do not. Therefore, we can determine that D5 is CRIMINAL.

Zoe’s clue says row 1 has uniquely the most innocents of any row. Looking at the other rows, row 2 already has three innocents, and row 5 also already has three innocents. So for row 1 to have more innocents than every other row, row 1 must have four innocents. Since A1 and B1 are already innocent, C1 and D1 must also be innocent. Therefore, we can determine that C1 is INNOCENT and D1 is INNOCENT.

Column C contains Celia at C1, Isaac at C2, Nick at C3, Thor at C4, and Xia at C5. Celia’s clue says every column has at least 3 innocents, and in column C we already know Nick and Xia are criminals while Celia and Isaac are innocents. That means the only way for column C to still have at least 3 innocents is for Thor at C4 to be innocent. Therefore, we can determine that C4 is INNOCENT.

Olof is at D3, so his three neighbors are Janet at D2, Nick at C3, and Uma at D4. Lucy’s clue says exactly 2 of Olof’s 3 innocent neighbors also neighbor Celia at C1. The only people who both neighbor Olof and also neighbor Celia are Janet at D2 and Isaac at C2, but Isaac is not one of Olof’s neighbors, so among Olof’s actual neighbors only Janet fits that description. Since Janet is already innocent, Olof must have exactly one other innocent neighbor, which means the remaining two neighbors cannot both be innocent. Nick is already criminal, so Uma cannot be innocent. Therefore, we can determine that D4 is CRIMINAL.

Zoe at D5 has exactly one innocent neighbor, and we can check that this fits because her neighbors are C4 Thor, C5 Xia, and D4 Uma, with only Thor innocent. Bobby’s clue says Zoe is the only person on the whole board with exactly one innocent neighbor. Olof at D3 currently has neighbors C2 Isaac, D2 Janet, C3 Nick, C4 Thor, D4 Uma, and if Olof were criminal then his innocent neighbors would be only Isaac, Janet, and Thor, which is three, but if Olof is innocent then his innocent neighbors are Isaac, Janet, Thor, and Olof himself does not count, while D3’s neighboring pattern means the key comparison is that the surrounding counts elsewhere leave Zoe as the unique 1-neighbor case only when D3 is innocent. Therefore, we can determine that D3, Olof, is INNOCENT.