Puzzle Pack #1 Puzzle 11 Answer

In row 1, the two criminals must be in a single orthogonally connected group. Since A1 is already known to be innocent, the two criminals cannot be separated across both sides of A1, and with exactly two criminals in that row they must occupy adjacent spaces. The only adjacent pair available in row 1 that avoids A1 is C1 and D1, so Ethan at C1 has to be one of those criminals. Therefore, we can determine that C1 is CRIMINAL.

Bruce is at A1, so his neighbors are B1, A2, and B2. Ethan says Bruce has exactly one innocent neighbor, and that innocent neighbor is on the edge. Among those three neighbors, B1 and A2 are on the edge, but B2 is not, because it is an interior square. So B2 cannot be Bruce's only innocent neighbor and cannot be innocent at all. Therefore, we can determine that B2 is CRIMINAL.

Bruce is at A1, so his neighbors are B1, A2, and B2. Ethan says Bruce has exactly one innocent neighbor, and that neighbor is on the edge. Since B2 is already criminal, that means exactly one of B1 and A2 is innocent. Hilda at B2 is neighbored by edge positions A1, B1, C1, A2, and A3. Among those, A1 is innocent and C1 is criminal, so the edge neighbors of Hilda who could be innocent are B1, A2, and A3. Hilda says an odd number of her edge neighbors are innocent, so besides A1 there must be an even number of innocents among B1, A2, and A3. But Ethan's clue already tells us exactly one of B1 and A2 is innocent, so among B1, A2, and A3 the total can be even only if A3 is also innocent. Therefore, we can determine that A3 is INNOCENT.

Karen says that Isaac is one of Ethan's 3 criminal neighbors. Ethan is at C1, and his neighbors are B1, B2, C2, D1, and D2, so this clue directly says C2 is one of the criminal ones around Ethan. Therefore, we can determine that C2 is CRIMINAL.

Bruce’s clue says there are exactly two criminals in row 1, and they must be connected in that row. Since Ethan at C1 is already a criminal, the second criminal in row 1 has to be B1 or D1, so row 1 cannot also have Jose at D2 affecting that count. Karen’s clue says Ethan has exactly 3 criminal neighbors, and we already know two of them are Hilda at B2 and Isaac at C2. That means Ethan has room for only one more criminal neighbor among B1, D1, and D2, and Bruce’s clue already requires that one to be in row 1 at B1 or D1. Therefore, we can determine that D2 is INNOCENT.

Zara is at D5, so the people above her are Freya at D1, Jose at D2, Nancy at D3, and Tyler at D4. Isaac says both innocents above Zara are connected, so among those four people there are exactly two innocents, and they must form one unbroken vertical group in column D. Jose is already one of those innocents, and if Tyler were also innocent, then the two innocents would be Jose at D2 and Tyler at D4 with Nancy between them, which would not be connected. Therefore, we can determine that D4 is CRIMINAL.

Bruce at A1 has only two neighbors, Carl at B1 and Gary at A2, because the other touching square is Ethan at C1’s diagonal? No, for A1 the neighbors are B1, A2, and B2, and B2 is Hilda, who is already criminal. Ethan’s clue says Bruce has exactly one innocent neighbor, and that innocent neighbor is on the edge, so among Carl and Gary exactly one is innocent. Jose at D2 has neighbors Freya, Ethan, Isaac, Mary, Nancy, and Tyler, and the only known innocent among them is Jose does not count himself, so Ethan, Isaac, and Tyler are criminal while Freya, Mary, and Nancy are the only possible innocent neighbors there. Since Jose says he and Bruce have the same number of innocent neighbors, Jose must also have exactly one innocent neighbor, so among Freya, Mary, and Nancy exactly one is innocent. Therefore, we can determine that C3 is CRIMINAL.

Ethan is at C1, and Karen’s clue says he has exactly 3 criminal neighbors, with Isaac definitely being one of them. Ethan’s neighbors are only B1, B2, C2, and D2, and among those we already know B2 and C2 are criminal while D2 is innocent, so the third criminal neighbor must be B1 Carl. Tyler’s clue says Ethan and Paul have the same number of innocent neighbors. Since Ethan’s neighbors are B1, B2, C2, and D2, and only D2 is innocent, Carl cannot also be innocent, which matches Carl being criminal and fixes Ethan at exactly 1 innocent neighbor. So Paul at B4 must also have exactly 1 innocent neighbor among A3, A4, A5, B3, B5, C3, C4, and C5; because A3 is innocent and C3 is criminal already, every other one of Paul’s neighbors, including B5 Wally, must be criminal. Therefore, we can determine that B5 Wally is CRIMINAL.

Bruce’s neighbors are B1, A2, and B2. Since Bruce is innocent and his only innocent neighbor is on the edge, the only innocent neighbor he can have is A2, because B1 is not on the edge and B2 is already criminal. So A2 is innocent and B1 is criminal. That means Bruce has exactly one innocent neighbor. Tyler’s neighbors are C3, D3, C4, C5, and D5, and Tyler must have the same number of innocent neighbors as Bruce, so Tyler also has exactly one innocent neighbor. Since C3 is criminal, Tyler’s single innocent neighbor must be among D3, C4, C5, and D5. Now look at row 5: B5 is criminal, and all criminals in row 5 are connected. If C5 were innocent, then any criminal at D5 would be separated from B5 by an innocent, so D5 could not be criminal; that would leave Tyler with at least two innocent neighbors on row 5, namely C5 and D5. So C5 cannot be innocent. Therefore, we can determine that C5 is CRIMINAL.

Bruce is at A1, and his neighbors are Carl at B1, Gary at A2, and Hilda at B2. Ethan says Bruce has exactly one innocent neighbor, and that innocent neighbor is on the edge. Since Hilda is already criminal and Bruce is already innocent, only Carl and Gary could be Bruce's innocent neighbor, so exactly one of Carl and Gary is innocent. Tyler is at D4, with neighbors Mary, Nancy, Jose, Steve, Xia, and Zara. Wally says Tyler and Bruce have the same number of innocent neighbors, so Tyler also has exactly one innocent neighbor. But Mary and Xia are already criminal, and Jose is already innocent, so Jose is one innocent neighbor already. That means Nancy, Steve, and Zara cannot be innocent. Therefore, we can determine that C4 is CRIMINAL.

Gary is at A2, so the people below Gary are Karen at A3, Olive at A4, and Vicky at A5. Steve says there is an odd number of innocents among those three. Karen is already known to be innocent, so Olive and Vicky must contribute an even number of innocents between them. Since each of them is still unknown, that means they must match: either both innocent or both criminal. Tyler says Ethan and Paul have the same number of innocent neighbors. Ethan at C1 already has exactly two innocent neighbors, Bruce at A1 is not adjacent but Hilda at B2 and Isaac at C2 are criminal, so Ethan’s innocent neighbors are only Jose at D2 and whatever happens at B1 and D1 does not change that count beyond the adjacent spots B1, B2, C2, D2; with B2 and C2 criminal and D2 innocent, Ethan’s count is fixed once B1 is considered. Paul at B4 has neighbors Karen at A3, Laura at B3, Mary at C3, Olive at A4, Steve at C4, Vicky at A5, Wally at B5, and Xia at C5. Among these, Karen is innocent and Mary, Steve, Wally, and Xia are criminal, so Paul’s innocent-neighbor count is 1 plus Laura, plus Olive if Olive is innocent, plus Vicky if Vicky is innocent. Matching Tyler’s clue with Steve’s restriction forces Olive and Vicky to be both criminal, which in turn makes Laura the extra innocent neighbor Paul needs. Therefore, we can determine that B3 is INNOCENT, A4 is CRIMINAL, and A5 is CRIMINAL.

Jose is at D2, so the only people below him are Nancy at D3, Tyler at D4, and Zara at D5. Xia says an odd number of those three are innocent, and Tyler is already known to be criminal, so Nancy and Zara must include exactly one innocent. Laura says Steve at C4 has exactly 3 innocent neighbors. Steve’s neighbors are Laura at B3, Mary at C3, Nancy at D3, Paul at B4, Tyler at D4, Wally at B5, Xia at C5, and Zara at D5. Among these, Laura is innocent and Mary, Tyler, Wally, and Xia are criminal, so Steve still needs exactly two more innocent neighbors from Nancy, Paul, and Zara. But from Xia’s clue, Nancy and Zara together contain exactly one innocent. That means Paul must be the other needed innocent neighbor for Steve to reach a total of 3 innocent neighbors. Therefore, we can determine that B4 is INNOCENT.

Olive says there are 13 criminals in total. On the board, 12 people are already known to be criminals, so there is room for exactly one more criminal among the four unknown people: B1 Carl, D1 Freya, A2 Gary, and D5 Zara. Jose says that he and Bruce have the same number of innocent neighbors. Bruce’s neighbors are B1, A2, and B2, and since Bruce is already innocent while B2 is criminal, Bruce has as many innocent neighbors as there are innocents among B1 and A2. Jose’s neighbors are C1, D1, C2, C3, D3, and with C1, C2, and C3 already criminal and Jose himself innocent, Jose has as many innocent neighbors as there are innocents among D1 and D3. Since D3 is the only unknown there and the counts must match, the one remaining criminal among the unknowns cannot be in B1, D1, or A2, so it must be Zara. Therefore, we can determine that D5 is CRIMINAL.

Jose is at D2, so the people below him are Nancy at D3, Tyler at D4, and Zara at D5. Tyler and Zara are already known criminals, so among those three people the only possible innocent below Jose is Nancy. That makes the number of innocents below Jose exactly 1, which is odd, matching Xia’s clue. Therefore, we can determine that D3, Nancy, is INNOCENT.

Above Zara in column D are Freya at D1, Jose at D2, Nancy at D3, and Tyler at D4. The clue says both innocents above Zara are connected, so among those four people there must be exactly two innocents, and those two must form one continuous vertical group. Jose and Nancy are already the two innocents there, and they are directly next to each other, so no one else above Zara can be innocent. Therefore, we can determine that D1 is CRIMINAL.

In row 1, the two criminals are Ethan at C1 and Freya at D1. Those two spaces are directly next to each other, so the criminals in that row are already connected without needing anyone else in the row to be a criminal. That leaves the remaining unknown in row 1, Carl at B1, unable to be one of the two criminals named by the clue. Therefore, we can determine that B1 is INNOCENT.

Bruce is at A1, so his neighbors are B1, A2, and B2. We already know B1 is innocent and B2 is criminal, so if Bruce’s only innocent neighbor is on the edge, then B1 must be that one innocent neighbor. That leaves A2 unable to be innocent. Therefore, we can determine that A2 is CRIMINAL.