Puzzle Pack #2 Puzzle 22 Answer

Below Alice in column A are Gary at A2, Kumar at A3, Ollie at A4, and Vera at A5. Hope is at B2, so among those four people, Hope’s neighbors are Gary, Kumar, and Ollie, while Vera is not Hope’s neighbor. Xavi says that only 1 of the 3 innocents below Alice is Hope’s neighbor. Since exactly three of those four people are innocent, and Vera is the only one who is not Hope’s neighbor, Vera must be one of the three innocents. That means the other two innocents among Hope’s neighbors must be Kumar and Ollie, so Ollie is innocent as well. Therefore, we can determine that A4 is INNOCENT and A5 is INNOCENT.

Ollie is at A4, and her neighbors are A3, B3, A5, and B5. Vera says that Kumar is one of Ollie’s 2 criminal neighbors, so Kumar must be a neighbor of Ollie and must be criminal. Kumar is at A3, which is indeed next to Ollie, so the clue applies directly to him. Therefore, we can determine that A3 is CRIMINAL.

Below Alice in column A are Gary at A2, Kumar at A3, Ollie at A4, and Vera at A5. We already know Ollie and Vera are innocent, and Kumar is criminal, so the clue’s “3 innocents below Alice” means Alice, Gary, Ollie, and Vera contain exactly 3 innocents, leaving only one of Alice or Gary as criminal. Hope at B2 is a neighbor of Alice at A1, Gary at A2, and Ollie at A4, but not Vera at A5. So among the innocents in that set, Hope’s neighbors are Alice and Ollie if Alice is innocent, or Gary and Ollie if Gary is innocent. Since the clue says only 1 of those 3 innocents is Hope’s neighbor, Gary cannot be criminal, because that would still leave two innocent neighbors of Hope. Therefore, we can determine that A2 is INNOCENT.

Cheryl is at B1, so the people below Cheryl are B2 Hope, B3 Logan, B4 Rose, and B5 Wanda. Ollie says exactly 2 of those 4 are innocent. Vera says Kumar is one of Ollie’s 2 criminal neighbors. Ollie is at A4, and Kumar at A3 is indeed one neighbor, so Ollie must have exactly one other criminal neighbor. Ollie’s neighbors are A3 Kumar, B3 Logan, B4 Rose, and A5 Vera, and Vera is already innocent, so among B3 and B4 exactly one is criminal and the other is innocent. That means in Cheryl’s column, B3 and B4 contribute exactly 1 innocent. Since there must be exactly 2 innocents total among B2, B3, B4, and B5, and B5 cannot be innocent because then B2 would also have to be innocent to make the count reach 2, leaving no room for the required single innocent among B3 and B4, Hope must be the remaining non-innocent in that column count. Therefore, we can determine that B2 is CRIMINAL.

Cheryl is at B1, so the people below her are B2, B3, B4, and B5. Ollie’s clue says exactly 2 of those 4 people are innocent. We already know B2 is criminal, so among B3, B4, and B5 exactly 2 are innocent and exactly 1 is criminal. Kumar’s clue says every column has at least 3 innocents. In column B, that means among Cheryl, B3, B4, and B5 there must be at least 3 innocents, because B2 is already criminal. Since B3, B4, and B5 contain only 2 innocents, Cheryl has to be the third innocent in that column. Therefore, we can determine that B1 is INNOCENT.

Xavi is at C5, so anyone above Xavi must be somewhere in column C. Hope’s eight neighbors are A1, B1, C1, A2, C2, A3, B3, and C3, and among those the ones above Xavi are C1, C2, and C3. Hope says exactly one of her four innocent neighbors is above Xavi, so exactly one of C1, C2, and C3 is innocent. Kumar says every column has at least 3 innocents. In column C, Xavi at C5 is already innocent, so among C1, C2, C3, and C4 there must be at least two more innocents. Since only one of C1, C2, and C3 can be innocent, the other needed innocent in column C has to be C4. Therefore, we can determine that C4 is INNOCENT.

Xavi is at C5, so anyone above Xavi must be somewhere in column C. Hope says she has exactly 4 innocent neighbors, and only 1 of those 4 is above Xavi. Hope’s neighbors are Cheryl at B1, Gary at A2, Isaac at C2, Kumar at A3, Logan at B3, and Mark at C3; among these, Cheryl and Gary are already innocent, Kumar is criminal, and the only neighbors above Xavi are Cheryl, Isaac, and Mark. Since only 1 innocent neighbor is above Xavi, Cheryl must be that one, so Isaac and Mark are criminal, and then Logan must be innocent to make Hope’s total of 4 innocent neighbors. Now use Cheryl’s clue. Ollie has criminal neighbors Kumar and Logan, so Ollie has 2 criminal neighbors. Therefore Jose must also have 2 criminal neighbors. Jose’s neighbors are C1, D1, Isaac, Mark, Nicole, Scott, and Tina; with Isaac and Mark criminal and Scott innocent, Jose already has exactly 2 criminal neighbors, so C1, D1, Nicole, and Tina must all be innocent. Therefore, we can determine that D1 is INNOCENT and D3 is INNOCENT.

Xavi is at C5, so everyone above Xavi is C1, C2, C3, and C4. Hope says that among her four innocent neighbors, only one is above Xavi; Hope’s innocent neighbors are B1, A2, and C3, so exactly one of Cheryl at B1 and Mark at C3 is innocent. Nicole says exactly one innocent above Xavi is neighboring Isaac, and the people above Xavi who neighbor Isaac are Cheryl at B1, Diane at C1, Mark at C3, and Scott at C4; since Scott is already innocent, that one innocent is already accounted for, so Cheryl, Diane, and Mark are not innocent. In particular, Mark is not innocent, so Hope’s statement forces Cheryl to be the one innocent neighbor above Xavi, and then Hope must also have a fourth innocent neighbor at C2. Therefore, we can determine that C2 is CRIMINAL.

Jose is at D2 and Ollie is at A4. From the known people around Ollie, Ollie has exactly one criminal neighbor already: Kumar at A3, while Gary, Logan, Rose, and Vera are the other neighbors. Cheryl’s clue says Jose and Ollie have the same number of criminal neighbors, so Jose must also have exactly one criminal neighbor. Around Jose, the known neighbors are Isaac at C2 and Nicole at D3, with Frank at D1 also innocent, so the only criminal neighbor Jose already has is Isaac; that means Logan at B3 cannot be criminal, because then Ollie and Jose would no longer match. Isaac’s clue fits this as well, since Jose then has more innocent neighbors than Vera. Therefore, we can determine that B3 is INNOCENT.

Hope’s neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Among those, the innocents we already know are B1, A2, and B3, and all three of them are above Xavi at C5 because rows 1, 2, and 3 are above row 5. Since Hope says only 1 of her 4 innocent neighbors is above Xavi, the fourth innocent neighbor cannot also be above Xavi, so none of A1, C1, or C3 can be that fourth innocent. Alice at A1 is one of those above-Xavi neighbors, so she cannot be innocent. Therefore, we can determine that A1 is CRIMINAL.

Hope’s neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. The four innocents among them are B1, A2, B3, and exactly one of C1 or C3, because A1, C2, and A3 are criminals. Hope says only 1 of those 4 innocent neighbors is above Xavi at C5. Everyone in rows 1 through 4 is above Xavi, so any innocent neighbor of Hope is above Xavi. That means exactly one of Hope’s neighbors can be an innocent, so the only way her statement works is that the fourth innocent neighbor must be Jose at D2 rather than C1 or C3. Alice’s clue fits this as well: Isaac’s neighbors are B1, C1, D1, B2, D2, B3, C3, and D3, and with Jose innocent he has exactly six innocent neighbors. Therefore, we can determine that D2 is INNOCENT.

Hope’s innocent neighbors are B1, A2, B3, and C3, and “above Xavi” means being somewhere in column C above C5. Among those four people, only C3 is above Xavi, so for Hope’s clue to say only 1 of those 4 neighboring people is innocent, C3 must be innocent. Then Jose has innocent neighbors C1, C3, D1, and D3, so Jose has either 3 or 4 innocent neighbors depending on C1. Zoe’s neighbors are C4, D4, and C5, and since C4 and C5 are already innocent, Zoe has either 2 or 3 innocent neighbors depending on D4. Because Jose and Zoe have the same number of innocent neighbors, Zoe cannot have only 2 while Jose has at least 3, so D4 must be innocent. Therefore, we can determine that D4 is INNOCENT.

Jose at D2 has criminal neighbors at C1, C2, and C3, while his other neighbors are innocent, so Jose has either 2 or 3 criminal neighbors depending on C1. Ollie at A4 has criminal neighbors at A3 and possibly B4, so for Cheryl’s clue to be true, Ollie must also have the same count as Jose; that only works if C1 is criminal and B4 is innocent, making both Jose and Ollie have 3 criminal neighbors. With C1 criminal and B4 innocent, Isaac at C2 has criminal neighbors A1, B2, C1, A3, and possibly C3, so Isaac has either 4 or 5 criminal neighbors depending on C3. Xavi at C5 then has criminal neighbors only at B4 and D5, but B4 is already innocent, so Tina’s clue says D5 must be innocent as well for Xavi to stay below Isaac’s count. Therefore, we can determine that D5 is INNOCENT.

Cheryl is at B1, so the people below her are Hope at B2, Logan at B3, Rose at B4, and Wanda at B5. Ollie says exactly 2 of those 4 are innocent, and Logan is already known innocent while Hope is already known criminal, so Rose and Wanda must split as one innocent and one criminal. Zoe says Vera and Frank have the same number of innocent neighbors. Vera at A5 has neighbors Ollie, Rose, and Wanda, so with Ollie innocent and exactly one of Rose and Wanda innocent, Vera has 2 innocent neighbors. Frank at D1 has neighbors Diane, Isaac, and Jose, and Jose is innocent while Isaac is criminal, so for Frank also to have 2 innocent neighbors, Diane must be innocent. Therefore, we can determine that C1 is INNOCENT.

Hope’s neighbors are Alice, Cheryl, Diane, Gary, Isaac, Kumar, Logan, and Mark. Among those, the innocents are Cheryl, Diane, Gary, and Logan, which matches her clue saying there are 4 innocent neighbors. Of those four innocents, the ones above Xavi at C5 are Cheryl at B1, Diane at C1, Gary at A2, and Logan at B3, so all 4 are above Xavi, not only 1. That means Hope cannot really have 4 innocent neighbors unless Mark is not innocent. Therefore, we can determine that C3, Mark, is CRIMINAL.

Scott is at C4, and his neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. Among those, the edge neighbors are D4, B5, C5, and D5, since those four positions are on the outer edge of the board. Mark says Scott has 4 innocent neighbors on the edges, so all four of those edge neighbors must be innocent. D4, C5, and D5 are already known innocent, so the fourth one, B5 Wanda, must also be innocent. Therefore, we can determine that B5 is INNOCENT.

Cheryl is at B1, so the people below Cheryl are Hope at B2, Logan at B3, Rose at B4, and Wanda at B5. Ollie’s clue says exactly 2 of those 4 people are innocents. We already know Hope is criminal, Logan is innocent, and Wanda is criminal, so among those four there is currently only one known innocent: Logan. That means Rose must be the second innocent required by the clue. Therefore, we can determine that B4 is INNOCENT.