Puzzle Pack #2 Puzzle 21 Answer

Paula is at B4, and her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Ruby says Katie is one of Paula's 5 innocent neighbors, so Katie must be a neighbor of Paula and must be innocent. Katie is at B3, which is indeed next to Paula. Therefore, we can determine that B3 is INNOCENT.

Column B contains Barnie at B1, Flora at B2, Katie at B3, Paula at B4, and Vince at B5. Katie says Barnie is one of 2 innocents in that column, and Katie herself is already known to be innocent, so her clue is true. Since Katie is in column B and is innocent, Barnie must be the other innocent counted there. Therefore, we can determine that B1 is INNOCENT.

Column B contains Barnie at B1, Flora at B2, Katie at B3, Paula at B4, and Vince at B5. Katie’s clue says Barnie is one of 2 innocents in that column, so column B has exactly two innocents total. We already know Barnie and Katie are innocent, which fills those two innocent spots completely. That leaves Flora, Paula, and Vince as the remaining people in column B, so none of them can be innocent. Therefore, we can determine that B2 is CRIMINAL, B5 is CRIMINAL, and B4 is CRIMINAL.

Paula is at B4, so her edge neighbors are A3, A4, A5, B5, C5, and C3. Among those, B5 is already known to be criminal. Ruby also says Katie is one of Paula’s five innocent neighbors, and Katie at B3 is indeed a neighbor of Paula but not on the edge, so Paula’s two innocent edge neighbors must be chosen from those six edge positions. Since one of those six, B5, cannot be innocent, the only way Paula can still have exactly five innocent neighbors in total with Katie included is for all five of the other edge positions to be innocent, including C3. Therefore, we can determine that C3 is INNOCENT.

Vince is at B5, so his edge neighbors are A4, A5, C5, and C4. Ruby at C4 is already innocent, and Flora’s clue says an odd number of those edge neighbors are innocent. That means among A4, A5, and C5, an even number are innocent. Ruby’s clue says Paula at B4 has exactly 5 innocent neighbors, and Katie at B3 is one of them. Paula’s neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. We already know B3, C3, and C4 are innocent, and B5 is criminal, so among A3, A4, A5, and C5 there must be exactly 2 innocents to make Paula’s total 5. Since A4, A5, and C5 already contain an even number of innocents, adding A3 must bring that group of four to exactly 2 innocents. That forces A3 to be innocent. Therefore, we can determine that A3 is INNOCENT.

Paula is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Ruby says exactly 5 of those neighbors are innocent, and we already know four of them: Jose at A3, Katie at B3, Logan at C3, and Ruby at C4. Jose also says there are exactly 3 innocents in column A, and exactly 2 of those are Paula's neighbors; since A3 is innocent and is Paula's neighbor, exactly one of A4 or A5 is also an innocent neighbor of Paula. That gives Paula exactly one more innocent neighbor among A4 and A5, so C5 cannot be innocent and must be the remaining non-innocent neighbor. Therefore, we can determine that C5 is CRIMINAL.

Wally’s clue says that Mark is one of the 4 innocents in column D, so it directly tells us Mark’s identity. Therefore, we can determine that D3 is INNOCENT.

Jose’s clue says column A contains exactly 3 innocents, and exactly 2 of those 3 are neighbors of Paula at B4. In column A, Paula’s neighbors are only Jose at A3, Nancy at A4, and Uma at A5, so those must be the 3 innocents in column A; that fixes A1 and A2 as criminals, and Nancy and Uma as innocents since Jose already is innocent. Wally’s clue says column D has 4 innocents, so with Mark already innocent that means D1, D2, D4, and D5 are all innocent. Now count innocents on the edge: B1, A3, D3, A4, D4, A5, D5, and the four edge cells in column D and row 5 just determined give 11 edge innocents unless C1 is also innocent, and Vince says the total must be odd. Therefore, we can determine that C1 is INNOCENT.

Jose says there are exactly 3 innocents in column A, so in column A the three innocents must be Jose plus exactly two of Alice, Emily, Nancy, and Uma. Barnie says only one column has exactly 3 innocents, and column B already has exactly 3 innocents because Barnie, Katie, and the criminal spots at B2, B4, and B5 leave only those two known innocents plus no room for a fourth. That means column A cannot also have exactly 3 innocents unless Barnie’s clue is satisfied some other way, so the count in column A fixes which nearby people can be innocent for Jose’s clue about Paula’s neighbors to work. Among the three innocents in column A, exactly 2 must be neighbors of Paula at B4, and the only people in column A who neighbor Paula are Jose at A3, Nancy at A4, and Uma at A5, while Alice at A1 and Emily at A2 do not. Since Jose is already one of the 3 innocents in column A, exactly one of Nancy and Uma must also be innocent, which leaves Alice and Emily as the other innocent in column A. That forces column C to contain Gabe along with the already innocent Carol, Logan, and Ruby, so column C cannot be the unique column with exactly 3 innocents unless Gabe is innocent. Therefore, we can determine that C2 is INNOCENT.

Column D contains Diane at D1, Isaac at D2, Mark at D3, Thor at D4, and Xavi at D5. Wally’s clue says Mark is one of 4 innocents in column D, and Mark is already known innocent, so exactly one person in that column is criminal and the other four are innocent. Gabe’s clue says there are exactly 3 innocents above Xavi. The people above Xavi are Diane, Isaac, Mark, and Thor, so among those four, exactly 3 are innocent. Since Mark is already innocent, that means those four include exactly 1 criminal. That matches column D having its single criminal somewhere above Xavi, which leaves Xavi as one of the 4 innocents in the column. Therefore, we can determine that D5 is INNOCENT.

In column D, Mark at D3 and Xavi at D5 are already innocent, so Wally’s clue means exactly two of the remaining three people in that column, D1, D2, and D4, are innocent. Xavi’s clue compares the innocent neighbors of Wally at C5 and Carol at C1. Carol’s neighbors are Barnie, Gabe, Flora, and Diane, and since Barnie and Gabe are innocent while Flora is criminal, Carol has either 2 or 3 innocent neighbors depending on Diane. Wally’s neighbors are Ruby, Thor, Vince, and Xavi, and since Ruby and Xavi are innocent while Vince is criminal, Wally has either 2 or 3 innocent neighbors depending on Thor, so Thor and Diane must have the same status. That means D1 and D4 match, and because exactly two of D1, D2, and D4 are innocent, D4 cannot be criminal alongside D1, so D4 must be innocent. Therefore, we can determine that D4 is INNOCENT.

Wally’s clue says there are exactly 4 innocents in column D. In that column, Mark at D3, Thor at D4, and Xavi at D5 are already innocent, so exactly one of D1 Diane or D2 Isaac must also be innocent. Thor’s clue says Carol at C1 and Uma at A5 have the same number of criminal neighbors. Carol’s neighbors are Barnie, Flora, Gabe, and Diane, and since Barnie and Gabe are innocent while Flora is criminal, Carol has 1 criminal neighbor if Diane is innocent or 2 if Diane is criminal. Uma’s neighbors are Nancy, Paula, and Vince, and since Paula and Vince are criminal, Uma has 2 criminal neighbors if Nancy is innocent or 3 if Nancy is criminal. Because Carol and Uma must have equal numbers of criminal neighbors, the only matching case is 2 and 2, which means Diane is criminal and Nancy is innocent. Therefore, we can determine that A4 is INNOCENT.

Paula is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Ruby says Katie is one of Paula's 5 innocent neighbors, so Paula must have exactly 5 innocent neighbors in total. Among those eight neighbors, A3 Jose, B3 Katie, C3 Logan, A4 Nancy, and C4 Ruby are already known innocent, while B5 Vince and C5 Wally are criminals. That leaves A5 Uma as the only remaining neighbor, and to keep Paula's innocent-neighbor count at exactly 5, Uma cannot be innocent. Therefore, we can determine that A5 is CRIMINAL.

Row 2 contains Emily at A2, Flora at B2, Gabe at C2, and Isaac at D2. Nancy’s clue says there are exactly 3 innocents in that row. Since Flora is already known to be criminal and Gabe is already known to be innocent, the only way for row 2 to have exactly 3 innocents is for both Emily and Isaac to be innocent as well. Therefore, we can determine that A2 is INNOCENT and D2 is INNOCENT.

In column A, the innocents are Emily at A2, Jose at A3, and Nancy at A4. Paula is at B4, and her neighbors are A3, A4, A5, B3, B5, C3, C4, and C5, so among those three column-A innocents she is next to Jose and Nancy, but not Emily. That makes exactly 2 of the 3 column-A innocents Paula's neighbors, so Jose's statement is true. Therefore, we can determine that A1, Alice, is CRIMINAL.

In column D, Isaac, Mark, Thor, and Xavi are already known to be innocent, which makes 4 innocents there if D1 were innocent too. Wally’s clue says that Mark is one of 4 innocents in column D, so Mark himself must be in column D. But Mark is actually at D3, so the 4 innocents in column D have to be the people at D2, D3, D4, and D5. That leaves D1 as the only person in column D who is not innocent. Therefore, we can determine that D1 is CRIMINAL.