Puzzle Pack #2 Puzzle 12 Answer

Nancy is at C3, so her six neighbors are B2 Gary, C2 Hank, D2 Jerry, B3 Logan, D3 Olive, and C4 Steve. Tina is at D4, and the people above Tina are D1 Emily, D2 Jerry, and D3 Olive. Among Nancy’s six neighboring innocents, exactly two are above Tina, so within Nancy’s neighbors the two innocent people in that “above Tina” group must be Jerry and Olive. Therefore, we can determine that D2 is INNOCENT and D3 is INNOCENT.

Wanda is at C5, so her neighbors are B4 Rob, C4 Steve, D4 Tina, B5 Vicky, and D5 Ziad, and the edge neighbors among those are Steve, Tina, Vicky, and Ziad. Jerry says an odd number of Wanda’s edge-neighboring innocents exist, while Olive says Wanda has exactly 3 innocent neighbors in total and only 1 of those 3 is below Emily, meaning only 1 innocent among Wanda’s neighbors can be in row 4 or 5 of column D, so among Tina and Ziad exactly 1 is innocent. Since the innocent edge neighbors must be odd in number, the other two edge positions next to Wanda, Steve and Vicky, must contribute an even total with Tina and Ziad fixed at 1 innocent between them; that forces exactly one of Steve and Vicky to be innocent. But Wanda has 3 innocent neighbors altogether, and with exactly 1 coming from Tina and Ziad, the remaining two innocent neighbors must be Rob and Steve, leaving Vicky as the only non-innocent among Rob, Steve, and Vicky. Therefore, we can determine that B4 is INNOCENT, C4 is INNOCENT, and B5 is CRIMINAL.

Steve is at C4, so his edge neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. Among those edge positions, the edge neighbors are D3, B4, D4, B5, C5, and D5, and Rob’s clue says exactly 2 of the 9 edge innocents are among them. Since D3 and B4 are already known innocents, none of D4, B5, C5, or D5 can be innocent. Wanda is at C5, so she is one of those four edge positions that cannot be innocent. Therefore, we can determine that C5 is CRIMINAL.

Wanda is at C5, so the edge neighbors around her are B4, C4, D4, and D5. Jerry says an odd number of those edge neighbors are innocent. We already know B4 and C4 are innocent, so among D4 and D5 exactly one must be innocent. Wanda also says there are exactly 4 innocents in column D. In that column, D2 and D3 are already innocent, so among D1, D4, and D5 there must be exactly two more innocents. Since exactly one of D4 and D5 is innocent, the other needed innocent in column D has to be D1. Therefore, we can determine that D1 is INNOCENT.

Steve is at C4, so his four innocent neighbors are Olive at D3, Pam at A4, Rob at B4, and Wanda is not innocent, so the fourth innocent neighbor must be Nancy at C3. Emily’s clue says that only one of those four innocents neighboring Steve is below Hank at C2. The people below Hank in column C are Nancy at C3, Steve at C4, and Wanda at C5, and among Steve’s innocent neighbors only Nancy fits that description. Therefore, we can determine that C3 is INNOCENT.

Wanda is at C5, so her neighboring edge positions are B4, C4, D4, and D5. Jerry says an odd number of the innocents on the edges neighbor Wanda. We already know B4 and C4 are innocent, so there are already 2 such innocents, and to make the total odd exactly one of D4 and D5 must also be innocent. Steve is at C4, and his four innocent neighbors are B4, C3, D3, and exactly one of D4 or D5. Emily says only 1 of those 4 innocents is below Hank, so Hank must be in row 3, because then the only innocent below him among those four is the one in row 5. That places Hank at C3, which is impossible because C3 is Nancy and already innocent, so Hank must instead be at B2, making the people below Hank in Steve’s innocent-neighbor set be B4 and exactly one of D4 or D5. For Emily’s clue to still say only 1 of those 4 is below Hank, Logan at B3 cannot be innocent, because if Logan were innocent then Steve would already have the four innocent neighbors B3, B4, C3, and D3, and both B4 and D3 would be below Hank. Therefore, we can determine that B3 is CRIMINAL.

Steve is at C4, so his neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. Among those, B3, B5, and C5 are already known criminals, and C3, D3, and B4 are known innocents, so the only undecided neighbors are D4 and D5. Rob says exactly 2 edge innocents are Steve's neighbors, and among Steve's neighbors the edge spaces are only B3, C3, D3, B4, D4, B5, C5, and D5; since C3, D3, and B4 are already three edge innocents there, D4 and D5 cannot also be innocent. That means the only innocent in column C must be C1 or C2, because C3, C4, and C5 are already fixed as innocent, innocent, and criminal. Logan says columns A and C contain the same number of criminals. In column A, A4 is innocent, so column A can have at most three criminals; but in column C, with C3 and C4 innocent and C5 criminal, having C1 criminal would force C2 innocent and leave only one criminal in column C, which does not match the needed balance from the current column counts. Therefore, we can determine that C1 is INNOCENT.

Steve is at C4, and his edge neighbors are D4 Tina, B5 Vicky, C5 Wanda, and D5 Ziad, because the other neighbors around him are not on the edge. Rob says exactly 2 of the 9 edge innocents are Steve's neighbors. Among those four edge neighbors, Vicky and Wanda are already criminals, so the only neighbors of Steve who could be edge innocents are Tina and Ziad. That means both D4 and D5 are innocent, giving column D exactly three innocents and therefore two criminals, since D1, D2, and D3 are already innocent. Nancy says columns A and D have the same number of criminals, so column A also has two criminals; but A4 is innocent, so among A1, A2, A3, and A5 there must be exactly two criminals. Looking at the top row, C1 and D1 are innocent, so for B1 to be innocent as well would leave only A1 as a possible criminal in that row, which would not fit the remaining column totals forced by the equal-count clue. Therefore, we can determine that B1 is CRIMINAL.

Steve is at C4, so his edge neighbors are B5, C5, D5, and D4. Among those, B5 and C5 are already known criminals, so the only edge neighbors of Steve who could be innocent are D4 and Ziad is not Steve's neighbor at all. Rob says exactly 2 of the 9 edge innocents are Steve's neighbors, so Steve must have exactly two innocent edge neighbors; with B5 and C5 not innocent, that forces D4 and C3? No, C3 is not on the edge, so the only way to reach two edge-innocent neighbors is that D4 is innocent and the count of edge innocents is fixed accordingly. Now use Logan's clue: column C contains Donna at C1, Hank at C2, Nancy at C3, Steve at C4, and Wanda at C5. Since C1, C3, and C4 are already innocent and C5 is criminal, column C can have either 1 criminal total if Hank is innocent or 2 criminals total if Hank is criminal. Column A currently has no known criminals at all, so to match column C's number of criminals, Hank cannot be criminal here. Therefore, we can determine that C2 is INNOCENT.

Above Uma in column A are Amy at A1, Frank at A2, Keith at A3, and Pam at A4. Donna’s clue says the number of innocents above Uma is odd, and among those four people Pam is already known to be innocent while the other three are not yet fixed. Rob’s clue settles that group: the edge innocents who are neighbors of Steve are exactly Pam, Rob, Jerry, Nancy, Tina, Vicky, Wanda, and Ziad around C4, so the remaining edge innocent must be Uma rather than anyone above her in column A. That makes the people above Uma contribute an odd total of innocents, matching Donna’s clue only when Uma herself is innocent. Therefore, we can determine that A5 is INNOCENT.

Wanda is at C5, so her neighboring edge positions are B4, C4, D4, B5, and D5. Among those, B4 Rob and C4 Steve are already innocent, B5 Vicky is criminal, and only D4 Tina and D5 Ziad are still unknown, so Jerry’s clue says the total number of innocents there must be odd. Since there are already exactly two known innocents among Wanda’s edge neighbors, exactly one of Tina or Ziad must also be innocent. Steve at C4 currently has criminal neighbors at B3 Logan, B5 Vicky, and C5 Wanda, with D5 Ziad and D4 Tina as the only unknown neighbors who could add more. Uma’s clue says Steve has the unique highest number of criminal neighbors, so Ziad cannot be criminal, because then Tina would also have four criminal neighbors from C5 Wanda, B5 Vicky, B3 Logan, and D5 Ziad, tying Steve instead of leaving him with the most. So Ziad must be innocent, which makes Tina criminal from Jerry’s clue. Now Tina at D4 is criminal, and her clue says all criminals above her in column D are connected. The only possible criminals above her are D2 Jerry and D3 Olive, but D3 Olive is innocent, which blocks any connected criminal chain reaching D2, so D2 cannot be criminal. With Jerry innocent, his clue that the number of innocents to his left in row 2 equals the number of criminals below him in column D gives 1 on the left already from Hank at C2, and below him there is now exactly 1 criminal at D4, so both unknowns to Jerry’s left, Frank at A2 and Gary at B2, must be criminal. Keith at A3 says the number of criminal neighbors of Gary equals the number of innocent neighbors of Vicky; Vicky has three innocent neighbors already at A4, C4, and A5, and Gary’s neighbors are A1, B1, C1, A2, C2, A3, and B3, where B1, A2, and B3 are criminal, so Gary already has three criminal neighbors before counting Keith. That forces Keith not to be criminal, so Keith must be innocent. Therefore, we can determine that A3 is INNOCENT.

Steve is on C4, so his neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. Among the edge innocents, the ones in those neighbor spots are C3, D3, B4, and possibly D4 and D5 if they were innocent; but Rob says exactly 2 of the 9 edge innocents are Steve's neighbors, so D4 and D5 cannot be innocent. That fixes Steve's innocent neighbors at B4, C3, and D3, so Steve has 3 innocent neighbors in total only if Gary at B2 is innocent, because his other neighbors B3, B5, C5, D4, and D5 are not innocent. Chris at B1 has neighbors A1, A2, B2, C1, and C2, and the known innocents among those are C1 and C2. Since Steve says he and Chris have the same number of innocent neighbors, Chris must also have 3 innocent neighbors, so one more of A1, A2, and B2 must be innocent. The only one forced by this comparison is Gary at B2, because that is the shared undecided point that raises Chris's count to match Steve's fixed total. Therefore, we can determine that B2 is INNOCENT.

Gary’s clue says every row must contain at least one criminal. In row 4, Pam at A4, Rob at B4, and Steve at C4 are already known to be innocent, so the only person left in that row who could satisfy the clue is Tina at D4. Therefore, we can determine that D4 is CRIMINAL.

Wanda is at C5, so her edge neighbors are Tina at D4, Vicky at B5, and Ziad at D5. Jerry says an odd number of innocents on the edges neighbor Wanda. Tina and Vicky are both already known criminals, so among those edge neighbors the only possible innocent is Ziad. That means the number of innocent edge neighbors is 1, which is odd, so Jerry’s clue forces Ziad to be innocent. Therefore, we can determine that D5 is INNOCENT.

Gary says each row has at least one criminal. In row 2, the only unknown person is Frank at A2, because Gary at B2, Hank at C2, and Jerry at D2 are already known to be innocent. So row 2 can only contain a criminal if Frank is that criminal. Therefore, we can determine that A2 is CRIMINAL.

Steve is at C4, and his neighbors are Logan, Nancy, Olive, Rob, Tina, Vicky, Wanda, and Ziad. Among those, Nancy, Olive, Rob, and Ziad are innocent, so Steve has 4 innocent neighbors. Chris is at B1, and his neighbors are Amy, Donna, Frank, Gary, Hank, and Logan. Donna, Gary, and Hank are innocent, Frank and Logan are criminal, so for Chris to also have 4 innocent neighbors, Amy must be innocent as well. Therefore, we can determine that A1 is INNOCENT.