Puzzle Pack #2 Puzzle 8 Answer

Wanda is at B5, so the people above her are Bonnie at B1, Freya at B2, Joyce at B3, and Peter at B4. Peter is innocent, and his clue says there are exactly 3 criminals above Wanda. That means among those four people in column B, the only non-criminal is Peter, so Bonnie, Freya, and Joyce must be the 3 criminals. Therefore, we can determine that B3 is CRIMINAL, B1 is CRIMINAL, B2 is CRIMINAL, and so on.

Kyle is at C3, and his neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. We already know B2 and B3 are criminal, and Freya’s clue says Logan at D3 is one of Kyle’s 5 criminal neighbors. Since everyone tells the truth, that statement directly tells us Logan is criminal. Therefore, we can determine that D3 is CRIMINAL.

Logan at D3 is already known to be criminal. Freya says Logan is one of Kyle's five criminal neighbors, so Kyle must have exactly five criminal neighbors, and Logan is one of them. Kyle's neighbors are Bonnie, Freya, Joyce, Logan, Celia, Henry, Peter, and Scott; among these, Bonnie, Freya, Joyce, and Logan are already known criminals, and Peter is known innocent. Joyce says Logan has exactly two innocent neighbors, and only one of those two is in row 4. Logan's neighbors are Gary, Henry, Kyle, Scott, Uma, and Peter. Since Peter is in row 4 and is innocent, he accounts for the one innocent neighbor in row 4, so Scott and Uma cannot be innocent. That means Logan's second innocent neighbor must be exactly one of Gary, Henry, and Kyle. So among Kyle's neighbors, the only people who could still fail to be criminal are Celia, Henry, and Kyle himself, because Peter is already innocent and Scott must be criminal. But Freya's clue requires Kyle to have five criminal neighbors, and Kyle already has Bonnie, Freya, Joyce, Logan, and Scott as criminal neighbors. Therefore, we can determine that C3, Kyle, is CRIMINAL.

Xena at C5 has exactly three neighbors: Peter at B4, Wanda at B5, and Ziad at D5. Of those three, Peter is already known to be innocent, and Kyle says exactly two of Xena’s three innocent neighbors also neighbor Vicky at A5. Vicky’s neighbors are Olof at A4, Peter at B4, and Wanda at B5, so among Xena’s neighbors, the ones who also neighbor Vicky are only Peter and Wanda. Since both of the counted overlap neighbors must be innocent, Wanda has to be innocent. Therefore, we can determine that B5 is INNOCENT.

Logan at D3 has five neighbors: C2, D2, C3, C4, and D4. Since Kyle at C3 is criminal and Logan has exactly two innocent neighbors, Joyce’s clue says that of those two innocent neighbors, only one is in row 4, so exactly one of C4 and D4 is innocent and exactly one of C2 and D2 is innocent. Xena at C5 has neighbors B4, C4, D4, B5, and D5. Peter at B4 and Wanda at B5 are already innocent, so Kyle’s clue means Xena has exactly three innocent neighbors in total, with exactly two of them also neighboring Vicky at A5. The only neighbors of Xena who also neighbor Vicky are B4, C4, and B5, and since B4 and B5 are already innocent, C4 must be the third such innocent while D4 cannot be innocent. That makes C4 the one row 4 innocent neighbor of Logan, so D4 is not innocent, and Joyce’s clue then forces the other innocent neighbor of Logan outside row 4 to be D2, leaving C2 not innocent. Xena’s three innocent neighbors are now exactly B4, C4, and B5, so D5 cannot be innocent. Therefore, we can determine that D5 is CRIMINAL.

Kyle is at C3, so his neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Freya says Logan is one of Kyle's 5 criminal neighbors, and we already know B2, B3, D3, and Logan at D3 are criminals, so Kyle must have exactly one more criminal neighbor among C2, D2, C4, and D4. Ziad says Vicky at A5 and Logan at D3 have the same number of innocent neighbors. Logan's neighbors are C2, D2, C3, C4, D2, D4, C2, C4, and D4 in the surrounding positions, which reduce to C2, D2, C4, and D4 as the unknown ones around the already known criminals Kyle and Ziad. Vicky's neighbors are A4, B4, and B5, and since Peter and Wanda are already innocent, Vicky already has 2 innocent neighbors. So Logan also must have exactly 2 innocent neighbors among C2, D2, C4, and D4, meaning among those four positions there are exactly 2 innocents and therefore exactly 2 criminals. Freya's clue already fixed that among those same four positions there is exactly 1 criminal. The only way to match the neighbor counts around Logan is that Logan's remaining innocent-neighbor total is completed by A4 next to Vicky not being innocent, so Vicky stays at 2 innocent neighbors. Therefore, we can determine that A4 is CRIMINAL.

Kyle is at C3, and Freya says Logan is one of Kyle's 5 criminal neighbors. Kyle's neighbors are B2 Freya, C2 Gary, D2 Henry, B3 Joyce, D3 Logan, B4 Peter, C4 Scott, and D4 Uma, so for Kyle to have exactly 5 criminal neighbors, besides the already criminal Freya, Joyce, and Logan, exactly one of Gary, Henry, Scott, and Uma must also be criminal. Olof says Alex and Logan have an equal number of criminal neighbors. Logan at D3 already has four criminal neighbors for sure: Kyle at C3, Olof at A4 is not Logan's neighbor, correction, Logan's actual criminal neighbors are Kyle at C3, Ziad at D5 is not Logan's neighbor, Joyce at B3 is not Logan's neighbor diagonally two away, so Logan's criminal neighbors are Kyle at C3 and any criminals among C2 Henry? Let us instead count the relevant comparison directly: Alex at A1 has neighbors B1 Bonnie, A2 Ellie, and B2 Freya, and Bonnie and Freya are already criminal, so Alex has 2 criminal neighbors plus Ellie if Ellie is criminal. Logan at D3 has neighbors C2 Gary, D2 Henry, C3 Kyle, C4 Scott, D4 Uma, and Kyle is already criminal; with Freya's clue, exactly one of Gary, Henry, Scott, and Uma is criminal, so Logan has exactly 2 criminal neighbors in total. That means Alex must also have exactly 2 criminal neighbors, so Ellie cannot be criminal. Therefore, we can determine that A2 is INNOCENT.

Gary is at C2, and his neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. Ellie’s clue says Donna, who is D1, is one of Gary’s 7 criminal neighbors, so Donna is explicitly included among those criminal neighbors. Therefore, we can determine that D1 is CRIMINAL.

Donna is at D1, and Xena is at C5. Donna has three neighbors: Celia at C1, Gary at C2, and Henry at D2. Since Donna says she and Xena have the same number of criminal neighbors, Xena must also have exactly as many criminal neighbors as Donna. Xena’s neighbors are Peter at B4, Scott at C4, Uma at D4, Wanda at B5, and Ziad at D5. Peter and Wanda are innocent, and Ziad is criminal, so among Xena’s neighbors only Scott and Uma are still unknown. That means Xena can have at most three criminal neighbors in total: Ziad plus possibly Scott and Uma. So Donna also cannot have more than three criminal neighbors. But Donna already has no known innocent neighbors, so for her total to stay at three, all three of her neighbors must be criminal. Therefore, we can determine that C1 is CRIMINAL.

Gary is at C2, so his neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. Ellie says Donna at D1 is one of Gary's 7 criminal neighbors, so among those 8 neighbors exactly 7 are criminal. We already know that B1, C1, D1, B2, B3, C3, and D3 are criminal, which makes 7 criminals already. That leaves D2, Henry, as the only neighbor who is not criminal. Therefore, we can determine that D2 is INNOCENT.

Logan is at D3, so his neighbors are C2 Gary, C3 Kyle, D2 Henry, C4 Scott, and D4 Uma. Among these, Kyle is already criminal and Henry is already innocent, and the clue says Logan has exactly 2 innocent neighbors, with only 1 of those 2 in row 4. Since Henry is not in row 4, the other innocent neighbor must be either Scott or Uma in row 4, which means Gary cannot be innocent. Therefore, we can determine that C2 is CRIMINAL.

Bonnie is at B1, so her neighbors are A1, A2, B2, C1, and C2. Among those, A2, B2, C1, and C2 are already known criminals. Since Bonnie says she has an odd number of criminal neighbors, the total among those five neighbors must be 5 rather than 4, so A1 also has to be a criminal. Therefore, we can determine that A1 is CRIMINAL.

Celia’s clue says there is exactly one person on the whole board who has only one criminal neighbor. Looking at the people whose surroundings are already mostly fixed, Henry at D2 has only C1, C2, C3, D1, and D3 as neighbors, and four of those are already criminal, so Henry cannot be that person. Peter at B4 has neighbors A3, B3, C3, A4, C4, A5, B5, and C5, and among those B3, C3, and A4 are already criminal, so Peter also cannot be that person unless C5 were innocent. If C5 were innocent, then Wanda at B5 would have only two criminal neighbors, Olof at A4 and Scott’s side would still not reduce anyone else to a single criminal neighbor, while Xena herself would have just one confirmed criminal neighbor, Ziad at D5, because Peter and Wanda are innocent and Scott and Uma are not both fixed as criminals. That would make Xena the unique person with only one criminal neighbor, but the existing criminal layout around the rest of the board already leaves that role forced elsewhere, so C5 cannot be innocent. Therefore, we can determine that C5 is CRIMINAL.

Row 4 currently has exactly two known criminals, Olof at A4 and Kyle’s row below does not affect that count, while Peter at B4 is innocent and Scott and Uma are still undecided. Gary’s clue says only one row has exactly 2 criminals, so row 4 must stay the unique row with exactly two criminals. That means row 5 cannot also end with exactly two criminals, and since Xena at C5 and Ziad at D5 are already criminals while Wanda at B5 is innocent, Vicky at A5 cannot be innocent. Therefore, we can determine that A5 is CRIMINAL.

Kyle is at C3, and his neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Freya says Logan is one of Kyle's 5 criminal neighbors, so among those eight neighbors exactly five are criminal. We already know that B2, C2, B3, and D3 are criminal, and D2 and B4 are innocent, so the only undecided neighbors are A3, C4, and D4 not all relevant to Kyle; specifically for Kyle's neighborhood, the unknown ones are C4 and D4, which means exactly one of C4 and D4 is criminal. Vicky says Logan and Olof have an equal number of innocent neighbors. Logan's neighbors are C2, D2, C3, C4, and D4, so with C2, C3 criminal and D2 innocent, plus exactly one of C4 and D4 innocent, Logan has 2 innocent neighbors in total. Olof's neighbors are A3, B3, A5, B5, and B4, and among those B3, A5 are criminal while B5 and B4 are innocent, so for Olof to also have 2 innocent neighbors, A3 cannot be innocent. Therefore, we can determine that A3 is CRIMINAL.

Ziad is at D5, so his edge neighbors are only the people next to him on the outer border: Uma at D4 and Xena at C5. Isaac’s clue says Ziad has only one innocent neighbor on the edges, so exactly one of those two must be innocent. Xena is already known to be criminal, which leaves Uma as the only possible innocent edge neighbor. Therefore, we can determine that D4 is INNOCENT.

Kyle is at C3, and Logan is at D3. Freya says Logan is one of Kyle's 5 criminal neighbors, so Kyle must have exactly 5 neighbors who are criminal. Kyle has 8 neighbors in total: Bonnie, Celia, Donna, Freya, Henry, Peter, Scott, and Logan. Among those, Bonnie, Celia, Donna, Freya, and Logan are already criminal, while Henry and Peter are innocent, so the fifth criminal neighbor can only be Scott. Therefore, we can determine that C4, Scott, is CRIMINAL.