Puzzle Pack #2 Puzzle 25 Answer

Eric is at A2, so his three neighbors are Adam at A1, Gary at B2, and Katie at A3. Pam is at B4, and among those three, only Katie and Gary also neighbor Pam; Adam does not. Jane’s clue says exactly 2 of Eric’s 3 innocent neighbors also neighbor Pam, so Eric must have exactly 3 innocent neighbors, and the two among them that also neighbor Pam must be Katie and Gary. That means Katie is innocent, and since all three of Eric’s neighbors are innocent, Linda at B3 is also innocent from the already-fixed neighborhood around Pam in this round. Therefore, we can determine that A3 is INNOCENT and B3 is INNOCENT.

Gary is at B2, so the people below him are B3, B4, and B5. Will is at C5, and among those three, both Pam at B4 and Vicky at B5 neighbor Will, while Linda at B3 does not. Linda’s clue says there are no innocents below Gary who neighbor Will. That means every person below Gary who does neighbor Will must be criminal. Therefore, we can determine that B4 is CRIMINAL and B5 is CRIMINAL.

Isaac is at C2, so his five neighboring innocents must be among A1, B1, C1, D1, A2, B2, D2, B3, C3, and D3, with Jane at D2, Katie at A3, and Linda at B3 already known innocent. “To the left of Nick” means everyone in row 3 to the left of D3, so among Isaac’s neighboring innocents the only possible people counted by the clue are Katie at A3, Linda at B3, and Mary at C3. Since the clue says exactly 2 of Isaac’s 5 neighboring innocents are to the left of Nick, and Katie and Linda already make 2 by themselves, Mary cannot be one of those five neighboring innocents. Therefore, we can determine that C3 is INNOCENT.

Katie says there are exactly 2 innocents in column B. In that column, B3 is already innocent and B5 is criminal, so exactly one of B1 or B2 must be innocent. Jane says exactly 2 of Eric's 3 innocent neighbors also neighbor Pam. Eric is at A2, and his neighbors are A1, B1, B2, A3, and B3; among these, the ones who also neighbor Pam at B4 are only A3 and B3. Since Eric has exactly 3 innocent neighbors, and exactly 2 of those also neighbor Pam, those two must be A3 and B3, so B1 and B2 are not innocent. That means the only two innocents in column B are B3 and not B1 or B2, so B1 is not innocent. Therefore, we can determine that A1 is CRIMINAL.

Isaac’s neighbors are Betty, Chad, Debra, Gary, Jane, Linda, Mary, and Ryan. Pam says exactly 5 of those 8 neighbors are innocent, and Adam says exactly 1 innocent neighboring Isaac is a cop. Among Isaac’s neighboring cops, Debra, Jane, and Nick, only Debra and Jane are actually neighbors of Isaac, and Jane is already known innocent, so Debra cannot also be innocent. Therefore, we can determine that Chad is CRIMINAL.

Linda’s neighbors are Katie, Gary, Mary, Pam, Vicky, Isaac, Ryan, and Sue, and the five innocent neighbors among them must include the three already known innocents Katie, Gary, and Mary. Pam’s neighbors are Gary, Mary, Isaac, Linda, Ryan, Vicky, Tyler, and Will, so among Linda’s innocent neighbors, the ones who also neighbor Pam are Gary, Mary, Isaac, and Ryan, while Katie does not. Vicky says exactly 3 of Linda’s 5 innocent neighbors also neighbor Pam. Since Gary and Mary are already innocent and do neighbor Pam, there is room for exactly one more such person among Isaac and Ryan. That means Isaac and Ryan cannot both be innocent, and because Linda must still have 5 innocent neighbors in total, Isaac must be one of them. Therefore, we can determine that C2 is INNOCENT.

Linda’s neighbors are A2, B2, C2, A3, C3, A4, B4, and C4, and the innocent ones already known among them are Isaac at C2, Katie at A3, and Mary at C3. Pam’s neighbors are A3, B3, C3, A4, C4, A5, B5, and C5, so among those three known innocent neighbors of Linda, only Katie and Mary also neighbor Pam, while Isaac does not. Vicky’s clue says exactly 3 of Linda’s 5 innocent neighbors also neighbor Pam, so Linda must have exactly two more innocent neighbors, and only one of those can also neighbor Pam; among the unknown neighbors A2, B2, A4, and C4, that means C4 must be innocent while the other extra innocent is either A2 or B2. Therefore, we can determine that C4 is INNOCENT.

Linda at B3 has five innocent neighbors: A2 Eric, A3 Katie, B2 Gary, C2 Isaac, and C3 Mary. Pam is at B4, and among those five, the ones who also neighbor Pam are A3 Katie, B2 Gary, C3 Mary, and A2 Eric, while A2 only counts if Olivia at A4 is innocent, because A2 and B4 are diagonal neighbors through Olivia’s position being part of Linda’s neighbor set. Since the clue says exactly three of Linda’s five innocent neighbors also neighbor Pam, A2 cannot be one of Linda’s innocent neighbors. Therefore, we can determine that A4 is CRIMINAL.

Isaac’s five neighbors are Betty, Chad, Gary, Linda, and Mary. Pam says exactly two of the innocents among those five are to the left of Nick, and since Chad is criminal while Linda and Mary are innocent, this fixes Betty and Gary as innocent and puts Nick in column D, so everyone in columns A, B, and C is to Nick’s left. That means Pam’s left-side innocent neighbors of Isaac are exactly Betty, Gary, Linda, and Mary, with exactly two of them counted by the clue, so the two innocents not counted there must be the two on the right of Nick’s line, which confirms the setup needed for Jane’s clue about Eric. Eric’s neighbors are Adam, Betty, Gary, Katie, and Linda, so his three innocent neighbors are Betty, Gary, and Katie. Among those three, Betty, Gary, and Katie all also neighbor Pam except Katie does not, because Pam’s neighbors are Katie, Linda, Olivia, Ryan, Tyler, Vicky, and Eric. So Jane’s clue says exactly two of Eric’s three innocent neighbors also neighbor Pam, which matches only if Eric himself is not one of the innocents around that area and is criminal. Now use Olivia’s clue. Column A already has Adam and Olivia as criminals, and with Eric now criminal it has three criminals; if Tyler were innocent, then no column would have exactly four criminals, but Olivia says only one column has exactly four criminals. So Tyler must also be criminal, making column A the one and only column with exactly four criminals. Therefore, we can determine that A5 is CRIMINAL and A2 is CRIMINAL.

Linda is at B3, and her five innocent neighbors are Isaac at C2, Jane at D2, Katie at A3, Mary at C3, and Ryan at C4. Pam is at B4, and among those five, the ones who also neighbor Pam are Katie, Mary, and Ryan, which is exactly three. For Vicky's statement to be true, Isaac and Jane must not also be innocent neighbors of Linda who neighbor Pam. Jane already is innocent, so the only way to keep the total at exactly three is for Gary at B2 not to add another innocent neighbor to Linda's count through that shared area. Therefore, we can determine that B2 is INNOCENT.

Eric at A2 has three innocent neighbors: Betty at B1, Gary at B2, and Katie at A3. Pam is at B4, and among those three, the ones who also neighbor Pam are Gary at B2 and Katie at A3, while Betty at B1 is not a neighbor of Pam. Since Jane says exactly 2 of Eric's 3 innocent neighbors also neighbor Pam, Eric's three innocent neighbors must be exactly those three people, so Betty has to be the third innocent neighbor of Eric. Therefore, we can determine that B1 is CRIMINAL.

Isaac is at C2, and his five innocent neighbors are Gary at B2, Jane at D2, Linda at B3, Mary at C3, and Ryan at C4. Nick is at D3, so “to the left of Nick” means those innocents in columns A, B, or C in Nick’s row, and among Isaac’s five innocent neighbors that is Gary, Linda, Mary, and Ryan, while Jane is not to Nick’s left. Pam’s clue says exactly 2 of those 5 innocents are to the left of Nick, so Ryan at C4 cannot stay innocent; that forces Sue at D4 to be criminal instead. Then look at column D: D2 is innocent, and with Sue now criminal the column has Debra unknown, Jane innocent, Nick unknown, Sue criminal, and Xavi unknown. Eric’s clue says there are more criminals than innocents in column D, so with 2 innocents already fixed there must be at least 3 criminals in that column. Since Sue gives only 1 criminal so far, both remaining undecided spots needed to reach 3 criminals here must be Debra and Xavi, and in particular Xavi is criminal. Therefore, we can determine that D4 is CRIMINAL and D5 is CRIMINAL.

Isaac is at C2, and his five innocent neighbors are Gary at B2, Jane at D2, Katie at A3, Linda at B3, and Mary at C3. Pam says exactly 2 of those 5 innocents are to the left of Nick, and since “left of Nick” means in Nick’s row, Nick must be at D3 so that the two innocents in row 3 to his left are Linda and Mary. That fixes Nick as innocent, because Pam’s clue is counting the innocents neighboring Isaac and needs Nick’s position to make that count work. Now use Mary’s clue about the edge. The edge currently has criminals at A1, B1, C1, A2, A4, D4, A5, B5, D5, which is 9 criminals, and the only unknown edge spaces are D1 and C5. Since the total number of edge criminals is odd, those two unknown edge spots must contain one criminal and one innocent. With D1 already fixed as innocent from the previous deduction, C5 must be the criminal. Therefore, we can determine that C5 is CRIMINAL.

The guards are Jane at D2, Mary at C3, and Vicky at B5. The clue says exactly one guard has an innocent directly above them. Mary has Isaac directly above her at C2, and Isaac is innocent, so Mary already fits the clue. Jane has Debra directly above her at D1, while Vicky has Pam directly above her at B4, and Pam is criminal, so Jane cannot also have an innocent above her. That means Debra at D1 is not innocent. Therefore, we can determine that D1 is CRIMINAL.

Isaac is at C2, and his five innocent neighbors are B2 Gary, D2 Jane, B3 Linda, C3 Mary, and one more among D3 Nick and C1 Chad. Pam’s clue says exactly 2 of those 5 innocent neighbors are to the left of Nick. Since “left of Nick” means in the same row as Nick, only the people in row 3 can count, and the innocent neighbors of Isaac in row 3 are Linda at B3 and Mary at C3. Those are exactly 2 people to Nick’s left only if Nick is at D3 and is also one of Isaac’s five innocent neighbors. Therefore, we can determine that D3 is INNOCENT.