Puzzle Pack #1 Puzzle 28 Answer

The corner people are A1, D1, A5, and D5, and the clue says there are exactly 2 innocent corners in total. It also says only 1 of those 2 innocent corners is in column D, so among the two corner spaces in column D, exactly one is innocent. Since D5 is already known to be innocent, D1 cannot also be innocent. Therefore, we can determine that D1 is CRIMINAL.

The corners are A1, D1, A5, and D5. Xena says there are exactly two innocent corners, and only one of those two is in column D. Since D5 is already known to be innocent and D1 is criminal, the only innocent corner in column D is D5, so there must be exactly one innocent corner in column A as well. Column A contains A1, A2, A3, A4, and A5, and Denis says there is only one innocent anywhere in that column. Because the single innocent in column A has to be one of the corners, it must be either A1 or A5. That means the non-corner spaces in column A cannot be innocent. Therefore, we can determine that A2 is CRIMINAL, A3 is CRIMINAL, and A4 is CRIMINAL.

Nicole’s clue is about row 4, which currently has Nicole at A4 as a criminal, while B4, C4, and D4 are still the only positions that could continue that row’s criminal group. Floyd’s clue says the number of criminal cooks equals the number of criminal sleuths, and since Nicole is a criminal sleuth while Alex is not known to be one, we already need at least one criminal cook. In row 4, the cooks are Ruth at C4 and Sam at D4, but Nicole’s connected-row clue means any criminal in row 4 must form one unbroken horizontal chain starting from A4. So to make Ruth at C4 a criminal cook, B4 must also be criminal, because A4 and C4 cannot be connected across row 4 otherwise. That gives us the needed criminal cook at C4 and forces B4 to be criminal as well. Therefore, we can determine that B4 is CRIMINAL and C4 is CRIMINAL.

Ruth’s clue is about the three cops: Megan at D3, Wanda at C5, and Xena at D5. Xena cannot count for the clue because nobody is directly above D5, so the two cops with a criminal directly above them must be Megan and Wanda. Wanda does fit, because Ruth at C4 is a criminal directly above C5, so Megan must also fit, which means the person directly above D3 is a criminal. The person directly above D3 is Isaac at D2, and Jane’s clue says there are exactly 2 innocents in column D; since Denis at D1 is already a criminal and Xena at D5 is already innocent, only one of D2, D3, and D4 can be innocent. If Isaac at D2 were that innocent, then both D3 and D4 would have to be criminals, so Megan would be a criminal cop with a criminal directly above her, which is exactly what Ruth’s clue requires. Therefore, we can determine that D3 is CRIMINAL.

Olive is at B4, so her six neighboring criminals must be among A3, B3, C3, A4, C4, and A5. The clue says exactly two of those neighboring criminals are in column A, and among the column A neighbors of Olive, A3 and A4 are already known criminals. That already makes the required two, so the remaining column A neighbor, A5, cannot be a criminal. Therefore, we can determine that A5 is INNOCENT.

The two innocents in corners must be chosen from A1, D1, A5, and D5. We already know D5 is innocent, while D1 is criminal and A5 is innocent, so the two innocent corners are A5 and D5. That means exactly one innocent corner is in column D, which is D5, so the remaining corner A1 cannot also be innocent. Therefore, we can determine that A1 is CRIMINAL.

The criminal sleuths are already fully known: Alex at A1 and Nicole at A4, so there are 2 criminal sleuths. The cooks are Ruth at C4 and Sam at D4, and Ruth is already known to be criminal. Floyd’s clue says the number of criminal cooks must match the number of criminal sleuths, so there must also be 2 criminal cooks. That forces Sam at D4 to be criminal as well. Therefore, we can determine that D4 is CRIMINAL.

Column D contains Denis at D1, Isaac at D2, Megan at D3, Sam at D4, and Xena at D5. Jane’s clue says there are exactly 2 innocents in that whole column. We already know Denis, Megan, and Sam are criminals, and Xena is innocent, so the only way for column D to contain exactly 2 innocents is for Isaac at D2 to be the second innocent. Therefore, we can determine that D2 is INNOCENT.

Olive says Ruth is one of 4 criminals in column C, so column C has exactly four criminals. Since Ruth at C4 is already criminal, the only undecided people in that column are Carol at C1, Hank at C2, Laura at C3, and Wanda at C5, and exactly three of those four must be criminals. Tom says all criminals below Carol are connected. The people below Carol in column C are Hank, Laura, Ruth, and Wanda. Because Ruth is criminal, the criminals in that group must form one unbroken vertical block in column C. That means if Wanda at C5 is one of the three needed criminals, then Laura at C3 must also be criminal to connect Wanda to Ruth; and if Wanda is not criminal, then the three needed criminals below Carol are Hank, Laura, and Ruth anyway, which still makes Laura criminal. Therefore, we can determine that C3, Laura, is CRIMINAL.

Olive is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. We already know six of those neighbors are criminals, and among those six, the ones in column A are A3 and A4. Megan’s clue says exactly 2 of Olive’s 6 criminal neighbors are in column A, so the other four criminal neighbors must all be outside column A, which forces B3 to be one of Olive’s criminal neighbors. Laura’s clue about row 5 is consistent with that count and does not change this result. Therefore, we can determine that B3 is CRIMINAL.

Olive says Ruth is one of 4 criminals in column C. In column C, Ruth at C4 is already criminal and Laura at C3 is also criminal, so to make 4 criminals in that column the only remaining two spaces, C1 and C5, must both be criminal. Kyle’s clue fits that as well, because the common neighbors of Denis at D1 and Isaac at D2 are C1, C2, and C3, and with C3 already criminal there must be exactly one innocent among C1 and C2. That means C1 cannot be forced innocent, so Olive’s clue is what fixes column C completely. Therefore, we can determine that C5 is CRIMINAL.

Olive is at B4, so her neighboring criminals are the six known criminals around her: A3, B3, C3, A4, C4, and C5. Among those six, exactly two are in column A, and those are A3 and A4. That means no other neighboring criminal of Olive can be in column A, so A5 cannot be one of Olive’s neighboring criminals; since A5 is Tom and he is already innocent, Olive must still have exactly six neighboring criminals among her eight neighbors. The only remaining unknown neighbor of Olive is B5, so B5 cannot be criminal and must be innocent. Therefore, we can determine that B5 is INNOCENT.

The relevant teachers are Bonnie at B1, Floyd at A2, and Gus at B2, because those are the only teachers on the board. Isaac’s clue says exactly two teachers have a criminal directly to their right. Floyd does, because Jane at A3 is not to his right; sorry, the teachers must be checked by row: Bonnie has Carol to her right at C1, Floyd has Gus to his right at B2, and Gus has Hank to his right at C2, so the teacher positions that can satisfy the clue are Bonnie, Floyd, and Gus according to who is immediately right of them. We already know Floyd is a teacher, and for Isaac’s count to reach exactly two, Gus must be one of the teachers with a criminal directly to the right, which means Hank is criminal; but Olive’s clue already fixes column C as having exactly four criminals, and with Laura, Ruth, and Wanda already criminal, only one of Carol, Hank, and C5’s status pattern can remain open, forcing Hank to be criminal and so making Gus one of the counted teachers. Therefore, we can determine that B2 is CRIMINAL.

Column C has Ruth at C4, Laura at C3, Hank at C2, Carol at C1, and Wanda at C5. Olive says Ruth is one of 4 criminals in column C, so exactly four of those five people in column C are criminals; since Laura, Ruth, and Wanda are already known criminals, that means exactly one of Hank and Carol is innocent. Gus says there are 5 innocents in total, and the board already has Isaac, Tom, Vince, and Xena as innocents, so there is room for only one more innocent anywhere. Because that one remaining innocent must be either Hank or Carol in column C, Bonnie cannot be innocent. Therefore, we can determine that B1 is CRIMINAL.

Carol at C1 and Xena at D5 must have the same number of criminal neighbors. Xena’s neighbors are C4 Ruth, C5 Wanda, D4 Sam, and all three are criminals, so Xena has 3 criminal neighbors. Carol’s neighbors are B1 Bonnie, B2 Gus, C2 Hank, D1 Denis, and D2 Isaac; among these, Bonnie, Gus, and Denis are criminals, Isaac is innocent, so Carol already has 3 criminal neighbors without Hank. To match Xena’s total of 3, Hank cannot be criminal. Therefore, we can determine that C2 is INNOCENT.

Olive says Ruth is one of 4 criminals in column C, so column C must contain exactly four criminals in total. In that column, Hank at C2 is already innocent, while Laura at C3, Ruth at C4, and Wanda at C5 are already criminal. That gives three criminals and one innocent in column C, so the only remaining person there, Carol at C1, has to be the fourth criminal. Therefore, we can determine that C1 Carol is CRIMINAL.