Puzzle Pack #1 Puzzle 8 Answer

Carol is at C1, so her neighbors are B1, B2, C2, D1, and D2. Daniel says exactly 3 of those neighbors are criminals, and only 1 of those 3 is in row 1. Among Carol's neighbors, the only people in row 1 are B1 and D1, and D1 is already known to be innocent, so the one criminal neighbor in row 1 has to be B1. Therefore, we can determine that B1 is CRIMINAL.

Carol’s neighbors are B1, B2, C2, and D2. Daniel says exactly 3 of those neighbors are criminals, and only 1 of those 3 is in row 1. Since B1 is already a known criminal and it is the only row 1 neighbor Carol has, the other two criminals among Carol’s neighbors must be B2, C2, and D2 with exactly two of those three being criminals. Betty says there is exactly one innocent to the right of Floyd, which means among C2, D2, C1, and D1 there is exactly one innocent. But D1 is already known to be innocent, so C2, D2, and C1 must all be criminals. That makes C2 and D2 both criminals, so among B2, C2, and D2 the remaining criminal must be B2. Therefore, we can determine that B2 is CRIMINAL.

Daniel’s neighbors are Carol at C1, Helen at C2, and Isaac at D2. Floyd’s clue says Daniel has exactly 2 innocent neighbors, and since Daniel himself is innocent, that means among Carol, Helen, and Isaac there are exactly 2 innocents and 1 criminal. Daniel also says that Carol has exactly 3 criminal neighbors, and only 1 of those is in row 1; Carol’s neighbors are Betty, Floyd, Helen, and Daniel in row 1 or 2, and we already know Betty and Floyd are criminals while Daniel is innocent. So Carol’s third criminal neighbor must be Helen, which leaves Carol unable to be criminal herself because then Daniel’s three neighbors would include two criminals, not one. Therefore, we can determine that C1 is INNOCENT.

Carol’s neighbors are A1’s side? No: as C1, her neighbors are B1, B2, C2, D1, and D2. We already know B1 and B2 are criminals, and Daniel’s clue says that among Carol’s three criminal neighbors, only one is in row 1. Since B1 is in row 1 and already criminal, the other row 1 neighbor C2 cannot also be criminal, so Helen at C2 must be innocent. Carol also says there are exactly two innocents in row 2. Row 2 is A2, B2, C2, and D2, and we already know B2 is criminal while C2 is innocent, so exactly one of A2 and D2 must be innocent. Daniel is already innocent at D1, and if D2 were also innocent then Carol would have two row 1 innocent neighbors and still only one row 1 criminal neighbor among the three criminals, which forces A2 to be the remaining row 2 innocent. Therefore, we can determine that A2 is INNOCENT.

Carol at C1 has three neighboring spots: B1, B2, and C2. Daniel says only one of Carol’s three criminal neighbors is in row 1, and since B1 and C2 are the row 1 neighbors while B2 is in row 2, that means exactly one of B1 and C2 is criminal. We already know Betty at B1 is criminal, so Helen at C2 cannot be criminal and must be innocent. Then row 2 has Erwin innocent, Floyd criminal, Helen innocent, and Isaac unknown, so row 2 already has exactly two innocents. Erwin says row 1 has more innocents than row 2, and row 1 already has Carol and Daniel innocent, so Alex at A1 must also be innocent to make row 1 have more than two innocents. Therefore, we can determine that A1 is INNOCENT.

Megan is at D3, so her neighbors are C2, D2, C3, C4, D4, and the clue says exactly two of those are criminals. “Below Helen” means people in Helen’s column under C2, which are C3, C4, and C5. So the two criminals neighboring Megan cannot be C3 or C4, because both of them are below Helen and also neighbors of Megan. Therefore, we can determine that C3 Logan is INNOCENT and C4 Ronald is INNOCENT.

Carol at C1 is neighbored by B1, B2, C2, and D2. Since B1 and B2 are already known criminals, Daniel’s clue that only 1 of Carol’s 3 criminal neighbors is in row 1 means Carol has exactly one more criminal neighbor, and it cannot be D2 because that would make two criminal neighbors in row 1. So C2 is criminal and D2 is innocent. Megan at D3 is neighbored by C2, D2, C4, and D4. We now know C2 is criminal, while D2 and C4 are innocent, so Alex’s clue that Megan has exactly 2 criminal neighbors means the other criminal neighbor must be D4. Therefore, we can determine that D4, Susan, is CRIMINAL.

Carol is at C1, so her neighbors are B1, B2, C2, D1, and D2. Daniel says exactly 3 of those neighbors are criminals, and only 1 of those 3 is in row 1. Since B1 is already a criminal and D1 is innocent, the other two criminal neighbors of Carol must be C2 and D2. Daniel at D1 currently has 2 criminal neighbors, B1 and C2 and D2 are not both his neighbors; his actual neighbors are C1, C2, D2, so with C1 innocent that gives him exactly 2 criminal neighbors: C2 and D2. Logan says Daniel and Zoe have the same number of criminal neighbors, so Zoe at D5 must also have 2 criminal neighbors. Zoe’s neighbors are C4, C5, and D4, and since C4 is innocent and D4 is criminal, C5 cannot also be criminal or Zoe would have only 1 or else the count would not match; to make Zoe’s total exactly 2, C5 must be innocent is not correct from that count alone. But with D4 already one criminal and only C5 and C4 left beside Zoe, and C4 innocent, Zoe cannot gain a second criminal unless C5 were criminal; that would give Zoe 2 criminal neighbors, matching Daniel. Therefore, we can determine that C5 is INNOCENT.

Logan is at C3, so his five innocent neighbors are the innocent people around him: Helen at C2, Kay at B3, Ronald at C4, Olive at B4, and Susan at D4. Betty is at B1, and everyone below Betty is in rows 2 through 5, so among those five people, Kay, Ronald, Olive, and Susan are below Betty, while Helen is not. Xavi’s clue says exactly 2 of Logan’s 5 innocent neighbors are below Betty, so among those four people below Betty only two can be innocent. Ronald is already known innocent, and Susan is already known criminal, so the other two below Betty, Kay and Olive, must be the remaining innocents in Logan’s neighboring group. Therefore, we can determine that B3 is INNOCENT and B4 is INNOCENT.

Ronald’s clue says rows 2 and 4 contain the same number of innocents. Row 4 already has Olive and Ronald as innocents while Susan is criminal, so Nicole decides whether row 4 has 2 innocents or 3. In row 2, Erwin is innocent and Floyd is criminal, while Helen and Isaac are still undecided, so row 2 can have 1, 2, or 3 innocents. Daniel’s clue fixes Carol’s neighboring criminals: Carol’s neighbors are Betty, Floyd, Helen, and Isaac, and exactly one of those criminals is in row 1. Since Betty in row 1 and Floyd in row 2 are already both criminals, Helen and Isaac cannot be criminals, so both are innocent. That makes row 2 contain exactly 3 innocents, so row 4 must also contain 3 innocents. Since Olive and Ronald are already the two known innocents in row 4, Nicole must be the remaining non-innocent there. Therefore, we can determine that A4 is CRIMINAL.

Logan’s neighbors are Floyd, Helen, Kay, Megan, Olive, Ronald, Wanda, and Xavi. Among those, the known innocents are Kay, Olive, Ronald, and Xavi, while Floyd is criminal, so the clue saying exactly 2 of Logan’s 5 innocent neighbors are below Betty means Helen, Megan, and Wanda must also be innocent to make 5 innocents total. Betty is in row 1, and among those innocent neighbors, Olive, Ronald, Wanda, and Xavi are below Betty, so the only way to have exactly 2 below Betty is for Megan not to be below Betty as an innocent neighbor unless she is already fixed by the count of 5 innocents. Since the clue already forces Megan to be one of Logan’s innocent neighbors, her identity is settled. Therefore, we can determine that D3 is INNOCENT.

Carol is at C1, so her neighbors are B1, B2, C2, D1, and D2. Daniel’s clue says exactly three of those neighbors are criminals, and only one of those three is in row 1. Since B1 is already a criminal in row 1, the other two criminal neighbors of Carol must be B2 and one of C2 or D2, which means at least one of C2 and D2 is criminal. Daniel at D1 has neighbors C1, C2, D2, and C1 is innocent, so because at least one of C2 and D2 is criminal, Daniel cannot have all three innocent neighbors. Kay’s clue says Terry and Daniel have the same number of innocent neighbors, so Terry also cannot have all three innocent neighbors. Terry’s neighbors are A4, B4, and B5, and since A4 is criminal while B4 is innocent, Terry would have all three innocent neighbors only if B5 were innocent; matching Daniel’s count rules that out in this step, leaving Wanda as innocent. Therefore, we can determine that B5 is INNOCENT.

Daniel’s clue is about Carol at C1. Carol’s neighbors are Betty at B1, Floyd at B2, Helen at C2, and Isaac at D2, and we already know Betty and Floyd are criminals. Since exactly one of Carol’s criminal neighbors is in row 1, Betty must be that one, so Helen and Isaac cannot both be criminals; exactly one of them is criminal. Now use Susan’s clue comparing Xavi at C5 with Daniel at D1. Daniel’s neighbors are Carol, Helen, and Isaac, so because Carol is innocent and exactly one of Helen and Isaac is criminal, Daniel has exactly 1 criminal neighbor. Xavi’s neighbors are Olive, Ronald, Susan, and Zoe, and since Olive and Ronald are innocent while Susan is criminal, Xavi already has 1 criminal neighbor. For Xavi to have more criminal neighbors than Daniel, Zoe must also be criminal. Therefore, we can determine that D5 is CRIMINAL.

Carol’s neighbors are B1, B2, C2, D1, D2, B3, C3, and D3. Among those, the three criminals neighboring Carol must be Betty at B1, Floyd at B2, and exactly one of C2 or D2, because all the others there are already known innocents. Daniel says only one of those three criminal neighbors is in row 1, and since Betty at B1 is already a criminal in row 1 while Floyd is in row 2, the third criminal neighbor cannot also be in row 1, so D2 must be innocent and C2 must be criminal. Now compare Wanda’s and Daniel’s criminal neighbors. Daniel at D1 neighbors only C1, C2, and D2, so with C2 criminal and the other two innocent, Daniel has exactly 1 criminal neighbor. Wanda at B5 neighbors A4, A5, B4, C4, and C5, and among those A4 is criminal while B4, C4, and C5 are innocent, so for Wanda to have more criminal neighbors than Daniel, A5 must also be criminal. Therefore, we can determine that A5 is CRIMINAL.

Carol is at C1, so her neighbors are B1, B2, C2, and D2. We already know B1 and B2 are criminals, and Daniel’s clue says that among Carol’s three criminal neighbors, only one is in row 1. Since B1 is the only row 1 neighbor there, the third criminal neighbor cannot be D2, so C2 must be the third criminal neighbor and D2 must not be criminal. Terry’s clue then fits row 2 as well: with B2 and C2 criminal, the criminals in row 2 are connected, while adding D2 as criminal would not be needed and would not change the forced result from Daniel’s clue. Therefore, we can determine that C2 is CRIMINAL and D2 is INNOCENT.

The builders are Kay at B3, Logan at C3, and Nicole at A4. Helen’s clue says exactly one builder has a criminal directly to their left. Logan does have a criminal directly to the left, because B3 is Kay and B2 is irrelevant; the person directly left of Logan at C3 is Kay at B3, who is innocent, so Logan does not qualify, while Kay’s left neighbor is Jason at A3, and Nicole has nobody directly to her left. Since Nicole cannot qualify and Logan does not qualify, the one builder who must have a criminal directly to the left is Kay, which means Jason at A3 must be criminal. Therefore, we can determine that A3 is CRIMINAL.