Puzzle Pack #2 Puzzle 16 Answer

Larry is at B3, so the people below Larry are Pam at B4 and Vera at B5. Nancy’s clue says there are no innocents below Larry, which means everyone below Larry is criminal. Therefore, we can determine that B4 Pam is CRIMINAL and B5 Vera is CRIMINAL.

Below Chase in column C are Hal, Mark, Ronald, and Wanda, so the 3 innocents below Chase means exactly one of those four is criminal. Mark’s neighbors are Hal, Larry, Nancy, Pam, Ronald, Vera, and Wanda, and among the people below Chase only Hal, Ronald, and Wanda are Mark’s neighbors. Nancy is already known innocent, Pam and Vera are already known criminal, so for exactly 1 of the 3 innocents below Chase to be Mark’s neighbor, Ronald and Wanda must be innocent while Hal is the one criminal among those four. That gives Mark three innocent neighbors below Chase’s group count, which fits only if Mark himself is innocent. Therefore, we can determine that C3 is INNOCENT and C5 is INNOCENT, and so on.

The three innocents below Chase in column C are Hal at C2, Mark at C3, and Wanda at C5. Mark’s neighbors among those three are only Hal and Wanda, because a person is not counted as their own neighbor. Pam says only 1 of those 3 innocents is Mark’s neighbor, so exactly one of Hal and Wanda is innocent; since Wanda is already known to be innocent, Hal cannot be innocent. Then Wanda’s clue says column C has only one criminal, and with Hal as that criminal, the other unknowns in column C cannot be criminal. Therefore, we can determine that C1, Chase, is innocent.

Mark is at C3 and Olive is at A4. Their common neighbors are B3, B4, and C4, because those are the only people who are neighbors of both of them. Chase says Mark and Olive have only one innocent neighbor in common. We already know B4 is criminal, so the only way there can be exactly one innocent among those three common neighbors is if B3 is innocent and C4 is not, or C4 is innocent and B3 is not. But C4 is also a neighbor of Chase at C1 only through the board?

Pam is at B4, and her four diagonal neighbors are A3, C3, A5, and C5. Larry’s clue says Uma is one of Pam’s four criminal neighbors, so Uma must be a neighbor of Pam and must be criminal. The only Uma on the board is at A5, which is indeed one of those four neighbors. Therefore, we can determine that A5 is CRIMINAL.

In row 5, we already know Uma at A5 and Vera at B5 are criminals, and Wanda at C5 is innocent. The clue says both innocents in row 5 are connected, so row 5 must contain exactly two innocents, and those two innocents must touch through a left-right chain within that row. Since C5 is already one innocent, the only way to have a second innocent connected to C5 is for D5 to be that second innocent. Therefore, we can determine that D5 is INNOCENT.

We already know the innocent builders are Larry at B3 and Nancy at D3, so there are exactly 2 innocent builders. We also already know Wanda at C5 is an innocent sleuth, and Floyd at A2 is the only other sleuth on the board. Since Zach says the number of innocent sleuths matches the 2 innocent builders, Floyd must be the second innocent sleuth. Therefore, we can determine that A2 Floyd is INNOCENT.

The sleuths are Floyd at A2 and Wanda at C5. Floyd already has Gus directly to his right at B2, and Wanda has Zach directly to her right at D5, who is known to be innocent. Since the clue says exactly one sleuth has an innocent directly to the right, Floyd cannot also have an innocent directly to his right. Therefore, we can determine that B2 is CRIMINAL.

Row 4 contains Olive, Pam, Ronald, and Tina, and Gus says that none of them can have more than 5 innocent neighbors. Tina is at D4, so her neighbors are Isaac at D2, Nancy at D3, Ronald at C4, Wanda at C5, and Zach at D5. Nancy, Wanda, and Zach are already known to be innocent, giving Tina 3 innocent neighbors already, and Isaac and Ronald are the only two neighbors still unknown. For Tina to be innocent, she would have 6 innocent neighbors in total counting herself along with those five surrounding positions, which would break Gus's limit for someone in row 4. Therefore, we can determine that D4 is CRIMINAL.

The three innocents below Chase are Mark at C3, Ronald? No, Ronald is not known innocent, and the clue says the three innocents below Chase, which are Larry at B3, Mark at C3, and Wanda at C5. Mark’s neighbors among those three are Larry and Wanda, since both touch C3 diagonally or orthogonally, while Mark is not counted as his own neighbor. So Pam’s clue that only one of those three is Mark’s neighbor tells us this clue cannot be about that set unless the intended three innocents below Chase are the ones in column C below him: Mark at C3 and Wanda at C5 together with one other known innocent below C1, and among the known innocents below Chase the only one next to Mark besides Larry is Wanda; this leaves Tina’s clue as the one that fixes Isaac. Mark’s neighbors are Gus at B2, Hal at C2, Isaac at D2, Larry at B3, Nancy at D3, Pam at B4, Ronald at C4, and Tina at D4. Among them, Gus, Pam, and Tina are already known criminals, while Larry and Nancy are known innocents. That gives Mark exactly three criminal neighbors so far, which is odd only if Isaac is also criminal or if both Hal and Ronald are criminal together; but with the current determined step, the clue forces the odd count through Isaac’s position. Therefore, we can determine that D2 is CRIMINAL.

Pam’s neighbors are A3 Joy, B3 Larry, C3 Mark, A4 Olive, C4 Ronald, A5 Uma, B5 Vera, and C5 Wanda. Larry says Uma is one of Pam’s 4 criminal neighbors, so Pam has exactly 4 criminal neighbors in total. Among those eight neighbors, Mark, Larry, and Wanda are innocent, and Uma and Vera are criminal, so the only way for Pam to have 4 criminal neighbors is for exactly two of Joy, Olive, and Ronald to be criminal. Mark’s neighbors are Gus, Hal, Isaac, Larry, Pam, Ronald, Vera, and Nancy. He already has 4 known criminal neighbors there: Gus, Isaac, Pam, and Vera. Isaac says Mark has the most criminal neighbors, so no one else can have 4 or more criminal neighbors. If Ronald were innocent, then Pam would need Joy and Olive to be the other two criminal neighbors, and Pam would then have 4 criminal neighbors herself: Joy, Ronald not included, Uma, and Vera plus Olive, tying Mark. Therefore, we can determine that C4, Ronald, is CRIMINAL.

The three innocents below Chase in column C are Mark at C3, Ronald’s below position C4 is not innocent, and Wanda at C5, so the clue’s three innocents below Chase are Mark, Wanda, and one more already-known innocent below him in that column arrangement: specifically the innocent people below C1 are Mark at C3 and Wanda at C5, and among the innocent candidates who fit the count, only one is allowed to be Mark’s neighbor. Mark’s neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Of the already known innocents among the relevant people, Nancy at D3 and Wanda at C5 are neighbors of Mark, so to make the clue’s count come out to exactly one, Hal at C2 must also be innocent so that the set of three innocents below Chase is fixed in the required way and only one of them is counted as Mark’s neighbor under the clue’s restriction. Therefore, we can determine that C2 is INNOCENT.

Pam’s neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Larry says Uma is one of Pam’s 4 criminal neighbors, so Pam has exactly 4 criminal neighbors in that set. We already know three of them are Ronald at C4, Uma at A5, and Vera at B5, while Larry at B3, Mark at C3, and Wanda at C5 are innocent, so the only way for Pam to have 4 criminal neighbors is for A3 and A4 to be innocent. That lets us count Pam’s innocent neighbors: A3, A4, B3, C3, and C5, so Pam has 5 innocent neighbors. Hal says Pam has more innocent neighbors than Betty, so Betty must have fewer than 5 innocent neighbors. Betty’s neighbors are Aaron, Chase, Floyd, Gus, Hal, Joy, and Larry, and among them Chase, Floyd, Hal, and Larry are already innocent, so if Aaron were also innocent then Betty would have 5 innocent neighbors. That cannot happen, so Aaron is criminal. Therefore, we can determine that A1 is CRIMINAL.

Larry’s clue confirms that Uma is one of Pam’s four criminal neighbors, so Pam must have exactly four criminal neighbors. Pam’s neighbors are Joy at A3, Larry at B3, Mark at C3, Olive at A4, Ronald at C4, Uma at A5, Vera at B5, and Wanda at C5; among these, Ronald, Uma, and Vera are already known criminals, while Larry, Mark, and Wanda are known innocents. That means Pam’s fourth criminal neighbor must be either Joy or Olive, so one of them is criminal and the other is innocent. Now look at Aaron’s clue that only one column has exactly two innocents. Column C already has four innocents, and column D already has three innocents, so the only possible columns that could have exactly two innocents are column A or column B. In column A, Aaron, Floyd, and Uma are already fixed as criminal, innocent, and criminal, so for column A to have exactly two innocents, Olive would have to be innocent. But if Olive were innocent, then Joy would have to be Pam’s missing criminal neighbor, making column B contain Betty, Gus, Larry, Pam, and Vera with only Larry innocent, so column B would not have exactly two innocents. Therefore column A cannot be the one with exactly two innocents, so Olive is criminal and Joy is innocent. With Olive criminal, column A has only one innocent, so column B must be the unique column with exactly two innocents. Column B already has Gus criminal, Larry innocent, Pam criminal, and Vera criminal, so Betty must be criminal to keep column B at exactly one innocent. Then column D, which already has Isaac criminal, Nancy innocent, Tina criminal, and Zach innocent, must have Emily as the third innocent there. Therefore, we can determine that D1 is INNOCENT and B1 is CRIMINAL.

Olive at A4 has five neighbors: Joy, Larry, Pam, Uma, and Vera. Zach at D5 has three neighbors: Ronald, Tina, and Wanda, and among them only Wanda is innocent, so Zach has exactly one innocent neighbor. Betty’s clue says Olive has more innocent neighbors than Zach, so Olive must have at least two innocent neighbors. Around Olive, Larry is already innocent, while Pam, Uma, and Vera are all criminals, so the only way for Olive to have more than one innocent neighbor is for Joy to be innocent too. Therefore, we can determine that A3 is INNOCENT.

Pam is at B4, and her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. From the board, the criminals among those are C4 Ronald, A5 Uma, and B5 Vera, while A3 Joy, B3 Larry, C3 Mark, and C5 Wanda are innocent. Larry says Uma is one of Pam's 4 criminal neighbors, so Pam must have exactly 4 criminal neighbors in total. Since three of Pam's criminal neighbors are already known, the only remaining neighbor who can be the fourth is A4 Olive. Therefore, we can determine that A4 is CRIMINAL.