Puzzle Pack #1 Puzzle 26 Answer

Cheryl is at B1, and her neighbors are A1, A2, B2, C1, and C2. Olive’s clue says Gabe is one of two or more innocents neighboring Cheryl, so Gabe must be an innocent neighbor of Cheryl, and B2 is indeed next to B1. Therefore, we can determine that B2 is INNOCENT.

Rose is at D4, and her neighbors are C3 Logan, D3 Megan, C4 Paul, C5 Xavi, and D5 Zoe. Gabe’s clue says Logan is one of Rose’s 3 innocent neighbors, which directly tells us that Logan is innocent. Therefore, we can determine that C3 is INNOCENT.

Logan’s clue says that Rose is one of the 7 innocents on the edges. Since everyone tells the truth, that directly tells us Rose is innocent. Therefore, we can determine that D4 is INNOCENT.

Rose is at D4, so her neighbors are C3 Logan, C4 Paul, C5 Xavi, D3 Megan, and D5 Zoe. Gabe says Logan is one of Rose's 3 innocent neighbors, so Rose has exactly 3 innocent neighbors in total; since Logan is already one of them and Rose herself is innocent, Paul's clue is the key to which others qualify. Paul is at C4, and his neighbors are B3 Katie, C3 Logan, D3 Megan, B4 Olive, D4 Rose, B5 Wally, C5 Xavi, and D5 Zoe. Rose says exactly 2 of Paul's 5 innocent neighbors also neighbor her. Among Paul's neighbors, the ones who also neighbor Rose are only Logan, Megan, Xavi, and Zoe, because Olive is not a neighbor of Rose. Logan is innocent, and Olive and Rose are also known innocents, so if Paul were not innocent then Paul's 5 innocent neighbors would already have to include Logan, Olive, and Rose plus exactly one of Megan, Xavi, or Zoe, which does not fit Rose having exactly 3 innocent neighbors including Logan. The count works only if Paul himself is one of the innocents around Rose and part of Paul's set of 5 innocent neighbors. Therefore, we can determine that C4 is INNOCENT.

The only guards are Isaac at D2 and Rose at D4. Paul’s clue says exactly one guard has a criminal directly below them, so among the two people below those guards, exactly one of D3 and D5 is criminal. Gabe’s clue says Logan at C3 is one of Rose’s three innocent neighbors, and Rose’s neighbors are C3, C4, C5, D3, and D5; since C4 and Rose herself are already innocent and C3 is named innocent in the clue, Rose’s only three innocent neighbors are C3, C4, and one of D3 or D5. That means both C5 and whichever of D3 or D5 is not innocent must be criminal. From Paul’s clue, exactly one of D3 and D5 is criminal, so C5 must be criminal in every case. Therefore, we can determine that C5 is CRIMINAL.

Rose is at D4, and her neighbors are C3 Logan, D3 Megan, C4 Paul, C5 Xavi, and D5 Zoe. Gabe says Logan is one of Rose's 3 innocent neighbors, and we already know Paul is innocent while Xavi is criminal, so Rose's other two neighbors, Megan and Zoe, must split as one innocent and one criminal. Xavi says Rose and Barb have the same number of criminal neighbors. Since Rose has five neighbors total and exactly three of them are innocent, Rose has exactly two criminal neighbors. Barb at A1 has only three neighbors: B1 Cheryl, A2 Floyd, and B2 Gabe, and Gabe is innocent. So for Barb to also have exactly two criminal neighbors, both Cheryl and Floyd must be criminal. Therefore, we can determine that A2 is CRIMINAL and B1 is CRIMINAL.

Rose is at D4, and her neighbors are C3 Logan, D3 Megan, C4 Paul, C5 Xavi, and D5 Zoe. We already know Logan and Paul are innocent, while Xavi is criminal, so for Rose to have exactly 3 innocent neighbors, Megan and Zoe cannot both be innocent. That means at least one of D3 or D5 is criminal. Floyd says there is an odd number of criminals below Emily at D1, so among D2, D3, D4, and D5 there must be an odd number of criminals. Rose at D4 is innocent, and since at least one of D3 or D5 is criminal, Isaac at D2 cannot also be criminal, or the total would become even. Therefore, we can determine that D2 is INNOCENT.

Isaac says there are exactly 4 innocents in column D. In that column, D2 is Isaac and D4 is Rose, and both are already known to be innocent. Rose’s neighbors are C3 Logan, D3 Megan, C4 Paul, C5 Xavi, D5 Zoe, and Logan is confirmed innocent by Gabe’s clue, while Paul is also already known innocent. That gives us the four innocents for column D as D2, D4, and the clue forces D1 Emily to be the remaining innocent in that column. Therefore, we can determine that D1 is INNOCENT.

Nancy at A4 is next to Floyd at A2, Julie at A3, Katie at B3, Olive at B4, and Terry at A5 and Wally at B5. Cheryl’s clue says exactly two of Nancy’s neighbors are criminals, and exactly one of those two is in column B. Since Floyd at A2 is already a known criminal and is in column A, Nancy must have exactly one criminal neighbor in column B, and no other neighbors in column A can be criminals. That means Julie at A3 and Terry at A5 are innocent, so Nancy is touching only one known criminal. Emily’s clue says Olive has exactly three criminal neighbors, and exactly one of them is in column A; around Olive, Floyd at A2 is already that one column A criminal, and Cheryl at B1 and Xavi at C5 are already two more criminal neighbors. So Olive’s three criminal neighbors are fixed as Floyd, Cheryl, and Xavi, which confirms Katie at B3 and Wally at B5 are innocent. All of Nancy’s neighbors are now innocent except Floyd, so Nancy has exactly one criminal neighbor. Therefore, we can determine that A4 is INNOCENT.

Nancy is at A4, so her neighboring criminals must be among A3, B3, and B4. Since B4 is innocent, Cheryl’s clue that Nancy has exactly 2 criminal neighbors tells us A3 and B3 are the two criminals, and among them only one is in column B, which fits B3 and leaves A3 in column A. Now look at Nancy’s clue that column A has more criminals than any other column. Column B already has at least the criminals B1 and B3, so for column A to have more criminals than column B, column A must have at least three criminals. We already have A2 and A3 as criminals, so A1 must also be criminal. Therefore, we can determine that A1 is CRIMINAL.

Barb’s clue is about all innocents above Xavi, which means the innocents in column C above C5: C4, C3, C2, and C1. We already know C4 and C3 are innocent, and they are directly next to each other, so for all innocents in that column to form one connected group, any other innocent there must connect to that pair through C2. Since C1 is not adjacent to C3 or C4, C1 could only join that connected innocent chain if C2 were also innocent. Therefore, we can determine that C2, Hal, is INNOCENT.

Emily’s neighbors are C1, C2, C2, C1, C2, C3, and D2 around D1, which means her actual neighbors are C1 Daniel, C2 Hal, C3 Logan, and D2 Isaac. Of those, Hal, Logan, and Isaac are already known innocents, so Emily has 3 innocent neighbors if Daniel is criminal, or 4 if Daniel is innocent. Wally’s neighbors are A4 Nancy, B4 Olive, C4 Paul, A5 Terry, C5 Xavi, and D5 Zoe. Among them, Nancy, Olive, and Paul are innocent and Xavi is criminal, so Wally’s number of innocent neighbors is 3 plus Zoe if Zoe is innocent, plus Terry if Terry is innocent. Hal says Wally and Emily have the same number of innocent neighbors. Since Emily can only have 3 or 4, Wally cannot have 5, so Terry and Zoe cannot both be innocent. And Wally also cannot have only 3, because then Emily would also need 3, which would make Daniel criminal, but that would leave no way for Wally to match Emily unless both Terry and Zoe were criminal, forcing Wally’s count below what the clue allows. The only matching count is 4, so Emily must have 4 innocent neighbors and Wally must also have 4. That means Daniel must be innocent, and Wally reaches 4 innocent neighbors only if Terry is criminal and Zoe is innocent. Therefore, we can determine that C1 is INNOCENT and A5 is CRIMINAL.

Nancy is at A4, so her neighbors are A3, B3, B4, A5, and B5. We already know B4 is innocent and A5 is criminal, and Cheryl’s clue says Nancy has exactly 2 criminal neighbors, with only 1 of those 2 in column B. That means Nancy must have one more criminal neighbor besides Terry, and it has to be in column B, so exactly one of B3 or B5 is criminal. The only other possible neighbor in column A is A3. If A3 were also criminal, then Nancy would already have two criminals in column A, which would leave no criminal in column B and break the clue. Therefore, we can determine that A3 is INNOCENT.

Rose is at D4, an edge position. Logan’s clue says Rose is one of the 7 innocents on the edges, so there are exactly 7 innocent edge people in total. Looking around the edge, the innocents already known are C1, D1, D2, A3, A4, D4, and C4. That is already 7 edge innocents, and every other edge spot besides those is occupied by someone already known to be criminal except B5 and D5. Since the total number of innocent edge people is exactly 7, neither B5 nor D5 can also be innocent. Therefore, we can determine that B5 is CRIMINAL.

Paul is at C4, and his five innocent neighbors are B3 Katie, C3 Logan, D3 Megan, B4 Olive, and D4 Rose. Rose is at D4, so among those five, the ones who also neighbor Rose are C3 Logan, D3 Megan, and C4 Paul’s other side does not matter here? Wait, only Paul’s innocent neighbors count, so the shared neighbors are C3 Logan, D3 Megan, and B4 Olive. Since the clue says exactly two of Paul’s five innocent neighbors also neighbor Rose, and Logan and Olive are already known innocents, Megan cannot be innocent. That leaves Katie as the remaining one among Paul’s five innocent neighbors, so Katie must be innocent. Therefore, we can determine that B3 is INNOCENT.

In row 3, Logan is at C3, so his neighbors in that row are Katie at B3 and Megan at D3. Katie’s clue says that exactly one innocent in row 3 is neighboring Logan. Katie herself is already known to be innocent, so she is that one innocent neighbor. That leaves Megan as Logan’s other row-3 neighbor, and Megan cannot also be innocent. Therefore, we can determine that D3 is CRIMINAL.

Rose is at D4, so her neighbors are C3 Logan, D3 Megan, C4 Paul, C5 Xavi, and D5 Zoe. Gabe says Logan is one of Rose's 3 innocent neighbors, so Rose must have exactly 3 innocent neighbors in total. Logan and Paul are already innocent, while Megan and Xavi are criminal, so the only way for Rose to have 3 innocent neighbors is for Zoe to be innocent as the third one. Therefore, we can determine that D5 is INNOCENT.