Puzzle Pack #2 Puzzle 18 Answer

Mark’s clue says that Janet is one of the 10 innocents on the edges. Janet is at D2, which is an edge position, so this clue directly includes her among the innocents. Therefore, we can determine that D2 Janet is INNOCENT.

Tina is at C4, so the people above her are C1, C2, and C3. Olive is at D3, and among those three people, the ones neighboring Olive are C2 and C3, while C1 is too far away. Janet’s clue says exactly 2 innocents above Tina are neighboring Olive, so those two neighboring people above Tina must both be innocents. Therefore, we can determine that C2 is INNOCENT and C3 is INNOCENT.

Evie is at C1, so the people below her in the same column are Isaac at C2, Nicole at C3, Tina at C4, and Xavi at C5. Isaac’s clue says that the two criminals below Evie are connected, so among those four people there are exactly two criminals, and they must form one continuous vertical group. Since Isaac at C2 and Nicole at C3 are already known innocents, the only possible pair is Tina at C4 and Xavi at C5, and they are adjacent so they are connected. Therefore, we can determine that C4 is CRIMINAL and C5 is CRIMINAL.

Column C contains Evie at C1, Isaac at C2, Nicole at C3, Tina at C4, and Xavi at C5. We already know Isaac and Nicole are innocent, and Tina and Xavi are criminal, so column C already has exactly 2 criminals unless Evie is also criminal. Since the clue says column C is the only column with exactly 2 criminals, Evie cannot be criminal, because that would make column C have 3 criminals instead of exactly 2. Therefore, we can determine that C1 is INNOCENT.

Vince is at A5, so his only two neighbors are Rohan at A4 and Wanda at B5. Evie’s clue says both of those neighbors are innocents, and exactly one of them is below Hazel. Hazel is at B2, so anyone in row 3, 4, or 5 is below her; both A4 and B5 are below Hazel. That means the only way the clue can fit is for just one of Vince’s neighboring innocents to be among those two positions, so A4 cannot be one of Vince’s neighboring innocents. Since the clue says Vince has exactly two neighboring innocents and A4 is one of only two neighbors Vince has, Rohan must be innocent. Therefore, we can determine that A4 is INNOCENT.

Uma at D4 is a neighbor of Tina at C4 and Xavi at C5, and both of those neighbors are criminals. Rohan’s clue says every person must have at least one innocent neighbor. Since Tina and Xavi already have Uma as a shared neighbor, Uma must be innocent so that they each have at least one innocent neighbor. Therefore, we can determine that D4 is INNOCENT.

Uma is at D4, and her neighbors are Nicole, Olive, Tina, Xavi, and Zed. Since Nicole and Uma herself are innocent, Tina and Xavi are criminal, and Olive and Zed are still unknown, Uma already has exactly 1 or 2 innocent neighbors. Vince is at A5, so his only neighbors are Rohan, Scott, and Wanda. Evie’s clue says Vince has exactly 2 innocent neighbors, because it refers to “the 2 innocents neighboring Vince.” Uma’s clue says Vince and Uma have the same number of innocent neighbors, so Uma must also have exactly 2 innocent neighbors. That forces exactly one of Olive and Zed to be innocent. Now use Xavi’s clue: column C has exactly 2 criminals, and we already know they are Tina at C4 and Xavi at C5, so everyone else in column C is innocent, including Evie at C1 and Nicole at C3 as already shown. Flora at D1 is a neighbor of both Evie and Janet, and Janet is innocent; with D2, D4, and C3 already innocent around the right side of the board, the only way to keep Uma’s neighborhood count at exactly 2 innocents is for Zed to be criminal, making Olive the one innocent among Olive and Zed. That fixes the top-right edge consistently with the innocent cluster around Janet and Evie, so Flora must be innocent as well. Therefore, we can determine that D1 is INNOCENT.

Vince is at A5, so the people above him are Austin at A1, Gus at A2, Luigi at A3, and Rohan at A4. Flora says exactly 3 of those 4 people are innocent, and Rohan is already known to be innocent, so among Austin, Gus, and Luigi exactly 2 are innocent and 1 is criminal. That means column A has exactly 1 criminal in A1–A4, and since Vince is also in column A, Xavi’s clue that column C is the only column with exactly 2 criminals means column A cannot end up with 2 criminals. So Vince cannot be criminal, because that would make column A have exactly 2 criminals. Therefore, we can determine that A5 is INNOCENT.

Vince’s innocent neighbors are Rohan at A4 and Wanda at B5, so Evie’s clue says exactly one of those two is below Hazel. Since “below Hazel” means in the same column as Hazel, Hazel must be in column A or column B. Vince’s clue says all innocents in column B are connected, and Mark at B3 is already innocent; if Hazel were also innocent while Scott at B4 were not, then the innocents in column B would be split apart instead of forming one connected group. So Scott has to be innocent to keep the innocents in column B connected. Therefore, we can determine that B4 is INNOCENT.

Vince is at A5, so his two innocent neighbors are Rohan at A4 and Scott at B4. Evie says only one of those two innocents is below Hazel, and “below” means somewhere lower in the same column. Rohan in column A is not in Hazel’s column, so he cannot be below Hazel; that means Scott must be the one below Hazel, which fixes Hazel at B2 or B3. Hazel cannot be at B3 because Mark at B3 is already innocent, so Hazel must be B2. Wanda at B5 is directly below Hazel in the same column, and Evie’s clue says only one of Vince’s two neighboring innocents is below Hazel. Since Scott at B4 is that one innocent below Hazel, Wanda cannot also be innocent. Therefore, we can determine that B5 is CRIMINAL.

Isaac is at C2, so his neighbors are Barb at B1, Evie at C1, Flora at D1, Hazel at B2, Janet at D2, Mark at B3, Nicole at C3, and Olive at D3. Among those, Evie, Flora, Janet, Mark, and Nicole are already known innocent, so Isaac already has 5 innocent neighbors, and he would have 6 if Barb is innocent. No one else can tie for the most, because the clue says Isaac has the most innocent neighbors uniquely. Scott at B4 already has 6 innocent neighbors around him: Luigi, Mark, Rohan, Vince, and Uma are innocent, and Hazel is also in his neighborhood. So Barb cannot be criminal, because that would let Hazel be innocent and keep Scott from reaching 6, while Isaac could still be at 5; but once we compare the only way to preserve Isaac as uniquely highest, Barb must be one of Isaac’s innocent neighbors so that Isaac reaches 6 and stays ahead. Therefore, we can determine that B1 is INNOCENT.

In column B, the known innocents are B1, B3, and B4, while B5 is criminal. Vince’s clue says all innocents in that column must form one continuous up-and-down group. Since B3 and B4 are already connected, B1 can join that same innocent group only if B2 is also innocent, because B2 is the only space between B1 and B3. Therefore, we can determine that B2 is INNOCENT.

Vince is at A5, so the people above him are Austin at A1, Gus at A2, Luigi at A3, and Rohan at A4. Flora says exactly 3 of those 4 are innocent, and since Rohan is already known to be innocent, exactly one of Austin, Gus, and Luigi must be criminal. That means among Olive’s three neighbors on the left side, Janet at D2 and Uma at D4 are innocent, and only one of Isaac at C2, Nicole at C3, and Tina at C4 is criminal because Isaac and Nicole are already innocent while Tina is criminal. So Olive has exactly 4 innocent neighbors if Olive is innocent, but only 3 innocent neighbors if Olive is criminal. Scott says Isaac has the most innocent neighbors. Isaac currently has 4 innocent neighbors around him, so nobody else can also have 4. Therefore Olive cannot be innocent-neighbored by 4 people, which forces Olive herself to be innocent rather than criminal. Therefore, we can determine that D3 is INNOCENT.

Uma is at D4 and Vince is at A5. Uma’s neighbors are Nicole, Olive, Tina, Xavi, and Zed; among those, Nicole, Olive, and Uma are not counted, so the known innocent neighbors are Nicole and Olive at D4’s upper-left and upper positions and Uma’s other neighbors include Tina and Xavi, who are criminals, leaving Zed as the only unknown. That means Uma has exactly 2 innocent neighbors if Zed is criminal, or 3 if Zed is innocent. Vince at A5 has only two neighbors, Rohan at A4 and Wanda at B5 and Scott at B4 diagonally, so his innocent neighbors are Rohan and Scott, while Wanda is criminal, giving Vince exactly 2 innocent neighbors. Since Uma says she and Vince have an equal number of innocent neighbors, Uma must also have 2 innocent neighbors, so Zed cannot be innocent. Therefore, we can determine that D5 is CRIMINAL.

Austin at A1 has only three neighbors: Barb at B1, Gus at A2, and Hazel at B2. Barb and Hazel are already known to be innocent, so Austin has either 2 or 3 innocent neighbors depending on Gus. Xavi at C5 has five neighbors: Tina at C4, Uma at D4, Scott at B4, Wanda at B5, and Zed at D5. Among them, only Scott and Uma are innocent, so Xavi has exactly 2 innocent neighbors. Zed says Austin has more innocent neighbors than Xavi, so Austin must have more than 2 innocent neighbors. That forces Gus to be innocent, making Austin's total 3. Therefore, we can determine that A2 is INNOCENT.

Rohan at A4 has neighbors A3, B3, A5, and B5. Since B3 and A5 are innocent and B5 is criminal, Rohan’s number of innocent neighbors depends on A3: he has 3 innocent neighbors if Luigi is innocent, but only 2 if Luigi is criminal. Olive at D3 has neighbors C2, D2, C3, C4, D4, C5, and D5. Among those, C2, D2, C3, and D4 are innocent, while C4, C5, and D5 are criminal, so Olive has exactly 4 innocent neighbors. Gus says Rohan and Olive have the same number of innocent neighbors, so Rohan must also have 4 innocent neighbors. That is only possible if Luigi is innocent. Therefore, we can determine that A3 is INNOCENT.

The edge spaces are A1, B1, C1, D1, A2, D2, A3, D3, A4, D4, A5, B5, C5, and D5, for 14 edge people in total. Mark’s clue says Janet is one of exactly 10 innocents on those edges, and on the edges we already know that B1, C1, D1, A2, D2, A3, D3, A4, D4, and A5 are innocent, which already makes 10. That leaves the four remaining edge people, A1, B5, C5, and D5, as not innocent; and since B5, C5, and D5 are already criminal, A1 must be the last criminal edge person. Therefore, we can determine that A1 is CRIMINAL.