Stratego Setup Double Logic Puzzle Walkthrough

Reviewing the Basics:

The board is 10 columns by 4 rows. Squares in different quizzes can be adjacent or share diagonals, etc. “Adjacent” includes diagonally; “next to” does not. All clues refer to the entire board (both quizzes) unless the phrasing “this side of the field” is used.

Here are the answers:

22222222 (8)
33333 (5)
4444 (4)
5555 (4)
6666 (4)
777 (3)
88 (2)
9
10
(S)py
(F)lag
BBBBBB (6 bombs)

Back to Part 1 (mostly blue)
Back to Part 2 (mostly red)

(Part 1: 2/20 – Part 2: 1/20)
B4 tells you that A4 or B3 or C4 is an 8. V4 tells you that no 8 can be in Row 4. So B3 is 8.

(Part 1: 2/20 – Part 2: 2/20)
No matter which side of the field is which, you can pinpoint the locations of the numbers mentioned in Y3. Four and six horizontal spaces from Y3 will either land you at C3 and E3 or E3 and C3; so you know that one of them is 3 and the other is 6. You also know from V4 that neither A4 nor C4 is a bomb, so B3 tells you that A3 is a (B)omb.

(Part 1: 2/20 – Part 2: 3/20)
You know from Y3 that E3 is a 3 or 6, so A3 tells you that E1 and E2 are both 2’s

(Part 1: 2/20 – Part 2: 5/20)
Five horizontal spaces from E2, regardless of which side is which, is Z2. So according to E2, Z2 is a 4.

(Part 1: 3/20 – Part 2: 5/20)
Again, you know from Y3 that C3 is a 3 or 6. Z2 tells you that both 8’s are next to a 7, so B2 is a 7.

(Part 1: 3/20 – Part 2: 6/20)
Look at B2’s clue. E1 tells you that all 4’s are on Part 1, since there are two 4’s there already. The only way for any piece to be next to three 4’s is if it is next to the two already known and one more. So “the piece that is next to three 4’s” is either  Y2 or Z3. B2 says it could be in Row 1 or 2 and be a bomb, and you know it’s not Row 1. So if “the piece that is next to three 4’s” is Y2, then it is bomb. Or it could be adjacent to four bombs, and not necessarily itself be a bomb. So check out Z3. Obviously Z3 can only be adjacent to four bombs if that side is the left side. Then Z3 could be adjacent to the bomb at A3, a bomb at A2, a bomb at Y2, and a bomb at Y4. Z4 wouldn’t work because it would have to be a 4. A4 wouldn’t work because of what V4 says. So the bottom line is whether “the piece that is next to three 4’s” is Y2 and is a bomb, or is z3 and might be a bomb but is definitely adjacent to four bombs – either way, Y2 is a (B)omb.

(Part 1: 4/20 – Part 2: 6/20)
If Z3 is “the piece that is next to three 4’s,” then according to B2, it is adjacent to four bombs, which would be, according to the step above, Y2, Y4, A2, and A3. It would also mean that Z4 is a 4. But according to Y2, bombs and 4’s cannot both show up more than once per column. therefore Z3 is not “the piece that is next to three 4’s,” so Y2 is, meaning that X2 or Y1 is a 4. E1 tells you that X2 is not a 4, so Y1 is a 4.

(Part 1: 5/20 – Part 2: 6/20)
According to B4, both B1 and W2 are either 2 or 6. According to E1, W2 cannot be a 6, so B1 and W2 are both 2.

(Part 1: 6/20 – Part 2: 7/20)
E1 tells you that each row has one 6. Since the 4’s are on the blue Part 1, you know the 6’s are all on the red Part 2. You also know from Y1 that neither C1 nor D1 is an even-numbered piece. This means the only place for the Row 1 6 to go is A1.

(Part 1: 6/20 – Part 2: 8/20)
A1 doesn’t tell you a whole lot, it might seem. Since it shares a column with A4, if E4 and A4 are the same, neither can be a 6. And actually, none of V2, W2, or X2 can be a 6 either, because of E1’s clue. So E4 cannot be a 6. But since Column B is already filled and has no 6, and since no two 6’s can share a column but each row on the red Part 2 must have one, and since there are four 6’s, the implication is that every row has one 6 AND every column (besides B) has one 6. With E4 ruled out, E3 has to be a 6.

(Part 1: 6/20 – Part 2: 9/20)
Based on Y3, you now know that the red Part 2 goes on the left and the blue Part 1 goes on the right. And therefore C3 is a 3.

Now that you know which side is which, you can determine that B1 must be referring to E4. That is the only cell that shares a diagonal with both B1 and W2. So E4 is a 3.

(Part 1: 6/20 – Part 2: 11/20)
Now you have to consider the number of 3’s. Since Only 4’s and 2’s can be repeated in a column, and since there are five 3’s and ten columns, five of the rows have exactly one 3 and five of the rows have no 3. Right now you only have two 3’s, but you can determine which columns do not have a 3.
Column A does not have a 3 because B1 says it is not adjacent to a 3 and B4 is only adjacent to one 3, which is at C3.
Column B is filled and does not have a 3.
Column D could have a 3.
W2 tells you that W4 is not a 3 or adjacent to one. And since W2 is adjacent to only one 3, that means that among Columns V, W, and X, there is only one 3.
Column Y could have a 3 in Y4.
Column Z cannot have a 3, according to E4.
So, no 3 in Columns A, B, or Z, and two more of V,W,X. This means you have identified five columns without a 3, so the other five must have a 3. Column D must have a 3 but it could go anywhere. Column Y must also have a 3, and it must be Y4.

(Part 1: 7/20 – Part 2: 11/20)
Y4 tells you that neither C4 nor D4 is an odd-numbered piece. This means neither can be a 7, but V4 tells you that Row 4 does have a 7. So it could be A4, W4, or X4. Not Z4 according to E4. So whether A4 or W4 or X4 is a 7, it is not next to a 6, since all 6’s are on the red Part 2. But Z2 says that two of the three 7’s are next to a 6. Now that you know the one in Row 4 is not, B2 must be next to a 6, and according to E1, it must be C2.

(Part 1: 7/20 – Part 2: 12/20)
According to E1, the final 6 must be D4.

(Part 1: 7/20 – Part 2: 13/20)
C4 is not a 7 because of Y4. D4 tells you that no other red cell can be a 7 except possibly A4. But A4 cannot be next to a 6. So you need to place a 7 somewhere that can be adjacent to a 6 to satisfy Z2, since the 7 in Row 4 cannot be next to a 6. The only place to put that 7 is V3.

(Part 1: 8/20 – Part 2: 13/20)
Y1 tells you that Z1 is not an even-numbered piece. E4 tells you that Z1 is not a 3, 7, or 9. V3 tells you that Z1 is not flag or spy. And C3 tells you that Z1 is not a 5. So Z1 is a (B)omb.

(Part 1: 9/20 – Part 2: 13/20)
A4 cannot be a 3 because of B4’s clue. So at this point you know, according to A1, that either V2 or X2 is a 3. Now z1 tells you that V2 cannot be a 3. So X2 is a 3.

(Part 1: 10/20 – Part 2: 13/20)
V4 told you that Row 4 contains two bombs that are not adjacent. This means that W4 or X4 or Z4 is a bomb. B3 told you that every bomb is adjacent to at least one other bomb. Y2 told you that only two pieces can appear more than once in the same column. You know that they are 2 and 4. So since there are six total bombs, the remaining two must be either W4 and X3 or X4 and W3. Now look at X2’s clue. If W3 is a 2, then the remaining two bombs would be W4 and X3. X4 cannot be a 7 according to Z1. Z4 cannot be a 7 according to E4. C4 cannot be a 7 according to Y4. So if W3 is a 2, then the 7 in Row 4 (V4 says there is one) would have to be A4. But X2 says “either…or” It can’t be both. So since W3 cannot be a 2, A4 is a 7.

(Part 1: 10/20 – Part 2: 14/20)
The one-of-a-kind pieces are 9, 10, spy, and flag. Since 9, spy, and flag are all adjacent to at least one of the other, but the 10 is not, you know that one of the former must be adjacent to both the other two. This means that somewhere on the board you need to have three open, unaccounted for spaces, at least one of them adjacent to both the other two. That doesn’t work in the W3-X4 square because you know that two of those are bombs. Even if the bombs were W4 and X3, V2-W3-X4 doesn’t work because the other 8 must be adjacent to the 7 in V3 (Z2+D4). V2 + two cells in the top row doesn’t work either, because of V3’s clue. The only place for those three to go is on the red Part 2, using D2 and D3. The third one could be C1 or C4, but not D1, because now you know that D1 is a 3, since that column had to have one.

Since the 9 must be C1 or D2 or D3 or C4, you know what V1 and X1 are. Being in the top row, they cannot be flag or spy, according to V3. They also cannot be even-numbered, according to Y1. The bombs are all accounted for, the remaing two being in W and X. 3’s and 7’s are all known. So all that’s left for both those cells is 5.

(Part 1: 12/20 – Part 2: 15/20)
At this point W1 could be 2 or 10. But according to C3, it must be a 2.

(Part 1: 13/20 – Part 2: 15/20)
If V2 is not an 8, then W3 would be an 8, since it must be adjacent to a 7 that is not already adjacent to an 8 (Z2+D4). If W3 is an 8, then W4 and X3 are bombs. Z4 cannot be a 4, since B2 indicates only one piece is next to three 4’s. That would make X4 the final 4, since they all must be in different rows but on one side of the field (E1). So now you know that whether V2 is an 8 or W3 is, X4 is a 4.

(Part 1: 14/20 – Part 2: 15/20)
Now you know where the last two bombs go. According to V4, Row 4 has two bombs. B3 told you that each one is adjacent to at least one other bomb, but not in the same column because of Y2. So the remaining two bombs are W4 and X3.

(Part 1: 16/20 – Part 2: 15/20)
Since V3 is a 7, neither V2 nor W3 can be a 5 (Z1). A2 cannot be a 5, according to C3. C4 cannot be a 5 according to Y3. D2 and D3 cannot be 5 because they must be 9 or flag or spy (A4). Z3 and Z4 cannot both be 5, because in order for one of them to be a 5, the other would have to be a 2 (C3). So one of them is a 5 and the other remaining 5 must be C1.

(Part 1: 16/20 – Part 2: 16/20)
One of Z3/Z4 could have been a 5, if the other were a 2 (C3). Now you know Z4 cannot be a 5, because that would require Z3 being a 2, but if Z3 is a 2, then Z4 cannot be a 5. Since Z4 cannot be a 5, or anything else odd (E4), or 8 or 10 (V4), or spy or flag (A4), everything else is all filled in except 2. Z4 is a 2.

(Part 1: 17/20 – Part 2: 16/20)
The only place for the remaining 5 to go is Z3.

(Part 1: 18/20 – Part 2: 16/20)
W3 cannot be a 2 because of X2. So V2 is adjacent to four 2’s. Therefore, according to Z3, V2 is the other 8.

(Part 1: 19/20 – Part 2: 16/20)
All that’s left is a 2, the 9, the spy, the flag, and the 10. Again, W3 cannot be a 2 because of X2. You established earlier, based on A4, that the 9, spy, and flag are in Columns C and D (now you know which three cells). So W3 must be the 10.

(Part 1: 20/20 – Part 2: 16/20)
D2, D3, and C4 being some combination of 9, spy, and flag, A2 must be the final 2.

(Part 1: 20/20 – Part 2: 17/20)
C4 is not “next to” to D3, so according to A2, the spy and 9 are D2 and D3 or vice versa. This means the flag is C4.

(Part 1: 20/20 – Part 2: 18/20)
The flag tells you where the 9 and spy go.

Back to Part 1 (mostly blue)
Back to Part 2 (mostly red)



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