# White & Black Bishop Logic Puzzle Walkthrough

A couple things to realize before you start:

1.) The black bishop can only be on certain squares and the white bishop can only be on certain squares because of the way they move: any amount of spaces diagonally. The little icons above and below the cell numbers actually indicate whether that cell was occupied by the white or black bishop.

2.) The answers are all something like “w01” or “b11.” Since you have to cover 28 additional squares with two bishops, the answer choices are W01-W14 and B01-B14.

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(2/30)
E1 tells you that E2 is W01.

(3/30)
W01 is in Row 1. W02 cannot be in Row 2 since it has to move diagonally from E2. W03 cannot be in Row 3 because A6 says W03 is in Column A and A3 is a black square. W04 could be in Row 4 if it is at C4, since W03 is in Column A. W05 cannot be in Row 5 because A6 says W05 is in Column E and E5 is a black square. So E2 is clearly referring either to W04 being in Row 4 or W06 being in Row 6. If W04 is in Row 4, it is in C4, since A6 says W03 is in Column A. So W03 would have to be A2. W05 would then have to be E6. But where would that leave W02? With W01 at E2 and W03 at A2, W02 would have to be at C4, but in this scenario W04 is at C4. So this scenario does not work, meaning W04 is not the time when the white bishop’s row number and turn number were the same. So it was W06 after all, meaning W06 must be in Row 6. If W06 were in E6, W05 could not be in Column E, but W05 is in Column E, according to A6. So W06 must be C6.

(4/30)
According to A6, W05 must be E4.

(5/30)
C6 tells you that A1 is B03.

(6/30)
Given the black bishop’s start B01 must be B5, C4, or D3. Of course whichever of those is B01 must share a diagonal with B02, which must share a diagonal with B03, which is at A1. E4 tells you that B01 is not C3. So of the three remaining possibilities for B01, none of them allows for B02 to be anywhere but C3. That is, since B01 cannot be C3, the only way for B02 to share a diagonal with both B01 and B03 is for Bo2 to be C3.

(7/30)
C3 tells you that A2 cannot be W03. Therefore, according to A6, W03 is in A4.

(8/30)
The only unused square that shares a diagonal with both A4 and E4 is C2. Since A4 is W03 and E4 is W05, C2 must be W04.

(9/30)
Since B03 is in the corner, B04 must be B2, D4, or E5. It cannot be B2, according to C3. It cannot be D4, according to A1. So B04 is E5.

W02 must share a diagonal with both A4 and E2, so it is either B5 or D1. Now you know, according to C2, that it is not D1. So B5 is W02.

(11/30)
W07 cannot go anywhere except D5.

D1 only shares a diagonal with one unused square: B3. This means that if anything comes after D1, it must be B3. Looking at the used squares that share a diagonal with D1, you see A4, C2, and E2. But none of these can come directly before D1 because W02, W04, and W05 are already placed. So since those used squares cannot come directly before D1, you know that the square that comes before D1 has to be B3. And since B3 was also the only square that could come after D1, you know that D1 has to be last. So D1 is W14 and B3 is W13.

(14/30)
D4 cannot be B05, according to C3, since D4 is adjacent to W05. If D6 were B05, then B06 would have to be A3, B4, or C5. According to B3, neither A3 nor B4 can be B06. And according to C3, C5 cannot be B06 either. So D6 cannot be B05. The only square available for B05, then, is B2.

B3 tells you that B01 must be at A5.

(16/30)
As just established, A3 cannot be B06. This leaves C1 and D4. If C1 were B06, then D2 would have to be B14, according to C3. B13 could not be B4 according to C3, so B13 would go in E3. Then, in order for A3 and B4 to be reached on two-digit numbered turns (B3), C5 would have to be B12. But then you see that B06 would not share a diagonal with B07, among other problems. So C1 cannot be B06, meaning D4 must be B06.

(17/30)
If E3 was reached on a two-digit numbered turn, it was reached on turn 13 (B13).
14 would not work because E3 is not adjacent to D1 (C3).
12 would not work because C1 and D2 would be B13 and B14, so A3 and B4 could not both be reached on two-digit numbered turns (B3).
11 and 10 would not work for a number of pretty obvious reasons.
So if E3 were B13, C1 and D1 would be B12 and B14, A3 and B4 would be B11 and B10, D6 would be B09, C5 would be B08, and B6 would be B07.
But if C5 is B08, then C4 could not be W08 because of C3. C4 being W08 is the only way D3 could be W09; otherwise it would be W10+, meaning that D3 would most definitely be reached on a two-digit-numbered turn if E3 was. So, looking at D4 again, the conclusion is that E3 was not reached on a two-digit numbered turn, making it B07, B08, or B09. But B09 at E3 does not work because it would require B13 at either A3 or B4, neither of which works, according to C3. But whether E3 is B07 or B08, B09 has to be C5 or B6 will be blocked off and unused.

(18/30)
Look at the “either…or” statement split between E5 and C5. E5’s part could be true as long as E3 is B08, not B07. C5’s part would require A2 to be W12, since no other square on the top half is a white square that shares a diagonal with W13 at B3. But if A2 were W12, you’d either end up with blocked off unused squares or you’d end up with W09 in C4, which does not work, according to C3. So it is E5 that is true, not C5, and so E3, having been established above as either B07 or B08, must be B08.

(19/30)
B07 must share a diagonal with both D4 and E3; therefore it goes in B6.

(20/30)
There are two ways to finish the white bishop’s route with W09 being adjacent to W08. Either C4 is W08 and D3 is W09 or A2 is W08 and B2 is W09. Either of these scenarios inevitably leads to W12 in E6.

(21/30)
E6 gives you enough to finish the black bishop’s route. B10 must now be at A3.

(22/30)
If B11 were at C1, D2 would have to be B12, but one of those two must be B14, according to C3. So B11, according to B6 not at B4 either, is at D6.

(23/30)
B4 must be B12, D2 must be B13, and C1 must be B14.

(26/30)
C1 tells you that C4 is not W08. So W08 must be at A2.

(27/30)
According to B6, W09 must be at B1.

(28/30)
D3 must be W10 and C4 must be W11.

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