Baseball Photo Logic Puzzle Walkthrough

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1.) In order for a row to have 3 or more consecutive pitchers, one of them must be in the center of the row. Therefore, according to A1, both C2 and C4 are P.

2.) C4 tells you that D3 is P.

2.) C4 tells you that neither D4 nor D5 is P. This means, according to A1, that both A4 and B4 are P.

2.) C4 tells you that neither D4 nor D5 is P. C2 tells you that every player is adjacent to at least one P. Therefore the P that is adjacent to E5 must be E4.

3.) At this point you know that row 5 has no P (A1). Therefore according to A4, either A3 or B3 is P. But according to E4, C4 is only adjacent to two P, so B3 cannot be P. Therefore A3 is P.

4.) The corner A3 is referring to is obviously either E1 or E5. But according to E4, C4 is only adjacent to one outfielder. If A3 were referring to E5, then both D4 and D5 would have to be outfielders. Therefore A3 is referring to E1. This means that D1, D2, and E2 must be some combination of P and two outfielders, whichever 2 are not in E1 itself. In other words, D2 and E2 cannot both be P. Therefore, according to A1, B2 must be P.

5.) D3 tells you that Column E has at least three outfielders. Therefore, according to B2, all other columns have either zero or one outfielder. According to step 4 above, D1 and/or D2 (but now you know it’s only one of them) is an outfielder, so D4 cannot be an outfielder. Therefore according to B4, D4 must be 1B.

6.) D4 tells you that at least three rows have at least three P and that there are 11 total P. Right now 9 P are known, and only row 4 has “mostly pitchers.” Row 1 would need two more P in order to have “mostly pitchers.” But that would bring the total up to 11, and then neither Row 2 nor Row 3 could have “mostly pitchers.” So only two rows would have three or more pitchers. Therefore the two remaining P must belong one in Row 2 and one in Row 3. According to E4, the final pitcher in Row 3 must be E3.

7.) According to D3, the rest of Column E are all outfielders. This means the pitcher that E1 is adjacent to (C2) must be in D1 or D2. But according to step 6 above, the final pitcher is in Row 2. Therefore D2 is P and there are no more pitchers.

8.) D4 is adjacent to four known cells, none of which is mentioned by D2. This means that the remaining cells adjacent to D4 are LF, CF, SS, SS. According to A3 and step 4 above, D1, E1, and E2 are all outfielders. So according to B2, D5 cannot be an outfielder. Therefore D5 is S.

9.) According to B2, every shortstop is adjacent to at least one other shortstop. So either C5 or E5 is S. But according to D3, everything left in Column E is an outfielder. Therefore C5 is S.

10.) D2 says “both shortstops,” so there are no more. D4 says there are 11 pitchers, so there are no more. B3 is adjacent to at least six P, so according to D5 it cannot be CA. According to C5, B3 cannot be 2B or 3B. E4 tells you that C4 is adjacent to only one outfielder, which you know to be C3 because of D2. Therefore the only thing B3 can be is 1B.

11.) You know that E2 is an outfielder. You know from D2 that C3 and E5 are LF and CF. You know from A3 that D1, E1, and E2 are some combination of LF, CF, RF. E3 says there are two of each outfielder. B3 says that B1 and E2 are the same, but they cannot be LF or CF or there would be three of them: those two and either C3 or E5. Therefore both B1 and E2 are RF.

12.) You know C3 is CF or LF, so according to E2, the (and apparently there is just one of each) 2B and 3B are in Column 1. D5 told you there are 2 CA. At this point you have all but 2 cells narrowed down: B5 and C1, and of course, both those cells are adjacent to three P. So B5 and C1 are both CA.

13.) According to B5, the two CF are in the same row or column. This means one of them must be E1. In fact, they must be in the same column because D1 and E1 have to be different (A3). So E1 and E5 are both CF.

14.) So C3 and D1 are both LF.

15.) Most of the “if, then” clues don’t help at all. They are all self-evident, except for C3. B3 and D4 are both adjacent to the LF at C3. Therefore A2 is 3B.

16.) So according to E2, A5 is 2B.

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