1.) C3 gives you A5 and E3.
2.) According to A5, an English word appears using the abbreviations. No combination of RM, RW, LM, and LW creates an English word, but using the A from C3 you can create “alm” and “arm.” According to E3, then, D3 must be RM.
3.) According to E3, D1 is right-handed. Therefore, according to C3, C1 is left-handed. According to D3, C1 is a man, so C1 is LM.
4.) D4 is adjacent to 8 cells. 3 of them are known, and the others, according to C1, are 4 RW and a LW/LM. Since Column D has no lefty (E3), D5 must be one of the RW.
5.) D5 tells you that all remaining corners are women. According to C3 and the fact that A3 is left-handed, you know that A1 and E1 are also left-handed. Therefore they are both LW.
6.) E1 tells you that Column E has only the two known lefties. D5 tells you that E5 is a woman. Therefore E5 is RW.
7.) E4, according to C1 is either left-handed or RW. It is not left-handed according to E1; therefore it is RW.
8.) According to C1, D4 is adjacent to two lefties, one still unknown. It is either C4 or C5. According to E5, Column B has a lefty. According to A1, Column B’s lefty can only be in B3. This means that A3 is an RM, according to E4.
9.) Having established that B3 is left-handed, the second lefty that is adjacent to D4 must be in C5, according to A1. And according to A3, everyone left in Row 5 must be male. Therefore C5 is LM. You also know, according to C1, that C4 is RW.
10.) According to A1, B5 is right-handed. According to A3, B5 is a man. Therefore B5 is RM.
11.) E5 and A1 told you that B3 is a lefty. A1 also tells you that no other box can be a lefty. Therefore, according to B5, B3 is LW
12.) B3 tells you that D4 is RM.
13.) D4’s clue refers to B3. A1 is two diagonal spaces away from the autobot only. The same is true of E1. E3 is two diagonal spaces away from two men, not one. Therefore B3 must be two diagonal spaces away from one man. That man has to be in D1, and according to C3, D1 is right handed. Therefore D1 is RM.
14.) Rows 3 and 5 do not have all four possibilities. Rows 2 and 4 cannot have a lefty, so they also do not have all four possibilities. Therefore Row 1 must have all four possibilities, according to C5. So B1 must be RW.
15.) The only column that might be symmetric is Column C. So, according to B1, C2 is not RW. Nor is it left-handed, according to A1. Therefore C2 is RM.
16.) You already knew that not every row has a lefty. But the only column that can be all female or all male is Column E, all female. Since all lefties are placed, C2 tells you that D2 is RW.
17.) The combination spoken of in E2 could be either 2 RW + 1 RM or 2 RM + 1 RW. This is because all lefties are already known, and because only one corner could be adjacent to 3 RW (A1) or 3 RM (A5). D1 tells you that you can only place up to 3 more men. If the combination E2 were speaking of were 2 RM + 1 RW, you would have to place four more men, too many for what D1 says. Therefore E2 means that three corners are adjacent to 2 RW + 1 RM. According to C4, A5 is the odd corner out. This means that D2 must be RW.
18.) D2 tells you that there are 12 men and 12 women, or 3 more men and 1 more woman. According to C4, A5 A4 and/or B4 is a man. This means that E2’s clue refers to A1, not A5. So either A2 or B2 must be RW and the other RM. Therefore, according to D1 and D2, both A4 and B4 are RM.
19.) B4 tells you that a column has more than three men. It must be A, so A2 is RM.
20.) B2 must be RW according to E2.