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Yesterday's 7x7 difficult (that's Feb 11, 2015) http://www.calcudoku.org/forum/viewtopic.php?f=16&t=663 
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Author:  clm [ Fri Feb 13, 2015 12:41 am ] 
Post subject:  Re: Yesterday's 7x7 difficult (that's Feb 11, 2015) 
tomas wrote: clm wrote: The parity rule is absolutely useful for this type of puzzles (the perfect reasoning of bram to decide that c7 = 4). But, in addition of this, many times it's interesting to see if a determined number can be placed or not in a determined line. For instance, another hint: Suposse that you make c3 =2, c4 = 7 then c5c6 = [1,5]. Since g5g6 = [1,6] and due to the three ones present in the three bottom rows, the 1 of column "b" would only be possible in cage "1" located in b3b4, that is, b3b4 = [1,2]. Why can the 1 of column b not appear in rows 1 or 2? clm wrote: And then, where do we place a 6 in column "b"?, there is not a place for it, I must confess I don't see that either. clm wrote: so cage "5" (c3c4) must necessarily be [1,6] and cage "4" (c5c6) = [3,7]. The 1 of column "b" could not go to either b1 or b2 because c1 or c2, respectiveley, would be a 2 and you already have a 2 in c3 in this hypothesis. The 6 could not be in cage "2" since it would produce a 4 (and there is a 4 in b5) and it can not go to b1 or b2 since c1 or c2, respectively, would be either a 5 or a 7 but both numbers are already present in column "c" in this hypothesis. 
Author:  tomas [ Fri Feb 13, 2015 12:42 am ] 
Post subject:  Re: Yesterday's 7x7 difficult (that's Feb 11, 2015) 
firefly wrote: Bram had the same first two steps that I was going to post. Combined parity for rows 56 lead to a5=[2,6], which combined with the 4 in b5 and the 1 in ef5 will block the evens in row 5, which makes g5=1. Thanks, firefly. I have a lot to learn, it seems. It's like learning a language, practice makes perfect. firefly wrote: The combined parity of columns bc lead to c7=4. The combined parity of rows 34 lead to d3=[2,6]. So that's because even though some cages extend beyond these rows per se, we know for example for a4, that it contains an odd number, right? And similarly for e3... firefly wrote: A few more incidental tricks: ef6 can't have a 7 due to the 6 in g6; de4 can't have a 7 because that would leave both a4 and c4 as 1. This means that the 7's in ac are locked in rows 4, 6, and 7, eliminating the 7s in b2, c2, and 5c, also meaning c6 can't be 3. Ah yes, quite nifty. firefly wrote: Since c6=[1,7] now, and since c34 has to be either [1,6] or [2,7], that blocks the 1's from c1 and c2. This leaves the puzzle looking like: I think honestly I might have brute forced the 4's in row 4 at this point, since it was such an obviously divergent binary option, but let me see if I can find a more elegant (ie logical) solution. Sure, have fun with that. I should get some sleep. More practice tomorrow... 
Author:  tomas [ Fri Feb 13, 2015 12:44 am ] 
Post subject:  Re: Yesterday's 7x7 difficult (that's Feb 11, 2015) 
clm wrote: tomas wrote: clm wrote: The parity rule is absolutely useful for this type of puzzles (the perfect reasoning of bram to decide that c7 = 4). But, in addition of this, many times it's interesting to see if a determined number can be placed or not in a determined line. For instance, another hint: Suposse that you make c3 =2, c4 = 7 then c5c6 = [1,5]. Since g5g6 = [1,6] and due to the three ones present in the three bottom rows, the 1 of column "b" would only be possible in cage "1" located in b3b4, that is, b3b4 = [1,2]. Why can the 1 of column b not appear in rows 1 or 2? clm wrote: And then, where do we place a 6 in column "b"?, there is not a place for it, I must confess I don't see that either. The 1 of column "b" could not go to either b1 or b2 because c1 or c2, respectiveley, would be a 2 and you already have a 2 in c3 in this hypothesis. The 6 could not be in cage "2" since it would produce a 4 (and there is a 4 in b5) and it can not go to b1 or b2 since c1 or c2, respectively, would be either a 5 or a 7 but both numbers are already present in column "c" in this hypothesis. Yes, you're right. I wasn't thinking through all the consequences of the original assumption. 
Author:  firefly [ Fri Feb 13, 2015 12:46 am ] 
Post subject:  Re: Yesterday's 7x7 difficult (that's Feb 11, 2015) 
tomas wrote: Why can the 1 of column b not appear in rows 1 or 2? clm is basically doing a TaE on the puzzle at that point. Assuming c3=2 means that neither c1 nor c2 can be 2, which means that neither b1 nor b2 can be 1. He just didn't spell out that step. tomas wrote: I must confess I don't see that either. Again, leading from c5=5 in his TaE, that eliminates the possibility of c1 or c2 being 5, which eliminates b1 or b2 being 6. 
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