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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 c5-c6 = [1,5]. Since g5-g6 = [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 b3-b4, that is, b3-b4 = [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-" (c3-c4) must necessarily be [1,6] and cage "4-" (c5-c6) = [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 5-6 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 b-c lead to c7=4.

The combined parity of rows 3-4 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 a-c 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:

Image

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. [smile] 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 c5-c6 = [1,5]. Since g5-g6 = [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 b3-b4, that is, b3-b4 = [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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