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Jan 2, 2014 8x8 difficult (modulo operator)
http://www.calcudoku.org/forum/viewtopic.php?f=16&t=528
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Author:  baldert  [ Thu Jan 02, 2014 7:34 pm ]
Post subject:  Jan 2, 2014 8x8 difficult (modulo operator)

Hi,
I can't seem to solve this puzzel and I hope you can help me! Usually i am able to solve an 8x8 within 15 minuten orr so but i'm struggling with this one for hours. Starting from the block in the middle working towards the right I've found all of the 1s and 8s. I can't find the next move other than trying (which feels like guessing)
Any tips?
Cheers

Edit: solved! Had to look for 4s

Author:  clm  [ Thu Jan 02, 2014 11:52 pm ]
Post subject:  Re: Jan 2, 2014 8x8 difficult (modulo operator)

baldert wrote:
Hi,
I can't seem to solve this puzzel and I hope you can help me! Usually i am able to solve an 8x8 within 15 minuten orr so but i'm struggling with this one for hours. Starting from the block in the middle working towards the right I've found all of the 1s and 8s. I can't find the next move other than trying (which feels like guessing)
Any tips?
Cheers

Edit: solved! Had to look for 4s


Hi, baldert (it's not yet 24:00 CET, one of this days Patrick is going to punish me, but since you are a new puzzler just commencing in 2014, I hope he will be kind this time). Once you have placed all 1s and 8s and completed row 8 with the shown candidates (the addition rule drives you to f8-g8 = [2,4] >>> d8 = 6, b8-c8 = [3,7]) you may alternately look for 7s, for instance, in column "a": if a6 = 7 >>> a5-b5 = [2,3] and that would make impossible the cage "2:" so a7 = 7, b7 = 4.

Now, in column "b", the 7 cann't go to b2 or b3 (it would require a 1 in the cage) so b8 = 7 (c8 = 3). Next: b2 = 6 (only place for a 6 in this column) and if you sum all you have till the moment in columns "a" and "b" you find: a1 + b3 = 5 (that is = 72 - 8 - 8 - 6 - 6 - 7 - 13 - 11 - 1 - 7) = [2,3] and necessarily a1 = 3, b3 = 2. From this point you quickly arrive to the solution (b5 = 3, h4 = 4, h5 = 2, "9+" = [3,6], h2 = 7, h1 = 5, g1 = 4, ...).

Image

Author:  baldert  [ Fri Jan 03, 2014 5:58 pm ]
Post subject:  Re: Jan 2, 2014 8x8 difficult (modulo operator)

Hi CLM,
thanks for your post. The picture you show corresponds to what I had at the moment I posted my question except for the e1=7 and [a4,b4]=[2,5].

I didn't see that a6=7 was not possible so a7=7. I worked around this and solved it eventually, but it would have solved me a lot of head aches [biggrin] Thanks for the explanation!

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