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Full Analytical Solution for Difficult KS on Friday 2/10/17
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firefly
Posted on: Tue Feb 14, 2017 6:38 am
Posts: 37 Joined: Mon May 05, 2014 4:59 am

Full Analytical Solution for Difficult KS on Friday 2/10/17
Here's a full analytical solution for the difficult killer sudoku on 2/10/2017: First, we know that g5 = 6 by using known values for columns hi. Then we quickly get h56 = [1,3], i45 = [2,8], and i67 = [5,9]. The 8 cell at h1 has to have a 1, so i1 = 1, which also makes h12 = [2,5] and h34 = [4,9]. g4 = 7 and g23 = [3,8] are the only options for the 18 cage, which also makes g1 = [4,9]. Known cage values in the a1c3 block mean that c1,3 = 6. c1 can't be 1 or 5, so c1,3 = [2,4]. This means that a23 = [1,6] and then a89 = [5,9]. With bracketing 4s in c1,c3,g1,h3, then c,h1 must be either [2,4] or [4,9]. Using known values for the d1f3 block, this means that d,f3 adds up to 5 or 12. The only usable possibilities are d,f3 = [3,2],[3,9],[5,7]. This also means that c4 = [5,9]. Using known values for the a4c6 block, a6,c4,c6 adds up to 15. c4 can't be 9 because c6 is blocked for [2,4] and a6 is blocked for [1,5]. This means that c4 = 5, leading to c3 = 2, d3 = 9. After some more clearing up, we have: There can't be an 8 in the 22 cage at e1, so the 14 cage at d2 = [1,5,8]. This means that f2 = 4 and e1,2 = [2,7], which also makes h1 = 5 and h2 = 2. Using known values for the a4c9 blocks, we find that d6,9 = [1,2]. Using known values for the d4f6 block, we find that e7 must be the same as d6. d6 and e7 can't be 1 because that would block the 1 in the 14 cage at d2, so d6 and e7 = 2, and d9 = 1. After some clearing, we get: From here on out, it's pretty easy. f45 = [1,9], then ef6 = [6,8], then c67 = [6,7], then a67 = [3,7], then ab1 = [8,3], then b78 = [1,4], etc. End result:




paulv66
Posted on: Tue Feb 14, 2017 10:26 am
Posts: 370 Joined: Tue Mar 01, 2016 10:03 pm

Re: Full Analytical Solution for Difficult KS on Friday 2/10
Many thanks for the detailed step by step analysis. I had a look at what I had saved before giving up/ running out of time last Friday and I had done all the steps up to your third diagram apart from the 5,9 in c4 and 3,5,9 in d3. I'm struggling to follow your reasoning in the final paragraph before the third diagram. How do you conclude that c,h1 must be 2,4 or 4,9?




paulv66
Posted on: Tue Feb 14, 2017 11:47 am
Posts: 370 Joined: Tue Mar 01, 2016 10:03 pm

Re: Full Analytical Solution for Difficult KS on Friday 2/10
paulv66 wrote: Many thanks for the detailed step by step analysis. I had a look at what I had saved before giving up/ running out of time last Friday and I had done all the steps up to your third diagram apart from the 5,9 in c4 and 3,5,9 in d3. I'm struggling to follow your reasoning in the final paragraph before the third diagram. How do you conclude that c,h1 must be 2,4 or 4,9? Sorry, I must have been half asleep. I had another look at it and it makes perfect sense. I think my problem on Friday (and the previous Friday) was that I left it too late to start doing the KS and then panicked when I was struggling to solve it. Hopefully I'll do better this week. Thanks again for the analysis. Paul




kozibrada
Posted on: Wed Feb 15, 2017 11:12 pm
Posts: 37 Joined: Sun Feb 03, 2013 3:25 am

Re: Full Analytical Solution for Difficult KS on Friday 2/10
My way to solve this puzzle was different from firefly. I helped myself by concentrating on placement of number 6 into the bottom left nonet. Probably more T&E, though. First I enter another pencil marks: i8–9 = [3,4] and b6 = [7,9] (relating to a sum of the rows 6–9 ( 170) and h6) So 6 in the bottom left nonet is located either in the cage "14" (b9 + c8–9), or "15": 1) "14": it’s remaining 8: a) [1,2,5]: d9 = 5 → a7 + c7 = 17 → impossible; b) [1,3,4]: the cage "5" must be [2,3] → d9 = 3 → i9 = 4 → c8 = 4 → duplicated/impossible2) Thus "15": correct. Then 6 in the middle left nonet is in b4; a4–5 = [2,4]; a1 = 8; c1–d1 = [4,6]; etc…






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