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Assassin 354 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1441 |
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Author: | Ed [ Sat Jul 14, 2018 7:57 pm ] |
Post subject: | Assassin 354 |
Attachment: a354.JPG [ 65.88 KiB | Viewed 6894 times ] Two key steps for me. If you miss either one you might find this as hard as what SudokuSolver (1.90) and JSudoku do (chains). Hopefully, as usual, you guys find different key steps. code: solution: Cheers Ed |
Author: | wellbeback [ Sun Jul 15, 2018 6:43 pm ] |
Post subject: | Re: Assassin 354 |
I enjoyed this puzzle very much. Congratulations on a gem Ed! Some edits to more fully explain the first key step. Second Edit. Added a simpler equivalent way to do my first key step. (Step 2). Assassin 354 WT: |
Author: | Andrew [ Tue Jul 24, 2018 4:19 am ] |
Post subject: | Re: Assassin 354 |
Thanks Ed for another fun Assassin. Looks like I found a very different way to solve it. Here is my walkthrough for Assassin 354: Rating Comment: |
Author: | Ed [ Thu Jul 26, 2018 6:16 am ] |
Post subject: | Re: Assassin 354 |
Another puzzle, another techniques post! Love how I'm still learning. I've managed to make a future Assassin using the techniques that wellbeback is using so might be handy to know! Glad I'm finally getting the hang of them. Anyway, I thought a slightly different explanation of wellbeback's step 1 and 2 might be useful. This is very close to his alternate explanation but just a little embellished. Attachment: WellStep2a.jpg [ 50.95 KiB | Viewed 6839 times ] 2. r3c6 sees all of n5 apart from r45c4 so must repeat there. 3. in other words, the 34(7) at r3c6 is also completely inside n5, ie, there is an (implied) 34(7) in n5 4. "45" on n5 -> the remaining two innies = 11, this is an implied 11(2) cage 5. that 11(2) must have 1 cell in one of r45c4 and one cell in one of r56c6. We don't know exactly which of those 4 cells has the 11(2), but that's not important. 6. the actual 11(2) cage at r89c6 sees both r45c6 -> the combinations in those two 11(2) cages must be different so they must have different numbers 7. since the 11(2)r89c6 also sees r3c6 (which repeats in r45c4, step 2) -> the two numbers at r45c4 must both be different to the 11(2)r89c6 8. so where are the two numbers in r89c6 going to fit in n5? 9. they can't both go in r456c5 since it is an 11(3) (hidden) cage so will go over the cage total 10. -> one of r89c6 must be in r6c4. Attachment: WellStep2b.jpg [ 69.92 KiB | Viewed 6839 times ] 11. since r6c4 is from (79) -> the 11(2)r89c6 must have 7 or 9 = {29/47} 12. further, since the (79) in 11(2)r89c6 is repeated in r6c4, this must also repeat in n2 in the 17(3) 13. from step 1. and step 12. -> the 17(3)n2 must have both 7&9 = {179} only Now to Andrew. Love step 3! I got the same thing but in a different way, but Andrew's is simpler. In general, I followed the same path. Thanks to Andrew for going through my WT and suggesting some improvements. A354 WT: Ed |
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