Calcudoku puzzle forum https://www.calcudoku.org/forum/ 

Re: 9 x 9 Tuesday May 3rd https://www.calcudoku.org/forum/viewtopic.php?f=16&t=788 
Page 1 of 2 
Author:  nicow [ Sat May 07, 2016 1:42 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
Admiration for Firefly! Firefly for president! (edit by pnm: this refers to this post: viewtopic.php?p=6103#p6103) Even harder is the 9×9 on august 18, 2015. 
Author:  pnm [ Sat May 07, 2016 2:00 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
nicow wrote: Even harder is the 9×9 on august 18, 2015. The same pattern, rated a little bit lower than the other two, and solved by 65 people on the first day, 23 on the second. So comparable, but maybe not harder (?) 
Author:  firefly [ Sat May 07, 2016 9:38 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
pnm wrote: nicow wrote: Even harder is the 9×9 on august 18, 2015. The same pattern, rated a little bit lower than the other two, and solved by 65 people on the first day, 23 on the second. So comparable, but maybe not harder (?) Any chance you could whitelist that date/puzzle on the calendar, Patrick? Maybe we can see if someone can draft an FAS solution to that one, too. I'll admit, I'm curious since it doesn't look like I solved it, though I often have severe time constraints to deal with. 
Author:  pnm [ Sat May 07, 2016 9:51 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
firefly wrote: Any chance you could whitelist that date/puzzle on the calendar, Patrick? Maybe this is easier: 
Author:  firefly [ Sat May 07, 2016 10:03 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
pnm wrote: firefly wrote: Any chance you could whitelist that date/puzzle on the calendar, Patrick? Maybe this is easier: Bah, except that I vastly prefer to solve your calcudoku puzzles in situ, as it were, with the keyboard and digital notes. :P Plus, it makes taking screenshots much more clean (for me, at least). Perhaps you could make it a "user" submitted puzzle for the day, sans solution and worth 0 points? Off to work for me now, so I'll check back in later. 
Author:  pnm [ Sat May 07, 2016 10:21 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
firefly wrote: Plus, it makes taking screenshots much more clean (for me, at least). For now you could do Right click > View Image, then Print ? 
Author:  clm [ Sun May 08, 2016 7:54 am ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
pnm wrote: firefly wrote: Any chance you could whitelist that date/puzzle on the calendar, Patrick? Maybe this is easier: ... This 9x9 is of similar level, in my opinion, quite difficult, due to the pattern type, as commented, I have solved it with the use of the parity rule in the first part (easier), the second part until completing all cells is some more difficult than the previous ones, probably due to the fact of having a majority of addition cages, very "compensated/balanced", that is, with similar amounts: 19, 22, 22, 23, 24, 27, 28. I cann't remember if when I solved this puzzle the first time, on August 18, 2015, I followed the same analysis, since I do not keep records, but I feel that probably yes. 
Author:  firefly [ Sun May 08, 2016 1:58 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
Full Analytical Solution (FAS) for the 9x9 on 8/18/2015: The empty puzzle: 1. Prefilled cells. 2. The 9: at 7d must be [9,1]. 3. The 45x at 6b must be [9,5]. 4. The 4: at 8b must be [2,8]. Through step 4: 5. Subtracting the known sums from rows 13 leaves 17 for the 2 at 2d, the 12x at 2f, and the two leftover cells at 3b and 3h. The leftovers must sum as least 3 and the 12x sums either 7 or 8, leaving a max value of 7 for the 2. With the 1 blocked in column d, that means the 2 is [2,4]. 6. The 12x cannot be [2,6] because this would block the 2s in rows 23, while forcing the leftovers at 3b and 3h to be [1,2], which is impossible. So the 12x at 2f must be [3,4] and the leftovers must be [1,3]. 7. Since both leftovers are part of 3 cages, the corresponding 4b and 4h cells must be [4,6]. 8. The 14+ at 4c must therefore be [9,5] due to the blocked 6s in row 4. 9. The 15+ at 2b must be [7,8] due to blocked 9s in columns bc. 10. The 15+ at 2g must therefore be [6,9] Through step 10: 11. Subtracting known sums from columns ac leaves 15 for the 3 at 3b and the leftover cell at 6c. The only values that work are a [3,6] for the 3 and a 6 for 6c. 12. The 3 at 3h is therefore [1,4]. 13. The 3 at 6c is also therefore [6,3] due to the 9 at 7d. 14. The 10+ at 6h must be [2,8] due to blocked 1s and 4s in column h as well as blocked 3s in rows 67. 15. Subtracting known sums from rows 46 leaves 18 for the 6 in 4f and the leftover cell at 6h. The only values that work are a [2,8] for the 6 and an 8 for 6h. Through step 15: 16. Subtracting known sums from columns df leaves 22 for the 1 at 7f and the leftover cells at 3f and 6f. The only values that work are an [8,7] for the 1, a 2 for 4f, and a 5 for 6f. 17. The 7 in row 7 must be in 7a, which also makes 7i a 4. 18. Subtracting known sums from rows 79 leaves 36 for the 3240x at 7i and the 1 at 8g. The only values for the 3240x that include a 4 sum at 27 and 29, which mean that the 1 must be 7 or 9, which means that 8g must be a 4. Through step 18: 19. 3a and 3i must both be [6,9]. 20. With the 9s, 8s, 6s, 5s, and 3s blocked in columns bc, the only choices for 9bc are [7,4,2,1]. 21. 9bc must contain a 7 because no other combination allows for usable values in 89a. ([1,2] leads to sum 18 which is impossible, [1,4] leads to sum 16, which is blocked by 7a, and [2,4] leads to sum 15, which is blocked by 7a and 3a.) 22. 8a cannot be a 6 because it allows for no usable values for 9a. (With two 7s and a 6 in the 28+ cage, 9a and the non7 in 9bc would sum 8: [1,7] is blocked by both 7s, [2,6] is blocked by the 6, and [4,4] is impossible.) 23. The only place for a 6 in row 8 is 8i. 24. The only option for the 3240x cage is now [5,4,9,6,3] which makes 8h 5 to balance of the sum of rows 79. 25. This forces a resolution of the cells in the 3240x at 7i, the 15+ at 2g, as well as 3a and 3i. Through step 25: 26. 8a cannot be a 3 because it allows for no usable values for 9a. (As in step 23, the remainders in the 28+ cage would sum 11: [1,10] is impossible, [2,9] is blocked by the 9 in row 9, and [4,7] is blocked by both 7s.) 27. 8a must therefore be 9 and 8e must be 3. 28. 9ac must be [1,4,7] due to the blocked 3 in row 9. 29. The remainder of [2,6,8] are easily resolved in 9df. Through step 29: 30. 1g must be a 2. 31. The only place for a 2 in column c is 8c, resolving the 4: at 8b, the 15+ at 2b, and 9c. 32. The only place for a 2 in column b is now 5b. 33. 1c must be a 4, which means 1b must be 1, further resolving 9ab. 34. 46a must be [3,4,2] and 12a must be [8,5]. Through step 34: 35. The rest of the puzzle is fairly simple. Final solution: 
Author:  pnm [ Sun May 08, 2016 2:58 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
Nice one :) Was this one easier/harder/the same? 
Author:  clm [ Sun May 08, 2016 3:25 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
firefly wrote: Full Analytical Solution (FAS) for the 9x9 on 8/18/2015: 16. Subtracting known sums from columns df leaves 22 for the 1 at 7f and the leftover cells at 3f and 6f. The only values that work are an [8,7] for the 1, a 2 for 4f, and a 5 for 6f. I followed practically the same way up this point (step 15): maximums and minimums, etc., except that, having determined via the parity rule, in the three bottom rows, that the 3240x cage is odd (so its sum is 27 or 29), g6 must be odd so f6 = 5, g6 = 1, and now the rest of the subtraction to the three central rows is 17 (instead of 22) what permits to more quicly see that f4 = 2 since f4 = 8 forces 45 in 1 at f7 which is impossible ... Very curiously, the green area that I first obtained is exactly the same as yours, with c8 undefined yet to complete a perfect simmetry. The rest is a little more difficult I think. 
Page 1 of 2  All times are UTC + 1 hour [ DST ] 
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group http://www.phpbb.com/ 