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9 x 9 Tuesday May 3rd https://www.calcudoku.org/forum/viewtopic.php?f=16&t=785 
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Author:  jaek [ Thu May 05, 2016 6:40 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
pnm wrote: Interestingly, this puzzle is maybe harder than the "world's hardest Calcudoku" (see http://www.calcudoku.org/hardest_logic_ ... #calcudoku and http://www.calcudoku.org/en/20130402/9/3) The one from April 2, 2013 was solved on time by 48 people (out of about 840 "monthly active users"). The one from this past Tuesday by 50 people (out of about 1050 monthly active users). I'll wait for clm's verdict Was there still just one day to turn in the solution on April 2, 2013? 
Author:  sjs34 [ Thu May 05, 2016 8:02 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
The puzzles as currently posted show 70 solutions on April 26 and 49 on April 5 but they they were both subscriber only puzzles. April 19 and April 12 each show 224 solutions and they were available to all. This week's was available to all and shows 77 solutions. The program rates this week's at 130. The previous 4 weeks were 120, 111.8, 102.3 and 125.2. Interestingly, Patrick's numbers say there were 50 solutions this week, the puzzle this morning said there were 74, and now the puzzle says there were 77. 
Author:  pnm [ Thu May 05, 2016 8:34 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
jaek wrote: Was there still just one day to turn in the solution on April 2, 2013? Ah, thanks for reminding me. An additional 20 people submitted the solution for this 9x9 yesterday (so on May 4th, 2016). A request came in for extending the deadline for Tuesday's 9x9 on August 21, 2014. I implemented the extension on June 4th, 2015 (amazingly responsive, I know ) So 70 / 1050 = 0.067 and for the one from 2013, 48 / 840 = 0.057 ... 
Author:  clm [ Thu May 05, 2016 11:35 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
jaek wrote: pnm wrote: Interestingly, this puzzle is maybe harder than the "world's hardest Calcudoku" (see http://www.calcudoku.org/hardest_logic_ ... #calcudoku and http://www.calcudoku.org/en/20130402/9/3) The one from April 2, 2013 was solved on time by 48 people (out of about 840 "monthly active users"). The one from this past Tuesday by 50 people (out of about 1050 monthly active users). I'll wait for clm's verdict Was there still just one day to turn in the solution on April 2, 2013? OK, according to the picklepep’s comment mentioned in the post viewtopic.php?f=3&t=436 , the rating of that puzzle (Apr 02, 2013) was also 130, exactly as this one (May 03, 2016). In my opinion both are of similar difficulty. This pattern in the 9x9’s seems to be the most difficult and it rarely appears in the site (btw, always on tuesday's). In this occasion I have not enough time to develop the full solution but I have some comments: I think that the probability of solving this puzzle only by TAE is so low that it would require a tremendous amount of time, so I thing that once more the analysis must be applied (parity rule, maximums and minimums, etc.). For instance: Considering that in an Lshape cage, like the 38+, a number cannot appear more than twice, we have five combinations: 99884, 99875, 99866, 99776 and 98876. We immediately eliminate the double 8 since we already have two 8’s in the three bottom rows, so we are left with: 99875, 99866 and 99776. With two 9’s in the cage, a 9 must be in b9 (in one of the branches of the L), since c3 = 9. The double 9, along with a third 9 in “7”, forces the cage “10+” to be [3,7] (considering that the cage “14+” in d2d3 must be [6,8]. Consequently, since we have a 7 in cage “168x”, two more 7’s are not possible in the three bottom rows and then 99776 must be eliminated. The combination 99866 must be eliminated too since b9 is occupied by a 9 and c7 = 6 so two 6’s cannot go in a7a8a9. Then “38+” = [5,7,8,9,9]. Combinations in other cages can be reduced accordingly, for instance, in cage “168x”: 7324, 7381, 7622 and 7641, the 7381 is suppressed due to 8’s (now we have three 8's in the three bottom rows) and the 7622 would produce three 2’s in cells d9e8f7 and, since “108x” has five combinations: 33621, 36611, 39221, 39411 (this one immediately suppressed due to 4’s in c5 and e7) and 69211, only the 36611 (without 2’s) would appear to be valid but now we would have three 6’s in the three central columns what enters in conflict with our initial hypothesis of 7622, so “168x” is left with 7324 or 7641, both of even parity (this fact will be used later). … 
Author:  firefly [ Fri May 06, 2016 9:35 am ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
Here's a Full Analytical Solution (FAS) for this puzzle, for those who are interested: The empty puzzle: 1) Prefilled boxes. 2) The 2x in 6h must be [1,2]. 3) The 8: in 8b must be [1,8]. 4) The 7 in 7f must be [2,9] due to the [1,8] in 8b. 5) The 14+ in 2d must be [6,8] due to the 5 and 9 in row 3. Through step 5: 6) The 38+ in 7a must have two 9s, because it can only have one 8 due to two 8s being spoken for in rows 79. (Largest otherwise would be [9,8,7,7,6], which is only 37.) 7) The 10+ in d7 must be [3,7] because of the [6,8] in 2d and the fact that the 9s are now spoken for in rows 78. 8) The 38+ in 7a must contain one 8 due to the fact that two of the three 7s in rows 79 are now spoken for. 9) The 38+ in 7a has a remaining balance of 12 after the two 9s and 8 have been established. The 9s and 8s are spoken for, and this can't be [6,6] due to the 6 in 7c (the symmetrical [9,6,8,6,9] would force the 8 to 9a and the 6 to 9c). Thus it is [9,9,8,7,5]. Through step 9: 10) Multiplying out the 38+ in 6a, the 168x in 8e, and the 180x in 7i, we get (9x9x8x7x5)x168x180 = 685843200. Factoring out the bottom row, we get 685843200/9! = 1890. We know that three of the remaining cells must be [5,7,9] because there are 5s in both the 38+ and the 180x, and 7s in both 38+ and 168x, and a second 9 in the 38+. Factoring these out gives us 1890/(5x7x9)=6. So those 5 cells must contain either [9,7,5,6,1] or [9,7,5,3,2]. 11) The 1 in 8g must contain a 4, because it's the only place remining for a 4 in row 8. 12) The 5 leftover cells from step 10 must now be [9,7,5,6,1] because otherwise there's no room for a 6 in row 8. 13) 7i must therefore be a 1, because the 1 in row 8 is already spoken for. 14) This in turn sets the numbers for the 2x in 6h, the 7 in 7f, and the second 9 in the 38+ in 7a. Through step 14: 15) The remaining cells in the 180x in 7i must be [6,5,3,2] because the 1s in columns gi are all spoken for. 16) The 168x in 8e must therefore be [7,6,4,1]. 17) 7b must be either 3 or 5, depending on the outcome of the [3,4,5] in the 1 in 8g. Through step 17: 18) Taking the sum of the 19+ in 1g, the 28+ in 4i, and the 180x in 7i, we get 19+28+17=64. Taking out the whole of column i, we get 6445=19. 19) The value of 5h must be 5 or higher due to the fact that the largest values for 46i is 24 and that the 4 in row 5 is spoken for in 5c. 20) The cells in 9gh cannot be [3,6] because that would force 1gh to be [2,3] in order to satisfy the sum of 19 from step 18, keeping in mind the minimum value of 5h from step 19 and the fact that the 1s are spoken for in columns gh. This would also mean that the 3s were spoken for in columns gh, which would force 8i to be 6, which would contradict the 6 already used in 9gh at the start of this step. 21) The cells in 9gh cannot be [2,6] because that would block the 2s in columns gh, making it impossible to satisfy the sum of 19 from step 18. (The minimum values are 2+6+5+3+4=20.) 22) The cells in 9gh cannot be [2,5] because it would be impossible to satisfy the sum of 19 from step 18 without having duplicate 5s in both 5h and 9h. 23) The cells in 9gh cannot be [2,3] because this would make 89i [5,6], which would make the largest value for 46i 4+8+9=21, leading to a minimum value of 8 in 5h due to the 7 being blocked by 5g. This makes the sum of 19 from step 18 impossible. (The minimum values are 2+3+8+3+4=20.) 24) Thus, the cells in 9gh must be [3,5], which means that 9i is 2 and 8i is 6, which in turns leads to more resolution of previous sets. Through step 24: 25) With the 6 in column i spoken for, the minimum value of 5h is now 6 (28[9,8,5]=6), and the minimum value of 1gh is still [2,3]. Both are needed to fulfil the sum of 19 from step 18, now that we know that 9gh are [3,5]. (3+5+6+2+3=19) 26) The 12x in 6fg must be [3,4] due to the 6 in 6b. 27) The 2 in 3h must be [7,9], which leads to many easy resolutions in sets at 2g, 2d, 4f, 6c, 7a, and 8b. Through step 27: 28) The 5 in 3b must be [2,7]. 29) The only place for a 5 in row 5 is 5b. 30) This leads to a 5 in 1c, necessary for the 240x. 31) This last 5 goes in 2f. 32) The only place for a 7 in column f is 1f. 33) This leads to a 7 in 2i. Through step 33: 34) The rest pretty much solves itself. Final solution: Steps 1824 were clearly the most difficult part to analyze of the whole puzzle. There were other ways to go at that point, but the ones I saw required too much TAE for my liking. Perhaps clm came up with a more elegant way of getting through that middle section. He mentioned parity in the previous post, and while I toyed with parity during my solve, I ultimately never really used it. So perhaps he was able to see something I missed. Edit: corrected a glitch in the order of steps 2930. 
Author:  pnm [ Fri May 06, 2016 10:07 am ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
firefly wrote: Here's a Full Analytical Solution (FAS) for this puzzle, for those who are interested: Thanks Very nice to see such an analysis from someone other than clm I saved the images of your post just in case "tinypic.com" disappears some time in the future.. 
Author:  firefly [ Fri May 06, 2016 10:38 am ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
pnm wrote: Thanks Very nice to see such an analysis from someone other than clm I saved the images of your post just in case "tinypic.com" disappears some time in the future.. You're very welcome. And thanks for providing us with such an enjoyable and fulfilling website! :) And to be honest, clm usually just beats me to it. I play somewhat sporadically, so by the time I see a post about solving strategy, clm has usually made any further input somewhat unnecessary. :P But you're right that this one was one of the hardest ever, which is largely what made me want to post an FAS guide to it. One of my first memories of this site is reading clm's treatment of the 4/2/13 puzzle. A few years ago I think I googled "hardest logic puzzles" and stumbled across your "10 hardest logic/number puzzles" article, which in turn led me to the calcudoku puzzle, clm's solution, and this site. It's been fun ever since! :) 
Author:  pnm [ Fri May 06, 2016 12:12 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
firefly wrote: But you're right that this one was one of the hardest ever, which is largely what made me want to post an FAS guide to it. One of my first memories of this site is reading clm's treatment of the 4/2/13 puzzle. Would an even harder puzzle be fun? Or maybe a book with about 50 of these?? 
Author:  sjs34 [ Fri May 06, 2016 1:48 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
Fiirefly, thank you for your very excellent, very clear, very helpful analysis. I hope you will do more in the future. Patrick, I think that a book of super hard puzzles would be a welcome extension for calcudoku.org although many of us would probably not solve many of them. If you did such a book, I would urge that you give it a ranking list in the style of the timed ranking lists. There are some puzzlers whose excellence I don't appreciate from existing ranking lists (e.g. firefly) because they are not completists. 
Author:  pnm [ Fri May 06, 2016 2:03 pm ] 
Post subject:  Re: 9 x 9 Tuesday May 3rd 
sjs34 wrote: Patrick, I think that a book of super hard puzzles would be a welcome extension for calcudoku.org although many of us would probably not solve many of them. If you did such a book, I would urge that you give it a ranking list in the style of the timed ranking lists. There are some puzzlers whose excellence I don't appreciate from existing ranking lists (e.g. firefly) because they are not completists. Interesting idea, thanks: number of Tuesday 9x9's solved ranking? 
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