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Killer Sudoku 06/01/2017 https://www.calcudoku.org/forum/viewtopic.php?f=16&t=855 
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Author:  paulv66 [ Sat Jan 07, 2017 3:57 pm ] 
Post subject:  Killer Sudoku 06/01/2017 
I found yesterday's difficult killer sudoku very hard to get to grips with. I managed to solve it eventually, but I'd say I spent around 3 hours in total working on it. I see that it's only been solved just over 100 times, which is much lower than usual for a killer sudoku. There were two 5s that could be input straight away, but after that I played around for ages before making any further progress. I finally managed to get my hooks into it by sorting out the options for column A and then solving row 9 when I determined that the cage in the bottom right corner had to be 126. It's certainly the most difficult killer sudoku I can remember seeing on this site. Did anyone manage to find a quick key to solving it? 
Author:  eclipsegirl [ Sat Jan 07, 2017 9:37 pm ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
I agree with you that The Jan 6 Difficult Killer Sudoku was tough. I think that is always true of Friday's Difficult Killer Sudoku, but I am not positive. It is like how Tues 9 x 9 is supposed to be the most difficult puzzle of the week. Althought I found this Tues 9 x 9 to be easier than expected 
Author:  sjs34 [ Sat Jan 07, 2017 9:54 pm ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
I had not previously noticed that the Friday killer sudokus were atypically difficult, but I did not complete last week's or this week's, so now I agree with you. 
Author:  pnm [ Sat Jan 07, 2017 10:38 pm ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
I checked and everyone is right: a while ago someone sent me email about "krazydad's" "insane" Killer Sudoku puzzles, claiming they were more difficult than the "most difficult Killer Sudoku" on this page: http://www.calcudoku.org/hardest_logic_ ... es/#killer I explained that I could always "dial up" the difficulty level, and proceeded to do so (first by analyzing the level of those "insane" puzzles, and then adjusting the generator so it'd create even harder puzzles). I forgot to remove the adjustments, so as a result, Friday's difficult Killer Sudoku was much harder than usual I'll turn it down a bit for future puzzles.. 
Author:  clm [ Sat Jan 07, 2017 11:08 pm ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
paulv66 wrote: I found yesterday's difficult killer sudoku very hard to get to grips with. I managed to solve it eventually, but I'd say I spent around 3 hours in total working on it. I see that it's only been solved just over 100 times, which is much lower than usual for a killer sudoku. There were two 5s that could be input straight away, but after that I played around for ages before making any further progress. I finally managed to get my hooks into it by sorting out the options for column A and then solving row 9 when I determined that the cage in the bottom right corner had to be 126. It's certainly the most difficult killer sudoku I can remember seeing on this site. Did anyone manage to find a quick key to solving it? Hi, paulv66, yes, also in my opinion the puzzle is quite difficult (of course there are KS’s much more harder). I have not enough time to provide the graphics (it would be useful if Patrick could simply add the proposed puzzle to this thread for a future use) but I’ll try to explain the way I have followed. First I placed (as you did) the 5’s in a9 and e6 (this last as an “outie” of the sum ( 180) of the 4 leftmost columns). Cage “17” in b7 is [8,9] so the 8 of column 1 must be in cage “13” (a1a2a3) considering that an 8 in a6 is not possible due to a7a8 = [2,4] what would make impossible the cage “10” in a9. To determine that c1 = 9 we depart from the fact that the sum of c1, b3 and c3 is 20, and having an 8 in cage “13” consequently the combinations contain a 9 so: [4,7,9] or [5,6,9]. A 9 is not possible in c3 because the rest of cage “20” would be 11 = [1,2,3,5] regardless of the sequence. A 4 (>>> c1 = 7) could not be in b3, and in case of 5 and 6, the 5 would go to b3 and the 6 to c1 >>> d1 = 9 and this situation is impossible because then the cage “22” in c4 could not be formed due to two simultaneous 9’s in c3 and d1. Then c1 = 9 >>> d1 = 6. The first consequence of this is that the 9 of the referred cage “22” is in d4 or d5; now looking to the nonet number 8 (that is, the 3x3 box in the middle of the three bottom rows), since the 9 could not be in f8 because in that case there is no way of placing a 9 in row 9 (g9 is unable due to repetition of 9’s in cage ”30”), then the cage “15” must contain the 9, that is: “15” = [6,9]. And now cage “9” in i8 is not [1,3,5] or [1,2,6] because in both cases the cage “10” in a9 could not be formed so cage “9” in i8 = [2,3,4] and then “10” in a9 is [1,4,5], but not [2,3,5], that is, i8 = 4 and h9i9 = [2,3]. Now we look at cage “22” in c4. The case [5,8,9] must be eliminated because an 8 in d4 or d5 would require the 8 of the nonet number 8 to go to f8 with similar consequences as in the case of the 9 previously studied, that is, an 8 could not be placed in row 9. So, cage “22” in c4 is [6,7,9] with c4 = 6 and d4d5 = [7,9]. Three consequences: 1) Cage “13” in a4 = [4,9] (>>> cage “13” in a1 = [2,3,8]); 2) cage “14” in column a is [1,6,7] with a6 = 1 and a7a8 = [6,7]; 3) a 7 could not go to f8 using the previous logic (a 7 could not be placed in row 9) so the Lshape cage “10” is [1,2,7]. From this point, clearly, g9 = 7 and d9 = 8. Now looking to row 7, d7 = 4 and obviously since d8 cann’t be a 5 >>> f8 = 5 and d8 = 3 >>> c8 = 2 >>> c7 = 3. Since we have left a total of 10 in the 5cell cage “22” necessarily c6 = 8 and d6 = 2 and d2d3 = [1,5]. Also b4 = 3 and b3c3 = [4,7] (we are left with 11 in cage “20”, but [5,6] would repeat a 5). Additionally, since “14” in b5 is [2,5,7] >>> b6 = 7, b5 =2, c5 = 5. Now b3 = 4, c3 = 7 (and b9 = 1, c9 =4). And since cage “12” is [1,5,6] >>> c2 = 1 (>>> d2 = 5, d3 = 1), b2 = 6, b1 = 5. I will continue a little bit: g7 + i7 = 11 = [5,6] then g7 = 6 (due to c6 = 8), i7 = 5 >>> i6 = 9. Also a7 = 7 and a8 = 6. And e8 = 7 with e7f7 = [1,2]. Due to the nonrepetition of 6’s in cage “19” h6 = 6 >>> f6g6 = [3,4] >>> g4 + g5 = 6 = [1,5], that is, g4 = 5, g5 = 1. Now we observe that h5 + i5 = 12 so = [4,8], that is: h5 = 4, i5 = 8 >>> a4 = 4, a5 = 9, d4 = 9 and d5 = 7. Also g6 = 3 and f6 = 4. Cage “9” in e5 = [3,6]. Additionally, h4i4 = [2,7], the rest of cage “32” being 23 (BTW with a 6 in i3 observing row 3) then: g1 + g2 = 6 = [2,4]. The 1 of nonet 9 is in h8, … . Last comment for Patrick: From time to time it's sane to have a difficult KS like this . 
Author:  pnm [ Sat Jan 07, 2017 11:12 pm ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
clm wrote: From time to time it's sane to have a difficult KS like this . It's maybe an idea then to make Friday's Killer Sudoku a bit harder than the rest Here is the puzzle for reference: 
Author:  bram [ Sat Jan 07, 2017 11:46 pm ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
pnm wrote: I forgot to remove the adjustments, so as a result, Friday's difficult Killer Sudoku was much harder than usual Well, that explains it, then! Yesterday's difficult killer is indeed one of those infuriating puzzles where there's a certain number of seemingly promising ideas but it seems like none of them can be pieced together or lead anywhere on their own. Today's diffkill, on the other hand, has a really nice flow to it I'll admit to having solved yesterday's puzzle by trial and error (cue chorus of boos ). Assuming the numbers in the "14" cage in the rightmost column to be [5,9], I explored both of the two possible positions of the number 5 in row 8. One led to a contradiction down the line while the other led to a correct solution. But I couldn't find a way to exclude the possibility of [6,8] in the "14" cage Afterwards I started over and (apparently) managed to solve the puzzle "analytically" after a long, tortuous series of inferences that I don't quite recall But now clm has provided a roadmap for all of us 
Author:  bram [ Sun Jan 08, 2017 12:11 am ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
paulv66 wrote: (…) and then solving row 9 when I determined that the cage in the bottom right corner had to be 126. clm wrote: And now cage “9” in i8 is not [1,3,5] or [1,2,6] because in both cases the cage “10” in a9 could not be formed so cage “9” in i8 = [2,3,4] and then “10” in a9 is [1,4,5], but not [2,3,5], that is, i8 = 4 and h9i9 = [2,3]. At any rate, after having looked a long time in vain for a quick key to this puzzle, I guess there's probably no such thing and the rest of us will just have to follow in clm's footsteps 
Author:  paulv66 [ Sun Jan 08, 2017 11:51 am ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
bram wrote: paulv66 wrote: (…) and then solving row 9 when I determined that the cage in the bottom right corner had to be 126. Sorry, I got that backwards. B9/C9 had to be 14 or 23. This meant that the cage in i9 had to be either 126 or had to have 4 in cell i8. As you surmised, I initially thought that i8 could not be 4 (not sure why) and I went off on a tangent with 126. I try to make sure each step is solid when solving a puzzle, but sometimes have to go down blind alleys before reaching the solution. Anyway, it was an interesting challenge and would have been the first puzzle in a while that I had been unable to solve. I did it mostly on a flight and then was able to submit it when I got to my destination and was able to connect to wifi less than an hour before the deadline. 
Author:  paulv66 [ Mon Jan 09, 2017 9:05 pm ] 
Post subject:  Re: Killer Sudoku 06/01/2017 
By the way bram, I notice you have scored 192 points for book puzzles over the past 7 days. As the highest scoring puzzles are worth 12 points each (apart from a couple of one off exceptions) and can only submit 2 book puzzles a day, I would have thought the maximum should be 12 x 2 x 7 = 168. How did you manage to score 192? 
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