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6x6 Difficult September 6  Help http://www.calcudoku.org/forum/viewtopic.php?f=16&t=478 
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Author:  pluto [ Sat Sep 07, 2013 8:00 pm ] 
Post subject:  6x6 Difficult September 6  Help 
I' m trying to solve this analytically butcannot get beyond determining that the 24x cage is odd. If anyone could get me started before the time runs out I'd be most appreciative. Thank you. 
Author:  pluto [ Sat Sep 07, 2013 8:16 pm ] 
Post subject:  Re: 6x6 Difficult September 6  Help 
Oops! Because I'm in Canada I may have got the date wrong  it could be the September 7 6X^ Difficult. Sorry about that. 
Author:  bram [ Sat Sep 07, 2013 8:28 pm ] 
Post subject:  Re: 6x6 Difficult September 6  Help 
I take it you mean today's difficult 6x6 because you want to solve it before the deadline. OK, so you have already found that "24x" + "6x" + "3:" = 23 and, given that the latter two are both odd, the former also must be. Now consider the possibility that "3:" contains [1,1,3]. What must "6x" then contain? And is it then still possible to satisfy the condition of "24x" while also satisfying "24x" + "6x" + "3:" = 23? If not, what does that tell you about "3:"? 
Author:  pluto [ Sat Sep 07, 2013 8:44 pm ] 
Post subject:  Re: 6x6 Difficult September 6  Help 
Thanks Bram, but but the sum could be 21 or 25, for example, so I don't see how it must be 23. 
Author:  bram [ Sat Sep 07, 2013 8:55 pm ] 
Post subject:  Re: 6x6 Difficult September 6  Help 
Take a look at the cages in the four middle columns whose sums are known: "120x" = 1 + 4 + 5 + 6 = 16 "180x" = 2 + 3 + 5 + 6 = 16 (and those whose sums are already indicated as requirements) Subtract those sums from the total of the four middle columns: 84  16  16  7  8  14 = 23 
Author:  kristina [ Sat Sep 07, 2013 8:58 pm ] 
Post subject:  Re: 6x6 Difficult September 6  Help 
start with cage 14+. 4 is possible to be there. the other one is in d1 or e1. leave cage 24x for end. thatt cage has maximum combinations. After 14+, move on 6x and 3: (use a pencil) and than go on 8+. You should calculate a lot. If you solve these cages, the others will apear automatically. I hope this helps to start solving problem 
Author:  pluto [ Sat Sep 07, 2013 9:03 pm ] 
Post subject:  Re: 6x6 Difficult September 6  Help 
Bram and Kristina: Thanks very much! 
Author:  bram [ Sat Sep 07, 2013 9:57 pm ] 
Post subject:  Re: 6x6 Difficult September 6  Help 
Pluto, in case you get stuck: The 24x cage is actually quite manageable, with only three possible combinations (when the arrangement of numbers in individual cells is left out of consideration): [1,4,6] (sum 11), [2,2,6] (sum 10, with the even sum immediately excluding this possibility) and [2,3,4] (sum 9). It doesn't require much calculation to continue along the path I proposed in my first answer: Having excluded the possibility [1,1,3] for the 3: cage, it follows that it must contain [1,2,6], the sum of which is 9, which leaves 23  9 = 14 as the total sum of the 6x and 24x cages. It is then easily determined which combinations must go in each of those two cages. Having done that, you look at the placement of the 4s in columns d and e like Kristina proposed. It is clear that none of them can be in the lower cages and that only one of them can be in the 14+ cage (because of the combination [2,3,4] in the 24x cage), which means that just one combination of numbers is possible in the 14+ cage and that the rest of columns d and e (and, after that, the rest of the puzzle) can easily be solved. 
Author:  pluto [ Sun Sep 08, 2013 5:19 am ] 
Post subject:  Re: 6x6 Difficult September 6  Help 
Thank you Bram, in particular, and Kristina also. I had to leave the puzzle because of family commitments, so I missed the deadline. However, I've just returned to it and it fell into place immediately after I followed your advice. Thank you very much! It is gratifying to know that it could be solved analytically, because that's the way I do the puzzles, and was the reason for me turning away from Sudoku, where trialanderror often seemed to be necessary, to Calcudoku. Thanks again. 
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