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Guessing for 7x7 difficult 21/04/2014?
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baldert
Posted on: Mon Apr 21, 2014 9:21 pm
Posts: 7 Joined: Wed Jan 01, 2014 12:02 am

Guessing for 7x7 difficult 21/04/2014?
Hi, I'm wondering how often people "guess" a number and see if it works out. I got stuck on the 7x7 difficult puzzle from today (21/04/2014), meaning i could not find the next logical step in the puzzle. How i solved it was: There are two options for the 2401^ cell ([c1, d1, c2]) and eventually i "guessed" one of the two options to be able to move on. It turned out to be the right guess and i solved the puzzle. If i would have got stuck after filling in the next 10 or so steps i would have tried the other option. Is this common practice? I usually try to avoid this way of solving a puzzle but i really couldn't find the next step in a different way..




beaker
Posted on: Tue Apr 22, 2014 2:10 am
Posts: 530 Location: Ladysmith, BC, Canada Joined: Fri May 13, 2011 1:37 am

Re: Guessing for 7x7 difficult 21/04/2014?
I did the correct guess also but am unaware of a definitive way of "knowing" which guess would be the correct one......by the way, the "guessing topic" has come up a number of times in past posts!!




bram
Posted on: Tue Apr 22, 2014 11:26 am
Posts: 193 Joined: Tue May 24, 2011 4:55 pm

Re: Guessing for 7x7 difficult 21/04/2014?
beaker wrote: I did the correct guess also but am unaware of a definitive way of "knowing" which guess would be the correct one 7x7 difficult 21/04/2014: The reason why [1, 4, 7] in [c1, d1, c2] is wrong: It means that in the upper row we have c1 = 1, d1 = 4, e1 = 5. Now [a1, b1] must contain either [2, 3] or [6, 7] with the "unused" pair going to [f1, g1]. But that would mean that either g2 = 9 (which is out of bounds in a 7x7) or g2 = 1 (which is impossible because [2a, 2b] = [1, 5]). Therefore, the correct option must be the other one: c1 = 7, d2 = 2, c2 = 2. Sorry if the notation isn't exactly correct, but you get the idea beaker wrote: by the way, the "guessing topic" has come up a number of times in past posts!! Indeed. I also turn to guessing when I can find no other way, but if I stumble upon the correct solution when trying out the first guess, I then often duly explore the other options (which may branch to "suboptions" ) to "disprove" them by getting stuck somewhere . Sometimes I will notice that some constraints appear in every option, which may set me on the path to an "analytical" solution. Another option is to vent your frustration with a specific puzzle in that section of the forum (sjs34's post describes exactly how I was feeling about that puzzle, too ) and hope that clm puts in a couple of hours to provide an illustrated analytical solution




clm
Posted on: Tue Apr 22, 2014 11:26 pm
Posts: 700 Joined: Fri May 13, 2011 6:51 pm

Re: Guessing for 7x7 difficult 21/04/2014?
bram wrote: ... The reason why [1, 4, 7] in [c1, d1, c2] is wrong: It means that in the upper row we have c1 = 1, d1 = 4, e1 = 5. Now [a1, b1] must contain either [2, 3] or [6, 7] with the "unused" pair going to [f1, g1]. But that would mean that either g2 = 9 (which is out of bounds in a 7x7) or g2 = 1 (which is impossible because [2a, 2b] = [1, 5]). Therefore, the correct option must be the other one: c1 = 7, d2 = 2, c2 = 2. Sorry if the notation isn't exactly correct, but you get the idea ... set me on the path to an "analytical" solution ... and hope that clm puts in a couple of hours to provide an illustrated analytical solution I think you know me well, bram, and you know that I always defend the "analytical" way according to my "theorem": "If a Calcudoku has a unique solution then it can be solved using only analytical means". I basically agree with yor first reasoning, but using the "parity rule" instead to see that if c2 = 7, c1 = 1, d1 = 4, f1 + g1 is odd then g2 would be odd = 3 (unique) but now it would not be possible to get a sum of 11 in f1g1 since [4,7] or [5,6] would be both invalid (first graphic). Once we have c1 = 7, c2 = 2, d1 = 2, a1b1 can only be [3,4], e2f2 = [3,4], f1g1 = [1,6] >>> g2 = 7, d2 = 6 and d3e3 = [1,3]. The 1's in f1g1 and/or f4g4 make "6+" (f3g3) = [2,4] and "84x" must be [2,6,7] being impossible [3,4,7] with the numbers in the shown positions: a4 = 2 >>> cage "2:" (a6a7) must be [3,6], etc. All this is shown in the next graphic. Next we place a 5 in c3. The following step is to determine that the 6 of row 4 must be in c4 since it cann't go to e4 as explained in the next graphic. And, if c5 = 3, "3mod" in f5g5 must be [4,7] (unique). This quickly drives to b5 = 2 and/or, consequently, e5 = 6, d5 = 5. The next graphics are self explanatory; we must remind that a "0mod" cage is always composed of numbers that are one multiple of the other or necessarily a 1 is inside, which in fact is a particular case of the previous, here e6 = 2 and e7 = 7.






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