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Re: Step by step solution of a 7x7 subtractions only https://www.calcudoku.org/forum/viewtopic.php?f=3&t=61 |
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Author: | jomapil [ Wed Sep 28, 2011 3:38 pm ] |
Post subject: | Re: Step by step solution of a 7x7 subtractions only |
Thanks for your good work, clm. It's a little complicated but enough elucidatory. I have learned so much with you. I'm waiting for your next help. Thank you very much, again. |
Author: | clm [ Thu Sep 29, 2011 10:13 am ] |
Post subject: | Re: Step by step solution of a 7x7 subtractions only |
jomapil wrote: Thanks for your good work, clm. It's a little complicated but enough elucidatory. I have learned so much with you. I'm waiting for your next help. Thank you very much, again. You are very welcome, thank you for appreciating it. It is a little complicated because it is difficult to show the variety of steps (many ways or alternatives...) that a solver is following with the pencil and the eraser. I will see if I can go to higher puzzles in the future, like 8x8's ... The main idea, though, is transmitting confidence in the sense that every puzzle can be completed by everyone (regardless of its difficulty), and that there is a lot of information inside the puzzle itself to arrive to the unique solution. You are in danger of becoming soon (as me) a "calcuholic" (this game is very addictive for people who love numbers, and there must be a few old members in a detoxification centre). |
Author: | jomapil [ Fri Sep 30, 2011 10:09 am ] |
Post subject: | Re: Step by step solution of a 7x7 subtractions only |
I have just two weeks I joined to calcudoku and I'm already a calkuholic. I only had some experience with the simpler kenken. But calcudoku has much more interest and much more variety. From the 12x12 and 10x10 puzzles to the exponentiation or bitwise OR... But the Modulo Puzzle is the more difficult on account of the many possibilities of each cage that become it very compact, confusing and heavy. It's the only special puzzle I never succeed to solve. |
Author: | clm [ Mon Oct 03, 2011 9:05 am ] |
Post subject: | Re: Step by step solution of a 7x7 subtractions only |
jomapil wrote: ... But the Modulo Puzzle is the more difficult on account of the many possibilities of each cage that become it very compact, confusing and heavy. It's the only special puzzle I never succeed to solve. Here is the full table: 7mod: 78 (1 combination) 6mod: 67 68 (2) 5mod: 56 57 58 (3) 4mod: 45 46 47 48 (4) 3mod: 34 35 36 37 38 and additionally 85 74 (7) 2mod: 23 24 25 26 27 28 and 86 83 75 64 53 (11) 1mod: 12 13 14 15 16 17 18 and 87 76 73 72 65 54 52 43 32 (16) 0mod: 12 13 14 15 16 17 18 and 84 82 63 62 42 (12). Total of possible combinations: 56. The main tips are: 1) A number x can never be inside a cage ymod if x < y, for instance, 2 can never go inside a cage 3mod, 4mod, 5mod, 6mod or 7mod. 2) In a cage 0mod or 1mod "any" number (from 1 to 8) can initially be present. 3) In cages 4mod, 5mod, 6mod and 7mod, the 4, 5, 6 and 7, respectively, must be present; the cages 1mod, 2mod and 3mod are possible without the respective 1, 2 or 3. The cage 0mod has pairs that are multiples or a 1 is inside, what in fact is a particular case of the multiples. There are a few combinations (if you forget any when solving, the full puzzle may collapse) but less than in the "bitwise OR". I am sure next thursday you will solve it. |
Author: | pharosian [ Tue Oct 04, 2011 3:40 pm ] |
Post subject: | Re: Step by step solution of a 7x7 subtractions only |
Thank you very much for posting those tips, clm! The term "mod" was taught to me in school as a "remainder" function when any number is divided by a smaller one. So I was really in the dark for a long time about how a valid entry in a "4mod" cage could be 5,4 -- I was convinced that had to be a 1mod. The stuff I read online was just confusing. This really makes it much clearer. |
Author: | clm [ Tue Oct 04, 2011 10:39 pm ] |
Post subject: | Re: Step by step solution of a 7x7 subtractions only |
pharosian wrote: Thank you very much for posting those tips, clm! The term "mod" was taught to me in school as a "remainder" function when any number is divided by a smaller one. So I was really in the dark for a long time about how a valid entry in a "4mod" cage could be 5,4 -- I was convinced that had to be a 1mod. The stuff I read online was just confusing. This really makes it much clearer. Very welcome, pharosian. Yes, it is always the "remainder", and this is what produces the confussion because in the "calcudoku" "any" of the two numbers in the cage can be the dividend and the other the divisor (changing the role, following the same philosophy of the "-" cages, etc., ...) so 4 / 7 gives a quotient of 0 and a remainder of 4 then 47 is a valid combination for 4mod, but the "natural" division (7 / 4) would produce the 3mod, in fact, obviously, 47 is a valid combination (among others) for 3mod. In the table I have shown first, for each "xmod", the pairs of numbers in the sequence that produce a quotient 0, followed by the natural divisions. In the case of the parity (your post under "Trial and error"): In today's 10x10 (Puzzle id: 368234), you will probably see quickly that cell b2 is even (2 upper rows = 110; we have 78 + 2 cages "2-" with an even total, so 110 - even = even); a 2 in b2 would produce a 4 in b3 and there is already a 4 inside "7+" in d3-e3; a 4 is not possible since there is already a 4 inside "5+" in e2-f2; a 6 is not possible because (for the "2-") the roommates, 4 or 8, are already present in row 3, so ... b2 = 8 (and b3 = 6 since b1 = 10) (I hope Patrick does not object for this little help). So, using the "parity rule", you have determined the value of a cell, in this case b2. |
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