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Searching for the operands https://www.calcudoku.org/forum/viewtopic.php?f=3&t=34 
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Author:  clm [ Mon Jun 13, 2011 8:51 pm ] 
Post subject:  Searching for the operands 
Basic strategies (perhaps trivial but may be useful to someone). “Third rule”: Searching for the operands (elementary or compound) (I bring here a set of recommendations for beginners, since the expert “calcudokers” undoubtedly know these ideas). I The first thing to do with a product is to find the “prime factors” (it’s an arithmetic subject; you may find a wider explanation in the Patrick Min’s Vols I and II), then proceed to find (and write down) all possible combinations with the elementary factors or the compound factors based on them. When obtaining the operands it is of the maximum importance not forgetting any, because if you do it the full puzzle may collapse (or you will loose a lot of additional time, even the expert “calcudokers” probably have experienced this terrible situation having to erase the full sheet, for instance, in a 12x12). For instance, “72x” in a 9x9 puzzle, in the cells let’s say A1A2B1 (triangle), produces the combinations 338, 364, 662, 924 and 981 with, respectively, sums of 14, 13, 14, 15 and 18 (these totals, their parities, etc., will be combined with the rest of the techniques to solve the puzzle). If you forget the 981, for instance, the full puzzle may crash. But the same “72x”, in a 12x12 puzzle, has additionally the combinations 1232 and 1261. Another example, in this case with the “modulo” operation (in an 8x8 puzzle). The “1mod” (2 cells) could be 12, 13, 14, 15, 16, 17, 18, 23, 34, 45, 56, 67, 78 (the last six with numbers that do not contain the 1), but also we have the 25, 27 and 37 (if you forget, for instance, any of these last three, you can have serious difficulties in finding the solution for the puzzle). II Now let’s think in a cage 3 in three cells, say A1A2B1, in a “from zero” puzzle (8x8). It is specially important not forgetting any of the combinations with the “0” (it’s easy to forget it, it’s not a natural number); combinations are: 003, 115, 227 (these three perfectly defined in position, usually the first to analyze), 014, 025, 036, 047, 126 and 137 (in the last six you can rotate the numbers); if you may eliminate, via analysis, the 003, 115, 227, 126 and 137, for instance, a 0 must be in). If the same cage were in a 12x12 standard puzzle, the combinations would be: 115, 227, 339, 4411 (these four perfectly defined in position), 126, 137, 148, 159, 1610, 1711, 1812, 238, 249, 2510, 2611, 2712, 3410, 3511, 3612 and 4512 (in the last 16 you can rotate the numbers), a total of 20 combinations, in which the “addition” value, always odd of course, ranges from 7 thru a maximum of 21. In general, the maximum “addition” value of an n cage, regardless of the number of cells, equals 2s – n (if the cage is 0, that is, if n = 0, that “addition” value is 2s, where s is the puzzle size [and also the value of the higher number in the puzzle]); it’s easy to see: if we call “r” the “addition” value of all the smaller numbers then s – r = n, r = s – n, then s + r [that is the sum of all numbers inside the cage] = s + (s  n) = 2s –n (here we have implicit the rule of the parity: 2s is even, then when n is even, the cage is of even parity, and when n is odd the cage is parity odd); example: the maximum “addition” value of a two cells cage 5, in a 9x9, is 2 x 9  5 = 13 (this is the case for the operands 9 and 4). If the “addition” value of a cage, V, has been previously determined by other means, now the “higher” number, m, inside the cage, equals (V+n)/2, since m – r = n, r = m –n, V = m + r = m + (m – n) = 2m – n = V, then m = (V+n)/2; example: a four cell cage 2, in a 9x9, where the total sum must be 14, then m = (14 + 2)/2 = 8, being the sum of all the smaller numbers 6 (these type of results are usually obtained quickly and mentally by expert “calcudokers”). Let’s now suppose that the cage (three cells), in the same position, is a 20+ (in a 9x9 puzzle). Combinations are: 992, 884, 776, 668, 983, 974, 965 and 875, the first four perfectly defined in position, in the other four the operands can be rotated. All combinations must be carefully analyzed along the puzzle resolution process. The first four would be very useful because they would “erase” the duplicated number in two rows and two columns simultaneously, unless they are impossible (if the duplicated number is already present in any other cell of the 2 rows or the 2 columns involved). III Let’s see now the case of the exponentiation. If we have a cage (three cells) “262144^” (in an 8x8 puzzle), once we obtain that 262144^ = 2 ^ 18 = 2 ^ (3x6) = (2 ^ 3) ^ 6 = 8 ^ 6 = 8 ^ ( 6 ^ 1) and also that 2 ^ 18 = 2 ^ (2x9) = (2 ^ 2) ^ 9 = 4 ^ 9 = 4 ^ (3x3) = 4 ^ (3 ^ 2), we arrive to the combinations 168, 234 [5 and 7 are not possible because they can not be either in the base or in the exponent but, for instance, in 32768^ = 2 ^ 15 = 2 ^ (3x5) = (2 ^ 3) ^ 5 = 8 ^ 5 = 8 ^ (5x1), 851is a valid combination although the 5 is not a divisor of 32768]. For the 1 ^ (three cells) all combinations of the form 1xy are possible (with any number no more than twice inside). In the operation “bitwise OR”, for instance, for a three cells cage with “7” (in an 8x8 puzzle), let’s say in cells A1A2B1, knowing that the operand 8 can not be in, all combinations are: the 421 and all of the type 7xy, 61x, 63x, 65x, 52x, 53x and 43x (where x and/or y take all values from 1 to 7) (of course any operand cann’t be present more than twice in the combination, situating in this case the duplicated operand in the opposite corners of the triangle). The number of possibilities (see the list, eliminating the redundant) is really big (56) (in cases like this it is not necessary to analyze all combinations, of course, only to clearly know how the operator works, the “” in a cage must modify our conventional thinking, being relevant the binary expression of the “numbers” 1 to 8 and having no meaning its “value”). 421 717 727 737 747 757 767 716 726 736 746 756 766 715 725 735 745 755 714 724 734 744 713 723 733 712 722 711 616 636 656 525 535 434 615 635 655 524 534 433 614 634 654 523 533 432 613 633 652 522 531 431 612 632 521 611 IV In summary:  Guessing a number in a big puzzle is not recommended (too many possibilities).  Use the analysis systematically.  Before starting (with practice it will not be necessary) find all possible combinations (with the elementary and/or compound operands) for the cages (products, divisions, n cages, n+ cages, module and exponentiation).  Do the same with the “well defined” cages “bitwise OR” (“6” is only made with 24, 26 and 46; “5” is only possible with 14, 15 and 45; ”3” is only made with 12, 13 and 23; “n“, with n higher that 8, contains the 8, being the operands 8 and n8 if the cage is of two cells; m can not be inside a cage “n” if m > n, for instance, 8 in cages “7”, “6”, “5” or “3”, 7 in cages “6”, “5” or “3”, 6 in cages “5” or “3”, and 5 in cages “3”, etc.).  Do the same with the “well defined“ cages “from 0” (if any at the beginning; in the case of “from 0” it is a common practice to write them later along the solution process, etc.).  Write the combinations down and erase them one by one as soon as, via analysis, you verify they would drive you to an impossibility (“reductio ad absurdum”).  Use the tools already provided (addition all numbers of a row or column, multiplication of all numbers of a row or column, parity, and the rest that will come in the future) 
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