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The "rule of quadrants"  6x6 difficult from 09May2012 http://www.calcudoku.org/forum/viewtopic.php?f=3&t=211 
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Author:  clm [ Fri May 11, 2012 8:55 pm ] 
Post subject:  Re: The "rule of quadrants"  6x6 difficult from 09May2012 
jomapil wrote: jomapil wrote: The "demonstration" of Clm or better his logical reasoning is a perfect justification for the "rule". But his observation for the greater than 6x6... even so I think it can be useful ( although in some cases can be of difficult use ) due also the applicability at two squares and two rectangles. Clm, you are right as always: In fact for greater than 6x6 it is very hard to profit this rule. However, the jotempe’s observation, which is a true condition, using the rectangles in addition to the squares, is very interesting and a new and original point of view and perhaps in some particular case (in a 7x7, … ) it could accelerate the solution. The case for a 6x6 of the type referred is a clear application since usually (as commented in the discussion already held in November in the thread “Something I’ve always been curious about”) the top left 3x3 box is solved first and then you know every number in the bottom right 3x3 box, for instance. Another point of view for the demonstration (a little bit more “abstract”) is as follows: Since the subset of numbers in the rectangle is the complement of the subset of numbers in the square (we can take any of the two squares for this purpose) to a well defined set of numbers (the full r rows or the full c columns involved) and this also happens with the other rectangle (for the same number of “lines”), necessarily both rectangles must contain the same subset of numbers. 
Author:  clm [ Fri May 11, 2012 9:10 pm ] 
Post subject:  Re: The "rule of quadrants"  6x6 difficult from 09May2012 
picklepep wrote: jomapil wrote: The "demonstration" of Clm or better his logical reasoning is a perfect justification for the "rule". But his observation for the greater than 6x6... even so I think it can be useful ( although in some cases can be of difficult use ) due also the applicability at two squares and two rectangles. I like your discussions  clm  because often times it will lead me to think about the puzzles slightly different. This happened when you were discussing the 7x7 only  op and brought up the fact that often times the 'odd' square can very easily be found to be even or odd. I have since applied this method in various other puzzles to eliminate half the possibilities! A very useful tool. While i usually solve the 6x6 pattern with other methods, I never really thought about how filling in a block of numbers neccitates that their be an isomorph opposite it. I believe that this can be a very powerful tool! Thank you for putting so much time into your analysis! It’s nice to listen puzzlers who enjoy these discussions. In the case of the parity, your comments give me the opportunity to make some additional observations on this parity property or rule, going a little beyond, though we are now in a different thread. First let’s remind that the parity of the full puzzle is even in a 2x2 (baby ), 3x3 (baby ), 4x4, 6x6, 7x7, 8x8, 10x10, 11x11, 12x12, 14x14, … . Puzzles like the 1x1 (baby ), 5x5, 9x9, 13x13, 17x17, … , that is when the size is a multiple of 4 plus a unit, are odd puzzles, more precisely the full sum respectively would be 1, 75, 405, 1183, 2601, … (the algebraic demonstration is at the end of the post). Consequently, if, for instance, we divide a 7x7 (even) by any two “axis” (one horizontal and another vertical lines), being any line even (28), and being even any number of lines, and being the rectangle and the square complementary, both must have the same parity. As the rectangles contain the same subset of numbers they have the same parity and consequently the four areas must always be of the same parity (all four even or all four odd). This is valid in a 3x3, 4x4, 7x7, 8x8, 11x11, 12x12, … , for “any” division including the case when the rectangles are “linear” that is leaving just a corner alone. (For the 6x6, 10x10, 14x14, ... , that is, puzzles even but with the line odd, in these particular cases 21, 55, 105, ... , the divison produces as before the four areas with the same parity but if the side of the squares is even and two areas even and two odd, opposite, when the side of the squares is odd; the 2x2 is trivial). But in the case of the 5x5, 9x9, 13x13, 17x17, … , the puzzle is odd, then such a division must have one of the areas odd (or three areas odd, only these two possibilities), so one of the squares must be odd (the rectangles, of the same parity, can be both even or both odd) and the two squares are always of different parity, again whichever is the division. This can be quickly seen in any 5x5. This conclusion, for the 5x5 and the 9x9, and the previous ones, for a 4x4, 6x6, 7x7, and 8x8, so in practical cases, perhaps can be applied some times. Now, let’ divide the puzzle, for instance a 7x7, into, i.e., 3 areas, by any arbritary lines, not necessarily straight lines. Since a 7x7 is even, either all areas are even or two of the areas are odd. In the case of a 5x5 or a 9x9, the arbitrary division into three areas would produce this result: at least one the areas is odd (or the three areas are odd). And those areas could be as small as a single cell. Furthermore, we can divide the puzzle in any number of “areas” (not cages), a particular case of which is the division into the calcudoku "cages". Those areas can have the cells connected or the cells can be arbitrarily distributed over the puzzle as long as we know they are part of a determined area. We can derive many different observations, depending on the total number of areas, the type of puzzle, etc.. For instance, in the 7x7 subtractions only (on wednesdays) the number of odd cages (including obviously the odd single cells) must be even.  Demonstration of the parity of the full puzzle: Since a line has a sum of L = s * (s + 1) / 2 (the sum of all numbers of an arithmetic progression, where s is the size of the puzzle), the full puzzle (s lines) has a sum of P = s * s * (s + 1) / 2 (this expression produces the results shown above). When s = 4n + 4 = 4 * (n + 1) (n = 0, 1, 2, 3, … , thus generating the types 4x4, 8x8, 12x12, 16x16, … ) then P = 16 * (n + 1) * (n +1) * (4n + 5) / 2 = 8 * (n + 1) * (n + 1) * (4n +5) which is always even. When s = 4n + 3 (puzzles 3x3, 7x7, 11x11, 15x15, … ) then P = (4n + 3) * (4n +3) * (4n + 4) / 2 = 2 * (4n + 3) * (4n + 3) * (n + 1) always even. When s = 4n +2 = 2 * (2n + 1) (puzzles 2x2, 6x6, 10x10, 14x14, … ) then P = 4 * (2n + 1) * (2n + 1) * (4n + 3) / 2 = 2 * (2n + 1) * (2n + 1) * (4n + 3) always even. But if s = 4n + 1 (puzzles 1x1, 5x5, 9x9, 13x13, 17x17, … ) then P = (4n + 1) * (4n + 1) * (4n + 2) / 2 = (16n^2 + 8n + 1) * (2n +1) = 32n^3 + 32n^2 + 10n + 1 which is always odd since all terms are even except the last. 
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