FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in          Games  Calendar Log in to check your private messages    Problems applying X-Colors Algorithm          Sudoku Programmers Forum Index -> Solving sudoku View previous topic :: View next topic   Author Message whizter Joined: 19 Jul 2007 Posts: 30 : Items Posted: Fri Jul 27, 2007 11:22 am    Post subject: Problems applying X-Colors Algorithm I tried to apply the X-Colors technique for Candidate 1 at this stage (I started coloring at r1c4): Code:    *--------------------------------------------------------------------*  | 7      6      4      | 13     8      2      | 9      13     5      |  | 5      3      1      | 6      47     9      | 2      478    48     |  | 2      9      8      | 57     457    13     | 17     13467  346    |  |----------------------+----------------------+----------------------|  | 16     8      37     | 9      2      5      | 17     13467  346    |  | 4      5      9      | 37     1      6      | 8      37     2      |  | 16     2      37     | 4      37     8      | 5      169    69     |  |----------------------+----------------------+----------------------|  | 39     4      2      | 1358   35     13     | 6      89     7      |  | 39     1      5      | 38     6      7      | 4      2      89     |  | 8      7      6      | 2      9      4      | 3      5      1      |  *--------------------------------------------------------------------* Now look at Block 6 and at X-Colors Rule 4.3, which says: "If ALL the cells of a house are peers of cells colored with the same color A, you can conclude that THE OTHER COLOR B IS TRUE." All Cells of Block 6 are peers of cells colored with color A, no cell is a peer of a Cell with color B. Now applying the rule, i would put 1 into r1c4 and r7c6. But this is wrong. Where's my mistake? This is what it looks like after applying steps 1-3: Back to top daj95376 Joined: 05 Feb 2006 Posts: 349 : Items Posted: Fri Jul 27, 2007 1:31 pm    Post subject: I don't have the rules for X-Colors in front of me, but here's what I see. [r1c8]=Green => [r46c8]<>Green => [r4c7],[r6c1]=Green => [r3c7],r4c1]=Blue The elimination cells should now be obvious. Note: Using X-Colors, I don't believe that you can conclude [r6c8]=Blue at this stage. Obviously, it is Blue after the eliminations. Last edited by daj95376 on Fri Jul 27, 2007 1:49 pm; edited 1 time in total Back to top whizter Joined: 19 Jul 2007 Posts: 30 : Items Posted: Fri Jul 27, 2007 1:48 pm    Post subject: I know that this situation can be solved in other ways, but I'm currently learning this X-Colors strategy and I don't know what's wrong with the rules or my understanding of the rules. They can be found here: http://www.sudopedia.org/wiki/X-Colors Back to top daj95376 Joined: 05 Feb 2006 Posts: 349 : Items Posted: Fri Jul 27, 2007 1:59 pm    Post subject: Re: Problems applying X-Colors Algorithm Quote: Now look at Block 6 and at X-Colors Rule 4.3, which says: "If ALL the cells of a house are peers of cells colored with the same color A, you can conclude that THE OTHER COLOR B IS TRUE." All Cells of Block 6 are peers of cells colored with color A, no cell is a peer of a Cell with color B. Now applying the rule, i would put 1 into r1c4 and r7c6. But this is wrong. Where's my mistake? To answer your question, cell [r4c7] is not a peer of Green ... it is Green. BTW, I believe that my coloring those two cells Blue are part of the X-Colors technique. Back to top whizter Joined: 19 Jul 2007 Posts: 30 : Items Posted: Fri Jul 27, 2007 2:13 pm    Post subject: Ok, now I understand why rule 4.3 doesn't apply here. But I don't understand your example, Step 1 of the rules says: "Select one pair of conjugate cells, color one of them with color A and the other with color B." I did that with r1c4 and r1c8. Step 2.1: "Find a not colored cell that is pair-conjugated of one cell already colored (with any color A or B)." How can you get to this using these rules? --> "[r1c8]=Green => [r46c8]<>Green" Back to top daj95376 Joined: 05 Feb 2006 Posts: 349 : Items Posted: Fri Jul 27, 2007 3:38 pm    Post subject: Quote: Ok, now I understand why rule 4.3 doesn't apply here. But I don't understand your example, Step 1 of the rules says: "Select one pair of conjugate cells, color one of them with color A and the other with color B." I did that with r1c4 and r1c8. Step 2.1: "Find a not colored cell that is pair-conjugated of one cell already colored (with any color A or B)." How can you get to this using these rules? --> "[r1c8]=Green => [r46c8]<>Green" Okay, lets take it from the top. (These are my steps and not those listed in Sudopedia) Step 1) Perform simple Colors until you can't proceed any further. Perform any eliminations allowed through simple Colors. Continue until no eliminations exist. Step 2) If a color sees all candidate cells in a house, then select the conjugate color and return to Step (1). Step 3) If all but one candidate cell in a house is seen by one color, then select that color for the exception cell. Continue until no exception cells can be found. (My "[r1c8]=Green => [r46c8]<>Green" applies here to get [r4c7]=Green. One more application of this step gets [r6c1]=Green.) Step 4) Return to Step (1) using the cells colored in Step (3). You performed Steps 1-3, but never took advantage of the coloring you performed. {Note: Edited several time to correct typos and clarify message.] Back to top whizter Joined: 19 Jul 2007 Posts: 30 : Items Posted: Sat Jul 28, 2007 8:23 am    Post subject: Quote: Step 2) If a color sees all candidate cells in a house, then select the conjugate color and return to Step (1). I didn't understand this, what exactly happens in this step? Is it meant like this? -> If all candidate Cells within a house of only uncolored candidate cells can be seen by Color A, but none of these Cells can be seen by Color B, assume that Color B is true. Except for that step, everything makes sense to me. Back to top daj95376 Joined: 05 Feb 2006 Posts: 349 : Items Posted: Sat Jul 28, 2007 3:15 pm    Post subject: whizter wrote: Quote: Step 2) If a color sees all candidate cells in a house, then select the conjugate color and return to Step (1). I didn't understand this, what exactly happens in this step? Is it meant like this? -> If all candidate Cells within a house of only uncolored candidate cells can be seen by Color A, but none of these Cells can be seen by Color B, assume that Color B is true. Except for that step, everything makes sense to me. First, my apologies for being too brief on this step. Here's an expanded (and more accurate) explanation. Step 2) If all of the candidates in a house are uncolored and one color sees all of them, then select the conjugate color as true and return to Step 1). Explanation: If color A meets the above if condition and is assumed true, then it will eliminate all of the candidates in this house and the puzzle can not have a valid solution. Therefore, color A is unacceptable. At this point, most people would say the candidate needs to be eliminated from all cells with color A. This is then followed by a bunch of Hidden Singles for the candiate in the cells with color B. I say, cut to the chase and just select this candidate in all color B cells; i.e. select the conjugate color. I hope this is less confusing! Back to top whizter Joined: 19 Jul 2007 Posts: 30 : Items Posted: Sat Jul 28, 2007 4:36 pm    Post subject: Thanks, that was understandable. 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