Calcudoku puzzle forumhttps://www.calcudoku.org/forum/ April 13 9x9https://www.calcudoku.org/forum/viewtopic.php?f=16&t=586 Page 1 of 1

 Author: sjs34  [ Tue Apr 15, 2014 3:14 am ] Post subject: April 13 9x9 Does anyone have an elegant solution to yesterday's 9x9? I ultimately was able to overpower it but my effort was ugly. Way too much trial and error.

 Author: clm  [ Tue Apr 15, 2014 2:28 pm ] Post subject: Re: April 13 9x9 sjs34 wrote:Does anyone have an elegant solution to yesterday's 9x9? I ultimately was able to overpower it but my effort was ugly. Way too much trial and error.I suppose you are referring to April's 14 9x9 since on Sunday it was not a 9x9.Well, as you well know, the full solution requires many steps and it’s a long process.What I do first is, as usual, to place the “immediate” numbers and candidates: "24x" = [3,8], i1 = 1, i2 = 8, a7 = 1, a8 = 9 (“27x” = [1,1,3,9] and 9 is not a submultiple of “336x” or “84x”), d9 = 1 (h9 cann’t be a 1 because a sum of 25 in g7-g8-h8 would require [7,9,9] or [8,8,9] both invalid), “18x” = [2,9], "42x” = [6,7] and in column “i”: “3-" = [6,9], “8+” = [3,5] and “5-" = [2,7].Next, looking to column “a”, I multiply 336 x 84 x 8 x 1 x 9 x 5 and divide this by 9! = 362880 (multiplication rule to column “a”) finding: b2 x b5 = 28 = [4,7]. This means that a 9 or a 5 cann’t be in “1-” (b3-b4) then b6-b7 = [5,9] >>> c6 = 2.The only possibilities for “336x” would be [3,4,4,7] and [2,4,6,7] since no 8’s are allowed. The first option is immediately suppressed because a1= 7, a2 = 3, b2 = 4 >>> c2 + “0-” (g2-h1-h2) = 16, that is, c2 even = 6 (unique) and “0-" had a sum of 10 with a 5 as the higher number but [1,4,5] or [2,3,5] are now invalid. Consequently “336x” = [2,4,6,7] and “84x” = [3,4,7].Now it’s easy to see that candidates for “0-” (c5-d5-d6) are [3,5,8]: No 1’s or 2’s or 9’s are allowed and [3,3,6], [4,4,8] are invalid (due to two 6’s in rows 5 and 6 and c1 = 4). So we are only left with the combinations [3,4,7] and [3,5,8], but [3,4,7] is not possible since this would force a4 = 7 (7's in rows 5 and 6) and a5-b5 = [3,4] against the hypothesis; consequently the "0-" cage is [3,5,8].Now let’s define the cage “11+” in f8-f9; this cage cann’t be [2,9] (f8 =2, f9 = 9) because of h9 = 6 what would make impossible to place a 2 in row 5, so “11+” = [5,6] >>> f8 = 5, f9 = 6, h9 = 9. This drives to h1 = 5 because e1 = 5 would necessarily require e2 = 9, f2 = 5 repeating a 5 in column “f”....(Patrick, it would help to add a drawing of this puzzle in the Forum for future reference, I am having problems now with ImageShack, which is now a subscribers site ... )Thank you. Reedit to add imageReedited to continue a little more.What comes now is very nice:Column “h”: A 2 is not possible in h8 because a sum of 15 would not be possible in g7-g8. And a 2 cann’t go to h2 (necessarily h3 = 1) or h3 (necessarily h2 =1) because in both cases the 6 of column “h” should go to h8 which is impossible because g7 + g8 should sum 11 which cann’t be obtained in those positions. And we can observe that h4 or h7 must be different than 2 (both producing an additional 6 in rows 5 and 6) so h5 = 2, h4 = 6.And now we repeat a similar reasoning this time using the possibilities of the 7 in column “h”. A 7 cann’t go to h8 because a sum of 10 would not be possible in g7-g8. And a 7 cann’t go to h2 (necessarily h3 = 1) or h3 (necessarily h2 = 1) because in both cases it’s not possible to place a 3 in column “h”, consequently “4-” (h6-h7) = [3,7] and the 4 of row 6 is only allowed in e6.Reedited again to continue a little bit more:Row 5: Since f8 = 5 we cann’t place a 5 in f5 so the 5 of row 5 must be in c5-d5, consequently d6 = 3 and this fulfills row 6 as shown. Additionally, using the addition rule applied to the three bottom rows (3 x 45 = 135 in total), we conclude that e7 + f7 = 12 >>> e7 = 8, f7 = 4 (unique).And from here: a4 = 7, b5 = 4 (we remind here that b2-b5 = [4,7]), a5 = 3, f5 = 9, e4 = 5, f4 = 2, i4 = 3, i3 = 5. Also b2 = 7, a1 = 2, d1 = 9, d2 = 2, e1 = 7 (e2 = 9, f2 = 3, f1 = 8, g1 = 3, f3 = 7, c2 = 5, a2 = 6, a3 = 4). And, besides that, b3 = 2, b4 = 1, b8 = 3, c8 = 1. And to complete the solution from this point is easy (next two graphics).

 Author: pnm  [ Tue Apr 15, 2014 3:17 pm ] Post subject: Re: April 13 9x9 clm wrote:(Patrick, it would help to add a drawing of this puzzle in the Forum for future reference, I am having problems now with ImageShack, which is now a subscribers site ... )The one to use now is http://www.imgur.com,free and much easier to use Edit: I've updated the instructions at viewtopic.php?f=3&t=38&p=269#p269

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