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March 13 9x9 cages 25+, 27+, 27+ https://www.calcudoku.org/forum/viewtopic.php?f=3&t=171 
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Author:  honkhonk [ Thu Mar 15, 2012 9:59 pm ] 
Post subject:  Re: March 13 9x9 cages 25+, 27+, 27+ 
picklepep wrote: arjen wrote: It might help if you bolded a few more things. The double 6's in the 23+ cage can be bolded as they can not be in any other position. You can also look at rows 6 and 7. You can figure out that the 5 and 1 in row 7 can only be in 2 spots, hence they are fixed that makes row 7 columnn 5 be 8, thus r8c1=8. I think there is more... I agree with Pickelepep, to solve the last cages you'll need to find and fix more numbers. Cages 23+ and 128+ all green: Since there is no '1' or '2' in 11664x, [b6,c6] = [1,2] There is no [2,4] in the 25+, 19+, 11664x , 630x cages >>> [a7,b7] = [2,4] >>> a8=8. From this and since d9=1, the '1' in column a can only be in a2 >>> a2=1. Because i4=2 >> b3=2 >> c4 =8 >> b6=1 >> c6 =2 >> a7=2 >> b7=4 , the cage 23+ is all green Going to the top 27+: Then looking at column a, a1 = 4 because in 23+ and 25+ there are no '4's Knowing a1=4, the last '4' to be placed must be in h2 >>> h2=4. Since we know b3=2 and i4=2 there can be no '2' in the lower 27+. In fact because g9=2 and h2=4 >>> h1 =2! Now we know the top 27+ is either [2,4,5,8,8] or [2,4,5,9,7] Because f1=5, the 5 of the top 27+ must be in [g2,g3] Going down to the lower half of the puzzle: In row 6 the '8' must be in [g6, h6] since the '8'in 128x is already placed in a8 In row 7 the '8' must be in e7 since the '8' in 128x is in row 8 and the '8'of 11664 in row 6! Because of the 10+, the 1 and the 11664 are sharing the '7', the '3' and the '6', c7 and g7 are [1,5]. We know there is a 5 in [g2,g3] >>> g7=1 >>> c7 = 5 From this follows c8 = 1, since 10080x has no '1' in it an column c needs a '1' Now we know [c2,c3] = [9,7]! Fixing the 14+: The '1' in the first row, must be placed in e1, since the cages 10080x and 27+upper contain no '1' e1=1 and b3=2 >>> e2=2 >>> e3=7 Leads to fixing [c2,c3] and the [1,9] cage: e3=7 leads to c2=7 >>> c3=9 >>> f3=1 >>> f4=9! But then in row 4 all we miss is an '1' and a '7' >>> [g4,h4] = [7,1] We know g7=1 >>> g4 = 7 >>> h5=1 >> h6=5 >>> i5=1! This will fix the 630x cage: [g2,g3] = 5 and h6=5 thus the 5 in 630x must be in i9 >>> i9=5 >>> e9=6 >>> e8=5 Because g1=6 and e9=6 the '6' in 630x must be in h8 >>> h8=6 We know g4=7, h8 and i9 are occupied >>> the '7' in 630x is in h9 >>> h9=7 This leaves g8=3 Go to the 11664x cage: Since h8=6 and g1=6 >>> i7=6 And since g8=3 >>> g6=8 >>> h6 =3 Go to the upper 27+ cage: We know column g is only missing a '9' and c3=9 >>> g2=9 >>> g3=5 >>> i1=7 This makes the lower 27+ cage [1,3,7,8,8]... Now the rest is easy.... I did not include any pictures, I hope you can follow. If there is need for pictures I'll try make some... :D 
Author:  arjen [ Fri Mar 16, 2012 6:56 pm ] 
Post subject:  Re: March 13 9x9 cages 25+, 27+, 27+ 
Thanks for the explanation. I totally overlooked the 2 sixs in 23+. After minutes staring at the numbers I gave up, that happens relatively rarely. 
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