Comments  Talk
... by: ajay
hi your puzzle solving methods very charming, I do like it .hence forward I will fallow your tricks to solve the puzzle. Thanku very much. I do fallow till I get to hard puzzle. byebye AJAy
... by: Ilma
Appreciation for this ifnromatoin is over 9000thank you!
... by: LizH
In your explanation of pointing pairs, the first example has 3s in columns 1,2,3,7, and 9 of row B; and you say the "pair" of 3s in columns 7 and 9 point to and eliminate the 3s in columns 1,2, and 3 of row B. Why not the other way round, with the triple in box 1 (columns 1,2, nd 3 of row B) pointing to and eliminating the 2s in column 7 and 9?
Andrew Stuart writes: Because box 1 has 3s elsewhere, namely in C2 and C3. It's not a pair/triple if there's other numbers in the same box.
... by: Ed
In practice, how do use coloring? I can't use a magic marker on my screen. When there are more than several cells that are colored, I loose track of what each number is colored.
Andrew Stuart writes: When printed, you should only need two colours. Or you can use a square and circle, or a circle and cross out. I agree it’s a little tough to manually do this on the screen. I'd like to add more features like that to help people use the solver for those types of strategies but I fear over complicating the user interface.
... by: Konrad M Kritzinger
Anthony, I don't think that you can look at the box in isolation. For the "pointing pairs" to work, there needs to be an absence of candidate 8s from the rest of the first column. If there are no other candidate 8s in that column then the column's 8 must be in the 5/8 or 7/8 cell. In that case the box cannot contain another 8 and the other four candidate 8s can be eliminated. If on the other hand there is another candidate 8 elsewhere in that first column, there would be no certainty that the subject box's 8 need be in the 5/8 or 7/8 cell, and "pointing pairs" would not work.
... by: Anthony
I have a question. Please see my diagram of a block below. On another website they say with a pointing pair such as the 5/8 in the top left corner and the 7/8 in the bottom left corner are pointing pairs. Therefore I can eliminate all of the other 8s in the block. I do not see why this rule holds true. I am probably missing something but I just don't see it. Why couldn't I put the 8 in the middle top box and then put the 5 in the top left box and the 7 in the bottom left box? I have looked at quite a few puzzles and the rule holds true but WHY??? Thanks for any help in this.
58 1589 1589 2 6 14789 78 4789 3
... by: Paleo3D
Much, much appreciated and very intelligent strategies...
... by: alvin
NICE SITE.. ive been to many sites ( videos, explanation, sodoku solver..) this site is the best of the best... thanks man. ^_^
... by: CS VIDYASAGAR
Another excelleng explaination of difficult concept. In Type I , digit 2 have no alternative as it has to be in cell B4 and B6 in the Centre Box. Where as in Right Box, digit 2 appear in Row B and Row C. Hence the Centre Box has priority to contain digit 2. Since 2 has to be either in cell B4 or B6, it has no other place to go in the same Row ie. Row B. So 2 from box Right in Row C can be removed. The concept is NO ALTERNATIVE FOR DIGIT IN ROW OR COLUMN IN A PARTICULAR BOX. Thanks for nice and simple elucidation of difficult concept.
... by: Patrick Barnaby
These pairs are easy to spot if you first look at a box then ask is there a single line in a box? But to spot the sevens and eitghts you have to see the two XWings first.
There is an XWing for sevens and an XWing for eights.
... by: Clell
It would have been much easier to understand if you had made the point of alternatives. I did not understand why the double 2s could not eliminate the triple 2s and vice versa until I figured out that what matters is whether or not there is an alternative choice. GAR.
I had looked at this and also could not see the relationship until I reread the above and saw there were no other alternatives in the pointing pair box that forced the others to be eliminated. thanks
... by: Greg
Moses  If A or B is a 9, then how can there be another 9 in the rest of those box of nine squares? You can therefore eliminate the possibility of a 9 in C, D or E.
... by: grosenthal08@cox.net
It would have been much easier to understand if you had made the point of alternatives. I did not understand why the double 2s could not eliminate the triple 2s and vice versa until I figured out that what matters is whether or not there is an alternative choice. GAR.
... by: Rick Aben
Shouldn't this strategy be called "intersection prevalence" (or something like that) instead of "intersection removal", because what you actually do (see for instance example nr 1) is keeping the two 2's in the intersection of row B and box 2 (since they do not appear in the remainder of box 2, so that one of them must be a solution in that box) and remove the 2's out of the remainder of row B (in box 3). The same counts for "type 2" (with a swap of box and rows). In that example the two 9's can be found in the intersection of row 4 en box 6 (the box on the right) and not in the remainder of row 4, which means the 9's in the remainder of box 6 can be eliminated. Ofcource, the solution is the same, the difference is in the logical explanation. Isn't it?
Andrew Stuart writes: The name stuck  in the early days. Not my invention but you are perhaps closer to the mark
... by: moses
please explain more on box line reduction and show some examples thanks

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