... 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 X-Wings first.

There is an X-Wing for sevens and an X-Wing 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: W Gerald

Why isn't your printed version the same as the site version with no changes in layout and only background and text colours reversed? The site version display is superb.

... 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?

... by: moses

please explain more on box line reduction and show some examples thanks