I'm creating a sudoku solver in c++, and I'm using your explanations as base to work from, while writing my own algorithms. (I tried reading your javascript code, but I didn't understand it at all, so I stopped trying)
So far I've built everything up to pointing pairs,and my project for tomorrow is Box Line Reduction.
Shall I just go on in order, or do you recommend specific steps to do next to be able to solve most puzzles?

(If you want to check it out you can find my code so far on github, using the same name as I'm using right now)

I find that the use of 'intersection removal' will solve puzzles ranked 'diabolical' (Telegraph) and 'super fiendish' (Times) so the solution process becomes purely mechanical. I've stopped doing them for that reason - a difficult killer sudoku is a much more rewarding puzzle to solve.

Incidentally, I had trouble sending this comment because I followed the instruction 'to the letter' and didn't enter the number. It should say " Please enter the characters you see"!

I'm curious about logistics. Example: our puzzle is on paper and we find pointing pairs. Do you use another piece of paper to indicate these pairs? Do you have a duplicate copy of the puzzle ? I cannot keep all these numbers and positions in my head. What are good ideas to track candidates, patterns, etc. Thanks

Look for a pair (or triple) of n's inside a box that is also part of a row (or column). Then look to see if it's a pointing pair (or triple). It's a pointing pair if there are no other n's inside the box (though there may be nothing for it to "point to" and eliminate). If the pair (or triple) isn't a pointing pair (or triple), then look to see if n does not appear outside the box on that row (or column). If n doesn't appear at all out there, then you have a box-line reduction.

I am learning Perl and I would like to learn some of the algorithms to solve for x-box, naked pairs etc. Is it OK if I use your source code? I do not know what source code you use and if it is applicable to Perl.
thanks
-eac

I really like your explanations here. I am just wondering if you sell books about these strategies? Thanks!

Finally an explanation with appropriate examples that make pointing / intersecting pairs understandable.

pcw - persevere! I'm nearly 80 and no brain box. But I enjoy tackling Sudoku and slowly improving my solving skills.

Yes, what he's asking for is something that mostly only Browning has ever done, I think (plus Mr Nagra himself, of course) - because that's how Brow'ingns mind worked. But them's the rules, at least it doesn't have to rhyme.

Perhaps I'm missing something... this article talks about pairs and triples, as if that's important.

I don't think this is the basis of the logic. What is important is that the digit does not occur in another unit. Maybe I should explain using the example given. In the first example, digit 3 can be removed from B1,B2 and B3. It is stated that this is because there is a pair of 3's in B7,B9. However, what is important here is that there is no 3 on A7,A8,A9,C7,C8,C9. The amount of 3's in B7,B8,B9 makes no difference, the same logic works for 1, 2 or 3 of them.

In other words: I believe explaining this technique using pairs and triples masks the real logic used.

Am I missing something?

In the first example on pointing pairs, isn't 9 in E2 and E3 also a pointing pair that results in eliminating 9 from E5, E7 and E9. Thus making 9 in D9 and F9 a pointing pair too.

Hi Andrew:

I just wanted to thank you for the excellent site.

You explanations and examples are brilliant!

You obviously put in a tremendous amount of time and effort creating this site. You are to be congratulated!

Thanks

Robert

I refer to your example of "box line reduction".

How has number 2 been eliminated as a candidate for cells A1 and B1?

There is no number 2 present in that box, or in row A or column 1

I assume the number 2 has already been eliminated from these cells at a previous stage, using the pointing pair of 2s in row A. However this is not made clear when the example is viewed in isolation.

I have recently started playing Sudoku and am totally addicted! This evening I came across your site as I was looking for an explanation of X -wing and Swordfish.

Thank you so much for clarifying both these strategies.

Playing the game is going to be so much more fun and I can comtinue to challenge myself!

Many thanks.

Just for information, those solving technics are sometime colled "Locked Candidates" or "Intersections", with two sub-technics called "pointing" and "claiming" (what you call Box/Line Reduction).

Thank you for your simple explanations. At long last the "fog" has cleared!

In Triple BLR, the 7s ocupy G4,G5,G6, then the 7 in H4,H5, and J5 must go.

To Gretchen: Look at box 3: Digit 3 appears to be limited to row 'B' (no 3's elsewhere in box 3). That means, either B7 or B9 MUST be set in box 3, otherwise the '3' would be missing in box 3 - which is impossible at all.

Now, from this fact, we can conclude: in row B, there may be no other '3' set. If there were some hints of Digit 3 elsewhere in row B (let alone box 3, this box remains untouched and we could not decide now, which of B7 or B9 would really be true, finally), all those hints in row B could now be removed (look at B1...6), so in this example: B123.

Now your question about "why do the 3s in box 1 do not mean anything". It's going like this: neither, the 3's in box 1 are restricted to row B (there are 3's in C2, C3), nor, the 3's in row B are restricted to box 1 (because of the 3's in box 3). Conclusion: you CANNOT conclude any reduction of pencilmarks out of the 3's in box1.

I hope I could help you, gretchen

Your first example of pointing pairs is confusing to me. Why do the pairs of 3s in box 3 constitute the pointing pair but the trio of 3s in box 1 don't? How do you determine those are the only candidates?

I am delighted to have found your excellent site. I am still at the intermediate level but have found your explanations totally understandable, not something I can say about some others!

Thank you.

Great web-site.,
I been playing Soduko for 3 yrs now, and still trying to figure out ways to solve them,
your web-site helps a lot with tips and hints.

here is a cool puzzle, and it looks good too !
it has 1 unique solution and is solvable.

this site is a real help for solving Sudoku

hi
your puzzle solving methods very charming, I do like it .hence forward I will fallow
your tricks to solve the puzzle. Thank-u very much. I do fallow till I get to hard puzzle.
bye-bye
AJAy

Appreciation for this information is over 9000-thank you!

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?

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.

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.

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

Much, much appreciated and very intelligent strategies...

NICE SITE.. ive been to many sites ( videos, explanation, sodoku solver..) this site is the best of the best... thanks man. ^_^

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.

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.

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

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.

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.

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?

In the first example of pointing pairs of 2 in B4-B6 leads to elimination of 2's in B7-B8-B9.
What prevents the choice of pointing triples of 2 in B7-B8-B9 or even the choice of pointing pairs of sny combination in that box, which would lead to elimination of B4-B6?