If any one number occurs twice or three times in just one unit (any row, column or box) then we can remove that number from the intersection of another unit. There are four types of intersection:

- A Pair or Triple in a box - if they are aligned on a row, n can be removed from the rest of the row.
- A Pair or Triple in a box - if they are aligned on a column, n can be removed from the rest of the column.
- A Pair or Triple on a row - if they are all in the same box, n can be removed from the rest of the box.
- A Pair or Triple on a column - if they are all in the same box, n can be removed from the rest of the box.

Looking at each box in turn there may be two or three occurrences of a particular number. If these numbers are aligned on a single row or column (as a pair or a triple) then we know that number MUST occur on that line. Therefore, if the number occurs anywhere else on the row or column outside the box WHICH THEY ARE ALIGNED ON then it can be removed. The pair or triple *points* along the line at any numbers which can be removed.

Here are two Pointing Pairs at the same time on this tough rated puzzle. The 3s in B7 and B9 are alone in box 3 and they are aligned on the row. So looking along the row we can remove all the 3s in Box 1. In a similar manner the 2s in G4 and G5 point along the row to G2.

Here are two Pointing Pairs at the same time on this tough rated puzzle. The 3s in B7 and B9 are alone in box 3 and they are aligned on the row. So looking along the row we can remove all the 3s in Box 1. In a similar manner the 2s in G4 and G5 point along the row to G2.

Now this is a rather special puzzle and a little extreme, but if we look at the whole board you can see I have highlighted a whole cluster of Pointing Pairs. It is obviously not necessary to spot everyone to progress the board but there are so many good examples it is worth looking at. The eliminations are highlighted in yellow. You should be able to see which eliminations belong to which Pointing Pair. A couple are consequences of a previous elimination.

Here is a Pointing Triple found in a tough grade puzzle containing many wholesome and nutritious Pointing Pairs. All the 3s are found in G6, H6 and J6 and nowhere else in box 8, so they point up the column to another 3 which can be removed.

This strategy involves careful comparison of rows and columns against the content of boxes (3 x 3 squares). If we find numbers in any row or column that are grouped together in just one box, we can exclude those numbers from the rest of the box. For example:

This Sudoku contains numerous Pointing Pairs and Box/Line Reductions and is worth stepping through from the start. Consider row A. The only 2s left are in A4 and A5 which means we should should check the rest of the box. 2 has to go somewhere on row A and it will be in one of those two cells. So we can eliminate 2 from B5, C4 and C5.

This Sudoku contains numerous Pointing Pairs and Box/Line Reductions and is worth stepping through from the start. Consider row A. The only 2s left are in A4 and A5 which means we should should check the rest of the box. 2 has to go somewhere on row A and it will be in one of those two cells. So we can eliminate 2 from B5, C4 and C5.

In the very next step of the same puzzle we have two 4s alone in column 8. That fixes 4 to be in either cell A8 or B8. We can remove the other 4s in box 3 and get our next solved cell: 2 in C7.

I've rolled two examples into one diagram here, I hope it wont be confusing. The one involving 6s follows on immediately from the first one found among the 3s (purely because it searches in order from 1 to 9). They are both 'triple' versions of Box/Line reduction.

In column 1 the 3s occupy G1, H1, J1 (shorthand = GHJ1) which are all in box 7. The solution is pinned to column 1 so the other 3s in the box must go.

The 6s are likewise pinned by column 2. You can see there are other 6s in column 3, in A3 and J3) which is why it's okay to remove the 6s in DEF3. You won't run out.

If you are a fan of Jigsaw Sudoku puzzles, you may want to read the articles on Double Pointing Pairs and Double Line/Box Reduction which extend the ideas here but are strategies possible only in the Jigsaw variant of the puzzle.

## Comments

Comments Talk## Tuesday 28-Oct-2014

## ... by: Mal

Thank you for your simple explanations. At long last the "fog" has cleared!## Sunday 27-Jul-2014

## ... by: 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?## Sunday 20-Jul-2014

## ... by: JG

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.

## Tuesday 10-Jun-2014

## ... by: Joseph Boronka

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.

## Thursday 19-Dec-2013

## ... by: George kunnappally

this site is a real help for solving Sudoku## Friday 11-Oct-2013

## ... by: ajay

hiyour 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

## Monday 18-Feb-2013

## ... by: Ilma

Appreciation for this information is over 9000-thank you!## Saturday 28-Jul-2012

## ... 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?## Thursday 12-Jul-2012

## ... 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.## Friday 4-May-2012

## ... 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.## Saturday 4-Feb-2012

## ... 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

## Wednesday 9-Mar-2011

## ... by: Paleo3D

Much, much appreciated and very intelligent strategies...## Wednesday 7-Apr-2010

## ... by: alvin

NICE SITE.. ive been to many sites ( videos, explanation, sodoku solver..) this site is the best of the best... thanks man. ^_^## Thursday 25-Feb-2010

## ... 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.

## Wednesday 20-Jan-2010

## ... 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.

## Friday 20-Nov-2009

## ... 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

## Friday 25-Sep-2009

## ... 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.## Thursday 17-Sep-2009

## ... 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.## Sunday 12-Jul-2009

## ... 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?

## Sunday 21-Jun-2009

## ... by: moses

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