Unique Rectangles takes advantage of the fact that published Sudokus have only one solution. If your Sudoku source does not guarantee this then this strategy will not work. But it is very powerful and there are quite a few interesting variants.

__Credits first__: The initial ideas for these strategies I lifted wholesale from **MadOverLord**'s description (11 Jun 05) on the forum at www.sudoku.com. (Sadly the thread is no longer hosted). Many others have since added to and improved them. I have provided my own examples. (please email me for other credit requests). I will stick to MadOverLord's nomenclature.

In Figure 1 we have two example rectangles formed by four cells each. The pattern in red marked A consists of four conjugate pairs of 4/5. They reside on two rows, two columns and two boxes. Such a group of four pairs is impossible in a Sudoku with one solution. The reason? Pick any cell with 4/5. If the cell solution was 4 then we quickly know what the other three cells are. But it would be equally possible to have 5 in that cell and the others would be the reverse. There are two solutions to any Sudoku with this deadly pattern. If you have achieved this state in your solution something has gone wrong.

The pattern ringed in green looks like a deadly pattern but there is a crucial difference. The 7/9 still resides on two rows and two columns, but instead of two boxes it is spread over four boxes. Now, such a situation is fine since you can't guarantee that swapping the 7 and 9 in an alternate manner will produce two valid Sudokus. One of them is the real solution, the other a mess. Why? Swapping the 7 and 9 around places them in different boxes and 1 to 9 must exist in each box only once. In the red example, swapping within the box does not change the content of that box.

The pattern ringed in green looks like a deadly pattern but there is a crucial difference. The 7/9 still resides on two rows and two columns, but instead of two boxes it is spread over four boxes. Now, such a situation is fine since you can't guarantee that swapping the 7 and 9 in an alternate manner will produce two valid Sudokus. One of them is the real solution, the other a mess. Why? Swapping the 7 and 9 around places them in different boxes and 1 to 9 must exist in each box only once. In the red example, swapping within the box does not change the content of that box.

For all Unique Rectangles we are going to look for *potential* deadly patterns and take advantage of them. A **Type 1 Unique Rectangle** is illustrated in Figure 2. The three yellow cells marked contain 2/9. The fourth corner marked in orange also contains 2/9 and two other candidates. If the 1/5 were removed from that cell we would have a Deadly Pattern. This cannot be allowed to happen so its safe to remove 2 and 9 from that cell.

The proof is pretty straightforward once you get your head around the basic idea. Assume D1 is 2. That forces D9 to be 9, F9 to be 2, and F1 to be 9. That's the deadly pattern; you can swap the 2's and 9's and the puzzle still can be filled in. So if the Sudoku is valid, D1 cannot be 2. The exact same logic applies if you assume D1 is 9.

So D1 can't be a 2, and can't be a 9

- it must be either 1 or 5.

The proof is pretty straightforward once you get your head around the basic idea. Assume D1 is 2. That forces D9 to be 9, F9 to be 2, and F1 to be 9. That's the deadly pattern; you can swap the 2's and 9's and the puzzle still can be filled in. So if the Sudoku is valid, D1 cannot be 2. The exact same logic applies if you assume D1 is 9.

So D1 can't be a 2, and can't be a 9

- it must be either 1 or 5.

In Figure 3 we have a similar pattern, but this time, A5 and A6 (cells which are also in the same box) both have a single extra possibility - in this case, 7.

To make subsequent discussion easier to follow, we will refer to the two squares that only have two possibilities as the floor squares (because they form the foundation of the Unique Rectangle); the other two squares, with extra possibilities shall be called the roof squares.

To make subsequent discussion easier to follow, we will refer to the two squares that only have two possibilities as the floor squares (because they form the foundation of the Unique Rectangle); the other two squares, with extra possibilities shall be called the roof squares.

In this **"Type-2 Unique Rectangle"**, one of the boxes contains the floor squares, and the other contains the roof squares. In order to avoid the deadly pattern, 7 must appear in either A5 or A6 (the roof squares). Therefore, it can be removed from all other cells in the units (row, column and box) that contain *both* of the **roof** cells (in this case, row 1 and box 2).

Now that you've gotten your head around the basic unique rectangle concept, the proof should come clear:

If neither A5 or A6 contains an 7, then they both become cells with possibilities 1/5. This results in the deadly pattern - so one of those cells must be the 7, and none of the other squares in the intersecting units can contain 7. So A3 and C6 can have 7 removed. This cracks the Sudoku.

Now that you've gotten your head around the basic unique rectangle concept, the proof should come clear:

If neither A5 or A6 contains an 7, then they both become cells with possibilities 1/5. This results in the deadly pattern - so one of those cells must be the 7, and none of the other squares in the intersecting units can contain 7. So A3 and C6 can have 7 removed. This cracks the Sudoku.

I couldn't resist adding this example which I found while looking for Empty Rectangles. It's as clear as day how the 8s in H4 and H6 combine with the [1,6] Deadly Pattern. 8's aligned with the brown cells are eliminated. I need say no more.

There is a second variant of Type-2 Unique Rectangles as illustrated In Figure 5.

In this puzzle, we have the same pattern of 4 squares in 2 boxes, 2 rows and 2 columns. The floor squares are A1 and A9, and the roof squares are B1 and B9. However, in this Unique Rectangle, each of the boxes contains one floor and one roof cell. This is perfectly fine, but it means that the only unit (row/column/box) that contains both of the roof cells is row 2, so that is the only unit that you can attempt to reduce; in this case, B7 cannot contain a 8. This is called at**"Type-2B Unique Rectangle"**.

You will need to untick 3D-Medusa to see this example.

In this puzzle, we have the same pattern of 4 squares in 2 boxes, 2 rows and 2 columns. The floor squares are A1 and A9, and the roof squares are B1 and B9. However, in this Unique Rectangle, each of the boxes contains one floor and one roof cell. This is perfectly fine, but it means that the only unit (row/column/box) that contains both of the roof cells is row 2, so that is the only unit that you can attempt to reduce; in this case, B7 cannot contain a 8. This is called at

You will need to untick 3D-Medusa to see this example.

HoDoKu records this as a Type 5 but since it's a variation on Type 2 I'm going to categorize it as 2C. It is fairly rare but easy to understand.

This variation again has the extra candidate but in cells diagonally opposite. You can see the extra 6 in cells B7 and H9 (marked as C and D). 6 Must exist in one of those two cells so any 6 not in the deadly rectangle pattern that can 'see' C and D can be removed.

Not shown in the example is the chance that A or B can also contain the extra candidate 6. Lets say B did - the eliminations hold since those cells can see B9. But if the third extra candidate was in H7 then we could not look in box 3.

This variation again has the extra candidate but in cells diagonally opposite. You can see the extra 6 in cells B7 and H9 (marked as C and D). 6 Must exist in one of those two cells so any 6 not in the deadly rectangle pattern that can 'see' C and D can be removed.

Not shown in the example is the chance that A or B can also contain the extra candidate 6. Lets say B did - the eliminations hold since those cells can see B9. But if the third extra candidate was in H7 then we could not look in box 3.

In this variant we have a floor with a pair as before but the roof contains two different candidates (occurring once or twice in each cell of the roof). In the Figure 7 the floor and roof contain [3/9] and the extra candidates in the roof are 1 and 7. Removing both the 1 and 7 from the roof would leave the deadly pattern so either the 1 in B7 must be a solution or the 7 in G7 is a solution. Knowing this does not get us as far as an elimination, however, but we can say that the 1 in B7 and the 7 in G7 act as a pseudo-cell in their own right. The clever bit is using this pseudo-cell with a bi-value cell containing the same candidates. Such a cell is circled on the board – J7. We effectively have a “locked set” like a Naked Pair. 1 will occur in B7 or J7 or 7 will occur in G7 or J7, we just don’t know which way round.

Such reasoning allows us to remove other 1s and 7s on the unit shared by the roof cells (but not the cell we’re using to create a locked set with). There are 1s and 7s in C7 and H7 which can be eliminated.

There is a compliment to Type 3, where the roof cells share the same box and that means the floor must be in a different box. Cells J4 and J6 form a pair with 6 and 7. These can see two cells in row E which also contain 6/7 as well as a 3 and an 8. We need to match this 3 and 8 in the roof cells with a bi-value cell to make a locked set (effectively a Hidden Pair). As we have roof cells in the same row AND box we can look in both row 5 to E1 and remove all other 3s and 8s in the row, plus F5 and eliminate 3 from D5.

I'd like to credit Hervé Gérard for the first example which he sent to me in November 2013 and later Tom Morrin for stating the box should be used as well.

I'd like to credit Hervé Gérard for the first example which he sent to me in November 2013 and later Tom Morrin for stating the box should be used as well.

So far Type 3 and 3b have worked with a pseudo-cell linking with a bi-value cell to make a pseudo pair. This is a valid locked set. Well, locked sets don't have to be pairs, they can involve three candidates over three cells.

In this example we have a deadly rectangle based on 4 and 8. The remaining candidates in the Roof are 6 and 9. Whether there are one or two of either 6 and 9 in the Roof is not important. We know 6 OR 9 has to appear in H23. Members of a Locked Set must all see each other, so we are looking at other cells that align with the roof, namely row H (and occasionally the box as well). For a Triple cell Locked Set we need two other cells that contain 6 and 9 and one other candidate not present in the Roof. H4 contains 5 and 6 and H8 5, 6 and 9. With our pseudo-cell that makes a Naked Triple. Naked Triple rules apply so 6 and 9 can be removed from H1.

All credit to David Hollenberg for working this insight that nicely expands the strategy. In a test of 5,000 puzzles with some kind of UR, a third more Type 3 and 3B were found with Triples.

All credit to David Hollenberg for working this insight that nicely expands the strategy. In a test of 5,000 puzzles with some kind of UR, a third more Type 3 and 3B were found with Triples.

- Cracking the Rectangle with Conjugate Pairs. (See also Extended UR Type 4.)

An interesting observation is that it is sometimes possible to remove one of the original pair of possibilities from the roof squares. Consider the following puzzle in Figure 10 which is a continuation of the Sudoku puzzle used in the first example.

Look closely at the roof squares, H1 and H2, but this time, don't look at their extra possibilities; look at the possibilities they share with the floor squares.

An interesting observation is that it is sometimes possible to remove one of the original pair of possibilities from the roof squares. Consider the following puzzle in Figure 10 which is a continuation of the Sudoku puzzle used in the first example.

Look closely at the roof squares, H1 and H2, but this time, don't look at their extra possibilities; look at the possibilities they share with the floor squares.

If you look carefully, you'll see that in box 7, the roof squares are the only squares that can contain a 7. This means that, no matter what, one of those squares must be 7 - and from this you can conclude that neither of the squares can contain a 9, since this would create the "deadly pattern"! So you can remove 9 from H1 and H2.

This is an example of a

And, as you might expect, there is a **Type-4B Unique Rectangle** variant, in which the floor squares are not in the same box, and you can only look for the conjugate pair in their common row or column. For example:

In this case, since 2 can only appear in row E in the roof squares, 5 can be removed from both of them.

As Type-4 Unique Rectangle solutions "destroy" the Unique Rectangle, it is usually best to look for them only after you've done any other possible Unique Rectangle reductions.

In this case, since 2 can only appear in row E in the roof squares, 5 can be removed from both of them.

As Type-4 Unique Rectangle solutions "destroy" the Unique Rectangle, it is usually best to look for them only after you've done any other possible Unique Rectangle reductions.

As of January 2021 we have a new Unique Rectangle elimination! Thanks to Ivar Agï¿½y from Norway who shared an example with me. Not to say that it might have been discovered elsewhere, I can't check, but please come forward if there are earlier references. Since it is vaguely related to Type 1 (in that it attacks candidates in the rectangle, not outside), and is quite simple to spot, the solver searches for it after Type 1 and before the others. However, I am obliged to call Type 5 since I don't want to re-number everything.

Any rectangle across two boxes that contains two candidates in all four cells might have two opposite corners containing only two candidates, like the {2,8} in E6 and F1. If one of those candidates is linked to the other corners with strong links (ie no other n in the two rows and two columns of the rectangle), it could create a Deadly Rectangle. We are looking for a diagonal Naked Pair - each cell can't see the other, but it is locked. If 2 (the weakly linked candidate in the pair) was ON in either E6 or F1 it would force 8 to be in the other corners. This is not allowed so 2 can be removed from the Naked Pair. Very cool.

Ivar's example can be found if you turn off 3D Medusa first.

They make good practice puzzles.

Created by Klaus Brenner

## Comments

## ... by: Karl Kirkman

## ... by: Karl Kirkman

Here is some nomenclature to help me describe my thoughts. Let the candidates in the floor cells of the rectangle be A and B. Then both of the roof cells (which comprise the pseudo-cell) also contain A and B as well as other candidates, at least one each. Let the set of all of these other candidates in the roof cells (combined) be R. (Note that R does not include A or B).

Consider naked groups. As long as you can find ‘n-1' cells such that when all of their candidates are combined with the candidates in R it gives you exactly ‘n’ distinct candidates, you have found a naked group. ‘n’ could be any number from 1 through whatever the unit will allow, so you can have naked singles on up. So let I be the set of ‘n-1' cells that are included in this group (not including the pseudo-cell). Let E be the set of cells that are excluded (not including the pseudo-cell). So the entire unit (row, column, or square – if the roof cells are in the same square) is made up of the sets E, I, and R. The ‘n’ candidates can be removed from all the cells in set E. But here is where it gets interesting. Suppose that at least one of the cells in set I includes candidate A (and that none of the cells in set I includes candidate B). So A is one of the ‘n’ candidates. Since all of the cells in set I and half of the pseudo-cell will be resolved to the ‘n’ candidates (including A), that means that the other half of the pseudo-cell will have to be resolved to B (no other candidates are left for it), so B can be removed from all of the cells in set E. Furthermore, to prevent the deadly rectangle, A cannot be the solution in either half of the pseudo-cell (since B is in one of the halves), so it can be removed from both halves of it. Finally, for the sake of completeness, observe that it is not possible for both A and B to be included by the cells in set I, because then there would be no possible solution for the ‘other’ half of the pseudo-cell (‘n’ candidates cannot be spread among ‘n+1' cells).

Now consider hidden groups. If you can find ‘n-1' cells such that when combined with the pseudo-cell, they are the only cells in the unit in which exactly ‘n’ distinct candidates occur, then you have a hidden group. ‘n’ could be any number from 1 through whatever the unit will allow, so you can have hidden singles on up. Let the sets of included cells (I) and excluded cells (E) be defined as above. All candidates other than the ‘n’ can be removed from all of the cells in set I. (If you find a group with n = 1, it doesn’t allow you to remove any candidates, but you also know there is no point in searching for larger groups). Now consider what happens if A (but not B) is a candidate in at least one of the cells in set I. As long as at least one candidate in R is not one of the ‘n’ candidates, we can go no further. But it gets interesting when every candidate in R is one of the ‘n’ candidates. This means that each of the ‘n-1' cells in set I and half of the pseudo-cell will be resolved to one of the ‘n’ candidates (including A), so the only choice left for the other half of the pseudo-cell is B, and therefore B can be removed from every cell in set E, and furthermore A can be removed from both halves of the pseudo-cell. (Note that with a hidden group, it is possible that both A and B can be included in the cells in set I, but only if not all candidates in R are included in the ‘n’ candidates, and that doesn't yield any benefit).

## ... by: Mika

## ... by: David Harkness

It seems type 3b should extend to naked quads and hidden pairs/triples/quads as well.

Cheers,

David

## ... by: Robert

a) B2 and B3 are in the same cage, and D2 and D3 are in the same cage, or

b) B2 and D2 are in the same cage, and B3 and D3 are in the same cage.

We could also have all four in the same cage, but this is precluded if we are using the "killer cage convention".

## ... by: Robert

So this idea that all regular Sudoku strategies also work on Killer Sudokus - I don't think it extends to "uniqueness" strategies, at least not without some additional assumptions on the cages.

This can be seen in Killer #5768 (June 22). Use the solver until it decides to remove "5" from cell J7. This is based on a "UR" strategy, and ultimately finds one solution. However, this is another solution that does have a "5" in J7 - the inference, that the 5 can be removed, is only valid in one of the two solutions.

## ... by: Steve Benoit

I think I've seen your type 5 Unique Rectangles before. It took me a while to remember where... Check out Sudoku Swami's "Unique Rectangles Part 2 / Sudoku Tutorial #21" on Youtube beginning at time-stamp 12:32. He refers to them as UR-Type 6.

Fantastic website by the way! I wish I had found it much sooner...

Steve

## ... by: Tapio Ranta-aho

I wonder if this can be a version of unique rectangles? This happened once and candidate 5 appeared to be correct, but I don't know that wasn't just a coincidence.

___ ___ ___ ___ ___ ___ ___ ___ ___

___ ___ ___ ___ ___ ___ ___ ___ ___

___ 12_ 12_ ___ ___ ___ ___ ___ ___

___ 12_ ___ ___ ___ 12_ ___ ___ ___

___ ___ ___ ___ ___ ___ ___ ___ ___

___ ___ 12_ ___ ___ 125 ___ ___ ___

Thanks,

Tapsa

Assuming the "12" pairs are Naked Pairs (no other candidates), this is virtually a "stretched" Type 1 UR.

Also @Andrew

I refer you to Jonathan Handojo's comment on 2021-06-9 finding that the Solver skips UR's and jumps to WXYZ, similar to my earlier comment on 2019-12-16.

I also found a sorta Type 5 in your daily puzzle for 2023-01-21 but not in the Solver order, which I will try to retrieve and send separately.

As always thanks for a fabulous website!

Ciao, Pieter

## ... by: Maggie Mc

## ... by: Anonymous

andit mentions that it is the same as two HUR type 1's.Seeing the website hasn't been updated for about 8-9 years, this type has been known for a long time already.

## ... by: Anonymous

## ... by: Pieter, Newtown, Oz

Re Type 5 URs

1. The output from the solver says, for example, "Uniqueness Type 5: removing 2 from E6 and F1 because if true it would create a Deadly Rectangle with E1 and F1". It should read "... would create a Deadly Rectangle with E1 and ***F6***".

2. I've noticed odd characters in some places in the text, such as "Thanks to Ivar Agï¿½y from Norway" in Type 5 above. Also in the Digit Forcing Chains strategy - Second Example "DIGIT FORCING CHAIN: because â€¦ -5[G9]+5[F9]-5[F1]+5[C1] ï¿½ï¿½-4[C1]+4[B1".

I find it curious that you "updated" version 2.08, twice, rather than naming them 2.09, 2.10?

Thanks as always

Ciao

Pieter

## ... by: alpreucil

Never been down except for server migrations - [Del]

## ... by: Tunner

## ... by: SG

Click on this link

## ... by: BobW

I believe the new type 5 that you posted (January 2021), has been known for some time.

It is described on the enjoysudoku forum, along with many other variations of UR's having two extra candidate cells, located diagonally. That post is dated April 2006. I ran across it while implementing UR's in my own solver last year.

http://forum.enjoysudoku.com/post26448.html#p26448

If I'm not mistaken, the one that you refer to as Type 5 is referred to as UR+2D/1SL (not to be confused with UR+2D or UR+2d).

I spent many hours going through that post and many others in the same thread. The logic for many of them is so convoluted that it made my head spin. It occurred to me that it would be much simpler just to work out the truth tables for them. One day, having nothing better to do, that's exactly what I did. I worked out the truth tables for every possible combination of UR's with 2, 3 or 4 extra candidate cells ("Guardian cells" in my notation) with every possible combination of strong links between the various cells. Summarizing only patterns that give eliminations of internal UR candidates, I found 8 different UR+2 patterns with adjacent guardians, 11 different UR+2 patterns with diagonal guardians, 18 different UR+3 patterns, and 5 different UR+4 patterns.

When I refer to the number of patterns found, I've excluded reflections and rotations of the basic patterns. I implemented the resulting rules using a lookup table in my solver.

There's not enough room here to give all of my results, but if you contact me by email, I'd be happy to share what I've done.

That sounds like thorough work, well done. I should take it all in and completely redo the UR. Big job though. - [Del]

## ... by: Jonathan Handojo

If you use the Solve Path with every checkbox ticked, you get a WXYZ-Wing evaluation. It seems to have skipped through the Unique Rectangles strategy when there's one in B7, B9, J7 and J9 that allows the eliminations: 8 from J7, 6 from J9, and 8 from B9. Either value you force into them forms the deadly pattern, so I solved the puzzle quite with ease after using this. Now I'm not too sure of the rules with this, but I only saw it through forcing.

## ... by: Pieter, Newtown, Oz

As always thanks for a fabulous website!

Solving your Daily Sudoku for 2019-11-15 I believe it has a Type 4 UR which the solver does not find.

After solving the basics we are left with this board and the solver next finds an XYZ-Wing.

If you backstep and turn XYZ and X-Cycles off, I would expect the solver to find what I believe to be a Type 4 UR in EJ89, with the floor in E8/9, and the conjugate pair of 4's in J8/9, eliminating the 5's in J8/9. However, the solver skips all URs and finds a WXYZ.

Is this correct/a bug?

Ciao, Pieter

## ... by: Niki

I ask because I just spotted one for the first time in an extreme jigsaw puzzle - it was a 3/3b (triple), and it made the rest of the puzzle collapse nicely. Despite the jigsaw shape, it still fell in two rows, two columns, and two boxes just like it ought. But I was hesitant to follow up on it, having never seen the Unique Rectangle strategy mentioned in conjunction with jigsaws.

- [Del]

## ... by: Pieter, Newtown, Oz

I am confused by the solvers handling of URs in this partially solved puzzle

I have spotted what I believe to be a clear Type 4 UR in CH45, on 1/9. However, the solver finds a Type 3b combining with the 5/8 in H6 and then the puzzle solves.

I tried unchecking URs and the solver finds a Hidden UR Type 1 and eliminates the 1 in H5. Why does it not eliminate the 1 in H4? Taking Steps to try again, even re-checking the UR box, the solver moves to a HUR in EF28.

A 1 in H4 would create the Deadly Pattern. Why isn't it eliminated? Bug?

Ciao, Pieter

## ... by: digituer

The pseudo-cell concept in Type 3 is great! I think based on this pseudo-cell concept, a hidden pair strategy can also be applied.

12 . . | . . 12AC

. . . | . . .

12 . . | . . 12AB

---------------------------

. . . | . . 12FE

If the 2 roof cells and that extra cell are the only three cells holding 12 in the unit, then FE can be removed (ABCFE can be any numbers).

## ... by: AAHoffman

A1=A2={2,4}

D1=D2={2,5}

G1=G2={4,5}

as a simple example. Solutions devolving to this can be eliminated.

Multiple floors can be bent and don't have to be in a straight line, i.e.,

A1=A2={2,4} (row)

D1=E2={2,4} (note: forms a diagonal that "bends" the structure)

D4=E4={2,4} (column)

Another possibility uses naked triples in three rows, i.e.,

{A1,A2,A3}={2,3,4} (in any combination (except naked single or pair), i.e., {23},{24},{34})

{D1,D2,D3}={2,3,4} (i.e., {24},{34},{23})

{G1,G2,G3}={2,3,4} (i.e., {34},{23},{24}).

All these patterns occur often enough to be useful.

By the way, not sure how nomenclatures are standardized but I have named these "meta" strategies instead of naked rectangles... because they are a rule dependent on a rule of the game. When three digits are involved as above, I call them "triple meta".

## ... by: Harold

In the following Sudoku I'm thinking of D3, D6, F3, F4, H4, H6. I know the 1 and 6 in F3 can be eliminated by other strategies but I find it an easier strategy than some of the others, (including XY chains),

Load here

H

## ... by: Barry

## ... by: Thinkist

I came to a clearer realization on why I dislike uniqueness strategies: they mess with the solution count if a sudoku happens to have multiple solutions, erroneously decimating otherwise valid solutions. I point you to an example of mine

here. It has 63 solutions, but 4 URs erroneously deduce it to 3, and then one use of the BUG strategy erroneously deduces it to 1. So, this makes the puzzle appear valid when in fact it should have 63 solutions all along and the solver should fail to complete correctly. The chaining, netting, and simple combinatoric strategies don't mess with the solution count. In those cases, simply because there is more than one solution, the solver will merely fail to complete correctly rather than throw up an error. I believe that any puzzle (or "puzzle", if you will) that throws up a contradiction should have 0 solutions from the start, not because of a uniqueness strategy.Re-reading your reply to my first comment, I'm glad that you've not decided to add a solution count before any logical solving takes place, as that would of course ruin the purpose of using the solver, which is to showcase a bottom-up approach to sudoku solving rather than cheat it with a brute-force top-down approach.

And for the record, I realize that any valid sudoku puzzle should have only a single solution. The point I'm trying to make is that uniqueness strategies aren't the best way to find it (which is why I still have them unchecked in the solver). In essence, they assume what you are trying to prove (via chaining, netting, and simple combinatoric strategies): that the puzzle has a unique solution.

The problem with a) is that I don’t know of a solver that claims to solve all puzzles (certainly I don't pretend mine is the most advanced and I have a long job queue of improvements) and the problem with b) is that I'm sure a non-valid puzzle could accidentally complete giving the impression of a single solution solved logically. Wish I had one to hand, but I strongly suppose it.

I think there are good reasons to set aside uniqueness strategies especially from a purist POV but I've always considered muli-solution puzzles to be broken or mis-defined puzzles. They look like ducks, walk like ducks, but are not in fact ducks. They are another class of object, to be filtered away. That's not to deny they are interesting in their own right but they are outside the scope of the solver (mine at least).

However there is a good practical reason for using logic rather than brute force for solution counting, as the attached graph tries to show. It’s a crude diagram I knocked up to illustrate a point in a previous correspondence, but there is a point at which brute force becomes much slower than logic and it’s a function of clue density.

- [Del]

## ... by: Master Shuriken

If you have, for example empty boxes in A1, A2, H1, J2, H9, J9 with 2 clues (1,2) in each box, this wouldn't be possible.

You could extend this *uniqueness chain* as long as you like through the puzzle and extend it to triples, too, (as long as you make sure that there would be 2 solutions if everything else was filled in).

Like the unique rectangle, it would tell you something which is not possible, and could help eliminate clues, etc.

## ... by: dshaps

Quote

If you look carefully, you'll see that in box 7, the roof squares are the only squares that can contain a 7. This means that, no matter what, one of those squares must be 7 - and from this you can conclude that neither of the squares can contain a 9, since this would create the "deadly pattern"! So you can remove 9 from H1and H2.

Unquote

How is it that deadly pattern will get created, if either of roof squares contains 9. Can you please elaborate?

## ... by: PCForrest

"In the 'Type 1 Unique Rectangles' example (Unique Rectangle Figure 2), can't the same explanation be used for the 2 and the 9 in D2? Logically, it doesn't make sense to eliminate the 2 and the 9 in D1, but not in D2. It seems that by eliminating the 2 and the 9 in D1, we move the deadly rectangle corner over to D2. Should we then use the same logic to eliminate the 2 and the 9 from D2? Or are we ignoring that square for the sake of the example?"

D2 does not a rectangle make, let alone a unique rectangle. To be valid, all four cells must occupy exactly 2 rows, 2 columns and 2 boxes. Logically, it makes perfect sense to eliminate both 2 and 9 from D1 but not from D2, for the simple reason D2 must take the value 2 in the final solution. If you eliminate the 2, you immediately invalidate the puzzle. The logic that applies to D1 does not apply to D2. We are ignoring D2 for the simple reason it plays no part in this unique rectangle.

## ... by: Sherman

For example, figure 7 in the Type3b UR discussion has strong links between all four 6's in the UR. If 7 is placed in cell E6, 6's must be placed in cells E4 and J6, forcing a 7 in J4 - a deadly pattern. So 7 cannot go in E6. Likewise, 7 cannot go in E4.

The HUR logic can also be applied to figure 6, removing a couple of 3's.

Type 4/4b can be analyzed with HUR, which results in the same eliminations as UR. So in some sense Type 4/4b is really part of the HUR logic.

When you find a Type 2/2b or 3/3b UR, it pays to look at the strong links in it to see if HUR can also be applied. Apply HUR after UR, as it destroys the deadly pattern.

## ... by: Dan, UK

how about this:

http://www.sudokuwiki.org/sudoku.htm?bd=000764381364010572187235000013082607000601035000090018200000159001020700070150820

my solver says:

Unique rectangle type 3 at r2c4, r2c6, r8c4 and r8c6 allows 4 and 6 to be eliminated from r8c1, r8c2

you can see a screenshots from my solver here:

http://danhen.xf.cz/pics/shared/su_ur3_01.jpg

http://danhen.xf.cz/pics/shared/su_ur3_02.jpg

explanation: floor if clearly defined. As the roof contains additional candidates defined as 346, we can search for TWO cells, which contains some of these candidates, but only them, as defined in naked triple rule. As we found two cells 46 and 346, we can consider this as triple (two found cells and two cells from the roof of unique rectangle) and eliminate candidates 346 from the other cells in the unit (as seen on screenshots).

HODOKU solver can recognize this 'triple-extended-type 3' as well :)

http://hodoku.sourceforge.net/en/show_example.php?file=u301&tech=Unique+Rectangle+Type+3

The same logic can be used for searching for 'quad', that means four additional candidates in the roof of rectangle, and we'll be looking for THREE cells in a unit, which contains some of those candidates, defined by the rule of naked quad. My solver should be able to solve such a problem, but unfortunatelly I didn't find any game to match this problem :( If ANYONE is able to share game with such a kind of unique rectangle, feel free to post it :)

Dan

## ... by: Random Sudoku Enthusiast

Logically, it doesn't make sense to eliminate the 2 and the 9 in D1, but not in D2.

It seems that by eliminating the 2 and the 9 in D1, we move the deadly rectangle corner over to D2. Should we then use the same logic to eliminate the 2 and the 9 from D2?

Or are we ignoring that square for the sake of the example?

## ... by: jaswant singh =sriganganagar INDIA

## ... by: DanielKlaus

there are 7 eliminations.

## ... by: JesseChisholm

Picture the group "B" in figure one (where "A" is the Deadly Pattern)

Add an expanded copy of group "B" such that:

* there are four boxes with a pair of 7/9 cells

* there are four rows with a pair of 7/9 cells

* there are four cols with a pair of 7/9 cells

These eight 7/9 cells together form a Nested Deadly Squares pattern.

-Jesse

## ... by: gerp124

I can accept that they're part of the game, and may be the only way to solve a board sometimes, but it seems like such an arbitrary rule. What's so special about singularity? Yes, that was intended as a joke, but I've never seen any aesthetic offense in a puzzle with more than one solution. Except that a single solution cannot then be published. But I've already made my case regarding published solutions :-)

## ... by: gerp124

And thanks very much for such a wonderful website- for years, until I discovered this website, I was completely befuddled as to the solutions that are always published- they are not needed, and they are not helpful. If you can finish, you don't need the solution, and if you _can't_ finish, you don't need the solution, rather, you need to know _how_ to get the solution. And this website has provided an wonderful practical education in reasoning via sudoku!

## ... by: gerp124

In the Sudoku of the Variety section of Minneapolis Star Tribune for Sunday has these values _given_:

2D=9

2E=7

7D=7

7E=9

Does it make a difference, that the values were _given_, as opposed to being values that needed _solving_? The entire puzzle constitutes the _solution_, doesn't it? And these values could move, yielding a second solution, true?

Here is the board as published, and thanks for in any thoughts.

http://www.sudokuwiki.org/sudoku.htm?bd=800002000004000009060908003090004700075801930006300010500403090900000200000700008

## ... by: Sonalita

A fundamental premise of Andrew's excellent solver (and any algorithm based solver) is that the puzzle MUST have a unique solution.

It is generally accepted that the definition of a valid sudoku puzzle includes the constraint that there must be a single solution, so your statement of "not believing in uniqueness strategies" makes no sense.

## ... by: Thinkist

I am thinking of adding a solution count to the start of the solve "take step" and stop any further progress until the puzzle is fixed. Mostly it is data entry and it causes confusion is people don’t check the solution count first.

- [Del]

## ... by: B.N.Hobson

## ... by: Anton Delprado

If we look at type 2 cells from this concept then the virtual cell has a single value and is "solved". This can then be used to eliminate that value in any cell that can see both cells. So type 2 are kinds of type 3 in a sense.

This can also be extended to naked multiples in the joined row/col. For example if roof cells are in the same row and have three non-deadly values that are shared in two other cells in the row then those values can be removed in other cells in the row.

Yes, I have a feeling it might be possible to generalize and extend UR types. I'm in touch with someone else who has a good idea about re-organizing the families under different rules. I think you are on a similar track

- [Del]

## ... by: Chuck Bruno - Virginia

## ... by: Alan Freberg

Here's one that I had on hand when I wrote my previous post, but had yet to logic it out. It's fairly straight forward, though. It appears to be a Type 5b as there are three candidates in the ceiling instead of two as in Type 3 or my previous examples.

In the Daily Nightmare for Dec. 29, 2005 the floor cells contain 3/4 in 2 B/D. The ceiling is in Col. 1 where the unsolved cells contain: 6/7/9 (A1), 3/4/7/9 (B1), 3/4/6/7 (D1), 4/7 (E1) and 6/7/9 (H1). The ceiling in cells B/D1 contain the Strong Links (if we can still call them that) 6/7/9. As you explain two cell tri-values on pg. 124, A1 and H1 will be reduced to a Naked Pair whichever of the three is the true value in the ceiling cells . Delete 7(E1).

Thanks again,

Alan Freberg

## ... by: Alan Freberg

A few months ago I purchased "The Book" with the intention of first reading it and then trying to solve all of Ruud's Daily Sudoku Nightmares. "The Book" was an excellent read. It illuminated several dark corners. There are still some dark corners left, but they are my failings and not yours.

One thing I'm always trying to do with a Sudoku is to find something new or somehow push the envelope past what I know. Having read your book lead me to something of that nature while working on the Nightmares.

My solutions for the Daily Nightmares for March 30, 2006, Feb. 15, 2006, Feb. 2, 2006, Dec, 19 2005 and Dec. 16, 2005 contain a pattern that I have not seen before. They are based on a combined application of Unique Rectangles and Aligned Pair Exclusion. At first I thought they were a variant of your Type 3 URs, but, since they allow for more deletions than Type 3 would, it appears that they are a new variation. I call them UR Type 5 or UR APE App(lication).

In the March 30 DN (please pardon my reverse chronological order, but if you choose to look them up it is the order in which you will find them.) the floor pair are 4/5 in Col. 5 A/C. The ceiling is in Col. 2 where the unsolved cells contain the following candidates: 4/5/6/(A2), 6/8 (B2), 4/5/9 (C2), 4/5/8/9 (E2) and 8/9 (J2). 6/9 are the Strong Link pair in the ceiling. According to the Type 3 rule there are no deletions possible. However, on pg. 122 of the chapter on APE the following rule is given--"Any two cells with only abc exclude combinations ab, ac and bc..." Applying this rule to this formation we find that the two cells B2 and J2 contain 6/8/9. However one looks at it, the values for the extra candidates in the ceiling cells are accounted for and 8/9 can be deleted from E2. Notice that 8(E2) is not one of the Strong Link candidates.

In the Feb. 15 DN the floor pair are 3/7 in Row E 1/2. The ceiling is in Row B. The unsolved cells in B contain: 1/3/7 (B1), 3/7/8 (B2), 8/9 (B4), 1/9 (B5) and 3/9 (B9). 1/8 are the Strong Link candidates in the ceiling and 1/8/9 in B4/5 fulfill the APE rule requirements. Delete 9 (B9).

In the Feb. 2 DN the floor pair are 7/8 in Col. 2 G/J. The ceiling is in Col. 6. The unsolved cells in Col. 6 contain: 3/6 (C6), 4/6/8 (D6), 4/6 (E6), 4/7/8 (G6) and 3/4/7/8 (H6). 3/4 are the strong links in the ceiling. 3/4/6 in C/E fulfill the APE rule requirements. Delete 4/6 (D6).

In the Dec. 19 DN the floor pair are 2/6 in Row H 2/3. The ceiling is in Row B. The unsolved cells in Box 1 are: 5/7 (A1), 1/5/7 (A3) and 6/7 (C2). 1/5 are the Strong Link pair in the ceiling with 1/5/7 in A 1/3 completing the APE requirements. Delete 7 (C2).

In the Dec. 16 DN the floor pair are 4/6 in Row D 1/2. The ceiling is in Row G. The unsolved cells in Row G contain: 4/6/7 (G3), 3/5/7 (G7) and 3/5 (G8). 5/7 are the Strong Link pair in the ceiling with 3/5/7 (G7/8) fulfilling the APE rule requirements. Delete 7 (G3).

If this is indeed a new application then you are the godfather of this baby as your explanation of the APE rule pointed the way for me.

Thank you very much,

Alan Freberg

## ... by: JCS

I am new to this site but I end to agree with Andi. There is no deadly pattern if R5C6 contains candidates 356.

Applying simple colouring technique to the puzzle will show that 2s can be removed from R2C8 R7C2 R8C7 and R8C9. That causes R2C8 to be a 9, R7C8 a 2, R7C3 a 9, R1C1 a 9, R1C5 a 4, R4C5 a 9, R4C6 a 7. At this stage that leaves 356 as candidates for R5C6. The puzzle can then easily be solved without applying that "unique rectangle" technique ending with a 6 in R5C6.

## ... by: csvidyasagar

1. You are absolutely correct. Type 4 and Type 4 B rely one logic that roof cell may contain a digit which is confined to that row or column or block and can not be removed. So the other digit can be easily removed from thase two cells in Roof Row. In Type 4 example, the Roof Row C has 1,5,6 in cell C1 and C3. Either digit 1 or 5 can be removed to ensure Deadly Pattern does not result leading to two solutions. But 5 is confined both in Row C and in Block Top Left or Block 1. So digit 1 can be removed from both cells in Roof Row C.

Have I confused you further ?

with regards,

CS Vidyasagar

## ... by: csvidyasagar

1. When you have three out of four corners in a unique rectangle have two digits (in this case 5,7), then the fourth corner can not have 5,7 as this will have all four corners having 5,7.This is a Deadly pattern as you can have two different solutions. But for a pure Sudoku you have to have only one or unique solution. That is possible only if you remove digits 5,7 which other three corner cells of Unique Rectangle have.

2. You aim must be to reduce maximum number of digits from cells containing multiple candidates. When you can remove 5,7 from 3,5,6,7, of cell R5 C6, you are reducing time to solve the puzzle but also reducing complexity. More candidates in ells mean more complexity and require more time to solve.

3. When you remove 5,7 from cell R5 C6, you see in column 6 you have digit 5 in cell R2C6 and digit 7 in cell R4C6. So the rule of all columns, rows and boxes to contain all digits from 1 to 9 is met.

4. I could only think of three reasons as to why you can remove 5,7 from cell R5 C6.

Have I confused you further ?

with regards,

CS Vidyasagar

## ... by: csvidyasagar

The four cells you have mentioned have the following pattern:

Col 1 Col 4

Row A 1,5,6 1,5,6

Row D 1, 5 1,5.

This is similar to Unique Rectangle Pattern 2B. That is floor cells are 1,5 and Roof cells are 1,5,6. Since these form a unique rectangle with 1,5 being common in all the four cells to avoid Deadly Pattern ( of all cells having 1,5), you have to keep 6 in Row A i.e. Roof cells. That means 1,5 will be removed later from Roof Cells in Row A.

But you can not have two 6s in Row A (Roof Row) as any row can have only one digit in one cell. So remove 6 from any other cell in Roof Row A or in the Block /Box of Roof cells.

Have I confused you further ?

with regards,

CS Vidyasagar

## ... by: Harmen Dijkstra

## ... by: Wiking 48

I'm quite new on this.

why do you bother with such a non-defined sudoku?

There are many more than 17 sulotions! Check for yourself, you see that 24, 42 in the left of your grid can be excanged.

Is there any idea to fight against non-defined?

Upper left corner can also be 1, 2 or 3 which gives a lot of solutions..

Down left corner in your example is 7 in one example which is quite common among the sulotions, 1 is more rare but possible...

## ... by: Harmen Dijkstra

. 9 . | 5 . . | 6 4 7

6 . 5 | . . 7 | . 9 3

. 7 . | . . . | 2 5 8

-----------------------

. 5 6 | . 7 . | 9 . .

. . 7 |9 . 5| 3 8 6

. . 3 |8 . 2| 4 7 5

-----------------------

. . . | . 5 .| 8 3 9

5 6 . | . . .| 7 2 .

. . . | . . 8| 5 6 4

If you filling this sudoku in the sudoku solver, the solution Count will say that there are 17 solutions. But if you try TAKE STEP then you will get only this sudoku (with Unique Rectangles):

2 9 1 | 5 8 3 | 6 4 7

6 8 5 | 2 4 7 | 1 9 3

3 7 4 | 1 9 6 | 2 5 8

-----------------------

8 5 6 | 3 7 4 | 9 1 2

4 2 7 | 9 1 5 | 3 8 6

9 1 3 | 8 6 2 | 4 7 5

-----------------------

7 4 2 | 6 5 1 | 8 3 9

5 6 8 | 4 3 9 | 7 2 1

1 3 9 | 7 2 8 | 5 6 4

But there are more solutions, for example:

1 9 2 | 5 8 3 | 6 4 7

6 8 5 | 2 4 7 | 1 9 3

3 7 4 | 6 9 1 | 2 5 8

-----------------------

8 5 6 | 3 7 4 | 9 1 2

2 4 7 | 9 1 5 | 3 8 6

9 1 3 | 8 6 2 | 4 7 5

-----------------------

4 2 1 | 7 5 6 | 8 3 9

5 6 8 | 4 3 9 | 7 2 1

7 3 9 | 1 2 8 | 5 6 4

The sudoku solver said that (because of Unique Rectangles) R7C4 has to be a 6, but in this example you see that you get an other solution with a 7.

Is this a fault in the sudoku solver?

## ... by: CS Vidyasagar

2. One has to understand the logic and solution becomes quite clear. You have seleted many good illustrations which have explained the logic clearly and unambiguosly.

3. Looking forward to more such illustrations.

Kudos for wonderful work you are doing.

with regards,

Vidyasagar

## ... by: Lea Hayes

Rectangle Cells:

A(1,5,6) B(1,5,6)

C(1,5) D(1,5)

Why remove 1 from both A and B and not either of the following:

A(5,6) B(1,5)

C(1,5) D(1,5)

or

A(1,5) B(5,6)

C(1,5) D(1,5)

## ... by: Lea Hayes

I do not understand how you came to the conclusion that 1 can be removed? Why not remove 6 instead?

## ... by: John_Ha

I too was puzzling over that but I think the answer is simple. See Fig 2. (5 and 7) are unsolved in Col 4 otherwise they wouldn't be candidates in R2C4. Similarly (5 and 7) are unsolved in Col 6. Similarly (5 and 7) are unsolved in Row 5. Because the cell at R5C4 is unsolved for 5 and 7, then the centre 3x3 must be unsolved for (5 and 7).

There is therefore no way before now that the cell at R5C6 could have had either 5 or 7 eliminated as candidates because there are no solved (5 or 7) that it can see. The 5 and 7 at R5C6 could not have been eliminated by something happening on R5, nor on C6, nor in its 3x3.

So, the cell at R5C6 MUST have BOTH 5 and 7 as candidates (as well as possibly others, in this case 3 and 6).

Now assume R5C6 is 5 - we get the deadly pattern. So R5C6 cannot be 5. Now assume R5C6 is 7 - we get the deadly pattern. So R5C6 cannot be 7.

So R5C6 cannot be 5 or 7 and both can be eliminated.

## ... by: Andi

But I do not see why you can eliminate 5 AND 7 for this reason from R5C6. IMHO, if in R5C6 3,5,6 would be possible, well then this would just imply R5C4 and R2C6 being 7 and R2C4 being 5. An "ordinary" solution. OTOH, if in R5C6 3,6,7 would be possible, R5C4 and R2C6 would be 5 and R2C4 would be 7. Quite straightforward, too. My point is: both possibilities would be valid. But I cannot spot a "deadly pattern" here because to remove the ambiguity in this sitatuation, all you can postulate is that *either* 5 *or* 7 must be removed from the possibilities in R5C6. With either number removed, the ambiguity is removed, thus no deadly pattern anymore. But I just don't see why you eliminate *both* 5 and 7 in R5C6?

Am I missing something?