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Extended Unique Rectangles

Unique Rectangles can be larger structures than simple 2 by 2.
In this extension I show how the solver can now detect 2 by 3.
Just as a Naked Pair or Hidden Pair can be extended to a Triple, so can the cells that form Unique Rectangles.

A Triple is any three cells with a total of three candidates left. That doesn't always mean
[1,2,3] + [1,2,3] + [1,2,3]
Which is exactly three candidates in three cells. Other Triples will have some digits missing, like
[1,2,3] + [1,2,3] + [2,3]
[1,2] + [1,2,3] + [1,2,3]
[1,2,3] + [1,2] + [2,3]
And the ultimate skinny triple will have two out of three in each cell:
[1,2] + [1,3] + [2,3]
But all of these have - in total - three candidates in three cells.

Type 1

Now, 2 by 3 is a bit of an odd shape. Other extensions, like X-Wing to Swordfish jump from 2 by 2 (four cells) to 3 by 3 (nine cells). We don't need to search for a nine-cell 3 by 3 although they do theoretically exist. Take a look at the diagram to the right. In the left hand side marked A I've drawn the full 3 by 3 Deadly Rectangle.

Note: it must occupy exactly three rows, three columns and three boxes.

To the right labelled B I've marked out a candidate distribution of 1s, 4s and 8s. For each row and each column you'll see a Triple. If we've arrived at this state we're in trouble because there are many ways to settle those cells in all combinations of 1, 4 and 8. The only thing that saves us are the extra candidates marked in green.

Here's how we go from the 3 by 3 ideal pattern to the much more likely 2 by 3 version: Take away any one of the columns with numbers and you still have a multiple solution problem. For example, if we took away column 8 (and ignored the green extra candidates for a moment), we could still fill column 7 with 1 + 8 + 4 (going down the rows), 4 + 1 + 8 and 4 + 8 + 1. Same for column 9. We could fill it with 4 + 1 + 8 or 4 + 8 + 1 or 8 + 4 + 1. If you take any column away it looks like the problem stays a problem. So conclusion - don't worry about looking for 3 by 3 Deadly Patterns - they will come out in the wash when you find 2 by 3.

Example Extended UR : Load Example or : From the Start

In this first example we have an Extended Deadly Pattern based on the three candidates {1,3,5}. The yellow cells mark the pattern. Notice this pattern is vertical like the theory example above - but as we are not looking for (and don't have) a 3 by 3, we need to look out for the pattern working horizontally as well. One of these is below.

The only type of UR I've implemented in the solver, so far, is Unique Rectangle Type 1.

This says that the odd cell out - the cell with an extra candidate or two - is the key. Those extra candidates must be part of the solution. If they didn't exist we've have a full Deadly Pattern and we need to avoid that at all costs. So 6 in C1 is the extra candidate - it must be the solution, so we can remove 1 and 5 from C1 and progress.

Example Extended UR 2 : Load Example or : From the Start

Here is a similar Extended Unique Rectangle based on the cells AEH78 (which just means cells on rows A, E and H and columns 7 and 8). The Triple is {4,7,9} and we have 1 and 6 - on cell E8. That's the get-out-of jail card avoiding the bigger Deadly Pattern. Take away the 4 and 7 from E8 and we're back into single solution territory.

Example Extended UR 3 : Load Example or : From the Start

Finally a horizontal example.

The Triple {3,7,8} sits over the two rows D and E. There is a mixture of bi-value and tri-value cells but we clearly have two triples with the same values very close to each other. We have to use the extra candidates provided. There is a lone 1 in D2 that gets us out of trouble.

Type 2

Type 2 extends the same idea found in Unique Rectangles to three boxes and three candidates. The Floor (yellow cells) is now four cells instead of two. Some of these cells can only have two candidates in them, but all four must have exactly three in total.

We can look in both the row and the column for the Roof cells (orange). We need to find two more cells that are in a different box and have exactly one extra candidate in both. In the example our Deadly Rectangle triple is {2/3/4}. The extra candidate in G23 is 6. The logic is that 6 must appear in either of G2 OR G3 since without it a deadly Rectangle on {2/3/4} would be created. Knowing this allows us to remove all 6s in the row and box.

If the orientation of the Type 2 was the other way round we'd eliminate in the box and column.

Many thanks to prompts in the comments below and to Ross Trusler for the example he found in the first puzzle in the Fireworks strategy. Turn off Fireworks to get it to show.

Type 4

We can extend the Type 4 Unique Rectangle in an interesting way as decribed in a comment below by Pieter. I'm going to use his example first.

Instead of a single locked pair of cells we can have two - if they are in different boxes and the roof is in a different box as well.

To see this example you do need to turn off the Tough and XY Chain strategies so the solver can find it. The double Locked Pairs on DF35 (locked by the 4's) which effectively solve to a 1/3 Locked Pair, which then become a UR Type 4 with D7/F7 locked by the 1's, eliminating the 3's. You can see there are no other 1s in column 7 or box 6.

I will admit, instances of the Extended Unique Rectangle are not common, but if you are just getting into Unique Rectangle you'll be looking for the very common Type 1. I think it's worth keeping an eye out for this formation at the same time. You will notice a similar cluster of common candidates and it's just possible - if you remember the three boxes spread rule - that you'll find one when you're looking for the 2 by 2 version.

Go back to Unique Rectangles

Continue to Hidden Unique Rectangles

Comments

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CommentsTalk

... by: martin

Friday 22-Nov-2024

I would think that type 1 and 2 extension are somewhat contrived and difficult to identify
The type 4 extension, however, is what I would call a transitive extension.
Theoretical at least, it applies to longer chains as well.
There is a couple of patterns of a similar nature (closed in themselves, hence deadly), that are easy to spot.
In either case, drop in a different candidate anywhere for a free out of jail card. Otherwise, you will have to admit the possible multiple solution.

A 90-degree bend:

(1,5) o o | o (1,5) o | o o o

(1,5) o o | (1,5) o o | o o o

o o o | o o o | o o o

- - - + - - - + - - -

o o o | (1,5) (1,5) o | o o o

o o o | o o o | o o o

o o o | o o o | o o o

and as already pointed out by BCorso, the permutation of 3 digits:

(1,5) o o | o (1,5) o | o o o

(1,9) o o | o (1,9) o | o o o

(5,9) o o | o (5,9) o | o o o

- - - + - - - + - - -
etc.

Of course the transitive combo remains intact as stated as EUR Type 4:

(1,5) o o | o (1,9) o | o o (5,9)

(1,5) o o | o (1,9) o | o o (5,9)

o o o | o o o | o o o

- - - + - - - + - - -

While these are out there and spotable, there are way more.

REPLY TO THIS POST

... by: Pieter, Newtown, Oz

Thursday 2-Feb-2023

@Tapio Ranta-aho.
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

REPLY TO THIS POST

... by: Anonymous

Tuesday 25-Oct-2022

Type 2 and Type 3 EURs are also possible.
Hidden EURs too.

Andrew Stuart writes:

As of Jan 2024, added Type 2. Will work on Type 3 next. Thanks!

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... by: Anonymous

Tuesday 25-Oct-2022

I think the term you're looking for is not "double locked pairs", it's "ALS".

Andrew Stuart writes:

I don't think there are enough surplus candidates (ie one extra) to be an ALS

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... by: Pieter, Newtown, Oz

Friday 27-May-2022

Hi Andrew
I believe I have found an Extended UR in your Daily Sudoku for 2022-05-7, not a 2x3, but a 3x2. After solving the basics (and turning various strategies off?) one can arrive at this board (as I did on paper):
bd=800000251001507830053812607145936782730285410008174503314628975080751300500000108

The EUR is in AJ456, however, the solver does not find it, I think cuz you haven't implemented this variety yet?

Also, I noticed several of my and others' earlier comments which you may not have seen or replied to yet.

Thanks and kudos as always!

Ciao, Pieter

Monday 11-Mar-2013

To bcorso,

I believe example 3 proves this theory to be incorrect. Unless I misunderstand what you're saying, the 38 pair on E2 and the 37 pairs on column 6 would imply that 7 & 8 are the only choices for the two cells in column 7. However, J7 eventually evaluates to be 7, meaning you really only had a choice between 3 and 8 for three of the cells in the extended UR.

To Andrew,

Great website! If I had an extra arm I'd give you three thumbs up!

REPLY TO THIS POST

... by: bcorso

Sunday 10-Feb-2013

Hey Andrew,

Regarding my previous post. I've come to the conclusion that the only deadly pattern available for a 3x2 extended UR is of the form

xy | xy
xz | xz
yz | yz

That is, to get a 3 x 2 deadly pattern each row must only contain 2 candidates, those 2 candidates must be the same for each row, and those 2 candidates can not be the same for any column. This leaves only permutations of the above patterns as possible deadly patterns.

Proof:
First off, each cell MUST only contain two candidates AND those two candidates MUST be the same for both deadly cells in that row. This is because a deadly pattern means that if ALL other cells are filled in on the board, there still exists multiple solutions. Thus, if all other cells in the rows are filled in, there are only two candidates that can be placed in the deadly cells of that row, and those candidates are the same for both of the cells.

Second, the pairs in the columns cannot be the same. This follows from the fact that if there were two pairs that matched, there would no solution (instead of multiple solutions). For example, if xz was repeated twice in the column:

xy | xy
xz | xz
xz | xz

Then there would be no solution because the second and third column would have to have the same solution, which is impossible.

REPLY TO THIS POST

... by: bcorso

Sunday 10-Feb-2013

Hey Andrew,

Please let me know if I am missing something important, but I see a flaw in the extended URs that you present here.

The way I understand it, a deadly solution only exists if the candidates in the "deadly" cells cannot see any other cell with those candidates (because the deadly cells cannot solve themselves, thus they need other cells to make further eliminations).

For example, in the 2x2 deadly URs it is deadly because the 4 cells contain a naked pair in the row, col, and box and the "naked pair" technique will remove those candidates from every cell those deadly cells can see. Thus, the naked pairs have multiple solutions because they cannot solve themselves.

Now, in the extended URs I image you would need the same circumstance, the deadly cells cannot see any other cells with their candidates, otherwise that cell can be used to break the deadly pattern. This brings me to your examples here. In both examples you only require a triple in the row OR col. but this obviously leaves cells (outside the deadly cells) with the same candidates which are in the same box and col OR row respectively.

I'm interested in hearing your solution, thanks Andrew.

REPLY TO THIS POST

... by: Dale Kloss

Sunday 16-Dec-2012

Help!!! I am having a problem with your first Extended Unique Rectangle example. Let me explain by tabulating the possible solutions of [C1,C3,E1,E3,G1,G3].

First the two Extended Deadly pattern solutions:

1 5 5 1
3 1 1 3
5 3 3 5

I don't see a problem with these as horizontally the 2 solutions are deadly, the patterns are deadly inside the 3 boxes, and they are deadly vertically. Since we are told the solution to the entire puzzle is unique they can be eliminated as possible solutions.

Now let's tabulate the other two possible solutions:

Pattern A: 1 5 Pattern B: 6 | (1,5)
5 3 (1,3,5) | (1,5)
3 1 (1,3,5) | (1,3,5)

This is where I have the problem. I don't see Pattern A as Deadly with respect to row E of the first two solutions nor Deadly with respect to row G of the first two solutions since the candidate sets are different. Nor is it Deadly with respect to Boxes 4 and 7 since the candidate sets are different.
Conclusion ? Either Pattern A or B destroys the Deadly pattern. We can't say that just the 6 destroys the pattern ?

One bright spot: By comparing Patterns A and B we see that we can eliminate the 5 in cell C1.

Andrew Stuart writes:

I've had another good look at this example.

Let's call the first two patterns you described as C and D as they are unnamed.

Pattern B is not deadly because it's actually going to be the solution, ie 6 in C1 and some combination of the other candidates in the other cells which will come out in the wash later.

Pattern A is deadly with both C and D. Now, A with C doesn't look swappable in isolation because they both share 1 and 5 in C1 and C3. But they do swap candidates in cells E1,E3 and G1,G3. So in isolation just thinking about A and C we are effectively using four cells out of 6 to prove a point which seems insufficient because C1 and C3 are not being swapped. However, the super-set pattern AC can be swapped with pattern D. So we have three patterns which in combination can create swappable cells. This is the neat thing about triples - we don't need every candidate to be swappable in all cells, just as long as each cell and each candidate can be swapped in at least one combination.

Another way to think about it: It would be easier to show the total swapability of all patterns if we had a 1,3 and 5 in all six cells. The fact we are missing a 3 in C1 and C3 and a 5 in E3 merely restricts the number of swappable patterns available to patterns A, C and D. The minimum number of swappable patterns is two and we would get that in slightly more rare situations where all cells contained 2 candidates (apart from the extra one candidate) similar to the second example.

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... by: Ehsan Vessal

Sunday 7-Oct-2012

Very cool Andrew! I just wanted to note that Peter Gordon in his Mensa Guide to solving Sudoku somewhat touches on what you are talking about here when he introduces what he calls "Gordonian Polygons". He has some examples here of not only rectangular shapes, but pentagonal and hexagonal shapes formed by possible candidates.

Andrew Stuart writes:

Interesting, I'll check into that

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