After noticing it recently in the puzzle below, I have started using a pattern which resembles a weaker but more common version of Type 5; in short, an elimination still is possible even if only a single strong link is present, as opposed to the full rectangle being linked strongly.

If a rectangle across two boxes has opposing corners with only the same two candidates X/Y, and if one candidate X has just one strong link to an adjacent corner, then X can be removed from the other corner. The reason is that placing X there would force adjacent double Y, but then the strong link demands opposite X, which is the ambiguous pattern we are trying to avoid.

This is an example puzzle, which reaches that scenario if XY-Chain and 3D Medusa are disabled:
000000025002753004900800000280010050030408010010020078000009007300586200690000000

It occurs in the rectangle AC37, and it becomes available immediately after A7 becomes 3/7 along with its opposite C3, also immediately before the solver finds other unique rectangles. From here, it suffices to note that 7 is strong in row C—that alone is enough to remove 7 from A3. Independently of that, because the other candidate 3 is strong in row A it can be eliminated from C7.

As it happens, 7 was also strong in column 7 (and 3 strong in column 3), but I emphasise that only one of those strong links are required for an elimination. The reduced need for strong links makes this partial pattern much more common; unlike the full Type 5 pattern (or the very useful Hidden Unique Rectangle pattern!), neither candidate is required to form an X-Wing.

Personally I imagine it as a "partial unique rectangle" or perhaps a "unique rectangle edge elimination" given that demanding uniqueness risks eliminating a candidate from one edge, but in the context here it is described very well as a lighter version of Type 5!

It seems like Type 5 Unique Rectangles is a variation of Type 4 (Not entirely sure).

Type 3 can be extended to hidden pairs/triples/quads similar to how it works with naked tuples by treating the floor cells as a single pseudo cell. Might this become type 3c?

In this example, the UR is in BC69. Since 347 form a hidden triple in C4, C5, and C69 (the pseudo cell), 5 and 6 may be eliminated from C4 and C5. It does not allow eliminating 56 from the floor cells, however.

+-----------------+------------------+--------------+
| 7    5   2      | 9     36   8     | 16  4   136  |
| 3    48  48     | 2     1    56    | 79  79  56   |
| 69   1   69     | 34567 3456 3567  | 2   8   356  |
+-----------------+------------------+--------------+
| 145  6   3      | 457   8    2     | 147 17  9    |
| 14   2   7      | 346   9    36    | 5   16  8    |
| 8    49  459    | 1     456  567   | 467 3   2    |
+-----------------+------------------+--------------+
| 2    7   1      | 356   356  9     | 8   56  4    |
| 4569 489 45689  | 56    2    1     | 3   569 7    |
| 569  3   569    | 8     7    4     | 169 2   156  |
+-----------------+------------------+--------------+

Load Sudoku

Unfortunately, the naked pair 56 in C1 and C3 eliminates the 6 from C6 and C9, breaking the UR.

Greetings, Andrew.

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

Cheers,
David

Just to follow up - in the first graphic on this page, the "deadly pattern" only remains deadly in Killer Sudoku if either

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".

I've just sent a detailed email, but the short summary here is - be careful applying the Unique Rectangle strategies to Killer Sudokus. It all depends on the particular "cages" the four points of the rectangle are in.

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.

Hi,
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

Hello .
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.

Thanks,
Tapsa

I played a game of brainium sudoku in which the four corners of a square had the same two numbers could only have those two numbers because they were the only spaces left in their respective rows. This should not be possible. I can send a screen shot. Can you explain?

HoDoKu already has a documentation of Type 5 (under the name Type 6), and it 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.

Isn't a Type 5 UR just two Type 1 HURs on the same spot?

Hi Andrew
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 Ag�y from Norway" in Type 5 above. Also in the Digit Forcing Chains strategy - Second Example "DIGIT FORCING CHAIN: because … -5[G9]+5[F9]-5[F1]+5[C1] ��-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

Andrew! Where is your Solver??? The wonderful one with the black background. I've used this solver for years to solve the puzzles that puzzled me, and I found it very user-friendly and helpful. NONE of the other solvers I've tried hold a candle to it. I want it back! Any reason why it has disappeared from the site??? Thanks! Al Preucil

Through out this section you repeatedly state that the "rectangle" consists of 2 rows, 2 columns, and 2 boxes. But, is it not true that the process is the same if there are 2 rows, 3 columns, and 2 boxes? Or 3 rows, 2 columns, and 2 boxes? I think the boxes are the important link. This may not be true for all types but it seems to work for type 1.

There is a Sudoku board I would like you to look at for another example of Type5.

Click on this link

Hi Andrew,
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.

Here's a puzzle that I purchased from one of your Diabolical packs (and I have been enjoying a lot of these. I hope it's fine with sharing one of those puzzles here.) load here

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.

Hi Andrew
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

I'm curious why the Unique Rectangle strategy isn't included in the Jigsaw Sudoku solver. Do other strategies there render it irrelevant?

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.

Hi Andrew
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

It's a pleasure reading your pages. I've learned a lot. Thanks Andrew!
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).

I often extend this strategy to 3, 4, or even more naked pairs forming multiple "floors" or a multi-story structure, i.e.,
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".

Using the type 1 example, couldn't the removal of just the 2 or just the 9 prevent a deadly pattern? Of course, you wouldn't necessarily know which one to eliminate at that point, but isn't it possible that another solving technique might eventually eliminate one of the primary candidates thus no longer making it part of a unique rectangle? I know the methodology works and that neither candidate will be part of the final solution, but just don't see the logic of being able to say that both candidates should be eliminated.

Hello again...

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.

You could have another strategy which is an extension of unique rectangles (extension of both 2x2 and 3x3).

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.

Under Type 4, following comment appears.

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?

@Random Sudoku Enthusiast

"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.

Two of the Unique Rectangle examples on this page can also have the Hidden Unique Rectangle logic applied to them, resulting in even more eliminations.

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.

type 3 can be extended to search for 'triples', even 'quads'. In your examples just a pairs are used - 17 in first one, 83 in second one.
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

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?

ref.sudoku puzzle creater UR type 4 =focuses on UR candidates -- U R of 3,9 see row J =I remove the 3,3 ofJ2 J3 OTHERWAY= J2,J3,J78 form subset triplet [157].hence 5 of J5 will be removed leaving there inJ5 the 2. In turn J4 will become 3. & in turn3 will be removed from J2andJ3...

I found a wonderful UR type 2 example in my own created sudokus:
there are 7 eliminations.

I didn't see the pattern on your site, but there is a "Nested Deadly Squares" that is quite rare, but could also be looked for and avoided with much the same techniques as Unique Rectangles.

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

OH! And while I have your attention, I would like to say that I dislike unique rectangles.

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 :-)

Thanks very much for the reply Andrew.

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!

I don't see much recent activity, but I hope this gets seen- I believe I have found a published contradiction, and so I'm wondering about the phrase above: 'published Sudokus have only one solution'.

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

@Thinkist
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.

Harmen Dijkstra has a fully valid point. This is why I don't believe in uniqueness strategies and have them unchecked in the solver. If a Sudoku has multiple solutions, so be it.

A word of caution. The Unique Rectangle doesn't work with Samurai Sudoku (five interlocking Sudoku puzzles). One section may have two answers, but one of them is wrong in relation to rest of the puzzle. i have recently had two cases where the Unique Rectangle gave the wrong answer.

There seems to be a lot of room for extension of type 3 or at least the concept of treating the two roof cells as a single virtual. This virtual cell has the union of values in the two roof cells excluding the deadly values.

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.

Isn't the example for type 3 also a type 4B, allowing us to remove the 4's from cells B2 and B9?

Hi Andrew,

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

Hi Andrew,

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

Andi and John_Ha

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.

Dear Denis,
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

Dear Andi,
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

Dear Lea,
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

Yes the point is, that there are indeed more solutions. So, if you use unique triangles, and get a solution, you will not know that your solution is a unique solution.

Hi Harmen Dijkstra,
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...

I made this sudoku:

. 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?

1. You have given very simple and lucid explanation of difficult concept of Unique Rectangles of four types. One comes across such situations in number of Sudoku puzzles of varying degrees of difficulty. The clear understanding of unique rectangles helps one to get out of log jam one is subjected to frequently in these interesting puzzles. I do not know whether any more explanation can be given to what you have done. Only one has to read your article may be twice but keep the most important lesson you brought out loud and clear that " DO NOT HAVE DEADLY PATTERN OF HAVING FOUR CONJUGATE PAIRS" as that leads to multple solutions. Sudoku permits only unique solution.

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

Just to clarify my question:

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)

Re: Type 4 Unique Rectangles - Cracking the Rectangle with Conjugate Pairs

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

Andi

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.

I do not understand this strategy. In the first example for Type 1 UR, I understand that if 3 or 6 would be removed from R5C6, you would end up with a deadly pattern. That's fine so far.

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?

Shouldn't your Type 4B example be "solved" using the Type 2A method? Using this method the 6 in R1C3 and R3C5 would be eliminated, forcing one of the "roof" cells to be a 6. I believe the end result is the same. This would be consistant with only using Type 4 when no other solution is possible.

Dennis