All the Unique Rectangles discussed on this page have had at least two bi-value cells (or a Naked Pair, if you like). An interesting situation emerges when we have only one bi-value cell and the Unique Rectangle is buried, or 'hidden', under lots of other candidates. The conditions have to be just right, but this formation is surprisingly common.

In Figure 1, the cells D3, D7, F3 and F7 (also known as DF37) form a rectangle with a potential deadly pattern on [1/6]. We want to avoid reducing this rectangle to 1/6 in all four cells. There is a bi-value cell at F7, but the other corners contain a clutter of other candidates.

The types of Unique Rectangles that we’ve examined so far simply won’t work here. However, all is not lost because something interesting is going on in the corner opposite to F7. 1 in D3 is part of two strong links in the row and column - that is, 1 occurs only twice in row D and column 3. Our 1 in this rectangle can have only one of two possibilities.

Looking at D3, lets make "Option A" the case where D3__is__ 1. The pairs or "strong links" remove 1 from in D7 and F3. Option B is where D3 __is not__ a 1. That would put a 1 in D7 and F3.

The types of Unique Rectangles that we’ve examined so far simply won’t work here. However, all is not lost because something interesting is going on in the corner opposite to F7. 1 in D3 is part of two strong links in the row and column - that is, 1 occurs only twice in row D and column 3. Our 1 in this rectangle can have only one of two possibilities.

Looking at D3, lets make "Option A" the case where D3

Let us conjecture that the solution to D3 is 6. That forces the 1s to take on option B *because they would be the only remaining 1s in that row and column (D7 and F3)*. The implication of this option is that the only candidate for F7 is 6. But, wait a moment - this is exactly the deadly pattern we need to avoid - two numbers in two rows, two columns and two boxes. Since none of these cells is a clue, the puzzle could have two solutions. We can swap the 1 and the 6 around. This makes option B unviable.

So surely we can fix the 1s in D7 and F3? Not quite. We haven’t taken into account the clutter of other clues in the corners of this rectangle. But we can say something about the 6 in D3. We’ve excluded it for the reason above, but option A also excludes the 6 in D3. Whichever way round F7 is, D3 cannot contain a 6, so we can remove it from that cell.

So surely we can fix the 1s in D7 and F3? Not quite. We haven’t taken into account the clutter of other clues in the corners of this rectangle. But we can say something about the 6 in D3. We’ve excluded it for the reason above, but option A also excludes the 6 in D3. Whichever way round F7 is, D3 cannot contain a 6, so we can remove it from that cell.

In Figure 2, the cells B1, B3, D1 and D3 (also known as BD13) form a rectangle with a potential deadly pattern on [6/8]. In **type 2** we have an identifiable Floor consisting of B1 and B3, both bi-value cells. And the Roof contains a clutter of other candidates. We still want to avoid reducing this rectangle to 6/8 in all four cells.

Type 2 starts with a Naked Pair, 8/6, and we're checking for which strong links exist on the pair candidates. The Naked Pair is obviously a double strong link on both numbers, but candidate 8 has a strong link to the Roof in column 1. From this we can start using the logic of Type 1.

Type 2 starts with a Naked Pair, 8/6, and we're checking for which strong links exist on the pair candidates. The Naked Pair is obviously a double strong link on both numbers, but candidate 8 has a strong link to the Roof in column 1. From this we can start using the logic of Type 1.

Is 6 viable for either Roof Cells? D1 is viable since D3 does not have to be a 8. However, D3 as 6 is trouble because it forces B3 to be 8, B1 to be 6 and because of the strong link, D1 becomes 8. A Deadly Pattern, so it is safe to remove 6 from D3. At first glance the formation looks symmetrical but its the other 6s and 8s (or lack of) in the columns that give us the implications of the Hidden Rectangle.

In common with Unique Rectangles there is a type B for this strategy, where the floor is across two boxes rather than in the same box.

The yellow cells are the floor with conjugate pairs 1/7. The Deadly Rectangle extends to E3 and H3 (orange) where 1 and 7 are also present among other candidates. The strong link on 1 between E2 and E3 means that a 7 in H3 would create a Deadly Rectangle.

Thanks to Jerry Foil of Virginia, USA for providing the first example. Interestingly this type of Hidden Unique Rectangle is almost twice as common as the Type 1 and three times as common as Type 2.

The yellow cells are the floor with conjugate pairs 1/7. The Deadly Rectangle extends to E3 and H3 (orange) where 1 and 7 are also present among other candidates. The strong link on 1 between E2 and E3 means that a 7 in H3 would create a Deadly Rectangle.

Thanks to Jerry Foil of Virginia, USA for providing the first example. Interestingly this type of Hidden Unique Rectangle is almost twice as common as the Type 1 and three times as common as Type 2.

Klaus Brenner in Germany has created some very beautiful Sudokus like the one on the right. It solves trivially up to the point where a Type 1, Type 2 and Type 2b are used consecutively. In the screen shot the Type 2B is shown with the floor in row H.

HIDDEN UNIQUE RECTANGLE Type 2b: removing 7 at J7 because of HJ47 and one strong link between H4 and J4 on 5

## Comments

Comments Talk## Tuesday 29-Oct-2019

## ... by: Pieter, Newtown, Australia

Hi AndrewAs always thanks for a fabulous website!

Solving your Daily Sudoku for 2019-09-28 I believe it has some Type 2b HUR's which the solver does not find.

After solving the basics, a UR, a 3DM, and a HUR, we are left with this board.

The solver then skips to an AIC, but I believe there is a HUR Type 2b (a double in fact) in HJ17.

The floor is a 7/8 pair in J1/J7.

The first HUR eliminates the 7 in H1 due to the strong link of 8's in H7/J7.

The second HUR eliminates the 8 in H7 due to the strong link of 7's in H1/J1.

Is this correct?

Ciao, Pieter

## Thursday 9-Oct-2014

## ... by: jaswant singh

hidden U R. type 1,UR of 1,6 is there in d3-d7,f3-f7. only f7 is clear bivalue cell -In row f, six 6 is available outside u r.- but one 1 is neither in row f nor in col 7 except in the u r .-hence other partner of 1 i.e. 6 will be removed from opposite corner -d3. the explanation in 5 paras is perhaps unnecessary ... now take U R type 2 U R of 6,8 in b1-b3 and d1-d3 .firstly take clear bivalue cell b1--six 6 is available in col one 1 at f1 and h1 but eight 8 is not available in row b nor in col 1 [except in u r ]. hence the other partner of 8 i.e. 6 will be removed from opp. corner d 3 now take clear bivalue cell in b3--8 and 6 both are avail. in col 3 hence no elimination from cell d 1 is possible j.s. sriganganagar india## Monday 31-Mar-2014

## ... by: Sheela Ravishankar

Hidden Rectangles - was confused till I read Avoidable Rectangles and came back to HUR. It really helped to understand HUR better.Also I have started appreciating the AIC technique better after reading your website. In fact I can solve almost all your Diabolic and Extreme Sudokus now.

Thanks for the good work and great help!!!

## Wednesday 16-Oct-2013

## ... by: J.S.SRAN sriganganagar INDIA

TYPE one == hidden unique rectangles ===The example does not contain FLOOR AND ROOF squares.Three cells contain more candidates and only one cell F7 is clear 1-6 the 1 has to be in F3 or F7 .so the other partner 6 will be removed from opp.corner D 3.## Saturday 14-Sep-2013

## ... by: Joris.engels@sdworx.com

I find the example in type 1 confusing, because the 1 in F7 is also part of 2 strong links, which blurs ones view "## Tuesday 10-Sep-2013

## ... by: Hassieb

I have so much trouble with Hidden Unique Rectangle. Why do we want to we want to avoid reducing this rectangle to 1/6 in all four cells. I have not had any trouble before with two values in four cells. Why 6 has to go from D3 ???## Sunday 28-Jul-2013

## ... by: TAU

A Type 1 HUR is also valid if it has two bi-valued corners, provided they are at diagonally opposite corners.## Wednesday 19-Sep-2012

## ... by: Jools

Andrew,Further to Sean's post, I've also found [at least] two examples of such a "diagonal" hidden non-unique rectangle that could be added to your solver. Both examples satisfy the existing NUR rules. You should be able to step through both grids in the solver until coming across an XY-wing, when they should become visible.

1st example

Type (i) 5/8 in F1 and G3; entering a 5 in F3 would force G1 to become one too - thus forming a "deadly" pattern, so the 5 in F3 can be removed.

Doing this enabled me to breeze through the rest of the puzzle (after also completing the XY-wing on C2-H2-G3 which removed the 5s in B3 and C3), because it immediately left just a single 5 in G3 for column 3.

Note that type (i) candidates do not have to form part of a N=2x2 pattern.

2nd example

This one actually has two types at once; the second removes two candidates!

Type (i) 2/6 in D7 and F5; entering a 2 in F7 would force D5 to become one too - thus forming a "deadly" pattern, so the 2 in F7 can be removed.

Type (ii) 6/7 in G6 and H9; entering a 7 in either G9 or H6 would force the other to be the same - again forming a "deadly" pattern, so both 7s in G9 and H6 can be removed.

Note that type (ii) candidates *do* form a N=2x2 pattern, which enables both corners to have a candidate removed.

Both examples come from my Nintendo DS Platinum Sudoku cartridge, from the 5,000,000 grids on "Pro" level. (I'm up to #3293 so far!)

## Tuesday 31-Jul-2012

## ... by: Sean Tremba

Andrew,I think I've come up with a variant on unique rectangles that doesn't seem to be covered by any of your HUR or UR strategy types. (You can certainly correct me if I'm wrong).

Suppose you have an "x-wing" formation on a given candidate, and you have identical bi-value cells diagonally opposed to each other in that formation. That is, you have two cells with xy as candidates at diagonal corners. The other two cells then must contain x with some other set of candidates (which may or may not include y). I maintain that you can then conclude that x is the solution to the two diagonal cells which have xy as candidates.

To prove this, assume that x is not the solution to these diagonally opposed cells. x is therefore solution to the other two cells in the "x wing" formation. Since the only other candidate for these cells is y, y must then be their solution. However, this leads to a deadly pattern, since you have a rectangle with y as the solution to the cells on one diagonal and x as the solution on the other; x and y could be reversed to give a second valid solution.

More concretely, (and also the example I found), suppose cells F1 and H2 are bivalue cells with 7 and 9 as candidates. In the sudoku I was working, cell F2 had candidates 1,7 and 9, and cell H1 had candidates 7,8 and 9. These cells were the only cells on rows F and H which contained 7's, and the only cells in columns 1 and 2 which contained 7's. The solution to cells F1 and H2 must then be 7. Otherwise, the solution to F1 and H2 would be 9, and the solution to F2 and H1 would be 7, which creates a deadly pattern.

## Wednesday 11-Jan-2012

## ... by: Chuck Bruno - Virginia

After importing the puzzle below and enabling only X-Wing, Y-Wing, Sword-Fish, Unique Rectangles, Hidden Unique Rectangles, and XYZ Wing/WXYZ Wing, stepping through the puzzle eventually yields a Type 1 Hidden Unique Rectangle at AC15 which enables the removal of 7 at C1. I can't see it, can you please help?Load puzzle

Thanks,

Chuck Bruno

## Sunday 10-May-2009

## ... by: John

I used this technique librally for several months with success until today when it kept giving me an incorrect solution. I just want to confirm that the floor has to be in the same box for hidden type 2's to work, right?