Sudoku Unique Rectangle Strategy with Examples

The other day, while driving I came upon a traffic signal. The red light was clearly lit, but so was the green. I was confused. And a little bit scared. A few more brown strands turned gray.

Suppose you are working hard on a puzzle. You get to the end. You see that no matter what, you cannot uniquely identify the solution. The puzzle has two (or more) valid solutions!

Puzzle solvers are enamored by the latter situation to the same extent drivers are endeared to the former.

For this reason, my opening blog codified this rule, practiced by all good puzzle creators:

  • Each Sudoku puzzle has a unique solution
This rule has some power attached to it. It is necessary, however that one has digested the previous techniques to apply this idea.

The simplest application of the unique solution rule is Unique Rectangles. However, there are some extensions of this idea that, to date, I have not found very useful. If a puzzle has no more than one solution, then

  • All possible partitions of the puzzle grid that have multiple solutions are forbidden
That seems fairly obvious. The specific application of this idea with Unique Rectangles follows:
  • The following configuration is forbidden
  • A single naked pair of candidates x,y exists in a box and a row
  • Another naked pair of candidates x,y exists in a different box and row
  • The column coordinates of both of these pairs are the same.
  • Therefor, any potential solving activity that would cause this forbidden configuration to occur is also forbidden.
  • Suppose the above configuration occurs
  • Then because of the naked pairs in two rows, two columns, two boxes
  • The puzzle is now partitioned into a large puzzle and a small subpuzzle
  • The small subpuzzle is never effected by the larger puzzle
  • The small subpuzzle has two solutions
Here is an example:

forbidden Unique Rectangle

Note here that we have pair 17 in rows 1,3 and columns a,i and boxes b2,h2. Therefor, nothing can happen to prefer a1=1 over a1=7. Then same is true for cells a3,i1,i3. Thus, the puzzle has at least two solutions that do not violate the standard rules.

The power in this technique, then, is that when one has an Almost Unique Rectangle, one can forbid something. I prefer the term forbidden rectangle over unique rectangle, but the nomenclature is fairly standard with this technique, so I guess it best to call this situation Unique Rectangle.

Here is an example of the simplest usage of this concept

Almost Unique Rectangle

In this example, clearly if a1<>9 then we would have a Unique Rectangle. Therefor, a1=9. This step could be written in a proof as:

  • AUR 17 at ai13 forbids a1=17

False Forbidden oops! Unique Rectangle

Not a Forbidden Rectangle

The example above reflects a common error. The situation above is not a Forbidden Rectangle. Note that although it looks very much like one, none of the pairs 23 share a common box. Thus, something could happen, say at cell c2, that would break the rectangle legally and not force two solutions.

Twelve Solutions

Triples with twelve solutions

There are many extensions of this idea to other shapes and configurations. I can say that except when trying to create a puzzle, I have never encountered any of them except BUG. The Bivalue Universal Grave may be treated in a later blog. A configuration like that shown above would also be forbidden, as it would have at least two solutions.

Deeply hidden Unique Rectangle

Almost Unique Rectangle Hidden Deeply

Can you find the Almost Unique Rectangle here? It is very well hidden.

If only life were as unambiguous as Sudoku!

Indicate which comments you would like to be able to see

The most pleasant of holidays to you, each and every one!!
is it
AUR at 39 at gi59, however no candidates can be eliminated?
If g8 is an '8' then you end up with the 39s at g9/i9 and g5/i5. Therefor, g8 cannot be an 8. Likewise h4 cannot be an 8. Therefor the 8s are in g5 and h7.
Ah - I see it now. Thank you kate!
Nicely done Kateblu - some mention of the cells gh7 could be warranted - but then a bit of vagueness is a good idea. One question though:
How is h4<>8?
Hmmm. My thinking was that if h4=8, then h7=5. However, Steve, as you point out, although we can eliminate a 4 in i9, but we are left with a 4 in g9, defeating the UAR. How did I get rid of the 4 in g9? I saw it yesterday but.... That's why this puzzle can drive me crazy.
The only AUR elimination that I could find was to forbid g8=8.
Steve, I think you are right. You know, this blog has really helped me. Now I am looking for AURs when easy eliminations are gone.
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23/May/07 9:25 AM
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23/May/07 9:28 AM
Hello Steve,
I saw the almost unique rectangle and understood why g8 can't be an 8, but my problem is : How to justify the elimination in this cell ? I mean, It's easy to see that, here, the 8 in g8 is eliminated but I'd like to deduce a rule that would work for every similar case of almost More...
27/Dec/07 2:50 AM
Hi sudoku addict!
An AUR basically says that one can not have two pairs prefectly aligned in exactly two columns, two rows, and wholly contained in
two boxes. Thus:
Consider the AUR 39 at gi59: One could knows: gi5=(naked pair 39) -- gi9=(hidden pair 39).
The key to make this easy, in More...
28/Dec/07 7:03 PM
I have lost familiarity with my own notation style from using the Eureka style of notation. More accurately, I think:

(g5=8) == (gi5=NP39) -- (gi9=HP39) == (g8=39)
forbids g8=8.

In Eureka form, but still using the grid chessboard grid notation:

(8)g5=(naked pair More...
28/Dec/07 7:08 PM
Perfect !

I began to study forbidding chains a few weeks ago, so I'm not yet very familiar with them. But your explanations are quite clear, thanks !

(I will spend more time searching for AURs...)
30/Dec/07 5:15 AM
Isnt G8<>5 for the same reason that G8<>8?
02/Jan/08 4:06 AM
Hi Ace!
I think not. The potential 8 at g5 prevents one from concluding g8<>5.
02/Jan/08 7:33 PM
Ok - got it !
I'll be caught up to you in about a month !
03/Jan/08 6:39 AM
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