... by: Anton Delprado

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.

... by: Chuck Bruno - Virginia

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

... by: Alan Freberg

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

... by: 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

... by: JCS

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.

... by: csvidyasagar

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

... by: csvidyasagar

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

... by: csvidyasagar

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

... by: Harmen Dijkstra

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.

... by: Wiking 48

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...

... by: Harmen Dijkstra

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?

... by: CS Vidyasagar

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

... by: Lea Hayes

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)

... by: Lea Hayes

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?

... by: John_Ha

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.

... by: Andi

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?

... by: Dennis Daft

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