## Almost Unique Rectangles: description and use in nice loops

Advanced methods and approaches for solving Sudoku puzzles

### Almost Unique Rectangles: description and use in nice loops

Almost Unique Rectangles: description and application in advanced nice loops

Thanks to the work of several people, we all know what unique rectangles are and how they can be used to make useful deductions. But, what can happen if we have, not a unique rectangle, but a set of cells that is very closely a unique rectangle? What I am about to say next may have been already “discovered” or used by someone else (if this is the case, I apologize), but even then let me share my thoughts.

Allowd me to define an almost unique rectangle (which, from now on, I will abbreviate as AUR) in the following way:

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`Almost Unique Rectangle (AUR): a set of four cells populated in such a way that if one or two of them does not have a specific candidate 'x', then those four cells form a unique rectangle.`

As we know, the presence of a unique rectangle will force some value in one of those cells or exclude a value from one or more cells. Let’s see three simple examples of AURs:

Code: Select all
` .  .   . | . . 12 | .   127 .  .  .   . | . . 127| .   124 .  . 178 178| . .  . | .   .   47    ----------+--------+------------  .  .   . | . .  . | .   .   .  .  .   . | . . 35 | 358 .   .  .  .   . | . . 357| 35  .   . ----------+--------+------------  . 179 179| . .  . | .   .   .  .  .   . | . .  . | .   .   .  .  .   . | . .  . | .   .   . `

Cells r3c23/r7c23 form what I call a type 1 AUR (we would still have this type of AUR if: one of the “9” were not present in r7c23, or if one of the “8” were not present in r3c23; instead of the two “9s” in r7c23 we have two “8s”): we can see that if “8” is not in both cells r3c2/r3c3 then uniqueness requires that “9” must be in one of the cells r7c2/r7c3, and vice-versa for “9” in cells r7c2/r7c3. This can be expressed by the link [r3c2|r3c3]=8|9=[r7c2|r7c3], where =8|9= is a link similar but not completely equivalent to the type of link that exists in an ALS having two of his values in just one cell each.
On the other hand, cells r5c67/r6c67 form what I call a type 2 AUR: in this case, if ”7” is not in r6c6 then uniqueness requires that “8” must be in r5c7; conversely, if “8” is not in r5c7, then “7” must be in r6c6. This is equivalent to the link [r6c6]=7|8=[r5c7]. This link is a very interesting one: please note that it relates directly two cells that are not in the same unit.
Finally, cells r12c6/r12c8 form what I call a type 3 AUR: if "7" is not in r2c6 then "4" or "7" must be in one of the cells r12c8, and these last two cells then form a naked pair with r3c9 (this bivalue cell must be present in order to this type of AUR be useful); if "4" and "7" are not present in cells r12c8 (and this can happen if one of these values occur in a cell of box 3 besides r12c8/r3c9), then "7" must be in r2c6. We can write this link as [r2c6]=7|4,7=[r1c8|r2c8].

As might be expected, the type of link that exists in an AUR can be included in a nice loop and used to make deductions. So, let’s see some examples of nice loops that make use of AURs.

I will begin with the type 1 AUR. Consider the following puzzle in the initial state (after the basic steps):

Code: Select all
` 167  2    13478| 468   3489 468  | 5     79   168  156  138  9    | 7     358  2568 | 246   246  1268  567  478  478  | 24568 4589 1    | 79    3    268    ----------------+-----------------+-----------------  4    15   126  | 13    7    58   | 1236  2568 9  3    59   16   | 458   2    4589 | 16    58   7  8    1579 127  | 13    6    59   | 123   25   4 ----------------+-----------------+-----------------  127  6    38   | 9     1458 24578| 247   247  235  1279 147  147  | 256   15   3    | 8     2679 256  279  38   5    | 2468  48   24678| 24679 1    236 `

If we look carefully to this grid, we can see two AURs of type 1: one in cells r3c2, r3c3, r8c2, r8c3, and the other in cells r2c7, r2c8, r7c7, r7c8. In the first one, if “8” is not in r3c2/r3c3 then “1” must be in one of the cells r8c2/r8c3, and vice-versa, so we have here the link [r8c2|r8c3]=1|8=[r3c2|r3c3]. We can make use of this link to eliminate an “8” from r2c2 and r3c5:

[r2c2]=3=[r1c3]-3-[r7c3]=3=[r7c9]=5=[r8c9]-5-[r8c5]-1-[r8c2|r8c3]=1|8=[r3c2|r3c3]-8-[r2c2]
=> r2c2<>8.

[r3c5]=9=[r1c5]=3=[r1c3]-3-[r7c3]=3=[r7c9]=5=[r8c9]-5-[r8c5]-1-[r8c2|r8c3]=1|8=[r3c2|r3c3]-8-[r3c5]
=> r3c5<>8.

Later in the solving of this grid it is also possible to use the other type 1 AUR to make a deduction. As an exercise, the reader may want to complete the solving of this puzzle.

Now let’s see an example of a type 2 AUR. Consider the following puzzle:

Code: Select all
` 3 1 . | 6 . . | . . .  . . 2 | . . . | . . .  . 8 . | . 9 . | 7 6 .    -------+-------+------  . . . | . . 5 | . . .  . 6 . | . 1 . | . 9 .  . . . | 4 . . | . . . -------+-------+------  . 9 8 | . 6 . | . 3 .  . . . | . . . | 5 . .  . . . | . . 7 | . 4 2 `

After all basic steps we get:

Code: Select all
` 3   1  9  | 6  7   48 | 248  25 458  6   7  2  | 38 5   348| 48   1  9  5   8  4  | 2  9   1  | 7    6  3    -----------+-----------+------------  278 4  137| 9  238 5  | 6    27 18  28  6  35 | 7  1   38 | 2348 9  458  9   23 157| 4  238 6  | 238  57 15 -----------+-----------+------------  4   9  8  | 5  6   2  | 1    3  7  27  23 37 | 1  4   9  | 5    8  6  1   5  6  | 38 38  7  | 9    4  2 `

Simple Sudoku continues the solving of this grid with two XY-Wings and then gets stuck. It is easy to find in this grid nice loops or perhaps other arguments that solve the puzzle. But please note the following: the cells r1c6, r1c7, r2c6, and r2c7 form an AUR of type 2, where it is easy to see that if “3” is not in r2c6, uniqueness requires that r1c7 = 2; conversely, if “2” is not in r1c7, then “3” must be in r2c6. So, we can consider the link [r2c6]=3|2=[r1c7] and write the following nice loop,

[r6c3]=1=[r4c3]-1-[r4c9]-8-[r6c7]=8=[r6c5]-8-[r5c6]-3-[r2c6]=3|2=[r1c7]-2-[r1c8]-5-[r6c8]-7-[r6c3]

which implies r6c3<>7 and that solve the puzzle.

Let’s look at another example (kindly provided by Bennys) of this type of AUR. The initial grid:

Code: Select all
` 7 . . | 6 . . | 3 . 1  . 4 6 | . . . | . 9 .  2 . . | . 4 . | . 8 .    -------+-------+------  . . . | 4 . 3 | . . 8  . . 4 | . 6 . | 1 . .  5 . . | 8 . 7 | . . . -------+-------+------  . 2 . | . 8 . | . . 3  . 7 . | . . . | 8 1 .  4 . 8 | . . 9 | . . 6 `

from where we get with the basic steps:

Code: Select all
` 7   589  59  | 6   259 258 | 3   4    1  18  4    6   | 35  357 18  | 257 9    25  2   1359 1359| 579 4   15  | 6   8    57    --------------+-------------+--------------  19  169  7   | 4   259 3   | 25  256  8  389 389  4   | 259 6   25  | 1   2357 257  5   36   2   | 8   1   7   | 49  36   49 --------------+-------------+--------------  169 2    159 | 157 8   46  | 49  57   3  369 7    359 | 235 235 46  | 8   1    49  4   135  8   | 135 357 9   | 257 25   6 `

In this grid we have a type 2 AUR in cells r4c7, r4c8, r9c7, r9c8, for which we have the link [r4c8]=6|7=[r9c7]. Thus, we can write the loop

[r4c8]=6|7=[r9c7]-7-[r7c8](-5-[r4c8])-5-[r9c8]-2-[r4c8]

which implies r4c8<>2,5 => r4c8=6 and that solve the puzzle.

A variation of the type 2 AUR exists when both candidates are in the same cell, which in this case function as a bivalue node. Let´s see an example (kindly provided by Ronk):

Code: Select all
` 368 36  68 | 5  2    1 | 79   4 79  1   4   2  | 3  79   79| 5    8 6  9   7   5  | 46 48   68| 3    2 1 ------------+-----------+-------------  36  369 4  | 8  679  2 | 79   1 5  2   5   69 | 1  4679 79| 48   3 4789  7   8   1  | 49 3    5 | 24   6 249 ------------+-----------+-------------  5   69  689| 2  1    4 | 68   7 3  468 1   3  | 7  5    68| 2468 9 248  468 2   7  | 69 89   3 | 1    5 48`

This is puzzle #272 from top1465. We have a modified type 2 AUR in cells r2c56/r5c56, which can be used in the following loop,

[r7c3]=8=[r7c7]-8-[r5c7]-4-[r5c5]-6-[r5c3]-9-[r7c3],

which implies r7c3<>9 and that solve the puzzle.

Let's see now an example of a type 3 AUR. The initial grid:

Code: Select all
` 1 . . | . . . | 7 . 2  . 3 . | 1 . . | . . .  . 9 . | . . . | . . .    -------+-------+------  . . . | 3 9 . | . 6 .  5 . 8 | . . . | 2 . .  . . . | . . . | . . . -------+-------+------  2 . . | . . 8 | 5 . .  . . . | 6 . . | . 9 .  . . . | . 4 . | . . . `

from which we get after some steps

Code: Select all
` 1   458  45 | 9  6    3  | 7  58 2  6   3    257| 1  2578 257| 9  58 4  78  9    257| 28 2578 4  | 6  1  3    -------------+------------+-----------  47  247  1  | 3  9    257| 48 6  58  5   67   8  | 4  17   167| 2  3  9  9   246  3  | 28 258  256| 14 7  15 -------------+------------+-----------  2   1    9  | 7  3    8  | 5  4  6  478 4578 457| 6  12   12 | 3  9  78  3   78   6  | 5  4    9  | 18 2  178 `

Here we have a type 3 AUR in cells r1c23/r8c23/r9c2 with the link
[r1c2]=8|7,8=[r8c2|r8c3] (note the presence of the bivalue "7,8" in r9c2).
This link can be used to show that r8c1<>7 in the following way:

[r8c1](-7-[r8c2|r8c3])-7-[r9c2]-8-[r8c2|r8c3]=7,8|8=[r1c2]-8-[r3c1]-7-[r8c1], => r8c1<>7.

To finish this post, allowd me to present an example of an almost unique loop. Consider the following puzzle,

Code: Select all
` . 1 . | . 7 . | . . .  . 2 6 | . . . | . 5 7  5 . . | . . . | . . 3    -------+-------+------  . . 8 | 2 . . | . . .  . 5 . | . . 4 | 3 . 2  4 . . | . 3 7 | . 8 . -------+-------+------  . . . | . 5 3 | 4 . .  . . 5 | . . . | . . 8  . 8 . | . . . | . 9 6 `

and allowd me to jump to the final step of my solution of this grid:

Code: Select all
` 389 1  39 | 35  7   25689| 2689 26 4  389 2  6  | 34  149 189  | 89   5  7  5   47 47 | 69  269 2689 | 2689 1  3    -----------+--------------+------------  679 3  8  | 2   169 1569 | 167  4  59  679 5  1  | 69  8   4    | 3    67 2  4   69 2  | 15  3   7    | 16   8  59 -----------+--------------+------------  26  69 79 | 8   5   3    | 4    27 1  12  47 5  | 147 69  69   | 27   3  8  13  8  347| 47  124 12   | 5    9  6 `

As can be seen, we have an almost unique loop in cells r3c2, r3c3, r8c2, r9c3, r8c4, and r9c4. Uniqueness requires that if r9c3 is not 3 then r8c4 must be 1, and if r8c4 is not 1 then r9c3 must be 3. Hence, we can consider the link [r9c3]=3|1=[r8c4] and write the loop

[r9c3]=3|1=[r8c4](-1-[r9c5])-1-[r9c6]-2-[r9c5](-4-[r9c3])-4-[r9c4]-7-[r9c3]

which implies r9c3<>4,7 => r9c3=3 and that solve the puzzle.

Carcul
Last edited by Carcul on Mon Apr 03, 2006 7:38 am, edited 8 times in total.
Carcul

Posts: 724
Joined: 04 November 2005

bennys

Posts: 156
Joined: 28 September 2005

### Re: Almost Unique Rectangles: description and use in nice lo

Carcul wrote:
Code: Select all
` 389 1  39 | 35  7   25689| 2689 26 4  389 2  6  | 34  149 189  | 89   5  7  5   47 47 | 69  269 2689 | 2689 1  3    -----------+--------------+------------  679 3  8  | 2   169 1569 | 167  4  59  679 5  1  | 69  8   4    | 3    67 2  4   69 2  | 15  3   7    | 16   8  59 -----------+--------------+------------  26  69 79 | 8   5   3    | 4    27 1  12  47 5  | 147 69  69   | 27   3  8  13  8  347| 47  124 12   | 5    9  6 `

As can be seen, we have an almost unique loop in cells r3c2, r3c3, r8c2, r9c3, r8c4, and r9c4. Uniqueness requires that if r9c3 is not 3 then r8c4 must be 1, and if r8c4 is not 1 then r9c3 must be 3. Hence, we can consider the link [r9c3]=3|1=[r8c4]

Well done, carcul. Just one comment; the notation of the AUR link is more meaningful like this:

[r9c3]-3-[AUR:r3c2|r3c3|r8c2|r9c3|r8c4|r9c4]-4|7-[r8c4]-1-
Jeff

Posts: 708
Joined: 01 August 2005

Thanks Bennys.

Jeff wrote:Just one comment; the notation of the AUR link is more meaningful like this:

[r9c3]-3-[AUR:r3c2|r3c3|r8c2|r9c3|r8c4|r9c4]-4|7-[r8c4]-1-

Yes, it is more meaningful to list all cells of the AUR (AULoop in this case), because it makes more clear where the link come from. However, I listed just the two cells involved in the link because its origin was clear from the explanation, and to not make the chain notation too complicated.

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Carcul:
A very useful analysis. I freely admit I would have dismissed the first example on the assumption that the 7s in R6c2 and R6c3 would "defend" against efforts to form the uniqueness contradiction. Obviously I would not be correct.

Your second uniqueness loop is reminiscent of ones I have seen before where there's a simple example of what I call a "withdrawal link". (By withdrawal I just mean that any placement of a 3 that "sees" R9c3 "withdraws" the 3 from the loop, and likewise any placement of a 1 that "sees" R8c4 withdraws the 1.) The (1,3) at R9c1 is a withdrawal point for the 3 that holds the 1 as the only other candidate. That creates a "one-step withdrawal link" in my parlance that eliminates the 1's at R9c5 and R9c6 - i.e., those are "withdrawal points" for the 1 that see R9c1, so placing a 1 there triggers the uniqueness contradiction by simultaneously withdrawing the 1 and forcing the 3 onto R9c1. More succinctly, R9c5 or R9c6=1=> R8c5<>1 and => R9c1=3 =>R9c3<>3. While this example involves two candidates, I have seen one recently that involves three.
Rod
Last edited by RodHagglund on Tue Jan 03, 2006 7:08 pm, edited 1 time in total.
RodHagglund

Posts: 9
Joined: 19 December 2005

RodHagglund wrote:Jeff: A very useful analysis.

Hi Rod, I think your post should address to Carcul who is the author of this advanced technique.
Jeff

Posts: 708
Joined: 01 August 2005

Jeff:
Quite correct. Message edited.
Rod
RodHagglund

Posts: 9
Joined: 19 December 2005

### Two more examples

Example 1

Consider the following grid:

Code: Select all
` 268    2689   1      | 4      5      26     | 79     789    3        356    3569   3456   | 136    7      8      | 2      149    149      7      238    348    | 13     9      12     | 5      148    6       ----------------------+----------------------+-------------------  68     68     9      | 7      1      3      | 4      5      2        4      13     2      | 9      6      5      | 8      13     7        135    7      35     | 2      8      4      | 6      139    19      ----------------------+----------------------+-------------------  1268   4      68     | 168    3      9      | 17     267    5        123568 123568 3568   | 168    4      7      | 19     269    89       9      168    7      | 5      2      16     | 3      46     48     `

We have a type 1 AUR in cells r2c89/r6c89 with the link
[r6c8]=3|4=[r2c8|r2c9]. We can use this link to write the following continuous nice loop:

[r6c8]=3|4=[r2c8|r2c9]-4-[r2c3]=4=[r3c3]=8=[r7c3|r8c3]-8-[r9c2]
=8=[r9c9]-8-[r8c9]-9-[r6c9]-1-[r5c8]-3-[r6c8],

which implies r7c1<>8, r8c1<>8, r8c2<>8, r2c9<>9, r6c8<>1.

Now, to solve the puzzle we only need the simple loop

[r9c2]=8=[r9c9]-8-[r8c9]-9-[r6c9]-1-[r5c8]-3-[r5c2]-1-[r9c2],
=> r9c2<>1.

Example 2

The initial grid (puzzle #87 of the top95):

Code: Select all
` . . . | 5 . 3 | . . .  . . . | . 6 . | 7 . .  5 . 8 | . . . | . 1 6    -------+-------+------  3 6 . | . 2 . | . . .  . . . | 4 . 1 | . . .  . . . | . 3 . | . . 5 -------+-------+------  6 7 . | . . . | 2 . 8  . . 4 | . 7 . | . . .  . . . | 2 . . | 5 . . `

After the basic techniques we get:

Code: Select all
`  7  49   6 | 5  1  3 | 48  489 2   14 1349 2 | 8  6  49| 7   5   349   5  349  8 | 7  49 2 | 34  1   6 ------------+---------+-------------   3  6    7 | 9  2  5 | 148 48  14   9  2    5 | 4  8  1 | 36  367 37   48 48   1 | 6  3  7 | 9   2   5 ------------+---------+-------------   6  7    39| 1  5  49| 2   34  8   2  5    4 | 3  7  8 | 16  69  19   18 18   39| 2  49 6 | 5   347 347`

Here we have a type 2 almost unique rectangle in cells r1c78/r4c78 with the link [r1c8]=9|1=[r4c7], and we can write

[r1c8]=9|1=[r4c7]-1-[r4c9](-4-[r4c8]-8-[r1c8])=1=[r8c9]-1-[r8c7]-6-[r5c7]-3-[r3c7]-4-[r1c8]

which implies r1c8<>4,8 => r1c8=9 and that solve the puzzle.

Carcul
Carcul

Posts: 724
Joined: 04 November 2005

### Re: Two more examples

Carcul wrote:Example 2

After the basic techniques we get:

Code: Select all
`  7  49   6 | 5  1  3 | 48  489 2   14 1349 2 | 8  6  49| 7   5   349   5  349  8 | 7  49 2 | 34  1   6 ------------+---------+-------------   3  6    7 | 9  2  5 | 148 48  14   9  2    5 | 4  8  1 | 36  367 37   48 48   1 | 6  3  7 | 9   2   5 ------------+---------+-------------   6  7    39| 1  5  49| 2   34  8   2  5    4 | 3  7  8 | 16  69  19   18 18   39| 2  49 6 | 5   347 347`

Here we have a type 2 almost unique rectangle in cells r1c78/r4c78 with the link [r1c8]=9|1=[r4c7], and we can write

[r1c8]=9|1=[r4c7]-1-[r4c9](-4-[r4c8]-8-[r1c8])=1=[r8c9]-1-[r8c7]-6-[r5c7]-3-[r3c7]-4-[r1c8]

which implies r1c8<>4,8 => r1c8=9 and that solve the puzzle.

Using a two-step approach with a much simpler nice loop for the type 2 AUR, we can write

[r4c8]=8=[r1c8]=9|1=[r4c7]-1-[r4c9]-4-[r4c8] => r4c8=8

Then after placing r4c8=8 and r1c7=8 we have a simple x-cycle in 4s and can write

[r3c2]-4-[r3c5]=4=[r2c6]-4-[r7c6]=4=[r7c8]-4-[r1c8]=4=[r1c2]-4-[r3c2] => r3c2<>4

and that solve the puzzle.

P.S. I'm sure you'll let me know if my '=' and '-' are still messed up.
ronk
2012 Supporter

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Location: Southeastern USA

Hi Ronk.

Ronk wrote:Using a two-step approach with a much simpler nice loop for the type 2 AUR, we can write

[r4c8]=8=[r1c8]=9|1=[r4c7]-1-[r4c9]-4-[r4c8] => r4c8=8

This nice loop is not correct, because it don't eliminate the "4" from r4c8.
If this loop were correct, then it would imply that if r4c8 is not "8", following the implications of the loop we would get r4c8 is not "4" either. This doesn´t happen in your loop: just make r4c8=4 and you will get no contradiction in the cells of the loop. It was just a coincidence that the elimination r4c8<>4 is correct.
When using the link in an AUR, it is not possible to propagate directly to this link with a strong link inside the AUR, as your loop demonstrates. Also, we must propagate from the link of an AUR to a cell not belonging to the AUR.

Carcul wrote:[r1c8]=9|1=[r4c7]-1-[r4c9](-4-[r4c8]-8-[r1c8])=1=[r8c9]-1-[r8c7]-6-[r5c7]-3-[r3c7]-4-[r1c8]

Now let's check the logic of this loop: we can consider that r1c8 is "9" or r1c8 is not "9" - but, if r1c8 is not "9", and if you follow the implications of the loop, you will get no candidate for r1c8. This is a contradiction, and so r1c8 must be "9", which solve the puzzle.

Ronk wrote:[r3c2]-4-[r3c5]=4=[r2c6]-4-[r7c6]=4=[r7c8]-4-[r1c8]=4=[r1c2]-4-[r3c2] => r3c2<>4

This X-cycle is correct and the notation is also correct. Well done. But keep in mind the following: more important than the rules for nice loop construction is the logic inside them, and it is always a good practice to check manualy the dedution made by a particular loop. Nice loops and their rules should not be used "blindly".

Keep the good work.

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Hi Ronk.

Here is an exercise for you (or others). In the thread "Nice loops for elementary level players - the x-cycle", look at the last reply (by Pep): there is in the grid posted more than one AUR, but two of them, if properly used, will (each one by itself) solve the puzzle.

Carcul
Last edited by Carcul on Mon Jan 09, 2006 7:33 am, edited 1 time in total.
Carcul

Posts: 724
Joined: 04 November 2005

Carcul wrote:It was just a coincidence that the elimination r4c8<>4 is correct.

I don't think it's coincidence at all. I started with implication chains and then assumed the deduction was expressible with a nice loop. Apparently that assumption was incorrect, and it's probably because I unknowingly treated r1c8 like it was a bivalued cell.

My starting implication chains are based on the fact that for this type 2 AUR example ... either r1c8=9 or r4c7=1 and therefore ...

r4c7=1 => r4c9<>1 => r4c9=4 => r4c8<>4 => r4c8=8
r4c7<>1 => r1c8=9 => r1c8<>8 => r4c8=8 => r4c8<>4

Where is the error in logic? Is is possible such a simple deduction is not expressible with a nice loop simpler than the one you posted? Tell me it isn't so!
ronk
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Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Carcul wrote:
Ronk wrote:Using a two-step approach with a much simpler nice loop for the type 2 AUR, we can write

[r4c8]=8=[r1c8]=9|1=[r4c7]-1-[r4c9]-4-[r4c8] => r4c8=8

This nice loop is not correct, because it don't eliminate the "4" from r4c8.

Don't give up, Ronk. I have high hope on you.
Jeff

Posts: 708
Joined: 01 August 2005

Hi Ronk and hi Jeff.

Ronk wrote:r4c7=1 => r4c9<>1 => r4c9=4 => r4c8<>4 => r4c8=8
r4c7<>1 => r1c8=9 => r1c8<>8 => r4c8=8 => r4c8<>4

This deduction is completely correct. However, I do not agree that this deduction can be expressed by your loop, and I think this is because of the link [r1c8]=9|1=[r4c7]. I have written in the original post about AURs that this link "is the same type of link that exists in an ALS having two of his values in just one cell each." (I have now edited the original post). I have now concluded that this is not completely true. Consider the following grid:

Code: Select all
` . . . | . .  .  | .   .   .  . . . | . .  .  | .   .   .  . . . | . 37 137| 378 389 .    -------+---------+----------  . . . | . .  .  | .   .   .  . . . | . .  .  | .   .   .  . . . | . .  14 | .   48  . -------+---------+----------  . . . | . .  .  | .   .   .  . . . | . .  .  | .   .   .  . . . | . .  .  | .   .   . `

and let's suppose that the only "8s" in column 8 are the ones showed. We have an ALS in row 3 for which we have the link [r3c8]=9|1=[r3c6]: if "9" is not in r3c8, then "1" must be in r3c6, and if "1" is not in r3c6 then "9" must be in r3c8 - so far, this is equivalent to the link in an AUR. However, for the ALS we also have the following: if another number besides "9" is in r3c8, then "1" must still be in r3c6 (and vice-versa for r3c6). This is not the case for the link in the AUR. Let's confirm.
In the grid above with the ALS we can write

[r6c8]=8=[r3c8]=9|1=[r3c6]-1-[r6c6]-4-[r6c8] => r6c8<>4.

If "8" is not in r6c8, then r3c8=8 => r3c6=1 => r6c6=4 => r6c8<>4.

In our grid:

Code: Select all
`  7  49   6 | 5  1  3 | 48  489 2   14 1349 2 | 8  6  49| 7   5   349   5  349  8 | 7  49 2 | 34  1   6 ------------+---------+-------------   3  6    7 | 9  2  5 | 148 48  14   9  2    5 | 4  8  1 | 36  367 37   48 48   1 | 6  3  7 | 9   2   5 ------------+---------+-------------   6  7    39| 1  5  49| 2   34  8   2  5    4 | 3  7  8 | 16  69  19   18 18   39| 2  49 6 | 5   347 347`

[r4c8]=8=[r1c8]=9|1=[r4c7]-1-[r4c9]-4-[r4c8] => r4c8=8

but if r4c8 is not "8" then r1c8=8 but in this case the link =9|1= doesn't imply that r4c7=1, and so we cannot prove that r4c8<>4.
So, I think we have two conclusions:

1: the link that exists in an AUR is not completely equivalent to a similar one in an ALS.

2: your deduction r4c8<>4 is correct, but in my opinion it is not correct to express it with your loop.

Ronk wrote:Is is possible such a simple deduction is not expressible with a nice loop simpler than the one you posted?

I think you cannot compare the loops because the deduction is not exactly the same.

Regards, Carcul
Last edited by Carcul on Mon Jan 09, 2006 9:09 am, edited 1 time in total.
Carcul

Posts: 724
Joined: 04 November 2005

Carcul wrote:
Code: Select all
` . . . | . . . | .  .   .  . . . | . . . | .  .   .  . . . | . . 13| 38 389 .    -------+-------+----------  . . . | . . . | .  .   .  . . . | . . . | .  .   .  . . . | . . 14| .  48  . -------+-------+----------  . . . | . . . | .  .   .  . . . | . . . | .  .   .  . . . | . . . | .  .   . `

In the grid above with the ALS we can write
[r6c8]=8=[r3c8]=9|1=[r3c6]-1-[r6c6]-4-[r6c8] => r6c8<>4

Hi Carcul, Correct me if I am wrong. Why not just use a simple nice loop for this one?

[r6c8]=8=[r3c8]-8-[r3c7]-3-[r3c6]-1-[r6c6]-4-[r6c8] => r6c8<>4.
Jeff

Posts: 708
Joined: 01 August 2005

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