The New Sudoku Players' Forum

by **Carcul** » Sat Jan 14, 2006 4:47 pm

I have already described three types of almost unique rectangles. However, there is a special case of the type 1 AURs that I think deserves a description by their own. So, allowd me to make the following definition:

Type E3 AUR: an almost unique rectangle whose four cells are populated by, and only with, the same three candidates.

The next grid provides an example:

Code: Select all: . . . | . . . | . . . . . . | . . . | . . . . . . | 458 458 . | . . . -------+-----------+------ . . . | . . . | . . . . . . | . . . | . . . . . . | . . . | . . . -------+-----------+------ . . . | 458 458 . | . . . . . . | . . . | . . . . . . | . . . | . . .

We can see that this type of AUR can be considered a “mixture” of three AURs, each one with the same “extra” candidate in the four cells (the “extra” candidate is the one whose absence would imply more than one solution to the puzzle). This is the reason for the “E3” designation: “3” stands for the three AURs embedded in the four cells, and “E” stands for equal and therefore exchangeable extra candidates in the cells. We can refer to each one of those AURs by writing between brakets the two numbers that would be left in each of the four cells if the extra candidate were absent from all of them: for example, considering the “8” as the extra candidate we are referring to the AUR (45).
In each one of these AURs we have a simple strong link for the extra candidate: considering the AUR (45), if “8” is not in both cells r7c45 then it must be in one of the cells r3c45, and vice-versa. This is equivalent to the link [r7c4|r7c5]=8=[r3c4|r3c5]. In the same way, for the AUR (48) we have [r7c4|r7c5]=5=[r3c4|r3c5], and for the AUR (58) we have [r7c4|r7c5]=4=[r3c4|r3c5]. So, and this is the interesting thing, we can use one of these links to make deductions, or use two or three of them in turn to make several deductions.

Let’s see two examples. Consider the following grid:

Code: Select all: 468 5 468 | 346 2 1 | 7 9 348 2 7 69 | 69 348 348 | 34 5 1 1 3 489 | 49 5 7 | 48 2 6 ----------------+----------------+---------------- 349 6 7 | 1 34 349 | 5 8 2 5 49 2 | 8 7 349 | 1 6 34 34 8 1 | 2 6 5 | 9 34 7 ----------------+----------------+---------------- 469 49 5 | 34 1 2 | 3468 7 348 7 2 346 | 5 348 348 | 346 1 9 48 1 348 | 7 9 6 | 2 34 5

We have a type E3 AUR in cells r2c56/r8c56. The AUR (48) can be used in the following way:

[r9c3]=3=[r8c3]-3-[r8c5|r8c6]=3=[r2c5|r2c6]-3-[r2c7]=3=[r7c7|r8c7]-3-[r9c8]=3=[r9c3], => r9c3=3 and that solve the puzzle.

Let’s see another example. The initial puzzle:

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

and after all basic steps we get:

Code: Select all: 1 245 9 | 6 3 245 | 7 25 8 235 8 235 | 7 1 9 | 25 4 6 6 245 7 | 245 8 245 | 3 1 9 ---------------+------------+------------ 2345 25 2345 | 245 9 7 | 6 8 1 245 7 1 | 3 6 8 | 245 9 25 9 6 8 | 1 45 245 | 245 3 7 ---------------+------------+------------ 245 3 6 | 9 45 1 | 8 7 245 8 1 245 | 45 7 6 | 9 25 3 7 9 45 | 8 2 3 | 1 6 45

From here, it is possible to solve the grid with standard techniques: coloring, a swordfish, locked candidates, and finally coloring again. However, a shorter solution can be made with the aid of the exchangeable properties of the type E3 AUR in cells r1c26/r3c26. First, we can consider the AUR (24):

[r3c4]=2=[r4c4]-2-[r4c2]-5-[r1c2|r3c2]=5=[r1c6|r3c6]-5-[r3c4], => r3c4<>5.

After eliminating the “5” from r6c6 due to locked candidates in box 2, we can now use the AUR (45):

[r6c5]=5=[r4c4]-5-[r4c2]-2-[r1c2|r3c2]=2=[r1c6|r3c6]-2-[r6c6]-4-[r6c5], => r6c5<>4,

which solve the puzzle. An even shorter solution could be reached from the grid above by using the AUR (45) in a different way:

[r4c2]-2-[r1c2|r3c2]=2=[r1c6|r3c6]-2-[r3c4]=2=[r4c4]-2-[r4c2], => r4c2<>2

and that solve the puzzle.

Carcul

by **ronk** » Sat Jan 14, 2006 7:53 pm

For your example #1:

Carcul wrote:
Code: Select all
468 5 468 | 346 2 1 | 7 9 348 2 7 69 | 69 348 348 | 34 5 1 1 3 489 | 49 5 7 | 48 2 6 ----------------+----------------+---------------- 349 6 7 | 1 34 349 | 5 8 2 5 49 2 | 8 7 349 | 1 6 34 34 8 1 | 2 6 5 | 9 34 7 ----------------+----------------+---------------- 469 49 5 | 34 1 2 | 3468 7 348 7 2 346 | 5 348 348 | 346 1 9 48 1 348 | 7 9 6 | 2 34 5

We have a type E3 AUR in cells r2c56/r8c56. The AUR (48) can be used in the following way:

[r9c3]=3=[r8c3]-3-[r8c5|r8c6]=3=[r2c5|r2c6]-3-[r2c7]=3=[r7c7|r8c7]-3-[r9c8]=3=[r9c3], => r9c3=3 and that solve the puzzle.

The AUR (38) yields a shorter and less complex nice loop:
[r7c7]-3-[r2c7]-4-[r2c5|r2c6]=4=[r8c5|r8c6]-4-[r7c4]-3-[r7c7] => r7c7<>3

Due to the symmetry of digits 3 and 4, the 3 and 4 may be interchanged to get another nice loop:
[r7c7]-4-[r2c7]-3-[r2c5|r2c6]=3=[r8c5|r8c6]-3-[r7c4]-4-[r7c7] => r7c7<>4

For your example #2:

Carcul wrote:An even shorter solution could be reached from the grid above by using the AUR (45) in a different way:

[r4c2]-2-[r1c2|r3c2]=2=[r1c6|r3c6]-2-[r3c4]=2=[r4c4]-2-[r4c2], => r4c2<>2

and that solve the puzzle.

With the AUR (45) "paired" as you have done, we also have a 2s "turbot fish":

Code: Select all: 1 a245 9 | 6 3 A245 | 7 25 8 235 8 235 | 7 1 9 | 25 4 6 6 a245 7 | 245 8 A245 | 3 1 9 ------------------+---------------+-------------- 2345 A25 2345 |A245 9 7 | 6 8 1 245 7 1 | 3 6 8 | 245 9 25 9 6 8 | 1 45 a245 | 245 3 7 ------------------+---------------+-------------- 245 3 6 | 9 45 1 | 8 7 245 8 1 245 | 45 7 6 | 9 25 3 7 9 45 | 8 2 3 | 1 6 45

[r4c2]-2-[r1c2|r3c2]=2=[r1c6|r3c6]-2-[r6c6]=2=[r4c4]-2-[r4c2] => r4c2<>2

and then because of the four strong sides of the turbot fish, r1c6<>2, r3c6<>2, and r4c4<>2. Simple coloring letters 'A' and 'a' are shown in the appropriate cells, and the 2s elimination cells are colored 'A'.

[edit: Oops, I just noticed your nice loop for example 2 is also a "turbot fish" with four strong sides, and the immediate implications following the identical elimination r4c2<>2 would obviously be the same. From a coloring perspective, just move the 'a' from r6c6 to r3c4.]

With 3 different AUR pairs (24, 25, 45) and two different pairings (row, col), there are six different ways this special case of AUR type 1 can be considered. Just lots of elimination possibilites, I think.

Great stuff, Carcul, keep up the good work!

Ron

by **Carcul** » Sun Jan 15, 2006 12:06 pm

Hi Ronk.

Ronk wrote:The AUR (38) yields a shorter and less complex nice loop:
[r7c7]-3-[r2c7]-4-[r2c5|r2c6]=4=[r8c5|r8c6]-4-[r7c4]-3-[r7c7] => r7c7<>3

My loop may be more complex (and I think "complexity" is a subjective issue), but immediately solves the puzzle, while your simpler one (and the next one) don't. Personaly, I prefer loops that allow a great step in the solution process, even if they are complex.

Ronk wrote:With 3 different AUR pairs (24, 25, 45) and two different pairings (row, col), there are six different ways this special case of AUR type 1 can be considered.

Good observation.

Ronk wrote:Great stuff, Carcul, keep up the good work!

Thanks.

Regards, Carcul

by **ronk** » Sun Jan 15, 2006 12:48 pm

Carcul wrote:My loop may be more complex (and I think "complexity" is a subjective issue), but immediately solves the puzzle, while your simpler one (and the next one) don't. Personaly, I prefer loops that allow a great step in the solution process, even if they are complex.

That is not logical to me because before one makes an elimination, one doesn't know the following steps will be simpler than for an alternate elimination. Maybe a Garry Kasporov or a Bobby Fischer type would know, but I don't. Do you?

by **ronk** » Sun Jan 15, 2006 1:58 pm

Hi Carcul, would you please help me out with the following nice loop dilemma?

This is your example #1 again:

Carcul wrote:
Code: Select all
468 5 468 | 346 2 1 | 7 9 348 2 7 69 | 69 348 348 | 34 5 1 1 3 489 | 49 5 7 | 48 2 6 ----------------+----------------+---------------- 349 6 7 | 1 34 349 | 5 8 2 5 49 2 | 8 7 349 | 1 6 34 34 8 1 | 2 6 5 | 9 34 7 ----------------+----------------+---------------- 469 49 5 | 34 1 2 | 3468 7 348 7 2 346 | 5 348 348 | 346 1 9 48 1 348 | 7 9 6 | 2 34 5

By inspection, we have:

r2c7=4 => r7c7<>4
r2c7=4 => r2c56<>4 => r8c56=4 => r7c4<>4
r7c4=4 => r7c7<>4
r7c4=4 => r8c56<>4 => r2c56=4 => r2c7<>4

Unorthodox as it is, either r2c7=4 or r7c4=4 implies r7c7<>4 ... and r2c7=4 implies r7c4<>4 and vice versa. It seems the nice loop ...

r7c7]-4-[r2c7]=4=[r2c5|r2c6]-4-[r8c5|r8c6]=4=[r7c4]-4-[r7c7] => r7c7<>4

... applies, as if r2c56=4, then r8c56<>4. But ... [r2c5|r2c6]-4-[r8c5|r8c6] ... is not valid AFAIK. Any thoughts?

TIA, Ron

by **Carcul** » Sun Jan 15, 2006 3:46 pm

Hi Ronk.

Ronk wrote:2. r2c7=4 => r2c56<>4 => r8c56=4 => r7c4<>4

This is not true, because r7c4 and r8c56 are not in the same row.

Ronk wrote:It seems the nice loop ...

r7c7]-4-[r2c7]=4=[r2c5|r2c6]-4-[r8c5|r8c6]=4=[r7c4]-4-[r7c7] => r7c7<>4

... applies, as if r2c56=4, then r8c56<>4. But ... [r2c5|r2c6]-4-[r8c5|r8c6] ... is not valid AFAIK.

That link is not valid, because if "4" is necessarily in one of the cells r2c56, then this fact does not eliminate the "4s" from r8c56: a "4" can be in one of r8c56. This type of link (in a E3 AUR) is not completely equal to a common strong link (even a grouped one), because, contrary to the last one, it cannot be used as a weak link. Let's check the logic of your loop above: if r7c7=4, then r2c7<>4, then one of r2c56 must be "4", and this doesn't eliminate both "4s" from r8c56.
I hope this helps.

Regards, Carcul

by **ronk** » Mon Jan 16, 2006 1:58 am

Carcul wrote:
Ronk wrote:2. r2c7=4 => r2c56<>4 => r8c56=4 => r7c4<>4

This is not true, because r7c4 and r8c56 are not in the same row.

But they are both in box 8.

That link is not valid, because if "4" is necessarily in one of the cells r2c56, then this fact does not eliminate the "4s" from r8c56: a "4" can be in one of r8c56.

For all the reasons you listed, I already knew my "nice loop" was incorrect. That's why I wrote "it seems" and "as if" and "is not valid".

Sorry for making the indirect request, as I should have asked directly if you could find the nice loop that fit my implication chains. But I have since progressed a little further. There is an xy-loop going through the AUR in cells r2c56 and r8c56. It's an AUR (48) in one direction and an AUR (38) in the opposite direction. The net effect of the loop is to make the bivalues r2c7=34 and r7c4=34 act as if they are were in the same group (unit). IOW, if r2c7=3, then r7c4=4 and vice versa.

I got this far ...

Code: Select all: 468 5 468 | 346 2 1 | 7 9 348 2 7 69 | 69 348 348 | 34 5 1 1 3 489 | 49 5 7 | 48 2 6 ----------------+----------------+---------------- 349 6 7 | 1 34 349 | 5 8 2 5 49 2 | 8 7 349 | 1 6 34 34 8 1 | 2 6 5 | 9 34 7 ----------------+----------------+---------------- 469 49 5 | 34 1 2 | 3468 7 348 7 2 346 | 5 348 348 | 346 1 9 48 1 348 | 7 9 6 | 2 34 5

-[r7c4]-4-[r8c5|r8c6]=4=[r2c5|r2c6]-4-[r2c7]-3-[r2c5|r2c6]=3=[r8c5|r8c6]-3-[r7c4]-

... although I'm unsure of the left and right ends.

After some likely touchup to the ends of the above xy-loop expression, the next step is to add the links to obtain both the eliminations r7c7<>3 and r7c7<>4. If that's doable without too much effort, would you please show me how?

[edit: On second thought, please ignore my request. It's just an academic exercise anyway, and likely one I'll never use again anyway.]

TIA, Ron

by **Jeff** » Mon Jan 16, 2006 3:42 am

ronk wrote:-[r7c4]-4-[r8c5|r8c6]=4=[r2c5|r2c6]-4-[r2c7]-3-[r2c5|r2c6]=3=[r8c5|r8c6]-3-[r7c4]-

After some likely touchup to the ends of the above xy-loop expression, ....

Hi Ronk, This is not an xy-loop?

It can be called one of the following:

AUR chain
AUR nice loop
AUR combination chain
Grouped AUR nice loop
Grouped AUR chain
Strong AUR nice loop
Strong grouped AUR nice loop
Continuous AUR chain
Continuous grouped AUR chain
Continuous AUR combination chain
Continuous AUR nice loop
Continuous strong AUR nice loop
Continuous strong grouped AUR nice loop

by **ronk** » Mon Jan 16, 2006 5:27 am

Jeff wrote:This is not an xy-loop?

It can be called one of the following:

[ronk edit: yada, yada, yada, yada]

Jeff, you have an irritating habit of nitpicking while avoiding the substance of matters ... smiley face notwithstanding ... and IMO of course.

by **Jeff** » Mon Jan 16, 2006 7:50 am

ronk wrote:Jeff, you have an irritating habit of nitpicking while avoiding the substance of matters ... smiley face notwithstanding ... and IMO of course.

I am just trying to leave the hard work to Carcul but make it looks like that I am busy.

by **Carcul** » Mon Jan 16, 2006 11:05 am

Hi Ronk.

Ronk wrote:But they are both in box 8.

You are right, of course, I was thinking in r7c7.

Ronk wrote:-[r7c4]-4-[r8c5|r8c6]=4=[r2c5|r2c6]-4-[r2c7]-3-[r2c5|r2c6]=3=[r8c5|r8c6]-3-[r7c4]-

Ronk, you are using the same cells twice in the same loop. According to your loop: r7c4=4 => r8c56=38 and r2c56=348 => r2c7<>4 => r2c7=3 => r2c56=48, from which I don't see any conclusion, but I might be wrong.

Ronk wrote:Sorry for making the indirect request, as I should have asked directly if you could find the nice loop that fit my implication chains.

Check this loop:

[r7c7|r7c9]-4-[r7c4]=4=[r8c5|r8c6]=3=[r2c5|r2c6]-3-[r2c7]=3=[r7c7|r8c7]-3-[r9c8]-4-[r7c7|r7c9]

which implies r7c7,r7c9<>4. I have used r7c7 twice, but the logic of the loop is OK. Jeff, I think you have here another good example of a grouped nice loop.

Regards, Carcul

by **Jeff** » Wed Jan 18, 2006 6:39 am

Carcul wrote:[r7c7|r7c9]-4-[r7c4]=4=[r8c5|r8c6]=3=[r2c5|r2c6]-3-[r2c7]=3=[r7c7|r8c7]-3-[r9c8]-4-[r7c7|r7c9]

which implies r7c7,r7c9<>4. I have used r7c7 twice, but the logic of the loop is OK. Jeff, I think you have here another good example of a grouped nice loop.

Thanks Carcul, I always try to use nice loops without overlaps as examples. I will surely watch out for the next opportunity.

by **ronk** » Thu Jan 19, 2006 2:37 pm

Carcul wrote:Check this loop:

[r7c7|r7c9]-4-[r7c4]=4=[r8c5|r8c6]=3=[r2c5|r2c6]-3-[r2c7]=3=[r7c7|r8c7]-3-[r9c8]-4-[r7c7|r7c9]

which implies r7c7,r7c9<>4. I have used r7c7 twice, but the logic of the loop is OK.

"Using r7c7 twice" certainly seems valid, since there is no contradiction between "r7c7<>4 and "r7c9<>4" and "r7c7=3 or r8c7=3".

Thanks for working that out, Ron

by **ronk** » Sat Feb 25, 2006 2:08 pm

Hi Carcul,

Here are a couple more examples of AUR type E3. The first is #1325 of the top1465.

Code: Select all: .3.|.8.|..4 ...|...|36. .6.|..1|..2 ---+---+--- ..3|..2|... ...|.38|15. ..8|...|.2. ---+---+--- 2..|.64|..9 .86|2..|... .7.|...|... 159 3 2 | 567 8 567 | 59 179 4 8 4 157 | 57 2 9 | 3 6 15 59 6 57 | 3 4 1 | 5789 789 2 ----------------+----------------+--------------- 157 159 3 | 1456 157 2 | 469 49 8 6 2 4 | 9 3 8 | 1 5 7 157 159 8 | 1456 157 56 | 469 2 3 ----------------+----------------+--------------- 2 15 15 | 78 6 4 | 78 3 9 34 8 6 | 2 9 37 | 457 147 15 34 7 9 | 158 15 35 | 2 48 6

The 2nd is #1391 of the same source.

Code: Select all: ...|.8.|.5. 93.|...|7.. ...|6.1|..2 ---+---+--- 698|.4.|5.. ...|...|.4. 5..|...|2.8 ---+---+--- 4..|.2.|.7. .69|...|... .52|...|9.. 12 1247 1467 | 379 8 379 | 1346 5 136 9 3 16 | 24 5 24 | 7 8 16 8 47 5 | 6 37 1 | 34 9 2 ----------------+----------------+--------------- 6 9 8 | 123 4 23 | 5 13 7 123 127 137 | 58 6 58 | 13 4 9 5 14 34 | 137 9 37 | 2 6 8 ----------------+----------------+--------------- 4 8 13 | 359 2 3569 | 136 7 1356 137 6 9 | 3457 137 3457 | 8 2 1345 137 5 2 | 48 137 468 | 9 13 46

Just now found them, so don't know whether eliminations may be made using the AURs.

Ron

by **Carcul** » Sat Feb 25, 2006 3:53 pm

Thanks Ronk.

In puzzle #1325 (the first one) I could use the AUR to show that r9c4<>8.
In the second puzzle, well, I leave the use of the AUR with you.

Have fun.

Regards, Carcul

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