The New Sudoku Players' Forum

by **Protector** » Mon Sep 17, 2007 6:47 pm

I'm stuck. Can't find the next step.

3X2XX4X7X
9741X583X
X687X3X4X
8XX216497
2XX347658
647958321
X8XXX926X
XX6XX271X
X2XXX1X83

Note: This thread has been split from this one. IMHO if anyone wants to request help for a seperate puzzle it's more appropriate to create a new thread. The poster of the original thread might not like his/her thread being hijacked.

udosuk, moderator

by **re'born** » Mon Sep 17, 2007 9:21 pm

Protector wrote:I'm stuck.Can't find the next step.

3X2XX4X7X
9741X583X
X687X3X4X
8XX216497
2XX347658
647958321
X8XXX926X
XX6XX271X
X2XXX1X83

From here

Code: Select all: .---------------.---------------.---------------. | 3 15 2 | 68 689 4 | 159 7 569 | | 9 7 4 | 1 26 5 | 8 3 26 | | 15 6 8 | 7 29 3 | 159 4 259 | :---------------+---------------+---------------: | 8 35 35 | 2 1 6 | 4 9 7 | | 2 19 19 | 3 4 7 | 6 5 8 | | 6 4 7 | 9 5 8 | 3 2 1 | :---------------+---------------+---------------: | 17 8 13 | 45 37 9 | 2 6 45 | | 45 359 6 | 458 38 2 | 7 1 459 | | 47 2 59 | 46 67 1 | 59 8 3 | '---------------'---------------'---------------'

there is an xy-chain that solves the puzzle

- 9 - [r9c3 - 5 - r4c3 - 3 - r4c2 - 5 - r1c2 - 1 - r5c2] - 9 -, => r5c3, r8c2 <>9.

by **Protector** » Tue Sep 18, 2007 9:09 am

Yes!!I was thinking thinking and thinking yestorday
And i found a some kind of Turbot fish or forcing chain. :!:

And efter that i was been able to continue solving.
Txs for help

by **udosuk** » Tue Sep 18, 2007 2:00 pm

Here is another way to solve from this position:

Code: Select all: *--------------------------------------------------* | 3 15 2 | 68 689 4 | 159 7 569 | | 9 7 4 | 1 26 5 | 8 3 26 | | 15 6 8 | 7 29 3 | 159 4 259 | |----------------+----------------+----------------| | 8 35 @35 | 2 1 6 | 4 9 7 | | 2 19 $19 | 3 4 7 | 6 5 8 | | 6 4 7 | 9 5 8 | 3 2 1 | |----------------+----------------+----------------| |#17 8 *13 | 45 37 9 | 2 6 45 | |#45 359 6 | 458 38 2 | 7 1 459 | |#47 2 *59 | 46 67 1 | 59 8 3 | *--------------------------------------------------*

APE (Aligned Pair Exclusion):

r79c3 can only have 4 different combinations: [15|19|35|39]
[19] conflicts r5c3
[35] conflicts r4c3
[15] conflicts r789c1
Therefore r79c3 must be [39], and singles solve the rest.

I keep looking for more elegant routes without chains, but couldn't find any.

Another route goes like this:

One of r4c23 must be 5
=> Either r1c2=1 or r9c3=9
=> r5c23 can't be [19]
=> r5c23=[91], and singles solve the rest.

Unfortunately I couldn't formulate this move into any familiar existing technique. It feels like a Y-wing/W-wing, but not quite.

by **re'born** » Tue Sep 18, 2007 3:13 pm

udosuk wrote:r79c3 can only have 4 different combinations: [15|19|35|39]
[19] conflicts r5c3
[35] conflicts r4c3
[15] conflicts r789c1
Therefore r79c3 must be [39], and singles solve the rest.

I guess the middle conflict is superfluous, as the first and last will imply r7c3<>1.

udosuk wrote:One of r4c23 must be 5
=> Either r1c2=1 or r9c3=9
=> r5c23 can't be [19]
=> r5c23=[91], and singles solve the rest.

Unfortunately I couldn't formulate this move into any familiar existing technique. It feels like a Y-wing/W-wing, but not quite.

How did you conclude that either r1c2=1 or r9c3=9? Certainly it is true, but the only way I see to accomplish it is by running through (in some order) part of the xy-chain I suggested above.

By the way, there is a semi-remote naked pair on the grid, {r8c1 = 5 = r7c4} => r8c4<>5, but it doesn't seem to advance the puzzle much.

by **Steve R** » Tue Sep 18, 2007 3:35 pm

When I spot an ineffective xy- or xyz-wing, I check the puzzle for transport.

In udosuk’s position

Code: Select all: *--------------------------------------------------* | 3 15 2 | 68 689 4 | 159 7 569 | | 9 7 4 | 1 26 5 | 8 3 26 | | 15 6 8 | 7 29 3 | 159 4 259 | |----------------+----------------+----------------| | 8 *35 35 | 2 1 6 | 4 9 7 | | 2 19 19 | 3 4 7 | 6 5 8 | | 6 4 7 | 9 5 8 | 3 2 1 | |----------------+----------------+----------------| | 17 8 13 | 45 37 9 | 2 6 45 | | 45 *359 6 | 458 38 2 | 7 1 459 | | 47 2 *59 | 46 67 1 | 59 8 3 | *--------------------------------------------------*

the xyz-wing is marked with asterisks.

As it stands it would eliminate 5 only from b7c2. Happily transport is offered by the conjugate 5s in b1. The effect is to move the elimination to b7c1 and the elimination of 5 from r8c1 also solves the puzzle.

A chain by any other name, I suppose.

Steve

by **udosuk** » Tue Sep 18, 2007 3:45 pm

re'born wrote:How did you conclude that either r1c2=1 or r9c3=9? Certainly it is true, but the only way I see to accomplish it is by running through (in some order) part of the xy-chain I suggested above.

One of r4c23 must be 5.
If r4c2=5, r1c2 can't be 5, must be 1.
If r4c3=5, r9c3 can't be 5, must be 9.
I know it's chain-like, but for some reason I find it more elegant than the actual chain interpretation.

BTW I understand the logic of you "transport" move Steve, but perhaps it's a bit too complicated for the beginners to understand (without clearer demonstrations).

by **Ruud** » Tue Sep 18, 2007 5:16 pm

udosuk wrote:I keep looking for more elegant routes without chains, but couldn't find any.

ALS-xz rule:
A = r789c1|r9c3 (14579)
B = r5c3 (19)
X (restricted common) = 9
Z = 1 (eliminates 1 from r7c3)

Code: Select all: *--------------------------------------------------* | 3 15 2 | 68 689 4 | 159 7 569 | | 9 7 4 | 1 26 5 | 8 3 26 | | 15 6 8 | 7 29 3 | 159 4 259 | |----------------+----------------+----------------| | 8 35 35 | 2 1 6 | 4 9 7 | | 2 19 B19 | 3 4 7 | 6 5 8 | | 6 4 7 | 9 5 8 | 3 2 1 | |----------------+----------------+----------------| |A17 8 -13 | 45 37 9 | 2 6 45 | |A45 359 6 | 458 38 2 | 7 1 459 | |A47 2 A59 | 46 67 1 | 59 8 3 | *--------------------------------------------------*

No chain, no monkey

Ruud

by **Sudtyro** » Wed Sep 19, 2007 12:26 pm

Ruud wrote:No chain, no monkey

AKA a VWXYZ-Wing with V=9 and Z=1.

But I guess they're really just chains:
(1=9)r5c3 - (9=4571)r789c1|r9c3 => r7c3 <> 1.

BTW, I'm not familiar with Steve R's "transport" technique, either. Can anyone elaborate a bit?

by **udosuk** » Thu Sep 20, 2007 5:09 pm

Ruud wrote:No chain, no monkey

Sudtyro wrote: AKA a VWXYZ-Wing with V=9 and Z=1.

But I guess they're really just chains:
(1=9)r5c3 - (9=4571)r789c1|r9c3 => r7c3 <> 1.

BTW, I'm not familiar with Steve R's "transport" technique, either. Can anyone elaborate a bit?

Great spotting Sudtyro. The VWXYZ-Wing interpretation allows r9c3 to be {14579} and the move still stands.

There is another VWXYZ-Wing, with r7c3 as the pivot, r789c1 and r4c3 as the wings which eliminates 5 from r9c3. In this case r7c3 could be {13457} for the move to stand.

Ruud, I wasn't considering ALS-xz at the beginning. It's just too powerful and easy. I was looking for something stylish like the VWXYZ-Wing. I know they are all interpretations of the same chain-logic but somehow the wings make me feel more elegant than chains (and ALS).

As for the "transport" concept, I can understand the logic but it's hard to express it systematically. I guess I better leave it to Steve to elaborate himself.

by **re'born** » Fri Sep 21, 2007 1:47 pm

udosuk wrote:As for the "transport" concept, I can understand the logic but it's hard to express it systematically. I guess I better leave it to Steve to elaborate himself.

Here is another example of it from the same grid:

Code: Select all: .---------------.---------------.---------------. | 3 15 2 | 68 689 4 | 159 7 569 | | 9 7 4 | 1 26 5 | 8 3 26 | | 15 6 8 | 7 29 3 | 159 4 259 | :---------------+---------------+---------------: | 8 35 35 | 2 1 6 | 4 9 7 | | 2 19 19 | 3 4 7 | 6 5 8 | | 6 4 7 | 9 5 8 | 3 2 1 | :---------------+---------------+---------------: | 17 8 13 | 45a 37 9 | 2 6 45- | | 45* 359 6 | 458A 38 2 | 7 1 459*| | 47 2 59 | 46 67 1 | 59* 8 3 | '---------------'---------------'---------------'

The *'d cells form an xyz-wing with no eliminations. Such an xyz-wing implies that

Code: Select all: r8c1=5 or r8c9=5 or r9c7=5.

Now, the 5 in r8c1 is weakly linked to the 5 in r8c4, which is strongly linked to the 5 in r7c4. Thus if r7c4<>5, then r8c1<>5. So in effect, we've transported one of the points (r8c1) of the xyz-wing to another point (r7c4) and we conclude that

Code: Select all: r7c4=5 or r8c9=5 or r9c7=5

As r7c9 is weakly linked to all 3 cells, we conclude r7c9<>5 (which as far as I can tell does little to advance the puzzle).
One way to classify the pattern systematically is as an nrcz-chain. This also suggests that one needn't stop at transporting away by one weak and one strong link, but by any nice alternating chain (starting with a weak link). It also suggests that one needn't restrict themselves to only transporting one endpoint. You can transport from both endpoints (or even from both endpoints and the middle piece :!:

).

Another interesting variation is if the two endpoints are connected by transport. Both examples above exhibit this property (thought the first exhibits it in both directions) and it allows you to consider an xyz-wing with only 2 cells

This line of thinking implies in the first example that not only is r8c1<>5, but also r1c2<>5.

[Edit: I'm not so sure any more that one can view it always as an nrcz-chain. Namely, if all three points are transported, it looks like a more complicated nrcz-net to me.]

by **re'born** » Fri Sep 21, 2007 2:28 pm

Here is another an interesting example of transporting xyz-wings.

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

After basics, we get to

Code: Select all: .---------------------.---------------------.---------------------. | 15 4 9 | 3 2 8 | 157 6 17 | | 2 6 158 | 19 7 4 | 135 189 389 | | 18 7 3 | 19 6 5 | 2 4 89 | :---------------------+---------------------+---------------------: | 6 8 7 | 2 4 39 | 13 19 5 | | 159 1359 125 | 6 58 7 | 4 89 2389 | | 4 359 25 | 58 1 39 | 6 7 2389 | :---------------------+---------------------+---------------------: | 1578 15A 6 | 4 58 2 | 9 3 17 | | 15789- 159- 158B | 58C 3 6 | 17 2 4 | | 3 2 4 | 7 9 1 | 8 5 6 | '---------------------'---------------------'---------------------'

where an xyz-wing eliminates 5 from r8c12. However, we can also transport both A and C to r5c5 and get:

Code: Select all: .---------------------.---------------------.---------------------. | 15 4 9 | 3 2 8 | 157 6 17 | | 2 6 158 | 19 7 4 | 135 189 389 | | 18 7 3 | 19 6 5 | 2 4 89 | :---------------------+---------------------+---------------------: | 6 8 7 | 2 4 39 | 13 19 5 | | 159 1359 125- | 6 58AC 7 | 4 89 2389 | | 4 359 25 | 58 1 39 | 6 7 2389 | :---------------------+---------------------+---------------------: | 1578 15 6 | 4 58 2 | 9 3 17 | | 15789 159 158B | 58 3 6 | 17 2 4 | | 3 2 4 | 7 9 1 | 8 5 6 | '---------------------'---------------------'---------------------'

allowing us to eliminate 5 from r5c3. Then we transport B to r1c1 and get:

Code: Select all: .---------------------.---------------------.---------------------. | 15B 4 9 | 3 2 8 | 157 6 17 | | 2 6 158 | 19 7 4 | 135 189 389 | | 18 7 3 | 19 6 5 | 2 4 89 | :---------------------+---------------------+---------------------: | 6 8 7 | 2 4 39 | 13 19 5 | | 159- 1359 125 | 6 58AC 7 | 4 89 2389 | | 4 359 25 | 58 1 39 | 6 7 2389 | :---------------------+---------------------+---------------------: | 1578 15 6 | 4 58 2 | 9 3 17 | | 15789 159 158 | 58 3 6 | 17 2 4 | | 3 2 4 | 7 9 1 | 8 5 6 | '---------------------'---------------------'---------------------'

allowing us eliminate 5 from r5c1.

by **Sudtyro** » Sun Sep 23, 2007 12:41 pm

Thanks, re’born, for the excellent “transport” examples and explanations.

Without knowing the technique’s name, I’ve actually been using something pretty much equivalent when working with a puzzle’s single-digit grids. I first look for the derived strong inferences in any (easy to spot) XYZ-Wings and then try to use those inferences to form grouped single-digit chains.

For example, the ALS’s in Steve R’s XYZ-Wing can be linked as
(5=9)r9c3 – (9=35)r48c2, or
(5=3)r4c2 – (3=95)r8c2|r9c3.
Hence, the derived strong inferences are
(5)r9c3 = (5)r48c2, and
(5)r4c2 = (5)r8c2|r9c3.

Looking now at the first tower of the 5’s grid (see below), one can then easily spot the two grouped AIC’s (effectively, grouped Turbot fish)
(5): r9c3 = r48c2 – r1c2 = r3c1 => r8c1 <> 5.
(5): r4c2 = r8c2|r9c3 – r8c1 = r3c1 => r1c2 <> 5.

Code: Select all: --------- . 5 . | . . . | 5 . . | --------+ . 5 5 | . . . | . . . | --------+ . . . | 5 5 . | . . 5 | ---------

You also pointed out the second elimination (r1c2 <> 5) in stating that “... it allows you to consider an xyz-wing with only 2 cells

.”

Hmmm...I think I’d better stick with the single-digit AIC’s!