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DonM wrote:
Luke451 wrote:
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`Ruud's Sunday Nightmare, 1/6/08.27....3.94...35..1...2........9...3..93.46..3...6........1...4..28...75.5....98. *--------------------------------------------------------------------* | 58     2      7      | 49     58     16     | 14     3      169    | | 9      4      68     | 167    78     3      | 5      126    12678  | | 1      368    3568   | 49     2      5678   | 478    469    6789   | |----------------------+----------------------+----------------------| | 2458  *1678   14568  | 1257   9      12578  | 12478  1245   3      | | 258   *178    9      | 3      578    4      | 6      125    1278   | | 3     *178    1458   | 1257   6      12578  | 12478  12459  12789  | |----------------------+----------------------+----------------------| | 78     9      38     | 2567   1      2567   | 23     26     4      | | 6     *13     2      | 8      4      9      | 13     7      5      | | 47     5      14     | 267    3      267    | 9      8      126    | *--------------------------------------------------------------------*`

Just for the heckuvit, as another Joe solver , this would be my first move targeting another hidden set:
ht(145)r469c3=(5)r3c3-r3c6=r1c5-(5=8)r1c1-(8=6)r2c3 => r4c3<>6 -> r4c2=6

I like the way your move not only winds its way back into the starting set but to a group within the set. I tend to move away from the starting set and forget to ever look back, leaving some fruit on the tree. Another good example of this was the handling of Extreme #122 in the Almost SdC tutorial.

The opening does look unusual, though. I assume the "express route" is a little shorthand you snuck in there, but I don't see where it is. As you wrote it makes sense well enough: if the ht isn't true, then the 5 in r3c3 is, then off to the groups.

I think I get it after all. The groups were not completely spelled out .

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

Luke451 wrote:
DonM wrote:
Luke451 wrote:
Code: Select all
`Ruud's Sunday Nightmare, 1/6/08.27....3.94...35..1...2........9...3..93.46..3...6........1...4..28...75.5....98. *--------------------------------------------------------------------* | 58     2      7      | 49     58     16     | 14     3      169    | | 9      4      68     | 167    78     3      | 5      126    12678  | | 1      368    3568   | 49     2      5678   | 478    469    6789   | |----------------------+----------------------+----------------------| | 2458  *1678   14568  | 1257   9      12578  | 12478  1245   3      | | 258   *178    9      | 3      578    4      | 6      125    1278   | | 3     *178    1458   | 1257   6      12578  | 12478  12459  12789  | |----------------------+----------------------+----------------------| | 78     9      38     | 2567   1      2567   | 23     26     4      | | 6     *13     2      | 8      4      9      | 13     7      5      | | 47     5      14     | 267    3      267    | 9      8      126    | *--------------------------------------------------------------------*`

Just for the heckuvit, as another Joe solver , this would be my first move targeting another hidden set:
ht(145)r469c3=(5)r3c3-r3c6=r1c5-(5=8)r1c1-(8=6)r2c3 => r4c3<>6 -> r4c2=6

I like the way your move not only winds its way back into the starting set but to a group within the set. I tend to move away from the starting set and forget to ever look back, leaving some fruit on the tree. Another good example of this was the handling of Extreme #122 in the Almost SdC tutorial.

The opening does look unusual, though. I assume the "express route" is a little shorthand you snuck in there, but I don't see where it is. As you wrote it makes sense well enough: if the ht isn't true, then the 5 in r3c3 is, then off to the groups.

Luke
You've fallen into a slight trap, which I would like to point out because it's important in the understanding of hidden sets : this statement of yours "if the ht isn't true, then the 5 in r3c3 is" is wrong.
not the hidden triple means precisely that : the combination 145 is excluded, but only that total combination (to achieve which there must necessarily be either a 6 or an 8 in the relevant cells). Combinations involving 5 remain at this stage possible (eg 156, 458 etc).
The necessary 6 or 8 then unite with r2c3 to form a pair 68 which places 3 in r7c3, all of which removes 368 from r3c3 to leave 5r3c3.
So Don was being very concise
aran

Posts: 334
Joined: 02 March 2007

DonM wrote:
Code: Select all
`Ruud's Sunday Nightmare, 1/6/08.27....3.94...35..1...2........9...3..93.46..3...6........1...4..28...75.5....98. *--------------------------------------------------------------------* | 58     2      7      | 49     58     16     | 14     3      169    | | 9      4      68     | 167    78     3      | 5      126    12678  | | 1      368    3568   | 49     2      5678   | 478    469    6789   | |----------------------+----------------------+----------------------| | 2458  *1678   14568  | 1257   9      12578  | 12478  1245   3      | | 258   *178    9      | 3      578    4      | 6      125    1278   | | 3     *178    1458   | 1257   6      12578  | 12478  12459  12789  | |----------------------+----------------------+----------------------| | 78     9      38     | 2567   1      2567   | 23     26     4      | | 6     *13     2      | 8      4      9      | 13     7      5      | | 47     5      14     | 267    3      267    | 9      8      126    | *--------------------------------------------------------------------*`

Just for the heckuvit, as another Joe solver , this would be my first move targeting another hidden set:
ht(145)r469c3=(5)r3c3-r3c6=r1c5-(5=8)r1c1-(8=6)r2c3 => r4c3<>6 -> r4c2=6

The way I interpret this move is, that if r3c3 is not 5 (i.e. one of {368}), then we'll have a hidden triple {145} @ r469c3. Of course that will prevent r4c3 from being 6. On the other hand, if r3c3 is 5, then through the chain we can conclude that r2c3 will be forced to be 6, again preventing r4c3 to be 6. So no matter what r4c3 can't be 6.
udosuk

Posts: 2698
Joined: 17 July 2005

aran wrote:You've fallen into a slight trap, which I would like to point out because it's important in the understanding of hidden sets : this statement of yours "if the ht isn't true, then the 5 in r3c3 is" is wrong.

Looks like I tried to pound a square peg into a round hole.

Isn't/is was a very poor choice of words. My understanding is that one node simply exercizes an inference on the next, so using words that suggest otherwise is indeed wrong.

I'll take another look at everyone's suggestions later. For now, three Super Bowl parties await!

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

Luke451 wrote:I'll take another look at everyone's suggestions later. For now, three Super Bowl parties await!

That's right. Forget the game; go for the parties!
DonM
2013 Supporter

Posts: 475
Joined: 13 January 2008

Luke451 wrote:
DonM wrote:this would be my first move targeting another hidden set:
ht(145)r469c3=(5)r3c3-r3c6=r1c5-(5=8)r1c1-(8=6)r2c3 => r4c3<>6 -> r4c2=6

As you wrote it makes sense well enough: if the ht isn't true, then the 5 in r3c3 is

Correct, it's equivalent to a finned swordfish, where either the fish is true or at least one of the fin cells is true. Indeed, although by itself it yields no eliminations, it is a finned swordfish in Denis Berthier's rn-space.

Ruud's SudoCue allows us to view rn-space without doing a lot of work. Paste these pencilmarks into SudoCue and select View->RN View. You'll see the '3' -- for the original c3 -- in three n1, n4, and n5 digit "columns" can be covered by three rows with a leftover fin at r3n5. However, viewing rn-space is totally unnecessary.
ronk
2012 Supporter

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Location: Southeastern USA

To recapitulate on
ht(145)r469c3=(5)r3c3

Approach 1 : ht(145)r469c3=68r246c3-(8=3)r7c3-(368=5)r3c3.
Approach 2 : noting that the bivalue 14 in r3c9 is conjugate with 14r46c3, this must definitely appear in all versions of r469c3, hence to avoid 145 there cannot be 5 in r46c3 hence 5r3c3.
Approach 3 : go into RN space, note that there is a finned swordfish as follows :
On “3” (representing Column 3)
R4N1 R6N1 R9N1
R4N4 R6N4 R9N4
R4N5 R6N5
Fin R3N5.
Then reason : not swordfish (ie not hidden triple back in RC space)=>Fin true=>C3R3N5 true ie 5r3c3.

In my reply to Don : I assumed that Approach 1 had been compressed, but Don may well have used Approach 2.
In my reply to Luke, I thought he had been snared by a logical trap.
But he may well have intended Approach 2.
If appropriate, best apologies

For the hand solver, taking a trip into new space is a labour of love, and as has been pointed out, is certainly not necessary in this instance.
Last edited by aran on Mon Feb 02, 2009 1:59 am, edited 1 time in total.
aran

Posts: 334
Joined: 02 March 2007

I don't understand the strong link claim for

Code: Select all
`ht(145)r469c3=(5)r3c3`

In particular: aran's Approach 1 appears to be a forcing network. His Approach 2 lacks a supporting chain. Going off into Denis Berthier spaces for Approach 3 seems unnecessary.

At best, I see a forcing network for [r46c3]<>68

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`[r3c3]<>5 [r237c3]=368                    => [r46c3]<>68                               [r2c3]=6                   /                    \[r3c5]= 5 [r1c1]=8                        => [r46c3]<>68                   \                    /                     [r7c1]<>8 [r7c3]=8`
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

aran wrote:To recapitulate on
ht(145)r469c3=(5)r3c3

Approach 1 : ht(145)r469c3=68r246c3-(68=3)r7c3-(368=5)r3c3.
Approach 2 : noting that the bivalue 14 in r3c9 is conjugate with 14r46c3, this must definitely appear in all versions of r469c3, hence to avoid 145 there cannot be 5 in r46c3 hence 5r3c3.
Approach 3 : go into RN space, note that there is a finned swordfish as follows :
On “3” (representing Column 3)
R4N1 R6N1 R9N1
R4N4 R6N4 R9N4
R4N5 R6N5
Fin R3N5.
Then reason : not swordfish (ie not hidden triple back in RC space)=>Fin true=>C3R3N5 true ie 5r3c3.

In my reply to Don : I assumed that Approach 1 had been compressed, but Don may well have used Approach 2.
In my reply to Luke, I thought he had been snared by a logical trap.
But he may well have intended Approach 2.
If appropriate, best apologies

For the hand solver, taking a trip into new space is a labour of love, and as has been pointed out, is certainly not necessary in this instance.
Save your apologies for those who deserve one, because in regards to me you were right in your original assessment. I had written, "The opening does look unusual, though," but plowed ahead anyway (I blame Super Bowl Anxiety Syndrome.)

I understand Don's entire move now, and can appreciate why the "express route" was taken. I also see what I interpret as the use of a little memory at the end of Approach 1.

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

daj95376 wrote:I don't understand the strong link claim for

Code: Select all
`ht(145)r469c3=(5)r3c3`

In particular: aran's Approach 1 appears to be a forcing network.

In fact it is neither forcing nor a network
IMHO you are confusing forcing with implying :
forcing means
this cell is either a or b
if it is a then that cell is c
if it is b then that cell is c
therefore that cell is c.
it does not mean
if this cell is a, then that cell is b : which is implication
it does not mean :
if this cell is a, which means that cell is b, which jointly mean that other cell is c : which is still implication
Implication derived from an unbranched chain is neither forcing not networked.
daj95376 wrote:His Approach 2 lacks a supporting chain.

145r469c3=ALS1468r469c3-5r46c3=5r3c3
daj95376 wrote:Going off into Denis Berthier spaces for Approach 3 seems unnecessary
I entirely agree : for the handsolver, it's a huge effort for no extra return.
aran

Posts: 334
Joined: 02 March 2007

daj95376 wrote:I don't understand the strong link claim for

Code: Select all
`ht(145)r469c3=(5)r3c3`

In hindsight that's a perfectly valid claim.

Because clearly we have the "grouped strong link":

Code: Select all
`(5)r46c3=(5)r3c3`

But in that state (5)r46c3 (5 locked @ r46c3 for c3) is equivalent to ht(145)r469c3 since {14} is locked @ r469c3 for c3.

So in words, that would mean: "On c3, either 5 is locked in r46c3, or 5 is placed in r3c3. If the former happens, then there is a hidden triple of {145} in r469c3. If the latter happens, then r3c3=5. So (exactly) one of the following 2 must happen: a hidden triple of {145} in r469c3, or r3c3=5."
udosuk

Posts: 2698
Joined: 17 July 2005

And then there's always the constraint set POV.

Base Sets (5): 145c3, r1c1 and r2c3

Cover sets (6): r469c3, 58b1, 6c3

Elimination: the double-covered 6r4c3

To no one in particular, do you see the analogous finned swordfish now For the graphic, an honorable mention to Allan Barker.

aran wrote:Approach 3 : go into RN space, note that there is a finned swordfish as follows :
On “3” (representing Column 3)
R4N1 R6N1 R9N1
R4N4 R6N4 R9N4
R4N5 R6N5
Fin R3N5.
Then reason : not swordfish (ie not hidden triple back in RC space)=>Fin true=>C3R3N5 true ie 5r3c3.

You are twisting words. My presentation was an analogy, not an "approach".
ronk
2012 Supporter

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Joined: 02 November 2005
Location: Southeastern USA

aran wrote:
daj95376 wrote:I don't understand the strong link claim for

Code: Select all
`ht(145)r469c3=(5)r3c3`

In particular: aran's Approach 1 appears to be a forcing network.

In fact it is neither forcing nor a network
IMHO you are confusing forcing with implying :
forcing means
this cell is either a or b
if it is a then that cell is c
if it is b then that cell is c
therefore that cell is c.
it does not mean
if this cell is a, then that cell is b : which is implication
it does not mean :
if this cell is a, which means that cell is b, which jointly mean that other cell is c : which is still implication
Implication derived from an unbranched chain is neither forcing not networked.

Well, I'm working from the concept that chains are based on only one condition causing another. Once you explain this link in terms of one condition causing another, then I'll retract my statement that it appears that you are using a forcing network.

Approach 1 : ht(145)r469c3=68r246c3-(8=3)r7c3-(368=5)r3c3.

And, if it takes an ALS to support your Approach 2, then you haven't convinced me of anything.

Bottom Line: This whole mess could be replaced by a simple (and real) chain

Code: Select all
`[r3c3]=3=[r7c3]=8=[r7c1]-8-[r1c1]-5-[r3c3]  =>  [r3c3]<>5(3)r3c3 = (3-8)r7c3 = (8)r7c1 - (8=5)r1c1 - (5)r3c3  =>  [r3c3]<>5`
daj95376
2014 Supporter

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Joined: 15 May 2006

daj95376 wrote:Well, I'm working from the concept that chains are based on only one condition causing another. Once you explain this link in terms of one condition causing another, then I'll retract my statement that it appears that you are using a forcing network

If a causes b, and b causes c, the a has caused b and c.
daj95376 wrote:And, if it takes an ALS to support your Approach 2, then you haven't convinced me of anything

I don't much like ALSs myself but on occasion do call upon them.
daj95376 wrote:Bottom Line: This whole mess

Mess, what mess ?....
aran

Posts: 334
Joined: 02 March 2007

ronk wrote:And then there's always the constraint set POV.

Base Sets (5): 145c3, r1c1 and r2c3
Cover sets (6): r469c3, 58b1, 6c3
Elimination: the double-covered 6r4c3

An alternative combining the delights of both worlds
Sets : r1c1, r2c3, r3c3, r7c3
Linksets : 5b1, 3c3, 6c3, 8c3.
=>Rank 0, except for the fact that 8r1c1 isn't covered.
But 8r1c1=>6r2c3=><6>r4c3.
And Rank 0=><6>r4c3
Thus <6>r4c3

ronk wrote:You are twisting words. My presentation was an analogy, not an "approach".

Good heavens, if that is important to you, I withdraw "approach" forthwith.
aran

Posts: 334
Joined: 02 March 2007

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