help with puzzle - what could be my next move

Post the puzzle or solving technique that's causing you trouble and someone will help

Postby Steve K » Wed Apr 09, 2008 9:39 am

1a) hpl 1589 => b45<>26,h5<>37,de2<>27,ef8<>67,a3g9<>8,c1i7<>9

1b) qntx159bdfh5=qntx159e2468 loop => both qnts exist =>
(qnt loop) => ai5<>1,ch5<>5,ci5<>9,e19<>1,e37<>5,e17<>9
alt: (p59=p1x)bh5-(1)df5=(1)e46-(p1x=p59)e28-(5)e46=(5)df5
1c) rsf or csf 9 => d2,h6<>9
1d)All puzzle conatining hp loops, (formally generalized as a simple interplay of 8 box sis), such that the four focal cells, in this case r5c28,r28c5, have these cells limited to the same 3 or fewer candidates, in this case 159, will have the EMZ property such that for example (8=9),(8-9)r56c2. (A true quantum cell) This puzzle is bit simpler than many hpl puzzles, as only depth 7 is required max to achieve the 16 relationships. Once (1) is determined, the other 15 follow the hp loop path around the grid.
One does not have to go as deep as the grid determination detail below, but given the symmetry, it takes almost no additional human effort:
[(9)'s r5c2&r8c2]-(9)r8c5&r5c8=loop15Box9-(15=8)r6c8&r2c4&(8)grid determined]=
[(9)'s r5c8&r3c4-(9)r2c5&r5c2=loop15Box1-(15=8)r2c6&r6c2&(8)grid] loop
=> r4c1,r1c4,r7c6,r6c7<>8 note we have also proven Z equiv EM.
Here is a sample chain: (915)r5c2,(815)r5c2, (1)Box 1, (5)Box1,
(915)r2c5,(9)r8c53,(9)r94c2 => (8=9)r56c2. By symmetry, one knows:
(8=9)r45c8. Then, (8=9)r56c2-(9=8)r45c8 =< r4c1,r6c7<>8. One can break this down into small chunks, or do it all at once. There are several ways to reach the same conclusions. There need be no T&E.
Probably from this point, if Denis could program in the 16 weak and strong links proven in 1d, the puzzle should solve with *t chains. These are new cells, and no inferences are required.
Code: Select all
 
 *-----------------------------------------------------------*
 | 9     126   237   | 124   2467  78    | 3467  378   5     |
 | 16    4     567   | 3     19    158   | 78    2     679   |
 | 237   256   8     | 2459  2467  2457  | 1     39    3467  |
 |-------------------+-------------------+-------------------|
 | 124   7     2459  | 6     1245  3     | 2458  18    249   |
 | 2346  59    2346  | 1245  8     124   | 2347  159   2347  |
 | 18    158   2345  | 7     245   9     | 2345  6     1234  |
 |-------------------+-------------------+-------------------|
 | 246   68    1     | 2458  2347  2457  | 9     57    367   |
 | 78    3     79    | 58    159   6     | 25    4     12    |
 | 5     269   24    | 149   2347  1247  | 367   137   8     |
 *-----------------------------------------------------------*

Above is the puzzle at the point my mind got tired. There should still be many steps, clearly not T&E, available. I have one more letter left in the alphabet.

Denis, it appears that your T&E is very, very deep. Piecewise shallow, but since many determinations build on each other, deep in total.
Edit: depth 7 required required for step 1d series. By symmetry, similar other depth 7 or less steps exist to prove all 16 relationships. All of those steps are purely AIC. None are T&E. If one chooses not to remember the newly proven pseudo cells, one can reprove whichever relationship one requires each time. Added one example of the series.
Last edited by Steve K on Thu Apr 10, 2008 8:52 am, edited 2 times in total.
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Postby denis_berthier » Wed Apr 09, 2008 10:40 am

Steve K wrote:Probably from this point, if Denis could program in the 16 weak and strong links proven in
1d, the puzzle should solve with *t chains.

As my approach is not based on weak and strong links, no chance I'd program them in my solver.
If you can provide a list of eliminations done (apart from the 15 done by your loop, already used in my solution), I could try to restart from this point.

Steve K wrote:Denis, it appears that your T&E is very, very deep. Piecewise shallow, but since many determinations build on each other, deep in total.

I fear you're confused by an imprecise vocabulary and you don't make a difference between the depth of recursive T&E and the length of chain patterns.
I'm using recursive T&E pruned by the nrczt-rules. As I need eliminate only one candidate by T&E, this is the shallowest form of T&E one can imagine.
Within this single elimination, I use long chains, but it is not "deep" in any sensible sense, the T&E local depth of this elimination is zero.
The global depth of T&E is thus 1.

Using nrczt rules to prune the search thus allows to have depth 1 instead of the depth 6 you announced when you use only singles to prune the search. (In any case, depth 6 is not optimised, it is very likely that depth 3 or even 2 is enough)
Notice that, when you announce depth 6, you're supposing that the pruning done by elementary contraints and singles don't count as additional depth.
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Postby ronk » Wed Apr 09, 2008 10:45 am

Pat, your post with the long string of underscores is very user-unfriendly. It changes every other paragraph -- those encapsuled in the HTML < P > tag -- into one long line, forcing us to use the horizontal scroll bar for other posts on the same page.

Please figure out some other way to present your data ... and edit the post accordingly. Thanks, Ron
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Postby denis_berthier » Wed Apr 09, 2008 10:48 am

ronk wrote:Please figure out some other way to present your data ... and edit the post accordingly. Thanks, Ron


I had to do this for my solver. Here are 2 easier formats.

9.......5
.4.3...2.
..8...1..
.7.6.3...
....8....
...7.9.6.
..1...9..
.3...6.4.
5.......8


900000005040300020008000100070603000000080000000709060001000900030006040500000008
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Postby tarek » Wed Apr 09, 2008 11:11 am

denis_berthier wrote:Apart from the usual permutations of rows, columns, numbers, floors and towers, it doesn't seem there are other possibilities for the SK-loop to appear but from this EM core.
Ah,

So it is puzzle-core specific rather than configuration/symmetry specific.

Thanx

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Postby ronk » Wed Apr 09, 2008 11:35 am

tarek wrote:the 4 boxes (containing 1 clue each) that flank the centre box can easily be transposed from a double diagonal to achieve a full rotational symmetry & vice versa (a II+ symmetry:D ).

Does this property have to do with the SK loop ?

Not sure I understand your question, but fully-rotationally-symmetric clues in the center box of ...
Code: Select all
 1 . . | . . . | . . 2
 . 3 . | 4 . . | . 5 .
 . . 6 | . . . | 7 . .
-------+-------+-------
 . . . | O O O | . . .
 . 4 . | O O O | . 3 .
 . . . | O O O | . . .
-------+-------+-------
 . . 7 | . . . | 6 . .
 . 5 . | . . 3 | . 4 .
 2 . . | . . . | . . 1

... would not produce a fully-rotationally-symmetric puzzle. The puzzle would, however, contain an "SK loop."

I don't know whether or not such a puzzle can also be minimal.
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Postby denis_berthier » Wed Apr 09, 2008 11:42 am

tarek wrote:
denis_berthier wrote:Apart from the usual permutations of rows, columns, numbers, floors and towers, it doesn't seem there are other possibilities for the SK-loop to appear but from this EM core.
Ah,

So it is puzzle-core specific rather than configuration/symmetry specific.

Thanx

tarek


It seems to be so. The only important data are:
- those in the four outer blocks
- plus 4 entries (one per central row crossing these blocks and one per central column these blocks) that eliminate candidates that would otherwise block the pattern
All other entries are only there to insure uniqueness of the solution.
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Postby tarek » Wed Apr 09, 2008 12:50 pm

ronk wrote:
tarek wrote:the 4 boxes (containing 1 clue each) that flank the centre box can easily be transposed from a double diagonal to achieve a full rotational symmetry & vice versa (a II+ symmetry:D ).

Does this property have to do with the SK loop ?

Not sure I understand your question, but fully-rotationally-symmetric clues in the center box of ...
Code: Select all
 1 . . | . . . | . . 2
 . 3 . | 4 . . | . 5 .
 . . 6 | . . . | 7 . .
-------+-------+-------
 . . . | O O O | . . .
 . 4 . | O O O | . 3 .
 . . . | O O O | . . .
-------+-------+-------
 . . 7 | . . . | 6 . .
 . 5 . | . . 3 | . 4 .
 2 . . | . . . | . . 1

... would not produce a fully-rotationally-symmetric puzzle. The puzzle would, however, contain an "SK loop."

I don't know whether or not such a puzzle can also be minimal.


You are right ronk. I was referring to the whole puzzle not just the centre box..... many fully rotational puzzles produced can be easily isomorphed into a double diagonal symmetry ..... As the SK loop was demonstrated for tarek-0014 (which is one those puzzles), the idea just struck me.

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Postby denis_berthier » Wed Apr 09, 2008 1:20 pm

tarek wrote:
ronk wrote:
tarek wrote:the 4 boxes (containing 1 clue each) that flank the centre box can easily be transposed from a double diagonal to achieve a full rotational symmetry & vice versa (a II+ symmetry:D ).

Does this property have to do with the SK loop ?

Not sure I understand your question, but fully-rotationally-symmetric clues in the center box of ...
Code: Select all
 1 . . | . . . | . . 2
 . 3 . | 4 . . | . 5 .
 . . 6 | . . . | 7 . .
-------+-------+-------
 . . . | O O O | . . .
 . 4 . | O O O | . 3 .
 . . . | O O O | . . .
-------+-------+-------
 . . 7 | . . . | 6 . .
 . 5 . | . . 3 | . 4 .
 2 . . | . . . | . . 1

... would not produce a fully-rotationally-symmetric puzzle. The puzzle would, however, contain an "SK loop."

I don't know whether or not such a puzzle can also be minimal.


You are right ronk. I was referring to the whole puzzle not just the centre box..... many fully rotational puzzles produced can be easily isomorphed into a double diagonal symmetry ..... As the SK loop was demonstrated for tarek-0014 (which is one those puzzles), the idea just struck me.

tarek

In this very special version of the EM core, moving the 4 anywhere in the 2 line within the 2nd block and similar moves for the other 4 and 3's would disrupt any symmetry but keep the SK-loop intact.
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Postby Steve K » Thu Apr 10, 2008 12:41 pm

Per Ronk, I am using the term Quantum (Q) below instead of Hybrid. Quantum is extended, though, to include non-uniqueness arguments.
step 1d: (from above)
AIC style is possible, but unwieldy. Below is a conceptual AIC chain. Note it is not strictly accurate, as AIC does not handle multiple simulataneous inferences with clarity. The portion in bold is where standard AIC fails only in notation: This is analagous to AIC's failure to handle even a standard full valued 3x3 swordfish. Of course, there are many ways to accurately notate the chain.
[(9=8)r56c2]=(pair15)r56c2-(1&5)r13c2=(pair15)r2c13-(1&5=9)r2c5-(9)r2c9=(9)r4c9-(9)r5c8=(9)r5c2
=> (9=8)r56c2

By symmetry, (9=8)r45c2,r8c45,r2c56
(9=8)r56c2-(8)r7c2=(8)r8c1-(8=9)r8c45-(9)r8c7=(9)r9c2 loop =>
(9-8)r56c2,r8c45 ,[(9=8)&(9-8)]r8c13,r79c2. Now, by symmetry:
Q cells (89) at r56c2,r45c8,r2c56,r8c45,r79c2,r8c13,r13c8,r2c79
Symmetrically also these complimentary Quantum cells:
(15) at r56c2,r45c8,r2c56,r8c45,r79c8,r8c79,r13c2,r2c13
Thus, 16 Quantum cells referred to as Z, or Z EM, as this relationship was first uncovered in EM.
Using the quantum cells:
(8=9)r56c2-(9=8)r45c8 => r4c1,r6c7<>8
(8=9)r8c45-(9=8)r2c56 => r1c4,r7c6<>8
Because of the excessive symmetry, once one sees the first relationship, the rest are easy finds. Thus, the conclusions are not difficult, nor are they T&E unless all steps are T&E.

Here is one way to unambigously notate the AIC style chain. The following is a pigeonhole matrix. The first column, a label column, simply notes the sis used in each row for convenience. Each row is a sis - at least one truth per row. Each column, except column 2, contain mutually exclusive items. (no more than one truth per column). As the matrix is nxn, excluding the label column - the first column is a proven sis. The proof is simple: at least n minus at most (n-1) = at least 1
Code: Select all
         
(9=8)r56c2                     
r5c2  9   1   5           
r6c2  8   1   5           
1B1      c2      r2         
5B1          c2       r2     
r2c5              1    5   9   
9c9                       r2  r4
9r5  c2                       c8

If one does not like to remember proven Quantum cells, each deduction that uses this relationship could repeat the information given above. That, of course, makes little logical sense.

A few more steps:
1e) (sf6)r259c137=[(6)r9c2=(6-9)r2c9=(9)r3c8-(9)r5c8=(9)r5c2]-[(9)r2c9&(9)r9c2]=Z[(8-6)r7c2&(8-6)r2c7]=(sf6)r257c139 => r1c3,r3c1<>6

Below, typical hidden pair & Z used in conjuction with a pair, Qp <=>Quantum pair, hidden or naked

1f) (hp46)r7c1,r9c3=(6)r79c2-(6)r13c2=(6)r2c13-(Qp689=Qp789)r2c79-(7)r13c8=(7)r79c8-(7)r8c79=(7)r8c13 =>r7c1,r9c3<>7

1g)Locked 7's r8c13 => r8c79<>7

1h) (Qnp: 2=5-1=2)r8c79 => r8c13,r9c7,r7c9<>2
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