Bands and low-clue puzzles

Everything about Sudoku that doesn't fit in one of the other sections

Re: Bands and low-clue puzzles

Postby champagne » Thu Feb 25, 2016 3:48 am

blue wrote:.

This is something I put in a PM to champagne, to show why the non-minimal cases are needed (at some point in the game):
(...) here is a 17 clue puzzle metioned in Mladen's thread, and a version with bands 2 and 3 filled in from the solution grid.
In the 54+4 puzzle, 2r2c9 is redundant, but (of course) it isn't redundant in the 17-clue puzzle.

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

. . . | . . . | . . .
. . . | . . 1 | . . 2     -> 2r2c9 is redundant
. . 3 | . . . | . 4 .
------+-------+------
6 3 1 | 4 9 8 | 5 2 7
8 7 2 | 1 6 5 | 3 9 4
4 5 9 | 3 7 2 | 1 8 6
------+-------+------
5 9 6 | 2 8 7 | 4 3 1
3 1 8 | 6 5 4 | 2 7 9
7 2 4 | 9 1 3 | 8 6 5



I have many comments to do to the last PM of blue, but i am so busy these days that it it on the wait state.

but more on that point

It is clear to me that having three valid bands does not give a sudoke, that means that we have usually several gangster triplets valids for the same three patterns.

One way to come to a unique solution is to add clues to fill all unavoidable sets.

Doing so, at the end, some bands will have a redudant clue.

The other possibility, likely more restrictive, is to use minimal subpatterns with more clues.

So Mladen is basically right (as Blue), redundant clues will come.

The open question is "in which process" and is is better to start with the file of minimal sub pattern.

As long as we don't know how to handle the situation, it is impossible to answer
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Re: Bands and low-clue puzzles

Postby blue » Thu Feb 25, 2016 5:51 pm

dobrichev wrote:The revised results
(...)

100% matching results, now.
Thank you.
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Re: Bands and low-clue puzzles

Postby coloin » Thu Feb 25, 2016 11:59 pm

champagne wrote:As long as we don't know how to handle the situation, it is impossible to answer

i sense pain here ...

In dobrichevs initial tables with crossing bands - he showed 3 17-puzzles with 8 clues in each band in both crossing directions.....

2 of the 3 puzzles had seven clues in total in one of the other crossing bands [CB]

Maybe this can provide a possible progress.

here is a 17-puzzle similar to those two published - there are 7 clues in B12347
Code: Select all
+---+---+---+                     +--+--+--+
|1..|2..|...|                     |.7|.9|.9|
|.3.|...|4..|                     +--+--+--+
|...|...|...|                     |10|.9|10|
+---+---+---+                     +--+--+--+
|...|.4.|.3.|                     |10|11|10|
|5..|...|..1|                     +--+--+--+    [85 clues / 5 = 17]
|...|89.|...|
+---+---+---+
|..4|...|86.|
|.7.|..1|...|
|...|5..|2..|
+---+---+---+   7 clues in B12347

Code: Select all
+---+---+---+                     +--+--+--+
|1..|2..|...|                     |.8|10|.9|
|.3.|...|.4.|                     +--+--+--+
|...|..1|...|                     |10|10|.9|
+---+---+---+                     +--+--+--+
|...|.43|7..|                     |.9|11|.9|
|5..|...|8..|                     +--+--+--+    [85 clues / 5 = 17]
|...|.6.|...|                     
+---+---+---+                     
|...|5..|1.2|                     
|.6.|7..|...|                     
|..4|...|...|                     
+---+---+---+ 8 clues in B12347

Code: Select all
+---+---+---+                     +--+--+--+
|12.|.3.|...|                     |.9|11|.9|
|...|...|.45|                     +--+--+--+
|...|6..|...|                     |.9|.9|.9|
+---+---+---+                     +--+--+--+
|6..|2..|7..|                     |.9|11|.9|
|...|9.5|...|                     +--+--+--+     [85 clues / 5 = 17]
|...|...|...|
+---+---+---+
|7..|.8.|1..|
|..5|..4|.9.|
|...|...|...|
+---+---+---+  9 clues in B12347 [there is no 10  but there are x2 11s]

There has to be at least a 10 as
Code: Select all
+--+--+--+
| 9| 9| 9|
+--+--+--+
| 9| 9| 9|
+--+--+--+
| 9| 9| 9|
+--+--+--+       81/5 < 17

from two of the 40 maximal patterns,
Code: Select all
+-----+-----+-----+    +-----+-----+-----+ 
|. . .|. . x|. . x|    |. . .|. . x|. x x| 
|. . .|. . x|. . x|    |. . .|. . x|. x x| 
|. . x|. . x|. . x|    |. . .|. . x|. x x| 
+-----+-----+-----+    +-----+-----+-----+ 
|. . .|x x x|x x x|    |. . .|x x x|x x x| 
|. . .|x x x|x x x|    |. . .|x x x|x x x| 
|x x x|x x x|x x x|    |x x x|x x x|x x x| 
+-----+-----+-----+    +-----+-----+-----+ 
|. . .|x x x|x x x|    |. . .|x x x|x x x| 
|. . .|x x x|x x x|    |x x x|x x x|x x x| 
|x x x|x x x|x x x|    |x x x|x x x|x x x| 
+-----+-----+-----+    +-----+-----+-----+ 

with no empty box six clues are necessary ....  like

+---+---+---+   +---+---+---+
|1..|...|...|   |1..|...|2..|
|.2.|...|..3|   |...|.3.|...|
|...|..4|...|   |...|..4|...|
+---+---+---+   +---+---+---+
|...|258|794|   |..5|492|138|
|...|761|235|   |...|386|547|
|..5|439|861|   |...|715|926|
+---+---+---+   +---+---+---+
|...|647|182|   |6..|928|714|
|...|893|657|   |...|643|859|
|.6.|125|349|   |...|157|362|
+---+---+---+   +---+---+---+

and with an empty box in B1 - 7 clues are necessary !!!!

+---+---+---+
|...|...|..1|
|...|..2|...|
|...|.3.|.4.|
+---+---+---+
|...|358|197|
|...|716|524|
|..5|249|386|
+---+---+---+
|...|597|632|
|.6.|184|975|
|7..|623|418|
+---+---+---+

with an empty box [not in Box 1] there are puzzles with 5 clues [but not 4] in a crossing band.
Code: Select all
+---+---+---+   +---+---+---+
|12.|...|...|   |1..|...|...|
|3..|...|...|   |.2.|...|...|
|...|..4|...|   |...|..3|..4|
+---+---+---+   +---+---+---+
|...|296|458|   |...|281|956|
|...|543|127|   |...|564|173|
|..5|781|936|   |.5.|739|248|
+---+---+---+   +---+---+---+
|...|639|712|   |...|328|795|
|...|178|549|   |...|976|481|
|...|425|683|   |...|145|632|
+---+---+---+   +---+---+---+


Pretty sure there are no 17s with 5 clues in one of the crossing bands, there may indeed be a 6 .... but I think unlikely.

So....

the number of valid patterns with 7 clues in two crossing bands [CBs] is calculable [with a complete B5689]
Completing the B12347 and extrapolating the other several thousand B5689s is also easy
Maybe it is possible to do this....... probably the 8 is a step too far ??
Last edited by coloin on Mon Jan 16, 2017 1:18 pm, edited 2 times in total.
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Re: Bands and low-clue puzzles

Postby dobrichev » Fri Feb 26, 2016 6:43 am

Hi Coloin,

Essentially you propose different approach. Instead of splitting the search space in 3 bands and exploit orthogonality, the space can be split into 2 pieces: B12347 + B5689.
Our experience in this area is
- all valid grids generation by fixing B123 then B4 then B5
- some constrains to patterns having at least one unique puzzle with B5689 all given (but not fixed to particular values!)
- investigations on fractal pattern where this approach gave result due to the extreme symmetry of the pattern

Enumerating all valid puzzles of size N for fixed givens (= pattern with values) in B12347 is achievable, but how many such combinations have to be examined?
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Re: Bands and low-clue puzzles

Postby dobrichev » Fri Feb 26, 2016 7:21 am

Clarification of the post for the maximal clues in band.
If we enumerate puzzles by band population by examining

B0+B1+B3
0+0+17
0+1+16
...
17+0+0

and apply constrains to the maximal number of givens per band to be examined, then we know that x+y+8 populations are valid and must be examined.
We know x+y+9 exemplars, but after rotation they become m+n+(<9), and would be found after x+y+(<=8) enumeration.
We know 3 exemplars having x+y+8 population in both directions.
We don't know whether x+y+(9+) in both directions isn't valid - we just haven't such exemplar.
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Re: Bands and low-clue puzzles

Postby coloin » Fri Feb 26, 2016 11:28 am

dobrichev wrote:Enumerating all valid puzzles of size N for fixed givens (= pattern with values) in B12347 is achievable, but how many such combinations have to be examined?

Probably wiser to explore feasibility before doing anything !
I am not sure how many essentially different B12347 and B5689 there are - it was examined a long time ago.
I think enumerating all patterns for the 7-clues-in-B12347 is very possible - just add a clue to each of the relevant maximal patterns in Serg's 40-patterns. I am expecting there to be fewer patterns than initially invisaged ....

As a preliminary
How easy is it for your program to find the 17-puzzle [ with a full B5689] from
Code: Select all
+---+---+---+                   
|1..|2..|...|                 
|.3.|...|4..|
|...|...|...|                     
+---+---+---+   
|...|.4.|.3.|
|5..|...|..1|   
|...|89.|...|
+---+---+---+
|..4|...|86.|
|.7.|..1|...|
|...|5..|2..|
+---+---+---+  ?


dobrichev wrote:We don't know whether x+y+(9+) in both directions isn't valid - we just haven't such exemplar.

There are options
Code: Select all
+--+--+--+
|..|..|..|
+--1--+.7+
|..|..|..|
+--+--+--+
|..7..|.2|
+--+--+--+    2 clues in B9 - no 17 - as only 1 clue in B1245

+--+--+--+
|..|..|..|
+--2--+.6+
|..|..|..|
+--+--+--+
|..6..|.3|
+--+--+--+    3 clues in B9 - probably easy to prove
 
+--+--+--+
|..|..|..|
+--3--+.5+
|..|..|..|
+--+--+--+
|..5..|.4|
+--+--+--+    4 clues in B9 - probably easy to prove too ?

+--+--+--+
|..|..|..|
+--4--+.4+
|..|..|..|
+--+--+--+
|..4..|.5|
+--+--+--+    5 clues in B9 there are only 6 17-puzzles with 5 clues in a box - extensively searched for - but not proven

              6 clues in B9 - also not proved - but almost certainly one needs 12 clues in B12345678


As an aside, in a 17-puzzle, there must always be 7 or less [ not 8] clues at least one of the 9 "B1245"
because adding up all the [9] B1245 in a 17-puzzle - the total has to come to 68 - so there has to be a B1245 with 7 - or less clues.
Code: Select all
+--+--+--+
|.7|.8|.7|
+--+--+--+
|.8|.8|.8|
+--+--+--+
|.7|.8|.7|
+--+--+--+     68/4 = 17


Looking through the 17-puzzles - typically 7 and 6 clues are to be found in one of the representative B1245s ..... not sure if there can be a 5[B1245] plus 12[B36789] .... hmmm
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Re: Bands and low-clue puzzles

Postby dobrichev » Fri Feb 26, 2016 6:54 pm

coloin wrote:How easy is it for your program to find the 17-puzzle [ with a full B5689] from
Code: Select all
+---+---+---+                   
|1..|2..|...|                 
|.3.|...|4..|
|...|...|...|                     
+---+---+---+   
|...|.4.|.3.|
|5..|...|..1|   
|...|89.|...|
+---+---+---+
|..4|...|86.|
|.7.|..1|...|
|...|5..|2..|
+---+---+---+  ?



Could you be more specific what from the sub-grids and masks below is given and what is the target?
Code: Select all
1..2.....
.3....4..
.........
....4..3.
5.......1
...89....
..4...86.
.7...1...
...5..2.. A

149256783
732918456
658437129
927145638
583762941
461893572
214379865
875621394
396584217 B

1..2.....
.3....4..
.........
...145638
5..762941
...893572
..4379865
.7.621394
...584217 C

?..?.....
.?....?..
.........
...145638
?..762941
...893572
..?379865
.?.621394
...584217 D

1..2.....
.3....4..
.........
...??????
5..??????
...??????
..4??????
.7.??????
...?????? E

?..?.....
.?....?..
.........
...??????
?..??????
...??????
..???????
.?.??????
...?????? F

C->A
D->A
E->A
F->A
D->C
E->C
F->C
F->E
???
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Re: Bands and low-clue puzzles

Postby coloin » Sun Feb 28, 2016 10:42 am

Sorry for not being totally clear.
The case would be:
Code: Select all
1..2.....
.3....4..
.........
...145638
5..762941
...893572
..4379865
.7.621394
...584217

to

1..2.....
.3....4..
.........
....4..3.
5.......1
...89....
..4...86.
.7...1...
...5..2..

Code: Select all
+---+---+---+
|...|...|.83|
|...|...|...|
|...|...|12.|
+---+---+---+
|...|...|.38|
|..3|..2|..1|
|..1|..3|..2|
+---+---+---+
|...|...|...|
|..5|6..|...|
|..6|5..|21.|
+---+---+---+ 4 unavoidable sets - need a clue ...



coloin wrote:Looking through the 17-puzzles - typically 7 and 6 clues are to be found in one of the representative B1245s ..... not sure if there can be a 5[B1245] plus 12[B36789] .... hmmm

it seems the 5 clues in B1245 is not rare .....
Code: Select all
+---+---+---+
|...|...|...|
|...|..1|..2|
|..3|...|.4.|
+---+---+---+
|...|2..|.3.|
|.1.|...|...|
|7..|...|8..|
+---+---+---+
|...|...|5..|
|..2|.6.|3..|
|.5.|.7.|18.|
+---+---+---+

5 clues in B1245 ...., count for the 4 boxes displayed in the disjoint box
Code: Select all
+--+--+--+
|.9|10|.7|
+--+--+--+
|.9|.9|.6|
+--+--+--+
|.6|.7|.5|
+--+--+--+     68/4 = 17

EDIT - this puzzle/pattern is unusual - in that it doesn't have a B5689 equivalent which has 8 clues / doesn't have a B12347 which has 9 clues !
Last edited by coloin on Thu Mar 03, 2016 12:42 am, edited 1 time in total.
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Re: Bands and low-clue puzzles

Postby m_b_metcalf » Mon Feb 29, 2016 7:25 pm

coloin wrote:As a preliminary
How easy is it for your program to find the 17-puzzle [ with a full B5689] from
Code: Select all
+---+---+---+                   
|1..|2..|...|                 
|.3.|...|4..|
|...|...|...|                     
+---+---+---+   
|...|.4.|.3.|
|5..|...|..1|   
|...|89.|...|
+---+---+---+
|..4|...|86.|
|.7.|..1|...|
|...|5..|2..|
+---+---+---+  ?


I remembered writing a recursive procedure, back in 2007, that could do this (although it's spectacularly inefficient). Here is its record of each new minimum:
Code: Select all
 1 . . 2 . . . . .
 . 3 . . . . 4 . .
 . . . . . . . . .
 . . . 1 4 5 6 3 8
 5 . . 7 6 2 9 4 1
 . . . 8 9 3 5 7 2
 . . 4 3 7 9 8 6 5
 . 7 . 6 2 1 3 9 4
 . . . 5 8 4 2 1 7

 Final count: 21
 1 . . 2 . . . . .
 . 3 . . . . 4 . .
 . . . . . . . . .
 . . . . 4 . . . 8
 5 . . . . . . . 1
 . . . . 9 3 . . .
 . . 4 . . . . . 5
 . 7 . . . 1 3 9 .
 . . . 5 8 . 2 1 7

 Final count: 20
 1 . . 2 . . . . .
 . 3 . . . . 4 . .
 . . . . . . . . .
 . . . . 4 . . . 8
 5 . . . . . . . 1
 . . . . 9 3 . . .
 . . 4 . . . . . 5
 . 7 . 6 . 1 . 9 .
 . . . 5 8 . 2 1 .

 Final count: 19
 1 . . 2 . . . . .
 . 3 . . . . 4 . .
 . . . . . . . . .
 . . . . 4 . . . 8
 5 . . . . . . . 1
 . . . . 9 3 . . .
 . . 4 . . . . 6 .
 . 7 . . . 1 3 9 .
 . . . 5 8 . 2 . .

 Final count: 18
 1 . . 2 . . . . .
 . 3 . . . . 4 . .
 . . . . . . . . .
 . . . . 4 . . . 8
 5 . . . . . . . 1
 . . . 8 9 3 . . .
 . . 4 . . . 8 6 .
 . 7 . . . 1 . . .
 . . . 5 . . 2 . .

 Final count: 17
 1 . . 2 . . . . .
 . 3 . . . . 4 . .
 . . . . . . . . .
 . . . . 4 . . 3 .
 5 . . . . . . . 1
 . . . 8 9 . . . .
 . . 4 . . . 8 6 .
 . 7 . . . 1 . . .
 . . . 5 . . 2 . .

The bad news it that it took 30 minutes to get there, but it was a good test that it works. But, clearly, it's not up to the job of carrying out research projects.

Regards,

Mike Metcalf
Last edited by m_b_metcalf on Wed Mar 02, 2016 11:05 am, edited 1 time in total.
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Re: Bands and low-clue puzzles

Postby dobrichev » Tue Mar 01, 2016 6:33 am

So, taking a valid puzzle, finding its solution grid, and searching for puzzles with the same number of givens having the same solution grid and same givens within one band and one stack, took 3266 seconds CPU time for the known 49157 17-clue puzzles.
Each puzzle is tested against all 3 * 3 = 9 combinations of "fixed" givens and non-givens in a band and stack. The grid preparation takes most of the time. Roughly 15 puzzles per second are processed and maybe "clearing" the rest of the combinations of 4 boxes wouldn't significantly affect the processing time.
It is a modification of old code and I am not sure whether there are no bugs. At least all known puzzles have been reproduced.

Update: It took 1577 seconds for the same puzzles when fixing only band 1 and stack 1, ~30 puzzles per second.
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Re: Bands and low-clue puzzles

Postby coloin » Wed Mar 02, 2016 10:48 pm

Indeed that is very impressive .... and I can see that analyzing the whole grid solution - then fixing clues - will speed up the checking process.
However despite that speed as ever we are still facing a massive task to find new 17s .....

I have looked at the B12347 pattern combinations which are valid.

5 clues - 34 patterns found
6 clues - at least 340 - probably a few more .....
7 clues [ no clue in B1] - 20 patterns found

Anyhow - it would seem that the number of valid patterns with 7 clues anywhere in B12347 is going to be significantly more ...

Also the number of ed ways to complete B12347 is considerably more than the 416 ^2. The number of ed ways to complete this pattern
Code: Select all
+---+---+---+
|127|456|389|
|348|912|567|
|56.|...|...|
+---+---+---+
|78.|...|...|
|91.|...|...|
|23.|...|...|
+---+---+---+
|45.|...|...|
|67.|...|...|
|89.|...|...|
+---+---+---+   [at least two clues are needed for this pattern, one of the two must be in B1]
is more than 126000 .....

Anyhow it seems there are way more possibilities than we can ever search .....
It would be interesting to see what proportion of B12347 is solvable in 5 clues, 6 clues, 7 clues [with an empty box 1].
I looked through many of the 17-puzzles but i could not find a puzzle with 6 clues in B12347 - but that is too big a search in any case.
There will be some B12347 which cant be completed in 7 clues - ie have 8 unavoidable clues .....

C
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Re: Bands and low-clue puzzles

Postby champagne » Thu Mar 03, 2016 2:47 am

Still too early to be affirmative, but I see, using a variation of the blue code based on valid puzzles in a band to search all 27 + X + 3 for new 17's (if any)

I should have a better idea in some weeks, when I'll be back home, unless meantime blue comes to a conclusion.
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Re: Bands and low-clue puzzles

Postby champagne » Thu Mar 10, 2016 10:03 am

One week later, I am more confident that the idea will work.

From blue enumerations, we know that the costly step, in his process to generate some 27+X+Y puzzles is the brute force, checking, in nearly all cases, that the puzzle has multiple solutions.

I made some tests on the gangster type
123456789124357689136278459
index 34 in blues table.

that gangster has no automorphism, so it is a very very long to process it against the ED X+Y patterns.

I made the test on the 3+5 file.
In blue's process, the elementary task is to process one of the morphs of the ED basic patterns, having cleared all possible redundancy seen.

The current process gives 4 282 308 calls but only 432 different patterns for the band having 3 clues (here after band3.

These 432 patterns for the band3 have been tested against all possible gangsters for the band3 derived from the gangster 34 in band1.

2 remarks at that point

I got 80136 band 3 (3 given) which is an average of 185.5 bands per pattern not so far from a maximum possible 6x6x6 = 216.
I finished that step with 29 459 760 valid band3 an average of 368 possible gangsters per band 3



No evidence here that we can here reduce significantly (the target should be a reduction of 80% or more) the number of calls to the brute force at a reasonnable price.

splitting the average 368 gangster per band3, I got the following frequency table

Code: Select all
gangsters; number of bands
<50  13786
<100 10162
<200 10463
<400 13398
<800 21926
>799 10401


I made some tests comparing the number of brute force calls using the pairs {gangster + band3} on one side, a full expansion of the band 2 for a given band3 on the other side.

When you pass 1000 possible gangsters for a given band, you have in that process to test the brute force for all or nearly all of the possible values for the band 2 (5 clues).

But we can do more.

We know that the puzzle will have multiple solutions if several of the gangsters possible for a given band3 (average 368, with the split of the frequency table) give the same solutiion in band 2(5 clues).

One special case is very interesting. If 2 gangsters possible in band 2 have exactly the same expansion, then all the corresponding expanded band 2 will lead to a non valid bands2;3 puzzle.


Let's take one example


Code: Select all
123 456 789 | 124 357 689 | 136 278 459 here is our gangster in band 1


Code: Select all
4           |             |
    1       |             | 
            |             |2            here is the band 3 clues giving several (381) valid bands


Code: Select all
1   1       |             |
1           |1            |
            |             |1            here a valid 5 clues pattern for that 3 clues band


and now 2 different gangsters for the band 2 complementary to valid band3 gangsters

Code: Select all
A   B   C   |D   E   F   |G   H   I   
689 237 145 |789 146 235 |578 469 123
689 237 145 |789 146 235 |578 469 123
689 237 145 |789 146 235 |578 469 123

A   B   C   |D   E   F   |G   H   I   
689 237 145 |789 146 235 |578 349 126
689 237 145 |789 146 235 |578 349 126
689 237 145 |789 146 235 |578 349 126


both will produce exactly the same set of 5 given.
for that entire set , the puzzle will have multiple solutions.


The key point here is that the set of given generated depends only on the columns pattern in the band with 5 clues.
Both gangsters have identical columns 1;2;4;7, this is enough to consider them as equivalent to expand the band with 5 clues.

and we see now what can be the "optimal process"

Shrink the list of (381) gangsters replacing each gangster by the smallest equivalent for the 5 clues pattern,
Flag all gangsters having duplicates and flag as multiples the corresponding sets of 5 clues given

expand the rest (if any) till the brute force check.

My expectation is that such a process in 27+5+3 would cut dramatically the runtime. The code should be ready for tests and validation when I am back home in a little more than a week.
champagne
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Re: Bands and low-clue puzzles

Postby coloin » Thu Mar 10, 2016 11:46 am

hi champagne - trying to follow -

There are only 44 different 27-clue gansters. [ and also 44 different 54-clue double gangsters ....... each 6-tuple in each column having many [6!] variations ...]

Is band 3 the 27 clue Band ? - if so its ok to use as it is the 3-tuple of clues which mean that there can only be one of six possibilities in any of the clues in bands 1 and 2.
So band 2 will have 5 clues
And band one will have 3 clues ....

the ed 416 bands are are always completable with 2-6 clues, a select few with only 3.
The pattern of 3 clues will be smallish - in band one - clue one and clue two have to be in different boxes and different rows [9x6], clue three can be anywhere - is this the 432 patterns ?
perhaps you could use this info to expand your idea ?
C
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Re: Bands and low-clue puzzles

Postby champagne » Thu Mar 10, 2016 3:30 pm

Hi coloin,

I try to decipher your post to see where we are in line and where we are not

As a preliminary statement, in the 27 + X + Y problem, we usually have X >= Y.
This is the way ED X+Y patterns have been expressed.

so here band 1 has 27 given (the gangster), band2 has 5 clues and band 3 has 3 clues.



coloin wrote:There are only 44 different 27-clue gansters. [ and also 44 different 54-clue double gangsters ....... each 6-tuple in each column having many [6!] variations ...]
C


That's ok but considering the band 3 valid bands, we have to consider that each box in band 1 can produce 56 different boxes 1 in band 3,
so, before any automorphism consideration, we have 56 x 56 x 56 different gangsters for band 3 (each of them having a complementary gangster for band 2.

Each of these gangsters can be valid for a set of 3 given in band 3.

This is the way I got the 29 459 760 valid band3



coloin wrote:The pattern of 3 clues will be smallish
clue one and clue two have to be in different boxes and different rows [9x6], clue three can be anywhere - is this the 432 patterns ?
C


Basically, this is true, except that each ED pattern has to be expanded to all morphs against the gangster. This is a point widely discussed in another thread.

The 432 patterns is the count of seen band 3 pattern having processed the 3+5 ED file using the blue code.

The blue code creates all morphs of an ED pattern (exchanging stacks and columns within a stack) and then reduces the count applying automorphisms and filtering all redundancies

Sorting the results, I came to 432 different patterns in band 3.



coloin wrote:the ed 416 bands are are always completable with 2-6 clues, a select few with only 3.


This is the mladen table shown above.

Here, I did not use the table, instead, I tried to solve each of the derived 56 x 56 x 56 gangsters (brute force in a band, a very fast process) for each set of given generated out of a band3 pattern.

hope this will help
champagne
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