The New Sudoku Players' Forum

by **Mauriès Robert** » Sun Dec 13, 2020 9:30 am

Hi all,
I offer you this puzzle to solve.
.....3.45.....497.7...583..9.7.8..6...6.3.4...8....7.9..389...7.591.....27.3.....

puzzle: Show

Robert

Website · by **denis_berthier** » Sun Dec 13, 2020 9:41 am

Solvable in W5 or in Z6, using only Subsets and reversible chains:

Hidden Text: Show

by **Cenoman** » Tue Dec 15, 2020 4:05 pm

Nine simple steps:

Hidden Text: Show

1. (8)r5c8 = (8-3)r8c8 = r6c8 - r6c1 = (3-58)r2c13 = (8)r2c9 => -8 r5c9
2. (3)r4c9 = (3-4)r8c9 = r9c9 - (4=62)r69c5 - (2=1)r4c6 => -1 r4c9

Hidden Text: Show

3. (8)r9c3 = r8c1 - r2c1 = r2c9 = -8 r9c9
4. Kraken cell (368)r2c1
(3)r2c1 - r6c1 = r6c8 - r8c8 = (3-4)r8c9 = (4)r9c9
(6)r2c1 - (6=81)b7p49
(8)r2c1 - (8=261)r235c9
=> -1 r9c9

Hidden Text: Show

5. (1=23)r38c8 - r6c8 = r6c1 - (3=681)r128c1 => -1 r1c7
6. (1=26)r47c6 - (6=41)r7c12 - r9c3 = (1)r9c7 => -1 r4c7
7. (1=26)r47c6 - r9c5 = r9c9 - (6=124)r3c389 - (4=21)b4p59 => -1 r4c2
8. (2=1)r5c2 - r5c9 = r3c9 - (1=24)r3c38 - (4=21)b4p59 => -2 r4c2

Hidden Text: Show

9. (8=1)r1c1 - r6c1 = r6c3 - (1=8)r9c3 => -8 r1c3, r8c1; ste

by **Mauriès Robert** » Thu Dec 17, 2020 5:49 pm

Hi all,
This puzzle can be solved in different ways, with only two conjugated sets of tracks (level TDP=2), with 3 chains of "reasonable" length or several shorter ones (Cenoman) or according to the principle of the simplest first (Berthier).
I present a resolution with 3 conjugated sets of tracks (JP) by developing each of the tracks in stages and by only looking for validations (intersection), which corresponds well to the way of doing it by hand.

1) JP(3r4)
P(3r4c9) : 3r4c9->3r8c8->8r5c8->...
P(3r4c2) : 3r4c2->(3r2c1->5r2c3)->8r2c9->8r5c8->... => r5c8=8 + 2 placements.

puzzle1: Show

If we continue the development of the two tracks, we also place the 4r7c1 :
P(3r4c9) : ...->3r6c1->4r7c1->...
P(3r4c2) : ...->3r6c8->[(12r45c9 and 3r6c8->2r8c8->2r6c5->2r5c2)->2r1c7]->2r3c3->4r3c2->4r7c1->...
=> r7c1=4

puzzle2: Show

2) JP(8B7)
P(8r8c1) : (8r8c1->8r2c9)->8r9c7->5r9c6->...
P(8r9c3) : 8r9c3->(1r7c2->1r5c9)->1r9c7->5r9c6->...
=>r9c6=5

puzzle3: Show

If we continue the development of the two tracks, we also place the 1r4c6 :
P(8r8c1) : ...->6r7c2->2r7c6->1r7c6->...
P(8r9c3) : ...->1r4c6->...
=> r4c6=1 + elimination by basic techniques.

puzzle4: Show

If we continue the development of the two tracks, we also place the 1r3c8 :
P(8r8c1) : ...->8r1c3->2r3c3->1r3c8->...
P(8r9c3) : ...->1r3c8->...
=> r3c8=1 + elimination by basic techniques.

puzzle5: Show

3) JP(4r3)
P(4r3c2) : 4r3c2->9r3c2->1r7c2->8r9c3->...
P(4r3c3) : 4r3c3->1r6c3->8r9c3->...
=> r9c3=8, stte.

puzzle6: Show

Robert

by **SpAce** » Fri Dec 18, 2020 1:32 am

Hi Robert,

Mauriès Robert wrote:I present a resolution with 3 conjugated sets of tracks (JP) by developing each of the tracks in stages and by only looking for validations (intersection), which corresponds well to the way of doing it by hand.

I agree. Your solution is pretty much how I would have solved it with GEM, except I would have accepted the btte-solution available with the second coloring.

1) JP(3r4)
P(3r4c9) : 3r4c9->3r8c8->8r5c8->...
P(3r4c2) : 3r4c2->(3r2c1->5r2c3)->8r2c9->8r5c8->... => r5c8=8 + 2 placements.
...
If we continue the development of the two tracks, we also place the 4r7c1 :
P(3r4c9) : ...->3r6c1->4r7c1->...
P(3r4c2) : ...->3r6c8->[(12r45c9 and 3r6c8->2r8c8->2r6c5->2r5c2)->2r1c7]->2r3c3->4r3c2->4r7c1->...
=> r7c1=4

Here's what that looks like with GEM:

Code: Select all: .---------------------------.--------------------.--------------------------. | †8‡1-6 '9:26-1 †1:28 | '6:29 7 3 | '2:68-1 4 5 | | †3:68-5 †6‡3-2 +5-28 | '2:6 1 4 | 9 7 '8:26 | | 7 '4:269-1 '2:4-1 | '9:26 5 8 | 3 '1:2 '6:12 | :---------------------------+--------------------+--------------------------: | 9 †3:124 7 | '4:5 8 '1:2 | '5:12 6 †2‡3-1 | | +5-1 '2:1 6 | 7 3 9 | 4 +8-125 '1:2-8 | | †1‡3-45 8 '4:12-5 | '5:4 '2:6 '6:12 | 7 †3:125 9 | :---------------------------+--------------------+--------------------------: | +4-16 '1:6-4 3 | 8 9 '2:56 | '6:125 '5:12 7 | | '6:8 5 9 | 1 '4:26 7 | '8:26 †2‡3-8 †3:2468 | | 2 7 '8:1 | 3 '6:4 '5:6 | '1:568 9 '4:168 | '---------------------------'--------------------'--------------------------'

Step 1. GEM (3R4) => +4 r7c1, +5 r2c3,r5c1, +8 r5c8; -1 r1c27,r3c23,r4c9, -2 r2c2, -6 r1c1

2) JP(8B7)
P(8r8c1) : (8r8c1->8r2c9)->8r9c7->5r9c6->...
P(8r9c3) : 8r9c3->(1r7c2->1r5c9)->1r9c7->5r9c6->...
=>r9c6=5
...
If we continue the development of the two tracks, we also place the 1r4c6 :
P(8r8c1) : ...->6r7c2->2r7c6->1r7c6->...
P(8r9c3) : ...->1r4c6->...
=> r4c6=1 + elimination by basic techniques.
...
If we continue the development of the two tracks, we also place the 1r3c8 :
P(8r8c1) : ...->8r1c3->2r3c3->1r3c8->...
P(8r9c3) : ...->1r3c8->...
=> r3c8=1 + elimination by basic techniques.

The same with GEM:

Code: Select all: .------------------------.-----------------------.---------------------------. | "1.8 "9.6-2 ‡8.12 | "6.29 7 3 | "2.68 4 5 | | ‡6.38 "3.6 5 | "2.6 1 4 | 9 7 "8.26 | | 7 "4.69-2 "2.4 | "9.26 5 8 | 3 +1-2 "6.2-1 | :------------------------+-----------------------+---------------------------: | 9 ‡2.34-1 7 | 45 8 +1-2 | "5.2-1 6 "3.2 | | 5 +†2-1 6 | 7 3 9 | 4 8 +1-2 | | "3.1 8 "4.1-2 | 45 "2.6 "6.2-1 | 7 .35-12 9 | :------------------------+-----------------------+---------------------------: | 4 †1‡6 3 | 8 9 "2.6-5 | ‡1.256 "5.2-1 7 | | †6‡8 5 9 | 1 4.2:6 7 | 26.8 "3.2 24.38-6 | | 2 7 †8‡1 | 3 46 +5-6 | †1‡8-56 9 46-18 | '------------------------'-----------------------'---------------------------'

Step 2. GEM (8B7) => +1 r3c8,r4c6,r5c9; +2 r5c2; +5 r9c6; btte

Or a contradiction (r6c8 empty) for parity ‡ => -8 r8c1 (etc); btte

(Placement of both 1r4c6 and 1r3c8 yields a btte solution, so I don't really see a reason for your third step.)

by **Mauriès Robert** » Fri Dec 18, 2020 3:39 pm

Hi SpAce,

SpAce wrote:Placement of both 1r4c6 and 1r3c8 yields a btte solution, so I don't really see a reason for your third step.)

As I wrote in my introduction, I have voluntarily limited myself to cross validation and you will have noticed that I have not proceeded to any elimination by cross colour, that is why I have done the 3rd step.
Robert

by **SpAce** » Sat Dec 19, 2020 1:46 am

Hi Robert,

Mauriès Robert wrote:As I wrote in my introduction, I have voluntarily limited myself to cross validation and you will have noticed that I have not proceeded to any elimination by cross colour, that is why I have done the 3rd step.

You're right. I saw what you wrote but failed to see it made a difference. Somehow I thought those placements gave btte directly, but now that I rechecked, the -1r1c2 elimination in my first step was actually significant. My mistake.

by **Mauriès Robert** » Sat Dec 19, 2020 12:57 pm

SpAce wrote:Hi Robert,

Mauriès Robert wrote:As I wrote in my introduction, I have voluntarily limited myself to cross validation and you will have noticed that I have not proceeded to any elimination by cross colour, that is why I have done the 3rd step.

You're right. I saw what you wrote but failed to see it made a difference. Somehow I thought those placements gave btte directly, but now that I rechecked, the -1r1c2 elimination in my first step was actually significant. My mistake.

I could have eliminated the 1r2c2 in the JP(3r4) phase by lengthening the two tracks a little and thus avoid the JP(4r3) phase, but I only wanted to be interested in validations with the shortest possible tracks.
Robert

Website · by **denis_berthier** » Sat Jan 16, 2021 5:52 am

Back to this old puzzle. I decided to try my recent Forcing-TE procedure on it.

After the trivial start:
***********************************************************************************************
*** SudoRules 20.1.s based on CSP-Rules 2.1.s, config = S
*** Using CLIPS 6.32-r779
***********************************************************************************************
singles ==> r9c8 = 9, r5c6 = 9, r5c4 = 7, r1c5 = 7, r2c5 = 1, r8c6 = 7
152 candidates, 743 csp-links and 743 links. Density = 6.47%
whip[1]: r7n4{c2 .} ==> r9c3 ≠ 4, r8c1 ≠ 4
whip[1]: c4n5{r6 .} ==> r6c6 ≠ 5, r4c6 ≠ 5
whip[1]: c4n4{r6 .} ==> r6c5 ≠ 4
whip[1]: b2n6{r3c4 .} ==> r6c4 ≠ 6
whip[1]: b2n2{r3c4 .} ==> r6c4 ≠ 2, r4c4 ≠ 2
PUZZLE 0 NOT SOLVED. 45 VALUES MISSING.

Code: Select all: CURRENT RESOLUTION STATE: 168 1269 128 269 7 3 1268 4 5 3568 236 258 26 1 4 9 7 268 7 12469 124 269 5 8 3 12 126 9 1234 7 45 8 12 125 6 123 15 12 6 7 3 9 4 1258 128 1345 8 1245 45 26 126 7 1235 9 146 146 3 8 9 256 1256 125 7 68 5 9 1 246 7 268 238 23468 2 7 18 3 46 56 1568 9 1468

two steps of FTE will be enough:

FORCING-T&E(S) applied to bivalue candidates n3r4c2 and n3r4c9 :
.... 45 values decided by n3r4c2 : n3r4c2 n3r2c1 n5r2c3 n8r2c9 n8r5c8 n5r5c1 n4r4c4 n5r6c4 n5r4c7 n5r7c8 n5r9c6 n3r6c8 n2r8c8 n1r3c8 n2r1c7 n6r3c9 n2r7c6 n1r4c6 n6r6c6 n2r6c5 n2r4c9 n1r5c9 n4r9c9 n6r9c5 n4r8c5 n3r8c9 n2r5c2 n6r2c2 n2r2c4 n9r3c4 n4r3c2 n1r7c2 n8r9c3 n1r9c7 n6r8c1 n8r8c7 n4r7c1 n1r6c1 n4r6c3 n8r1c1 n1r1c3 n6r7c7 n9r1c2 n2r3c3 n6r1c4
.... 9 values decided by n3r4c9 : n3r4c9 n3r2c2 n3r6c1 n4r7c1 n3r8c8 n8r5c8 n5r5c1 n5r2c3 n1r1c1
===> 4 values decided in both cases: n5r2c3 n8r5c8 n5r5c1 n4r7c1
===> 22 candidates eliminated in both cases: n6r1c1 n1r1c2 n1r1c7 n5r2c1 n2r2c2 n2r2c3 n8r2c3 n1r3c2 n1r3c3 n1r4c9 n1r5c1 n1r5c8 n2r5c8 n5r5c8 n8r5c9 n4r6c1 n5r6c1 n5r6c3 n1r7c1 n6r7c1 n4r7c2 n8r8c8

Code: Select all: CURRENT RESOLUTION STATE: 18 269 128 269 7 3 268 4 5 368 36 5 26 1 4 9 7 268 7 2469 24 269 5 8 3 12 126 9 1234 7 45 8 12 125 6 23 5 12 6 7 3 9 4 8 12 13 8 124 45 26 126 7 1235 9 4 16 3 8 9 256 1256 125 7 68 5 9 1 246 7 268 23 23468 2 7 18 3 46 56 1568 9 1468

FORCING-T&E(S) applied to bivalue candidates n6r8c1 and n8r8c1 :
.... 41 values decided by n6r8c1 : n6r8c1 n1r7c2 n8r9c3 n2r5c2 n1r5c9 n1r3c8 n1r4c6 n1r9c7 n5r9c6 n8r8c7 n8r2c9 n3r2c1 n1r6c1 n4r6c3 n5r6c4 n3r6c8 n2r8c8 n4r8c5 n6r9c5 n4r9c9 n2r7c6 n6r6c6 n2r6c5 n3r8c9 n5r7c8 n6r7c7 n2r1c7 n5r4c7 n6r3c9 n1r1c3 n2r4c9 n4r4c4 n3r4c2 n2r3c3 n9r3c4 n4r3c2 n6r1c4 n2r2c4 n9r1c2 n8r1c1 n6r2c2
.... 35 values decided by n8r8c1 : n8r8c1 n1r1c1 n3r6c1 n6r2c1 n2r2c4 n8r2c9 n3r2c2 n1r9c3 n6r7c2 n8r1c3 n3r4c9 n3r8c8 n8r9c7 n5r9c6 n2r7c6 n1r4c6 n6r6c6 n2r6c5 n4r6c3 n5r6c4 n1r6c8 n5r7c8 n1r7c7 n2r5c9 n1r5c2 n5r4c7 n2r3c8 n6r1c7 n2r8c7 n1r3c9 n9r1c4 n6r3c4 n2r1c2 n4r4c4 n9r3c2
===> 12 values decided in both cases: n1r4c6 n5r9c6 n8r2c9 n4r6c3 n5r6c4 n2r7c6 n6r6c6 n2r6c5 n5r7c8 n5r4c7 n4r4c4 n2r2c4
===> 43 candidates eliminated in both cases: n6r1c2 n2r1c3 n2r1c4 n8r1c7 n8r2c1 n6r2c4 n2r2c9 n6r2c9 n2r3c2 n6r3c2 n4r3c3 n2r3c4 n2r3c9 n1r4c2 n2r4c2 n4r4c2 n5r4c4 n2r4c6 n1r4c7 n2r4c7 n1r6c3 n2r6c3 n4r6c4 n6r6c5 n1r6c6 n2r6c6 n2r6c8 n5r6c8 n5r7c6 n6r7c6 n2r7c7 n5r7c7 n1r7c8 n2r7c8 n2r8c5 n6r8c7 n2r8c9 n8r8c9 n6r9c6 n5r9c7 n6r9c7 n1r9c9 n8r9c9

Code: Select all: CURRENT RESOLUTION STATE: 18 29 18 69 7 3 26 4 5 36 36 5 2 1 4 9 7 8 7 49 2 69 5 8 3 12 16 9 3 7 4 8 1 5 6 23 5 12 6 7 3 9 4 8 12 13 8 4 5 2 6 7 13 9 4 16 3 8 9 2 16 5 7 68 5 9 1 46 7 28 23 346 2 7 18 3 46 5 18 9 46

stte

Note that the bivalue pairs at the start of each FTE procedure are fund automatically by SudoRules.

by **DEFISE** » Sat Jan 16, 2021 11:38 am

Hi Denis,
I checked your resolution with forcing T&E procedure.
I agree with the first pair but for the second pair I found that only the first 9 values can be decided by n6r8c1 :
n6r8c1 n1r7c2 n8r9c3 n2r5c2 n1r5c9 n1r3c8 n1r4c6 n1r9c7 n5r9c6
Then I have to use 2 whips [1] to get the other values.
In summary the mere T&E (Singles) is not enough. T&E (Singles+ Whip [1]) is necessary.

Website · by **denis_berthier** » Sat Jan 16, 2021 11:47 am

DEFISE wrote:Hi Denis,
I checked your resolution with forcing T&E procedure.
I agree with the first pair but for the second pair I found that only the first 9 values can be decided by n6r8c1 :
n6r8c1 n1r7c2 n8r9c3 n2r5c2 n1r5c9 n1r3c8 n1r4c6 n1r9c7 n5r9c6
Then I have to use 2 whips [1] to get the other values.
In summary the mere T&E (Singles) is not enough. T&E (Singles+ Whip [1]) is necessary.

Isn't it what you do in conjugated tracks: take all the implications by Subsets?

by **DEFISE** » Sat Jan 16, 2021 12:19 pm

denis_berthier wrote:Isn't it what you do in conjugated tracks: take all the implications by Subsets?

You haven't talked about tracks before. I thought that for you T&E meant T&E (Singles).
On the other hand, to develop a track, I can very well use only singles. It’s configurable.

Website · by **denis_berthier** » Sat Jan 16, 2021 12:25 pm

DEFISE wrote:
denis_berthier wrote:Isn't it what you do in conjugated tracks: take all the implications by Subsets?

You haven't talked about tracks before. I thought that for you T&E meant T&E (Singles).
On the other hand, to develop a track, I can very well use only singles. It’s configurable.

When I defined both T&E and Forcing-T&E, I clearly said they can be defined for any resolution theory with the confluence property, which is the case for Subsets.
In order to avoid any ambiguity, I'll modify the name.

by **DEFISE** » Sat Jan 16, 2021 1:30 pm

denis_berthier wrote:When I defined both T&E and Forcing-T&E, I clearly said they can be defined for any resolution theory with the confluence property, which is the case for Subsets.
In order to avoid any ambiguity, I'll modify the name.

In fact, this notion of forcing-T & E with a pair does not interest me too much.
I prefer to use the contradiction, it seems to me more effective.
I'm talking about as a computer solver of course.
I don't know if you have been able to demonstrate that a puzzle solvable by T&E (R) is also solvable by Forcing T&E (R) ?
I think it is wrong and the converse is true.

Website · by **denis_berthier** » Sat Jan 16, 2021 2:19 pm

DEFISE wrote:In fact, this notion of forcing-T & E with a pair does not interest me too much.

I'm not very interested either.
I wanted to show that, if Robert took seriously his definition of conjugated tracks as sets (and not as sequences), this is exactly what he would obtain.
Additionally, if one wants to optimise the number of steps, it's easier to do it with T&E or Forcing-T&E than with braids or forcing braids.
We have recently seen several solutions doing this. It's very easy (for a program). But optimising the number of steps without keeping the length of whips/braids/... under control has no interest to me.

DEFISE wrote:I prefer to use the contradiction, it seems to me more effective.
I don't know if you have been able to demonstrate that a puzzle solvable by T&E (R) is also solvable by Forcing T&E (R) ?
I think it is wrong and the converse is true.

It's obviously easier to use a single braids (or T&E procedure) than two in parallel.
For the implications, I don't know. Maybe the two are false. On one side, not every candidate belongs to a bivalue pair. On the other side, FTE allows several eliminations.
It may also depend on what you do when a branch of FTE meets a contradiction (the proper thing would be to delete its root).
It should be easy to check if all the puzzles in T&E(1) can be solved by FTE, but I'm currently busy on other things.

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