The New Sudoku Players' Forum

Website · by **denis_berthier** » Sat Dec 05, 2020 4:32 am

(The first time I saw the spiral pattern is here: http://forum.enjoysudoku.com/sudokus-with-an-original-rare-shape-t4147-8.html, but the example is solvable by Singles.
I don't know if the spiral pattern has minimal puzzles (I couldn't find one in one hour of computation).

Code: Select all: ———XXXX—— ——X————X— —X——XX——X —X—X——X—X —X————X—X ——X——X——X X——XX——X— —X————X—— ——XXXX———

But it has lots of non-minimal ones Here's a relatively easy one with SER 8.3 (the hardest however in the 10,000 I generated with gsf's program) :

Code: Select all: +-------+-------+-------+ ! . . . ! 8 6 5 ! 4 . . ! ! . . 6 ! . . . ! . 5 . ! ! . 4 . ! . 1 3 ! . . 6 ! +-------+-------+-------+ ! . 2 . ! 6 . . ! 5 . 8 ! ! . 6 . ! . . . ! 3 . 1 ! ! . . 1 ! . . 8 ! . . 4 ! +-------+-------+-------+ ! 6 . . ! 3 7 . ! . 4 . ! ! . 9 . ! . . . ! 6 . . ! ! . . 2 ! 9 5 6 ! . . . ! +-------+-------+-------+ ...8654....6....5..4..13..6.2.6..5.8.6....3.1..1..8..46..37..4..9....6....2956... # 1060 FNBTXY C30 30 givens, SER = 8.3, W = 4

by **Ajò Dimonios** » Sat Dec 05, 2020 3:13 pm

Resolution in two steps.

After basics

Code: Select all: +------------------+-------------+-----------------+ | 12379 37 379 | 8 6 5 | 4 12379 2379 | | 12378 378 6 | 247 249 79 | 1278 5 237 | | 25789 4 5789 | 27 1 3 | 2789 2789 6 | +------------------+-------------+-----------------+ | 379 2 3479 | 6 349 1 | 5 79 8 | | 5789 6 45789 | 2457 249 79 | 3 279 1 | | 3579 357 1 | 257 239 8 | 279 6 4 | +------------------+-------------+-----------------+ | 6 158 58 | 3 7 2 | 189 4 59 | | 357 9 357 | 1 8 4 | 6 237 2357 | | 4 1378 2 | 9 5 6 | 178 1378 37 | +------------------+-------------+-----------------+

Common candidate eliminations of two conjugated tracks P(5r7c9) and P’(5r7c9).
P(5r7c9)=>{r7c9=5;r7c7=9;r7c2=1;r7c3=8;r8c8=3;r8c9=2;r9c2=3;r9c9=7;r1c2=7;r1c3=3;r1c9=9;r2c2=8;r2c4=4;r2c9=3;r5c1=8;r5c4=5;r6c2=5}
P’(5r7c9)=>{r7c9=9;r8c8=2;r8c9=5;r6c7=2;r3c7=9;r3c8=8}=>-79r1c8;-8r2c7;-8r3c13;-9r3c8;-79r5c1;-7r5c4;-2r6c5;-9r6c7;-7r8c89;-3r8c9;-7r9c28

Code: Select all: +------------------+-------------+-----------------+ | 12379 37 379 | 8 6 5 | 4 123 2379 | | 12378 378 6 | 247 249 79 | 127 5 237 | | 2579 4 579 | 27 1 3 | 2789 278 6 | +------------------+-------------+-----------------+ | 379 2 3479 | 6 349 1 | 5 79 8 | | 58 6 45789 | 245 249 79 | 3 279 1 | | 3579 357 1 | 257 39 8 | 27 6 4 | +------------------+-------------+-----------------+ | 6 158 58 | 3 7 2 | 189 4 59 | | 357 9 357 | 1 8 4 | 6 23 257 | | 4 138 2 | 9 5 6 | 178 138 37 | +------------------+-------------+-----------------+

Common candidate eliminations of two conjugated tracks P(2r3c4) backdoor and P’(2r3c4).
P’(2r3c4)=>{r3c4=7;r2c6=9;r4c8=7;r5c6=7;r5c8=9;r6c4=5;r6c7=2}
P (2r3c4)=>{r3c4=2;r2c6=7;r4c8=9;r5c6=9;r5c8=7;r6c4=7;r6c7=9…..solution}=>-7r3c13;-7r2c4;-2r2c79;-2r3c7;-7r3c8;-7r4c1;-79r5c3;-9r5c5;-2r5c8;-5r6c1;-2r6c4;-7r6c7;-5r7c3;-5r8c1=>Stte

Paolo

by **DEFISE** » Sat Dec 05, 2020 5:38 pm

Here is my resolution with 5 whips of length <= 6

Hidden Text: Show

Website · by **denis_berthier** » Sun Dec 06, 2020 3:43 am

Ajò Dimonios wrote:Common candidate eliminations of two conjugated tracks P(5r7c9) and P’(5r7c9).
P(5r7c9)=>{r7c9=5;r7c7=9;r7c2=1;r7c3=8;r8c8=3;r8c9=2;r9c2=3;r9c9=7;r1c2=7;r1c3=3;r1c9=9;r2c2=8;r2c4=4;r2c9=3;r5c1=8;r5c4=5;r6c2=5}
P’(5r7c9)=>{r7c9=9;r8c8=2;r8c9=5;r6c7=2;r3c7=9;r3c8=8}=>-79r1c8;-8r2c7;-8r3c13;-9r3c8;-79r5c1;-7r5c4;-2r6c5;-9r6c7;-7r8c89;-3r8c9;-7r9c28

Total length: 17+6+1 = 24

Ajò Dimonios wrote:Common candidate eliminations of two conjugated tracks P(2r3c4) backdoor and P’(2r3c4).
P’(2r3c4)=>{r3c4=7;r2c6=9;r4c8=7;r5c6=7;r5c8=9;r6c4=5;r6c7=2}
P (2r3c4)=>{r3c4=2;r2c6=7;r4c8=9;r5c6=9;r5c8=7;r6c4=7;r6c7=9…..solution}

Total length: 7+7+1

(The +1 is for the conjugacy at the start.)

The puzzle can be solved with chains of length 4.

Website · by **denis_berthier** » Sun Dec 06, 2020 3:53 am

After the trivial part:

Hidden Text: Show

whips of length ≤ 4 are enough

Hidden Text: Show

Alternatively, using only reversible chains, length 5 is enough

Hidden Text: Show

by **Mauriès Robert** » Sun Dec 06, 2020 11:19 am

Hi Denis,
Your answer to Paolo (Ajo Domonios) as it stands makes no sense to me. It means that it would be easier to solve the puzzle in 20 strings from n≤4 (I only count those with n>1 in your resolution). If this is true for a programmed computer solver, it is certainly not the case for a manual solver like Paolo and me.
Indeed, TDP is based on examining the possibilities of developing conjuguated tracks from pairs (of candidates or sets of candidates). The interest of the manual solver is therefore to first choose pairs that allow for the development of both tracks. Consequently, it is no more complicated (nor less rapid) for a manual solver to develop the tracks completely to increase its possibilities of elimination than it is to develop them partially (which would be equivalent to your short chains). Don't forget that a manual solver works visually on the puzzle with markings.
On the other hand, to make me do this from time to time, I can tell you that for a manual solver, the search for short chains of contradiction is more complicated than the exploitation of conjuguated tracks. It is complicated for the manual solver to find the Z candidates which allow you to develop chains of contradiction (your chains), whereas it is easy for your programme.
Robert

Website · by **denis_berthier** » Sun Dec 06, 2020 11:43 am

Mauriès Robert wrote:Your answer to Paolo (Ajo Domonios) as it stands makes no sense to me. It means that it would be easier to solve the puzzle in 20 strings from n≤4 (I only count those with n>1 in your resolution).

It makes no sense, but it means something !!!

Mauriès Robert wrote:If this is true for a programmed computer solver, it is certainly not the case for a manual solver like Paolo and me.
Indeed, TDP is based on examining the possibilities of developing conjuguated tracks from pairs (of candidates or sets of candidates). The interest of the manual solver is therefore to first choose pairs that allow for the development of both tracks. Consequently, it is no more complicated (nor less rapid) for a manual solver to develop the tracks completely to increase its possibilities of elimination than it is to develop them partially (which would be equivalent to your short chains). Don't forget that a manual solver works visually on the puzzle with markings.

I've heard the "manual solver" argument ad nauseam for years and I have no desire to talk about it again. If you're happy with your double (or sometimes triple) T&E procedures, that's fine for me.

Mauriès Robert wrote:On the other hand, to make me do this from time to time, I can tell you that for a manual solver, the search for short chains of contradiction is more complicated than the exploitation of conjuguated tracks. It is complicated for the manual solver to find the Z candidates which allow you to develop chains of contradiction (your chains), whereas it is easy for your programme.

The problem is probably you haven't yet understood that the start of a chain is not a target, but a partial-whip[1] pattern - which is not more complex than intersections - and that the continuity condition plays a major role in simplifying the extensions.

by **Ajò Dimonios** » Sun Dec 06, 2020 12:00 pm

Hi Denis

Denis wrote:
The puzzle can be solved with chains of length 4.

Surely if we interpret the difficulty of a puzzle, the length of the longest whip present in the resolution, I agree with you. But if we also consider the number of whips and other logical steps, we should multiply the length of these by the number of whips. For a human solver it is not the length of the single logical chain that determines the difficulty but the total number of single simple logical steps that lead to the solution. It must also be considered that once a mechanism has been triggered, albeit long, it is a question of applying simple rules (singles, allignaments and only rarely naked and hidden), while, always for a human solver, the search for another whip valid after having concluded positively one certainly has a certain difficulty.

Paolo

P.S.
I could divide the two steps of my resolution into n steps by eliminating, instead of 17 candidates at once, one candidate at a time, clearly always using the same logic. I don't think that in this case, the breaking up of the method makes it easier to solve the puzzle

Website · by **denis_berthier** » Sun Dec 06, 2020 12:24 pm

Ajò Dimonios wrote:Hi Denis
Denis wrote:
The puzzle can be solved with chains of length 4.
Surely if we interpret the difficulty of a puzzle, the length of the longest whip present in the resolution, I agree with you. But if we also consider the number of whips and other logical steps, we should multiply the length of these by the number of whips. For a human solver it is not the length of the single logical chain that determines the difficulty but the total number of single simple logical steps that lead to the solution. It must also be considered that once a mechanism has been triggered, albeit long, it is a question of applying simple rules (singles, allignaments and only rarely naked and hidden), while, always for a human solver, the search for another whip valid after having concluded positively one certainly has a certain difficulty.

Hi Paolo
For the main points, see my answer to Robert.

SudoRules applies simplest-first search. I always said that a human solver would not be so systematic.
SudoRues doesn't try to minimise the number of steps. Some steps are not necessary. I'm sure François would find a simpler path with whips no longer than 4.

As for your idea of defining difficulty as a total number of steps, it makes no sense. All the elementary steps are obvious. But they are not all useful. So, using useful steps / total possible steps would be more realistic.

However, there are three problems with definitions based on the total number of steps:
- they allow no theoretical analyses. That's why the theoretical content of Robert's approach is close to null.
- they can't take every step into account in consistent ways. If you don't count singles (let alone elementary eliminations) appearing at the top level, why would you count them when you use them within your double T&E procedure?
- how do you count the cost of comparing the two or three branches of T&E? (especially as you don't keep track of the eliminations)

by **m_b_metcalf** » Sun Dec 06, 2020 12:42 pm

elsewhere, denis_berthier wrote:BTW, could you try it for minimal puzzles on the spiral pattern? I couldn't find any.

No luck. The least number of independently redundant clues I find is three:

Code: Select all: . . . 4 8 7 5 . . . . 4 . . . . 6 . . 8 . . 5 3 . . 2 . 6 . 2 . . 3 . 8 . 9 . . . . 2 . 7 . . 7 . . 8 . . 6 8 . . 7 6 . . 2 . . 4 . . . . 6 . . . . 6 3 2 5 . . . 6 9 is redundant 7 4 is redundant 8 7 is redundant

by **Mauriès Robert** » Sun Dec 06, 2020 2:01 pm

Hi Denis,
In response to your comments on my intervention and that of Paolo, I would say that it must have been a long time since you solved a puzzle manually. It is SudoRules that gives you the resolutions and it is therefore easy to criticize other people's resolutions!
In reality what does your solver do (apart from the subsets) that a human can't do otherwise than by spending a lot of time (the puzzle Spiral 8.3 after simplifications by basic techniques counts 146 candidates of which 69 lead to contradiction with singles) :
1) It looks for candidates whose validation leads, with singles, to contradiction with n sequences n≤1, whip[1] .
2) Once he has identified all these whips[1], he looks for candidates whose validation leads with singles to contradiction with n sequences n≤2, whip[2] or chain[2]. Then he starts the loop again by looking again for the whip [1], then the whip [2] or chain[2].
3) Once the searches of 1) and 2) have been exhausted, it looks for candidates whose validation leads with simple to n sequences n≤3, whip [3] or chain[3]. Etc...
It is also possible to do it with tracks by making a programme. Besides François, unless he contradicts me, developed his first routine like that.
Robert

Website · by **denis_berthier** » Sun Dec 06, 2020 2:43 pm

Mauriès Robert wrote:In reality what does your solver do (apart from the subsets) that a human can't do otherwise than by spending a lot of time (the puzzle Spiral 8.3 after simplifications by basic techniques counts 146 candidates of which 69 lead to contradiction with singles) :
1) It looks for candidates whose validation leads, with singles, to contradiction with n sequences n≤1, whip[1] .
2) Once he has identified all these whips[1], he looks for candidates whose validation leads with singles to contradiction with n sequences n≤2, whip[2] or chain[2]. Then he starts the loop again by looking again for the whip [1], then the whip [2] or chain[2].
3) Once the searches of 1) and 2) have been exhausted, it looks for candidates whose validation leads with simple to n sequences n≤3, whip [3] or chain[3]. Etc...

It's absolutely not the way SudoRules works: 1) it keeps the partial-chains obtained at the previous steps. and 2) it re-starts at level 0 after each elimination.
And, once more, a manual solver doesn't have to use the simplest-first strategy. If 69 candidates out of 146 lead to a contradiction, then the manual solver has a high chance of finding one by starting from anywhere. He has still a much larger chance if he starts from a partial-whip[1].

Mauriès Robert wrote:It is also possible to do it with tracks by making a programme. Besides François, unless he contradicts me, developed his first routine like that.

NO. As tracks are sets of candidates and don't use the continuity property, you have no way of knowing in advance that a track will be interpretable as a whip. François must have written a specific procedure for whips and a different one for braids.

Tracks are basically T&E, conjugate tracks are double T&E with comparison of the two results. It's easy to do by hand and by computer. There's no shame about it: world champs of Sudoku use T&E.

by **Ajò Dimonios** » Sun Dec 06, 2020 3:56 pm

Hi Denis

Denis wrote:
Tracks are basically T&E, conjugate tracks are double T&E with comparison of the two results. It's easy to do by hand and by computer. There's no shame about it: world champs of Sudoku use T&E.

I thought that the T&E technique should always produce a contradiction. In the case that indicates as double T&E there is no contradiction, the two tracks P (5r7c9) and P '(5r7c9) of the case in question do not produce contradiction. After the 17 eliminations I am unable to decide which of the two is the one that produces a contradiction.

Paolo

Website · by **denis_berthier** » Sun Dec 06, 2020 4:01 pm

Ajò Dimonios wrote:Hi Denis
Denis wrote:
Tracks are basically T&E, conjugate tracks are double T&E with comparison of the two results. It's easy to do by hand and by computer. There's no shame about it: world champs of Sudoku use T&E.

I thought that the T&E technique should always produce a contradiction. In the case that indicates as double T&E there is no contradiction, the two tracks P (5r7c9) and P '(5r7c9) of the case in question do not produce contradiction. After the 17 eliminations I am unable to decide which of the two is the one that produces a contradiction.

Yes, that's why I call it double-T&E: you apply the procedure to two complementary candidates and (instead of eliminating one) you compare the two outcomes; whenever they assert or delete a candidate, you can assert or delete it at the top level. That's mere reasoning by cases.

by **m_b_metcalf** » Sun Dec 06, 2020 4:19 pm

m_b_metcalf wrote:The least number of independently redundant clues I find is three:

Now reduced to just one:

Code: Select all: . . . 4 8 1 5 . . . . 4 . . . . 6 . . 8 . . 5 3 . . 2 . 1 . 2 . . 3 . 8 . 9 . . . . 2 . 7 . . 7 . . 8 . . 6 8 . . 7 4 . . 3 . . 4 . . . . 7 . . . . 1 3 2 5 . . . r9c4 is redundant

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