The New Sudoku Players' Forum

[Ajò Dimonios]

Hi Eleven

Ajò Dimonios wrote:
Having an extra clue cannot make the scheme more difficult because it is in contradiction with the fact that it is always possible to adopt a method of solving the difficulty identical to that of the scheme with a clue in less (always with single solution) simply using the elimination of the candidate as the first resolution move.

Eleven wrote:
No.
If you remove a clue from a unique puzzle, you cannot know, if it still has a single solution. You would have to prove it, before you use the UR. But to prove it, you must solve it without uniqueness techniques.
So the puzzle with the extra clue is definitely harder.

I show you with a simple chain that what you say is not correct. We start from the fact that both schemes allow only one solution (it is an essential feature to be able to use the methods of uniqueness. Consequently, nobody forbids me to write the following chain on puzzle n2: 9r3c6 = {UR Type 3 (43 + 36) r35c46} => - 3 r7c4 => 9r3c6; stte. If 9r3c6 is false it implies that the UR type3 is true and consequently that r7c4 = 3 is false and that r3c6 = 9 is true which leads to the solution. Therefore the two schemes they have the same difficulty.

Paolo

[eleven]

Eleven wrote:
If you remove a clue from a unique puzzle, you cannot know, if it still has a single solution.

... We start from the fact that both schemes allow only one solution

Did you read my post ?

[eleven]

Here is another link to add to the discussion from Andrew Stuart about Avoidable Rectangles:
https://www.sudokuwiki.org/Avoidable_Rectangles
Here you need to know that the puzzle has a single solution & that the information you are taking comes from a solved cell and not from a given clue

The avoidable rectangles are a subset of what many call hidden unique rectangles.
Any digit, which forces such an unavoidable set (one of the possible solutions of a uniqueness pattern) in cells without givens can be eliminated, if you have a solved cell or not.

[champagne]

I show you with a simple chain that what you say is not correct. We start from the fact that both schemes allow only one solution (it is an essential feature to be able to use the methods of uniqueness. Consequently, nobody forbids me to write the following chain on puzzle n2: 9r3c6 = {UR Type 3 (43 + 36) r35c46} => - 3 r7c4 => 9r3c6; stte. If 9r3c6 is false it implies that the UR type3 is true and consequently that r7c4 = 3 is false and that r3c6 = 9 is true which leads to the solution. Therefore the two schemes they have the same difficulty.

Paolo

To tell it in another way, you can't write
9r3c6 = {UR Type 3 (43 + 36) r35c46} => - 3 r7c4 => 9r3c6;
in a UR, none of the cells is a given.

Here is the problem.

[Leren]

Ajò Dimonios wrote: ... We start from the fact that both schemes allow only one solution (it is an essential feature to be able to use the methods of uniqueness. Consequently, nobody forbids me to write the following chain on puzzle n2: 9r3c6 = {UR Type 3 (43 + 36) r35c46} => - 3 r7c4 => 9r3c6; stte ...

You appear to be arguing in a circle here. Your scheme only "works" on Puzzle n2 (the one with the clue 9 in r3c6) because you know that we got there by adding the 9 to puzzle n1 (the one without a clue in r3c6).

But if you are just given puzzle n2 you are not supposed to know this. In general, if you are given a puzzle and remove a clue from it, the most likely result is that the puzzle would have more than one solution. If you then play a uniqueness move on the modified puzzle you might find a single solution, but it would only be one of the multiple solutions that the puzzle has.

Leren

[Ajò Dimonios]

Hi Eleven , Hi Leren.

I believe that, since they are used to solve uniqueness methods, in order to have a correct measure of the difficulties in solving the two puzzles 1 and 2, it is essential that the initial conditions of the two puzzles are provided by the puzzle creator with similar conditions. In practice, the information provided by the creator of the schemes relating to the single solution is not similar. The unique solution of puzzle n1 also implies the unique solution of puzzle n2 (the n1 contains in its solution the 9 in cell r3c6), while the information relating to puzzle n2 does not automatically imply the unique resolution on puzzle n1. This different information from the creator of the puzzle must necessarily be provided to the player by the creator of the games to have a fair comparison between the two difficulties of resolution. In practice, the creator of the games, to make lawful the use of uniqueness methods, must provide identical information relating to the uniqueness of the two resolutions in such a way that puzzle n1 implies n2 and puzzle n2 implies n1. If this information is not provided by the creator of the game it is clearly the responsibility of the player, this makes the comparison not measurable because the uniqueness of puzzle n1 is much more powerful than the uniqueness of puzzle n2, it also contains within it the demonstration of the implication uniqueness puzzle n1 => uniqueness puzzle n2. This that I have described is precisely one of the problems that for me make the resolutions performed with the methods of uniqueness not comparable with respect to the resolutions that also contain the demonstration of uniqueness.

Paolo

[Ajò Dimonios]

Hi Champagne.

Champagne wrote:
To tell it in another way, you can't write
9r3c6 = {UR Type 3 (43 + 36) r35c46} => - 3 r7c4 => 9r3c6;
in a UR, none of the cells is a given.

Here is the problem.

It certainly is not spelled correctly. I intended to translate the logical information that is present between the two puzzles n1 and n 2. In practice, if it is provided by the creator of the two puzzles that the uniqueness of puzzle n1 implies that of n2 and that that of puzzle n2 implies that of n1, there is a strong inference between the two schemes and consequently the use of the method of uniqueness of the scheme n1 implies the insertion of 9 in r3c9 and consequently this resolution of n1 also belongs to n2.
Paolo

[StrmCkr]

Paolo

What she ia refering to is n1 and n2 are independent of each other if you havent verified singular solutions for both puzzles
As removing any clue results in muti solutions or it could still be zero or 1. Not to mentiom the fact n1 and n2 could have divergent solutions with 18 cells minimalistic diffrent from that 1 added clue. Or in other words 63/81 identical solved cells
If they arent verified.

if you only had n2 puzzle removing clues could make it a muti solution puzzle and a unique argument could make the muti solution puzzle singular

You would have to run a verification program to ensure it has 1 solution, then remove superfluous clues after words in thia case the 9 would be extra resulting in puzzle n1 ie verified n2 is a subset of n1.

Which allows the uniquness argument as it has 1 solution already verified other wise the uniqueness argument results in 0, 1 or still muti solutions .

Now if you try n1 puzzle and remove other clues that arent superfluous you are left with muti solutions and again verification
Of 1 solution else uniqieness arguments result in 1,0 or muti solutions for n1.

This is a very old debate from uniquness arguments shortcutting larger logic constructs by removing data.

This also happens with other typea of logic being valid befor insertion of clues, but often leaves shorter n amout of chains to solve the largers ones eliminations. (2 movea instead of 1) with lower ratings but more steps.
In rare cases some singular eliminations arent covered by ahortwr chains and need higher order logic of the same technique
Often subverted by a diffrent techniqie again lower rated step but usually more of them.

Cheers
Strmckr

[Ajò Dimonios]

HI StmrCkr

I fully agree with what you reported in your post. However, it remains not precisely defined what is meant by solving a puzzle. In practice it is not clear whether a resolution must also report the verification of uniqueness or the latter is exclusively the responsibility of the creator of the puzzle. Most of the resolutions reported include verification of uniqueness except clearly those that use uniqueness methods.

Ciao
Paolo

[eleven]

But you are aware of the fact, that none of all the basic and advanced and very advanced sudoku solution techniques can solve a multisolution puzzle, are you ?
So what sense should such a puzzle have for a player ?

[tarek]

This is another reason that puzzles are best presented as minimal. If you are given a pencilmark grid, the original (minimal) puzzle should be also supplied.

tarek

[Ajò Dimonios]

Robert Mauriès has proposed on his site https://www.assistant-sudoku.com/ different puzzles with multiple solutions. The site shows the resolutions that players have obtained using the TDP method. In particular see puzzles resolutions 600 and 630.

[champagne]

Robert Mauriès has proposed on his site https://www.assistant-sudoku.com/ different puzzles with multiple solutions. The site shows the resolutions that players have obtained using the TDP method. In particular see puzzles resolutions 600 and 630.

As written before applying a uniqueness rule to a puzzle with multiple solutions can lead to one of the solutions.
When I read the TDP description, I noticed some implicit uses of the uniqueness property. It is likely the case for these puzzles.
As noticed eleven, no logical rule can give you an elimination in the area where are the multiple solutions.

EDIT: I had a look to the puzzle 600.
It is clear that the final step is to count valid solutions, having solved what can be solved.
By chance, this puzzle has only 5 solutions.
In this forum, this is not called a logical elimination

I have plenty of code sequences counting the solutions of a puzzle. This is the best approach to extract unavoidable sets from a solution grid. But this is another story.

[dobrichev]

But you are aware of the fact, that none of all the basic and advanced and very advanced sudoku solution techniques can solve a multisolution puzzle, are you ?
So what sense should such a puzzle have for a player ?

Agree, but the process for determining that the puzzle has multiple or no solutions still can be rated.

For example this "puzzle" looks like multi-solution but it isn't.

Code: Select all

....12.3
4.134..2.
..423..1.
..21.34..
.3..1...4
24..2...1
3.34....2
11...234.
.2...413..

....12.34.134..2...423..1...21.34...3..1...424..2...13.34....211...234..2...413..


[Ajò Dimonios]

HI Champagne

Champagne wrote:
As written before applying a uniqueness rule to a puzzle with multiple solutions can lead to one of the solutions.
When I read the TDP description, I noticed some implicit uses of the uniqueness property. It is likely the case for these puzzles.
As noticed eleven, no logical rule can give you an elimination in the area where are the multiple solutions.

EDIT: I had a look to the puzzle 600.
It is clear that the final step is to count valid solutions, having solved what can be solved.
By chance, this puzzle has only 5 solutions.
In this forum, this is not called a logical elimination

I have plenty of code sequences counting the solutions of a puzzle. This is the best approach to extract unavoidable sets from a solution grid. But this is another story.

I disagree with the fact that the solutions found are illogical. The solutions are fully logical. They are not accepted on the site simply because they are chains of contradiction that use the basic technique. When a logical chain proves that a candidate leads to a contradiction, this contradiction is valid for all solutions of the puzzle, not just one in particular. All deletions made with an AIC are also valid for all solutions.

Paolo