To

coloin, thank you for confirmation about imam bayildi variation and recommendation of 3 puzzles.

coloin wrote:I can confirm that the 5th puzzle is a non-minimal version of the 4th

***tknottet omitted a code.***

In your list the 5th puzzle has a superfluos 1@r1c1 [would be the single 6@r2c9 in the above 4th puzzle]

Confusingly the non minimal version was the original imam bayildi coined by eleven it seems

The 4th puzzle is minimal - is elev;2;6;22; in the published list of 9 puzzles with SE 11.9

I don't think you should forget about the Easter Monster puzzle where the sk-loop was first found or the Fata_Morgana or the puzzles with high Suexratt

***tknottet omitted 3 codes.***

These may or may not score highly in your program - i am suspecting the suexrat9-10364 will

I'll remove the 5th puzzle(non minimal, I named New_Imam_bayildi) from my list.

Very bad news for me: the rating is 22.1194 for the 4th and 22.1706 for 5th.

As I wrote yesterday:

The puzzle obtained by applying T on the minimal Imam_bayildi is same as 23 clue version, in comparison of big numbers after some transposition and number permutation. So their ratings must be same.

I will check it in detail then report here later.

As for 3 puzzles which you recommended, I will rate them one by one.

To

eleven(please read later), thank you for information.

eleven wrote:In simple chains the length is probably the number of strong links (e.g. 3 for an xy-wing), in forcing chains their sums.

But i don't know, how it is done for nets (multipath chains). The rating must be the higher the more paths/links/cells are used.

If e.g. you try a number and get a contradiction applying your rules, this normally is equivalent to a more or less complex net (when numbers are forced by 2 or more earlier forced numbers).

These nets already have a high rating.

It is more surprise for me, when SE and your rating correlate, than when they differ. I agree with Coloin, that you probably will get a high rating for Suexrat9-10364 (suexrat counts the nodes needed to find a solution with dancing links, this process is repeated then, i think 10000 times).

But just because of the differences it would be nice to have both ratings for a puzzle, because they show different properties.

I will try to make my rating to be somewhat useful for this forum 's activity.

As a part of it, I will propose the measurement(s) ( for example, summation of numbers of the paths/links/cells which are used when my rating tool solve the puzzle ) to be reported here.

To

denis_berthier, thank you very much for your effort to clarify the detail of my rating.

denis_berthier wrote:I know the devil is in the details, but let me skip them for a while and concentrate on the potential meaning of your new rating, R.

I'll use the same notations as in my first post.

Starting from some puzzle P and applying repeatedly the rules in T until quiescence, one gets a puzzle (with candidates) P1 = T(P).

I forgot to mention it, but this is meaningful only if T has the confluence property (and I didn't check whether this is the case for the particular T you are using). Otherwise, depending in which order rules in T are applied, you may reach different puzzles P1.

Contrary to most of the known ratings, R is not intended as a rating of the hardest step in a resolution path.

Is it a rating of a full resolution path? Indeed, it can be considered neither as a rating of some resolution path with a particular property (e.g. simplest in some sense) nor as the mean rating of all the possible resolution paths (even modulo T). What is it then?

First, R(P) is defined as R(P1) (and similarly for each recursive step), which could be expressed as "the rating R is defined modulo T". But no candidate is ever eliminated for real from P1. So, what we get is:

R(P) is the mean rating, modulo T, of the first elimination step, wrt the strategy consisting of trying randomly each candidate at each level of T&E [restricted by your * condition (as notated in my first post)].

Do you agree with this definition?

To use it as supplement of the answers to your questions, I described the detail of my rating in below.

About the order of rules, it depends on P. But once Pis fixed, the order does not vary. So I have no question about uniqueness of P1. I described the order of rules in [3].

As for the relation of resolution paths, my intention is the ｓtatistical expectation over all possible paths. I described it in [6].

As the definition which you mentioned at the last line, I cannot understand why you limit to "the first" elimination. My intention is expectation(in other word: mean) of total number of times T is applied until the puzzle is solved. The description in [6] is intended to explain not only "all possible path" but also "until resolved".

detail of the rating[1]Purpose

This rating Is intended to rate the mean difficulty of a puzzle when one follows a specific resolution model which is described in [4], choosing randomly when the model allows several possibilities.

[2]Notation and Conceptual Definition

(1)P:puzzle:Not only the target puzzle to be rated, but also each of intermediate puzzles is denoted as P.

To distinguish puzzles, variation of notation (for example P1) may be used.

(2)C:candidate:For efficiency reasons, only restricted candidates concern. The restriction rule is described in [5}.

To distinguish candidates, variation of notation (for example C_S or C_C) may be used.

(3)T:a specific set of rules:The application manner on P is described in [3].

(4)R:rating:R(P) and R(C) is defined as follows.

R(P) is defined as ｓtatistical expectation of "how many times T is applied from first quiescence until determined solved/unsolvable"

R(C) is ｓtatistical expectation of "how many times T is applied from the candidate is assumed as the big number of the cell until determined whether the puzzle P1 is solved/unsolvable".

R(C) is used in the situation where starting from some puzzle P and applying T until quiescence, one gets a puzzle P1, then C in P1 is assumed as the big number of the cell, then T is applied in recursive manner.

[3]Rule Applying Manner

In this rating, T consists 15 rules.

The rules are arranged in a sequence, from r1 to r15.

Each rule is applied in order.

Every time a rule is applied successfully, next application starts from r1.

When r15 ends in failure, it means quiescence.

For your reference, examples of rules I found English name are r1(Full House), r7(Locked Candidates), r8(Naked Pair/Triple/Quad) and r9(Hidden Pair/Triple/Quad).

[4]Resolution Model

Let P be a puzzle

1)apply repeatedly to P all the rules in T until quiescence, thus obtaining a puzzle P1 (with both decided values and candidates)

2)if P1 is solved, the procedure ends.

3)if P1 displays a contradiction, backtrack to 4-3).

4) otherwise,

4-1)randomly select a cell from cells with the smallest number of candidates in P1

4-2)randomly select a candidate (say C) in the cell

4-3)create a new puzzle P1(C) by adding C as a value and then put it into P and start recursively from 1)

4-4)when backtrack occurs in 3) or 4-5), try another candidate (say C) in the same cell then restart from 4-3)

4-5)if all candidate fails, backtrack to 4-3) of the previous recursive procedure

[5]Description of Rating Procedures

Let P be a puzzle

1)apply repeatedly to P all the rules in T until quiescence, thus obtaining a puzzle P1 (with both decided values and candidates)

2)if P1 is solved, define R(P) as 0.

3)if P1 displays a contradiction, define R(P) as 0.

4) otherwise, define R(P) as the weighted average of the R ratings of the (yet undecided) candidates in P1, where the weight for each candidate C in P1 is the inverse of the number of candidates in the same cell as C, (*** for efficiency reasons, this is limited to candidates C in cells with the smallest number of candidates in P1, say n);

5) for each candidate C in P1 [restricted by ***], create a new puzzle P1(C) by adding C as a value and then applying repeatedly all the rules in T until quiescence; define the rating R of C as follows:

6) if P1(C) is solved then R(C) = R(P1(C))+1.

7) if P1(C) displays a contradiction, R(C) is computed each case as follows.

-In case a candidate in the same cell as C is (directly or after applying rules recursively) determined as the big number of the cell (name the candidate as C_S):

R(C) = R(P1(C))+1 + R(C_S) + (1+R(P1(C_C))/2

where C_C is the other candidate in the same cell as C (in case of n=2 no C_C exists, in case of n=3 only one C_C exists)

-In the other case:

R(C) = sum(R(P1(Ci))+1)

whether Ci is every candidate in the same cell as C including C itself, then the summation is calculated over all candidates.

8) otherwise, R(P1(C)) is computed by applying recursively the above procedure.

After the procedure, R(C) is calculated by the equation in 2_5) or 2_6) for each case.

[6}A Simple Example

In case that

there is only one 2-candidate cell (name candidates C1 and C2) in P1

P1(C1) is 'otherwise', then there is only one 2-candidate cell (name candidates C3 and C4) in P1(C1)

(P1(C1) becomes new P1 for C3 and C4)

p1(C3) displays a contradiction

P1(C4) is solved

P1(C2) is 'otherwise', then there is only one 2-candidate cell (name candidates C5and C6) in P1(C2)

(P1(C2) becomes new P1 for C5 and C6)

p1(C5) displays a contradiction

P1(C6) displays a contradiction

Following 6 solving paths are possible

1)P>T>P1>selectC1>T>P1(C1)>selectC3>T>P1(C3)(Failure)>BT>C4remained>T>P1(C4)(Solved)

2)P>T>P1>selectC1>T>P1(C1)>selectC4>T>P1(C4)(Solved)

3)P>T>P1>selectC2>T>P1(C1)>selectC5>T>P1(C5)(Failure)>BT>C6remained>T>P1(C6)(Failure)>

BT>BT>C1remained>T>P1(C1)>selectC3>T>P1(C3)(Failure)>BT>C4remained>T>P1(C4)(Solved)

4)P>T>P1>selectC2>T>P1(C1)>selectC5>T>P1(C5)(Failure)>BT>C6remained>T>P1(C6)(Failure)>

BT>BT>C1remained>T>P1(C1)>selectC4>T>P1(C4)(Solved)

5)P>T>P1>selectC2>T>P1(C1)>selectC6>T>P1(C5)(Failure)>BT>selectC5>T>P1(C6)(Failure)>

BT>BT>C1remained>T>P1(C1)>selectC3>T>P1(C3)(Failure)>BT>selectC4>T>P1(C4)(Solved)

6)P>T>P1>selectC2>T>P1(C1)>selectC6>T>P1(C5)(Failure)>BT>selectC5>T>P1(C6)(Failure)>

BT>BT>C1remained>T>P1(C1)>selectC4>T>P1(C4)(Solved)

(BT denotes backtracking.)

This rating adopts the number of times T is applied as puzzle's difficulty. It is intended to calculate the expectation of that number over all possible paths.

Probability of each path is 1/2**NS, where NS is number of "selectCx".

So 1)1/4, 2)1/4, 3)1/8, 4)1/8, 5)1/8, 6)1/8

Number of T for each (T in P>T>P1 is excluded by definition)

1)3, 2)2, 3)6, 4)5, 5)6, 6)5

Expectation should be 3/4+2/4+6/8+5/8+6/8+5/8=

4According to rating procedure

R(P1(C4))=0 from [5]2)

R(P1(C3))=R(P1(C5))=R(P1(C6))=0 from [5]3)

R(C4)=1 from [5]6)

R(C3)=1+1=2 from [5]7)above case

So R(P1(C1))=(1+2)/2=3/2 from [5]4)weights(1/2for both) are omitted

Then R(C1)=5/2 from [5]6)

R(C5)=R(C6)=1+1=2 from [5]7)the other case

So R(P1(C2))=(2+2)/2=2 from [5]4)weights(1/2for both) are omitted

Then R(C2)=2+1+5/2=11/2 from [5]7)above case

Then R(P)=(R(C1)+R(C2))/2=(5/2+11/2)/2=

4 from [5]4)weights(1/2for both) are omitted