## Ulterior Puzzles

Everything about Sudoku that doesn't fit in one of the other sections
thanx Ruud, I fixed a number of bugs, The Log that you provided coupled with the ratings of RW's #3 & #4 puzzles were helpful.

I've now replicated your results for all RWs puzzles.....

I have added the Fruitfulness formula to the solver so RW's puzzles have the following
Code: Select all
`       Steps  B1    Fruitful Ratio #1     15     5     2        0.4 #2     17     6     2        0.333 #3     12     11    6        0.545 #4     15     8     3        0.375 #5     9      7     4        0.571 #6     10     6     4        0.666`
where B1 is the number of steps where Locking candidates was possible.

Ocean's beautiful 27 Stepper & 25 Steppes
Code: Select all
`       Steps  B1    Fruitful Ratio        27     13    2        0.153        25     8     4        0.5`

& my latest 2 puzzles
Code: Select all
`       Steps  B1    Fruitful Ratio #1     19     9     5        0.555 #2     22     7     2        0.285`

How about multiplying the ratio by the number of Steps to give us an overall score (ulterior motives )???

The clear winner until now would be Ocean's 25 Stepper Scoring 12.5 (RW that should make you happy)

tarek

tarek

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tarek wrote:The clear winner until now would be Ocean's 25 Stepper Scoring 12.5 (RW that should make you happy)

Yes, that makes me very happy.

I'm not quite sure if the amount of fruitful pointing pairs should be compared to the amount of possible pointing pairs. Why should puzzles with more possible (non fruitful) pointing pairs be punished with less points? Looking at my #1 and Ocean's 27:

Code: Select all
`        Steps  B1    Fruitful Ratio  Ulterior scoreMy #1   15     5     2        0.4    6Ocean's 27     13    2        0.153  4,131`

Doesn't seem right in any way if a 15 stepper with 2 fruitful pointing pairs can score more than a 27 stepper with 2... IMO it would be better to either compare them to the amount of total steps or not to compare them to anything at all (just add the amount to the steps, possibly multiplied with some constant that you find suitable).

PS. Not that I expect it to be a winner, but it would still be nice to see some ratings on my #7.

RW
RW
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This is unofficial (awaiting the Threadmaster to confirm), but your #7 RW scored a 14, StepCount:0604030305080401020406050302, B1 possible steps: 5; Fruitful B1 steps: 3........

I'm not sure if the fruitful steps have any role really, it would be nice to see puzzles with fruitful step counts closer to the total number of steps though......

As you head towards low stepper ulteriors that gap should be smaller, but at increased step count, it should get more difficult to have that small gap & hence IMO puzzles with more Fruitfuls steps should score more.

tarek

tarek

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Time for some low steppers.....

8 Steps
Fruitful B1 Steps 6
Total B1 Steps 7
Ratio .857
Not minimal
180 Symmetrically minimal

Code: Select all
` . . . | 5 . . | . 3 .   . . 1 | . . . | 9 . 8   . 9 . | 1 . . | . 5 .  -------+-------+------  4 . 7 | . 1 . | . . .   . . . | 6 9 5 | . . .   . . . | . 7 . | 1 . 2  -------+-------+------  . 4 . | . . 3 | . 9 .   5 . 8 | . . . | 6 . .   . 7 . | . . 1 | . . .  `

tarek

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Ruud, can you point to the definitive message in this thread that describes how ulterior steps are counted?
thanks
gsf
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gsf wrote:Ruud, can you point to the definitive message in this thread that describes how ulterior steps are counted?
thanks

Here it is:

Ruud wrote:
The majority of support goes to ab's method. Since this is a democratic thread, we'll let the majority decide before the majority changes its mind.

Here's the technique:

1. Find all locked pairs (type 1), but pretend they do not exist.
2. Make eliminations based on the locked pairs found in step 1.
3. Find all hidden singles, but pretend they do not exist.
4. Place the hidden singles found in step 3 and perform eliminations.
5. Increment step counter and repeat from step 1.

Conditions:

a. The puzzle must be completely solved using the above steps (duuh)
b. The FN test must produce an (invalid) result.

ab

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Joined: 06 September 2005

ab wrote:Here it is:
1. Find all locked pairs (type 1), but pretend they do not exist.
2. Make eliminations based on the locked pairs found in step 1.
3. Find all hidden singles, but pretend they do not exist.
4. Place the hidden singles found in step 3 and perform eliminations.
5. Increment step counter and repeat from step 1.

thanks
now I need a clarification from the RW#4 trace
I get a discrepancy at step 9
step 8 gets the puzzle here
Code: Select all
`  9    4    5  | 167  167  16  |  3    2    8 67    2   67  |  8    9    3  |  4    5    1  8    1    3  |  5    2    4  |  7    6    9---------------+---------------+--------------- 36    7    9  |  4  1356  16  | 156   8    2 346   8   46  |  9  1356   2  | 156  13    7  1    5    2  | 36    8    7  |  9   34   46---------------+---------------+--------------- 47    3   147 |  2   146   5  |  8    9   46  2    6    8  | 137 1347   9  |  1   14    5  5    9   14  | 16   146   8  |  2    7    3`

and the step 9 log is
Code: Select all
`Step 9Digit 1 locked in B9R8Eliminate 1 in R8C4Eliminate 1 in R8C5Single in B9D6Place 6 in R7C9`

but it missed the lone 6 in box9 which can eliminate [69]^6
(all candidates (6) in one row/col within a box (9))
commit that move and it opens up a hidden single [64]=6 for step 9
gsf
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gsf wrote:but it missed the lone 6 in box9 which can eliminate [69]^6
(all candidates (6) in one row/col within a box (9))
commit that move and it opens up a hidden single [64]=6 for step 9

Personally I'd agree with you here, but it depends on how people have programmed their algorithm for looking for locked candidates. If they've assumed they'd need to find 2 or 3 candidates in a line then they'd miss this. In the normal course of solving puzzles this situation wouldn't arise, because you'd place the 6. Maybe it explains why I've been unable to match Ruud's step counts so far
ab

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ab wrote:Personally I'd agree with you here, but it depends on how people have programmed their algorithm for looking for locked candidates. If they've assumed they'd need to find 2 or 3 candidates in a line then they'd miss this. In the normal course of solving puzzles this situation wouldn't arise, because you'd place the 6. Maybe it explains why I've been unable to match Ruud's step counts so far

time to do some 3rd party comparisons
are the current ulterior puzzles in one spot or spread out?
gsf
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You are right gsf in saying that a box line elimination is possible. however, I actually wouldn't go for a box-line elimination if there is an obvious hidden single there.....I would place the hidden single only

[If two deductions can be made based on one cell then I would go for the more obvious or easier deduction first, if that was the hidden single then the step ends there. If it was the box-line elimination then you are correct in your assumption].

I think that the setting should be at >1 per line per box because I consider the hidden single EASIER than box-line elimination.

tarek

tarek

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### re: RW ulterior puzzle #3

RW (2006.May.29) wrote:[ 6 ulterior puzzles ]

#3
Code: Select all
` 9 1 . | . 3 . | . 2 8  4 . . | . . 7 | . . 3  . . . | 8 . 5 | . . . -------+-------+------ . 4 3 | . . . | 8 . .  6 . . | . . . | . . 5  . . 9 | . . . | 4 7 . -------+-------+------ . . . | 2 . 9 | . . .  1 . . | 3 . . | . . 4  8 3 . | . 4 . | . 6 9 `

nice puzzle — does seem to need 4 exclusions from box to line.

the odd thing is, when i first solved it i was freely using both types of box-line interaction, as i usually do — and i found i had solved the puzzle with 3 exclusions from line to box — which is the type Ruud has dis-allowed for the present Topic — so i went looking for the equivalent exclusions of the other type, and i seem to need one more ( 4 ).

by the way, the Pappocom rating for puzzles #3 and #5 is Hard;
the other 4 puzzles are Medium.
Last edited by Pat on Sun Jun 04, 2006 8:25 am, edited 1 time in total.

Pat

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You are solving it Pat in the method which this thread does not employ (we employ ab's method, Hidden & box-line at the same time).... (Pat's method, is exhausting hidden singles before going to type one box-line eliminations)......

The puzzle you mentioned solves in 12 steps(ab's method), 11 out of these steps you can make box-line eliminations out of which 6 should put more placements than if left for hidden singles alone......

tarek

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### only 2 steps of "hidden singles"

RW (2006.May.29) wrote:In this diagram 'X' represents any number except 'A':

Code: Select all
`X X X|X . .|. . .. X .|. . .|. . .X . X|X . .|X . .-----+-----+-----. X X|. . .|X . X. . .|. . .|X . .. . .|. A .|. . .-----+-----+-----. X .|. . .|. . .X X X|X X .|. . .. . .|. . .|. A .`

All digits 'A' can be solved without pointing pairs in 5 steps:

Code: Select all
`X X X|X . .|5 . .. X .|5 . .|. . .X 4 X|X . .|X . .-----+-----+-----2 X X|. . .|X . X. . .|. . .|X . 1. . .|. A .|. . .-----+-----+-----. X 3|. . .|. . .X X X|X X 4|. . .. . .|. . .|. A .`

- no, only 2 steps of "hidden singles"
1. r4 r8 box6
2. r1 r7 c2 c4
Code: Select all
`X X X|X . .|2 . . . X .|2 . .|. . . X 2 X|X . .|X . . -----+-----+----- 1 X X|. . .|X . X . . .|. . .|X . 1 . . .|. A .|. . . -----+-----+----- . X 2|. . .|. . . X X X|X X 1|. . . . . .|. . .|. A . `

Pat

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Pat wrote:- no, only 2 steps of "hidden singles"

Oops... I think I mentioned in another post that I spot hidden singles within boxes a lot easier than hidden singles in rows. Apparently this is a good example of that...

RW
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ab wrote:Here it is:
1. Find all locked pairs (type 1), but pretend they do not exist.
2. Make eliminations based on the locked pairs found in step 1.
3. Find all hidden singles, but pretend they do not exist.
4. Place the hidden singles found in step 3 and perform eliminations.
5. Increment step counter and repeat from step 1.

tarek wrote:You are right gsf in saying that a box line elimination is possible.
however, I actually wouldn't go for a box-line elimination if there is
an obvious hidden single there.....I would place the hidden single only

[If two deductions can be made based on one cell then I would go for
the more obvious or easier deduction first, if that was the hidden
single then the step ends there. If it was the box-line elimination
then you are correct in your assumption].

I think that the setting should be at >1 per line per box because I
consider the hidden single EASIER than box-line elimination.

the point I hope is not to do what any particular user would do
but to do what the rules say
ab's ulterior rules don't have provisions for "more obvious"
Code: Select all
`        (1) locked pairs        (2) commit        (3) hidden singles        (4) commit        (5) increment step count`

tarek wrote:You are solving it Pat in the method which this thread does not employ
(we employ ab's method, Hidden & box-line at the same time).... (Pat's
method, is exhausting hidden singles before going to type one box-line
eliminations)......

The puzzle you mentioned solves in 12 steps(ab's method), 11 out of
these steps you can make box-line eliminations out of which 6 should
put more placements than if left for hidden singles alone......

even though its been spelled out there is still confusion
ab's rules are not exacty "Hidden & box-line at the same time"

I've was confused too until I came up with an expression syntax
(that can be input to my solver) that defines what a "step" is

(you can substitute any constraint names you like in the following
the operations on the constraints are what's important)

define a move to be one or more placements or eliminations
moves are identified by applying a constraint for a puzzle position
moves are committed by actually making the placements or eliminations
a concatenation of constraint names specifies a left to right order on constraint application

Code: Select all
`C1...Ci...Cn`

means:
Code: Select all
`  repeat:        apply the constraints in order C1 ... Cn        for the first Ci that has at least one move                identify all Ci moves                commit all Ci moves in one batch                increment the step-count                goto repeat        end for        if no Ci has moves then either                the puzzle is solvable in step-count steps or                the puzzle is not solvable (constrained) by the given constraints        end if`

let {...} treat groups of constraints as one constraint
Code: Select all
`{C1...Cn}`

means
Code: Select all
`        identify all C1 moves        ...        identify all Cn moves        commit all C1...Cn moves in one batch`

(all constraints in the group are applied, not all need produce moves)

finally, let : within a group denote "commit":
Code: Select all
`{C1:C2C3}`

means
Code: Select all
`        identify all C1 moves        commit all C1 moves        identify all C2 moves        identify all C3 moves        commit all C2 and C3 moves`

the inferior and ulterior threads involve these constraints
Code: Select all
`        F       naked singles        N       hidden singles        B2      locked pairs (box-line)`

so the ulterior step count expression (ab's rules) is
Code: Select all
`        {B2:N}`

and the inferior step count expression is
Code: Select all
`        {FN}`

the {FN} expression captures all of the posted inferior counts
{B2:N} captures the ulterior counts except the cases where F and/or N may also be identfied by B2
for reproducability it seems that B2 should identify singles if those singles match the
B2 constraints
gsf
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