The New Sudoku Players' Forum

[ronk]

For example
(4,3)!8 > (8,3)8 > (8,9)!8 > (2,9)8 > (4,9)4 > (4,3)!4
shows that r4c3 is not 4.

Since (4,3) has five candidates, and (4,3)!8 and (4,3)!4 is not a contradiction, how does the above forcing chain show anything?

Think about it. If (4,3)!8 implies (4,3)!4, then (4,3)!4.

If one "reverses" the argument to ... (4,3)4 > (4,9)!4 > (2,9)4 > (8,9)8 > (8,3)!8 > (4,3)8 ... then the contradiction is obvious implying r4c3 is not 4. But as you expressed, I don't see it.

[vidarino]

Think about it. If (4,3)!8 implies (4,3)!4, then (4,3)!4.

If one "reverses" the argument to ... (4,3)4 > (4,9)!4 > (2,9)4 > (8,9)8 > (8,3)!8 > (4,3)8 ... then the contradiction is obvious implying r4c3 is not 4. But as you expressed, I don't see it.

Hmm, I also thought it was pretty obvious.

(4,3)8 > (4,3)!4 (necessarily!)
(4,3)!8 > ...chain... > (4,3)!4

So regardless if the cell is 8 or not, it's not 4.

[ronk]

(4,3)8 > (4,3)!4 (necessarily!)
(4,3)!8 > ...chain... > (4,3)!4

So regardless if the cell is 8 or not, it's not 4.

Ahh so, the unstated implication stream! I guess that highlights one "nice" (pun intended) thing about nice loops, i.e., both implication streams are embedded in the nice loop expression.

Thanks, Ron

[tso]

Still, there are a few simple forcing chains. For example
(4,3)!8 > (8,3)8 > (8,9)!8 > (2,9)8 > (4,9)4 > (4,3)!4
shows that r4c3 is not 4.

That certainly is a pretty simple bilocation chain. Not sure why Susser doesn't find this. With "trevor's tables" engaged at default settings, it solve the puzzle using the tables three times.

How does Susser do on these two?

Code: Select all

2..5..6...9.....3...1..9..44..9..7...5..4..8...7..1..6...1..3...2..7..9...8..3..5
4..6..3...1..2..6...8..7..19..8..5...4..5..1......2..75.....6...3..8..4......9..5

TIA, Ron

In the first, without tabling engaged, Susser makes a single deduction, placing a '9' at r1c9. Even with tables engaged with default settings, Susser makes no further placements and few exclusions.

Second puzzle -- without tabling engaged, Susser makes no placements, just two exclusions by Nishio. With default tabling, a few more exclusions, but still, not a single placement.

Therefore, when restricted to the things that Susser can find, these two are much harder.

[gsf]

How does Susser do on these two?

Code: Select all

2..5..6...9.....3...1..9..44..9..7...5..4..8...7..1..6...1..3...2..7..9...8..3..5
4..6..3...1..2..6...8..7..19..8..5...4..5..1......2..75.....6...3..8..4......9..5

these are 1 constrained up to singles and box claims,
but each has only 1 backdoor

Code: Select all

[3,2]=8
[6,4]=4


with only 1 backdoor backtrackers can easily go astray
for my technique based solver on #2
breadth first: 93 guesses
depth first: 25 guesses

the speed solver using forward checking solved both without backtracking

I streamed through 37,315,135 random 1-constrained puzzles and 238 had only 1 backdoor
these 2 induced the most backtrack guessing, but not as much as #2 above

Code: Select all

1....6.8......9..378...35......4...8...2.....69...8...56.......842.3.9.........15
.23.......5....1.37...2.....3.9..64.........78.4..5....1...45......3...29...17...


[Jeff]

Ahh so, the unstated implication stream! I guess that highlights one "nice" (pun intended) thing about nice loops, i.e., both implication streams are embedded in the nice loop expression.

Glad you have observed the informative nature of the nice loop notation.

[Jeff]

That certainly is a pretty simple bilocation chain. Not sure why Susser doesn't find this. With "trevor's tables" engaged at default settings, it solve the puzzle using the tables three times.

Hi Tso, The version of Susser that I have downloaded was pretty old. However, the kind of forcing chain mentioned in the solver are in fact xy-chains. Nevertheless, Susser would have included advanced colouring which would have identified same deductions as bilocation chains do.

[Myth Jellies]

If you play tag between Vidar's POM solver, and Simple Sudoku, they eventually reach the solution. (Vidar's POM solver times out...put last result into Simple Sudoku and solve until it has no hints...put that back into POM solver...repeat as needed.)

[vidarino]

If you play tag between Vidar's POM solver, and Simple Sudoku, they eventually reach the solution. (Vidar's POM solver times out...put last result into Simple Sudoku and solve until it has no hints...put that back into POM solver...repeat as needed.)

I'm running the POM solver on my home computer as well, which is faster than the web-server, and it actually solves it in one go.

It took 86 steps and a wide variation of the equation-juggling techniques, if I recall correctly.

Vidar

[ronk]

Glad you have observed the informative nature of the nice loop notation.

I'm not sure if aeb's expression was meant as an implication stream or an equivalent to a nice loop. If the former, I certainly like the notation:

(4,3)!8 > (8,3)8 > (8,9)!8 > (2,9)8 > (4,9)4 > (4,3)!4 ... better than the ...

r4c3<>8 => r8c3=8 => r8c9<>8 => r2c9=8 => r4c9=4 => r4c3<>4 ... I've been using.

Ron

[maria45]

Well, I solved this by hand and each forcing chain in my head as Carcul wished, but used 16 forcing chains, 1 contradiction and 1 uniqueness (which could have been avoided, but came in handy). Somehow this feels more awkward than Carculs megachain. But I like my opening move with the first forcing chain:

e9=1 or e9=9, d3=9, h3=8, k7=8, b9=8, d9=4, d8=5, f8=1, f7=7, h7=1, gh9=37, a9=1 > b9e7f7g9h9!=1

Greetings, Maria

[ronk]

How does Susser do on these two?

Code: Select all

2..5..6...9.....3...1..9..44..9..7...5..4..8...7..1..6...1..3...2..7..9...8..3..5
4..6..3...1..2..6...8..7..19..8..5...4..5..1......2..75.....6...3..8..4......9..5

In the first, without tabling engaged, Susser makes a single deduction, placing a '9' at r1c9. Even with tables engaged with default settings, Susser makes no further placements and few exclusions.

Second puzzle -- without tabling engaged, Susser makes no placements, just two exclusions by Nishio. With default tabling, a few more exclusions, but still, not a single placement.

tso, I believe your implementation of tabling has a blind spot. I implemented Nick70's Bifurcating Implication Chains ... albeit in the forward direction ... and it solves both of those puzzles. At that link, Nick70 expressed a belief, which I share, that the two methods were equivalent.

FWIW the solver has to make 8 and 22 guesses, respectively. (Multi-coloring enabled. No forcing chains other than the trivial xy-wing and xyz-wing.)

Ron

[tso]

tso, I believe your implementation of tabling has a blind spot.

The implementation isn't mine, it's Robert Woodhead's.

[tso]

Well, I solved this by hand and each forcing chain in my head as Carcul wished, but used 16 forcing chains, 1 contradiction and 1 uniqueness (which could have been avoided, but came in handy). Somehow this feels more awkward than Carculs megachain. But I like my opening move with the first forcing chain:

e9=1 or e9=9, d3=9, h3=8, k7=8, b9=8, d9=4, d8=5, f8=1, f7=7, h7=1, gh9=37, a9=1 > b9e7f7g9h9!=1

Greetings, Maria

This is not a "forcing chain" in any recognized sense. Drawing the implications on the grid, I do not get a loop with each node connected to two others, but instead a network in which some nodes connect to 3 or 4 others.

Most of the links to not lead to the next link listed without the help of one, two or even three unspecified others. To be specific,

('!->' is 'does NOT imply without the help of one or more other links')

d9=4!->d8=5
d8=5!->f8=1
f8=1!->f7=7
f7=7!->h7=1
h7=1!->gh9=37
gh9=37!->a9=1
Plus, (e9=1 OR a9=1) does not imply e7f7!=1

In fact, the order of the listed links is partially arbitrary.

Contratulations on being able to find these in your head, and they're perfectly logical deductions, but it will continue to confuse people if you call them forcing chains.

Digressing, why would you want to avoid a UR?

[aeb]

Digressing, why would you want to avoid a UR?

If I do not use UR I know at the end: "this is the unique solution".
If I do use UR, then I only know "the number of solutions is odd, and this is one of them", a weaker result.