The New Sudoku Players' Forum

[Shinhan]

I wanna try the forcing chains, so I need some very hard sudokus. Simple Sudoku's Extremes require only swordfishes and colouring (which I think I have pretty good grasp of right now). WebSudoku's (on which I started) Evils are even easier. I tried one Times Fiendish, wasnt too hard. I heard about Super Fiendish but cant find those.
I have several other sudoku generators, but which one do you knows makes hardest sudoku. Of course it needs to be solvable without T&E

[angusj]

I wanna try the forcing chains, so I need some very hard sudokus.

Try these: http://angusj.com/sudoku/unlimited_puzzles.zip

[Pi]

the papocom software has "very hard" puzzles which are a lot harder than the times "feindish" which is either "hard" or "medium"

[Shinhan]

There are 100 unlimited. nice, thanks.
And I'll have to try papocom very hard's too. See what they need for solving.

[tso]

Pappocom's Very Hard puzzles NEVER require forcing chains or colors (They max out at X-Wings.) -- though I suppose in some cases the tactic *might* be applied. However, the Very Hards are generally more difficult to solve without computer assistance than nearly any printed source (newspapers, magazines, books, etc). In fact, before the recent explosion of popularity of Sudoku in the last couple of years, very few puzzle published anywhere were as difficult as Pappocom's Very Hards. I've checked though lots of old Japanese magazines -- they are never rated higher than HARD by Pappocom.

ASTRAWARE'S Free Sudoku of the Day comes in six difficulty levels. They keep a two week archive here. They spell out exactly which tactics are required for each puzzle and give a pictorial step by step solution to each and every puzzle in the archive. Try the Fiendish and Diabolical.

Sudoku Susser does not create puzzles but it will import them VERY easily from an vast variety of sources. You can then see which tatics the solver uses on the puzzle to see if you want to try it your self.

SOLO, from Simon Tatham's Portable Puzzle Collection, at the Advanced level will often require FC and at Extreme level will usually require them. At the Unreasonable level, even FC may not be enough to solve the puzzles. Especially interesting is the ability to make 6x6 puzzles that can be solved only with forcing chains -- a great way to learn the technique.

SadMan Sudoku will create puzzles that specifically require the techniques you want -- but it can take a long time to do so. You can have it automatically solve any puzzle and then examine the log to see if forcing chains were used.

Simple Sudoku does not create puzzles that need forcing chains, but you can load a puzzle that you've found elsewhere, press and hold f11. If it solves it completely, the puzzle doesn't need forcing chains. If it gets stuck, the puzzle *might* be solved with forcing chains -- or it could require a more creative solution.

[r.e.s.]

SOLO, from Simon Tatham's Portable Puzzle Collection, at the Advanced level will often require FC and at Extreme level will usually require them. At the Unreasonable level, even FC may not be enough to solve the puzzles. Especially interesting is the ability to make 6x6 puzzles that can be solved only with forcing chains -- a great way to learn the technique.

Indeed, the Unreasonable 6x6s are great for learning to use FC, both single- and multi-valued (i.e. with graph edges not all the same digit). Of the couple dozen or so Unreasonable 6x6s (with "2-way" rotational symmetry) that I tried, one of them is apparently unsolvable by FC:

Code: Select all

 . . . | . 4 .
 . 1 . | 3 . 5
-------+------
 . . . | 2 . .
 . . 3 | . . .
-------+------
 6 . 2 | . 5 .
 . 5 . | . . .

(for anyone who'd like to try).

[tso]

I couldn't find a solution using *simple* forcing chains, but that's the advantage of working on these smaller puzzles. Complex chains can be more easily seen. I doubt very much this is the simplest way to solve this one.

Starting grid:

Code: Select all

 . . . | . 4 .
 . 1 . | 3 . 5
 ------+------
 . . . | 2 . .
 . . 3 | . . .
 ------+------
 6 . 2 | . 5 .
 . 5 . | . . .

Not much progress after standard tactics:

Code: Select all

 . . . | . 4 .
 . 1 . | 3 . 5
 ------+------
 . . . | 2 . .
 . . 3 | 5 . .
 ------+------
 6 . 2 | . 5 .
 . 5 . | . . .

235   236    56     16     4   126
 24     1    46      3    26     5

145    46  1456      2   136   346
124   246     3      5    16    46

  6    34     2     14     5    13
134     5    14     46   236   236

[EDIT: first forcing chain I posted in the solution was unnecessary and I removed it.]

[RE-EDIT: Turns out, my first chain wasn't unnecessary! I'm putting it back in.]

r1c3=5 => r1c1<>5 => r3c1=5
r1c3=6 => r1c4=1 => r5c4=4 => r5c2=3 => r6c1<>3 => r246c1=[24][124][14] (forming a naked triple) => r3c1=5
Therefore, r3c1=5

Code: Select all

 . . 5 | . 4 .
 . 1 . | 3 . 5
 ------+------
 5 . . | 2 . .
 . . 3 | 5 . .
 ------+------
 6 . 2 | . 5 .
 . 5 . | . . .

 23   236     5     16     4   126
 24     1    46      3    26     5

  5    46   146      2   136   346
124   246     3      5    16    46

  6    34     2     14     5    13
134     5    14     46   236   236

r4c5=6 => r6c5<>6
r4c5=1 => r3c5<>1 => r3c3=1 => r6c3=4 => r6c4=6 => r6c5<>6
Therefore, r6c5<>6

r4c5=6 => r2c5=2
r4c5=1 => r3c5<>1 => r3c3=1 => r6c3=4 => r2c3=6 => r2c5=2
Therefore, => r2c5=2


And the rest is elementary:

Code: Select all

3 2 5 | 1 4 6
4 1 6 | 3 2 5
------+------
5 4 1 | 2 6 3
2 6 3 | 5 1 4
------+------
6 3 2 | 4 5 1
1 5 4 | 6 3 2

[dukuso]

isn't it faster with backtracking ?

[tso]

isn't it faster with backtracking ?

Sure. You could say that for nearly every other puzzle we've ever analyzed on in this forum. Guessing and/or trial and error and/or bactracking are often the quickest way to the solution with all but the easiest puzzles. When I'm in a puzzle solving competition with a time limit my strategy is often to progress the puzzle to the point where guessing will finish off the puzzle quickly.

[Sue De Coq]

tso,

Your chains feature the implication:

r3c5<>1 => r3c3=1

which I don't understand, because there are three candidate positions for the value 1 in Row 3.

[udosuk]

I agree with Sue...

r4c5=1 => r3c5<>1 => r3c3=1 or r3c1=1

is the best I could work to... how tso eliminate r3c1=1 I don't understand...

[Sue De Coq]

Here are the forced chains I found:

I. Consider r1c6

r1c6=1 => r5c6<>1 => r5c6=3 => r5c2<>3 => r5c2=4 => r4c2<>4
r1c6=2 => r1c2<>2 => r4c2=2 => r4c2<>4
r1c6=6 => r4c6<>6 => r4c6=4 => r4c6=4 => r4c2<>4

so r4c2<>4 (I'm not certain this chain is strictly necessary)

II. Consider r6c1

r6c1=1 => r4c1<>1 => r4c5=1 => r4c5<>6
r6c1=3 => r6c5<>3 => r3c5=3 => r3c5<>1 => r4c5=1 => r4c5<>6
r6c1=4 => r2c1<>4 => r2c1=2 => r2c5<>2 => r2c5=6 => r4c5<>6

so r4c5<>6, which leaves 1 as the sole candidate for r4c5.

Standard techniques now allow the folowing deductions:

The values 1 and 5 occupy the cells r3c1 and r3c3 in some order.
- The moves r3c1=4, r3c3=4 and r3c3=6 have been eliminated.
The value 6 in Box 1 must lie in Column 3.
- The move r1c2=6 has been eliminated.
The values 1, 3 and 5 occupy the cells r1c1, r3c1 and r6c1 in some order.
- The moves r1c1=2 and r6c1=4 have been eliminated.

whereupon another forced chain appears:

III. Assume r1c4<>1

r1c4<>1 => r1c6=1 => r1c6<>2 => r1c2=2 => r1c2<>3 => r5c2=3
r1c4<>1 => r1c6=1 => r5c6<>1 => r5c6=3 => r5c2<>3

Since the two deductions are contradictory, the original assumption must have been false, so r1c4=1.

The remainder of the puzzle is straightforward to solve.

[tso]

tso,

Your chains feature the implication:

r3c5<>1 => r3c3=1

which I don't understand, because there are three candidate positions for the value 1 in Row 3.

I agree with Sue...

r4c5=1 => r3c5<>1 => r3c3=1 or r3c1=1

is the best I could work to... how tso eliminate r3c1=1 I don't understand...

You are both correct. I originally had another step in the solution -- thought it wasn't needed and edited it out. I've edited it back into the orignal post.

It isn't a simple forcing chain -- but it is the kind of implication that can be followed more easily in a 6x6 than in a 9x9.

r1c3=5 => r1c1<>5 => r3c1=5
r1c3=6 => r1c4=1 => r5c4=4 => r5c2=3 => r6c1<>3 => r246c1=[24][124][14] (forming a naked triple) => r3c1=5
Therefore, r3c1=5

The second implication chain has a step that uses a naked triple as a link. As long as each step is implied by the previous one alone (as opposed to a forcing *net*), it is no more difficult to follow along in your head, and therefore no less valid a forcing chain.