The New Sudoku Players' Forum

[monkey89]

Code: Select all

4 . . | 6 3 7 | . . 5
1 5 9 | 2 8 4 | 7 6 3
7 6 3 | 1 9 5 | 4 2 8
------+-------+-------
5 3 . | 9 7 6 | . 8 4
8 . . | 5 2 1 | 3 . 6
6 . . | 3 4 8 | . 5 .
------+-------+-------
2 . 5 | . 6 9 | 8 3 .
3 . 6 | 8 1 2 | 5 . 9
9 . . | . 5 3 | 6 . 2


Simple Sudoku did a multi-colors, which I had a hard enough time getting, but after that it said it had no hint. Any strategies you know of to solve this without trial and error?

Thanks,
-Monkey

[re'born]

monkey89,

The simplest way I can see is the following:

If (5,3)7 then (5,8)9. The 4 cells (4,3), (4,7) and (6,3), (6,7) would then form a deadly pattern. Therefore, (5,3)!7. The same argument shows that (5,2)!7 and the puzzle solves easily from there.

[Cec]

"..If (5,3)7 then (5,8)9.... The 4 cells (4,3), (4,7) and (6,3), (6,7) would then form a deadly pattern. Therefore, (5,3)!7...."

Am I correct in interpreting this terminology as follows:
(5,3)7 means If 7 were in row5 column 3 and
(5,3)!7 means 7 can't be in row5 column 3
Cec

[re'born]

You got it.

[Cec]

Thanks rep'nA

Cec

[foxglove]

+7@9/2=>-7@8/2<=>+4@8/2<=>-4@8/8<=>+4@9/8=>-4@9/4<=>+7@9/4=>-7@9/2!
the same pattern, which is so frequent that deserves a name, removes 7@9/3

Now a more complicated loop solves the problem:
!+7@8/8=>-7@5/8<=>+7@6/9=>-7@6/3<=>+7@5/3<=>-4@5/3<=>+4@5/2=>-4@8/2<=>+7@8/2<=>-7@8/8!


best wishes

[QBasicMac]

I would like to have a contest whereby my competetor (not an autistic savant) is restricted to no T&E and I am allowed to use T&E when I run out of practical stuff like X-Wings and Hidden Triples (or just get tired of looking).

Who can solve the most such puzzles as this in an hour?

(Given the limitation that one must prove the puzzle has one and only one solution)

T&E solves this puzzle immediately.

And makes it fun in the bargain!

Mac

[tso]

I would like to have a contest whereby my competetor (not an autistic savant) is restricted to no T&E and I am allowed to use T&E when I run out of practical stuff like X-Wings and Hidden Triples (or just get tired of looking).

Who can solve the most such puzzles as this in an hour?

(Given the limitation that one must prove the puzzle has one and only one solution)

T&E solves this puzzle immediately.

And makes it fun in the bargain!

Mac

Yikes! So many things wrong with this.

The results of this contest would have *everything* to do with the choice of puzzles, and prove *nothing*.

1) We *all* agree that if time to to complete is the most important issue, guessing will often cut the time down subtantially in all but the easiest puzzles.

2) However, given the right set of puzzles, the premature guesser may well be at a *huge* disadvantage. No, I'm not taking about all-singles puzzles, but puzzles that do not easily resolve to mostly bivalue cells, where all or most guesses lead to a dead end -- not a contradiction, just a situation not particularly more simplified. The guesser is forced to use multiple, embedded guess, a tactic that can easily be as difficult to manage and complex to understand as a bilocation chain -- one that would have solved the puzzle and would have been found if you hadn't been flipping coins.

3) You are handicapping the logician but not the magician -- shouldn't you be restricted to just singles and maybe pairs?

4) BUG will solve a puzzle faster than guessing.
Many UR are just as easy to spot and faster than guessing.

But of course, neither of these tactics would be allowed by the logician, as she wouldn't be able to prove the puzzle was unique.

5) Fun is subjective. I hate Sudoku.