Q: Are zero 2-candidate cells ever possible?

Everything about Sudoku that doesn't fit in one of the other sections

Q: Are zero 2-candidate cells ever possible?

Postby Elroy » Mon Mar 13, 2006 4:08 pm

When all possible single-candidate cells (singles) have been resolved without final solution, a puzzle typically leaves doubles (cells of 2 possible values each) which are then logically resolved by some technique of "trial-reduction" logic:
Q: Would it ever be possible, in solveable puzzles with a unique solution, immediately after all possible singles are resolved, that NO two-candidate cells (doubles) remain? -- meaning all yet-unsolved cells are minimally triple-candidates which must be first resolved by some form of "trial-reduction" logic before any doubles can be resolved?

Thanks, can't find anything touching on this in tips/techniques,etc

Elroy
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Postby tso » Mon Mar 13, 2006 5:29 pm

Elroy wrote:When all possible single-candidate cells (singles) have been resolved without final solution, a puzzle typically leaves doubles (cells of 2 possible values each)...


No it doesn't. In fact, this is impossible. Perhaps you mean something like:

"After all naked and hidden singles are resolved, some puzzles reach a point where most (but not all) of the remaining cells are bivalue (have only two candidates)..."



which are then logically resolved by some technique of "trial-reduction" logic:


That an incredible leap you've taken there. (For those who've read me rant about the notion that some puzzles "require" trial and error or that some tactics are merely a facade for trial and error in multiple other threads -- behold -- here we have an example of someone who assumes EVERYthing beyond singles is "trial-reduction" -- a term I've no choice but to believe means "trial-and-error".)

Let's replace this part of your premise with: "."


Q: Would it ever be possible, in solveable puzzles with a unique solution, immediately after all possible singles are resolved, that NO two-candidate cells (doubles) remain? -- meaning all yet-unsolved cells are minimally triple-candidates


Good question -- though your question incorrectly implies that some puzzles have a unique solution and are unsolvable.



which must be first resolved by some form of "trial-reduction" logic before any doubles can be resolved?


Again, no such animal.


To answer your question, probably yes. I don't have an example. Here's one that has only one bivalue cell. I'm sure someone will post an example with none.



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

+-------------------+-------------------+-------------------+
| 8     125   259   | 4     12379 12357 | 6     1359  139   |
| 2459  3     2459  | 1259  6     1258  | 1589  7     1489  |
| 4569  1456  7     | 1359  1389  1358  | 1589  13459 2     |
+-------------------+-------------------+-------------------+
| 1     2458  24589 | 6     278   2578  | 3     459   479   |
| 2569  7     2569  | 1235  4     1235  | 1259  8     19    |
| 245   2458  3     | 1257  1278  9     | 1257  145   6     |
+-------------------+-------------------+-------------------+
| 23567 256   256   | 8     12379 12367 | 4     1369  1379  |
| 3467  9     468   | 137   5     13467 | 178   2     1378  |
| 23467 2468  1     | 2379  2379  23467 | 789   369   5     |
+-------------------+-------------------+-------------------+
tso
 
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Postby Elroy » Thu Mar 16, 2006 1:03 am

(wish I could've gotten back sooner):
Yes, thanks, tso, that's a much cleaner way to ask my exact topic-question -- AND you also EXPANDED the issue by zeroing in on my very reason for asking it..
Topic-question rephrased:
"Q: Is it ever possible, in a solveable puzzle with a unique solution, after all unsolved cells have been reduced to a minimum possible number of candidate-values via cross-elimination (includes Swordfish and X-wing eliminations, etc) and mathematical deduction, that there might be no BI-valued candidates? -- meaning all cells are either naked|hidden singles else cells with 3-or-more candidate values?":?:
-
And the next few lines, if answered, will indeed be MY biggest learning-curve in SuDoku, regardless how it's answered:
You responded with this example; True, it's not completely void of BI-value candidates (there's one in col9) but it may well suit the core-issue underlying my topic-question:
__8__ _125_ _259_:__4__ 12379 12357:__6__ _1359 _139_
_2459 __3__ _2459:_1259 __6__ _1258:_1589 __7__ _1489
_4569 _1456 __7__:_1359 _1389 _1358:_1589 13459 __2__
.
__1__ _2458 24589:__6__ _278_ _2578:__3__ _459_ _479_
_2569 __7__ _2569:_1235 __4__ _1235:_1259 __8__ __19_
__245 _2458 __3__:_1257 _1278 __9__:_1257 _145_ __6__
.
23567 _256_ _256_:____8 12379 12367:__4__ _1369 _1379
_3467 __9__ _468_:__137 __5__ 13467:_178_ __2__ _1378
23467 _2468 __1__:_2379 _2379 23467:_789_ _369_ __5__
.
Could you PLEASE(please) reveal, in this example,
1) The very first single cell-solution
2) Each sequential observation that got you there (as though doing manually on paper):?:
.
******
And, a core issue which you bring up and in fact underlies the reason for my topic question, is what I cautiously nomered "trial-reduction" (yes I knew of your expoundings about "t&e" so as a novice I tried to avoid any focus apart from main subject)
-- By "trial-reduction" I mean that, after I'm sure all candidates have been completely resolved/reduced by all other means, I find no recourse but to look for a most-logically prospective initial cell and mentally fixate one of it's values as a naked single, noting the chain of canditate-solutions resulting in other cells: If a chain of results gets too long or complex to retain mentally, I may have to record each result; If then such "trial" dead-ends with no further possible candidate-resolvements, I must obviously "erase" that recorded effort and eliminate that initial candidate;
-- If this is what you deem "t&e" then perhaps some of us are deemed to live with it..:(
-- Anyway, my first choices of a prospective "trial" cell would be from those with the fewest candidates, such as the "19" in your example above (OhPLEASE address my request posted beneath it);
***
The reason I'd asked is obvious to a paper-solver: Any puzzle that can't be logically reduced to anything less than tri-value candidates suggests more effort to solve - It'd certainly be helpful to know beforehand if such puzzle was even valid (and unique).

Thanks again, can hardly wait for some reply

Elroy
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Postby tso » Thu Mar 16, 2006 2:39 am

(Please see this post on how to make legible diagrams. Is there some reason you replaced the standard legible diagram?)

Elroy wrote:Topic-question rephrased:
"Q: Is it ever possible, in a solveable puzzle with a unique solution,


This is redundant -- A Sudoku is solvable if and only if it has a unique solution.

...after all unsolved cells have been reduced to a minimum possible number of candidate-values via cross-elimination (includes Swordfish and X-wing eliminations, etc) and mathematical deduction, that there might be no BI-valued candidates? -- meaning all cells are either naked|hidden singles else cells with 3-or-more candidate values?":?:


Of course not. There's no there there. "Mathematical deduction" would include all possible deductions, including those still undiscovered, and would solve every puzzle. You've GOT to pick a specific place to stop making deducutions. As it is, the line is arbitrary, depending only on the skill set of the solver.

If you were to restrict solving to ONLY hidden and naked singles -- or any other specific subset of logical methods, then you have a question that might be answered.

A) Is it possible that after filling all singles, each unsolved cell will have at least 3 candidates? Most likely, yes.

B) Is this possible after all singles, naked and hidden subsets, [fill in all the tactics you know here], and Swordfish? Most likely, yes.

C) Is this *still* possible after all currently known humanly-implemental non-bruteforce tactics have been applied? I don't know. IF any puzzles were left unsolved after all known tactics -- then it would probably just as likely as case (A) or (B) for all remaining cells to be tri-value. But this case will require agreement on which tactics are allowable, which are really human-implementable, which are really bruteforce.

Could you PLEASE(please) reveal, in this example,
1) The very first single cell-solution
2) Each sequential observation that got you there (as though doing manually on paper):?:


Uh ... nope. This puzzle is beyond my skill level. It is impervious to most known tactics. Many computer solvers can find no deductions at all. This is an important point -- just because I cannot solve the puzzle does note make it unsolvable. I cannot do a double back flip -- but it is certainly possible. The 4 minute mile was thought to be impossible at one time, now it's *required*.

I've been stumped by 100's of puzzles in my life, logic puzzles, word puzzles, mechanical puzzles -- yet I never made the leap that because I couldn't do it, it was impossible or required guessing.

I can't remember if anyone has given a solution for this puzzle or not. You are unlikely to come across something like this unless you go out of your way to look for it.



... is what I cautiously nomered "trial-reduction"...


Yes, this is trial and error -- a perfectly logical tactic to use. I rant that trial and error is never *required*, not that it shouldn't be used. For a puzzle like this one, T&E may well be very challenging to apply by hand. I don't think I ever made that distinction before -- it's one thing to get stuck half-way through a hardish puzzle and take a guess in one of a dozen bivalue cells, any of which will solve the puzzle -- thereby skipping past your chance to make a brilliant and rewarding deduction -- and quite another to try to solve a nasty puzzle like this one that might require multiple embeddied conjectures. You might find it just as rewarding. Ah, but how do you know which is which?


The reason I'd asked is obvious to a paper-solver: Any puzzle that can't be logically reduced to anything less than tri-value candidates suggests more effort to solve - It'd certainly be helpful to know beforehand if such puzzle was even valid (and unique).


Again, if the puzzle has a unique solution, it can be solved completely, at least in theory. Certainly it will be easier to construct a faulty puzzle that has muliple solutions that will leave mostly polyvalue cells after any specific set of tactics are applied. However, there is no litmus test short of actually solving that will tell you that the puzzle is faulty or that it requires tactics beyond what you intend to use. Many software apps will tell you if a puzzle is valid without giving you any hints (unless you ask) on how to go about reaching that solution.
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Postby vidarino » Thu Mar 16, 2006 9:40 am

tso wrote:I can't remember if anyone has given a solution for this puzzle or not. You are unlikely to come across something like this unless you go out of your way to look for it.


Hey, that's my puzzle!:)

I dubbed it "Monster #2" back in the day, and there are a couple of solutions for it here, but they are all very very (very) complex.

Vidar
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Postby Elroy » Thu Mar 16, 2006 4:42 pm

Thanks, Tso, indeed thank you. I was right -- your response indeed constitutes my greatest learning-curve ever in SuDoku, very satisfying..
And thank you also Vidar for that puzzle Tso found to use as example..

Elroy 8^] (no suitable Emoticon to express my gratitude for your time/effort)
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