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25 April 2009 @ 12:21 pm
A fitting sudoku?  
So, here is the Guinness puzzle - Scanraid assures me, as I felt, that it is not-sledgehammer solvable, or reasonable to do in a fast time without bifurcation (guessing). Do it in under 10x the 3:06 time without guessing (let alone the 10 minutes round time) and I'll be amazingly impressed. If anyone ever asks why I hate sudoku competitions but like puzzles, this is a good example. I feel puzzles should be human-solvable, fun, and certainly in a competition serve a purpose of choosing the best, not the luckiest. Sudoku are too easily both computer-generated and made ridiculous, at least in a competition sense, and I've unfortunately stopped bothering to try to always solve logically because I know the rest of the world, in catching up to me, took up bifurcation. This "puzzle" should soundly prove the flaws of "fiendish" puzzles at a sudoku championship if the community of solvers didn't see them before now.

 
 
( 15 comments — Leave a comment )
byronosaurusrex on April 25th, 2009 09:36 pm (UTC)
It's been 20 minutes since I started the sudoku....

And I have a grand total of one logically-placed digit.

This is a bifurcator's sudoku; decisively not a competition-caliber puzzle, and not something I would set my watch by, let alone a world record.

(Please delete the previous comment if possible; I posted anonymously by accident.)
lardarsegreg on April 26th, 2009 12:16 am (UTC)
Sad thing is, I think I know which digit it is...

I forget exactly what it means to be sledgehammer solvable. I know what it's referring to, but I guess I don't understand how its methods translate to "moves" that I would understand normally.
motrismotris on April 26th, 2009 06:21 am (UTC)
I can't say this definitely but generally sledgehammer complete includes an X-wing but not a Y-wing so you can sort of draw the dividing line on Scanraid steps at that point. This goes well below that line and scores a "record" 614 or so for a truly Diabolical Scanraid grade.
lardarsegreg on April 26th, 2009 06:59 am (UTC)
I so wish I followed sudoku enough to know all of the jargon.

Also: Am I the only person who thinks of Peter Gabriel when the sledgehammer solver is mentioned?
Adam R. Woodzotmeister on April 29th, 2009 09:27 pm (UTC)
"Also: Am I the only person who thinks of Peter Gabriel when the sledgehammer solver is mentioned?"

No. - ZM
lunchboylunchboy on April 26th, 2009 02:20 pm (UTC)
This is a seriously un-fun sudoku.
motrismotris on April 26th, 2009 02:47 pm (UTC)
The new terminology is "sudoku". If even the organizers cannot be put in a room for 1 day to prove the uniqueness of the answer without computers, it is an epic fail.
zundevilzundevil on April 28th, 2009 04:36 am (UTC)
Can we use a term like "quadrifurcation" for this? I don't even think a single guess will get you anywhere -- it probably branches out a few times.

I was going to ask the guy how he solved it so quickly, but I think I knew his answer already...
(Anonymous) on April 28th, 2009 05:04 pm (UTC)
Hmmm clearly the fitting thing to do here is to use the template, make sure that each number is used 3 times as a given, and make it as easy as posible so that this record has a chance of being discreditted.

Although not taking anything away from the belgian chappy. His final standing proves he was a very strong solver, and even making a guess there finishing solving in just over 3 minutes is still stunningly good.

Tom.C
(Anonymous) on April 30th, 2009 04:13 pm (UTC)
Only one guess required
After placing the 1 in R9C2, Vincent Bertrand may have noted that R9C5 could only be 3/7 and R7C5 could only be 3/8. Guess 3 in R9C5 and there is a complete logical path to the solution.

This has been bugging me since the championship. I didn't believe that Vincent would have made a cluster of guesses and now I am fairly certain he only needed to make one. There may well be other places in the grid where a single guess breaks the puzzle.

That said, I think everyone would prefer World Championship puzzles which could be solved logically.

David McNeill.
motrismotris on April 30th, 2009 04:16 pm (UTC)
Re: Only one guess required
Thanks for this observation. I focused my attention on Box 4 in the championship, and no 1 guess was to be found there. Computer graders could certainly be tweaked to check for these bifurcation "backdoors" but still, there MUST be a logical route that can be seen in the same amount of time.
(Anonymous) on May 1st, 2009 04:02 pm (UTC)
I know Vincent, he's a very strong sudoku player. Perhaps a future world champion. He's a very fast player and he improved a lot with variants this year !

Fred76
(Anonymous) on December 16th, 2010 12:47 am (UTC)
sudoku
I'm not sure where others get their terminology from (I guess pages like this), but I've completely, naturally, and on my own, generated my own set of terms since my accident - I was in a hit-n-run accident that left me in a coma (with a Traumatic Brain Injury). Now, by utilizing sudoku puzzles as brain fodder, I can do easy's in 3 minutes or less (some under 2), medium's in 5-10 (some under 4), and hard's in 8-15 (some under 7). The "very hard" or "challenger's" can take up to 15-40 (but sometimes I'll drop a sub-10). Anyone use the terms: "lockers", "linears" ('spears'), "50/50 splits" (I believe you all call them 'bifurcations'), or "there can be only one's"? Also: 'all-considered', '2-fer', 'rule of 2/3+', or 'imbalance method'? Just wondering if others [are as weird as me] had developed their own, as well.
(Anonymous) on March 24th, 2012 09:49 pm (UTC)
Solution
For posterity, here's the shortest logical solution I composed in a few hours. The tricky steps here are two graph eliminations (which actually locate seven removable contradictory loops), and two additional removable contradictions.

There's a fascinating pattern to the 6s and 4s that allows significant savings, and it's simplest to start with them all in one go. I graphed both patterns and found several cells that result in contradictions that prohibit placement of all 6 unseen copies of the digit: 47<>6, 48<>6, 57<>6, 14<>4, 46<>4, 47<>4, 64<>4. These can also be found by beginning with the cell and proving the contradictory loop, but graphing is simpler. So given these, you can start immediately with 47=1 (NOT 92=1!).

I don't regard visually dismissable dead ends as bifurcation. So a genius might then hit visually upon the important trail that if 53=8, then 81=8, 46=8, 51=6, 64=6, 96=6, and col 9 lacks a 6. So instead 53=7, 69=7, 57=4, 48=8, and then you can run the 6s and continue.

The last 22 cells in this path present a nontrivial fork. This is resolved by removing the contradictory loop (found by graphing bivalues): if 84=3, then 83=5, 73=9, 33=1, 35=7, and cell 95 lacks a number; so 84=4, and the rest follows.

As long as I have the reader's attention, here are a few unused trails left as exercises. 1. From the initial grid, prove that if 36=2, then 91=5, and if 36=7, then 91=2, so 91<>3. 2. From the initial grid, prove that whether 95=3 or 95=7, it follows that 11=3. 3. Given 53=7, 47<>6, 57<>6 as above, prove 18=4 and 89=9. 4. When only 22 empty cells remain, prove by graphing that 21<>3.
motrismotris on March 25th, 2012 08:04 am (UTC)
Re: Solution
Intriguing. I've never given this puzzle this much thought since the 10 disatisifying minutes with it in Slovakia. But it's good to see what small inroads exist.

FWIW, I consider "bifurcation" to be anything I have to write on paper and can't hold in my head so I like your term "visually dismissable" as it seems to be the same. For me, this normally means 5 or 6 placements of one kind of digit (sometimes called "simple coloring") or as many as 3 placements with 2 or three different numbers. Your third paragraph starting with "if 53=8" is probably a bit too much for me, but maybe not too much for this "genius" of which you speak.
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