Log in

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.

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.

(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 !

(Anonymous) on December 16th, 2010 12:47 am (UTC)
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)
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.