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Tough Sudoku for 1/November/2005


Choose a number, and place it in the grid above.

  1 2 3 4 5 6 7 8 9
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Sudoku Solving Techniques

Submitted by: sudoku lover

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Yes! I am first
I would of put something up here but these is no puzzle ready yet.. :)
Must be something wrong with your tunnel, Chaos! I can see the numbers gleaming in the dark here.
BTW I arrived here by accident from october 26; simply no control over these time/space switches; no tunnel in sight - I just jump.

Going back now, I hope.
Chaos: If you see no puzzle (the form but it's blank), just refresh. When no puzzle is really ready, I get taken back to the real day's puzzle.
same as yesterday's, up to minor changes.

Common pattern : easy to 27 filled, then some eliminations (involving a 10-FC closing in 11-agon) make it easy to 50 filled, then some eliminations (involving two 8-FCs) make it easy to unique solution.
Sorry :
Common pattern : easy to 27 filled, then some usual eliminations make it easy to 29 filled, then some eliminations (involving a 4-FC closing in pentagon and a a 10-FC closing in 11-agon) make it easy to 50 filled, then some eliminations (involving two 8-FCs) make it easy to unique solution.
Cryptic hints :
1) easy to 27 filled.
2) Looking at 7 in col Cf, at 9 in block Bb2, at 7 in col Cb, at 57 in c4c6, eliminate some possibles. Then easy to 29 filled.
3) Looking at 4 in Be5, eliminate some possibles. Looking at 6 in col Ch, eliminate a8=6 (beware, pentagon).
4) Looking at More...
5) Looking at g6, eliminate i2=9 (heptagon). Looking at b3, eliminate e7=3 (nonagon). Looking at 8s in row R1, eliminate h8=6(nonagon).Now easy to unique solution.
More details (complete proof, FC proof, equivalence with 09/20's puzzle) in sandbox. (don't know where it is? google 'sudoku sandbox')
13:16 with one very lucky, stab in the dark guess that just so happened to be right. Brain not functioning today :((
I love one's like this and yesterdays - makes a welcome change from the more straight forward ones we'd been getting for a while.
Did it (eventually) from 29 (somehow) easy to fill with an FC then 40 filled, FC then 55 filled FC to solution.
Took me more steps than gb - so still room to learn!
Hi gb and all. I have found the last few puzzles here excruciatingly difficult. I can follow your proofs, gb, but when I try to come up with the moves myself, it can take forever. What I don't get is, for instance, though I understand in step 4 that looking at the 8s in R7 eliminates a4=1, how More...
Deb : it's not sure at all your solution took more steps than mine, but to compare, you have to write down yours...
27 mins
gb - I made a mistake, I started with 29 not 30 easy filled - have just worked through yours to that point and we were the same.
From there I looked at a4=1/6 and c5=1/6 and tried a4=1, c5=6. This falied in bb5, so i put in
a4=6, c5=1 which took me to 39 filled.
Then looked at a7=4,a3=3 this also failed in bb8 so I put in a7=3,a3=4 and this took me to 55 filled.
Then e3=1, e2=6 failed so put in e3=6, e2=1 and this lead to final solution.
I really need to document my steps carefully - because I am betting I made a mistake writing it down for you!
Hi Blep, no secret in how to find a 11-agon. You just need :
1) to be desperately stuck after trying anything else.
2) to be desperately willing to prove that this f... puzzle MUST be solvable without guessing.
3) to have time enough and enough and enough...
When I found it on sept 20th More...
Anyway, I ask again for contribution : any solution avoiding such a long chain would be MOST appreciated AND immediatly put in the sandbox with inventor's name for eternal fame!
ps : eternal may be a bit optimistic, as in a sandbox one only build sandcastles.
to search hard?)
Awful time again, got ticked off and ended up guessing. Somedays I just don't have the patience that others (gb) does. Good luck to everyone else though.
Blep : I realize that my above answer was sort of escaping. In fact, on sept20th, I hadn't got yet the FC principle, and I must say that now with this tool, it's easier. An approach is to take paper and pen and write down some of the thick links you can see on the grid (eg at beginning of step 4 : More...
You don't need to look at the grid for the thin links as they're obvious on the notation : eg(d8=8)--(e7=8) because two 2s can't be in Be8.
Then make your collection of FCs grow, until one of them closes in a oddagon : the closing vertex can be eliminated.
Woof woof.
12.26 what a slog. Down Snowie.
19.23 for rather cool dog. I did make a guess but usually find it easier on paper where I can pencil in possibilities then rub them out as I eliminate them. Also circle the 'only 2 possibles' which cant be done on the computer screen.
Love the pic - looks similar to my pup!

I find solving tough on paper easier, maybe I'm just showing my age.
6:57, these seemed easy today, same pics as awhile ago.
3:53 and off to bed again {{{{yawn}}}}
Was bored so did it again, down to 8.53. No, not as good as Mr. Yawn from Spokane, but what the hey! Fireworks were good, sprogs enjoyed themselves. See you tomorrow, maybe...

Must rush through the tough; myself and Sister Dolmades are having a choir boy tonight.
I am free tonight Reverend. I will bring my 'parson's nose'.
Well... first time I've been beaten since I first solved a tough one. Well, I did get a solution, but I couldn't prove it was unique.
I got to 29 straightforwardly enough, then filled in all the possibles, which didn't help at all. I tried a few of the 'choice of 2' trial and errors but More...
Var is mein Regina?
You are extremely pedantic, Pedantic.
(I got a/the solution with one unproven T and E.)
Can't even understand gb's notation today never mind the proof. :/
Race Day today and the smart money is on Mr. Ed. A horse is a horse, of course, of course and no-body talks to a horse of course. Unless of course the name of the horse is the famous Mr. Ed. Ridden by I.M. Wilbur. 100 to 1.
pedantic : no wonder the 'A solver' gets stuck at 29 filled, since every tip it uses is a special case of FC of length 2,4, or 6, meanwhile today's proof seems to need a FC of length 10 (until s.o. finds a shorter one) to go over the 29 filled. For Andrei's notation : google 'andrei's notation'.
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