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Tough Sudoku for 28/August/2006


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Submitted by: Gath

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Maen. Am I first? 5:55
Easy one today - simple eliminations to solve
Good Maen to all! Great picture Keyan! Are those baby herons that we can see?
Perhaps Bruce and I are doing different puzzles, or perhaps I am addled.

Solution to tough sudoku of 08 28 06:

1) Start at 22 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
2) Hidden pair 45 at i26 forbids i2=239 and i6=237 UP 29

3a) Locked 2's More...
I agree with you Steve. My proof is worse than yours:

1) Start 22, Unique Possibilities to (UP) 26.
2) Naked triplet i7=239, i8=39, i9=239 forbids g7=239, g9=239, h8=3, i2=239, i3=239, i6=23. UP 29.
3) Hidden pair 58 at a78 forbids a7=369, a8=3469, c7=5, c8=5.
4) g7=6 == h8=6 -- More...
maEn all!

Yes, those are the baby herons. I was quite surprised to see the nest. Golden Gate Park is an urban park in San Francisco.

This was the easiest tough I've done - no guesses for me.
Another tough one.
To anyone interested, I have posted a less deep proof for the august 14, 2006 tough - I borrowed some of Clark's work, as noted. Also, to those who wish to delve into the archives, there are a few unproven puzzles in there: February, 2006 - the 15th, 17th, 20th, 21st.

Finally, for an example of a Sue... elimination, see July 28th puzzle I think one can be done there.
Neat picture!!
if these are supposed to be solutions they are harder to decypher than the sudoku... and they are tough enough!
Hi Gt!
There is a link at the bottom of the page(and top) 'proofs' - that can lead to an explanation of the terminology used in the proofs.
Too much hurrying posted proof today:

Two errors:

3a) Locked 2's at g46 forbids g2379=2

6c) fc on 4's: e9 == e56 -- f4 == a4 -- c56 == c89 forbids b9=4 UP 81

At 3a, an obvious dyslexic typo, corrected above.
At 6c, a cutting board rewrite error, also corrected above.
BTW - If one combines Clark's and mine today, one can get a better proof. One way to do it, after Clark's step 6, insert my step 6c) as written the second time, UP 81.

One can clean up the presentation of Clark's step 5, perhaps as follows:
d2=1 == d5=1 -- h5=1 == h5=3 -- h2=3 == More...
Clark's proof and mine combined: the depth 6 step is Clark's.

Another proof to tough sudoku of 08 28 06:

1) Start at 22 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
2) Hidden pair 45 at i26 forbids i2=239 and i6=237 UP 29

3a) Locked 2's at g46 More...
interesting photo
have a great day/night one and all

What does the rating formula give if you go with the naked triplet in step 2 (and lose 3b and 3c) instead of the hidden pair?
My apologies – right answer, wrong reasons.
1) Start 22 UP 26
2) G7= 6,8 G9=1,8 H8=1,6,8 forbids I789=1,6,8 UP 29
At this point I was either inspired or lucky & said I6=4 which solved the puzzle with simple eliminations. Having arrived at the “answer” I didn’t go back & check the More...
Hi Clark! A difference of about .01:
Hidden pair plus three locked sets: .03 +3(.01)
versus triplet .07.
This exposes, of course, one of the limitations of the formula, which only uses sets per step.
The main idea of the formula is comparisons - rough ones - especially when using deep More...
I have a more comlicated formula, which would make some adjustments to correct some limitations, but it is just as arbitrary, and it is much harder to use. I am perfectly happy to take suggestions on a better formula.

For now, each non-native elimination step adds to oeverall proof More...
Super hard, am stuck and would take me days to figure out proofs but not giving up, I will not succumb.
Oct. 1, 2007.
02/Oct/07 2:28 AM

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