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Tough Sudoku for 4/March/2006


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

Submitted by: sudoku lover

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Hi everyone, cryptic hints for today's :

1) easy fillings (ef) to 27 filled.
2) Eliminate a6=3 (depth 4, causes too much 9s in row 9), then ef to 40 filled
3) Eliminate a1=2 (depth 2, start with 2s in Row7), then ef to the end.

Total cost : 6 sets, max depth 4.

Interesting, as none of the 2 eliminations needed is pure fc.
5:47 with one very lucky guess. When I look at those photos I want to go travelling.
Good Maen one and all in Sudoku World

Thought and Fact for the Day

gb from France is GOOD BRAINED from France.
I envy your intelligence.
Thanks, GB, that info helped when I got stuck! At step 2, I looked at i4=39, eliminated i4=3 because that left no 9's in Bh8; then a quad in col. b (don't remember exact numbers desolee, 2's,4's and something else,) , giving b4=3,b5=9; then I needed your step 3 to solve the rest. Merci beaucoup
to IVOR DUNNET where are you from, your posting never says.
Hi all!
Here is what I did today, pared down a bit:
1) start at 23 filled. Easy fillings to 27.
2) Hidden pair 28 at G45 forbids 39 at g45.
3) b9=3 == i9=3 -- i4=3 == b4=3 (--b9=3) forbids b8,b5,i3=3
4) Ef to 29. Colors 9's: i4 == i9 -- f9 ==f6 forbids g6=9.
5) EF to 40. Odd More...
Wow GB!
I cannot imagine getting this one down to just 6 sets! Very nice!
My step 5 detailed a bit:
(all on 2's)
If e21 then (e1==e2--h2==h1).
Therefor, a7 or e1 or h1 must occur.
Perhaps I should have gauged this as depth 4, but I use only 3 native strong links, but I do use 1 additional implied strong link. Comments on proper depth of this step are appreciated.
Also, about that step: Is reverse order more clear?
e9-f7==a7. Still conclude {h1,e1,a7} is a strong set that forbids a1. I only bring about this discussion, as it seems similar to gb's moving strong sets idea. Move the three strong sets:
{h1,h2},{e1,e2,e9},{f7,a7} into new strong set{h1,e1,a7}. I think I have that correct.
An alternative to gb's solution: ef to 27; g6i4=39 elim. g45=39; b4i4b9i9=3 elim. b5b8i3=3; easy to 29; i2i3=16 elim. g1g2i7=1; easy to 30; fc i4=9==i9=9--f9=9==f6=9 elim. g6=9; easy to 40; a1b23=125 elim. a3c23=125; easy to 41; c45=2 elim. c9=2; fc a7=2==f7=2--f3=2==b3=2 elim.
b9=2; easy to end. More sets involved but the depth is 3!
Sorry, when posting the above comment, I haven't yet seen Steve's comments. His solution seems very close and also is of depth 3.
5:14 with 2 guesses. Wow
Then there's that other time-honoured cheating option ... have a wonderful weekend, everyone!
Hi Steve,

I've not still studied your step 5 because you can replace it by my step 3 (also monochromic on 2s but 2sets-deep) and finish with efs.

Your solution becomes then gauged 8 sets, depth 2 !
Steve, Are you my brother-in-law?
Incidentally, Hi andrei ! Seems a interesting discussion is to come about proofs measuring...

Since I practice it, I find this bidimensional measuring (total sets/max depth) a good indicator of the type of a given solution, eg :

*(ts=20/md=2): solution long (at least 10 steps!) but easy
*(ts=8/md=8): solution very short (a single step!) but hard to find
But I can't always decide how to turn it in a 1dim measure. Practically today :
mine was (ts=6/md=4)
Steve's is (ts=8/md=2)

On one hand I tend to prefer Steve's. The economy of 2 sets my solution makes seems to be too much expensively balanced by the non-obviousness of depth4 in step More...
BTW I didn't write yet a solution, only hints. So here it is
2) Look at the 4 strong sets :
A(f56=3), B(g6=39),C(i49=9),D(f569=9).
Whether f5=9 (giving f6=3 by A), f6=9 (giving g6=3 by B) or f9=9 (giving i4=9 by C, g6=3 by B), a6 can't be 3.
3)fc in 2s : a7==f7--f3==abc3 fobids a1=2.
5:56 - is this a repeat picture? Good Maen, all, and have a great weekend!
gb and Steve: I don't have a clear preference on which complexity measure (the total number of strong sets involved, or the depth) should have priority in deciding which solution is 'better.' Note also that there is a third equally natural measure: the total number of candidates involved in all the steps ('atomic complexity'). So I'd rather keep both (or even all three) measures.
Lee - to the best of my knowledge I have no brother in law named Lee.
I had lots of locked pairs and need only 2 fc's of depth 2 each. One eliminated i9=9 and the other eliminated f7=2. There was also one X-wing.
1. Easy to 27.
2. X-wing in 3's => b58<>3, f5=3
3. fc in 9's: f9==f6--g6==i4 => i9<>9 Fill to 41 taking note of several locked More...
After I posted, I read Steve's very similar solution.
DJ: your solution is of depth 3 because of the triple in your step 4 (it actually seems very close to mine and Steve's).
Wow, 1 Second! I just hit the 'show solution' button, and there it was!
TO Judie/Maryland - we would love to here how you got there in under 6 min.!!!!!!
used a combo of solutions. i started showing d5 is not 9 gives a triplet in that row. didn;t write the rest down but got done pretty easily. starting to 'see' these eliminations easier.
sorry anne thats only a non de plume, i only post for my own benefit when checking archives
Thanks gb for your cryptic hints. I was stuck at 40. Where would I be without you - still tearing my hair at midnight, I expect!!
For today's puzzle, I like your solution better GB, because it just seems more elegant. Of course, there can be no objective measure of elegance! As far as objective complexity - I would gauge that all the solutions posted today are of about the same degree - all relatively easy!
But, then that is not objective either... I guess what I mean to say is that when a puzzle is not hard, objectively measuring complexity is perhaps not very important. It is probably more important to the individual solver what he can see - and that too will not only vary from person to person, but with someone like me - it will vary from day to day also!
mAen to all. That looks like a temple in India. Is that where the picture is from? And for all those telling everyone else how they solved this, are you really having any fun doing these?
My opinion is that depth is a good indicator when any solution needs a given minimum depth.

There are many ways of solving today's, but none can stick at depth one, and some (like Steve's revisited with my last step instead of his) can stick at depth 2.
I've not still completely checked More...
I agree with Steve. For a puzzle like this where there are so many easy ways to eliminate candidates, all the posted solutions are pretty direct. (Mine was a mix of Steve's and DJ's.) It's when GB or Andrei post solutions involving 3 sets/max depth 4, and I have a solution that is 8 sets/max depth 7 (or no solution at all) that I am amazed at the beauty of the 'elegant' proof.
t about it).

When you're done with wordoku, Gath, would you mind considering adding a preview before posting here, or shall we forever be cut in our literary inspiration ?
HI Gb! - I sent a couple of addendums to my proof of 2/28 - reduced it to depth 4 (just one step needed revisiting). Hopefully you receive those emails also. Thanks again!
Sorry Steve, I only found now your addenda. Depth 4 ! It's a dream come true... I have to look at this ! Congratulations.
13:10 With a few guesses.
I think it's up to the rest of us who aren't as technical to have the fun for them.
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