Tough Sudoku

A clown fish? 5:25

Possible proof of not tough enough sudoku of 09 13 06:

1) Start at 22 filled - the given puzzle. Unique Possibilities to 27 filled. (UP 27).
2) fc on 1's: i3 == b3 -- a1 == a8 forbids i8 UP 81
Sets: 2 Depth 2 :Rating .03.
Site in serious danger of losing reputation for having good puzzles, methinks!

Haven't been here in awhile but finished all 4. Not my best time though.Practice does make a difference.

Very colorful!

I guess I'm not sure how these proofs are done...which way is A-I and which way is 1-8?

A1 is a given so how does it fit into ==A8?

Hi Fred - notation is similar to algebraic chess board notation. Columns are a thru i, left to right. Rows are 1 thru 9, bottom to top.

Explanation of == and -- available through 'proofs' link on this page.

Shorthand here when looking at just one type of candidate: I did not write down then '=1' 's.

WAAAY too many guesses - Yes, Jeannette, practice does help!

Same happens at a2=2==d2=2 forbids a7=2 UP81

I would surmise the mentally challenged were distressed by the sometimes arcane chatter and baffling - possibly evil symbolism employed by the provers at this site - and lobbied or otherwise blackmailed the web-master into de-clawing the toughs

Is that teardrops i detect? I can share that,i have lost a SON.

Agian, the definition of what defines a tough is not its difficulty but its tool requirement.

Easys are all unique numbers to solve
Mediums require cross hatching and row/colunm ownership
Hards require candidate elimination
Toughs require chains.

chaining is the hardest technique to learn, thus 'tough'.

I believe the catigory you are looking for is often called 'fiendish' and is not seprate from tough here.

15:24

Zoel - although I appreciate your definitions, and albeit they are generally accepted in many circles, to claim that tool requirements is the only measure of definition for puzzle labels is perhaps more the result of the difficulty in assigning proper lables then any other cause.

There can be no doubt that to call the recent spat of puzzles here 'tough' is a misnomer, without respect to the techniques required. A puzzle that requires some chaining can still be fairly easy, as the chains can be easy to find and or abundant in number. On the other hand, a puzzle that lends itself easily to a good guess can be very difficult to prove.
Historically, at this site, there has been a keen interest in proving puzzles. Given this interest, it is natural to assign puzzles that are difficult to prove as being difficult - with no respect to the techniques actually used. One can, if one wishes to, label a technique to any elimination. The possible list of individual techniques, though certainly finite, is doubtfully humanly finite.

Hi Pat!
The conspiracy theory you present may require too much depth to prove.

When the puzzles get easy, I muse over techniques that one may never find the occasion to use:

Possible explanation of Sue De Coq two sector disjoint subsets:

Consider the following hypothetical situation in a puzzle:
S1: a123=56789 - 5 candidates in three cells, in the same column, in the same box
S2: a4 =56 - 2 candidates, one cell, common column with S1, candidates also in S1
S3: b1= 78 - 2 candidates, one cell, common box with S1, candidates also in S1, but not in S2

This type of configuration can be represented in many ways - pairs of wrap around ALS chains, for example. The most transparent way that I know of to illustrate the power of this technique is with one of Andrei's matrices:

In this matrix, all the rows are strong sets. All the columns are mutually weak. The end result will be each column will be proven strong:

a1=5 a1=6 a1=7 a1=8 a1=9
a2=5 a2=6 a2=7 a2=8 a2=9
a3=5 a3=6 a3=7 a3=8 a3=9
a4=5 a4=6 0000 0000 0000
0000 0000 b1=7 b1=8 0000

Since we have 5 truths minimum, one for each row, and we have no more than one truth in each column, it of course follows that we have exactly five truths, one in each column:
Thus:
a1234=5 is strong forbidding a56789=5.
a1234=6 is strong forbidding a56789=6.
a123b1=7 is strong forbidding b23c123=7.
a123b1=8 is strong forbidding b23c123=8.
a123=9 is strong forbidding a56789, b23,c123=9.
A very nice technique, albeit rarely can it be used in this form. Note that some candidates could be missing, and that the technique can work as a 4 strong set (using 4 cells) idea also.

This technique can be used in a chain: Suppose that instead of b1=78, one had the ability to use a chain that showed, for example, b1=7 == b2=8 . It seems that one could still use this idea.

Natrual extension of this idea to Hidden sets, as there always is such an extension:

A similar idea can be used with Almost Hidden Sets. A very close example of this occurs in the puzzle of 2/21/06, my step 5a, but not the exact idea.
Consider the following hypothetical situation in a puzzle:
a1=123456789
a2=123456789
a3=123456789
a4=123456 789
S1:b1=1234,c1=1234
S2: 123 is limited in column a to 4 locations: a1234 - in other words, we have the almost triple 123 at a1234

This idea is again best shown with a matrix, the same rules of interpretation apply as before: For the sake of making the matrix more readable, let the group of cells a123=x1
a4=1 x1=1 0000 0000 0000
a4=2 0000 x1=2 0000 0000
a4=3 0000 0000 x1=3 0000
0000 b1=1 b1=2 b1=3 b1=4
0000 c1=1 c1=2 c1=3 c1=4

Again, we can assert that there is exactly one truth in each column, thus:
a4=123 is a strong set forbidding a4=456789
a123b1c1=1 is a strong set forbidding b23c23=1
a123b1c1=2 is a strong set forbidding b23c23=2
a123b1c1=3 is a strong set forbidding b23c23=3
b1c1=4 is a strong set forbidding: adefghi1=4 and forbidding a123,b23,c23=4

Of course, one may have to chain in order to get some of these sets as strong - the

Continued:

Of course, one may have to chain in order to get some of these sets as strong - the step 5a mentioned above is a rather strained example of the same - and it did not get the wrap around quality - only some of the eliminations applied (unless I missed something - which is of course possible, as it was the first time I used the idea).

Please forgive the likely typo's. Please ask if something makes no sense at all - as it may just not make any sense!

Hopefully, it is understood that some elements in each matrix could be missing without losing the technique:

For example:

a1=5 a1=6 a1=7 a1=8 a1=9
a2=5 0000 0000 a2=8 0000
0000 0000 a3=7 0000 a3=9
a4=5 a4=6 0000 0000 0000
0000 0000 b1=7 b1=8 0000

would still have the same potential eliminations as the first matrix.

In similar fashion:

a4=1 a12=1 00000 00000 0000
a4=2 00000 a23=2 00000 0000
a4=3 00000 00000 a13=3 0000
0000 00000 b1=2_ b1=3_ b1=4
0000 c1=1_ 00000 00000 c1=4

would also have the same eliminations in the second matrix.

Good mAen to all! Way long long on this one today!! Isn't she/he cute?

21:12

Morning.
12:20.

to Steve:
you may just have put the theory to the test.
the enemy is circling in a frenzy at this very moment, praying and cursing the heathen Sudoku
prover. I expect Homeland Security will pull the
site within a day!

I repeat, GOOD MAEN to every ONE !

Steve,
I saw your comment about the tough of 2/17/06 a few days ago. Your proof was a little beyond me so I posted one using only 'simple' or single number fc's. For the safety of us all, you may want to take a look. :-)

there's nothing to comment about the picture..So i suppose you all did it just to show off to everyone 'how good' you were in solving the easy puzzle.This type of action is very disgusting. Please stop this childish behaviour

pretty easy, 8:07 after setup including the time when I left to go put a show on TIVO for my daughter and forgot to stop the timer!

THat is certainly a freaky picture.

been doing it,doing it,and doing it..so many interruptions..cant concentrate..but seems to be a straight forward one..finished at last to see this fountain..
cheers

18.22 with 7 wrong guess

Oct. 6, 2007.

SE=6.6

18:30.