Tough Sudoku

The first to post today, good mAen to everybody!!!

The island looks like a baseball cap..wonder where it is

14:36 too long What rock is it?

I found this one really tough. Had to make 4 guesses.

Cool picture.

Hey, cms from nyc -- is that you?

it is I! I usually can't do this till too late for anyone to care!

I agree MB. Had to make 2 guesses,and both were bad.

I guess I am warmed up enough to attempt the ones in the newspaper now.

Maen to all. This one lived up to its name. Found it to be very tough, but got through it. Have a nice day/night all!

This puzzle is objectively much easier then the one from yesterday, or most of the recent ones.

A possible appraoch follows:

1) Start at 22 filled - the given puzzle. Unique Possibilities to 27. (UP 27).
2) Pair 57 at ab1 forbids 5 from a2,b3,d1 and forbids 7 from a2,b3,dgh1. UP 31.
3) i5=1 == a5=1 -- c6=1 == g9=1 forbids i9=1
4) i3=2 == b3=3 -- c2=2 == c9=2 forbids i9=2. UP 32.
5) The ALS {c6=136, c4=36} establishes at depth 2: g6=1 -- c2=6.
The chain at additional depth 2 for a total of 4:
g6=2 == g2=2 -- c2=2 == c2=6 forbids g6=1. UP 48.
6) d2=8 == d2=6 -- d6=6 == d6=5 -- f4=5 == f4=8 forbids f2=8 and forbids d5=8. UP 49.
7) Locked 3's at ef5 forbid 3 from e6.
8) f2=9 == f2=6 -- d2=6 == d6=6 -- e6=6 == e6=9 forbids f5=9. UP 51.
9) a8=1 == ef8=1 -- e7=1 == e7=7 -- d8=7 == d8=8 forbids a8=8. UP 81.

Maximum depth 4 at step5.
Total sets 2 + 2 + 2 + 4 + 3 + 1 + 3 + 3 = 20.

Only step 5 of my proof is a little bit tricky. The establishment of non-native weak links is a mainstay of the proofs that I like to present. One can present the proof differently, but I find the non-native weak link approach very straighforward. In this case, it is clear that c2=6 forbids g6=1 using the ALS, and that c2=2 forbids g6=1 using the chain. This type of combination of the ALS idea and the forbidding chain can often make a challenging puzzle much easier.

The rest of the steps are typical short forbidding chains. Step 9 is perhaps a little bit tricky, as it uses the strong 1's in row 8 in a slightly atypical fashion. A lot of documentation on sudoku solving methods leave the approaches I used in steps 5 and 9 out, because they fail to generalize forbidding chains enough.

To me, a forbidding chain is of this sort:
f -- A == B ........ -- C == D -- f.
f is what is forbidden. All of the letters are:
Any statement which is either true or false, but not both at the same time.

This generalized approach has not yet failed for me to solve any puzzle I have seen - but I am still a relative novice, so they are likely puzzles out there which cannot use this approach.

Sorry about the poor grammar above!
... so it is likely there are puzzles which cannot use this approach.

good mAen 11.23

Steve - that hurts my blonde head. It's still only Sunday morning.

17:15 with two guesses

21:05 Much easier than yesterday's.

Had to guess as usual. No good at tough. Is that the Bass Rock in east Lothian, Scotland?

Unified Theory of Sudoku Solving:
Let A,B,C be any statement that will be exactly one of {True,False}.
The chain: C -- A == B -- C will solve every sudoku.

Note: The statements A,B,C include, but are not limited to:
Forbidding chains which meet the parameters (for example, A == B -- C), proven theorems involving n tuples, coloring, X wings, Swordfish, etc (all really still the same thing as the theorem to start with), statements which are known to be only one of true or false (always true or always false), statements which rely on extraneous rules (unique rectangles, for example), chains which have multi-faceted trees ( still just a subset of the theorem).

I believe that it can be rigously proven that given such a general definition of a forbidding chain, that all sudokus which have a solution are solvable with this technique.

If gb, Andrei, or some other mathematician with qualifications better than mine wishes to help write such a proof to this theorem, kindly let me know.

11:46 with a couple of guesses! Good mAen!

Took 2 or 3 guesses on this one. Whew! I think I need a nap now. One day I will be doing toughs w/o guessing . . . I hope!

Step 5 of my proof presented as a subset of the Unified Theory of Suddoku solving:
Consider the statement:{c6=36 and c4=36} replace this statement with:

{c6=6 == c6=3 -- c4=3 == c4=6}
now we have:
c6=1 == {c6=6 == c6=3 -- c4=3 == c4=6} -- c2=6 == c2=2 -- g2=2 == g6=2 forbids g6=1.

Hopefully this example clears up what I mean by this unified theory.

BTW - this theory is nothing more than what gb teaches in his proof link below, except that I have generalized the 'arguments' or 'statements' a bit more widely. X cyles, Y cxycles, forbidding chains as they are commonly understood, ALS techniques, gaurdians, illegal-odd agons, singletons in a container, locked sets, etc, etc, all are provable using this one theorem. I know of no technique that cannot be proven using this theorem, although many of the techniques can be broadened to greater usefullness through the application of this theorem.

Steve, I would have to agree with Violette. Perhaps my rocket scientist husband could sit down with me and explain the intricasies of triangulating. I am very visual and would have to copy your solution and work it step by step...If I could understand the language. Thank God for 'guesses'

Hi Sarahoz!
The language is explained in the link below - proofs. Also, I typically print out the puzzle when I look for eliminations, as it helps in the visualization process. I often draw the chains on the paper to ensure that I really understand it. I think, therefor, that the process I describe above is very visual. Once one takes the time to understand the lingo, it is not something one has to be a rocket scientist to understand. I am basically just a construction worker, and it is quite within the abilities of most human beings.

Hi Steve. I've never been able to get my head around forbidding chains. For instance, let's go back to #3 of your proof today. You say i5=1, what happened to the 3, why couldn't i5 just as well = 3? If so then i9 = 1. Where's the proof that i5 doesn't = 3? What am I missing here?

48:27. Bleah.

Is this Ayres Rock? Clear maEn to all.

This was a regular Tough, finally! 8:34 total minutes, 3:13 to set up, and 5:21 to solve.

Hi Kaz - I do not say i5=1
I say i5=1 == something
This means i5=1 OR something
In the case of today,since the one's in row 5 are limited to two locations, either i5=1 OR a5=1
This is of course ALWAYS true for today's puzzle
== means OR. -- means NOT arg OR Not arg.
It is explained quite fully in gb's link - proofs - below.

A chain that says:
A == B -- C == D means
A OR B
not B OR not C
C OR D.

The linked arg's provide the power. It is not a supposition type proof at all

I noticed that you consider forbidding chains to be guessing in a previous comment - perhaps study them first, understand them, then revisit the opinion. One cannot prove that any technique is not guessing, as one cannot really define guessing well... But, I can tell you that when I solve a puzzle with a forbidding chain, it does not feel like guessing. I can see that something cannot be just as clearly as one can see it with a locked set, etc. type elimination.

Further note:
If anyone chooses to understand forbidding chains, my proposed unified theory simple expands the list of statements that one uses in the chains to include ANY statement that is either true or false.

Thus, in the proof step three, it does not matter whether or not i5=1. It only matters that i5=1 be exactly one of TRUE or FALSE. So, in the answer of what happened to the 3, I do not know, and it does not matter.

7.39 not bad for tough one. Not Ayers Rock! Not easily recogniseable.

Slightly more explaining:
A == B -- C == D means:
A OR B, NOT B or NOT C, C OR D.
Now, A is true or false.

If A is true, then whatever is forbidden by A is forbidden.
If A is false, then B must be true.
If B is true, then C must be false.
If C is false, then D must be true.
Now, one can see that A == D meaning
A is true OR D is true (perhaps both are true!)
In any event, that which is mutually forbidden by A and D is certainly false.

That is the essence of a forbidding chain.

Final note:
Often, it is misconstrued that A == B means exactly one is true. That may often be the case, but - the proof only requires that at least one of A, B be true - that both A and B cannot be false.
Similarily, A -- B only requires that at least one of A and B be false. It may be that exactly one is false, it may be that both are false, but it is only required that at least one be false (or, at most one is true).
That is why we say:
A -- B means A forbids B. Since it is a logical or statement, B forbids A is simultaneously valid.

Hi Kaz again!
I think that if you follow my proof closely, you will see that I did not bother to do whatever you did to limit the one's in column i to just two locations, as that did not help. If you follow the proof precisely as written, then you should be able to see exactly what I did. That is why I write full proofs - to let the reader have exactly the same grid as I did when I made a step.

Hi Everyone,

Just added another requested feature. There is a new checkbox 'Grey out Used Numbers' which will indicate when you have used up all of a number.

Let me know what you think.

,
Gath

Gath, Love the new feature!

I would love the new feature, I think, especially in Easy and Medium where possibles aren't needed, but it doesn't seem to happen. I don't see anything changing.

forgot to turn the timer on till after set up which is usually about 3mins plus 4.44 to finish
have a great day/night one and all

Good maeN all

I truly believe Steve from Ohio deserves his own special comments page :o)

No offense meant

Steve, I really appreciate your comments, especially your explanation yesterday. That one really had me stumped. Dave, I'm working on yours too.