The New Sudoku Players' Forum

[Carcul]

It is a pleasure to announce that Bill Richter and Gurth Bruins have found the correct solution for Steps 1 and 2. Check them, along with my original solution, in the Sudoku Discussions Forum.

Carcul

[gurth]

Hi everybody : I think this is my first post on your forum. I was the one who found the solution to step 1 of this riddle. ( See www.sudoku.org.uk - Forums - Eureka - The Riddle of Sho). I am now posting my solution to Steps 2, 3 and 4.
On your forum, because fortuitiously the Discussions Forum (www.sudoku.org.uk) doesn't weem to be available this morning.

Solution of the Riddle, after my successful solution of Step 1:

(1) It is wrong to suppose that step 2 has been solved, there has so far been no indication of how Sho arrived to his invalid grid (I love quoting). But now I have the answer to that.

Long before Sho started scratching his head, he tawt he taw a puddy tat. (Creia que estaba viendo a un gato precioso). ( to cats I worship and attribute all superhuman qualities of soul). He tawt he taw that a certain cell C was a certain candidate C. (He thought C*C). That "allowed" him to eliminate a few candidates....but at that point an almost unmentionable catastrophe occurred. Tsho was taken horribly short! A terrible call of nature necessitated a rapid retreat to the retreat (retrete). There, while wrestling with a tardy flow of excretory materials, doubtless caused by excessive consumption of Brazilian coffee (or possibly any coffee), it dawned upon Sho that the cat of his visions was but an insubstantial phantom. Caused by a logical error. So when he got back to the other job, he promptly restored the original contents of cell C. But forgot in the process to restore the other candidates he had eliminated! Leading by natural degrees to the aforesaid head-scratching.

(2) So much for step 2. Let's proceed more seriously from there.

What we now realize is that the debacle of the 6 missing candidates (at least one of which we know must be true) could have been caused by one action: the erroneous placement of a C in cell C. Whether or not that WAS the true cause, we deduce that C"C - we eliminate C from C.

(3) Now time to drop a major bombshell. The fact that the givens are perfectly symmetrical means that the solution is perfectly symmetrical. Take it from me that that is an axiom. At this point I shall give a decent definition of axiom : that which is obvious without proof. Anybody not happy with that? Apply to me for private lessons in logic.

So we can place a 5*55, and then all the other 5s are easily placed. Also, where we eliminated C"C, we can now also eliminate (10-C) at the opposite (corresponding) cell.

(4) Now just apply my system of Conditional Equivalence Marking. (To see how I do that, study my solution to U19). This will lead simply and inexorably to the falsification of the Red World so coloured in my "Sudoku of Sho (not the Riddle)". And the solution of the puzzle as well as Riddle.

(5) For later reference, and in case of disputes and arguments, I will note some of the landmarks along the route : A Blue 2*26, combined with a Red 2*92, eliminate 2*96. The "mirror" (symmetrical counterpart) process also occurs, naturally, eliminating 8*14. And these things lead to the presence of both a Red 1 and a Red 2 in cell 16, which falsifies Red (proves Red False, and Blue True).

(6) Censored.

(7) Oh, I forgot to mention which was candidate C and which was Cell C. I leave you to work that one out for yourself. Another
Riddle! To help you figure this one out, I give you this additional clue : 16715. Another clue : the Riddle of Sho required reading the mind, not of Sho, but of Carcul.

[ravel]

The fact that the givens are perfectly symmetrical means that the solution is perfectly symmetrical. Take it from me that that is an axiom. At this point I shall give a decent definition of axiom : that which is obvious without proof. Anybody not happy with that?

Yes, me Of course this is no legitimate approach (i could prove much nonsense by taking as axiom, what intuitively seems to be right).
But i think it is an interesting question, that maybe could be proved in an elegant way. If someone has a proof, i would appreciate to see it (in an own thread).

When i remember right, Carcul said that the solution to this riddle has nothing to do with the symmetrical properties. But i will reserve my holidays for yours

[aeb]

The fact that the givens are perfectly symmetrical means that the solution is perfectly symmetrical. Take it from me that that is an axiom. At this point I shall give a decent definition of axiom : that which is obvious without proof. Anybody not happy with that?

Yes, me Of course this is no legitimate approach (i could prove much nonsense by taking as axiom, what intuitively seems to be right).
But i think it is an interesting question, that maybe could be proved in an elegant way. If someone has a proof, i would appreciate to see it (in an own thread).

gurth confuses matters by calling "axiom" what is a trivial theorem.
ravel, if the givens show a symmetry and you complete the grid and apply the symmetry to the solution then you get a solution again; if the solution was unique then it must have every symmetry that the original puzzle had

[keith]

if the solution was unique then it must have every symmetry that the original puzzle had

Can we use this property to solve the puzzle?

Keith

[ravel]

ravel, if the givens show a symmetry and you complete the grid and apply the symmetry to the solution then you get a solution again; if the solution was unique then it must have every symmetry that the original puzzle had

Maybe i am missing something trivial, but i still cant see it. To reformulate the question:
A perfectly symmetrical grid is one where the sum of the values in (row i,column j) and (row 10-i, column 10-j) is 10 for all cells. Is it true, that any unique sudoku which only has perfectly symmetrical givens, has a perfectly symmetrical solution ? (Why cannot not be there a unique solution, where not all missing numbers are perfectly symmetrical ?)

Can we use this property to solve the puzzle?

If it is true, we can - and this puzzles becomes easy, because then r5c5 must always be 5. You come here and can solve it with simple coloring in 2 and 8 to eliminate them from r1c4 (and r9c6).

Code: Select all

14   29   3   |  289   7     128  |  6    48   5         
14   5    26  |  3     16    128  |  48   9    7         
67   8    79  |  5     69    4    |  2    3    1   
--------------------------------------------------     
68   4    5   |  1     2     36   |  9    7    38         
28   69   79  |  47    5     36   |  13   14   28         
27   3    1   |  47    8     9    |  5    6    24 
--------------------------------------------------       
9    7    8   |  6     14    5    |  13   2    34         
3    1    26  |  289   49    7    |  48   5    69         
5    26   4   |  289   3     128  |  7    18   69

[ravel]

[Edit:] Counter-example deleted, it was not symmetrical.

I am pretty sure now, that it is true:

If it is unique, there must be a proof for all cells, that only a unique number can be placed there. But that means for symmetrical reasons, that there also must be a proof, that there "partner numbers" must be placed in the opposite cell (just replace all row/column and candidate numbers x by 10-x). So the solution always has to be perfectly symmetrical either.

[Myth Jellies]

Another question: Do the nonets need to share in this symmetry? You might be able to find a jigsaw sudoku with symmetrical clues that does not turn out to have a symmetrical solution, for example.

[gurth]

ravel, I see that aeb's argument did not convince you, and you expressed your doubts with an intelligent question. Let's see if I can help you to understand.

Please do the following thought experiment. Imagine a man solving a puzzle with symmetrical givens. When he solves his first cell C, he uses some set of the givens, call it S, and a technique, call it T. Now instead of doing that, he could just as well have used the set of givens symmetrically opposed to S, and the same technique T, to solve the cell symmetrically opposed to C.

And in fact, he could in this way have solved both cells, one after the other,
arriving at a perfectly symmetrical partial solution.

And moreover, as he proceeds further with his solving process, he can in the same way preserve the symmetry of his solution at all stages of its development. Right to the end. So he can only arrive at a symmetrical solution, and as that solution is unique, there can be no asymmetrical solution.

Does that help you?

Please note also: I did not define as axiom "what intuitively seems to be right." That is only YOUR idea. I said "that which is obvious without proof." Something very different, and a perfectly true definition of "axiom".
Some people say that an axiom can't be proved, but what they mean (or should mean) is that any proof of an axiom is at least as complicated and difficult to understand as the axiom itself, so therefore it is superfluous as a proof. Yet minds work in different ways, and what might be obvious to one may not be to another. There is in fact no rigorous definition possible of the (meaning of the) term "proof" itself. That which is universally accepted? No. A proof of the famous 4-Colour Theorem was "universally accepted" for 11 years, until somebody pointed out that it was false!

[ravel]

Thanks gurth, for the elaborate response.

As you can see above, i had already found a proof based on a similar argument.

My understanding of axioms seems to differ from yours: axioms ar not only obviously true (in the sense that there is common agreement that they are not contradicting other axioms), but they also cannot be proven without introducing another axiom. It was always the goal to keep the set of axioms as small as possible. For sudoku questions no new axioms are needed, all is covered by mathematical/logical axioms.

Anyway, we have a new, very simple and very exotic solution technique, which i want to call gurths symmetrical placement (if you dont mind):
When all givens are perfectly symmetrical, then r5c5=5.
[Added:]
Note that the perfect symmetry might not be obvious like in this scrambled version of the puzzle:

Code: Select all

*-----------------------*
| . . . | . . . | 8 . 7 |
| . . . | 6 . . | . 2 . |
| . 1 . | . . 5 | . . 9 |
|-------+-------+-------|
| . 5 . | . 4 . | 2 3 . |
| . . . | . . . | . . . |
| . 6 9 | . 1 . | . 8 . |
|-------+-------+-------|
| 2 . . | 8 . . | . 4 . |
| . 9 . | . . 3 | . . . |
| 7 . 5 | . . . | . . . |
*-----------------------*
r5c5=7


[Edit:] So to generalize it:

Code: Select all

If a puzzle is rotational symmetrical and all of the given numbers always have the same number in the opposite cell, then r5c5 must be either the number that is given in the opposite cells too or - if only 8 of the numbers are given - the number that is missing.


To come back to this thread's topic:
Carcul, now when the effort to solve your riddle even brought a new, most elegant way to solve the puzzle, i think it is time to reveal the solution to the Riddle of Sho

[gurth]

Thanks, ravel, I would be quite delighted naturally to have anything so elegant called after me, so of course I don't mind, though I don't imagine many people wll follow your suggestion.

When it comes to arguments about the meanings of words, I am usually willing to agree with anything remotely reasonable, and as I think you reasonable, I will agree to anything you say.

I had noticed that you already seemed to understand, but as I thought the message I had already composed (my access to internet is a bit limited) was interesting, I posted it anyway.

[ravel]

I had noticed that you already seemed to understand, but as I thought the message I had already composed (my access to internet is a bit limited) was interesting, I posted it anyway.

My problem was, that i had not realized, what aeb meant with "... and apply the symmetry to the solution", so i had to find the argument myself. Your explanation is the easiest to understand, i think.

I posted another puzzle for this technique there.

[ronk]

I posted another puzzle for this technique there.

Gurth's theory in color

[udosuk]

Thanks to gurth and ravel's inspiring discussion I understood another intriguing technique... This would probably come useful for a couple sudoku variant puzzles I posted here, since both solutions are perfectly symmetrical...

As a matter of fact, if I add 2 more clues to the first one:

Code: Select all

Only 2 of the 9 digits can repeat diagonally

....4...9
.........
..3.9.8..
.........
.........
.........
..2.1.7..
.........
1...6....

Then we could apply the gurths symmetrical placement on it and with the diagonal property (1 & 9 are the 2 digits allowed to repeat diagonally) perhaps this puzzle could be solved elegantly without T&E...