Tough Sudoku for 17/April/2006

First here too?

'METHINKS THOU DOTH PROTEST TOO MUCH'

Maen to all, and once again, Happy Easter from this side of the world. Meandering river in very flat land. Where was this picture taken? Still waiting for those captions to show up, or did the poll vote down showing captions on the pictures?

This is timely. That is the DALY river NT - taken about a month ago meandering out into the Timor Sea south west of Darwin. This is down stream from Katherine and about here the waters (that flooded Katherine last week) will peak tonight. The community of Daly River or Nauiyu is on alert to see if the town needs to be evacuated. The air strip is under water. I was talking tonight to a resident who hopes to fly in there tomorrow. I was taking a photo at the Daly River crossing (currently 13.4 meters under water) about a month ago and noticed a little croc - with just his eyes and snout above the water sittin about a meter from the bank. Glad I didnt go too close. If you want to check out a web site http://www.travelnt.com/regions/katherin e/region_daly_river.htm

A River where? Australia - water is mud colourerd - or not. I wish someone would tell us what or where these pictures are from.

Just read fi's posting. The Daly River, how I would love to go up North, I don't feel like a 'real' Australian because I am 62 and have never seen this part of my country.

7:44

21:13. No guesses today!

The old river just keeps rolling along.

Thanks for the great info on the pic. I checked out the website too. Your part of the world fascinates me. Would love to come there sometime.

Hi kaz!
Thanks for your opinion.
Here is the crux of the problem though:
The rules of the game Sudoku are mute about the number of solutions to a puzzle. The technique, unique rectangles, assumes no more than 1 solution. Therefor, the technique is only valid if, and only if, the rules of the game are changed. I did not make the rules!

To change the rules of the game, call out that one is doing that before starting the game... usually that is the courtesy afforded those playing!

If one is going to allow this significant rule change, then there is whole new line of reasoning that opens up. I am fairly certain that this new rule would serve to significantly trivialize the game!

maeN to all. Can solve without guessing

15:17 The flat land looks boring (I'm used to hilly terrain), but the photos & info on the website make me want to visit Australia...right after Scotland & Ireland!

Perseverance was the motto for today. Good maen to all.

should have times myself... that went reather quickly for a 'tough' one...

12:11

1) Start at 22 filled - the given puzzle. Unique possibilities to 26. (UP 26).
2) Locked 9's at e23 forbid 9 from e456.
3) Naked triple 379 at hgi4 forbids: 9 from bc4 and g5, 3 from e4,gh5, and h6, 7 from bde4 and h56. UP 29.
4) Hidden pair 34 at e78 forbids: 3 at e56, 4 from e23, 567 from e7, 1567 from e8.
5) Locked 5's at ef9 forbid 5 from agh9.
6) Locked 7's at d789 forbid 7 from ef9.
7) Y wing d7=67, i7=79, g9=69 forbids 6 from def9 and h7.
8) Locked 6's at d89 forbid 6 from d4. UP 40.
9) Locked 5's at c12 forbid 5 from c8.
10) this step involves multi chaining involving multiple weak link connections. Non-standard fc, but just as valid.
One could alternately forbid the 7 from e6 using a pair of ALS's and one strong link connecting them:
e2=7 == e6=7 -- f5=7 == d5=7 -- c23=7, c8=7.
c8=7 == c8=6 -- c12=6
c1=76 == c1=5 -- c2=5
c2=567 == c2=9 forbids e2=9. UP 81.
Alternate step 10:
Consider the ALS c1=567, c2=5679, c8=76 - 4 candidates, 3 cells.
Consider the ALS e2=79.
Nines are weak across the set, thus the 7's are strong in one set or the other.
The chain c5=7 == f5=7 forbids e6=7 as we now have proven:
e2=7 == f5=7.

Not checked for typo's, nor reduced for efficiency. For those for whom it is Easter:
Happy Easter!

BTW - the beauty of the ALS alternate step above is that one need not forbid the 5 from c8. Hopefully this is clear to all!

There is absolutely no place like that in my 'neighborhood' or even remotely close. We do not get to see sky like that, although you can see 'big sky' in places like Kansas. Would love, love, love to see Australia in person.

I had assumed that these particular puzzles only had ONE solution. In other words, were unique. I have never seen otherwise.

8:46 total time, 3:55 setup, 4:51 solving.

Good mAen to everyone.

Hi Steve. I thought you might be interested in this, which I found on 'Wikipedia'....
It is possible to set starting grids with more than one solution and to set grids with no solution, but such are not considered proper Sudoku puzzles; as in most other pure-logic puzzles, a unique solution is expected.

you're right kaz, true sudoku have only one solution. too bad it took me nearly 45 minutes to find this one, sheesh. 5 guess moves and only two were right. maybe i should stick to easy/medium/hard. ha.

35:42

This is my first posting of proof. I'm not sure yet about fc. If wrong, pls let me know. 22->26
1) y wing (g9=96, i7=97, d7=67) elim def=6
2) locked 5s at ef6 forbids 5 at abc6
3) locked 8s at e56 forbids 8 at e123 up to 34
4) triplet (i7=97, h9=76, g9=96) forbids h7=7
5) triplet (g4=93, h4=37, i4=97) forbids 7 at h45
6) locked 2s at ab6 forbids h6=2 up to 40
7) then comes fs
i7=9==i7=7--h9=7==h9=6--h3=6==h3=4--a3=4 ==a3=9 forbids ab7=9, h2=6 and ef3=4 up to 81

10.02
great pic
thanks fi for the info , hope there is not too much damage when the water recedes.
have a great day/night one and all

this also looks a little like the Athabasca river in NE alberta,but not enough trees along banks.
maEn all

Here I am in Perth. Don't know if I have time to solve the puzzle today, but it seems as though Steve has the proof well in hand.

Hi DJ!
Hope your vacation is going well!

Hi Miko!
Thanks for posting your proof. I am trying to follow it, and I will look back later. At this time, I have some trouble with your first step - I still have 569 at g9 after UP 26. Let me know, if you please, what you did to eliminate the 5 there so I can follow the rest of your proof.

Hi, Steve!
Sorry, I skipped the first part.
Because of the locked pairs 43 at e78, we get locked 5s at ef9, which eliminates 5 at g9. Thank you for reading my proof.

Hi Miko again:
Assuming that at 40 filled you and I had the same grid - Ihave reviewed your step 7. It is a legitimate fc, but you have overreached what it forbids, I think...
I see that i7=9 == a3=9 forbids a7=9. How are the other eliminations listed justified?
The chain is definately not a wrap around chain, as i7=9 does not forbid a3=9 natively. Therefor, I cannot agrre that the chain forbids h2=6 nor ef3=4. We probably do not have the same grid, as I no longer have a 4 at e3 at 40 filled. Finally, I do not see how 9 is forbidden from b7.

I could be wrong...
The way I see it, the fc you present is valid only for forbidding 9 from a7, as you have proven i7=9 == a3=9. You have not proven with the chain h3=4 == a3=4 nor h9=6 == h3=6. Although one may be able to prove those things...

Wrap around chains are great things, but they need the following set up:

A == B --- C == D -- A.
Then we would have A == D and also:
C == D -- A == B and therefor: C == B.
The critical point being that the chain is continous with the weak and strong links. (Of course, one can generalize to chain of any length)

You have:
I -- A == B -- C == D -- E == F -- G == H -- I.
There is no connection between a3=9 and i7=9 to complete the wrap around. Hopefully this explanation is helpful.

Hi Kaz
I am well aware that a unique solution is expected (but, not required!). But, it is not one of the three rules. Therefor, to assume a unique solution is clearly circular reasoning. Like I said before, a shortcut, but not rigourous. In fact, the set of possible techniques one can derive by adding that one rule - that the puzzle has exactly one solution - is quite large and not at all limited to the unique rectangle idea. You may, of course, choose to operate under that assumption - but then you are clearly adding a rule. Click rules at the bottom of this page - exactly three rules are listed.

One more note - I frankly do not care if we are to agree that there is an unwritten rule - that there is exactly one solution. The technique listed at the site you referred to kaz - I talked about in passing a long time ago here on this site. It is an interesting idea, and does indeed have merit, but only if we add a rule. As in all logic - all the rules should be explicitly listed. Implicit rules are never good enough.

At any rate, I stand behind my objections to the examples listed at the site scanraid. The figure one example of the unique rectangle forbids numbers from the wrong cell for the wrong reason. Luckily, the numbers forbidden do not invalidate the puzzle, I suppose. The argument used in that example is spurious and illogical. All this is true even assuming the validity of adding the rule. One can make a valid elimination using the new rule in that example, but not from the cell suggested. That is one the reasons for my warning - the examples are ill considered, and in some cases fallacious.

Perhaps you should contact Andrew Stuart with your comments. I'm sure he'd be most interested.
I was using figure 2 of his Unique Rectangles, which didn't seem spurious to me. Interesting how we all interpret rules and strategies, as is obvious just on this site alone.

Kaz - I have contacted him and told him of my concerns. The figure two idea is fine, if you add a rule. The figure one example is fallacious, as I noted

Thank you, Steve, for your explanation of a wrap around chain. You often use wrap around chains and sometimes with multiple ends. I thought I was doing the same thing, without understanding a wrap around chain! So the eliminations other than a7=9 are not justified. (b7=9 elimination is a simple error.)

You are most welcome, miko!

To anyone who has been forced to forbear the exchange between kaz and me, I am sorry.

Finally, I have mispoken. If one is to visit the scanraid site and study the unique rectangle idea there, the example he provides for type I unique rectangles is his figure 2. That is the example to which I had objected previously regarding that technique. I am removing my objection.
That example argues that one can forbid both a 5 and 7 from a cell that if it was limited to only a 5 and 7 would form a unique rectangle. In fact, one can do exactly that assuming the rule that there is one solution.
Sorry about my error.

Steve, there is no need to apologise. Such exchanges are what this site (at tough level) is all about. I'm sure there are many like myself, who quietly work through the info adding to our overall understanding of sudoku, the logic behind it and the various techniques. Some like Miko, even dare to join in the sharing of proofs and learn in the process. Thanks