Need help with this...

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Need help with this...

Postby EmilyS » Fri Apr 21, 2006 12:24 pm

I've been doing these for a while and never had any problems until a friend gave me this puzzle:

Code: Select all
...|.79|3..
1..|.3.|...
..8|.26|..9
-----------
4.1|...|..6
.59|...|..7
...|...|15.
-----------
2..|3..|6..
...|2..|..5
.4.|..5|.8.


I've searched this and other sudokusites for days to find a technique that could help me, but I still can't solve more than 3 numbers. Can this puzzle be solved?

thanks -E
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Postby Hud » Fri Apr 21, 2006 1:01 pm

If it can be solved, it must require advanced techniques. Pappocom's software doesn't allow it to be dubbed.
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Postby ravel » Fri Apr 21, 2006 1:04 pm

Of course it can be solved, like every unique sudoku. But it is a toughie. I am sure, Carcul will like it (4 brute force steps in my program) and provide a solution. If you want to understand it then, you are advised to read this or this before:)
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Postby EmilyS » Fri Apr 21, 2006 1:09 pm

Yeah, I've read them already, but probably didn't understand all of them. I'll read them again more carefully.

-E
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Postby QBasicMac » Fri Apr 21, 2006 2:36 pm

I have a list of 10 numbers
10, 2, 4, 5, 9, 8, 6, x, y, z

What are x, y, and z?

Answer: 7, 3 and 1
They are 1-10 in Eurolengo
un, du, tre, fier, fyf, sis, seven, oko, nun, dies
arranged alphabetically
dies, du, fier, fyf, nun, oko, sis, seven, tre, un

Moral of this story?

Many puzzles have solutions, but it is a waste of time to work on them. Simply move on to something that is more fun.

If I had to solve this, I would use simple T&E and not bog down forever trying to find Eurolengo.

Mac

P.S. If you are asking if there is a satisfying continuation such as a hidden triple or x-wing, I sure can't spot one.

Edit:
Uh, never mind about "simple T&E". Even that is tough.

I took cell r2c7 which had candidates 24578.

Cases 2, 4 and 8 led to impossible cases, leaving only 5 and 7 which I got stuck on and skipped.

Going back to 5 and 7, I had to use T&E to analyze them!

I took my own advice and gave up.
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Postby RW » Fri Apr 21, 2006 8:10 pm

EmilyS wrote:a friend gave me this puzzle:


I'd reconsider the use of the word friend. This puzzle don't seem to be an act of friendship... it really is devilish! But what could be more fun than beating the devil?

QBasicMac wrote:Many puzzles have solutions, but it is a waste of time to work on them. Simply move on to something that is more fun... ..I took my own advice and gave up.


I wouldn't give up quite yet. So you removed candidates 2, 4 and 8 from r2c7. I suppose you ended up with something like this:

Code: Select all
 *--------------------------------------------------------------------*
 | 56     26     246    | 1458   7      9      | 3      1246   1248   |
 | 1      9      2467   | 458    3      48     | 57     2467   248    |
 | 357    37     8      | 145    2      6      | 457    147    9      |
 |----------------------+----------------------+----------------------|
 | 4      2378   1      | 78     5      2378   | 289    239    6      |
 | 368    5      9      | 468    1468   12348  | 248    234    7      |
 | 3678   23678  2367   | 46789  4689   23478  | 1      5      2348   |
 |----------------------+----------------------+----------------------|
 | 2      178    5      | 3      1489   1478   | 6      1479   14     |
 | 36789  13678  367    | 2      14689  1478   | 479    13479  5      |
 | 3679   4      367    | 679    169    5      | 279    8      123    |
 *--------------------------------------------------------------------*


First, let's make some uniqueness reductions:

If r1c8=4 => r2c3=4 => r3c4=4 => r1c4=1 => r3c8=1 => UR in r13c48 => r1c8<>4

If r1c4=4 => r1c23=(26) => r1c8=1 => r3c4=1 => r2c4=5 => r2c7=7 => r3c8=4 => UR in r13c48 => r1c4<>4

Then the next move is simple:
If r1c9=4 => r23c7=(57) => r1c23=(26) => r1c8=1 => r3c8=empty cell => r1c9<>4

Solved a single: r1c3=4

Doesn't really help solving the rest of the puzzle, but brings you one step closer to the solution. I'll have a look at possibilities to proceed sometime later in the weekend.

Regards RW
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Postby Unk » Fri Apr 21, 2006 8:33 pm

In row 7, the only 5 is in column 3. Cell 73 is 5.
In column 5, the only 5 is in row 4. Cell 45 is 5.
In row 2, the only 9 is in column 2. Cell 22 is 9.
In row 6, 9 appears only in block 22. Cell 44 is not 9.
In block 12, 1 appears only in column 4. Cells 44 and 94 are not 1.

From here, I have to guess. Here are the 5 rules I use, given a grid of possible values for each cell.

Rule 1: If exactly one digit appears in a cell’s list, then that digit must be the digit for the cell.

Rule 2: If in a group, a digit appears in exactly one cell, then that digit must be the digit for the cell.

Rule 3: If in group A, a digit appears only in the intersection of group A and another group B, then the digit does not appear elsewhere in group B.

Rule 4: If in a group, the union of N cells is exactly N digits, then none of those digits appear elsewhere in the group.

Rule 5: If in a group, N digits appear in exactly N cells, then no other digits appear in those cells.

Hope this helps. I wrote a program for PC which solves every puzzle, even tracks good and bad guesses. You'd be amazed how simple some "Diabolical" puzzls are. Interested?
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Postby ravel » Fri Apr 21, 2006 8:41 pm

Emily,

i am interested, if your friend created the puzzle himself or where he has it from. There are not so much ultra hard puzzles around. I only know 4 ones, that seem to be harder and a few ones with about the same difficulty (but there exists no objective measure, any good solution can show that a puzzle is easier than assumed).
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Postby tarek » Fri Apr 21, 2006 9:19 pm

The above puzzles required contradiction elimination by my solver to crack.......

difficult indeed...

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Postby lunababy_moonchild » Fri Apr 21, 2006 9:20 pm

Unk wrote:My program is posted at www.eop.mu.edu/lawrence/Sudoku.zip and may be downloaded and shared at will. I didn't write it for profit, just the challenge.

Save the zipped file in a local folder (like Sudoku, perhaps), unzip it and follow the directions in README.TXT.

Difficulty is based on which rules are used, and bumped by 1 if guessing is necessary.

You can set it to: scroll toward an almost immediate solution or to wait for you to push ENTER after seeing each step; let you do the guessing or guess automatially. If you choose the right settings, it may still feel as if you're solving the puzzle yourself.

Originally, I just wanted a program to do the bookkeeping for me, but I then taught it how to use the 5 basic rules and decided to make it guess for me, since all the thrill of solving the puzzles by hand had waned in the process of programming.

Now the only interesting result is how misleading and intimidating the provided levels of difficulty often are.

If people know of other rules that aren't some amalgam of intuition and guessing, can communicate them to me, and I can program them, it would be greatly appreciated.

I split this post from the thread because I thought it wasn't relevant. I now see that it is relevant, this is the only way I can 'put it back'. Apologies for the confusion - Luna
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Postby Unk » Fri Apr 21, 2006 9:23 pm

Thanks for replacing the post. For a while I was afraid I had violated some clause of some agreement. I mean no harm. The program doesn't do anything more than what a non-expert would do, so I don't think I'm bursting any bubbles, stealing any thunder, stepping on any toes, ...
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Postby lunababy_moonchild » Fri Apr 21, 2006 9:24 pm

You are not, please see your PM for an explanation:D

Luna
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Postby re'born » Fri Apr 21, 2006 9:29 pm

RW wrote:Then the next move is simple:
If r1c9=4 => r23c7=(57) => r1c23=(26) => r1c8=1 => r3c8=empty cell => r1c9<>4

Solved a single: r1c3=4


If one doesn't want to assume Mac's reductions (which I presume were based on trial & error), we can still make this "simple" reduction (using RW's first 2 deductions which do not depend on Mac's work).

(2,3)4 > (2,6)8 >> (2,4)5 >> (1,4)1 > (1, [28])<26> > (1,3)4, a contradiction. Therefore, (2,3)!4 and so (1,3)4.

Incidentally, there is also a fiined x-wing that makes the elimination (9,1)!3.
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Postby RW » Sat Apr 22, 2006 9:27 am

rep'nA wrote:If one doesn't want to assume Mac's reductions (which I presume were based on trial & error), we can still make this "simple" reduction (using RW's first 2 deductions which do not depend on Mac's work).


I think the second of my reductions would need some extra steps to remove 2, 4 and 8 from r2c7, otherwise it works fine:

If r1c4=4 => r1c23=(26) => r1c8=1 => r3c4=1 => r2c3=4 => r2c4=5 => r2c6=8 => r2c9=2 => r2c7=7 => r3c8=4 => UR in r13c48 => r1c4<>4

Then the last reduction could also be:
If r1c9=4 => r1c23=(26) => r1c8=1 => r2c3=4 => r2c6=8 => r2c4=5 => r1c4= empty cell => r1c9<>4

But that's essentially the same as your move, so it doesn't matter which we use. What matters is that r1c3=4, which I think is very well proved by now.

I had another look at the puzzle this morning. I was hoping to find a solution that didn't need 10+ step trails, but that seems to be very hard. Solving the next number required two shorter trails and one 12 step (unfortunately these also continue from Macs already made T&E reductions, maybe I should go back and find something that doesn't depend on them):

First a quite short trail:

If r9c7=7 => r2c7=5 => r3c7=4 => r9c9=2 => r2c9=8 => r2c6=4 => r2c4= empty cell => r9c7<>7

Then I can smell an uniqueness reduction, so let's do that:

If r4c7=2 => r9c9=2 => r8c8=3 => r5c8=4 => r5c7=8 => r5c4=6 => r5c1=3 => r9c4=7 => r9c1=6 => BUG-lite in r1c28, r2c38, r6c23 => r4c7<>2

[Edit: missing step in previous trail r9c7=9]

The next step is quite ugly, but I've been working this same area for a while and wish to see some result:

If r9c7=9 => r9c9=2 => r9c5=1 => r4c7=8 => r6c9=3 => r5c78=(24) => r5c45=(68) => r5c1=3 => r4c4=7 => r4c2=2 => r1c2=6 => r1c1=5 => r3c1=7 => r9c1=6 => r9c4= empty cell => r9c7<>9

[Edit: Typos fixed]

=> r9c7=2

A small step for a man and unfortunately feels like an even smaller step in terms of solving the puzzle... it's not very easy to get an opening in this evil creation.

RW
Last edited by RW on Sat Apr 22, 2006 6:46 am, edited 1 time in total.
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Postby EmilyS » Sat Apr 22, 2006 9:46 am

i am interested, if your friend created the puzzle himself or where he has it from.


He made it himself. He's making puzzles every now and then (by hand, I think) and one day he just showed me this and said it was his "most fiendish creation".

Thanks RW for solving those numbers. If the computers need T&E to solve this puzzle, then maybe we should allow ourselves to do so also. But I sure like your short trails better than the long one. I especially liked your BUG-lite move. I've used uniqueness technique myself, but I cannot possibly understand how you could "smell" the BUG-lite in your reduction.

-E
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