Very very hard?

Advanced methods and approaches for solving Sudoku puzzles

Very very hard?

Postby joolslee » Tue Jun 07, 2005 10:49 pm

Either I am missing the obvious with this V. Hard puzzle or it requires a technique I've not yet got to grips with ... any help to move me on would be appreciated. Below is original grid then the same with the few cells I've managed to fill so far ...

* 9 * 7 * 3 * * *
* * 1 5 9 * * * *
3 * 7 * * * * * *
* 8 6 * * * 1 * 7
* * 3 * * * 6 * *
7 * 5 * * * 8 9 *
* * * * * * 3 * 1
* * * * 7 4 9 * *
* * * 9 * 8 * 5 *


* 9 * 7 * 3 * * *
* * 1 5 9 * * * *
3 * 7 * * * * * 9
* 8 6 * 5 9 1 * 7
9 * 3 * * 7 6 * 5
7 * 5 * * * 8 9 *
* * 9 * * 5 3 * 1
* * * * 7 4 9 * *
* * * 9 * 8 * 5 *

Even my pencil marks have got pencil marks ... I think they're breeding!!
Last edited by joolslee on Tue Jun 07, 2005 7:02 pm, edited 1 time in total.
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Re: Very very hard?

Postby whohe » Tue Jun 07, 2005 10:57 pm

joolslee wrote:Either I am missing the obvious with this V. Hard puzzle or it requires a technique I've not yet got to grips with ... any help to move me on would be appreciated. Below is original grid then the same with the few cells I've managed to fill so far ...
Code: Select all
* 9 *  7 * 3  * * *             * 9 *  7 * 3  * * * 
* * 1  5 9 *  * * *             * * 1  5 9 *  * * *
3 * 7  * * *  * * *             3 * 7  * * *  * * 9

* 8 6  * * *  1 * 7             * 8 6  * 5 9  1 * 7
* * 3  * * *  6 * *             9 * 3  * * 7  6 * 5
7 * 5  * * *  8 9 *             7 * 5  * * *  8 9 *

* * *  * * *  3 * 1             * * 9  * * 5  3 * 1
* * *  * 7 4  9 * *             * * *  * 7 4  9 * *
* * *  9 * 8  * 5 *             * * *  9 * 8  * 5 *

Even my pencil marks have got pencil marks ... I think they're breeding!!


Hi joolslee -

Use the "code" tags, otherwise it's going to get very confusing!
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Postby joolslee » Tue Jun 07, 2005 11:03 pm

Thanks Whohe ... was trying to save space but it didn't work so edited it now
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Postby simes » Wed Jun 08, 2005 6:12 am

Code: Select all
Note: Cell coordinates are in (row,column) format with (1,1) at top-left and (9,9) at bottom-right.
block 4: column 2 must contain 1, removing 1 from candidates for cell(s) (8,2)(9,2)
row 7: disjoint subset 26 in cells (7,4)(7,5) - updating candidates for cell(s) (7,1)(7,2)(7,8)
row 8: disjoint subset 268 in cells (8,3)(8,8)(8,9) - updating candidates for cell(s) (8,1)(8,2)(8,4)
block 7: disjoint subset 248 in cells (7,1)(8,3)(9,3) - updating candidates for cell(s) (7,2)(9,1)(9,2)
block 8: disjoint subset 26 in cells (7,4)(7,5) - updating candidates for cell(s) (9,5)
(7,2) = 7 : only possible value for this cell
etc
Last edited by simes on Sun Dec 11, 2011 2:36 pm, edited 1 time in total.
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Postby joolslee » Wed Jun 08, 2005 9:30 am

simes wrote:
Code: Select all
block 7: disjoint subset 248 in cells (7,1)(8,3)(9,3) - updating candidates for cell(s) (7,2)(9,1)(9,2)


Simes
http://www.simes.clara.co.uk/programs/sudoku.htm


Thank you Simes. I had overlooked a couple of those strategies however the block 7 hint above is one I can't figure out logically. My candidates for that block are ...

4/8 - 4/7 - 9
1/5/8 - 3/5 - 2/8
1/2/4/6 - 2/3/4/6/7 - 2/4

I'm failing to see how 4/8, 2/8 and 2/4 in those three cells enables me to delete 2s 4s and 8s from the other cells. Could you explain please?
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Postby abailes » Wed Jun 08, 2005 9:54 am

joolslee,

There are only three numbers (2, 4 and 8) possible in three cells. Simply, if you put any of those three numbers in a different cell in that box then there would not be enough of 2, 4 and 8 to go around in the three cells listed.

Try it! I know people frown on trial and error in these forums, but see what happens if you put a 2 in (say) cell (7,3). Then your codes would look like:

48 47 9
158 35 8
2 3467 4

So, you put in the 4 in (9,3) and the 8 in (9,2). Then - what do you put in (7,1)??

There are three numbers and three cells. They cannot possibly go anywhere else.

You will end up with

48 47 9
15 35 28
16 367 24

Which is Simes' solver's logical step.

Andrew
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Postby joolslee » Wed Jun 08, 2005 7:19 pm

Thank you for explaining that so clearly Abailes - I understand how that figures now.
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Postby simes » Wed Jun 08, 2005 9:08 pm

Sorry Jools - I didn't explain enough. Thanks to abailes for doing it for me!
Last edited by simes on Sun Dec 11, 2011 2:37 pm, edited 1 time in total.
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slowww person here

Postby azn_devil29 » Fri Jun 10, 2005 9:34 am

You guyz are thinking way too fast...:(
But yeh, I still dont get whats going on...
I get up to this part

48 47 9
15 35 28
16 367 24

now wat?
Sorry guyz, slow learner over here:(
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Postby abailes » Fri Jun 10, 2005 9:39 am

This is going to hurt, but bear with it...:D

That step doesn't give you a number, it is just a necessary step to do the next piece of logic!

Follow the solver on:

"...block 8: disjoint subset 26 in cells (7,4)(7,5) - updating candidates for cell(s) (9,5)
(7,2) = 7 : only possible value for this cell"

I haven't done the puzzle, but I assume after the 7 goes in (7,2), it is relatively easy. Let us know if it isn't.
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Postby azn_devil29 » Fri Jun 10, 2005 10:05 am

i still dont get it...
how can (7,2) = 2?
:s *cries
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Postby azn_devil29 » Fri Jun 10, 2005 10:06 am

i mean how can (7,2) = 7?
Can't 7 be in (9,2) as well?
:(
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Re: slowww person here

Postby Animator » Fri Jun 10, 2005 10:09 am

48 47 9
15 35 28
16 367 24


That's a copy of Andrew's post but unfortunally it is incorrect.

His post explains that the numbers 2, 4 and 8 form a group of three numbers. This allows you to remove them from all other cells.

This also allows you to do something Andrew forgot... remove the 4 as candidate from r7c2 (or r2c1 in the small grid posted above). This leaving only the 7 as a candidate.


Try to fill in the number 4 in r7c2 and see what happens... you will have an empty cell.

(Andrew, just for the record, most people here do not frown on trial and error when it is used to explain a pattern.)
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Postby azn_devil29 » Fri Jun 10, 2005 10:18 am

ohh yehh!!!
thanks animator
I just picked up a new technique... lolz
never thought about that before:)
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Postby abailes » Fri Jun 10, 2005 10:19 am

Whoops. That shows the value of actually doing the puzzles you are talking about, instead of just chatting confidently about somebody else's puzzle.

Sorry all.

Mind you, I feel a tiny bit better about being able to make the smallest of corrections to Animator's last post. The cell you remove the additional 4 from is r1c2 in the small grid, not r2c1.
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