Very hard help please?

Advanced methods and approaches for solving Sudoku puzzles

Very hard help please?

Postby paulf2127 » Fri Jun 03, 2005 5:51 pm

What a difference!

Starting grid;

1 4 x = 6 x x = x 8 x
x x x = 1 9 x = x x x
x x 7 = x x x = 5 x x
==============
x x x = x x x = x 6 2
6 x x = 7 x 4 = x x 9
5 8 x = x x x = x x x
==============
x x 2 = x x x = 3 x x
x x x = x 7 9 = x x x
x 5 x = x x 1 = x 9 4

After 2 hours got to ;

1 4 x = 6 x x = 9 8 x
x x x = 1 9 x = x x x
x x 7 = x x x = 5 x x
==============
x 7 9 = x x x = x 6 2
6 2 x = 7 x 4 = x 5 9
5 8 x = 9 x x = x x x
==============
x x 2 = x x x = 3 7 x
x x x = x 7 9 = x x x
7 5 x = x x 1 = x 9 4

Absolutely stumped. Can't see any x wings that are any use. Obviously the 9's would be x-wings but don't help any. Have checked my pencil marks but n movement.

A little nudge please.

Paul F
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Postby paulf2127 » Fri Jun 03, 2005 6:03 pm

Hold up, I think I have movement. Belay the last.
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Postby Animator » Fri Jun 03, 2005 7:10 pm

Column 9 is intresting, and when you removed some candidates then row 7 gets intresting...
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Postby paulf2127 » Fri Jun 03, 2005 7:52 pm

Sorry animator, thanks for your help but I managed this one. I had what alcoholics call 'a moment of clarity' moments after posting.....
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Postby Guest » Sun Jun 05, 2005 4:19 pm

paulf, if you finished it, mind posting the answer on?? I wanna check mine
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Postby scrose » Sun Jun 05, 2005 5:35 pm

There are those who might not want the answer spoiled. Instead of having the solution posted here, why don't you try plugging the clues into one of the many online solvers to discover the answer for yourself.
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help

Postby Bjorn » Tue Jun 07, 2005 11:43 am

Im stuck in

--------------------
=14*=6**=98*=
=***=19*=***=
=**7=***=5**=
--------------------
=**9=***=*62=
=62*=7*4=*59=
=58*=***=***=
--------------------
=**2=***=3**=
=***=*79=***=
=*5*=**1=*94=
--------------------

Hints plz.

Greetings

/Björn
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Re: help

Postby whohe » Tue Jun 07, 2005 1:55 pm

Bjorn wrote:Im stuck in

--------------------
=14*=6**=98*=
=***=19*=***=
=**7=***=5**=
--------------------
=**9=***=*62=
=62*=7*4=*59=
=58*=***=***=
--------------------
=**2=***=3**=
=***=*79=***=
=*5*=**1=*94=
--------------------

Hints plz.

Greetings

/Björn


You're gonna kick yourself - look at the 9's and the centre block
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Postby Bjorn » Tue Jun 07, 2005 2:20 pm

hm, yes


its these pencilmarks that messes things up for me. I obviously stops thinking. Thank you!
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Postby Bjorn » Tue Jun 07, 2005 4:12 pm

Even thouht it gives me the 9 in the middle block I wont get any further.
Right now I'm searching doubles and triplets to try eliminate some possibles but have done that for a while with no progress and yet again I ask for help...

Greetings

/Björn
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Postby whohe » Tue Jun 07, 2005 4:42 pm

Bjorn wrote:Even thouht it gives me the 9 in the middle block I wont get any further.
Right now I'm searching doubles and triplets to try eliminate some possibles but have done that for a while with no progress and yet again I ask for help...

Greetings

/Björn


Hi Bjorn -

Well it is described as a hard one!

I think you're going to need to think a little more about doubles & triplets and what they can mean.

The simplest way I can think of descibing this generalised technique is as follows -

If you have n candidates appearing in n cells within any row, column or block, then no other cell in that row, column or block can contain any of those candidates.

This is simplest to understand when you have 2 candidates in 2 cells.

Clearly if the only possible candidates for 2 cells in a row/column/block are a and b, then neither a nor b can appear in any other cell in that row/colum/block.

Things get a little more tricky for n >2 as there are more permutations amongst which the candidates can be placed in cells, but the technique still applies.

Consider the candidates a, b and c dispersed across 3 cells.

abc|abc|abc

Clearly if the candidates for the three cells are in all cases a,b and c then it is obvious that no other cell can contain those candidates.

However, for this techique to work, it doesn't matter whether or not the three candidates each appear in all of the cells.

The three cells could contain -

ab|bc|ba

The technique still applies - none of a, b or c can appear in any other cell in the row/column/block that is under investigation.

Exteding to 4 candidates, you could have:

bd|bcd|abc|abd

Same thing applies. It's just a single rule - regardless of how it looks.

n candidates in n cells within a row/column/block - those candidates cannot appear in any other cell within that row/column/block.

What this enables us to do is to eliminate those candidates from other cells in the row/column/block.

For example -

For a group of 6 cells, you may have:

bd|bcd|abc|abd|abedf|abceg

From the first 4 cells, using just the one rule, you can eliminate a, b, c and d from the last two cells, leaving

bd|bcd|abc|abd|ef|eg
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Postby Bjorn » Tue Jun 07, 2005 5:02 pm

I agree with the conclution in your example, but cant see how you from your first 4 cells can eliminate in the last 2. I look in the last 2 and find triple efg (not occuring in first 4), thus eliminating others from theese. Obviously we get the same result so the question is rather how to identify these patterns in a complex grid, this is my problem. Some I have found (ie 23 in the middle block) but obviously its not enough. Concerning more advanced solvning methods (xwing,samuraj T&E etc) I have not learnt how to identify and apply these yet.
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Postby whohe » Tue Jun 07, 2005 5:21 pm

Bjorn wrote:I agree with the conclution in your example, but cant see how you from your first 4 cells can eliminate in the last 2. I look in the last 2 and find triple efg (not occuring in first 4), thus eliminating others from theese. Obviously we get the same result so the question is rather how to identify these patterns in a complex grid, this is my problem. Some I have found (ie 23 in the middle block) but obviously its not enough. Concerning more advanced solvning methods (xwing,samuraj T&E etc) I have not learnt how to identify and apply these yet.


It's because you have n candidates appearing in n cells. It doesn't matter where else they appear - you can literally ignore them, they're effectively a red herring.

efg occuring in 2 cells doesn't allow any elimination because you have 3 candidates but in only 2 cells (don't forget of course that there will be other cells in this group, but we don't have to consider them).

"n" must = "n"!
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Postby Bjorn » Tue Jun 07, 2005 5:32 pm

Hehe, we look at different sides of the same coin I think. I agree with you now (but in the subset of our 6 cells, my reasoning is just as valid ;). The question is still, how can I "easily" identify this pattern?
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Postby Animator » Tue Jun 07, 2005 5:44 pm

you never need a subset of 6 numbers/cells.

the maximum you'll ever need is 4 numbers/cells.

Why? The answer is simple, there are always two groups.

Either n cells share the same numbers (and only these numbers),

or there are m cells that are the only valid candidates for m numbers.


A little example of that: in column 9 of this puzzles there are two groups, one of 2 numbers, and one of 4 numbers.

The 'n'-example: the group: 1, 3, 6, 7. 4 numbers, 4 cells. All of these four cells have share (only) these numbers.

The 'm' example: the group 5, 8. There are only two cells that can have these as candidate. Ofcourse these cells has extra candidates aswell, but they are just irrelevant.
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