Hard to me !!

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Hard to me !!

Postby danlm » Thu Feb 22, 2007 8:03 pm

I am not so good with the sudoku !!:(
Who can explain how to find the next step?
Puzzle is:
Code: Select all
4  6  5   | 13  7   8    | 2   39 139
8  7  9   | 136 36  2    | 35  4  135
2  1  3   | 5   49  49   | 7   8  6
----------+--------------+-------------
3  5  48  | 689 689 1    | 49  2  7
6  2  78  | 4   589 579  | 1   39 389
1  9  478 | 78  2   3    | 45  6  58
----------+--------------+-------------
7  8  1   | 2   349 469  | 369 5  39
59 3  2   | 789 589 5679 | 689 1  4
59 4  6   | 38  1   59   | 38  7  2


Thanks
danlm
 
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Joined: 03 April 2006

Postby udosuk » Thu Feb 22, 2007 8:14 pm

Code: Select all
 *-----------------------------------------------------------*
 | 4     6     5     | 13    7     8     | 2     39    139   |
 | 8     7     9     | 136  *36    2     |-35    4     135   |
 | 2     1     3     | 5     49    49    | 7     8     6     |
 |-------------------+-------------------+-------------------|
 | 3     5     48    | 689   689   1     | 49    2     7     |
 | 6     2     78    | 4     589   579   | 1     39    389   |
 | 1     9     478   | 78    2     3     | 45    6     58    |
 |-------------------+-------------------+-------------------|
 | 7     8     1     | 2    *349   469   | 369   5     39    |
 | 59    3     2     | 789   589   5679  | 689   1     4     |
 | 59    4     6     |*38    1     59    |*38    7     2     |
 *-----------------------------------------------------------*

You can use a move called "skyscraper/turbot fish/simple colours"...

Basically, when you see a candidate appearing exactly twice on a row/column/block, it is called a "strong link"...

A strong link has the property that exactly 1 of the 2 cells involved must contain the said candidate...

In this puzzle there are strong links of 3s on r9 (r9c4+r9c7) and c5 (r2c5+r7c5)...

Now notice r9c4 and r7c5 are linked in the same block, so at most one of them can be 3...

Consequently, at least one of the other ends of the strong links (r9c7 & r2c5) must contain a 3...

Therefore all cells simultaneously linked to these 2 cells must not contain 3s...

That include r2c7, hence we can eliminate the 3 from r2c7, and conclude that r2c7=5...

The rest of the puzzle is solved with singles...
udosuk
 
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Joined: 17 July 2005

Postby danlm » Thu Feb 22, 2007 11:01 pm

I am impressed and admiring!

It' is masterly!

And better, I very understood!!

Thank you.

Do you know a web page showing other exemple of this method?

DanLM
danlm
 
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Joined: 03 April 2006

Postby re'born » Thu Feb 22, 2007 11:24 pm

danlm wrote:I am impressed and admiring!

It' is masterly!

And better, I very understood!!

Thank you.

Do you know a web page showing other exemple of this method?

DanLM


Try Harvard's Strong Links for Beginners. By the way, this pattern is not actually a Skyscraper, but I'll leave it to you to discover which of the three basic turbot fish patterns it is.
re'born
 
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Joined: 31 May 2007

Postby danlm » Fri Feb 23, 2007 1:03 pm

Thanks a lot !!

I learnt ...

DanLM
danlm
 
Posts: 17
Joined: 03 April 2006

Postby Carcul » Fri Feb 23, 2007 1:42 pm

Alternatively:

Code: Select all
 *-----------------------------------------------------------*
 | 4     6     5     | 13    7     8     | 2     39    139   |
 | 8     7     9     | 136   36    2     | 35    4     135   |
 | 2     1     3     | 5     49    49    | 7     8     6     |
 |-------------------+-------------------+-------------------|
 | 3     5     48    | 689   689   1     | 49    2     7     |
 | 6     2     78    | 4     589   579   | 1     39    389   |
 | 1     9     478   | 78    2     3     | 45    6     58    |
 |-------------------+-------------------+-------------------|
 | 7     8     1     | 2     349   469   | 369   5     39    |
 | 59    3     2     | 789   589   5679  | 689   1     4     |
 | 59    4     6     | 38    1     59    | 38    7     2     |
 *-----------------------------------------------------------*

[r5c9]=8|1=[r1c9](=9=[r1c8]=3=[r5c8]-3-[r5c9])-1-[r1c4]-3-[r9c4]=3=
=[r9c7]-3-[r7c9]-9-[r5c9], => r5c9=8.

Other alternatives:

1) [r9c4]-8-[r8c5]=8=[r5c56|r6c4]-8-[r4c45]-9-[r4c7]=9=[r7c7]-9-[r7c9]-
-[r9c7]-8-[r9c4], => r9c4<>8.

2) [r7c9]=9=[r7c7]-9-[r4c7]-4-[r6c7]-5-[r6c9]-8-[r5c9]=8|1=[r1c9]-1-
-[r1c4]-3-[r9c4]=3=[r9c7]-3-[r7c9], => r7c9<>3.

Carcul
Last edited by Carcul on Fri Feb 23, 2007 10:04 am, edited 1 time in total.
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Postby udosuk » Fri Feb 23, 2007 1:58 pm

rep'nA wrote:... By the way, this pattern is not actually a Skyscraper, but I'll leave it to you to discover which of the three basic turbot fish patterns it is.

Thanks rep'nA for the clarification... I have to admit I've never really studied hard the correct names of these moves... Well next time I'll just call them all "turbot fishes"...
udosuk
 
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Postby _m_k » Fri Feb 23, 2007 5:06 pm

r1c4=1 or 3
Claim: r1c4=3 =>(r8c5=5 and r9c6=5), a contradiction.
Hence, r1c4=1, and the puzzle is solved using only elimination and hidden singles.

Proof of Claim:
r1c4=3 => r9c4=8 => r6c4=7 => r8c4=9 => (r8c5=5 and r9c6=5).

Note 1. I got this from my computer program.
Note 2. Some people prefer not to use this simplest method because they consider it a kind of trial & error.

M.K.
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Postby re'born » Fri Feb 23, 2007 5:23 pm

_m_k wrote:r1c4=1 or 3
Claim: r1c4=3 =>(r8c5=5 and r9c6=5), a contradiction.
Hence, r1c4=1, and the puzzle is solved using only elimination and hidden singles.

Proof of Claim:
r1c4=3 => r9c4=8 => r6c4=7 => r8c4=9 => (r8c5=5 and r9c6=5).

Note 1. I got this from my computer program.
Note 2. Some people prefer not to use this simplest method because they consider it a kind of trial & error.

M.K.


One doesn't have consider it as trial and error. Consider the cells r68c4, r8c5 and r9c6. The possible candidates are {5,7,8,9} with max. multiplicities (1,1,2,1); 5 candidates for 4 cells. However, if r9c4 = 8, then there can be no 8's in our set, leaving us with only three candidates for the 4 cells, an impossibility. Thus r9c4<>8, solving the puzzle.

Edit: This looks like a messed up Sue de Coq, but I suppose all one should say is that it is the ALS xz-rule in action: A= {r6c4}, B={r8c45, r9c6}, x=7, z=8.
re'born
 
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