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hollydoll Guest

Posted: Tue Aug 30, 2005 2:53 pm Post subject: august 28 


Where do I go from here? Thanks.
8xx x57 xxx
x3x x4x xx2
xx6 x21 xx5
1xx 78x 59x
xx9 2x4 8xx
x8x x95 x23
4xx 57x 2xx
2xx x6x x5x
xxx 4x2 xx7 

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Guest

Posted: Tue Aug 30, 2005 5:20 pm Post subject: 


You can place a 4 in square 3 (but I wont tell you where).
Then you can place an 8 in square 7.
After that, I'm stuck too. I know what the next hint is, but I can't see why.
Anyone tell me? 

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chobans
Joined: 21 Aug 2005 Posts: 39

Posted: Tue Aug 30, 2005 6:20 pm Post subject: 


After you place 8 in box 7 (r7c1r9c3) at r9c3, if you notice, in column 3, 5 can't go anywhere else except at r2c3. After you place 5 there, it'll eliminate the possibility of 5 at r2c1, leaving only 7 and 9 as possibilities. These are the same two numbers that's possible at r3c1. So you can eliminate 7 and 9 from the other cells in box 1 (r1c1r3c3), meaning r3c2 can only be a 4.
Let me know if this isn't enough to solve from here. 

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Hollydoll Guest

Posted: Tue Aug 30, 2005 7:19 pm Post subject: Aug 28 


How do you know where to put an 8 in the 7th box? And as far as where the 4 goes in the third box  I don't get that either. More help is needed. Thanks 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Tue Aug 30, 2005 8:48 pm Post subject: 


Hi Hollydoll,
You have to learn and use the techniques:
"Column on 3x3 Block interaction"
"Row on 3x3 Block interaction"
"3x3 Block on Row/Column interaction"
you can take a look at:
http://www.simes.clara.co.uk/programs/sudokutechniques.htm
for this.
Than there will be no more problems to solve it.
Or, you will have to work it out by yourself (the hard way).
If you would like a particular help for this Sudoku (that other participants have also refused till now), we can provide it. Please, just tell us.
P.S. I know that you can do it!!!
see u, 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Tue Aug 30, 2005 9:03 pm Post subject: Re: Aug 28 


Hollydoll wrote:  How do you know where to put an 8 in the 7th box? And as far as where the 4 goes in the third box  I don't get that either. More help is needed. Thanks 
Start by looking at the third box (intersection of r13 with c79). Because of the "8"s in row 1 and in column 7, you can see that the "8" in this box has to go in column 8  either in r2c8, or in r3c8. This implies that the "8" in box 9 must go in column 9  specifically, in either r7c9 or r8c9.
That leaves only one place to put an "8" in row 9  it has to go in r9c3. It can't fit in the first two columns because of the "8"s in r1c1 and r6c2; it can't fit in column 5 because of the "8" in r4c5; and it can't fit in column 7 or column 8 because we know the "8" in the ninth box has to be in either row 7 or row 8  not in row 9.
I'm sort of in a hurry right now ... what I do see about the "4" in box 3 is that it has to go in column 8. There are only 2 possible spots for a "4" in the ninth box (r8c7 or r8c9) and there are only two spots in the 6th box (r4c9 and r6c7) where the "4" can go  so the only way a "4" can fit in the 8th column is if it's in one of the first three rows. This gives us two numbers {4,8} that have to fit in the three cells r1c8, r2c8, r3c8  I don't see what the third one has to be, right off  maybe you can figure that out.
Good luck! dcb 

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Bruno Guest

Posted: Wed Aug 31, 2005 10:50 am Post subject: 


Thanks for this useful techniques !
But, i can't understand the Swordfish one
Coul'd you help me please ? 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Wed Aug 31, 2005 1:03 pm Post subject: Re: Swordfish 


There's a good discussion of "swordfish" under the Other Puzzles category in this forum, right here.
Briefly, the Swordfish works by identifying 3 (or more) rows or columns in which a particular value can only appear in two cells. We then get a connected loop of {n or ~n} pairs. We may not know which state the loop is in, but we can be sure that it is in one of two states, and that fact may help us to make inferences about the rest of the puzzle.
I'll post another message later, with an example. Be sure to check out the link above  it will help you understand. dcb 

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Bruno Guest

Posted: Thu Sep 01, 2005 12:46 pm Post subject: 


Thank you for the link and the sumary of the explication
now i understand the "loop" technique
I'll check your example later ! 

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Guest

Posted: Thu Sep 01, 2005 1:37 pm Post subject: Xwing, Swordfish, Jellyfish and Squirmbag ... 


Hi,
It's me again.
He a very nice and abstract description of several very advanced techniques, that I found:
Look for N columns (2 for Xwing, 3 for the Swordfish, 4 for a Jellyfish, 5 for a Squirmbag) with only two candidate cells for ONE given digit. If these fall on exactly N common rows, and each of those rows has at least 2 candidate cells, then all N rows can be cleared of that digit (except in the defining cells!).
I am still learning, ....
see u, 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Thu Sep 01, 2005 1:41 pm Post subject: 


Previous mesage was from me,
someone_somewhere.
I just was under deep cover ...
see u,
P.S. for the ones from you that know japoneze:
I am "inbijiburu" and you can reach me at inbijiburu@gmail.com
P.P.S. please no japooneze emails. 

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Going Mad... Guest

Posted: Sat Sep 03, 2005 3:21 pm Post subject: Why the Number 6 in Box 4? 


Doing this and now stuck. The hint says the number 6 in Box 4 but can't see why that the next logical step after doing number 4 in Box 3. Can anyone help explain please? 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sat Sep 03, 2005 6:13 pm Post subject: 


give us your exact position in the universe, so we can guide you ...
it's hard to imagine at what coordinates you are stucked pilot.
SOS without any additional information will not bring help to you.
see u, 

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Going Mad Guest

Posted: Sat Sep 03, 2005 7:07 pm Post subject: 


Sorry! Ok  here we go...
8xx x57 x4x
x35 x4x xx2
x46 x21 xx5
1xx 78x 59x
xx9 2x4 8xx
x8x x95 x23
4xx 57x 2xx
2xx x6x x5x
xx8 4x2 xx7
According to the hint, it's 6 in r6c1  but I can't figure out why!
Thanks!!! It's been driving me mad for hours! 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sat Sep 03, 2005 10:57 pm Post subject: Column on Block interaction is the clue 


Hi, Holly!
I'm starting from the position in your latest post.
You need to place one more number in the puzzle before you can decide to put the "6" at r6c1. As it stands, you have {1, 2, 4, 8} in column 1, and {2, 3, 5, 8, 9} in row 6, so either a "6" or a "7" is possible in this square right now. Let's look for one other number.
Look first at the 3x3 box in the lower left corner (the intersection of rows 7, 8, & 9 with columns 1, 2, & 3). From the position of the "5"s in rows 8 and 9 it's clear that the only way to fit a "5" in this box is to place it in r9c1 or in r9c2. In other words, it has to go in column 1 or in column 2  it can't fit in column 3.
Now turn your attention to the first 3x3 box in the middle row (the intersection of rows 4, 5, & 6 with columns 1, 2, & 3). It's clear that "5" can only fit in this box if it goes in row 5  in r5c1 or in r5c2, to be precise. So we know that the "5" in this box has to fit in column 1 or in column 2  it can't fit in column 3.
The only way to fit a "5" in column 3, then, is if it goes in the first box (the intersection of rows 1, 2, & 3 with columns 1, 2, & 3). And since the first row already has a "5" at r5c5, the only place to put the "5" in that first box is at r2c3.
This process  deducing that a particular number has to fit in a single column within a particular box  is the "Column on Block" interaction others have written about.
Anyway, with five numbers placed in the first box, we see that the only four remaining as possibilities are {1, 2, 7, 9}. And since "1" and "2" already appear in the first column, we see that the pair {7, 9} must appear in r2c1 & r3c1, in some order. Similarly, the pair {1, 2} must appear in the cells r1c2 & r1c3, in some order.
Now the deduction that "6" goes in r6c1 is immediate, since the possibility of a "7" going there has been eliminated  we know that the "7" in the first column has to fit in either r2c1 or in r3c1.
From here the puzzle falls into place pretty rapidly ... the next step is clearly a "1" in r6c4, and it's pretty much downhill from there.
I hope this helps with the "Column on Block" interaction concept. I'll try to write something on your other recent post soon. dcb 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sun Sep 04, 2005 6:03 am Post subject: 


Hi,
David has pointed out correct, how to move one.
The buttom line is that you have to know how to apply 3 techniques:
 Row on 3x3 Block interaction,
 Column on 3x3 Block interaction and
 3x3 Block on Row/Column interaction
The point you was stucked is a good exercise for learning them.
Here the short description for a possible to get home:
1 not in r1c7, it is in r2c7 or r2c8 (Row on 3x3 Block interaction)
1 not in r1c9, it is in r2c7 or r2c8 (Row on 3x3 Block interaction)
8 not in r7c8, it is in r7c9 or r8c9 (Column on 3x3 Block interaction)
7 not in r5c1, it is in r2c1 or r3c1 of the 3x3 (3x3 Block on Row/Column interaction)
7 not in r6c1, it is in r2c1 or r3c1 of the 3x3 (3x3 Block on Row/Column interaction)
3 not in r8c4, it is in r1c4 or r3c4 of the 3x3 (3x3 Block on Row/Column interaction)
6 in r6c1  Sole Candidate
P.S. I can't beat the literar talent of David.
see u, 

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Going Mad Guest

Posted: Sun Sep 04, 2005 9:38 am Post subject: 


THANKS!!!!
Just finished it!! 

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