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techschool Guest

Posted: Thu Jan 19, 2006 5:57 pm Post subject: 19Jan2006 classic logic question 


I successfully made the first 17 moves, but fail to see the logic in the 18th move; it looks like a guess to me (which the authors say should not happen). I do not believe they are mistaken, but that I can not see it. 

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

Posted: Thu Jan 19, 2006 7:02 pm Post subject: 


Hi,
I suppose you mean this puzzle:
Code:  . . . . . . 3 . 8
. . 6 . . 4 . 7 5
. . 2 . 9 . . . .
. 6 . 9 5 . 7 . 2
. 7 . . . . . 6 .
1 . 5 . 2 6 . 9 .
. . . . 3 . 6 . .
6 3 . 4 . . 2 . .
9 . 7 . . . . . . 
and after 17 moves, you should find out:
 4 it is in r1c1 or r1c3, so you can eliminate it from r3c1 and r3c2
 8 it is in r7c3 or r8c3, so you can eliminate it from r7c1, r7c2 and r9c2
 1 is in r3c7 or r3c8 of the 3x3, so you can exclude it from r3c2 and r3c4
finally you have to find out the:
 Hidden Pair 7 9 in r7c6 and r7c9, so can exclude 2, 5 and 8 from r7c6
and you can set 2 in r7c2 and r9c6 ....
and the rest should not be a problem.
hope I could help you, and you don't refer to a different puzzle ...
see u, 

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techschool Guest

Posted: Thu Jan 19, 2006 7:53 pm Post subject: yes that puzzle 


I must be dumb, for I do not see the elimination here: " 8 it is in r7c3 or r8c3, so you can eliminate it from r7c1, r7c2 and r9c2" (after 17 moves I have one 8 located in r1c9 and therefore can only eliminate 8s from r1 or c9. 

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

Posted: Thu Jan 19, 2006 7:56 pm Post subject: This was an interesting puzzle ... 


I worked through this puzzle without making any extraneous pencil marks, and got hung up at exactly the point our guest is talking about.
Code:  . 9 . . . . 3 2 8
3 . 6 2 . 4 9 7 5
. . 2 . 9 . . . 6
. 6 3 9 5 1 7 . 2
2 7 9 . 4 . 5 6 1
1 . 5 7 2 6 . 9 3
. . . . 3 . 6 . .
6 3 . 4 . . 2 . .
9 . 7 . . . . 3 4 
At this point I should have been able to spot the second piece of the {7, 9} pair in row 7, but for some reason I didn't see it. It's obvious enough now, because of the {7, 9} in the bottom left 3x3 box, the {7, 9} in column 4, and the {7, 9} in column 8. But I did find another way to proceed beyond this point.
What struck me was the fact that this puzzle could only have a _unique_ solution if the "7" in column 5 lies in the bottom center 3x3 box. If there were a "7" at r1c5 then there would be a {7, 9} pair in r7c6 & r8c6  this would match up with the {7, 9} pair in r7c9 & r8c9, making at least two solutions possible.
Armed with confidence in Samgj's devotion to unique solutions, I placed a "7" at r8c5 and went merrily on my way. dcb 

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techschool Guest

Posted: Thu Jan 19, 2006 8:22 pm Post subject: Sounds like guessing 


I follow "At this point I should have been able to spot the second piece of the {7, 9} pair in row 7, but for some reason I didn't see it. It's obvious enough now, because of the {7, 9} in the bottom left 3x3 box, the {7, 9} in column 4, and the {7, 9} in column 8. But I did find another way to proceed beyond this point.
What struck me was the fact that this puzzle could only have a _unique_ solution if the "7" in column 5 lies in the bottom center 3x3 box. If there were a "7" at r1c5 then there would be a {7, 9} pair in r7c6 & r8c6  this would match up with the {7, 9} pair in r7c9 & r8c9, making at least two solutions possible.
Armed with confidence in Samgj's devotion to unique solutions, I placed a "7" at r8c5 and went merrily on my way. dcb" but disagree with the logic because r1c5; r1c6; and r3c6 could also contain a 7 and I see a {7, 9} at r7c6; r7c9; r8c6; and r8c9. I think the problem lies in the fact that there is one 8 located in a corner box which leaves too many variables to solve the puzzle without guessing. 

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

Posted: Thu Jan 19, 2006 9:20 pm Post subject: It isn't guessing ... 


OK, I guess this hasn't been explained clearly enough yet. I suppose that the position you arrived at is the one illustrated in my previous post.
Code:  . 9 . . . . 3 2 8
3 . 6 2 . 4 9 7 5
. . 2 . 9 . . . 6
. 6 3 9 5 1 7 . 2
2 7 9 . 4 . 5 6 1
1 . 5 7 2 6 . 9 3
. . . . 3 . 6 . .
6 3 . 4 . . 2 . .
9 . 7 . . . . 3 4 
Let's analyze the cells in row 7.
 Neither "7" nor "9" can appear in r7c1, r7c2, or r7c3 because of the "9" at r9c1 and the "7" at r9c3.
 Neither "7" nor "9" can appear at r7c4 because of the "9" at r4c4 and the "7" at r6c4.
 Neither "7" nor "9" can appear at r7c8 because of the "7" at r2c8 and the "9" at r6c8.
 The only two cells in row 7 which can accomodate either a "7" or a "9" are r7c6 & r7c9. Therefore the pair {7, 9} must occupy these two cells in some order.
 There cannot be a "2" at r7c1 because of the "2" at r5c1.
 There cannot be a "2" at r7c3 because of the "2" at r3c3.
 There cannot be a "2" at r7c4 because of the "2" at r2c4.
 There cannot be a "2" at r7c8 because of the "2" at r8c7.
 There cannot be a "2" at either r7c6 or r7c9 because of the aforementioned {7, 9} pair.
 Therefore r7c2 = 2  it's the only possible place that a "2" can fit in row 7.
With a "2" in r7c2 the following moves are forced:
r6c2 = 4 (Unique in column)
r6c7 = 8 (Sole candidate)
r4c8 = 4 (Sole candidate)
r4c1 = 8 (Sole candidate)
r3c8 = 1 (Sole candidate)
r3c7 = 4 (Sole candidate)
r9c7 = 1 (Sole candidate)
r2c2 = 1 (Unique in column)
r2c5 = 8 (Sole candidate)
...
and the rest of the puzzle unwinds in pretty short order.
That's one approach. Now for the other one.
 There has to be a "7" in column 5 somewhere. The only possibilities are r1c5 & r8c5  the rest of the cells in column 5 are either occupied, or they're blocked (by "7"s in row 2 and in row 9).
 The only possibilities at r7c9 and r8c9 are {7, 9}. This is clear because {1, 2, 3, 4, 5, 6, 8) are already entered in column 9.
 Suppose that r1c5 = 7. Then we can make the following inferences in the bottom middle 3x3 box.
* Neither "7" nor "9" can appear in r9c4, r9c5, or r9c6, because of the {7, 9} in the bottom left 3x3 box.
* Neither "7" nor "9" can appear in r7c4 because of the {7, 9} appearing in r6c4 & r4c4.
* Neither "7" nor "9" can appear in r8c5 because of the "7" assumed at r1c5 and the "9" already appearing at r3c5.
 Therefore, if r1c5 = 7, the {7, 9} pair must lie in r7c6 & r8c6, because those are the only two cells in the bottom center 3x3 box where either a "7" or a "9" can possibly appear.
 But then the puzzle must have at least two solutions, because we can either have r7c6 = 7, r7c9 = 9, r8c6 = 9, r8c9 = 7, or, alternatively, we can have r7c6 = 9, r7c9 = 7, r8c6 = 7, r8c9 = 9.
 Therefore, if we assume the puzzle has a unique solution, we must have r8c5 = 7.
Again, the puzzle is easily solved from this point, with r8c5 = 7. dcb 

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

Posted: Thu Jan 19, 2006 10:20 pm Post subject: Re: yes that puzzle 


techschool wrote:  ... I do not see the elimination here: " 8 it is in r7c3 or r8c3, so you can eliminate it from r7c1, r7c2 and r9c2" (after 17 moves I have one 8 located in r1c9 and therefore can only eliminate 8s from r1 or c9. 
Once again it will help to look at the actual position.
Code:  . 9 . . . . 3 2 8
3 . 6 2 . 4 9 7 5
. . 2 . 9 . . . 6
. 6 3 9 5 1 7 . 2
2 7 9 . 4 . 5 6 1
1 . 5 7 2 6 . 9 3
. . . . 3 . 6 . .
6 3 . 4 . . 2 . .
9 . 7 . . . . 3 4 
 There must be an "8" in column 3.
 It cannot lie at r1c3 because of the "8" at r1c9.
 It cannot lie in rows 2, 3, 4, 5, 6, or 9 because those cells are already filled.
 Therefore the "8" in column 3 must lie in r7c3 or r8c3, as asserted. dcb 

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techschool Guest

Posted: Fri Jan 20, 2006 2:40 pm Post subject: not enough patience 


Thank you David, I see it now! Knew there was something I was missing. My patience was in short supply yesterday. 

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Guest

Posted: Fri Jan 20, 2006 4:51 pm Post subject: 


im stuck at the same place david bryant is at with his first post. i did not read any further because i want to solve this without any help or guessing. 

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Booya
Joined: 24 Jan 2006 Posts: 1

Posted: Tue Jan 24, 2006 4:36 pm Post subject: 


WOW, I was stuck at the EXACT same point too...
I too missed the hidden pair 7,9 That led to the unique 2's. Isn't it funny how logic works?
________
weed vaporizer
Last edited by Booya on Fri Feb 04, 2011 2:51 am; edited 1 time in total 

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alanr555
Joined: 01 Aug 2005 Posts: 198 Location: Bideford Devon EX39

Posted: Tue Jan 24, 2006 11:52 pm Post subject: Re: yes that puzzle 


> There must be an "8" in column 3.
> It must lie in r7c3 or r8c3, as asserted.
This also is the conclusion of Mandatory Pairs.
However, what broke the back of this one for me was
not the 79 in row 7 but (after setting the Candidate Profiles!)
the 1458 quadruple in row 7. I supposed that it being "naked"
made the difference!
Having spotted the quadruple, it was the subsequent eliminations
which exposed the 79 pair.
+++
I treat each row or column as an exercise in congruence. This
requires that the row be determinable as a collection of subsets
of which any with just one member will be the resolved cells and
can safely be ignored.
In this case we started with
(3)(6) (1245789)
The 3,6 are the resolved cells and so can be ignored.
Spotting the quadruple transforms the subsets to
(3)(6) (1458) (279)
but then the eliminations expose the 2 in r7c2
and the subsets become
(2)(3)(6) (1458) (79)
The congruence rule requires that any subset must contain
exactly as many digits as the number of cells that it covers.
This is a valuable check on eliminations. Here, in what was
initially the (279) subset the cell profiles were
12458
5789
79
The latter 79 had been reduced previously because of the
mutual reception between r7c9 and r8c9. Mutual Reception
"ties together" the two cells, thereby excluding interlopers.
Clearly the above profiles contain more than the three
distinct digits implied by the length of the subset (279)
and so any digits that occur in the other subset must be
eliminated  meaning removal of 1458 wherever they
occur. This left 2, 79, 79  which IS a congruent set!
However it is trivial to spot the further division into
(2) (79) whilst maintaining congruence.
In practice, I ignore the "singlemember" subsets when
testing for congruence but in theory they should be
included so that the "line" profile always contains exactly
nine members.
The congruence principle applies equally to rows, columns
and regions. Indeed, one technique is inspection of the
derived profiles to ensure that congruence is maintained
after resolving one or more cells. If INcongruence is found
then further eliminations are required!
Alan Rayner BS23 2QT 

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