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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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 Posted: Wed Nov 23, 2005 11:19 pm Post subject: Nov 23 USA Today puzzle, need help! |
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This is a four star out of five puzzle, so I didn't think it would be very hard, but it has me stumped.
120 340 500
000 000 000
560 200 003
007 050 800
201 000 906
003 060 200
900 008 025
000 000 000
002 097 041
I got to this point:
120 346 570
000 005 062
560 270 003
697 152 834
201 700 956
053 060 217
906 408 025
005 620 000
002 597 641
In the spaces that have not been solved, I have only pairs, yet I still can't figure out the next move. Any suggestions? I will be busy over the next couple of days with Thanksgiving so if I seem rude by not responding it's just that I have to cook for 22 people! This is the link to the puzzle if you want to print it out. It's the Nov. 23 puzzle. Thanks!
http://puzzles.usatoday.com/sudoku/ |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Thu Nov 24, 2005 8:01 am Post subject: |
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Hi,
here is a possible solution:
8 in r3c7 - Sole Candidate
5 in r3c1 3 in r3c6 - Sole Candidate
5 in r1c8 3 in r2c3 2 in r3c4 8 in r4c9 9 in r4c2 7 in r7c7 6 in r9c8 - Unique Horizontal
4 in r1c2 6 in r1c6 6 in r2c7 4 in r3c8 4 in r4c5 4 in r6c9 4 in r8c3 2 in r9c9 - Unique Horizontal
9 in r3c3 3 in r4c7 2 in r7c3 4 in r7c6 6 in r8c1 - Unique Horizontal
6 in r6c3 8 in r8c2 - Unique Horizontal
1 in r6c1 9 in r1c5 1 in r8c7 9 in r2c8 1 in r2c9 - Unique Vertical
2 in r4c1 8 in r9c5 8 in r2c6 1 in r5c8 - Unique Vertical
8 in r1c1 7 in r5c6 2 in r6c8 - Unique Vertical
7 in r2c1 7 in r6c2 7 in r1c4 2 in r5c5 5 in r9c6 7 in r4c8 - Unique Vertical
3 in r5c2 5 in r7c2 3 in r9c4 5 in r5c4 3 in r6c5 5 in r8c9 - Unique Vertical
1 in r9c2 1 in r7c4 9 in r7c9 - Unique Vertical
9 in r8c4 - Unique Vertical
see u,
P.S. and I almost forgot:
| Code: | Final SuDoku Table
8 4 1 7 9 6 2 5 3
7 2 3 4 5 8 6 9 1
5 6 9 2 1 3 8 4 7
2 9 5 6 4 1 3 7 8
4 3 8 5 2 7 9 1 6
1 7 6 8 3 9 5 2 4
3 5 2 1 6 4 7 8 9
6 8 4 9 7 2 1 3 5
9 1 7 3 8 5 4 6 2 |
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sicnic
Joined: 24 Nov 2005
Posts: 3
Location: Somwhere in Massachusetts
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| Posted: Thu Nov 24, 2005 11:53 pm Post subject: |
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Try this!
5 4 1 7 9 8 2 6 3
7 2 3 4 5 6 8 9 1
8 6 9 2 1 3 4 5 7
2 9 5 6 4 1 3 7 8
4 3 8 5 2 7 9 1 6
1 7 6 8 3 9 5 2 4
3 5 2 1 6 4 7 8 9
6 8 4 9 7 2 1 3 5
9 1 7 2 8 5 6 4 2 |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Nov 25, 2005 12:27 am Post subject: That's not a solution. |
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| sicnic wrote: | | Code: | 5 4 1 7 9 8 2 6 3
7 2 3 4 5 6 8 9 1
8 6 9 2 1 3 4 5 7
2 9 5 6 4 1 3 7 8
4 3 8 5 2 7 9 1 6
1 7 6 8 3 9 5 2 4
3 5 2 1 6 4 7 8 9
6 8 4 9 7 2 1 3 5
9 1 7 2 8 5 6 4 2 |
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This is invalid. There are two "2"s in column 4, and there are two "2"s in row 9.
On top of that, this isn't even the puzzle Louise asked about, anyway. dcb |
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sicnic
Joined: 24 Nov 2005
Posts: 3
Location: Somwhere in Massachusetts
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| Posted: Fri Nov 25, 2005 1:02 am Post subject: |
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Sorry, that 2 in column 4 should be a "3'" and what do u mean that that's not the one she was talking about?
See below
5 4 1 7 9 8 2 6 3
7 2 3 4 5 6 8 9 1
8 6 9 2 1 3 4 5 7
2 9 5 6 4 1 3 7 8
4 3 8 5 2 7 9 1 6
1 7 6 8 3 9 5 2 4
3 5 2 1 6 4 7 8 9
6 8 4 9 7 2 1 3 5
9 1 7 3 8 5 6 4 2 |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Nov 25, 2005 2:05 am Post subject: This is the puzzle Louise asked about |
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| sicnic wrote: | | Sorry, that 2 in column 4 should be a "3'" and what do u mean that that's not the one she was talking about? |
Please refer to the first message in this thread. Here's the puzzle Louise was asking about:
| Code: | 120 340 500
000 000 000
560 200 003
007 050 800
201 000 906
003 060 200
900 008 025
000 000 000
002 097 041 |
As you can see, the solution you've presented doesn't answer the question Louise was asking. dcb 
Last edited by David Bryant on Fri Nov 25, 2005 2:15 am; edited 1 time in total |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Nov 25, 2005 2:13 am Post subject: That is a tough one, Louise! |
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Hi, Louise!
I hope you had a good Thanksgiving with those 22 people. That's a big enough crowd to polish off the whole turkey!
Here's one way to crack your puzzle. I had to resort to "forcing chains".
In r5c5, the only possibilities are {3, 8}. You can rule out the possibility "3" as follows:
r5c5 = 3 ==> r6c4 = 8 ==> r6c1 = 4 ==> r8c1 = 7 ==> r7c2 = 1 ==> r7c5 = 3
But this is a contradiction; therefore r5c5 = 8, and the rest of the puzzle is fairly simple. dcb |
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Sun Nov 27, 2005 6:50 pm Post subject: Re: That is a tough one, Louise! |
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| Code: |
> Here's one way to crack your puzzle. I had to resort to "forcing chains".
> In r5c5, the only possibilities are {3, 8}. You can rule out the
> possibility "3" as follows:
> r5c5 = 3 ==> r6c4 = 8 ==> r6c1 = 4 ==> r8c1 = 7 ==> r7c2 = 1
> ==> r7c5 = 3
> But this is a contradiction; therefore r5c5 = 8, and the rest of the
> puzzle is fairly simple.
Whilst this is excellent work, I still do not like the "reductio adabsurdum"
usage (ie try a value and find a contradiction).
However there IS a method using forcing chains that gives the answer
without relying upon 'contradictions'. The above solution is what gave
me the hint to find it.
1) The nature of the remaining non-resolved cells suggests strongly
that a forcing chain should be possible.
2) From previous work it appears that a pair of forcing chains CAN lead
to a SINGLE value for a cell for EACH possible value of another cell.
The challenge is to find where this occurs.
3) The previous work suggests that the 'final' link in each chain must
be to a cell containing XY where the linkages are from different cells
with one link being "equals X" with the other "Not-Y"
4) In theory it should then be possible to work back on possible chains
so as to find a point of intersection of the chains and that cell then
becomes the "start point" of two chains (one for each possible value)
that lead to a single value of the end cell.
5) Here we have been given the start cell (r5c5) which has values (3,8)
My endeavour was to find the 'end cell" - and then to explore the
chains in order to complete the demonstration.
6) Column 1 is very interesting. (37),(48),(47),(38).
Every one of those cells has two binary links for each of its candidates.
Thus for example using (38) as the target - it may be linked from
37 as 'equals-3' or from (48) as 'not-8' both resulting in '3'.
Immediately we have the possibility of two chains giving the same
result in the target cell. All this is necessary is to trace back the
chains to a common cell such that each chain has a different value
in that common cell.
7) Here we can do that from the (47) cell in r8c1:
r8c1=4
r2c1=7 (not-7)
r2c2=3 (not-3)
r7c2=7 (not-7)
Remember this is looking at the chain backwards and so the link
will be from r7c7 to r2c2. This is a one-way link as the presence
of 7 in r7c2 implies 'not-7' for TWO other cells and proving just
one of them to be 'not-7' would NOT prove either to be '7'.
r7c5=1 (not-1)
r5c5=3 (not-3)
The rules in parenthesis are the FORWARD rules
(ie read-up the page!)
r8c1=4
r6c1=8 (equals 4)
r6c6=4 (not-4)
r5c6=3 (equals 4)
r5c5=8 (equals 3)
Again reading up the page gives the forward chain.
8) Thus it is demonstrated that whether r5c5=3 or r5c5=8 the value
in r8c1 MUST be '4'.
I regard this is as being POSITIVE logic in that a unique value has
been set for a cell (r8c1) rather than setting r5c5 on the basis of
a contradiction down the line.
9) Once r8c1 has been set the rest falls like the proverbial stack
of dominos.
10) Using data from the 'contradiction' chains we would have
r5c5=3, r6c4=8, r6c1=4, r8c1=7
which does indeed contradict the chain above which leads to
value '4' in r8c1 but this chain is consistent with r5c5=8 leading
to r8c1=4. Thus the demonstration of a single value for a
target cell WITHOUT using contradiction CANNOT rely upon
the same chain as is used for the contradiction. Why not?
Why does my chain (r5c5, r7c5, r7c2, r2c1,r8c1) lead to a different
result for r8c1 than (r5c5, r6c4, r6c1,r8c1)?
If we understand this, we may have some clues as to how to
select the routes for the forcing chains!!
++++++
The question then is how to spot such things.
It would be useful to know how r5c5 came to be selected as the start
point for the "contradiction" exercise as this might provide some clues
as to how the chain start/end points can be identified.
What can be said about the "end point" is that the cell must have
binary links to TWO other cells for at least one (but preferably two)
of its candidate values.
In the case above, it was possible to demonstate two chains coming
to the same end point with value 4 but one would need also to check
chains ending with 7 in r8c1.
Here the trail back goes via r6c1 (not-8) or r2c1 (equal-7)
The former has three routes back to r5c5 - all ending with value 3 - but
the important thing is the route via r2c1 which then goes back to r2c2
- which must have value 7. This is NO binary chain resulting in a forced
value 7 for that cell (as 7 occurs THREE times in column 2) and so only
the (not-3) link is valid. This goes back to r9c2 (value 3) and r9c1
(value 8) before reaching r6c1 value 4 - which is where the other chain
reached after just one backward step. As the the value of r6c1 is the
same for each chain then r8c1=7 does not comply with the rules for
finding the two chains with the attributes stated.
++
The essence of using this technique is to identify the start/end points
such that both X and Y in the start point lead to the same value Z in
the end point.
Identification of potential Z points is relatively easy. My method was to
highlight all the cells two binary links for at least one candidate. In this
case there were 19 such cells but several of these were within triplets
and triplets are linked by the fact that resolving one solves them all
so that the binary links could not be independent. Inspection revealed
that column 1 is "perfect" for meeting the conditions and that col2 may
be worth investigating - although the 147 in r8c2 means that some links
will be one-way only. As only ONE cell obeying the rules is needed, I
took the easiest choice of using col1.
Then there was a choice of four cells and a lot of work on plotting
chains for each one. At this point I cheated and started from r5c5
to find where it led in col1 - cheating because I had read the earlier
posting referreing to that cell!
Thus I am very keen to know why r5c5 was selected as the start point
for the 'find a contradiction' exercise. It seems definitely to have a
relevance also for at least one 'forcing chains' resolution.
+++
Applying this to the general scenario
Is it better to find a potential end-point (as outlined above) or is there
a way to identify a good start-point (as with the r5c5)?
Until we have some discernment on that point, we are left with a
technique that involves a horrendous amount of work (mostly with no
productive result?) in order to unearth the really valuable relationship.
Alan Rayner BS23 2QT
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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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| Posted: Tue Nov 29, 2005 7:39 pm Post subject: |
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Thanks David, Alan, Someone, and sicnic for your help! That was an interesting puzzle. Good explanation David.
"Until we have some discernment on that point, we are left with a
technique that involves a horrendous amount of work (mostly with no
productive result?) in order to unearth the really valuable relationship.
Alan Rayner BS23 2QT"
Alan, are you talking about sudoku or marriage?
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Wed Nov 30, 2005 2:36 am Post subject: |
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| Code: |
> "Until we have some discernment on that point, we are left with a
> technique that involves a horrendous amount of work (mostly with no
> productive result?) in order to unearth the really valuable relationship. "
> Are you talking about sudoku or marriage?
You may well ask that question, I could not possibly comment!!
Well perhaps I could say that the preparation of enormous amounts of
food for Thanksgiving could hardly be said to be without productive
result. Additionally many marriages produce(!) children, do they not??
However I agree about the horrendous amount of work!
PS: I was married in late 1991 but psychological baggage brought
in from a previous relationship led to a parting in early 1992 (she
could not bear to be in my company because the comparisons with
her previously ingrained concepts of marriage were incompatible
with my views - and her aspirations - concerning the autonomy of
the partners in marriage. She expected to be controlled and I
declined to control her.). Thus I have very limited experience on
which to draw when considering the perils and rewards of marriage.
Alan Rayner BS23 2QT
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