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Posted: Thu Nov 17, 2005 11:32 am Post subject: November 16 - V.Hard |
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For the benefit of those interested in the Mandatory Pairs method, I
can say that this one was solvable without resorting to a full scan
of candidate profiles. However, I did need to record the "missing"
profile against most of the rows and columns in order to limit the
amount of mental processing of data.
This was a tough one - and took some time to solve - but was all the
more satisfying for that. I find it a challenge to solve the hard and very
hard puzzles without resort to candidate profiles and the "advanced"
techniques that rely on spotting patterns from those profiles - although
it must be admitted that not every puzzle is solvable in this way. Using just Mandatory Pairs plus the "missing" profile for some of the lines can
provide sufficient "visible" information to allow the mind to concentrate
on the logic rather than having to hold onto all the "preliminary" logic
that would result in the information recorded by the mandatory pairs.
Alan Rayner BS23 2QT
(one of the warmer parts of Britain)
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PoppaPoppa
Joined: 06 Nov 2005 Posts: 21 Location: Arkansas USA
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Posted: Fri Nov 18, 2005 3:35 am Post subject: Nov 16 - very hard indeed |
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Well Alan I attempted the use of mandatory pairs and still got stuck on Nov 16
Here is where I am
456830900
039400800
008009643
060008400
500000008
841900026
082600000
304285760
605090280
I used the draw feature and asked for a hint; this resulted in a 2 at 5,2 but I cannot see why the 2 is required here; put another way why is the 9 excuded? (the 2-9 is a mandatory pair at 4,1 and 5,2 ) Help! |
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geoff h
Joined: 07 Aug 2005 Posts: 58 Location: Sydney
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Posted: Fri Nov 18, 2005 5:30 am Post subject: |
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Hi there,
You've done well so far. What you need to do now is consider Row 5. There is a triplet of 1,3,7 in this row - cells r5c3, r5c4 and r5c7. Therefore you can delete all other 1s, 3s and 7s from all other cells in Row 5. This then leaves Nr 9 as the only possibility for cell r5c8, which immediately means Nr 2 is the only possibility for r5c2 as given in the hint.
Those triplets and quadruplets are always very, very useful!!
Cheers. |
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guest Guest
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Posted: Fri Nov 18, 2005 10:28 am Post subject: Nov 16th. |
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I do not see how you can have 137 triplets in Cells R5C3 and R5C7. |
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guest Guest
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Posted: Fri Nov 18, 2005 10:44 am Post subject: Nov. 16th |
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Sorry. I was looking at an incomplete grid in which the 4 at R8C3 had not been entered. My fault. |
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newToSudoku Guest
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Posted: Mon Nov 21, 2005 5:02 am Post subject: |
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Hi,
I am new to Sudoku.
Can you explain how 3 goes into r8c1.
Any help is appreciated |
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lolumens
Joined: 25 Aug 2005 Posts: 5
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Posted: Thu Nov 24, 2005 1:07 pm Post subject: Nov. 16 |
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I struggled with this one too. My real problem was neglecting to recognize 9 as a possibility in one of the cells.
See if any of this helps.
3 is not a possibility for r7c6 or r7c1 because quadrant 9 needs r7c7 or r7c8 to be a 3.
Either r4c3 or r5c3 must be a 7 for column 3 so you can eliminate 7 as other possibilities in cells of quadrant 4.
That leaves you with 4 in r6c2.
Eliminate 4 as possibilities in the cells of quadrant 4 and you have naked pair of 3 and 7 in r3c3 and r4c3.
Now you can eliminate 3 as a possibility for r8c3 leaving 4.
This leaves you with 3 in r8c1 as the only possibility for a 3 in quadrant 7.
This should open up the puzzle for you and the only other tough one I found after this was a hidden pair in row 5. |
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zaks
Joined: 25 Nov 2005 Posts: 13
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Posted: Fri Nov 25, 2005 9:54 pm Post subject: Re: November 16 - V.Hard |
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Solution of Nov16 (hard) sudoku:
1. 8g8 * (* means that move can be done from the beginning)
2. 6i4 * 3. 6a1 * 4. 6c9 Now cells b4,f4 make pair 4,7, hence next 3 moves:
5. 5e4 6. 3g4 7. 8a4 8. 8c7 Now cells a7,b7,e7 make triple 1,2,7.
9. 5d7 10. 3i7 11. 3h3 12. 1g5 13. 5g5 14. 2g1 15. 4a9
16. 8d8 17. 8e2 18. 3d2 19. 3a2 20. 4c2 21. 9h5
22. 3c6 23. 1d6 24. 2e6 25. 9a6
26. 7c5 27. 3d5 28. 7d1 29. 1b1 30. 9b2
31. 7a3 32. 7b7 33. 2b5 34. 4b4 35. 4c1
36. 3f1 37. 9i3 38. 1i2 39. 7f4 40. 1e7
41. 2a7 42. 1a8 43. 1f3 44. 4e3 45. 4f5
46. 6e5 47. 6f8 48. 2f9 49. 7e8 50. 2i8
51. 5h8 52. 5i6 53. 7h6 54. 1h9 55. 7i9 x!
Hope i made not too many misprints.
Enjoy, zaks
PS Is there any solver-program which presents also moves not only final position?! |
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haggisking Guest
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Posted: Sat Nov 26, 2005 5:54 pm Post subject: Homebrew solver with log |
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My program which I coded myself logs details on how it solved the puzzle. Here is what it made of Nov 16 based on the grid provided by PoppaPoppa.
Before you read the log here is how it is formatted. The first number is the step number. The second field is the action taken. The third field is the date and time. The fourth field and beyond is what you want. The fourth field is the Column, which is 0 based (i.e the first col = 0 last col = . The fifth field is the row, again 0 based. Finally we have the number inserted into that square.
For example step 11 we inserted value 7 in r4c2.
The program makes what I call logical leaps. This is where there is more than one possible move forward. When we do this you will see the text "Logical Leap Required <Time> Forward". If it turns out that the move was a mistake, you will see later on that we make a "Logical Leap Required. Error in puzzle detected <Time> Backward"
The file is tab seperated so you could load it into excel for easy viewing.
I hope this helps.
Beginning Puzzle Solving
1 Computing options for all squares 26/11/2005 17:43:08
Checking Status of puzzle
2 Logical Leap Required 26/11/2005 17:43:08 Forward
3 Candidate for logical leap located 26/11/2005 17:43:08 7 0 1
4 Computing options for all squares 26/11/2005 17:43:08
Checking Status of puzzle
5 Logical Leap Required 26/11/2005 17:43:08 Forward
6 Candidate for logical leap located 26/11/2005 17:43:08 5 0 2
7 Computing options for all squares 26/11/2005 17:43:08
Checking Status of puzzle
8 Logical Leap Required 26/11/2005 17:43:08 Forward
9 Candidate for logical leap located 26/11/2005 17:43:08 2 3 3
10 Computing options for all squares 26/11/2005 17:43:08
11 Found square with one option 26/11/2005 17:43:08 2 4 7
Checking Status of puzzle
12 Computing options for all squares 26/11/2005 17:43:08
Checking Status of puzzle
13 Logical Leap Required 26/11/2005 17:43:08 Forward
14 Candidate for logical leap located 26/11/2005 17:43:08 0 1 1
15 Computing options for all squares 26/11/2005 17:43:08
Checking Status of puzzle
16 Logical Leap Required 26/11/2005 17:43:08 Forward
17 Candidate for logical leap located 26/11/2005 17:43:08 4 1 6
18 Computing options for all squares 26/11/2005 17:43:08
19 Found square with one option 26/11/2005 17:43:08 5 1 7
Checking Status of puzzle
20 Computing options for all squares 26/11/2005 17:43:08
21 Found square with one option 26/11/2005 17:43:08 5 5 3
Checking Status of puzzle
22 Computing options for all squares 26/11/2005 17:43:08
23 Found square with one option 26/11/2005 17:43:08 3 4 1
24 Found square with one option 26/11/2005 17:43:08 6 5 5
25 Found square with one option 26/11/2005 17:43:08 4 6 1
Checking Status of puzzle
26 Logical Leap Required. Error in puzzle detected. 26/11/2005 17:43:08 Backward
27 Moving back to status at previous logical leap 26/11/2005 17:43:08
28 New forward move located begining leap forward 26/11/2005 17:43:08 4 1 7
29 Computing options for all squares 26/11/2005 17:43:08
Checking Status of puzzle
30 Logical Leap Required 26/11/2005 17:43:08 Forward
31 Candidate for logical leap located 26/11/2005 17:43:08 5 1 6
32 Computing options for all squares 26/11/2005 17:43:08
33 Found square with one option 26/11/2005 17:43:08 4 5 5
Checking Status of puzzle
34 Computing options for all squares 26/11/2005 17:43:08
35 Found square with one option 26/11/2005 17:43:08 4 2 1
36 Found square with one option 26/11/2005 17:43:08 3 3 1
37 Found square with one option 26/11/2005 17:43:08 6 5 3
Checking Status of puzzle
38 Computing options for all squares 26/11/2005 17:43:08
39 Found square with one option 26/11/2005 17:43:08 1 8 1
Checking Status of puzzle
40 Computing options for all squares 26/11/2005 17:43:08
41 Found square with one option 26/11/2005 17:43:08 8 7 1
Checking Status of puzzle
42 Computing options for all squares 26/11/2005 17:43:08
Checking Status of puzzle
Puzzle Solved |
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alanr555
Joined: 01 Aug 2005 Posts: 198 Location: Bideford Devon EX39
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Posted: Mon Nov 28, 2005 11:19 am Post subject: Re: Nov 16 - very hard indeed |
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Code: |
> I attempted the use of mandatory pairs and still got stuck on Nov 16.
> Here is where I am
456830900
039400800
008009643
060008400
500000008
841900026
082600000
304285760
605090280
> I used the draw feature and asked for a hint; this resulted in a 2 at
> 5,2 but I cannot see why the 2 is required here; put another way why
> is the 9 excuded? (the 2-9 is a mandatory pair at 4,1 and 5,2 ) Help!
Thank you for the interest in Mandatory Pairs.
This puzzle seems to be easier than the V.Hard of Nov 23rd and should
not need resort to candidate profiles if M/Ps are used fully.
Although M/P exclusion is a common technique, it is not the only one.
Very often one gets an 'external' reason to make a value impossible.
Here the external factor is that r5c2 must be 2 rather than that
r5c2 cannot be 9 or r4c1 cannot be 2.
M/Ps can hold a lot of information about the position in a region and
SOME about the lines (rows/cols) in which they are situated BUT any
puzzle above the simple level MUST rely on information from sources
external to the region to be completely resolved. The M/P methodology
allows one to look at those 'external' factors without losing sight of
what has already been discovered inside the region.
+++
From the situation quoted above. The following subscripts should be
present on the grid. It is important to check that all pairs have been
identified as far as possible as else one could omit the critical one
which leads to the solution. This is a case in point.
2 in r1c9 and r2c9
6 in r2c5 and r2c6
5 in r2c8 and r2c9
5 in r3c4 and r3c5
9 in r4c1 and r5c2
2 in r4c1 and r5c2
3 in r4c3 and r5c3
7 in r4c3 and r5c3
4 in r5c5 and r5c6
6 in r5c5 and r5c6
7 in r6c5 and r6c6
7 in r7c1 and r9c2
9 in r7c1 and r8c2
3 in r7c7 and r7c8
4 in r7c9 and r9c9
3 in r9c4 and r9c6
I have never used the hint facility but in this case I agree with it that
the next step is r5c2=2 as quoted. Why?
The answer is by inspection of row 5.
The unallocated cells are r5c2, r5c3, r5c4, r5c5,r5c6,r5c7,r5c8
but they hold a lot of hidden information.
Try each cell for '9' and the row does not resolve
but try the row for '2' and the answer emerges.
Columns 3,4,7,8 already contain a 2.
This leaves r5c2, r5c5, r5c6 as possible places for the '2' in the row.
Now the power of M/P comes into play!
r5c5 and r5c6 form a "Mutual Reception" with values 4 and 6.
Thus they cannot be the place for the 2.
Cell r5c2 becomes the sole location for the '2' - after which the
puzzle resolves easily using Mandatory Pairs.
Why look at row 5 in particular?
I chose to take a closer look after deriving the 'missing' profile and
noting that there was a mutual reception present. This pattern can give
M/Ps an advatage as the computer solver regards the identification of
the hidden pair as a more advanced technique whereas M/Ps uses it
as part of its standard repertoire. Thus it was a 'hunch' to look there -
but one that paid off on this occasion!
The resolution depended upon having the 46 mutual reception recorded
and emphasises the point that one never knows which M/P will prove
useful - or when it will prove so. The technique is like orientation in
new territory - one leaves marks to record that one has been there
before and gradually the accumulation of such marks aids one's
picture of the whole territory without having to carry all the detail
in one's head.
+++
Apologies for the delay in responding to this one. It sometimes takes
me a few days to get a puzzle into my filing system and I work on
the puzzles in a different location to my computer equipment.
This one is a vindication that Mandatory pairs can, sometimes, resolve
even a puzzle graded as 'very hard' - but not always as evidenced by
Nov 23rd where I needed to look at the candidate profiles.
Alan Rayner BS23 2QT
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