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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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 Posted: Sun Dec 04, 2005 1:05 pm Post subject: December 4th - V.Hard |
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This puzzle reveals the handicap with using Mandatory Pairs.
The method does not highlight relationships "at a distance".
I needed to move to Candidate Profiles with this one - and
that revealed the critical relationship.
r1c2 and r7c2 form a pair which can be used to eliminate
possible values from other cells in column 2 - leading to
the resolution of one cell in that column which, in turn, then
reveals a (mandatory) pair in columns 7 and 9.
I am sure that it would be POSSIBLE to discern the relationship
in column 2 without proceeding to list out the candidates but
one would have to notice that each of two digits has the same
two cells as their only possible locations in that column. In my
view the human mind is unlikely to spot that relationship but
I am willing to be persuaded otherwise if I have missed something
that is demonstrable very simply.
+++
The moral of all this is that where the grading program finds "pairs"
it clocks up points towards being hard or very hard BUT it does not
distinguish between pairs which are within a region and those
which 'straddle' regions. The M/P technique will, usually (and with a
competent human solver), have found the pairs within the regions
and so the 'hardness' does not exist BUT if the pairs are of cells
in different regions the M/P will not find them (or at least will not
record them!).
One correspondent suggested that M/P should be renamed as
"Mandatory Box-wise Pairs" and there is certainly some merit
in that title (apart from its length!).
I have never claimed that M/Ps is a solution to all problems. Part
of the challenge is to know WHEN to move on to "Missing Profiles"
and then "Candidate Profiles". With this puzzle, I made the move
and feel justified in doing so. There have been times with other
puzzles when I have made such a move and then found that I
need not have done as I had just failed to exhaust the potential
inherent in the M/P methodology.
I managed to resolve 13 cells before resorting to compiling the
candidate profiles. Otheres may be able to do better.
Alan Rayner BS23 2QT
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sun Dec 04, 2005 2:16 pm Post subject: |
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Hi Alan,
Probable after finding the 13 digits, if you "see" the Hidden Pair of digits
2 and 4 in the cells r2c2 and r5c2 (in column), you can eliminate digit 9 from r2c2 and from r5c2 and this will lead you to:
9 in r4c2 - as Sole Candidate.
From here one ... it could be easy(er).
see u, |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sun Dec 04, 2005 4:32 pm Post subject: "Seeing" hidden pairs |
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| AlanR555 wrote: | | I am sure that it would be POSSIBLE to discern the relationship in column 2 without proceeding to list out the candidates but one would have to notice that each of two digits has the same two cells as their only possible locations in that column. In my view the human mind is unlikely to spot that relationship but I am willing to be persuaded otherwise if I have missed something that is demonstrable very simply. |
I did spot the "hidden pair" {2, 4} fairly early on. Here's how I saw it.
In the initial setup, there are 5 empty cells in column 2. I immediately noticed the {2, 4} pair in row 4, and in row 7. I also saw a "2" in row 1. "Aha!" I thought -- "if I only had a '4' in row 1, I could enter the pair {2, 4} at r2c2 & r2c4."
And then I noticed that I had to place a "4" at r1c6, because that's the only way it can fit in the top center 3x3 box. dcb
PS I always look for hidden pairs in Samgj's puzzles, because he seems to be quite fond of them. |
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bmassey
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| Posted: Sun Dec 04, 2005 5:11 pm Post subject: New Sudokukoo is missing something |
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I have yet to solve a very hard puzzle like Dec 4. I'm missing some tools.
I used the Draw tool to get me moving, and I don't understand how the computer picked the hint.
Why is the "9" in r7c6 an obvious choice when there could be 9's in any other space of c6?
I've profiled every square with superscripts looking for patterns.
Hope you can help.
Best,
Brian Massey
Austin, TX USA |
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zaks
Joined: 25 Nov 2005
Posts: 13
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| Posted: Sun Dec 04, 2005 6:05 pm Post subject: Re: December 4th - V.Hard |
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| alanr555 wrote: | | Code: | I managed to resolve 13 cells before resorting to compiling the
candidate profiles. Otheres may be able to do better.
Alan Rayner BS23 2QT
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Alan,
i could do only 8 after which the position is
000 804 012
000 051 380
081 326 000
100 240 070
000 030 000
070 065 003
000 410 020
059 680 000
810 508 000
then i had to use candidates, but how do you find another 5 cells, and what is your full solution?
thanks, zaks |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sun Dec 04, 2005 6:42 pm Post subject: |
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Hi,
You have to get a couple of steps back, or restart from the beginning because you have: Nr=8 in r9c1 and r9c6 !
see u, |
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Guest
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| Posted: Sun Dec 04, 2005 6:53 pm Post subject: |
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| someone_somewhere wrote: | Hi,
You have to get a couple of steps back, or restart from the beginning because you have: Nr=8 in r9c1 and r9c6 !
see u, |
sorry, the position'd be of course as this:
000 804 012
000 051 380
081 326 000
100 240 070
000 030 000
070 065 003
000 410 020
059 680 000
810 500 000
zaks |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sun Dec 04, 2005 8:41 pm Post subject: zaks' question |
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You need to concentrate on the bottom 1/3 of the puzzle, and look at the "7"s. Your next move is r9c5 = 7 -- here's why.
-- Because of the {2, 4] pair in row 7, you can see that the pair {2, 4} must occupy the two cells r8c1 & r9c3 in the bottom left 3x3 box. This implies a "hidden triplet" {3, 6, 7} lying in r7c1, r7c2, & r7c3.
-- You already know that the candidates for r9c5 are {7, 9} -- that's obvious from the numbers you've already filled in for column 5. Now, because the "3" in row 7 must lie in the bottom left 3x3 box, you can infer that there's a hidden pair {2, 3} lying in r8c6 & r9c6. So the pair {7, 9} must lie in r7c6 & r9c5, in the bottom center 3x3 box.
-- But r7c6 can't be a "7" because of the hidden triplet -- the "7" in row 7 has to go in r7c1, r7c2, or r7c3. So the "7" in the bottom center 3x3 box has to appear at r9c5. Then r7c6 = 9, and r1c5 = 9, and you ought to be in pretty good shape from there. dcb |
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Mon Dec 05, 2005 1:23 am Post subject: Re: New Sudokukoo is missing something |
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> Why is the "9" in r7c6 an obvious choice when there could be 9's
> in any other space of c6?
Given where you are, I do not know (but see below!)
One of the attributes of Sudoku is that almost all the time the
SEQUENCE of emplacements will differ between solvers. To
that extent there is NOT a SINGLE solution - although the FINAL
set 0f 81 placements will (on this site!!) be unique.
The '9' in r7c6 was one that I placed before resorting to the
candidate profiles BUT I had more on the grid than is shewn
in the question.
++
When looking at region 8 (centre bottom region) there are three
initial values and 1 in r7c5 plus 5 in r9c4 have been derived.
I had also '7' in r9c5 and "reservations" on r8c6 and r9c6.
a) In region 7 the '4' must be in r8c1 or r9c3
b) In region 7 the '2' must be in r8c1 or r9c3
Those two apply because row 7 contains them and that leaves only
two cells remaining in the region.
c) r8c1 and r9c3 having a 'pair' exclude any other values and so
in region 7 both row 8 and row 9 are complete. Thus the missing
values (3,6,7) must be in row 7.
d) r3c5 must be '2' (row 1 and col 4 precluded)
e) Thus the '2' in region 8 must be in col 6
f) Further it must be in row 8 or row 9.
g) Because of (c) above the '3' in region 8 must be in row 8 or 9.
h) It cannot be in r9c4 (already set) and so must be in r8c6 or r9c6
i) Thus r8c6 and r9c6 form a pair and cannot accept any other digit.
j) The '7' in region 8 cannot be in row 7 (as per 'c' above) and the
only remaining cell for the '7' is r9c5.
k) Thus r7c6 must be 9 - as this is the only unallocated digit for
the whole region (given that 2 and 3 are allocated to r8c6 and
r9c6 - even though we do not yet know in which order).
+++
This may seem tortured logic. I did not keep notes on the sequence
of my placements and so I have had to try and reconstruct what I
was doing in the early hours of the morning!
The reconstruction has taken much longer than the original working!!!
Alan Rayner BS23 2QT
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Mon Dec 05, 2005 1:41 am Post subject: Re: "Seeing" hidden pairs |
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> I did spot the "hidden pair" {2, 4} fairly early on. Here's how I saw it.
I did NOT spot it. In fact I found the OTHER pair in col 2 first!
> In the initial setup, there are 5 empty cells in column 2. I immediately
> noticed the {2, 4} pair in row 4, and in row 7. I also saw a "2" in row
> 1. "Aha!" I thought -- "if I only had a '4' in row 1, I could enter the pair
> {2, 4} at r2c2 & r2c4."
This is a mark of a "higher" level of command of Sudoku.
It involves seeing overall relationships in 'apparently' disparate
parts of the puzzle.
a) Noticing the (24) pairs in rows 4 and 7
b) Noticing the potential for another (24) pair in row 1
c) Noticing that these three rows correspond with blanks in column 2
d) Knowing that such a pattern could assist with the remaining
blanks in col 2.
e) Finding the 4 in r1c6 (well THAT bit is easy!!!)
f) Placing the (24) pair as a result.
I certainly had no inkling of a,b,c,or d.
It is situations like this that tend to confirm that there is a particular
mindset suited to Sudoku solutions. Perhaps it also applies to some
computer programmers and could explain why I never progressed
from being a systems analyst who could dabble with programming
into being a "real" programmer. In latter days, I found myself
unable even to 'read' the code - let alone write it - and so the days
of my usefulness were definitely numbered.
+++
> PS I always look for hidden pairs in Samgj's puzzles, because
> he seems to be quite fond of them.
Possibly that is why the Mandatory Pairs method often does quite
well with his puzzles - except when the pairs are widely spaced!
Alan Rayner BS23 2QT
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Mon Dec 05, 2005 2:01 am Post subject: Re: December 4th - V.Hard |
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| Code: |
> I managed to resolve 13 cells before resorting to compiling the
> candidate profiles. Others may be able to do better.
> I could do only 8 after which the position is
000 804 012
000 051 380
081 326 000
100 240 070
000 030 000
070 065 003
000 410 020
059 680 000
810 500 000
> then i had to use candidates, but how do you find another 5
> cells, and what is your full solution?
+++
It is not my practice to give full solutions. They are boring unless
there is a definite educational point. Usually the latter can be
achieved by concentrating on the 'crunch' point(s).
However my thirteen cells were
r1c4, r1c5, r1c6, r2c4, r3c3, r3c4, r3c5,
r4c6, r5c6, r7c5, r7c6, r9c4, r9c5.
A separate post under this topic describes how some of the
other derivations were made.
There were also some important 'mutual receptions' which aided
the compilation of the candidate profiles.
Col 4 - (19) pair
Col 6 - (23) pair
Row 7 - (58) pair
Row 8 - (17) pair
+++
Subsequently I have seen the posts by others drawing attention to
the (24) pair in column 2 and I have commented on that separately.
Resolving the '9' in r4c2 as a result of finding the (24) pair in col2
would certainly have unlocked a lot (perhaps all) of the puzzle
and allowed the M/P method to be continued.
However, how many of us would have spotted the (24) pair in the
way that was described in that other post?
Alan Rayner BS23 2QT
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Guest
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| Posted: Mon Dec 05, 2005 7:35 am Post subject: |
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| holy cow, david.. thanks a bunch. I was seriously stuck until I read your hint. |
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PoppaPoppa
Joined: 06 Nov 2005
Posts: 21
Location: Arkansas USA
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| Posted: Tue Dec 06, 2005 11:20 pm Post subject: Re "missing something" from bmassey (Dec 4 puzzle) |
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| Alan's explanation was correct but I think a shorter version is this: the mandatory pair of (2,4) forces the 3, 6 and 7 into row 7 of region 7. Therefore, there CAN BE NO 3 or 7 in row 7 of region 8. If you look at all possible entries for the cell at r7c6, the only remaining candidate is 9. |
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Guest
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| Posted: Wed Dec 07, 2005 1:33 pm Post subject: Re: Re "missing something" from bmassey (Dec 4 puz |
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| PoppaPoppa wrote: | | Alan's explanation was correct but I think a shorter version is this: the mandatory pair of (2,4) forces the 3, 6 and 7 into row 7 of region 7. Therefore, there CAN BE NO 3 or 7 in row 7 of region 8. If you look at all possible entries for the cell at r7c6, the only remaining candidate is 9. |
That's good. Making it shorter should make it more obvious.
To make it maybe not shorter, but even less detailed, use lots of hindsight:
note that the interaction of box 7 and row 7 pivots around the shared triplet r7c1..3.
The rest of box 7 must be the same as the rest of row 7.
Try fitting the rest of the box into the rest of the row.
{1} is already there but {5,8,9} must be accomodated in the still empty cells; i.e. {5,8,9} is the candidate profile for these cells.
Of these values, only {9} will fit in r7c6.
This explanation bypasses the {2,4} and {3,6,7} steps.
Such intermediate spinoff is useful later on, of course.
You get the pair by fitting the rest of the row {4,1,2} into the rest of the box, i.e. going in the other direction.
And having done r7c6 we are left with {5,8}, yet another pair (shared with box 9). |
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franko
Guest
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| Posted: Thu Dec 08, 2005 4:50 am Post subject: Dec 4 very hard |
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| I have followed the discussion/logic to this point, but am still stuck. Does it come down to trial and error beyond this point? |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Thu Dec 08, 2005 7:32 am Post subject: |
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Hi,
Let me help you.
In order to solove it you will need "only" the clasic techniques:
- Sole Candidate in Cell
- Unique digit in Line
- Unique digit in Column
- Unique digit in 3x3 Block
- Row on 3x3 Block interaction
- Hidden Pair in Row and
- Hidden Pair in Column
Now let's get to work.
The first 13 digits are easy to solve:
| Code: | 1 in r3c3 - Unique Horizontal
5 in r9c4 1 in r7c5 4 in r1c6 - Unique Vertical
2 in r3c5 8 in r1c4 - Unique in 3x3 block
3 in r3c4 - Unique in 3x3 block
6 not in r7c7, Hidden Pair 5 8 in r7c7 and r7c9 (in Row)
6 not in r7c9, Hidden Pair 5 8 in r7c7 and r7c9 (in Row)
7 not in r7c7, Hidden Pair 5 8 in r7c7 and r7c9 (in Row)
7 not in r7c9, Hidden Pair 5 8 in r7c7 and r7c9 (in Row)
9 not in r7c7, Hidden Pair 5 8 in r7c7 and r7c9 (in Row)
9 not in r7c9, Hidden Pair 5 8 in r7c7 and r7c9 (in Row)
9 in r7c6 - Unique Horizontal
9 in r1c5 - Unique Vertical
7 in r9c5 7 in r5c6 - Unique Vertical
7 in r2c4 8 in r4c6 - Unique Vertical |
Now we got to the difficult moment, described before.
We could continue like this:
| Code: | 3 not in r8c1, it is in r7c1 or r7c2 or r7c3 (Row on 3x3 Block interaction)
3 not in r9c3, it is in r7c1 or r7c2 or r7c3 (Row on 3x3 Block interaction)
6 not in r9c3, it is in r7c1 or r7c2 or r7c3 (Row on 3x3 Block interaction)
7 not in r8c1, it is in r7c1 or r7c3 (Row on 3x3 Block interaction)
4 not in r8c7, Hidden Pair 1 7 in r8c7 and r8c9 (in Row)
4 not in r8c9, Hidden Pair 1 7 in r8c7 and r8c9 (in Row)
6 not in r2c2, Hidden Pair 2 4 in r2c2 and r5c2 (in Column)
6 not in r5c2, Hidden Pair 2 4 in r2c2 and r5c2 (in Column)
9 not in r2c2, Hidden Pair 2 4 in r2c2 and r5c2 (in Column)
9 not in r5c2, Hidden Pair 2 4 in r2c2 and r5c2 (in Column)
3 not in r4c2, Nacked Pair 3 6 in r1c2 and r7c2 (same Column)
6 not in r4c2, Nacked Pair 3 6 in r1c2 and r7c2 (same Column)
9 in r4c2 - Sole Candidate
3 in r4c3 - Unique Horizontal |
Or you could try the M/P described by Alan (in Thomas Mann style  )
And from here on, the rest is easy:
| Code: | 5 not in r5c7, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
5 not in r5c8, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
5 not in r5c9, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
6 not in r5c7, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
6 not in r5c8, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
6 not in r5c9, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
5 in r3c8 6 in r9c8 - Unique Vertical
3 in r8c8 - Unique Vertical
3 in r9c6 - Unique Vertical
2 in r8c6 - Unique Vertical
2 in r9c3 - Unique in 3x3 block
4 in r8c1 - Unique in 3x3 block
2 in r6c1 - Sole Candidate
4 in r5c2 - Sole Candidate
2 in r2c2 9 in r5c8 8 in r6c3 - Sole Candidate
1 in r5c4 4 in r6c8 - Sole Candidate
8 in r5c9 9 in r6c4 1 in r6c7 - Sole Candidate
2 in r5c7 5 in r7c9 7 in r8c7 - Sole Candidate
6 in r1c7 6 in r4c9 8 in r7c7 1 in r8c9 - Sole Candidate
3 in r1c2 5 in r4c7 - Sole Candidate
6 in r7c2 - Sole Candidate
7 in r7c3 - Sole Candidate
5 in r1c3 3 in r7c1 - Sole Candidate
7 in r1c1 6 in r5c3 - Sole Candidate
4 in r2c3 9 in r3c1 5 in r5c1 - Sole Candidate
6 in r2c1 9 in r2c9 4 in r3c7 - Sole Candidate
7 in r3c9 9 in r9c7 4 in r9c9 - Sole Candidate |
Any other technique that leads you to the result:
| Code: | 7 3 5 8 9 4 6 1 2
6 2 4 7 5 1 3 8 9
9 8 1 3 2 6 4 5 7
1 9 3 2 4 8 5 7 6
5 4 6 1 3 7 2 9 8
2 7 8 9 6 5 1 4 3
3 6 7 4 1 9 8 2 5
4 5 9 6 8 2 7 3 1
8 1 2 5 7 3 9 6 4 |
is also ok.
hope I could help you,
see u, |
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geoff h
Joined: 07 Aug 2005
Posts: 58
Location: Sydney
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| Posted: Thu Dec 08, 2005 7:39 am Post subject: |
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Hi Franko,
No, you certainly don't need trial and error for these puzzles. Consider the following;
1. First look at Column 2. There is a pair of 3,6 in r1c2 and r7c2. Therefore, you can eliminate all other 3s and 6s from Column 2. Therefore, you can easily place Nr 9 in r4c2.
2. This then really opens up the puzzle. You can then delete all other 9s from Row 4 which leaves a pair of 5,6 in r4c7 and r4c9. This then means you can place Nr 3 in r4c3.
You should be right from there.
Cheers, |
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Thu Dec 08, 2005 4:39 pm Post subject: Re: Dec 4 very hard |
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> I have followed the discussion/logic to this point, but
> am still stuck. Does it come down to trial and error
> beyond this point?
NO!
With puzzles set on THIS site, it NEVER comes to that!!!!
Alan Rayner BS23 2QT |
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