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Nenthorn
Joined: 13 Nov 2005
Posts: 11
Location: Scotland
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 Posted: Sun Nov 13, 2005 12:27 pm Post subject: 12nov2005 hint |
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I am absolutely stumped by this one. So far I am at:
391 687 425
854 231 769
762 459 100
937 000 600
206 000 007
005 760 090
023 870 906
570 306 240
600 020 070
The hint of 5 at r7c8 I find inexplicable. Can someone help me? |
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stomer
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| Posted: Sun Nov 13, 2005 12:49 pm Post subject: sudoku 12nov; hint of 5 |
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I am stuck at the same place. I have been trying to find the logic behind this hint but I am not able to. Ther should be a page at the web site that explains the logic..regards |
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dext100
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| Posted: Sun Nov 13, 2005 1:04 pm Post subject: 11/12/05 |
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| I hope someone come to the rescue soon. We are all stuck at the same spot. |
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stomer
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| Posted: Sun Nov 13, 2005 1:18 pm Post subject: Re: 11/12/05 |
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| dext100 wrote: | | I hope someone come to the rescue soon. We are all stuck at the same spot. |
At one point I thought: One possible way to enter the number 5 was to try and eliminate number 1 in the same box. So I satrted to work with the number one, but I could not find anything. Have you tried this? |
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Jimmy Chan
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| Posted: Sun Nov 13, 2005 2:17 pm Post subject: Re: 12 Nov |
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First: r7c8 must either be 1 or 5, the reason is trivial.
Second: both r4c9 and r6c9 cannot be 1 because they must either be 2 or 4.
Third: Therefore in the sixth block, 1 must appear in either r4c8 or r5c8.
Fourth: Therefore 1 can be eliminated from r7c8 and leave the 5 there. |
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Sun Nov 13, 2005 4:14 pm Post subject: |
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For once, a "very hard" puzzle certainly lived up to its name. Often
"hard" or even "medium" have seemed harder.
However, I did manage to solve it using the "Mandatory Pairs" method
coupled with logic and without using candidate profiles or "advanced"
techniques. It did take about an hour to work through everything.
The seemd seldom to be anything that gave a foothold for placing the
'1' digits and, as I recall, it was the '5' digit that eventually provided
the break-through.
One of the virtues of Mandatory Pairs is that very often the method
leads to the resolution of several cells directly from resolving one - but
that one may well not be the one that one is targeting!
Aside from the recording of "Mandatory Pairs" (see below for a guide
to what they are) the principal techniques used were
a) "Counting" to ascertain what was "missing" in a row, column or region.
b) Recording of the "missing" numbers in a row or column by means of
a "digit string" outside the grid - when the number of missing values
was down to five or less.
c) Separation of the "missing" values into subsets where it was possible
to allocate them to particular subgroups of cells. This latter applied
particularly to columns 6 (being 238 in rows 456 and 45 in rows 7
and 9) and 9 (being 138 in rows 389 and 24 in rows 4 and 6).
Within mandatory pairs, the useful techniques were
a) "Mutual reception" (the same two digits in two cells within a region)
disbars any other digit from those two cells - and resolves the partner
of any 'interloper' already there. A very powerful scenario.
b) "Mutual reception" also contributes both digits to the count of "placed"
digits when considering the OTHER missing digits.
(eg a region with 1234 each placed in a cell has five 'empty' cells -
which must hold 56789 in some combination. If two of those cells
each have a '56' marked then one knows immediately that the
other three cells have 789 within them in some order. If later the
'7' is resolved, the '89' immediately become a mutual reception pair
and are likely to lead to toher useful inferences.).
c) Where a mandatory pair lies within a row or column, that row/column
is closed to any other occurrence of that digit. This restricts further
placing of such digit to the OTHER rows (or columns) that interesect
the region in which the pair is situated.
d) If two mandatory pairs (for the same digit) occupy separate regions
but the same two rows (or columns) the occurrence of that digit in
the third region sharing those rows/columns MUST be in the row (or
column) that is NOT occupied by the original pairs.
+++
A "Mandatory Pair" is a mark to indicate that a specific digit MUST
occur in one of only TWO positions within a region and CANNOT appear
in any other cell within that region.
By convention the digit involved is written at the bottom left corner of
the cell. As the solution progresses, these marks will be expunged BUT
they may well be replaced by other digits as logic later requires. At NO
time should the number of occurrences of any digit be other than two
or zero - except in the middle of writing a new pair (physically one has
to be written first!) or removing a pair (ditto for erasures!!).
Example
--8 --- ---
--- --8 ---
--- --- a4c
Clearly either 'a' or 'c' must be 8 and no other cell in that region can
be 8. Therefore the cells 'a' and 'c' are marked with a small '8' in the
bottom left corner to indicate that they are paired in this way.
We do not know (until more information is processed) in which of the
two cells the '8' will occur BUT (the power of the method!) as soon as
it is proved that digit '8' is not in 'a' then it MUST be in 'c' (and vice
versa; if not in 'c' then in 'a')
If we find that 'a' cannot be 8 then the procedure is to erase the '8' in
cell 'a' and then IMMEDIATELY to resolve cell 'c' by writing a big '8'
in it and resolving the subscript '8'. If cell 'c' happened to have say '68'
as its (subscript) pair markings, then there is an immediate route to
resolving digit '6' (erase the '6' subscript from cell 'c' and move to its
partner - the other cell in the same region with subscript '6' - so that
digit '6' can be resolved in the partner's cell).
The key to success is using logic to identify as many pairs as possible
- and then to eliminate them! However, elimination does depend upon
successful finding in the first place.
Taking this puzzle as an example.
First moves:
1) Set r7c1 and r8c1 as a pair on 5.
2) Set r4c3 as a resolved 7 (no pair needed!)
3) Set r1c1 and r2c1 as a pair on 3.
Much later r2c5 becomes resolved as 3.
That cell is in line with r2c1 and so r2c1 cannot be 3 - but it
contains a subscript '3'. Thus its partner MUST be resolved.
Accordingly r1c1 is set to '3' and the pair marks '3' in both
r1c1 and r2c1 will have been erased.
It is true that Mandatory Pairs will not resolve the "super-difficult"
class of puzzle but, in my view, it provides an opportunity to use
logical thinking without having to hold whole huge trains of logical
deductions in ones head and without reducing solvability to the
ability to spot patterns in candidate profiles (or more likely the
avoidance of errors in writing and erasing). Using Mandatory Pairs
would seem to be less error-prone than using candidate profiles
but it would be difficult to prove that one either way.
In some puzzles, I have needed to xx from Mandatory pairs to the
full candidate profiles in order to reach a solution. My personal
challenge is to reduce the frequency of needing to do this - and to
look for ways to identify further pairs, or to demonstrate that such
pairs are not to be found. Even when the xx has been made, it is
still possible to use "pairs" data in conjunction with the profiles - and
even to xx back to "Mandatory Pairs" after the 'crunch' point.
On the subject of this topic (hint?), what I can say is that I have never
needed to use the [hint] facility. In my earlier days of acquaintance
with Sudoku, I used a 'step' solver when I got stuck - but as a learning
tool, as it produces a commentary for each step. That commentary
includes details of the "rule" used to make an elimination from the
candidate profile. However, I tired of poring through candidate profiles
and developed the "Mandatory Pairs" as retaining the 'challenge' of
the game, whilst removing the chore of setting up the profiles.
Remember, the Pairs need to be found and recorded as and when they
are located. The setting up of candidate profiles is a more mechanistic
task that has be COMPLETED before use can be made of it. The Pairs
come and go - like electrons colliding and exchanging with electrons
in neighbouring atoms - and any of them can be used as soon as they
have been recorded.
I wish you well if you decide to explore Mandatory Pairs further. I am
sure that it could be developed - but I am too close to it to be able to
see what is there rather than what I assume to be there.
Alan Rayner BS23 2QT
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Nenthorn
Joined: 13 Nov 2005
Posts: 11
Location: Scotland
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| Posted: Sun Nov 13, 2005 11:26 pm Post subject: Re: 12 Nov |
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| Jimmy Chan wrote: | First: r7c8 must either be 1 or 5, the reason is trivial.
Second: both r4c9 and r6c9 cannot be 1 because they must either be 2 or 4.
Third: Therefore in the sixth block, 1 must appear in either r4c8 or r5c8.
Fourth: Therefore 1 can be eliminated from r7c8 and leave the 5 there. |
Thanks Jimmy. I'm kicking myself for not noticing the 2/4 pairs. You're a star. |
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