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Sarah
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 Posted: Fri Mar 10, 2006 7:23 pm Post subject: Still stuck on Mar.9 |
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I can't even get as far as the previous poster ...
I've only got:
5?69?7??4
?9?5?467?
74????195
46??59??1
359???426
8??64???9
9?4?????8
?8549????
6??8?594?
I don't see the logic in determining the 5's in rows 6&7, or the 7 in r9c8.
Any guidance?
Thanks,
Sarah |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Mar 10, 2006 9:29 pm Post subject: |
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Look at the possibilities in C8. There is a pair, which will enable you to solve the two <5>'s.
If you need a further clue, there is a triple (and a hidden pair) in block 5, which will lead to resolving C4 into two pairs.
Keith |
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Guest
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| Posted: Sun Mar 12, 2006 6:13 pm Post subject: |
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Sarah:
I assume you were able to follow Keith's logic, but just in case you couldn't find the pair, r1c8 and r4c8 are <3,8>. Therefore, r6c8 must be a 5, which means that r7c7 is also a 5.
- Emily |
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Sun Mar 12, 2006 11:44 pm Post subject: |
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I have already posted a solution on the other heading
and so I will confine this posting to the steps in reaching
the position at the start of the other topic.
There are THREE cells to be resolved r6c8, r7c7, r8c9
++
Starting from the position initially posted, I used Mandatory
Pairs to record additional information before proceeding to
derivation of the candidates. Normally I would have the M/P
data marked already but some had been eliminated on my
own solution which I completed on 10th.
The markings include
Box1- 8 in col3
Box2- 1 in col5, 6 in row3
Box3- 8 in row 1, 2 askew
Box4- 1 in row 6
Box5- Triple in row 5 (pairs for 1,7,8)
Mutual Reception (23)
Box6- 8 in row4
(NB 7 in col 7 is NOT included yet)
Box7- No pairs at this stage
Box8- No pairs at this stage
Box9- ONE Mutual Receptions (16)
(NB The other on 23 NOT derived yet)
To continue:
a) Row/Box Interesect on '5' requires the
'5' in box 9 to be in row 7. There are
two possibilities r7c7 and r7c8.
b) R7c8 is NOT available for the '5' as it
is in "Mutual Reception" with r8c8 with (16).
This is one of the techniques with Mandatory
Pairs:- If two cells are in Mutual Reception,
no 'interloper' is allowed.
c) Thus r7c7 MUST be '5'
++
d) Slice/Dice on '5' gives r6c8 to be '5' as well.
++
e) The '7' in box 6 MUST be in col 7 as col 9 is
full and col 8 already has a '7'
f) This means that ONLY col 9 is available for
the '7' in box 9 - meaning r8c9 or r9c9.
g) Looking at row 8 there is a '7' already in
col1 and col 6 plus there WILL be a '7'
in col 6 as per 'e' above.
h) This means that of the unresolved cells in
row 8 ONLY r8c9 is a possibility for the '7'
Thus r8c9 is '7' by "Sole Position"!
++++
"Sole Position" is one of the harder things for a
human solver to spot as it involves eliminating
the impossible rather than being guided directly
to the solution by the logic of the situation as
would be the case with "sole candidates".
In fact it was this "sole position" that I missed in
my original working and which led me to resort to
candidate profiles. I had a (237) triple in box 9
rather than the (23) Mutual Reception that arises
once the '7' in r8c9 has been resolved.
Sole position is easy for computer solvers and
somewhat easier for humans if the candidate
profiles have been derived. It is the only occurrence
of the digit in a whole row/col/region - easily spotted
if one maintains a record of how often a digit appears
in each such row/col/region BUT how many human
solvers would keep such a record in short term memory?
++++
Once these three cells have been resolved the remainder
of the puzzle CAN be solved using Mandatory Pairs. The
process is described in the other topic heading for March 9th.
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For further information on this, it is possible to view
the following link:
Mandatory Pairs - October Notes.
Alan Rayner BS23 2QT |
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