So maybe this is some form of colouring, or forcing chains or something and
my problem is unfamiliarity with the naming conventions. At any rate, it is
rare that I get a problem hard enough to need it, but Ruud makes them all the
time.
The puzzle from Dec 13 2005 shows it off fairly well, and in an isolated
enough fashion that there isn't a lot of irrelevant mess. Of course
first you have to solve things well enough that only this technique is
left ...
So: first you do the obvious stuff and get to here:
Code: Select all
.-----------------------.-----------------------.-----------------------.
|7 6 1 |8 3 2 |4 9 5 |
|4 9 8 |1 6 5 |2 37 37 |
|3 5 2 |4 9 7 |6 1 8 |
:-----------------------+-----------------------+-----------------------:
|9 378 367 |63 4 1 |378 5 2 |
|568 4 356 |2 7 9 |1 368 36 |
|2 1 367 |36 5 8 |37 4 9 |
:-----------------------+-----------------------+-----------------------:
|1 37 9 |57 8 6 |537 2 4 |
|568 2 356 |57 1 4 |9 3678 367 |
|568 78 4 |9 2 3 |578 678 1 |
'-----------------------'-----------------------'-----------------------'
Then cells (1,5) (1,7) (8,5) (8,7) constrain the 8s. (I am still learning
the jargon for solving these things, and I think this is called an
X-wing.) This means you can eliminate the 8s from (1,1) and (1,8 )
Also cells (3,4) (3,6) (4,4) (4,6) constrain the 6s. Thus you can eliminate
6s from (3,5) (3,8 ). Furthermore, eliminating the 6 from (3,8 ) allows you
to eliminate the 6 from (1,5) because the 6 in column 1 must provide the
6 for the bottom left house.
(3,5) are naked pairs in column 3. Thus the 7 cannot be in (2,4)
This is great news because now cells (3,4) (3,6) (7,4) (7,6) constrain the
7s. You can eliminate them from (7,7) and (7,9).
So now you get:
Code: Select all
.-----------------------.-----------------------.-----------------------.
|7 6 1 |8 3 2 |4 9 5 |
|4 9 8 |1 6 5 |2 37 37 |
|3 5 2 |4 9 7 |6 1 8 |
:-----------------------+-----------------------+-----------------------:
|9 38 67 |63 4 1 |378 5 2 |
|58 4 35 |2 7 9 |1 368 36 |
|2 1 67 |36 5 8 |37 4 9 |
:-----------------------+-----------------------+-----------------------:
|1 37 9 |57 8 6 |35 2 4 |
|568 2 35 |57 1 4 |9 3678 367 |
|56 78 4 |9 2 3 |58 67 1 |
'-----------------------'-----------------------'-----------------------'
then maybe I haven't been understanding the discussion.
Make a list of each number, and how well solved they are. 1,2,4 and 9 are
completely solved, and therefore useless to us.
3, 5, 6, and 7 are far from being solved.
But 8 is in a very useful state. Every house has only 2 possible cell
locations for the 8. It is completely constrained -- either one half
of them are all 8s, or the other half is. So I write this on paper like
this: (And here is where I am wondering if what I am doing is colouring,
but I think not.)
Code: Select all
.-----------------------.-----------------------.-----------------------.
|7 6 1 |8 3 2 |4 9 5 |
|4 9 8 |1 6 5 |2 37 37 |
|3 5 2 |4 9 7 |6 1 8 |
:-----------------------+-----------------------+-----------------------:
|9 38- 67 |63 4 1 |378+ 5 2 |
|58+ 4 35 |2 7 9 |1 368- 36 |
|2 1 67 |36 5 8 |37 4 9 |
:-----------------------+-----------------------+-----------------------:
|1 37 9 |57 8 6 |35 2 4 |
|568- 2 35 |57 1 4 |9 3678+ 367 |
|56 78+ 4 |9 2 3 |58- 67 1 |
'-----------------------'-----------------------'-----------------------'
Now we need to find a relationship between the 8s and one of the other
unsolved numbers, the 3s, 5s, 6s, or 7s. You need to start in a house
which has the 8 paired with one of those numbers, and which only has
2 cell possibilities in that house, for that number.
We don't have any (6,8 ) candidates. So we cannot try it with them.
We could try to find (3,8 ) or (7,8 ) ones, but the fact that both
the bottom rightmost house and the middle leftmost house have (8,5)
candidates makes 5 the most attractive subject.
Let us try with 5s. Using the same notation, starting with the bottom
rightmost house, so (7,9) we get:
Code: Select all
.-----------------------.-----------------------.-----------------------.
|7 6 1 |8 3 2 |4 9 5 |
|4 9 8 |1 6 5 |2 37 37 |
|3 5 2 |4 9 7 |6 1 8 |
:-----------------------+-----------------------+-----------------------:
|9 38- 67 |63 4 1 |378+ 5 2 |
|58+ 4 35 |2 7 9 |1 368- 36 |
|2 1 67 |36 5 8 |37 4 9 |
:-----------------------+-----------------------+-----------------------:
|1 37 9 |5+7 8 6 |35- 2 4 |
|568- 2 35 |5-7 1 4 |9 3678+ 367 |
|5-6* 78+ 4 |9 2 3 |5+8- 67 1 |
'-----------------------'-----------------------'-----------------------'
be 5-,6
And this is perfect. Because we have another (5,8+) in (1,4). That
5 has got to be a - as well.
Thus column 1 has two 5-s, which means that '-' must mean 'IS NOT'
So we can solve for the 5s, and the 8s. Sometimes you prove that '+'
is 'is not'.
Sometimes you are not as fortunate, and what you connect up is a
- with a +. Even this is of value. Pretend for a moment that (1,4)
was (5, 8- ) Then at this point we would have a 5+ and a 5- in
column 1, which would allow us to remove all other 5s from that
column (in this case the one at (1,8 ))
Is this just forcing chains/colouring or am I onto something new?
Laura