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New Technique? (2005-12-13 Nightmare)

 
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lac
Hooked
Hooked


Joined: 02 Jan 2006
Posts: 43
Location: Göteborg, Sweden

PostPosted: Wed Jan 04, 2006 1:36 pm    Post subject: New Technique? (2005-12-13 Nightmare) Reply with quote

I think I have a new technique. But I am still learning the terminology.
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. Very Happy

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:

.-----------------------.-----------------------.-----------------------.
|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       |
'-----------------------'-----------------------'-----------------------'

Using column, row notation:

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:

.-----------------------.-----------------------.-----------------------.
|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       |
'-----------------------'-----------------------'-----------------------'


Ok, here comes the bit I haven't seen anybody else discuss doing. But
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:

.-----------------------.-----------------------.-----------------------.
|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       |
'-----------------------'-----------------------'-----------------------'


So either all the 8+s are 8s, or all the 8-s are.

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:

.-----------------------.-----------------------.-----------------------.
|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       |
'-----------------------'-----------------------'-----------------------'


Here we are at (1,9) (marked with a *) which we have just found to
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
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David Bryant
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Joined: 20 Jan 2006
Posts: 86
Location: Denver, Colorado

PostPosted: Thu Jan 26, 2006 4:34 pm    Post subject: "Coloring" and "double-implication chains&quo Reply with quote

Laura wrote:
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.)

This step in your explanation is the technique most people call coloring. In this particular case the simple coloring doesn't gain you any new information, but it does emphasize the fact that there are only two possible ways to put the rest of the "8"s into this puzzle. I tend to call your "+"s and "-"s blue and green, but that's just because I've gotten used to Angus Johnson's Simple Sudoku program.
Laura wrote:
... 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'.

Excellent description! The average guy might understand it a little better if you first pointed out that you either have to have a "5" in col 1 row 5, and an "8" in col 7 row 9, or else you have to have an "8" in col 1 row 5, and a "5" in col 7 row 9. So the only legal combinations of symbols are "5+8-" or its opposite, "5-8+". This probably seems obvious to you, Laura, but some of the rest of us might have to think about it for a while. Smile

Thank you for a very clear and edifying example. I think this technique is logically equivalent to the "double-implication chains" my friend someone_somewhere (over on the DailySudoku forum) is so fond of. dcb
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