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Puzzle: DB021906 - Once more, with feeling!

 
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Sun Feb 26, 2006 12:10 am    Post subject: Puzzle: DB021906 - Once more, with feeling! Reply with quote
This puzzle was buried in the thread on 4-star constellations, and drew no responses. So I thought I'd try again.

Code:


Puzzle: DB021906
+-------+-------+-------+
| 3 . . | . . . | 6 7 . |
| . 5 . | 6 . . | 9 3 . |
| . . . | 5 7 . | . . 4 |
+-------+-------+-------+
| . . . | 9 . 4 | . . . |
| . . 6 | . . . | 8 . . |
| . . . | 3 . 7 | . . . |
+-------+-------+-------+
| 8 . . | . 9 6 | . . . |
| . 6 4 | . . 5 | . 2 . |
| . 1 3 | . . . | . . 6 |
+-------+-------+-------+



Using basic techniques, I got to this point:

Code:



+-------------------+-------------------+-------------------+
| 3     248   128   | 48    124   9     | 6     7     5     |
| 1247  5     1278  | 6     124   28    | 9     3     128   |
| 6     289   1289  | 5     7     3     | 12    18    4     |
+-------------------+-------------------+-------------------+
| 1257  2378  12578 | 9     68    4     | 12357 156   1237  |
| 47    3479  6     | 2     5     1     | 8     49    379   |
| 1245  2489  12589 | 3     68    7     | 1245  14569 129   |
+-------------------+-------------------+-------------------+
| 8     27    257   | 147   9     6     | 13457 145   137   |
| 79    6     4     | 178   3     5     | 17    2     89    |
| 579   1     3     | 478   24    28    | 457   89    6     |
+-------------------+-------------------+-------------------+



The possibilities of <8> in the last four columns seemed fishy, and I found the following (constellation or double implication chain?):

R2C9 cannot be <8>. If we assume R2C9 is <8>, then R2C6 = <2>, R8C9 = <9>, and we find the contradiction R9C6 = <8> and R9C8 = <8>.

I was impressed, for the rest of the solution is straightforward.

However, there is an interesting cycle in the above situation. It involves the squares R5C1, R8C1, R8C9, R9C8, R5C8, and back to R5C1. So we see:

R5C1 or R8C1 must be <7>.
R8C1 or R8C9 must be <9>.
R8C9 or R9C8 must be <8>.
R9C8 or R5C8 must be <9>.
R5C8 or R5C1 must be <4>.

and then:

R2C1 cannot be <7>.
R4C1 cannot be <7>.
R9C1 cannot be <7>.
R6C8 cannot be <9>.
R5C2 cannot be <4>.

Cute as this may be, it does not immediately solve the puzzle: The constellation is still shining!

Best wishes,

Keith
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
PostPosted: Sun Feb 26, 2006 7:36 pm    Post subject: I thought this one was easy ... Reply with quote
Hi, Keith!

I might have written back about this puzzle last week, but didn't bother because there was a straightforward "coloring" solution. This is the same position you illustrated above.
Code:
+-------------------+-------------------+-------------------+
| 3     248   128   | 48    124   9     | 6     7     5     |
| 1247  5     1278  | 6     124   28~   | 9     3     128+  |
| 6     289   1289  | 5     7     3     | 12    18-   4     |
+-------------------+-------------------+-------------------+
| 1257  2378  12578 | 9     68    4     | 12357 156   1237  |
| 47    3479  6     | 2     5     1     | 8     49    379   |
| 1245  2489  12589 | 3     68    7     | 1245  14569 129   |
+-------------------+-------------------+-------------------+
| 8     27    257   | 147   9     6     | 13457 145   137   |
| 79    6     4     | 178   3     5     | 17    2     8-9   |
| 579   1     3     | 478   24    28=   | 457   8+9   6     |
+-------------------+-------------------+-------------------+

-- There are only two places to fit an "8" in column 6, at r2c6 or r9c6.

-- There are also just two places for an "8" in column 8, and in column 9.

-- The way these line up, if r2c9 = 8 then r9c8 = 8 also.

-- But this is impossible, because it would leave no way to fit an "8" in column 6.

-- We conclude that r3c8 = 8, and also that r8c9 = 8.

Your other note, about the "cycle," is pretty interesting ... I'll have to study that more closely. dcb
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