Finishing Off

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Finishing Off

Postby Yogi » Sat Apr 23, 2016 9:19 am

Occasionally I find that I get stumped with a puzzle right near the end, where the remaining numbers to complete it are largely part of an extended bi-value chain. Always wanting a result, I naturally fall back on brute force with the mighty pen (never a computer). With only two possible options, one of them has to be right, but I always feel I should be able to just see straight away what will work and what will not. Here's a couple of examples:
067351024103820765205706013310472506724605130056130247400513672672900351531267400
073800125281035604504021380325408710710350248408172503030287451157000832842513900
Both these have about 20 numbers to go. The first has a chain of mainly 8s or 9s, but is r1c1 8 or 9? I searched every way I knew but could not see that either of these was excluded by any of the available variables. Similarly, the 2nd example has mostly options of 6 or 9. Has anyone got a quick-scan technique that would easily disprove one of them?
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Re: Finishing Off

Postby JC Van Hay » Sat Apr 23, 2016 10:17 am

Code: Select all
+----------------+-------------+------------+
| 8-9   6     7  | 3  5   1    | 89  2   4  |
| 1     4(9)  3  | 8  2   4(9) | 7   6   5  |
| 2     489   5  | 7  49  6    | 89  1   3  |
+----------------+-------------+------------+
| 3     1     89 | 4  7   2    | 5   89  6  |
| 7     2     4  | 6  89  5    | 1   3   89 |
| 8(9)  5     6  | 1  3   8(9) | 2   4   7  |
+----------------+-------------+------------+
| 4     89    89 | 5  1   3    | 6   7   2  |
| 6     7     2  | 9  48  48   | 3   5   1  |
| 5     3     1  | 2  6   7    | 4   89  89 |
+----------------+-------------+------------+
Skyscraper : {9R26} -> -9r1c1; stte
Code: Select all
+--------------+---------------+--------------+
| 69  7     3  | 8    469  469 | 1  2    5    |
| 2   8     1  | 79   3    5   | 6  79   4    |
| 5   6(9)  4  | 679  2    1   | 3  8    7(9) |
+--------------+---------------+--------------+
| 3   2     5  | 4    69   8   | 7  1    6(9) |
| 7   1     69 | 3    5    69  | 2  4    8    |
| 4   6(9)  8  | 1    7    2   | 5  6-9  3    |
+--------------+---------------+--------------+
| 69  3     69 | 2    8    7   | 4  5    1    |
| 1   5     7  | 69   469  469 | 8  3    2    |
| 8   4     2  | 5    1    3   | 9  67   67   |
+--------------+---------------+--------------+
Skyscraper : {9C29} -> -9r6c8; stte
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Re: Finishing Off

Postby Kozo Kataya » Sun Apr 24, 2016 1:53 am

Yogi wrote:Has anyone got a quick-scan technique that would easily disprove one of them?

i used to select as shown below.
Code: Select all
*-----------------------------*
|-9       |    9  9 |         |
|         | 9       |   *9    |
|   *9    |-9       |      -9 |
|---------+---------+---------|
|         |    9    |       9 |
|      *9 |       9 |         |
|   -9    |         |   -9    |
|---------+---------+---------|
| 9    -9 |         |         |
|         | 9  9  9 |         |
|         |         | 9       |
*-----------------------------*

if r3c2 = 9, r1c1, r3c49 and r6c2 > -9 then r2c8 and r5c3 = 9
This results r6c8 and r7c3 > -9. So, r6 lost possible candidates of 9
9r1c1 is correct because 9r3c2 > -9r6c28 = contradiction
is this quick-scan or not ?
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Re: Finishing Off

Postby pjb » Sun Apr 24, 2016 12:30 pm

Alternatively for the first, all the unsolved cells have 2 candidates except one which has three. This allows you to apply the BUG+1 technique. This is described on many sudoku websites, the following is a good one: http://hodoku.sourceforge.net/en/techniques.php

Phil
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Re: Finishing Off

Postby Yogi » Mon Apr 25, 2016 1:51 am

Thanx for the input guys. I was out of my depth with the other techniques but JC's Skyscrapers were magic. Now let's see if I've actually got it!
I take it that {9R26} means the skyscraper encompassing the 9-candidate cells in Rows 2 and 6, and -9r1c1 means r1c1 can't be a 9 because it can see both of the 9-candidate cells in the sloping end of the side-lying skyscraper, and one of those must be a 9. With r1c1 now fixed at 8 the puzzle is solved. I suppose the right lingo is 8r1c1 or r1c1 -> 8
The 2nd case is equally simple once spotted. With {9C29} we have the upside-down skyscraper drawn by the 9-candidate cells in Columns 2 and 9. This eliminates 9 from r6c8 which can see both cells in the sloping edge of the skyscraper. Now, with the values in r6c2 and 8 fixed this puzzle is also quickly solved.
I'm not sure what stte means. Maybe it's short for SORTED!
One general observation I drew from all this is that the first puzzle had only candidates 4, 8 and 9 to play with and the second had only 6, 7 and 9 (4 was not important) which made looking for skyscrapers a lot easier.

Regards!
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Re: Finishing Off

Postby JasonLion » Mon Apr 25, 2016 2:27 am

"stte" means Singles To The End.

Which means nearly the same thing as Sorted :)
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