Historical 57

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Historical 57

Postby Leren » Thu Jan 02, 2020 10:10 am

Code: Select all
 *-----------*
 |2..|..4|9.5|
 |...|...|...|
 |.49|.3.|21.|
 |---+---+---|
 |1.5|..9|4..|
 |8..|...|..3|
 |..6|3..|7.2|
 |---+---+---|
 |.72|.6.|34.|
 |...|...|...|
 |6.4|9..|..1|
 *-----------*
2....49.5..........49.3.21.1.5..94..8.......3..63..7.2.72.6.34..........6.49....1
Leren
 
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Re: Historical 57

Postby Mauriès Robert » Thu Jan 02, 2020 12:10 pm

Hi
Resolution with TDP :
P(7r89c8) : 7r89c8->7r3c9->5r3c1->9r7c1
P(7r8c9) : 7r8c9->(7r9c5->7r4c4)->7r1c8->3r1c3->3r8c1->9r7c1
=>r7c1=9, stte
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Re: Historical 57

Postby Ngisa » Thu Jan 02, 2020 10:03 pm

Code: Select all
+---------------------+----------------------------+----------------------+
| 2      168     a138 |j1678       18       4      | 9      k3678    5    |
| 57-3   1568    a138 | 125678     9        125678 | 68     L3678    4    |
|c57     4        9   | 5678       3        5678   | 2       1      d678  |
+---------------------+----------------------------+----------------------+
| 1      3        5   |i27        h27       9      | 4       68      68   |
| 8      2        7   | 456        45       56     | 1       9       3    |
| 4      9        6   | 3          18       18     | 7       5       2    |
+---------------------+----------------------------+----------------------+
|b59     7        2   | 158        6        158    | 3       4       89   |
|b359    158     b138 | 24578      24578    2578   | 568     2678   e6789 |
| 6      58       4   | 9         g27       3      | 58     f27      1    |
+---------------------+----------------------------+----------------------+

(3)r12c3 = (3,9|5)b7p641 - (5=7)r3c1 - r3c9 = r8c9 - r9c8 = r9c5 - r4c5 = r4c4 - r1c4 = (7-3)r1c8 = (3)r2c8 => - 3 r2c1; stte

Clement
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Re: Historical 57

Postby Wecoc » Fri Jan 03, 2020 12:28 am

Code: Select all
+---------------+-------------------+-----------------+
| 2   168  g138 | i1678  18  4      | 9   h3678  5    |
|a357 1568  138 |  12568 9  b125678 | 68   368   4    |
| 57  4     9   |  568   3   5678   | 2    1     678  |
+---------------+-------------------+-----------------+
| 1   3     5   |  27    27  9      | 4    68    68   |
| 8   2     7   |  456   45  56     | 1    9     3    |
| 4   9     6   |  3     18  18     | 7    5     2    |
+---------------+-------------------+-----------------+
| 59  7     2   |  158   6   158    | 3    4     89   |
|e359 158  f138 |  2458  45 c2578   | 568  268  d6789 |
| 6   58    4   |  9     27  3      | 58   27    1    |
+---------------+-------------------+-----------------+

(AIC) r2c1-7 == r2c6-7 -- r8c6-7 == r8c9-7 -- r8c9-9 == r8c1-9 -- r8c1-3 == r8c3-3 -- r1c3-3 == r1c8-3 -- r1c8-7 == r1c4-7 <> r2c6-7; stte

If someone who doesn't understand this nomenclature reads this, I might fool them to think I know what I'm doing.
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Re: Historical 57

Postby eleven » Fri Jan 03, 2020 10:25 am

You need a swordfish first to eliminate the 7 in r2c48.
Then you can shorten the chain and start with r8c6-7 == r8c9-7...


Code: Select all
 *----------------------------------------------------------------------*
 |  2     168    138   |  1678     18   4        |  9    d3678   5      |
 | b357   1568   138   |  125678   9    125678   |  68   c3678   4      |
 | b57    4      9     |  5678     3    5678     |  2     1     a678    |
 |---------------------+-------------------------+----------------------|
 |  1     3      5     |  27       27   9        |  4     68    a68     |
 |  8     2      7     |  456      45   56       |  1     9      3      |
 |  4     9      6     |  3        18   18       |  7     5      2      |
 |---------------------+-------------------------+----------------------|
 | b59    7      2     |  158      6    158      |  3     4     a89     |
 |  359   158    138   |  24578    45   2578     |  568   2678   6789   |
 |  6     58     4     |  9        27   3        |  58    27     1      |
 *----------------------------------------------------------------------*

(7=689)r347c9 - (9=573)r237c1 - r3c8 = 3r1c8 => -7r1c8, stte
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Re: Historical 57

Postby SpAce » Fri Jan 03, 2020 6:26 pm

Hi Wecoc,

Wecoc wrote:(AIC) r2c1-7 == r2c6-7 -- r8c6-7 == r8c9-7 -- r8c9-9 == r8c1-9 -- r8c1-3 == r8c3-3 -- r1c3-3 == r1c8-3 -- r1c8-7 == r1c4-7 => r2c6 <> 7; stte

If someone who doesn't understand this nomenclature reads this, I might fool them to think I know what I'm doing.

Actually, I'd say that someone who can figure out your notation might be fooled to think you (kind of) know what you're doing. The logic is correct, after all (if eleven's points are noted). But why would you want to obscure your logic with such a verbose and confusing notation when it could be written much more simply with Eureka? Also, your "conclusion" makes no sense. Why not at least write it normally: => r2c6 <> 7.

As eleven said, we could cut the first two nodes as redundant so we'd get this comparison:

Yours (shortened and with the conclusion fixed):
Code: Select all
r8c6-7 == r8c9-7 -- r8c9-9 == r8c1-9 -- r8c1-3 == r8c3-3 -- r1c3-3 == r1c8-3 -- r1c8-7 == r1c4-7 => r2c6 <> 7

Eureka:
Code: Select all
(7)r8c6 = (7-9)r8c9 = (9-3)r8c1 = r8c3 - r1c3 = (3-7)r1c8 = (7)r1c4 => -7 r2c6

or:
Code: Select all
7r8c6 = (7-9)r8c9 = (9-3)r8c1 = r8c3 - r1c3 = (3-7)r1c8 = 7r1c4 => -7 r2c6

You should see that either form of Eureka is much shorter and easier to read. Plus it's standard.

If you *really* want to use your notation (I don't understand why), you should at the very least avoid using '-' for two different purposes. For example, this would already be much more understandable:

Code: Select all
r8c6:7 == r8c9:7 -- r8c9:9 == r8c1:9 -- r8c1:3 == r8c3:3 -- r1c3:3 == r1c8:3 -- r1c8:7 == r1c4:7 => r2c6 <> 7

Now anyone who understands Eureka can easily translate that. However, that begs the question why not write it in Eureka in the first place. Since you seem to understand the idea, it shouldn't be much of a leap for you.

Also, writing 'r8c9-7' is not only confusing because it looks like a weak link but because it would be translated as 'r8c9 - r8c7' in 3D Eureka. For example, your chain could be further shortened like this:

3D Eureka:
Code: Select all
(7)r8c6 = (7-9)r8c9 = (9-3)r8c1 = r8-1c3 = (3-7)r1c8 = (7)r1c4 => -7 r2c6

or:
Code: Select all
7r8c6 = (7-9)r8c9 = (9-3)r8c1 = r8-1c3 = (3-7)r1c8 = 7r1c4 => -7 r2c6

(Note that no one but me considers the 3D notation acceptable.)

Btw, this would be the most compacted Eureka expression I could come up with:

Code: Select all
793r8c691 = r8-1c3 = 37r1c84 => -7 r2c6

(Don't try that at home, lol.)
-SpAce-: Show
Code: Select all
   *             |    |               |    |    *
        *        |=()=|    /  _  \    |=()=|               *
            *    |    |   |-=( )=-|   |    |      *
     *                     \  ¯  /                   *   

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."
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Re: Historical 57

Postby Wecoc » Fri Jan 03, 2020 7:08 pm

SpAce wrote:(Note that no one but me considers the 3D notation acceptable.)

At least you consider yours acceptable, I can't say the same :lol:

SpAce wrote:But why would you want to obscure your logic with such a verbose and confusing notation when it could be written much more simply with Eureka?

I agree, I'm still learning how to use Eureka properly.
At least, now I can understand it... kind of. At first for me it was like you guys were talking chinese.

eleven wrote:You need a swordfish first to eliminate the 7 in r2c48.

Just to be sure, the Swordfish (in rows) would be this one, right?
r1c48 - r4c45 - r9c58 => -7 r238c4,r28c8
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Re: Historical 57

Postby eleven » Fri Jan 03, 2020 9:51 pm

Yes (r149c458). In this case you can get all eliminations also with 3 strong links (only two 7's in each row).
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