leftover

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Re: leftover

Postby SpAce » Thu Jun 27, 2019 9:32 pm

eleven wrote:i just critisized, that by your abbreviation very useful information (for quick understanding), was hidden.

Ok. Really cool solutions, btw!

I borrowed your BUG-Lite for the second puzzle:

Code: Select all
.--------------------.----------------.------------------.
|  e1-7    28    134 |  5     9    6  |  f3(7)  48   12  |
|   15679  1567  145 |  2     3    8  |   79    456  156 |
|  d569    28    35  |  1     7    4  | ef39    568  256 |
:--------------------+----------------+------------------:
|   8      4     9   |  3     6    5  |   1     2    7   |
|  c13+5   15    7   |  49+   18+  2  |   6     39+  48+ |
|  c13+6   16    2   | c49+7  18+  79 |   5     39+  48+ |
:--------------------+----------------+------------------:
| ad5[7]   9     8   | b67    2    3  |   4     1    56  |
|   4      3     6   |  8     5    1  |   2     7    9   |
|   2      157   15  |  679   4    79 |   8     56   3   |
'--------------------'----------------'------------------'

BUG-Lite (13489)r56c14589

(7)r7c1 = r7c4 - (7==5|6)r6c4,r56c1 - (56=7|9)r73c1 - (79)r1c1,r3c7 = (37)r31c7 => -7 r1c1; stte

----

To be honest, i did not read Leren's 2nd kraken to the end.

Thanks for admitting :D Here's that "kraken" translated into something sensible:

Code: Select all
"Kraken Column 9 Digit 2"

(2)r1c9 - (1)r1c9
||        ||
||        (1-7)r1c1 = (7)r1c7
||        ||
||        (1)r1c3 - r9c3 = (1-7)r9c8 = r7c1 - r1c1 = (7)r1c7
||
(2)r3c9 - (5)r3c9
          ||
          (5)r3c1 - (5=7)r7c1 - r1c1 = (7)r1c7
          ||
          (5-3)r3c3 = (3)r3c7
          ||
          (5)r3c8 - r9c8 = r9c23 - (5=7)r7c1 - r1c1 = (7)r1c7

=> -3 r1c7; stte

Now that it's actually krakenized, does the given name make any more sense? No, it does not, because there are actually two krakens but not with the 2s (which have a basic strong link). Furthermore, trivial analysis would have revealed a much simpler elimination. Here's how I'd name and write that:

Code: Select all
Double Kraken 5R3 & 1R1:

(57)r37c1
||
(53)r3c37 - (3=7)r1c7
||
(5)r3c8 - r9c8 = (57)r7c91
||
(52)r31c9 - (1)r1c9
            ||
            (1)r1c3 - (1=57)b7p91
            ||
            (1)r1c1

=> -7 r1c1; stte

for those who prefer simple nodes: Show
Code: Select all
(5)r3c1 - (5=7)r7c1
||
(5-3)r3c3 = r3c7 - (3=7)r1c7
||
(5)r3c8 - r9c8 = r7c9 - (5=7)r7c1
||
(5-2)r3c9 = (2-1)r1c9
              ||
              (1)r1c3 - (1=5)r9c3 - (5=7)r7c1
              ||
              (1)r1c1

8x8 TM: Show
Code: Select all
 7r7c1 5r7c1
       5r7c9 5r9c8
       5r9c3       1r9c3
 1r1c1             1r1c3 1r1c9
                         2r1c9 2r3c9
       5r3c1 5r3c8             5r3c9 5r3c3
                                     3r3c3 3r3c7
 7r1c7                                     3r1c7
------------------------------------------------
-7r1c1

So... while there was nothing wrong with the logic, the naming was incorrect for several reasons and the presentation was absolutely horrible. But, like I said, even bad examples serve a purpose :)

Edit: Oops! It appears that my simplification is an exact copy of Cenoman's hidden solution! It's a pure coincidence as I hadn't even seen that! I would have of course mentioned it immediately had I known. Sorry about that!
Last edited by SpAce on Fri Jun 28, 2019 4:49 pm, edited 1 time in total.
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Re: leftover

Postby Cenoman » Fri Jun 28, 2019 1:42 pm

eleven wrote:In both grids there's a complex invalid pattern:
Code: Select all
 *-----------------------------------------------------*
 |  .      .     34   | .    .   .   |  .   48    .    |
 |  .      .     45   | .    .   .   |  .   46    .    |
 |  69     28    35   | .    .   .   |  39  68    25   |
 *--------------------+--------------+-----------------*

(row 3 gives the links 3r3c3=6r3c8 and 5r3c3=8r3c8, so 5r3c3 -> 6r3c8 -> 4r2c8 -> 5r2c3 and 3r3c3 -> 3r1c3)

5r23c8 == 1r12c3 - (1=5)r9c3 => -5r9c8, stte


Wow !
That left me speechless...
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