Fear Me Not 9.0

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Fear Me Not 9.0

Postby AnotherLife » Tue Dec 07, 2021 9:26 am

This puzzle (Berthier 2412) is rated SER 9.0. Will you press the 'autosolve' button or try to solve it by human means?
Code: Select all
|12.|.5.|...|
|.5.|7.9|...|
|..9|.2.|5..|
|---+---+---|
|.17|.48|...|
|3..|6..|...|
|...|..2|4.1|
|---+---+---|
|5..|..4|...|
|.78|...|95.|
|...|..7|81.|

12..5.....5.7.9.....9.2.5...17.48...3..6..........24.15....4....78...95......781.

P.S.
Hidden Text: Show
I ask you to abstain from emotional outbursts. I may stop making posts on the forum if I come to the conclusion that my work is not enough appreciated.
Bogdan
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Re: Fear Me Not 9.0

Postby P.O. » Tue Dec 07, 2021 11:44 am

Hi AnotherLife, i am not a human solver i use algorithms; i understand the expression 'human means' as meaning a pattern that a human can find as oppose to a pattern that only a algorithm can find; but you can't know that a algorithm was used or not for the patterns you class as 'human'; of course remains the question of defining the class of human patterns;
concerning this puzzle i have always seen its rating as an anomaly or an error as it is easily solved with a shortest-chain first strategy, only 12 chains of which six are intersections;
here a 4-chains solution in the first state of the grid after intersections:
c3n2{r9 r5} - r5c7{n2 n7} - r1{c6c7}{n3n6} - r1c3{n3n6 n4} => r9c3 <> 4
r1c4{n4 n8} - r7c4{n8 n2} - c7n2{r7 r4r5} - c8n2{r4r5 r2} - r2n4{c8 c1c3} => r1c3 <> 4
b1n8{r3c2 r2c1} - c5n8{r2 r7} - r7c4{n8 n2} - b9n2{r7c7r7c8r7c9 r8c9r9c9} - r2n2{c9 c8} - r2n4{c8 c3} => r3c2 <> 4
r2n4{c8 c1c3} - r1{c3c6}{n3n6} - r1c7{n3n6 n7} - r5c7{n7 n2} - r5c3{n2 n4} - c2n4{r5 r9} - c1n4{r8r9 r2} => r2c3 <> 4
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Re: Fear Me Not 9.0

Postby totuan » Tue Dec 07, 2021 12:18 pm

Code: Select all
 *-----------------------------------------------------------------------------*
 | 1       2       36-4    |&48      5       36      | 367     346789  36789   |
 |*468     5       346     | 7       368     9       | 1       23468   2368    |
 | 7       3468    9       | 148     2       136     | 5       3468    368     |
 |-------------------------+-------------------------+-------------------------|
 |*269     1       7       | 39      4       8       | 236     2369    5       |
 | 3       489    &24      | 6       1       5       | 27      2789    2789    |
 | 689     689     5       | 39      7       2       | 4       3689    1       |
 |-------------------------+-------------------------+-------------------------|
 | 5       369     1       |&28      3689    4       |*2367   *2367   *2367    |
 |#246     7       8       | 12      36      136     | 9       5      #2346    |
 |#2469    3469    2346    | 5       369     7       | 8       1      #2346    |
 *-----------------------------------------------------------------------------*

UR(24)r89c19 => (2)r7c789=(2|4)r24c1
01: (4=8)r1c4-(8=2)r7c4-(2)r7c789=UR(24)r89c19=(2|4)r24c1-(2=4)r5c3 => r1c3<>4, some singles
Code: Select all
 *--------------------------------------------------------------------*
 | 1      2      36     | 48     5      36     | 7      489    89     |
 |*48     5      36     | 7      368    9      | 1     *23468 *2368   |
 | 7      48     9      | 148    2      136    | 5      3468   368    |
 |----------------------+----------------------+----------------------|
 | 2      1      7      | 39     4      8      | 36     369    5      |
 | 3      89     4      | 6      1      5      | 2      789    789    |
 | 689    689    5      | 39     7      2      | 4      389    1      |
 |----------------------+----------------------+----------------------|
 | 5      369    1      | 28     3689   4      | 36     2367   2367   |
 |*46     7      8      | 12     36     136    | 9      5     *346-2  |
 | 469    3469   2      | 5      369    7      | 8      1      346    |
 *--------------------------------------------------------------------*

02: (4)r8c9=r8c1-r2c1=(4-2)r2c8=r2c9 => r8c9<>2, stte

Thanks for the puzzle.
totuan
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Re: Fear Me Not 9.0

Postby eleven » Tue Dec 07, 2021 1:35 pm

Very nice, but i am not happy with the notation (2|4)r24c1-(2=4)r5c3.
The 4r2c1 directly eliminates 4r1c3, but does not imply 4r5c3, does it ?

Reuse of the same piece of chain:
Code: Select all
 *------------------------------------------------------------------------*
 |  1      2     a346    | i48    5     a36    |  7-36   4789-36  789-36  |
 | b468    5     b346    |  7     368    9     |  1     c23468   c2368    |
 |  7      3468   9      | h148   2      136   |  5      3468     368     |
 |-----------------------+---------------------+--------------------------|
 |  269    1      7      |  39    4      8     |  236    2369     5       |
 |  3      489    24     |  6     1      5     |  27     2789     2789    |
 |  689    689    5      |  39    7      2     |  4      3689     1       |
 |-----------------------+---------------------+--------------------------|
 |  5      369    1      | f28    3689   4     | e2367  e2367    d2367    |
 |  246    7      8      | g12    36     136   |  9      5       d2346    |
 |  2469   3469   2346   |  5     369    7     |  8      1       d2346    |
 *------------------------------------------------------------------------*

(36=4)r1c36 - 4r2c13 = 42r2c89 - r789c9 = r7c78 - r7c4 = (2-1)r8c4 = (1-4)r3c4 = r1c4 - (4=36)r1c36 => -36r1c789, -4r1c3

4 singles
Code: Select all
+----------------------+----------------------+----------------------+
| 1      2      36     | 48     5      36     | 7      489    89     |
|b468    5      36     | 7      368    9      | 1     c23468 c2368   |
| 7      3468   9      | 148    2      136    | 5      3468   368    |
+----------------------+----------------------+----------------------+
| 2      1      7      | 39     4      8      | 36     369    5      |
| 3      89     4      | 6      1      5      | 2      789    789    |
| 689    689    5      | 39     7      2      | 4      3689   1      |
+----------------------+----------------------+----------------------+
| 5      369    1      |f28     3689   4      | 36    e2367  d2367   |
|a46     7      8      |g12    a36     1-36   | 9      5     d2346   |
| 469    3469   2      | 5      369    7      | 8      1      346    |
+----------------------+----------------------+----------------------+

(36=4)r8c51 - 4r2c1 = 42r2c89 - r78c9 = r7c8 - r7c4 = (2-1)r8c4 = 1r8c6 => -36r8c6, stte
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Re: Fear Me Not 9.0

Postby totuan » Tue Dec 07, 2021 2:53 pm

eleven wrote:Very nice, but i am not happy with the notation (2|4)r24c1-(2=4)r5c3.
The 4r2c1 directly eliminates 4r1c3, but does not imply 4r5c3, does it ?

Yes. I meant, pause at (2|4) r24c1 to "kill" 4r1c3 the first time and (2=4) r5c3 to "kill" 4r1c3 the second time. So, 4r1c3 gets "killed" twice :D

totuan
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Re: Fear Me Not 9.0

Postby yzfwsf » Tue Dec 07, 2021 5:33 pm

2 chains and Naked pair,Locked Candidates.
Hidden Text: Show
Cell Forcing Chain: Each candidate in r2c5 true in turn will all lead r1c3<>4
(3-8)r2c5 = r7c5 - (8=2)r7c4 - r7c78 = r789c9 - r2c9 = (2-4)r2c8 = r2c13 - 4r1c3
(6-8)r2c5 = r7c5 - (8=2)r7c4 - r7c78 = r789c9 - r2c9 = (2-4)r2c8 = r2c13 - 4r1c3
8r2c5 - (8=4)r1c4 - 4r1c3
Naked Pair: in r1c3,r1c6 => r1c7<>36,r1c8<>36,r1c9<>36,
Naked Single: r1c7=7
Naked Single: r5c7=2
Hidden Single: 2 in b4 => r4c1=2
Hidden Single: 2 in b7 => r9c3=2
Naked Single: r5c3=4
Locked Candidates 2 (Claiming): 6 in r4 => r6c8<>6
Locked Candidates 1 (Pointing): 3 in b7 => r3c2<>3
Locked Candidates 2 (Claiming): 6 in c3 => r2c1<>6,r3c2<>6
AIC Type 2: 2r2c9 = (2-4)r2c8 = r2c1 - r8c1 = 4r8c9 => r8c9<>2
stte
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Re: Fear Me Not 9.0

Postby AnotherLife » Tue Dec 07, 2021 5:48 pm

P.O. wrote:Hi AnotherLife, i am not a human solver i use algorithms; i understand the expression 'human means' as meaning a pattern that a human can find as oppose to a pattern that only a algorithm can find; but you can't know that a algorithm was used or not for the patterns you class as 'human'; of course remains the question of defining the class of human patterns;
concerning this puzzle i have always seen its rating as an anomaly or an error as it is easily solved with a shortest-chain first strategy, only 12 chains of which six are intersections;
here a 4-chains solution in the first state of the grid after intersections:
c3n2{r9 r5} - r5c7{n2 n7} - r1{c6c7}{n3n6} - r1c3{n3n6 n4} => r9c3 <> 4
r1c4{n4 n8} - r7c4{n8 n2} - c7n2{r7 r4r5} - c8n2{r4r5 r2} - r2n4{c8 c1c3} => r1c3 <> 4
b1n8{r3c2 r2c1} - c5n8{r2 r7} - r7c4{n8 n2} - b9n2{r7c7r7c8r7c9 r8c9r9c9} - r2n2{c9 c8} - r2n4{c8 c3} => r3c2 <> 4
r2n4{c8 c1c3} - r1{c3c6}{n3n6} - r1c7{n3n6 n7} - r5c7{n7 n2} - r5c3{n2 n4} - c2n4{r5 r9} - c1n4{r8r9 r2} => r2c3 <> 4

Hi P.O.,
You see, the difference between human solutions and solutions found by programs is not only in the different choice of the methods but in the different order of steps. For example, after your steps 1 and 2 any human would spot a naked pair in r1,
Code: Select all
.-----------------.----------------.----------------------.
| 1     2     36* | 48   5     36* | 7-36  4789-36 789-36 |
| 468   5     346 | 7    368   9   | 1     23468   2368   |
| 7     3468  9   | 148  2     136 | 5     3468    368    |
:-----------------+----------------+----------------------:
| 269   1     7   | 39   4     8   | 236   2369    5      |
| 3     489   24  | 6    1     5   | 27    2789    2789   |
| 689   689   5   | 39   7     2   | 4     3689    1      |
:-----------------+----------------+----------------------:
| 5     369   1   | 28   3689  4   | 2367  2367    2367   |
| 246   7     8   | 12   36    136 | 9     5       2346   |
| 2469  3469  236 | 5    369   7   | 8     1       2346   |
'-----------------'----------------'----------------------'

which leads to 5 singles and intersections, and then he might apply your step 3 leading to a sequence of singles. So there in no need in step 4 from a human point of view.

As to the SER of this puzzle, I don’t consider it as an anomaly or an error. Rather, this is a drawback of the SE ratings that they don’t take account of group- and ALS-nodes in AICs and forcing chains, and if we restrict ourselves to such chains, we will need dynamic forcing chains to solve this puzzle, so SER 9.0 will be a correct estimate.

To Eleven and Totuan
Thanks for your solutions! I thought that ALS-nodes were needed to solve this puzzle, but it comes to be that group nodes suffice. Now I will have to make corrections to my recent post.. Unfortunately, HoDoKu can reach -4 r1c3 only via forcing chains, so it is a bug in the program. Let me write my variant, which is based on your solutions.
Code: Select all
.----------------- -.----------------.---------------------.
| 1     2     36-4  | g48  5     36  | 367   346789  36789 |
| a468  5     a346  | 7    368   9   | 1     b23468  c2368 |
| 7     3468  9     | 148  2     136 | 5     3468    368   |
:-------------------+----------------+---------------------:
| 269   1     7     | 39   4     8   | 236   2369    5     |
| 3     489   24    | 6    1     5   | 27    2789    2789  |
| 689   689   5     | 39   7     2   | 4     3689    1     |
:-------------------+----------------+---------------------:
| 5     369   1     | f28   3689  4  | e2367 e2367   d2367 |
| 246   7     8     | 12   36    136 | 9     5       d2346 |
| 2469  3469  2346  | 5    369   7   | 8     1       d2346 |
'-------------------'----------------'---------------------'

1. AIC with groups: (4)r2c13 = (4-2)r2c8 = r2c9 - r789c9 = r7c78 - (2=8)r7c4 - (8=4)r1c4 => -4 r1c3; naked pair, 5 singles, lcste

Code: Select all
.---------------.----------------.-------------------.
| 1    2     36 | 48   5     36  | 7   489    89     |
| c48  5     36 | 7    368   9   | 1   d23468 e2368  |
| 7    48    9  | 148  2     136 | 5   3468   368    |
:---------------+----------------+-------------------:
| 2    1     7  | 39   4     8   | 36  369    5      |
| 3    89    4  | 6    1     5   | 2   789    789    |
| 689  689   5  | 39   7     2   | 4   389    1      |
:---------------+----------------+-------------------:
| 5    369   1  | 28   3689  4   | 36  2367   2367   |
| b46  7     8  | 12   36    136 | 9   5      a346-2 |
| 469  3469  2  | 5    369   7   | 8   1      346    |
'---------------'----------------'-------------------'

2. Basic AIC: (4)r8c9 = r8c1 - r2c1 = (4-2)r2c8 = r2c9 => -2 r8c9; ste

P.S. Denis, thanks for the puzzle.
Bogdan
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