17December18

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17December18

Postby Yogi » Tue Dec 25, 2018 10:42 pm

......73.....5........4...........29.7.....5..3.8........3.74..5.....1..2..6.....

17Dec18.png
17Dec18.png (16.96 KiB) Viewed 153 times

Want to try this with Pencil & Paper only?
Merry Christmas!
Yogi
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Re: 17December18

Postby SpAce » Wed Dec 26, 2018 6:50 am

Pure p&p as requested:

Original: Show
Code: Select all
.----------------------.----------------.-----------------.
| f1(4)8  5    fe1489  |  2    6  e189  | 7   3      148  |
|  37     189   f14689 |  179  5   1389 | 2  b149    1468 |
|  37     2      1689  |  179  4   1389 | 5   19     168  |
:----------------------+----------------+-----------------:
|  6     18      148   |  5    7   14   | 3   2      9    |
|  9     7       2     | c14   3   6    | 8   5     b14   |
|  1-4   3       5     |  8    9   2    | 6  a1(4)   7    |
:----------------------+----------------+-----------------:
|  18    189     189   |  3    2   7    | 4   6      5    |
|  5     6       3     | c49   8  d49   | 1   7      2    |
|  2     4       7     |  6    1   5    | 9   8      3    |
'----------------------'----------------'-----------------'

4r6c8 = 4(r2c8*&r5c9) - 4r(5=8)c4 - (4=9)r8c6 - 9r1c(6=3) - 4b1p(3*6=1) => -4 r6c1; stte

Code: Select all
.----------------------.----------------.-----------------.
| e1(4)8  5    ed1489  |  2    6  d189  | 7   3     e148  |
|  37     189    14689 |  179  5   1389 | 2   149    1468 |
|  37     2      1689  |  179  4   1389 | 5   19     168  |
:----------------------+----------------+-----------------:
|  6     18      148   |  5    7   14   | 3   2      9    |
|  9     7       2     | b14   3   6    | 8   5     a14   |
|  1-4   3       5     |  8    9   2    | 6  a1(4)   7    |
:----------------------+----------------+-----------------:
|  18    189     189   |  3    2   7    | 4   6      5    |
|  5     6       3     | b49   8  c49   | 1   7      2    |
|  2     4       7     |  6    1   5    | 9   8      3    |
'----------------------'----------------'-----------------'

4b6p(8=6*) - 4r(5=8)c4 - (4=9)r8c6 - 9r1c(6=3) - 4r1c(3*9=1) => -4 r6c1; stte

In 2D notation: Show
(4)r6c8 = (4*)r5c9 - r5c4 = r8c4 - (4=9)r8c6 - r1c6 = r1c3 - (4)r1c3,*r1c9 = (4)r1c1 => -4 r6c1; stte

As Kraken Row 4r1: Show
4r1c1
||
(4-9)r1c3 = r1c6 - (9=4)r8c6 - 4r(8=5)c4 - 4b6p(6=8)
||
4r1c9 - 4b6p(6=8)

=> -4 r6c1; stte

Added. Inspired by eleven's solution:

Code: Select all
.------------------.-----------------.-----------------.
| b148  5    1489  |  2     6  b189  | 7   3     b148  |
|  37   189  14689 | a179   5   1389 | 2   149    1468 |
|  37   2    1689  | a179   4   1389 | 5   19     168  |
:------------------+-----------------+-----------------:
|  6    18   148   |  5     7   14   | 3   2      9    |
|  9    7    2     | a1(4)  3   6    | 8   5     c1-4  |
| c14   3    5     |  8     9   2    | 6  d1(4)   7    |
:------------------+-----------------+-----------------:
|  18   189  189   |  3     2   7    | 4   6      5    |
|  5    6    3     |  49    8   49   | 1   7      2    |
|  2    4    7     |  6     1   5    | 9   8      3    |
'------------------'-----------------'-----------------'

(4=179)r423c4 - (9=184)r1c619 - 4(r1c6&r5c9) = 4r6c8 => -4 r5c9

17December18.JPG
17December18.JPG (21.22 KiB) Viewed 144 times
Last edited by SpAce on Thu Dec 27, 2018 6:46 am, edited 2 times in total.
Code: Select all
   *             |    |               |    |    *
        *        |=()=|    /  _  \    |=()=|               *
            *    |    |   |-=( )=-|   |    |      *
     *                     \  ¯  /                   *   
SpAce
 
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Re: 17December18

Postby eleven » Wed Dec 26, 2018 10:33 am

Code: Select all
 +-------+-------+-------+
 | . 5 . | 2 6 . | 7 3 . |
 | . . . | . 5 . | 2 . . |
 | . 2 . | . 4 . | 5 . . |
 +-------+-------+-------+
 | 6 . . | 5 7 . | 3 2 9 |
 | 9 7 2 | . 3 6 | 8 5 . |
 | . 3 5 | 8 9 2 | 6 . 7 |
 +-------+-------+-------+
 | . . . | 3 2 7 | 4 6 5 |
 | 5 6 3 | . 8 . | 1 7 2 |
 | 2 4 7 | 6 1 5 | 9 8 3 |
 +-------+-------+-------+

ab = 14 or 41 (hh=37)
Code: Select all
 +-------+-------+-------+
 |b8 5 . | 2 6 . | 7 3 b8|
 | h . . | . 5 . | 2 . . |
 | h 2 . | . 4 . | 5 . . |
 +-------+-------+-------+
 | 6 . . | 5 7 a | 3 2 9 |
 | 9 7 2 | b 3 6 | 8 5 a |
 | a 3 5 | 8 9 2 | 6 b 7 |
 +-------+-------+-------+
 | . . . | 3 2 7 | 4 6 5 |
 | 5 6 3 | . 8 . | 1 7 2 |
 | 2 4 7 | 6 1 5 | 9 8 3 |
 +-------+-------+-------+

pair b8 in r1c19: -8r1c36
b in r1c19, a in r4c6: -abr1c6
=> r1c6=9
eleven
 
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Re: 17December18

Postby SpAce » Thu Dec 27, 2018 6:37 am

Very cool logic, eleven! I tried to reformulate it as an AIC, but it doesn't quite capture the elegance:

(18=4)r1c19 - r1c6&r5c9 = (4-1)r6c8 = r5c9&r6c1 - r5c4,r1c19 = (1)r4c6&(48)r1c19 => -18 r1c6; stte
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17December18

Postby Yogi » Thu Dec 27, 2018 7:59 am

Mine looks most like the Kracken Logic, a two-case verity:
1r6c1 => 1r1c3
||
4r6c1 => 4r1c3, so 9r1c6 stte
Yogi
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Re: 17December18

Postby SpAce » Thu Dec 27, 2018 8:52 am

Yogi wrote:Mine looks most like the Kracken Logic, a two-case verity:
1r6c1 => 1r1c3
||
4r6c1 => 4r1c3, so 9r1c6 stte

That works. It would be helpful to show the full paths, though. I might write it like this:

(1)r6c1* - r6c8 = r5c9^ - r5c4 = r4c6 - r1c6*1^9 = (1)r1c3
||
(4)r6c1* - r6c8 = r5c9 - r1c9*1 = (4)r1c3

=> -89 r1c3; stte
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17December18

Postby Yogi » Wed Jan 02, 2019 8:16 am

.5.26.73.....5.2...2..4.5..6..57.329972.3685..358926.7...327465563.8.172247615983
OK, taking it from here, let’s try a little translation. My solution could have been written more fully as:
1r6c1 => 1r5c9,1r4c6 => 1r1c3 (1 is now excluded from all other cells in Row1)
||
4r6c1 => 4r5c9,4r4c6 => 4r1c3 (the same exclusions now apply for candidate 4)
So r1c3 <> 9 => 9r1c6, stte. This could be read literally as meaning:
“The only two possible candidates for r6c1 both predict that same candidate in r1c3.
Therefore r1c3 cannot be 9 and so the 9 in Row1 must be at r1c6. This allows the remaining cells to be solved in singles.”
I hadn’t noticed the exclusion of 8 at r1c3, as the definite placement of 9 at r1c6 solved the puzzle for me.
SpAce described the eliminations in a chain (not sure about * and ^)
(1)r6c1* - r6c8 = r5c9^ - r5c4 = r4c6 - r1c6*1^9 = (1)r1c3
||
(4)r6c1* - r6c8 = r5c9 - r1c9*1 = (4)r1c3
=> -89 r1c3; stte. Is it possible to read this chain as a statement in English which tells the same story?
Yogi
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Re: 17December18

Postby SpAce » Wed Jan 02, 2019 11:00 am

Yogi wrote:OK, taking it from here, let’s try a little translation. My solution could have been written more fully as:
1r6c1 => 1r5c9,1r4c6 => 1r1c3
||
4r6c1 => 4r5c9,4r4c6 => 4r1c3

Almost. The latter implications are not independently correct as written because they depend on the original premise. Also, 4r4c6 is irrelevant so it should not be listed. If you insist on using the forcing chain style, I'd write it like this:

1r6c1 -> (1r6c1 & 1r5c9 & 1r4c6) -> 1r1c3 -> 9r1c6
||
4r6c1 -> (4r6c1 & 4r5c9) -> 4r1c3 -> 9r1c6

=> 9r1c6; stte

Either way, I personally think that style is hard to follow, because it jumps over the negation steps (and often others too, as in here as well).

I hadn’t noticed the exclusion of 8 at r1c3, as the definite placement of 9 at r1c6 solved the puzzle for me.

You're right that it's not necessary for the solution, but it's a good practice to list all eliminations following from a piece of logic. It wouldn't show in your style which has a placement as the final conclusion.

SpAce described the eliminations in a chain (not sure about * and ^)

They're memory markers and necessary here because the chains aren't linear (they branch, which makes them nets). I used one in my own (first) solution too. The markers are there basically for the same reason why the original premise had to be repeated in your forcing chain to make the logic work as intended.

(1)r6c1* - r6c8 = r5c9^ - r5c4 = r4c6 - r1c6*1^9 = (1)r1c3
||
(4)r6c1* - r6c8 = r5c9 - r1c9*1 = (4)r1c3
=> -89 r1c3; stte. Is it possible to read this chain as a statement in English which tells the same story?

Yes, but I don't like such plain English statements because they're long, hard to follow, and often ambiguous. Chains and nets describe the logic more concisely and accurately. Here's the same thing using a full net diagram:

Code: Select all
(1)r6c1 --------------------------------- (1)r1c1
||      \                                  ||
||       \- r6c8 = r5c9 ----------------- (1)r1c9
||                      \                  ||
||                       \- r5c4 = r4c6 - (1)r1c6
||                                         ||
||                                        (1)r1c3
(4)r6c1 ----------------- (4)r1c1
        \                  ||         
         \- r6c8 = r5c9 - (4)r1c9
                           ||
                          (4)r1c3

=> -89 r1c3; stte

That should make it apparent that the logic is actually quite complex. The way you originally presented your solution hid that fact.
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