December 10, 2019

Post puzzles for others to solve here.

December 10, 2019

Postby ArkieTech » Tue Dec 10, 2019 11:41 am

Code: Select all
 *-----------*
 |...|...|.2.|
 |2..|..6|1.3|
 |.5.|4..|...|
 |---+---+---|
 |1..|.7.|.9.|
 |.8.|925|.1.|
 |.4.|.1.|..5|
 |---+---+---|
 |...|..1|.6.|
 |6.4|5..|..9|
 |.7.|...|...|
 *-----------*

.......2.2....61.3.5.4.....1...7..9..8.925.1..4..1...5.....1.6.6.45....9.7.......


Play/Print this puzzle online
dan
User avatar
ArkieTech
 
Posts: 3355
Joined: 29 May 2006
Location: NW Arkansas USA

Re: December 10, 2019

Postby SpAce » Tue Dec 10, 2019 1:48 pm

Code: Select all
.-----------------.----------------.--------------------.
| 4    36  b368#7 | 1     9   378* |  5        2    78* |
| 2    9    78    | 78*   5   6    |  1        4    3   |
| 378  5    1     | 4     38  2    |  9        78   6   |
:-----------------+----------------+--------------------:
| 1    36   5     | 368   7   4    |  368      9    2   |
| 37   8   b367   | 9     2   5    | c6(3)     1    4   |
| 9    4    2     | 368   1   38   |  3678     378  5   |
:-----------------+----------------+--------------------:
| 5    2    9     | 378*  4   1    | a8[#7]-3  6    78* |
| 6    1    4     | 5     38  378  |  2        378  9   |
| 38   7    38    | 2     6   9    |  4        5    1   |
'-----------------'----------------'--------------------'

5-link Oddagon+2 (7)r17,c49,b2

(7)r7c7 == (7,6)r15c3 - (6=3)r5c7 => -3 r7c7; stte
-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."
User avatar
SpAce
 
Posts: 2671
Joined: 22 May 2017

Re: December 10, 2019

Postby Cenoman » Tue Dec 10, 2019 11:37 pm

SpAce's single step is unrivaled !
So, why not two steps of simple coloring:
Code: Select all
 +--------------------+-------------------+--------------------+
 |  4     36   3678   |  1     9    378*  |  5      2     78*  |
 |  2     9    78     |  78    5    6     |  1      4     3    |
 |  378   5    1      |  4     38   2     |  9      78    6    |
 +--------------------+-------------------+--------------------+
 |  1     36   5      |  368   7    4     |  368    9     2    |
 |  37    8    367    |  9     2    5     |  36     1     4    |
 |  9     4    2      |  368   1    38    |  3678   378   5    |
 +--------------------+-------------------+--------------------+
 |  5     2    9      |  38-7  4    1     |  378    6     78*  |
 |  6     1    4      |  5     38   378*  |  2      38-7  9    |
 |  38    7    38     |  2     6    9     |  4      5     1    |
 +--------------------+-------------------+--------------------+

1. (7)r8c6 = r1c6 - r1c9 = r7c9 => -7 r7c4, r8c8

Code: Select all
 +-----------------+------------------+--------------------+
 |  4    36*  67   |  1     9    38*  |  5      2     78   |
 |  2    9    8    |  7     5    6    |  1      4     3    |
 |  37   5    1    |  4     38   2    |  9      78    6    |
 +-----------------+------------------+--------------------+
 |  1    36*  5    |  68-3  7    4    |  368    9     2    |
 |  37   8    67   |  9     2    5    |  36     1     4    |
 |  9    4    2    |  68-3  1    38*  |  3678   378*  5    |
 +-----------------+------------------+--------------------+
 |  5    2    9    |  38*   4    1    |  378*   6     78   |
 |  6    1    4    |  5     38   7    |  2      38*   9    |
 |  8    7    3    |  2     6    9    |  4      5     1    |
 +-----------------+------------------+--------------------+

2. (3)r7c4 = r7c7 - r8c8 = r6c8^ - r6c6 = r1c6 - r1c2 = r4c2 => -3 r6c4^, -3 r4c4; ste

Maybe could it be written as a one stepper, considering the second step as an almost SC of the 3s at the starting point ??
[(3)r7c4 = r7c7 - r8c8 = r6c8^ - r6c6# = r1c6 - r1c2 = r4c2] = (3)r8c6 - [(7)r8c6 = r1c6 - r1c9 = r7c9 - r7c4 = (7)r8c6] => -3 r6c4^, -3 r4c4; ste
So bad if no one likes that...
Cenoman
Cenoman
 
Posts: 2975
Joined: 21 November 2016
Location: France

Re: December 10, 2019

Postby SpAce » Wed Dec 11, 2019 3:22 pm

Hi Cenoman!

Cenoman wrote:SpAce's single step is unrivaled !

Thanks! I didn't even realize how lucky that was until I started wondering why no other solutions showed up. Looks like it was the only easy one-stepper available. (There's one with a bivalue oddagon too, but it's pretty much the same).

What follows is my typical nitpicking... Hope you don't mind! :)

So, why not two steps of simple coloring

I like your steps but I'd rather call them X-Chains as presented. I know there's historical variance, but the de facto standard of Simple Coloring considers only a single cluster of conjugate pairs, i.e. situations where you'd use just one tag pair in full tagging. I guess that's one reason why it's called "Simple", the other being that it uses just single digits.

Thus (standard) Simple Coloring is not a synonym for all X-Chains, and I think it's best kept that way as we don't really need two names for exactly the same thing. In this case it's available for the 3s in the second step but not for the 7s in the first. If you want to present it as a standard coloring move (instead of a Skyscraper), you could use Multi-Coloring with two coloring clusters (tag pairs aA and bB):

Code: Select all
.-----------------.------------------.------------------.
|  4    36   3678 |  1     9   b38-7 | 5      2     a78 |
|  2    9   b8-7  | B7-8   5    6    | 1      4      3  |
| a378  5    1    |  4     38   2    | 9     A78     6  |
:-----------------+------------------+------------------:
|  1    36   5    |  368   7    4    | 368    9      2  |
| A37   8   a367  |  9     2    5    | 36     1      4  |
|  9    4    2    |  368   1    38   | 3678   378    5  |
:-----------------+------------------+------------------:
|  5    2    9    | b38-7  4    1    | 378    6     A78 |
|  6    1    4    |  5     38  B7-38 | 2     b38-7   9  |
|  38   7    38   |  2     6    9    | 4      5      1  |
'-----------------'------------------'------------------'

Step 1. Multi-Coloring (7s): b sees both a and A => b false, B true => +7 r2c4,r8c6; -7 r1c6,r2c3,r7c4,r8c8

The second step can be done with Simple Coloring (one tag pair, aA):

Code: Select all
.----------------.-------------------.-------------------.
|  4    a3-6  67 |  1      9    A8-3 |  5       2     78 |
|  2     9    8  |  7      5     6   |  1       4     3  |
| A7-3   5    1  |  4     a3-8   2   |  9       78    6  |
:----------------+-------------------+-------------------:
|  1    A6-3  5  |  68-3   7     4   |  368     9     2  |
| a3-7   8    67 |  9      2     5   | A6-3     1     4  |
|  9     4    2  |  68-3   1    a3-8 |  678-3  A78-3  5  |
:----------------+-------------------+-------------------:
|  5     2    9  | a3-8    4     1   | A78-3    6     78 |
|  6     1    4  |  5     A8-3   7   |  2      a3-8   9  |
|  8     7    3  |  2      6     9   |  4       5     1  |
'----------------'-------------------'-------------------'

Step 2. Simple Coloring (3s):

Wrap: duplicate 'A's (contradictions) in c7, b6 => A false (7 elims), a true (6 placements)
Traps (see both a and A): -3 r4c46, r6c7

Maybe could it be written as a one stepper, considering the second step as an almost SC of the 3s at the starting point ??
[(3)r7c4 = r7c7 - r8c8 = r6c8^ - r6c6# = r1c6 - r1c2 = r4c2] = (3)r8c6 - [(7)r8c6 = r1c6 - r1c9 = r7c9 - r7c4 = (7)r8c6] => -3 r6c4^, -3 r4c4; ste

I think that works, but the chain should be appended with "= contradiction". Otherwise it ends with a weak link, and the conclusion is hard to understand. Another way to make it more understandable (at least for me) is to reverse it:

[(7)r8c6 == (7)r8c6] - (3)r8c6 = [(3)r7c4 == (3)r6c8&r4c2] => +7 r8c6, -3 r46c4; stte

However, in that case I still see it effectively as two steps, because it starts with an independently valid AIC (resulting in +7r8c6), followed by a weak link that (truly) kills 3r8c6, resulting in the second AIC whose conclusion is what we really want. Thus, it's not really a normal single AIC with a derived strong link between two end points (or it is, but the first end point is an implied contradiction). Unfortunately I can't see another way to do it except with redundant backpedaling:

[(3)r7c4 == (3)r6c8&r4c2] = (3-7)r8c6 == (7-3)r8c6 = [(3)r7c4 == (3)r6c8&r4c2] => -3 r46c4; stte
User avatar
SpAce
 
Posts: 2671
Joined: 22 May 2017

Re: December 10, 2019

Postby Cenoman » Wed Dec 11, 2019 10:50 pm

Hi SpAce,

No objection to your comments.
X-chain rather than simple coloring: my mistake.
As regards the trial chain you state:
I think that works, but the chain should be appended with "= contradiction". Otherwise it ends with a weak link...

Yes, it ends (or starts in your version) with a weak link, on one side of which there is a node equal to True.
I saw it as a symmetry of a strong link, on one side of which there is a node equal to False (e.g. a DP), that is commonly practised.

But ok, let's forget it. Anyhow, it is nothing but disguised two-steps...
SpAce wrote:Looks like it was the only easy one-stepper available.

I preferred the two-stepper, but in one step, ste finish, there was:
Almost kite (*) and almost M-wing (#), having a common elimination and conflicting spoilers:
Code: Select all
 +--------------------+-------------------+--------------------+
 |  4     36  b3678   |  1     9  ia378*# |  5      2   ia78*# |
 |  2     9    78     | a78*   5    6     |  1      4     3    |
 |  378   5    1      |  4     38   2     |  9      78    6    |
 +--------------------+-------------------+--------------------+
 |  1     36   5      | g368   7    4     |  368    9     2    |
 |  37    8   c367    |  9     2    5     | d36     1     4    |
 |  9     4    2      | g368   1   h38    |  3678   378   5    |
 +--------------------+-------------------+--------------------+
 |  5     2    9      |fa378*  4    1     | e378    6     8-7  |
 |  6     1    4      |  5     38  i378#  |  2     i378#  9    |
 |  38    7    38     |  2     6    9     |  4      5     1    |
 +--------------------+-------------------+--------------------+

[(7)r7c4 = r2c4 - r1c6 = r1c9] = (7-6)r1c3 = r5c3 - (6=3)r5c7 - r7c7 = r7c4 - r46c4 = (3-8)r6c6 = [(7=8)r1c9 - r1c6 = (8-7)r8c6 = (7)r8c8] => -7 r7c9; ste
As a net:
Hidden Text: Show
Code: Select all
(7)r1c9@
 ||
(7)r1c6-r2c4=(7)r7c4@
 ||
(7-6)r1c3 = r5c3 - (6=3)r5c7 - r7c7 = r7c4 - r46c4 = (3-8)r6c6
                                                        ||
                                                       (8)r1c6 - (8=7)r1c9@
                                                        ||
                                                       (8-7)r8c6 = (7)r8c8@
------------
=> -7 r7c9; ste
Cenoman
Cenoman
 
Posts: 2975
Joined: 21 November 2016
Location: France

Re: December 10, 2019

Postby SpAce » Thu Dec 12, 2019 2:42 am

Hi Cenoman,

Cenoman wrote:X-chain rather than simple coloring: my mistake.

I wouldn't call it a mistake. Like I said, there are multiple definitions for Simple Coloring, so there's no single truth here. (Besides, you wrote it in lower case so it could mean any coloring technique that is simple! :) ). I just personally think the Sudopedia definition (also used by Hodoku, SudokuWiki, etc) is the most useful one, because it's the most restrictive. That way it lives up to its name (simple) and leaves room for more powerful coloring methods with different names. I like tight definitions and clear boundaries because they promote the least ambiguous communication.

As regards the trial chain you state:
I think that works, but the chain should be appended with "= contradiction". Otherwise it ends with a weak link...

Yes, it ends (or starts in your version) with a weak link, on one side of which there is a node equal to True.
I saw it as a symmetry of a strong link, on one side of which there is a node equal to False (e.g. a DP), that is commonly practised.

That's what I thought you meant. It just took me a while to figure out, which is why I thought it would probably be clearer if you stated the falsehood of the right hand side explicitly. That way it can be read as a normal AIC with a derived strong link between the left hand end point (the nested X-Chain of 3s) and a contradiction at the other end point, forcing the former to be true and the conclusion valid.

Generally I like all AICs to start and end with a strong link because then all of them can be interpreted the same way, even when used as a contradiction chain (i.e. one end point is a known false). I have to spend extra brain cycles if there's a deviation from that norm :) The one deviation I've learned to like is Dan's style, which I've dubbed "compact kraken". It starts with a known truth and a weak link (e.g. your chain reversed), making it easy to read because the truth state of every node is exactly what the preceding link implies. It works nicely when the alternative is a much longer and redundant backpedaling AIC (as in this case).

I preferred the two-stepper, but in one step, ste finish, there was:
Almost kite (*) and almost M-wing (#), having a common elimination and conflicting spoilers:

Very nice! I like that!
User avatar
SpAce
 
Posts: 2671
Joined: 22 May 2017


Return to Puzzles

cron