3D Medusa Help on a specific puzzle

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Re: 3D Medusa Help on a specific puzzle

Postby JasonLion » Sun Apr 14, 2013 1:25 am

David P Bird wrote:Did he change his method later or is it because different authors have interpreted the term in their own ways working not from the original but from someone else's description I wonder.
Going back and reading everything I can find from 2005, I have come to agree with you. This post seems especially telling if read closely. In the top right example of the previous post Jeff has shown an X-Cycle with two weak links and Bob then says:
Bob Hanson wrote:On the right we see that not all the connections in the chain have to be strong. We could have some "weak edges" anywhere we have "TRUE implies FALSE".
His wording isn't ideal, but the idea is clearly there.

Lower in the same post he is pushing his 3D Medusa proposal and refers back to the undated link I gave earlier, http://www.stolaf.edu/people/hansonr/sudoku/top95-analysis.htm. That pushes the date on that page back to no later than Dec 2005.

Reading only the descriptive text at his top95 analysis leads you to a Sudopedia like definition of 3D Medusa, which is a mistake I made when I mentioned the page above. But as you pointed out, working through the examples makes it quite clear that he is including weak links bridging clusters from different pairs of colors.

Curiously, the Sudopedia page was written by Rudd in Nov 2006 (as can be seen at the WayBack Machine's copy of the Sudopedia update log for the page) with no mention of bridging. Yet in Oct 2006 Rudd had a description of 3D Medusa at his own site that explained bridging clusters as part of 3D Medusa.

Summing up, two factors jump out at me. First, many people who have learned 3D Medusa after 2007 will have gone by the Sudopedia description or another site's description copied from Sudopedia. Second, back in 2005 Bob Hansen clearly understood and give examples using weak links to bridge clusters, yet if you only read his descriptive text you could easily mistake that and miss out on bridging entirely.
Last edited by JasonLion on Mon May 13, 2013 2:18 am, edited 2 times in total.
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Re: 3D Medusa Help on a specific puzzle

Postby JasonLion » Sun Apr 14, 2013 1:49 am

Going back to the puzzle that started this thread, we can indeed make some progress using 3D Medusa with bridging. Consider this coloring of the top floor of the puzzle.
MedusaPuzzle.png
MedusaPuzzle.png (8.84 KiB) Viewed 747 times

Here we have two small clusters, yellow and green on the left, blue and pink on the right. (4R13C2 is a conjugate pair.) The two clusters are bridged by 9 in row 3 and by 4 in row 1. Anything which sees both blue and green can be eliminated, clearing 4R1C46 (and 9R3C3), leaving a naked pair in R1C46, which in turn places a couple of digits with a few more simple steps.

This is actually a good example of the power of 3D Medusa. Without it, the next step is going to be something fairly complex.
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Re: 3D Medusa Help on a specific puzzle

Postby David P Bird » Sun Apr 14, 2013 10:14 am

Jason, When I replied to Richard I admit I hadn't seen (or perhaps hadn''t remembered) the Medusa elimination inference that arises from showing a combination of two colours from different colour pairs can't coexist.

Expressing this as strong and weak links between colours we get the chain:
(Yellow)r3c2 = (Green)r1c2 - (Blue=Red)r1c9

This proves that Yellow and Red can't both be false as Green and Blue can't both be true.
Any candidate weakly linked to both Yellow and Red candidates therefore must be false.
I therefore think your deduction is well reasoned, but somehow selects the wrong pair of killer colours.

Using candidates this translates to
(4)r3c2 = (4)r1c2 - (4=9)r1c9

Such translations should always be available from Medusa coloured grids to give AICs when just 2 colours pairs are involved. With more than two colour pairs, I can envisage that branched paths may be required in some instances though.

All in all I think that once a Medusa grid has been multi-coloured there's still an awful lot of work to be done to identify the eliminations.
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Re: 3D Medusa Help on a specific puzzle

Postby daj95376 » Sun Apr 14, 2013 5:34 pm

Well, I admit to "missing the boat" on Jason's coloring above. His eliminations are correct, but I can't follow his logic. So, I resorted to assuming that his eliminations were true and looked for contradictions that could be explained by his coloring. Still nothing!

Code: Select all
 after basics
 +--------------------------------------------------------------------------------+
 |  1       479     2679    |  246     8       246     |  5       3       49      |
 |  8       5       3       |  124     9       7       |  124     6       14      |
 |  29      49      269     |  12346   1235    123456  |  1249    8       7       |
 |--------------------------+--------------------------+--------------------------|
 |  459     3       589     |  1269    125     12569   |  14689   7       14569   |
 |  6       79      1       |  789     4       589     |  3       59      2       |
 |  459     2       5789    |  13679   135     13569   |  14689   1459    14569   |
 |--------------------------+--------------------------+--------------------------|
 |  7       1       259     |  23489   23      23489   |  469     2459    34569   |
 |  239     8       29      |  5       6       12349   |  7       1249    1349    |
 |  2359    6       4       |  1239    7       1239    |  19      1259    8       |
 +--------------------------------------------------------------------------------+
 # 133 eliminations remain

Code: Select all
 assume r1c4=4  =>  r1c29<>4, r3c2=4, r1c9=9, r3c27<>9 ... but no contradiction
 *--------------------------------------------------------------------*
 | 1      7      267    | 4      8      26     | 5      3      9      |
 | 8      5      3      | 12     9      7      | 124    6      14     |
 | 29     4      269    | 1236   1235   12356  | 12     8      7      |
 |----------------------+----------------------+----------------------|

Code: Select all
 assume r3c3=9  =>  r1c29<>4, r3c2=4, r1c9=9, r3c27<>9 ... but no contradiction
 *--------------------------------------------------------------------*
 | 1      7      267    | 246    8      246    | 5      3      9      |
 | 8      5      3      | 124    9      7      | 124    6      14     |
 | 2      4      9      | 1236   1235   12356  | 12     8      7      |
 |----------------------+----------------------+----------------------|

All that I could conclude: if r1c46=4 is true, then r3c13=9 is true ... and vice versa.
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Re: 3D Medusa Help on a specific puzzle

Postby David P Bird » Sun Apr 14, 2013 9:11 pm

Daj, With two pairs of colours there are 4 possible combinations one of which will be true
One of those can be eliminated to leave 3 remaining.
But if the eliminated one is considered to hold at least one truth, 66% of the time that assumption will be good and the right eliminations will be made.
Do you want to gamble or not?
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Re: 3D Medusa Help on a specific puzzle

Postby JasonLion » Mon Apr 15, 2013 1:19 am

David P Bird wrote:This proves that Yellow and Red can't both be false as Green and Blue can't both be true.
Sigh, that is at least the third mistake I have made in this topic. Yes, you are correct, and there are no eliminations from the coloring I showed. The yellow=red strong link is what I was after, but I got the colors swapped and saw the elimination I was looking for even though it isn't there. In this case that renders the pattern useless. Hopefully one can see that in some other situation it might have been more valuable.

Here is what I think is a much more powerful example (without bridging):
MedusaWrap.png
MedusaWrap.png (16.13 KiB) Viewed 718 times

In R5C7(blue) there are two candidates colored yellow, which contradicts yellow. All yellow candidates can be eliminated.

Code: Select all
93.64.1..4.187.639.7.139.42..97...1418.42..9.34.951...71329.45.8645179232953.4..1
9    3    28   | 6    4    25   | 1    78   578
4    25   1    | 8    7    25   | 6    3    9   
56   7    68   | 1    3    9    | 58   4    2   
---------------+----------------+---------------
56   25   9    | 7    68   368  | 23   1    4   
1    8    67   | 4    2    36   | 357  9    567
3    4    267  | 9    5    1    | 278  678  678
---------------+----------------+---------------
7    1    3    | 2    9    68   | 4    5    68 
8    6    4    | 5    1    7    | 9    2    3   
2    9    5    | 3    68   4    | 78   678  1

This is from my program, so hopefully no more manual mistakes. What I am less clear on is what approach someone else might use to solve this same (somewhat contrived) puzzle some other way. I also have no problem admitting that such "large" deductions from Medusa coloring are quite rare.

On the other hand, I can search/select for puzzles with entertaining moves like this one and people seem to like to play them. We sometimes forget that there are two different biased sets going on. We each have a set of techniques we like to use, and we use them on puzzles selected to exhibit specific sets of techniques. By changing the puzzle selection process, I can make techniques like 3D Medusa more interesting to use than they would be in some differently biased set of puzzles.
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Re: 3D Medusa Help on a specific puzzle

Postby David P Bird » Mon Apr 15, 2013 1:34 pm

JasonLion wrote:On the other hand, I can search/select for puzzles with entertaining moves like this one and people seem to like to play them.

Players have a fondness for Unique Rectangles as they're easy spot. In your demo puzzle there is (78)r16c89 which can be disrupted either by (8)r1c3 or (6)r6c89 which will tempt them.

(6=78)r6c89 –[UR]- (78)r1c89 = (8)r1c3 – (8=6)r3c6 => r6c7 <> 6 (& so r5c9 <> 6)

but this still leaves work to do.

Your deduction can be shown to arise from an impossible conjugate loop which proves the yellow candidates must be false.

(3')r5c7 ~ (3"~2')r4c7 ~ (2"~5')r4c2 ~ (5")r4c1 ~ (5')r3c1 ~ (5")r3c7 ~ (5')r5c7

This can be presented as an AIC by tucking (5-3)r5c7 into the body of the chain:

(5)r3c7 = (5-3)r5c7 = (3-2)r4c7 = (2-5)r4c2 = (5)r4c1 => r3c1 <> 5

What has emerged from our dialogue is confusion over the names for different colouring systems. What you're using is not 3D Medusa but what I called < Equivalence Marking > in 2006. Later that year I then produced the more advanced Graded Equivalence Marking.

Note that in the example I gave Ruud, a conjugate link though group node has been used to get from one candidate to another equivalent one, but the group node couldn't be marked.
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Re: 3D Medusa Help on a specific puzzle

Postby RichardGoodrich » Sat Apr 20, 2013 7:18 am

Gentlemen,

I had an incorrect markup on that puzzle = thinking incorrectly I could link BV cells in some way - but that does NOT compute!
Since then I have studied Bob Hanson's explanations and others and I now see that to get opposite colors there must be ONLY one other presence of the digit in question in the house whether it is BV or Bi-Location. I started working on another Unfair HoDuKu #1692 and what I did was to deliberately try to solve it after some basic "slicing and dicing" Using HoDoKu's coloring of the presence of various digits. There were still hidden subsets in it and such. Medusa 3D helped me discover some of them and other digit eliminations and then I tried to progress to more chain colors when I discovered the Sudoku Bridge. I have NOT seen a lot on that subject, but your discussion has helped immensely. I will work on it some more and post my results. I used Bob Hansen's Sudoku Assistant to analyze it and I was not quite sure of his notation. He had a table of strong links that showed 21 strong links. I was not sure how to relate that to a 3D Medusa Chain notation. I have read one place that the Eureka Notation (DIC's) Double Implication Chain was a way to document 3D Medusa, but that seemed TOO restrictive to me. Well I am up late and need to crash!

I will post my work after I have a chance to work on it tomorrow and I will eventually come back to this puzzle and undo my bad assumptions and see what happens!

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Re: 3D Medusa Help on a specific puzzle

Postby David P Bird » Wed May 01, 2013 9:41 am

Richard, here is the puzzle you published on your blog

Preliminary steps in brief:
Hidden and naked singles assigned
In box 5 (18)r56c4 & (25)r45c6 are hidden pairs
Box/line eliminations (1)r3c3,r5c9, (2)r7c7, (7)r13c2
(2) becomes single in r7c9
(17)r3c18 becomes a hidden pair

..8..1..3..5.87..4...52.8.6.1....6.8..6...4..4.7....2.6.3.18...9..65.3..8..2..5..
Code: Select all
 *-------------------*-------------------*-------------------*
 | 2"7'  2/9\  <8>   | 4     6     <1>   | 27"9/ 5     <3>   |
 | 1"23/ 6     <5>   | 3\9/  <8>   <7>   | 129   19    <4>   |
 | 1'7"  3\49  49    | <5>   <2>   3/9\  | <8>   1"7'  <6>   |
 *-------------------*-------------------*-------------------*
 | 235   <1>   29    | 3/7'9 4     25    | <6>   37"9  <8>   |
 | 235   2389  <6>   | 18    37"9  25    | <4>   1379  7'9"  |
 | <4>   389   <7>   | 18    39    6     | 19    <2>   5     |
 *-------------------*-------------------*-------------------*
 | <6>   5     <3>   | 7"9'  <1>   <8>   | 7'9"  4     2     |
 | <9>   27    12    | <6>   <5>   4     | <3>   8     17    |
 | <8>   47    14    | <2>   3/7'9 3\9/  | <5>   6     179'  |
 *-------------------*-------------------*-------------------*

The pencil marked grid shows two pairs of Medusa colours
(') and (") used as 1st pair of colours from (2"7')r1c1
(\) and (/) used as 2nd pair from (3\9/)r2c4
Note that one of (9)r2c78 must carry a (\) colour, although which one it will be isn't known, (9/)r1c7 can be marked as it's the only other 9 in box 3

We now know that (7"9/)r1c7 can't be true together and nor can (3/7')r4c4.
One of (7') and (7") must be true so all the digits coloured (/) are false.
A Eureka AIC to prove (9)r1c7 is false is
(7)r1c7 = (7)r3c8 - (7)r3c1 - (7=2)r1c1 - (2=9)r1c2 => r1c7 <> 9
[Added That's wrong, it should have been:
(7)r1c7 = (7-9)r7c7 = (9)r7c4 - (9)r2c5 = (9)r2c78 => r2c7 <> 9
A longer alternative that avoids the group node (9)r2c78 is:
(1)r2c1 = (1-7)r3c1 = (7)r1c1 - (7)r1c8 = (7)r8c7 - (7)r8c4 = (7-3)r4c4 = (3)r2c4 => r2c1 <> 3 ]


The = symbols link two Booleans that must contain a true condition, and the - symbols link two Booleans that must contain a false condition.
To check out how this works follow the links from left to right considering (7)r1c7 to be false. This shows that then (9)r1c2 must be true. If it is now considered false and the chain is followed backwards it will prove that then (7)r1c7 must be true. They therefore can't both be false, and as they are both incompatible with (9)r1c7, it must be false.

I would recommend learning the Eureka notation as it's the most commonly used one and is easier to error check.

I've ignored a couple of fish eliminations that are also available, and I leave you to read up on these for yourself.

As a comment, it appears you're working backwards from the more difficult techniques towards the simpler ones :!:
Last edited by David P Bird on Sun May 12, 2013 3:39 pm, edited 1 time in total.
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Re: 3D Medusa Help on a specific puzzle

Postby RichardGoodrich » Fri May 03, 2013 3:17 am

JasonLion wrote:Sigh, I keep mixing up several coloring techniques that don't have names (that I have been exploring recently) with 3D Medusa. This is the second time I made that mistake. Sorry.

What I colored is (hopefully) an AIC net represented with coloring. It has a great deal in common with 3D Medusa, but the elimination rules aren't really the same.


Jason,

I think AIC and 3D Medusa are exactly the same thing! That is my current understanding. What I don't get is the assistance on Eureka notation.

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Re: 3D Medusa Help on a specific puzzle

Postby JasonLion » Fri May 03, 2013 2:13 pm

First, keep in mind that 3D Medusa has two or three inter-related definitions, which can confuse things.

In simple 3D Medusa (one pair of colors) all links are conjugate pairs, thus all links are both strong and weak. Not all AIC can be represented in this format, since AIC can include links which are only weak. All 3D Medusa eliminations of this kind can be converted to one or more AICs. Color traps are fairly straightforward AIC, but Color wraps (contradicting one color of a pair) generally require multiple AIC to perform all of the eliminations.

In 3D Medusa with bridging (two or more pairs of colors) the bridges are weak links, while all other links are both strong and weak. This covers significantly more AIC chains, but still not all of them. Depending on the flavor of AIC you are using, AIC can include strong links that are not also weak links. AIC can also include grouped cells, which 3D Medusa has no way of representing. Again, all 3D Medusa eliminations of this kind can be represented as AIC eliminations (perhaps with multiple chains).

A couple of good places to start learning Eureka notation are:
http://www.sudocue.net/eureka-notation.php
http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=6708
http://sudopedia.enjoysudoku.com/Eureka.html

Do you have a more specific question about Eureka notation?
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Re: 3D Medusa Help on a specific puzzle

Postby daj95376 » Sat May 04, 2013 8:28 am

I'm having a problem with the claim that 3D Medusa can be expressed as an AIC. Consider the W-Wing in the grid below.

Code: Select all
 Here's the W-Wing marked with Blue/Green for simple 3D Medusa coloring
 +--------------------------------------------------------------+
 |  5     1     3     |  7     24    8     |  2469  269   269   |
 |  2     7     4     |  9     6     3     |  8     1     5     |
 |  8     9     6     |  5     24    1     |  24    3     7     |
 |--------------------+--------------------+--------------------|
 |  4    B23G   7     |  8     1     6-2   |  2369  5     2369  |
 | B39    5     29    |  263G  7     4     |  1     26    8     |
 |  1     6     8     | G23B   5     9     |  7     4     23    |
 |--------------------+--------------------+--------------------|
 |  3679  23    29    |  1     8     67    |  5     679   4     |
 |  679   8     1     |  4     39    5     |  2369  2679  2369  |
 |  679   4     5     |  26    39    267   |  369   8     1     |
 +--------------------------------------------------------------+
 # 53 eliminations remain

 Blue/Green on 2:3 in bivalue    cell  r4c2
 Green/Blue on   3 in bilocation cells r4c2:r5c1
 Blue/Green on   3 in bilocation cells      r5c1:r5c4
 Green/Blue on   3 in bilocation cells           r5c4:r6c4
 Blue/Green on 3:2 in bivalue    cell                 r6c4

Code: Select all
 Here's the W-Wing marked as an AIC
 +--------------------------------------------------------------+
 |  5     1     3     |  7     24    8     |  2469  269   269   |
 |  2     7     4     |  9     6     3     |  8     1     5     |
 |  8     9     6     |  5     24    1     |  24    3     7     |
 |--------------------+--------------------+--------------------|
 |  4    a23    7     |  8     1     6-2   |  2369  5     2369  |
 | b39    5     29    | c236   7     4     |  1     26    8     |
 |  1     6     8     | d23    5     9     |  7     4     23    |
 |--------------------+--------------------+--------------------|
 |  3679  23    29    |  1     8     67    |  5     679   4     |
 |  679   8     1     |  4     39    5     |  2369  2679  2369  |
 |  679   4     5     |  26    39    267   |  369   8     1     |
 +--------------------------------------------------------------+
 # 53 eliminations remain

 (2=3)r4c2 - (3)r5c1 = (3)r5c4 - (3=2)r6c4  =>  r4c6<>2

There's nothing in the AIC notation that relays the fact that the weak inferences are also strong links. IOW, the AIC is not really expressing all of the information present/necessary in the 3D Medusa coloring. Thus, I don't see it as equivalent.

Ironically, this W-Wing matches Keith's original definition where every inference is supported by a strong link. (He was big on strong links!)
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Re: 3D Medusa Help on a specific puzzle

Postby David P Bird » Sat May 04, 2013 9:23 am

daj, by calling "Equivalence Marking" "Simple Medusa" you're perpetuating a misnomer, but I'm getting to the point that I'm past caring.

Equivalence Marking follows conjugate chains not AICs. It just happens that using alternating linking symbols helps follow which are the odd nodes and which are the even ones - ie the ones which must all be true or false together. In comparison in an AIC two consecutive nodes can both be false when they're weakly linked or both be true if they're strongly linked.

You're mixing up sheep and goats.
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Re: 3D Medusa Help on a specific puzzle

Postby JasonLion » Sat May 04, 2013 11:55 am

A 3D Medusa taken as a whole can not be expressed as an AIC. However, any specific elimination found by 3D Medusa can always be expressed as an AIC. Information is lost in the process of converting the 3D Medusa expression of an elimination into the AIC form, since AIC only represent links that are either strong or weak, while 3D Medusa works with conjugate (strong&weak) links and links that are only weak. None the less, an AIC can always be selected so that the information that is lost is not relevant to the specific elimination being represented.
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Re: 3D Medusa Help on a specific puzzle

Postby daj95376 » Sat May 11, 2013 7:52 pm

David P Bird wrote:Richard, here is the puzzle you published on your blog

Preliminary steps in brief:
Hidden and naked singles assigned
In box 5 (18)r56c4 & (25)r45c6 are hidden pairs
Box/line eliminations (1)r3c3,r5c9, (2)r7c7, (7)r13c2
(2) becomes single in r7c9
(17)r3c18 becomes a hidden pair

..8..1..3..5.87..4...52.8.6.1....6.8..6...4..4.7....2.6.3.18...9..65.3..8..2..5..
Code: Select all
 *-------------------*-------------------*-------------------*
 | 2"7'  2/9\  <8>   | 4     6     <1>   | 27"9/ 5     <3>   |
 | 1"23/ 6     <5>   | 3\9/  <8>   <7>   | 129   19    <4>   |
 | 1'7"  3\49  49    | <5>   <2>   3/9\  | <8>   1"7'  <6>   |
 *-------------------*-------------------*-------------------*
 | 235   <1>   29    | 3/7'9 4     25    | <6>   37"9  <8>   |
 | 235   2389  <6>   | 18    37"9  25    | <4>   1379  7'9"  |
 | <4>   389   <7>   | 18    39    6     | 19    <2>   5     |
 *-------------------*-------------------*-------------------*
 | <6>   5     <3>   | 7"9'  <1>   <8>   | 7'9"  4     2     |
 | <9>   27    12    | <6>   <5>   4     | <3>   8     17    |
 | <8>   47    14    | <2>   3/7'9 3\9/  | <5>   6     179'  |
 *-------------------*-------------------*-------------------*

The pencil marked grid shows two pairs of Medusa colours
(') and (") used as 1st pair of colours from (2"7')r1c1
(\) and (/) used as 2nd pair from (3\9/)r2c4
Note that one of (9)r2c78 must carry a (\) colour, although which one it will be isn't known, (9/)r1c7 can be marked as it's the only other 9 in box 3

We now know that (7"9/)r1c7 can't be true together and nor can (3/7')r4c4.
One of (7') and (7") must be true so all the digits coloured (/) are false.
A Eureka AIC to prove (9)r1c7 is false is
(7)r1c7 = (7)r3c8 - (7)r3c1 - (7=2)r1c1 - (2=9)r1c2 => r1c7 <> 9

The = symbols link two Booleans that must contain a true condition, and the - symbols link two Booleans that must contain a false condition.
To check out how this works follow the links from left to right considering (7)r1c7 to be false. This shows that then (9)r1c2 must be true. If it is now considered false and the chain is followed backwards it will prove that then (7)r1c7 must be true. They therefore can't both be false, and as they are both incompatible with (9)r1c7, it must be false.

I would recommend learning the Eureka notation as it's the most commonly used one and is easier to error check.

I've ignored a couple of fish eliminations that are also available, and I leave you to read up on these for yourself.

As a comment, it appears you're working backwards from the more difficult techniques towards the simpler ones :!:

I think you should review your AIC chain !!!
daj95376
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