Getting in touch with "Red" Ed Russell

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Getting in touch with "Red" Ed Russell

Postby peterchayward » Mon Dec 07, 2020 10:34 pm

Hey!

My name is Peter - I run the company which is about to publish the Cracking the Cryptic book. There’s one puzzle that we’d really like to include, which was devised by “Red” Ed Russell, who used to frequent these forums.

I was wondering if anyone had any way to get in contact with him, so we could get his permission to include the puzzle in the book.

I've been in touch with some old associates of his, but no one has any leads.

Please let me know if you know how to reach him!
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Re: Getting in touch with "Red" Ed Russell

Postby m_b_metcalf » Tue Dec 08, 2020 8:25 am

I can only suggest you try to send him a private message:

http://forum.enjoysudoku.com/member256.html

I think he lives in the UK.

HTH

Mike

P.S. If you post the puzzle here, someone can make you an essentially same copy (isomorph) with a different appearance.
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Re: Getting in touch with "Red" Ed Russell

Postby peterchayward » Tue Dec 08, 2020 9:05 am

How does one message someone? I can't see that option anywhere
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Re: Getting in touch with "Red" Ed Russell

Postby rjamil » Tue Dec 08, 2020 8:18 pm

Welcome peterchayward,

You can just click above mentioned Mike post member256 link, type subject and body message then press Submit button.

R. Jamil
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Re: Getting in touch with "Red" Ed Russell

Postby eleven » Tue Dec 08, 2020 11:13 pm

I would try to ask co-author Frazer Jarvis for a contact address:
a.f.jarvis@sheffield.ac.uk
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Re: Getting in touch with "Red" Ed Russell

Postby mith » Wed Dec 09, 2020 2:48 am

m_b_metcalf wrote:I can only suggest you try to send him a private message:

http://forum.enjoysudoku.com/member256.html

I think he lives in the UK.

HTH

Mike

P.S. If you post the puzzle here, someone can make you an essentially same copy (isomorph) with a different appearance.


There will be new puzzles in the book, but videos of the existing puzzles are already on youtube, and links will be included in the book, so this wouldn't work (and would feel a bit sketchy anyway :)).
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Re: Getting in touch with "Red" Ed Russell

Postby coloin » Wed Dec 09, 2020 10:31 am

I am not aware of many actual sudoku puzzles posted by red ed
His contributions were very much based on the theoretical combiotronics rather than solving sudoku puzzles.
He was nearly always right.
I also dont think he would care too much ..... although am sure he would be naturally pleased to get credit.
Hopefully he will be amused by this epitaph !

What cryptic puzzle was it !!!! ??
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Re: Getting in touch with "Red" Ed Russell

Postby peterchayward » Thu Dec 10, 2020 1:02 am

eleven wrote:I would try to ask co-author Frazer Jarvis for a contact address:
a.f.jarvis@sheffield.ac.uk

This was my first move! Nothing there, alas.

The puzzle in question was this one: https://www.youtube.com/watch?v=f-o2RxvdZvw

I'll send him a PM! Thanks for the suggestion
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Re: Getting in touch with "Red" Ed Russell

Postby m_b_metcalf » Thu Dec 10, 2020 8:43 am

peterchayward wrote:The puzzle in question was this one: https://www.youtube.com/watch?v=f-o2RxvdZvw

This is, indeed, a rare rating. ER 5.4 is played in only about one Patterns Game in six, and ED=5.4/5.4/3.2 in particular has been played only once, by gsf in, coincidently, Game 54.

The pattern seems to 5.4 friendly, see last puzzle.

Regards,

Mike

Code: Select all
 . . . . . . . 1 .
 2 1 . . . 3 4 8 .
 . 3 9 8 . . 2 . .
 . 6 . 3 . 4 9 . .
 . . . . . . . . .
 . . 1 6 . 7 . 4 .
 . . 8 . . 2 1 7 .
 . 2 6 7 . . . 9 8   Red Ed
 . 9 . . . . . . .   ED=5.4/5.4/3.2

  2 2  is redundant
  2 8  is redundant
  3 4  is redundant
  4 4  is redundant
  4 6  is redundant
  4 7  is redundant
  6 3  is redundant
  6 4  is redundant
  6 6  is redundant
  7 6  is redundant
  8 2  is redundant
  8 8  is redundant


Code: Select all
008090500060080030300405001003000700740000068006000300900208007070040050004070200 #  71    54 54 32 - gsf


Code: Select all
 . . . . . . . 1 .
 2 3 . . . 4 5 6 .
 . 4 1 7 . . 3 . .
 . 1 . 2 . 8 6 . .
 . . . . . . . . .
 . . 9 5 . 3 . 7 .
 . . 2 . . 1 7 8 .
 . 9 5 8 . . . 2 3  Minimal, values of givens almost symmetrical (not r4c6 r6c4).
 . 7 . . . . . . .  ED=5.4/2.3/2.3 
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Re: Getting in touch with "Red" Ed Russell

Postby coloin » Thu Dec 10, 2020 1:27 pm

m_b_metcalf wrote:
Code: Select all
  2 2  is redundant
  2 8  is redundant
  3 4  is redundant
  4 4  is redundant
  4 6  is redundant
  4 7  is redundant
  6 3  is redundant
  6 4  is redundant
  6 6  is redundant
  7 6  is redundant
  8 2  is redundant
  8 8  is redundant

best seen in this image - symetrical redundant clues - black cells in the image
red ed puzz.jpg
red ed puzz.jpg (97.13 KiB) Viewed 249 times

The automorhic qualities of the grid solution obviously attracted red ed
index of bands is 139 139 314 , 139 139 314
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Re: Getting in touch with "Red" Ed Russell

Postby eleven » Mon Dec 14, 2020 9:57 pm

btw this is the original post of the puzzle.
And some members will get a smile, how it was solved in the video.
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Re: Getting in touch with "Red" Ed Russell

Postby SpAce » Tue Dec 15, 2020 1:21 am

eleven wrote:And some members will get a smile, how it was solved in the video.

How was it solved in the video? I only watched as far as when Gurth's theorem was mentioned and r5c5 was solved. After that it got boring, as there seemed to be no further application of Gurth's (I don't know if there was later). Is there a neat way to solve this with it? Here's mine, but it's not very neat:

Code: Select all
.--------------------------.----------------------.----------------------.
|  4568-7  458-7   a57'4   | 2459   245679   569  |  3567   1     35679  |
|  2       1      af57     | 59     5679     3    |  4      8     5679   |
| b456-7   3        9      | 8     c14567   d156  |  2      56    567    |
:--------------------------+----------------------+----------------------:
|  578     6        25-7   | 3      1258     4    |  9      25    1257   |
|  345789  4578     2345-7 | 1259   12589    1589 |  35678  2356  123567 |
|  3589    58       1      | 6      2589     7    |  358    4     235    |
:--------------------------+----------------------+----------------------:
|  345     45       8      | 459    34569    2    |  1      7     3456   |
|  1345    2        6      | 7      1345    e15   | f35     9     8      |
|  13457   9        345-7  | 145    134568   1568 |  356    2356  23456  |
'--------------------------'----------------------'----------------------'

(75=4)r12c3 - r3c1 = (4-1)r3c5 = r3c6 - (1=5)r8c6 - (5==3,7)r8c7,r2c3 => +7 r12c3; 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."
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Re: Getting in touch with "Red" Ed Russell

Postby eleven » Tue Dec 15, 2020 1:23 pm

If you click to a later state, you will see, that he did not find the x-wings.

For a symmetry solution, note that the puzzle has quarter symmetry (as Red Ed mentioned).
If you turn it by 90°, and change the digits according to the cycles (1,8,9,2),(3,4,7,6), you will get the same puzzle.

So e. g. we have
Either 1r3c6 or r3c67=56 => r3c9=7 => - 7r45c9 => (symmetry) 4r3c5
In both cases r3c5 cannot be 1.

(1=567)r3c78 - r45c9 = 7r5c7 & 4r3c5 => -1r3c5, stte
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Re: Getting in touch with "Red" Ed Russell

Postby ghfick » Tue Dec 15, 2020 10:37 pm

Actually, Simon does find the X-Wings. He does not use them until he can make a placement. I thought he was quite entertaining and very honest through the whole solving. He uses the Snyder marking on the fly rather than a complete mark up of candidates before solving.
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Re: Getting in touch with "Red" Ed Russell

Postby SpAce » Wed Dec 16, 2020 12:05 am

eleven wrote:For a symmetry solution, note that the puzzle has quarter symmetry (as Red Ed mentioned).
If you turn it by 90°, and change the digits according to the cycles (1,8,9,2),(3,4,7,6), you will get the same puzzle.
...
(1=567)r3c78 - r45c9 = 7r5c7 & 4r3c5 => -1r3c5, stte

Very nice! Thanks for that. It took me a while to understand it because I didn't originally see that cyclical symmetry. So, in this case the cycle is (7,6,3,4)r5c7,r7c5,r5c3,r3c5, thus 7r5c7 -> 4r3c5? Clever. I only used the basic digit pairing in my solution. This is much nicer, and shorter too.
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