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ESYudkowsky's profile
Eliezer Yudkowsky
Eliezer Yudkowsky
Eliezer Yudkowsky
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@ESYudkowsky

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Eliezer YudkowskyVerified account

@ESYudkowsky

Ours is the era of inadequate AI alignment theory. Any other facts about this era are relatively unimportant, but sometimes I tweet about them anyway.

Joined June 2014

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    Eliezer Yudkowsky‏Verified account @ESYudkowsky 13 Nov 2017

    Nerdsnipe: Consider the 4x4 analogue of Sudoku. Treat permutations of the 4 symbols as equivalent, and rotations and reflections as equivalent. Try to solve this problem *entirely in your head* (e.g. without pencil and paper): Is there more than one distinct solution?

    6:05 PM - 13 Nov 2017
    • 5 Retweets
    • 13 Likes
    • Deniz Yuret Paul Pajo #Insulin4All {#HODL} Irving Rivas Torsten Arndt Andrey Kormilitzin Mario⚡️Gibney mere_mortise bruce falck Nick Peck
    7 replies 5 retweets 13 likes
      1. New conversation
      2. Harry Altman‏ @Sniffnoy 13 Nov 2017
        Replying to @ESYudkowsky

        Btw, the transforms you list aren't the only maps preserving the puzzle structure. You should allow more equivalences IMO. :)

        1 reply 0 retweets 0 likes
      3. Harry Altman‏ @Sniffnoy 13 Nov 2017
        Replying to @Sniffnoy @ESYudkowsky

        As best I can tell the problem's full symmetry group is of order 3072, not a mere 192 :)

        1 reply 0 retweets 0 likes
      4. Eliezer Yudkowsky‏Verified account @ESYudkowsky 13 Nov 2017
        Replying to @Sniffnoy

        Hm. *Thinks.* Still working in my head, I believe my proof generalizes, & there remains more than one solution taking into account the extra 4 symmetries and 16x isomorphisms.

        1 reply 0 retweets 0 likes
      5. Harry Altman‏ @Sniffnoy 13 Nov 2017
        Replying to @ESYudkowsky

        I can confirm there are at least 2 distinct solutions even with this (dunno if there are more). Did that with pencil and paper though!

        1 reply 0 retweets 0 likes
      6. Harry Altman‏ @Sniffnoy 13 Nov 2017
        Replying to @Sniffnoy @ESYudkowsky

        I guess it was a pretty successful nerdsnipe after all, even if not as originally intended...

        0 replies 0 retweets 0 likes
      7. End of conversation
      1. New conversation
      2. plutoo‏ @qlutoo 13 Nov 2017
        Replying to @ESYudkowsky

        Take solution where n:th row is 1 2 3 4 cyclicly shifted by n. Observe every row and column is either increasing/decreasing by 1 (property A). Now swap first two columns. All columns still satisfy A. Rows no longer has satsify A. Rotation/reflection maintain the A property. 1/

        2 replies 0 retweets 2 likes
      3. plutoo‏ @qlutoo 13 Nov 2017
        Replying to @qlutoo @ESYudkowsky

        You can't "fix" the A property by permutation because if you fix row, column will break. Thus we have found at least two solutions. [I've yet to verify on paper] 2/

        1 reply 0 retweets 2 likes
      4. plutoo‏ @qlutoo 13 Nov 2017
        Replying to @qlutoo @ESYudkowsky

        lgtm

        1 reply 0 retweets 0 likes
      5. halloween is dumb except for candy‏ @7inchesofAUTISM 13 Nov 2017
        Replying to @qlutoo @ESYudkowsky

        Well? I'm curious. Is that correct?

        1 reply 0 retweets 0 likes
      6. plutoo‏ @qlutoo 13 Nov 2017
        Replying to @7inchesofAUTISM @ESYudkowsky

        Think so

        1 reply 0 retweets 0 likes
      7. halloween is dumb except for candy‏ @7inchesofAUTISM 13 Nov 2017
        Replying to @qlutoo @ESYudkowsky

        I just went by "analogue of Sudoku". If you have 1234 going counterclockwise around a 4×4 grid and spiraling inward, you can reverse that spiral to be clockwise and still solve the puzzle.

        1 reply 0 retweets 0 likes
      8. plutoo‏ @qlutoo 13 Nov 2017
        Replying to @7inchesofAUTISM @ESYudkowsky

        Beautiful. Rotation of a spiral remains a spiral. And reflection of a spiral remains a spiral. And symbol permutation of a spiral remains a spiral. Now if you can find a solution that's not a spiral you have proven there are at least two distinct solutions. :)

        2 replies 0 retweets 0 likes
      9. 2 more replies
      1. New conversation
      2. A. Ham‏ @mysterylover321 13 Nov 2017
        Replying to @ESYudkowsky

        Doesn't "treat permutations of the 4 symbols as equivalent" mean every solution is equivalent by definition?

        1 reply 0 retweets 0 likes
      3. Arthur MILCHIOR‏ @ArthurMilchior 13 Nov 2017
        Replying to @mysterylover321 @ESYudkowsky

        No. On a line "abab" would be equivalent to "baba", but not to "baab", so permutation does not means everything is equivalent to everything.

        1 reply 0 retweets 1 like
      4. A. Ham‏ @mysterylover321 13 Nov 2017
        Replying to @ArthurMilchior @ESYudkowsky

        Don't you need 4 *unique* symbols per row/column, i.e. "abcd" not "abab"?

        1 reply 0 retweets 0 likes
      5. Arthur MILCHIOR‏ @ArthurMilchior 13 Nov 2017
        Replying to @mysterylover321 @ESYudkowsky

        I was trying to illustrate permutation and not sudoku here. Better exampe abcd bcda is a permutation of bcda cdab (a always sent to b) But not of cdab dabc Since here a is sent to b and c on 1st and 2 nd line

        0 replies 0 retweets 1 like
      6. End of conversation
      1. Harry Altman‏ @Sniffnoy 13 Nov 2017
        Replying to @ESYudkowsky

        That's just not much of a nerdsnipe because I just quickly conclude I can't do that in my head.

        0 replies 0 retweets 1 like
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      1. Subbak‏ @Subb4k 14 Nov 2017
        Replying to @ESYudkowsky

        Yes. There is one solution where a symbol is never diagonally adjacent to itself and one where it always is (to 2 other copies). There are other solutions but harder to compute in your head.

        0 replies 0 retweets 0 likes
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      1. New conversation
      2. Torsten Arndt‏ @ivanpoweronoff 14 Nov 2017
        Replying to @ESYudkowsky

        You sniped me well! Head: def. more than one. Paper: I know the number. I wont spoil. Pic gives solution. Thanks Eliezer!pic.twitter.com/wqftUOM4ha

        1 reply 0 retweets 0 likes
      3. Torsten Arndt‏ @ivanpoweronoff 14 Nov 2017
        Replying to @ivanpoweronoff @ESYudkowsky

        Well. Made a mistake. Heres the correct one.pic.twitter.com/BGSbN8VJG4

        0 replies 0 retweets 0 likes
      4. End of conversation

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