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oe1cxw's profile
Claire Xen 🏳️‍⚧️🏳️‍🌈🧙🏻‍♀️ BLM 🏴🚩
Claire Xen 🏳️‍⚧️🏳️‍🌈🧙🏻‍♀️ BLM 🏴🚩
Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩
@oe1cxw

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Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩

@oe1cxw

Neurodiverse trans geek girl. Yosys, RISC-V, SAT/SMT.

She/her/hers
clairexen.net
Joined September 2014

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    1. Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩‏ @oe1cxw 19 Apr 2019
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      Given a set of unsigned integers. How can I find the bit-permutation so that the maximum number is minimal?

      3 replies 1 retweet 3 likes
    2. Pepijn de Vos‏ @pepijndevos 19 Apr 2019
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      Replying to @oe1cxw

      Take all the one bits and store them breadth first starting from lsb. Maximum is 2^(ceil(onebits/integers)+1)-1 I think

      1 reply 0 retweets 1 like
    3. Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩‏ @oe1cxw 19 Apr 2019
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      Replying to @pepijndevos

      Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩 Retweeted Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩

      What is a "one bit"? A position where there is at least one bit set? So what if there are 0s and 1s at each position? See also this tweet:https://twitter.com/oe1cxw/status/1119255354829832194 …

      Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩 added,

      Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩 @oe1cxw
      Replying to @tedyapo
      That method would suggest that for the set 001, 100, 110 the order does not matter, because there are zeros and ones in each position. But obviously the order does matter, because swapping the middle and LSB bit will reduce the max from 110 to 101.
      1 reply 0 retweets 0 likes
    4. Pepijn de Vos‏ @pepijndevos 19 Apr 2019
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      Replying to @oe1cxw

      Ah, I misunderstood the problem. You want to apply the same permutation to all integers. The other case is more trivial. I'll think about it a bit more.

      1 reply 0 retweets 0 likes
    5. Pepijn de Vos‏ @pepijndevos 19 Apr 2019
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      Replying to @pepijndevos @oe1cxw

      Obviously start with msb zero columns, then take the column with the least ones. The one rows are the "danger set", pick the column with the least ones in the danger set. Intersect the danger set and repeat. At every column, everything outside the danger set is lower.

      1 reply 0 retweets 0 likes
      Claire Xen  🏳️‍⚧️ 🏳️‍🌈 🧙🏻‍♀️ BLM  🏴 🚩‏ @oe1cxw 19 Apr 2019
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      Replying to @pepijndevos

      Ad column with least ones: In the example above the two columns that must be swapped have the same number of ones in them.

      8:54 AM - 19 Apr 2019
      1 reply 0 retweets 0 likes
        1. New conversation
        2. Pepijn de Vos‏ @pepijndevos 19 Apr 2019
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          Replying to @oe1cxw

          So you'd have to recursively try them all. If on average your danger set halves with each column, I think you still get an acceptable time complexity. BRB, writing code... *nerd sniped*

          1 reply 0 retweets 0 likes
        3. Pepijn de Vos‏ @pepijndevos 20 Apr 2019
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          Replying to @pepijndevos @oe1cxw

          This is what I was going for https://gist.github.com/pepijndevos/76de3cd3187bd16889ea95302a2e05f4 …

          0 replies 0 retweets 0 likes
        4. End of conversation

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