Re: [Gendispatch] New Version Notification for draft-knodel-nomcom-gender-representation-00.txt

Martin Thomson <mt@lowentropy.net> Thu, 23 November 2023 00:57 UTC

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Date: Thu, 23 Nov 2023 11:56:48 +1100
From: Martin Thomson <mt@lowentropy.net>
To: Eric Rescorla <ekr@rtfm.com>, Christian Huitema <huitema@huitema.net>
Cc: Mallory Knodel <mknodel@cdt.org>, Vittorio Bertola <vittorio.bertola@open-xchange.com>, gendispatch@ietf.org
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Archived-At: <https://mailarchive.ietf.org/arch/msg/gendispatch/fGDYfmz43FJCQ9QjIJaHOo14MDQ>
Subject: Re: [Gendispatch] New Version Notification for draft-knodel-nomcom-gender-representation-00.txt
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On Thu, Nov 23, 2023, at 07:05, Eric Rescorla wrote:
> Just so I understand what you are proposing in practice. Suppose we have four
> values: A, B, C, and PNTS, then the selection algorithm would be:
>
> exclude = None
> while selected_total < 10 {
>   candidate = draw_random()
>   if selected(gender(x)) == 8 {
>      exclude = gender(x)
>   }
>   if exclude != None {
>     if exclude == gender(x)  {
>        skip;
>     }
>     // This avoids giving pnts a preference
>     if gender(x) == pnts {
>        skip;
>     } 
>
>    select x;
> }

I know how to do code review!  (You missed a closing brace.)

A perhaps better approach is to count PNTS toward all genders, which is probably more robust.  Otherwise, an all-male NomCom can be selected.  With a limit of eight, if two male volunteers select PNTS, the count won't reach that limit, as long as the two male PNTS options aren't the last two selected (which is a bit weird).

while selected.len < 10 {
  candidate = draw();
  // insert affiliation check here
  if candidate.gender == PNTS {
    if genders[A] == limit || genders[B] == limit || genders[B] == limit {
      skip;
    }
    genders[A]++; genders[B]++; genders[C]++;
  } else {
    if genders[candidate.gender] == limit {
      skip;
    }
    genders[candidate.gender]++;
  }
  select candidate;
}