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Comment by merlincorey | Hacker News Reader

The Python problem can be solved with `itertools.tee`[0]

    from itertools import tee
    squares_min, squares_max = tee(x * x for x in range(10))
    print(min(squares_min))
    print(max(squares_max))

[0] https://docs.python.org/3/library/itertools.html#itertools.t...

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Unfortunately, tee's are backed by a queue and upon each read of a "fresh" value from one of the branches of the tee, the shared queue is appended to. There is no opportunity to remove an element from the queue until all branches of the tee have consumed it, so this will end up using O(n) memory as squares_min is completely consumed before squares_max is accessed. (Alternatingly calling next on the branches of the tee will solve the issue.) See the "Roughly equivalent to" and immediately previous text in your link.

    squares = [x * x for x in range(10)]
    print(min(squares))
    print(max(squares))

this one works and looks good.

It will have memory impact though with larger range.

A better approach might be:

    squares = lambda: (x * x for x in range(10))
    print(min(squares))
    print(max(squares()))

Yes, your solution is in the original post, under the heading 'Python solutions':

  def squares():
      return (x * x for x in range(10))

It also uses O(n) memory.

We need a language that can schedule and fuse those folds over an infinite sequence concurrently. Joining a set of futures over an infinite sequence fuses folds

I'm not sure adding more magic to languages is necessary when the alternative of using explicit generators is not too difficult.

Do generators compose?

Sure.

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