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Here's a line of code from a handbook written by Stephen Wolfram, which turns out to be very complicated for me.

If[#1 > 2, 2 #0[#1 - #0[#1 - 2]], 1] & /@ Range[50]

The output is:

{1, 1, 2, 4, 4, 2, 4, 4, 8, 4, 4, 8, 16, 4, 2, 16, 4, 2, 4, 4, 8, 4, 4, 8, 16, 4, 8, 16, 16, 8, 4, 16, 32, 4, 4, 32, 64, 4, 2, 64, 4, 2, 4, 4, 8, 4, 4, 8, 16, 4}

I am confused about the Slot 0(#0) here, or how could I break down the code and understand it?

Shawn
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1 Answers1

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#0 refers to the function itself. This is consistent with the "0th" argument being the head of an expression.

Example:

Print[#0] &[]

(* prints Print[#0]& *)

In practice, this is useful for writing recursive functions. This is what it is used for in your example. The example could be rephrased as

f[x_] := If[x > 2, 2 f[x - f[x - 2]], 1]
Szabolcs
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    I'm going to have to remember this trick the next time I'm over on PCG Stackexchange. – Michael Seifert Feb 22 '19 at 13:34
  • @MichaelSeifert The problem is that we still need to add a stopping condition for the recursion, and with this method it needs to be done with an If[...]. That will sometimes make the code longer than the pattern matching version. – Szabolcs Feb 22 '19 at 13:38
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    @MichaelSeifert Well, actually a Fibonacci is shorter with the pure function method: http://i.stack.imgur.com/Fg38W.png – Szabolcs Feb 22 '19 at 13:40