What's the meaning of #0?
9
$begingroup$
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?
pure-function
asked 17 hours ago
ShawnShawn
482
New contributor
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
9
$begingroup$
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?
pure-function
asked 17 hours ago
ShawnShawn
482
New contributor
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Here is a dedicated Q/A on this topic.
$endgroup$
– Leonid Shifrin
9 hours ago
add a comment |
9
9
9
2
$begingroup$
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?
pure-function
asked 17 hours ago
ShawnShawn
482
New contributor
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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?
pure-function
pure-function
asked 17 hours ago
ShawnShawn
482
New contributor
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
ShawnShawn
482
New contributor
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
ShawnShawn
482
New contributor
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
ShawnShawn
482
asked 17 hours ago
ShawnShawn
482
482
New contributor
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Shawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Here is a dedicated Q/A on this topic.
$endgroup$
– Leonid Shifrin
9 hours ago
add a comment |
$begingroup$
Here is a dedicated Q/A on this topic.
$endgroup$
– Leonid Shifrin
9 hours ago
$begingroup$
Here is a dedicated Q/A on this topic.
$endgroup$
– Leonid Shifrin
9 hours ago
$begingroup$
Here is a dedicated Q/A on this topic.
$endgroup$
– Leonid Shifrin
9 hours ago
add a comment |
1 Answer
1
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oldest
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10
$begingroup$
#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]
answered 17 hours ago
SzabolcsSzabolcs
161k14438936
$endgroup$
3
$begingroup$
I'm going to have to remember this trick the next time I'm over on PCG Stackexchange.
$endgroup$
– Michael Seifert
12 hours ago
$begingroup$
@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 anIf[...]. That will sometimes make the code longer than the pattern matching version.
$endgroup$
– Szabolcs
12 hours ago
1
$begingroup$
@MichaelSeifert Well, actually a Fibonacci is shorter with the pure function method: i.stack.imgur.com/Fg38W.png
$endgroup$
– Szabolcs
12 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
10
$begingroup$
#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]
answered 17 hours ago
SzabolcsSzabolcs
161k14438936
$endgroup$
3
$begingroup$
I'm going to have to remember this trick the next time I'm over on PCG Stackexchange.
$endgroup$
– Michael Seifert
12 hours ago
$begingroup$
@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 anIf[...]. That will sometimes make the code longer than the pattern matching version.
$endgroup$
– Szabolcs
12 hours ago
1
$begingroup$
@MichaelSeifert Well, actually a Fibonacci is shorter with the pure function method: i.stack.imgur.com/Fg38W.png
$endgroup$
– Szabolcs
12 hours ago
add a comment |
10
$begingroup$
#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]
answered 17 hours ago
SzabolcsSzabolcs
161k14438936
$endgroup$
3
$begingroup$
I'm going to have to remember this trick the next time I'm over on PCG Stackexchange.
$endgroup$
– Michael Seifert
12 hours ago
$begingroup$
@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 anIf[...]. That will sometimes make the code longer than the pattern matching version.
$endgroup$
– Szabolcs
12 hours ago
1
$begingroup$
@MichaelSeifert Well, actually a Fibonacci is shorter with the pure function method: i.stack.imgur.com/Fg38W.png
$endgroup$
– Szabolcs
12 hours ago
add a comment |
10
10
10
$begingroup$
#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]
answered 17 hours ago
SzabolcsSzabolcs
161k14438936
$endgroup$
#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]
answered 17 hours ago
SzabolcsSzabolcs
161k14438936
answered 17 hours ago
SzabolcsSzabolcs
161k14438936
answered 17 hours ago
SzabolcsSzabolcs
161k14438936
answered 17 hours ago
SzabolcsSzabolcs
161k14438936
161k14438936
3
$begingroup$
I'm going to have to remember this trick the next time I'm over on PCG Stackexchange.
$endgroup$
– Michael Seifert
12 hours ago
$begingroup$
@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 anIf[...]. That will sometimes make the code longer than the pattern matching version.
$endgroup$
– Szabolcs
12 hours ago
1
$begingroup$
@MichaelSeifert Well, actually a Fibonacci is shorter with the pure function method: i.stack.imgur.com/Fg38W.png
$endgroup$
– Szabolcs
12 hours ago
add a comment |
3
$begingroup$
I'm going to have to remember this trick the next time I'm over on PCG Stackexchange.
$endgroup$
– Michael Seifert
12 hours ago
$begingroup$
@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 anIf[...]. That will sometimes make the code longer than the pattern matching version.
$endgroup$
– Szabolcs
12 hours ago
1
$begingroup$
@MichaelSeifert Well, actually a Fibonacci is shorter with the pure function method: i.stack.imgur.com/Fg38W.png
$endgroup$
– Szabolcs
12 hours ago
3
3
$begingroup$
I'm going to have to remember this trick the next time I'm over on PCG Stackexchange.
$endgroup$
– Michael Seifert
12 hours ago
$begingroup$
I'm going to have to remember this trick the next time I'm over on PCG Stackexchange.
$endgroup$
– Michael Seifert
12 hours ago
$begingroup$
@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.$endgroup$
– Szabolcs
12 hours ago
$begingroup$
@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.$endgroup$
– Szabolcs
12 hours ago
1
1
$begingroup$
@MichaelSeifert Well, actually a Fibonacci is shorter with the pure function method: i.stack.imgur.com/Fg38W.png
$endgroup$
– Szabolcs
12 hours ago
$begingroup$
@MichaelSeifert Well, actually a Fibonacci is shorter with the pure function method: i.stack.imgur.com/Fg38W.png
$endgroup$
– Szabolcs
12 hours ago
add a comment |
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$begingroup$
Here is a dedicated Q/A on this topic.
$endgroup$
– Leonid Shifrin
9 hours ago