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I'm using this example to study Mathematics.Example Demo. In this example, before generating recursion tree, she flats the function. But I have a difficultly to understand how flatten works, especially the 3rd line of the code below. Could someone teach me? Thank you very much.
f[___, 0] := {};
f[___, 1] := {};
f[m___, p_] := {{m, p} -> {m, p, p - 1}, {m, p} -> {m, p, p - 2}, //How to understand this line?
f[m, p, p - 1], f[m, p, p - 2]}
j[n_] := f[n] // Flatten
To understand j[n__]:= f[n] //Flatten we need to understand f[n]. As an example, take f[3].
MMA applies more specific definitions first. In this case this will be f[m___, p_], where m is empty. Therefore, we replace f[3] by
{{3} -> {3, 2}, {3} -> {3, 1}, f[3,2],f[3,1]}.
f[3, 1] is, by definition, an empty list.
f[3, 2] is, by the pattern: f[m___, p_],
{{3, 2} -> {3, 2, 1}, {3, 2} -> {3, 2, 0}, f[3, 2, 1], f3, 2, 0]}.
f[3, 2, 1] and f3, 2, 0] are by definition empty lists, so this is:
{{3, 2} -> {3, 2, 1}, {3, 2} -> {3, 2, 0}, {}, {}}
Inserting f[3, 2] and f[3, 1] in the list above:
{{3} -> {3, 2}, {3} -> {3, 1}, {{3, 2} -> {3, 2, 1}, {3, 2} -> {3, 2, 0}, {}, {}}, {}}
Flatten will take away the empty lists. So we finally get:
{{3} -> {3, 2}, {3} -> {3, 1}, {3, 2} -> {3, 2, 1}, {3, 2} -> {3, 2, 0}}
{{m, p} -> {m, p, p - 1}, {m, p} -> {m, p, p - 2}. If MMA find a part that matches the pattern, it replaces it by the replacement. Look pattern matching up in the help. E.g. _ means "anything". You may name this, e.g. p_. __means one or more "anythings" e.t.c
$\endgroup$
– Daniel Huber
Feb 16 at 19:33
