Understanding Recursion to generate permutations

Question

I find recursion, apart from very straight forward ones like factorial, very difficult to understand. The following snippet prints all permutations of a string. Can anyone h

Answer 1

PaulR has the right suggestion. You have to run through the code by "hand" (using whatever tools you want - debuggers, paper, logging function calls and variables at certain points) until you understand it. For an explanation of the code I'll refer you to quasiverse's excellent answer.

Perhaps this visualization of the call graph with a slightly smaller string makes it more obvious how it works:

The graph was made with graphviz.

// x.dot
// dot x.dot -Tpng -o x.png
digraph x {
rankdir=LR
size="16,10"

node [label="permute(\"ABC\", 0, 2)"] n0;
 node [label="permute(\"ABC\", 1, 2)"] n1;
  node [label="permute(\"ABC\", 2, 2)"] n2;
  node [label="permute(\"ACB\", 2, 2)"] n3;
 node [label="permute(\"BAC\", 1, 2)"] n4;
  node [label="permute(\"BAC\", 2, 2)"] n5;
  node [label="permute(\"BCA\", 2, 2)"] n6;
 node [label="permute(\"CBA\", 1, 2)"] n7;
  node [label="permute(\"CBA\", 2, 2)"] n8;
  node [label="permute(\"CAB\", 2, 2)"] n9;

n0 -> n1 [label="swap(0, 0)"];
n0 -> n4 [label="swap(0, 1)"];
n0 -> n7 [label="swap(0, 2)"];

n1 -> n2 [label="swap(1, 1)"];
n1 -> n3 [label="swap(1, 2)"];

n4 -> n5 [label="swap(1, 1)"];
n4 -> n6 [label="swap(1, 2)"];

n7 -> n8 [label="swap(1, 1)"];
n7 -> n9 [label="swap(1, 2)"];
}