algorithm - the logic behind bash power set function -

the function (output power set of given input)

p() { [ $# -eq 0 ] && echo || (shift; p "$@") |         while read r ; echo -e "$1 $r\n$r"; done } 

test input

p $(echo -e "1 2 3") 

test output

1 2 3 2 3 1 3 3 1 2 2 1 

i have difficulty grasping recursion in following code. tried understand placing variables inside of code denote level of recursion , order of execution, still puzzled.

here things can tell far:

1. the subshell's output not shown on final output, gets redirected read command via pipe
2. the echo command appends new line of output

the order of execution see is:

1. p (1 2 3) -> 1 followed combination of output below\n combination of output below
2. p (2 3) -> 2 3\n3\n
3. p (3) -> 3
4. p () ->

so think should have p(2) instead of p(3) on execution #3, how happen? since shift goes in 1 direction.

if use "p(1 2 3 4)" input, part shows "1 2 3" in output confuses me.

the use of -e in echo command seems me pure obfuscation, since have been written:

p() { [ $# -eq 0 ] && echo || (shift; p "$@") |       while read r ;         echo $1 $r         echo $r       done     } 

in other words, "for every set in power set of first argument (shift; p "$@"), output both set , without first argument."

the bash function works setting chain of subshells, each 1 reading next one, this, each box subshell , below it, i've shown output reads each line of input: (i used "" make "nothing" visible. => means "call"; <- means "read".)

+---------+     +-------+     +-------+     +-------+ | p 1 2 3 | ==> | p 2 3 | ==> |  p 3  | ==> |   p   | +---------+     +-------+     +-------+     +-------+   1 2 3 "" <--+-- 2 3 "" <---+-- 3 "" <-----+-- ""   2 3 ""   <-/              /              /   1 3 ""   <--+-- 3 ""   <-/              /   3 ""     <-/                           /   1 2 ""   <--+-- 2 ""   <---+-- ""   <-/   2 ""     <-/              /   1 ""     <--+-- ""     <-/   ""       <-/