Pick-flatten a list

Question

Consider the process of "picking" a nested list. Picking is defined as follows:

• If the argument is a list, take an element from the list at random (uniformly), and pick from that.
• If the argument is not a list, simply return it.

An example implementation in Python:

import random
def pick(obj):
    if isinstance(obj, list):
        return pick(random.choice(obj))
    else:
        return obj

For simplicity, we assume that nested lists only contain integers or further nested lists.

Given any list, it is possible to create a flattened version which is indistinguishable by pick, i.e. picking from it yields the same results, with the same probability.

For example, "pick-flattening" the list

[1, 2, [3, 4, 5]]

yields the list

[1, 1, 1, 2, 2, 2, 3, 4, 5]

. The reason simply flattening is invalid is because elements of sub-lists have a lower probability of being chosen, e.g. in the list [1, [2, 3]] the 1 has a 2/4 = 1/2 chance of being chosen while 3 and 4 both have a 1/4 chance each.

Also note that picking from a singleton list is equivalent to picking from its element, and that picking from an empty list has no meaning.

The Challenge

Given a nested list of nonnegative integers, return a flattened list of nonnegative integers from which picking yields the same results with the same probability.

This is code-golf, so the shortest valid answer (measured in bytes) wins.

Specifications

• The inputs [2, 3, 4], [2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4], and [2, [3, 3], [[4]]] are equivalent (i.e. they should give equivalent results).
• The outputs [2, 2, 2, 2, 3, 3, 3, 3] and [2, 3] are equivalent (i.e. either one could be output).
• You can assume only numbers in the inclusive range 1-100 will be present in the lists.
• You can assume the top-level input will be a list, i.e. 2 is not a valid input.
• You can use any reasonable representation of nested lists, for example:
  [1, [2, 3]], 1 {2 3}, "[ 1 [ 2 3 ] ]", etc.
• Instead of a list, you can output a multiset or a mapping, or, since only numbers in the range 1-100 are allowed, a length-100 list of integers representing quantities.

Test Cases

Note that the listed outputs are only one valid possibility; see specifications for what constitutes a valid input or output.

format:
input -> output
[3]                          -> [3]
[1, [1, 1]]                  -> [1]
[1, [2, 3]]                  -> [1, 1, 2, 3]
[2, 3, [4, [5, 5, 6], 6, 7]] -> [2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7]
[[1, 1, 2], [2, 3, 3]]       -> [1, 2, 3]
[[1, 1, 2], [2, 3, 3, 3]]    -> [1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3]

Answer 1

Wolfram Language (Mathematica), 41 20 bytes

Flatten@*Tuples//@#&

Try it online! Ignore the many warnings, it all works out in the end.

How it works

For a list of depth 2 such as {{1,2},{3},{4,5,6}}, Tuples will generate the list {{1,3,4},{1,3,5},{1,3,6},{2,3,4},{2,3,5},{2,3,6}} corresponding to all the ways to pick an element from {1,2} and pick an element from {3} and pick an element from {4,5,6}.

If we Flatten this, then we get all the elements with the correct frequencies, because picking an element from one of {1,2}, {3} or {4,5,6} is equivalent to picking an element from all of them, then choosing which one to keep.

We use //@ to apply this at all levels of the input. In the process, Mathematica complains a lot, because it's turning atoms such as 17 into Tuples[17], which is really not supposed to be a thing. But these simplify to the right result later on (Tuples is happy to treat Tuples[17] as a list of length 1, even if it has a head other than List), so the complaining is irrelevant.

Answer 2

Python 2, 105 102 99 bytes

g=lambda y=[],*z:[w+[n]for n in y for w in g(*z)]or[y]
f=lambda x:x<[]and[x]or sum(g(*map(f,x)),[])

Try it online!

Answer 3

Jelly, 9 8 bytes

߀Œp$¬¡F

Try it online!

How it works

߀Œp$¬¡F  Main link. Argument: x (array or positive integer)

     ¬    Compute elementwise logical NOT of x: a non-empty array for a non-empty array, 0 for a positive integer.
      ¡   Apply the link to the left once if ¬ returned a non-empty
          array, zero timed if it returned 0.
    $     Monadic chain:
߀            Map the main link over x.
  Œp          Take the Cartesian product.
       F  Flatten the result.

Answer 4

Jelly, 10 bytes

L€P⁸ṁ€ẎµÐL

Try it online!

Answer 5

Python 2, 128 bytes

def f(l,p=0):m=reduce(int.__mul__,[i*0<[]or len(i)for i in l]);return p*(p==l)or f(sum([(([i],i)[i*0>0]*m)[:m]for i in l],[]),l)

Try it online!

Port of my Jelly answer.

-12 thanks to Jonathan Frech.

Answer 6

C (gcc), 234 223 bytes

h[9][101];o[101];n[9];l;L;e;main(x){for(;(x=scanf("%d",&e))>=0;x?++h[l][e],++n[l]:(e=getchar())-'['?e-']'?0:--l:++l>L&&++L);for(e=1,l=L+1;l--;){for(x=101;--x;o[x]+=e*h[l][x]);e*=n[l];}while(o[x]--?printf("%d ",x):++x<101);}

Try it online!

Explanation:

h[9][101];  // <- number occurences per nesting level
o[101];     // <- number occurences in "flattened" array
n[9];       // <- number of entries per nesting level
l;          // <- current nesting level
L;          // <- max nesting level
e;          // <- multi-purpose temporary
main(x){    // x: multi-purpose temporary
    for(;
            // while not EOF try reading number
            (x=scanf("%d",&e))>=0;

            // number was read?
            x

                // then increment occurence and # entries in level
                ?++h[l][e],++n[l]

                // else read any character ... if not [
                :(e=getchar())-'['

                    // if not ]
                    ?e-']'

                        // do nothing
                        ?0

                        // else decrement nesting level
                        :--l

                    // else increment nesting level and adjust max level
                    :++l>L&&++L);

    // init factor in e to 1, iterate over nesting level from innermost
    for(e=1,l=L+1;l--;){

        // iterate over all numbers
        for(x=101;
                --x;

                // add factor times occurence on current level to output
                o[x]+=e*h[l][x]);

        // multiply factor by number of entries on current level
        e*=n[l];
    }

    // iterate over all numbers and output count times
    while(o[x]--?printf("%d ",x):++x<101);
}

Answer 7

Python 2, 144 139 bytes

def f(A,p):[F.append((len(A)*p,a))if a*0<[]else f(a,len(A)*p)for a in A]
F=[];f(input(),1);R=[]
for v in F:R+=max(F)[0]/v[0]*[v[1]]
print R

Try it online!

Answer 8

JavaScript (ES6), 132 131 bytes

f=A=>(_=(a,m)=>[].concat(...a.map(m)),n=1,A=A.map(a=>a.map?f(a):[a]),_(A,a=>n*=a.length),_(A,a=>_(a.map(x=>Array(n/a.length).fill(x)))))

f=A=>(_=(a,m)=>[].concat(...a.map(m)),n=1,A=A.map(a=>a.map?f(a):[a]),_(A,a=>n*=a.length),_(A,a=>_(a,x=>Array(n/a.length).fill(x))))

console.log( f( [3] ) )
console.log( f( [1, [1, 1]] ) )
console.log( f( [1, [2, 3]] ) )
console.log( f( [2, 3, [4, [5, 5, 6], 6, 7]] ) )
console.log( f( [[1, 1, 2], [2, 3, 3]] ) )
console.log( f( [[1, 1, 2], [2, 3, 3, 3]] ) )
console.log( f( [1, 2, [3, 4, 5], [6, 7]] ) )