Index finite lists of positive integers

Index finite lists of positive integers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

It is known that the set of finite lists of positive integers is countable. This means that there exists a subset of natural integers which can be mapped to the set of finite lists of positive integers. The purpose of this task is to implement such a mapping :

• write a function rank which assigns an integer to any finite, arbitrarily long list of arbitrary large integers.

• write a function unrank which is rank 's inverse function.

Demonstrate your solution by picking a random-length list of random positive integers, turn it into an integer and get the list back.

D

This solution isn't efficient.

Translation of: Python

<lang d>import std.stdio, std.algorithm, std.array, std.conv, std.bigint;

BigInt rank(T)(in T[] x) pure /*nothrow @safe*/ {

   return BigInt("0x" ~ x.map!text.join("F"));

}

BigInt[] unrank(BigInt n) pure /*nothrow @safe*/ {

   string s;
   while (n) {
       s = "0123456789ABCDEF"[n % 16] ~ s;
       n /= 16;
   }
   return s.split("F").map!BigInt.array;

}

void main() {

   immutable s = [1, 2, 3, 10, 100, 987654321];
   s.writeln;
   s.rank.writeln;
   s.rank.unrank.writeln;

}</lang>

Output:

[1, 2, 3, 10, 100, 987654321]
37699814998383067155219233
[1, 2, 3, 10, 100, 987654321]

J

Implementation:

<lang j>scrunch=:3 :0

 n=.1x+>./y
 #.(1#~##:n),0,n,&#:n#.y

)

hcnurcs=:3 :0

 b=.#:y
 m=.b i.0
 n=.#.m{.(m+1)}.b
 n #.inv#.(1+2*m)}.b

)</lang>

Example use:

<lang J> scrunch 4 5 7 9 0 8 8 7 4 8 8 4 1 4314664669630761

  hcnurcs 4314664669630761

4 5 7 9 0 8 8 7 4 8 8 4 1</lang>

Explanation. We treat the sequence as an n digit number in base m where n is the length of the list and m is 1+the largest value in the list. (This is equivalent to treating it as a polynomial in m with coefficients which are the values of the list.) In other words 4 5 7 9 0 8 8 7 4 8 8 4 1 becomes 4579088748841. Now we just need to encode the base (10, in this case). To do that we treat this number as a sequence of bits and prepend it with 1 1 1 1 0 1 0 1 0. This is a sequence of '1's whose length matches the number of bits needed to represent the base of our polynomial, followed by a 0 followed by the length of our polynomial.

To extract the original list we reverse this process: Find the position of the first zero, that's the size of our base, extract the base and then use that to find the coefficients of our polynomial, which is or original list.

Whether this is an efficient representation or not depends, of course, on the nature of the list being represented.

Perl 6

<lang perl6>sub expand(Int $n is copy, Int $dimension where * > 1) {

   my @reversed-digits = gather while $n > 0 {

take $n % $dimension; $n div= $dimension;

   }
   return map {

reduce * * $dimension + *, 0, 0, reverse @reversed-digits[$_, * + $dimension ... *]

   }, ^$dimension;

}

sub compress(*@n is copy where @n > 1) {

   my $dimension = @n.elems;
   reduce * * $dimension + *, 0, 0
   reverse gather while @n.any > 0 {

(state $i = 0) %= $dimension; take @n[$i] % $dimension; @n[$i] div= $dimension; $i++;

   }

}

sub rank(@n) { compress compress(@n), +@n } sub unrank(Int $n) { expand |expand($n, 2) }

say my @list = (^10).roll((2..20).pick); say my $rank = rank @list; say unrank $rank;</lang>

Output:

4 5 7 9 0 8 8 7 4 8 8 4 1
406578125236287223374090483
4 5 7 9 0 8 8 7 4 8 8 4 1

Python

<lang python>def rank(x): return int('a'.join(map(str, x)), 11)

def unrank(n): s = while n: s,n = "0123456789a"[n%11] + s, n//11 return map(int, out.split('a'))

l = [1, 2, 3, 10, 100, 987654321] print l n = rank(l) print n l = unrank(n) print l</lang>

Output:

[1, 2, 3, 10, 100, 987654321]
14307647611639042485573
[1, 2, 3, 10, 100, 987654321]

Tcl

Works with: Tcl version 8.6

Inspired by the D solution. <lang tcl>package require Tcl 8.6

proc rank {integers} {

   join [lmap i $integers {format %llo $i}] 8

}

proc unrank {codedValue} {

   lmap i [split $codedValue 8] {scan $i %llo}

}</lang> Demonstrating: <lang tcl>set s {1 2 3 10 100 987654321 135792468107264516704251 7} puts "prior: $s" set c [rank $s] puts "encoded: $c" set t [unrank $c] puts "after: $t"</lang>

Output:

prior: 1 2 3 10 100 987654321 135792468107264516704251 7
encoded: 1828381281448726746426183460251416730347660304377387
after: 1 2 3 10 100 987654321 135792468107264516704251 7