Index finite lists of positive integers

From Rosetta Code
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.


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

There are many ways to do this. Feel free to choose any one you like.


Extra credit:

Make rank as a bijection and show unrank(n) for n varying from 0 to 10.

D

This solution isn't efficient.

Translation of: Python
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;
}
Output:
[1, 2, 3, 10, 100, 987654321]
37699814998383067155219233
[1, 2, 3, 10, 100, 987654321]

Go

Bijective

A list element n is encoded as a 1 followed by n 0's. Element encodings are concatenated to form a single integer rank. An advantage of this encoding is that no special case is required to handle the empty list.

package main
 
import (
"fmt"
"math/big"
)
 
func rank(l []uint) (r big.Int) {
for _, n := range l {
r.Lsh(&r, n+1)
r.SetBit(&r, int(n), 1)
}
return
}
 
func unrank(n big.Int) (l []uint) {
m := new(big.Int).Set(&n)
for a := m.BitLen(); a > 0; {
m.SetBit(m, a-1, 0)
b := m.BitLen()
l = append(l, uint(a-b-1))
a = b
}
return
}
 
func main() {
var b big.Int
for i := 0; i <= 10; i++ {
b.SetInt64(int64(i))
u := unrank(b)
r := rank(u)
fmt.Println(i, u, &r)
}
b.SetString("12345678901234567890", 10)
u := unrank(b)
r := rank(u)
fmt.Printf("\n%v\n%d\n%d\n", &b, u, &r)
}
Output:
0 [] 0
1 [0] 1
2 [1] 2
3 [0 0] 3
4 [2] 4
5 [1 0] 5
6 [0 1] 6
7 [0 0 0] 7
8 [3] 8
9 [2 0] 9
10 [1 1] 10

12345678901234567890
[1 1 1 0 1 1 1 2 1 1 2 0 3 0 2 0 0 1 1 0 3 0 0 0 0 4 1 1 0 1 2 1]
12345678901234567890

Alternative

A bit of a hack to make a base 11 number then interpret it as base 16, just because that's easiest. Not bijective. Practical though for small lists of large numbers.

package main
 
import (
"fmt"
"math/big"
"math/rand"
"strings"
"time"
)
 
// Prepend base 10 representation with an "a" and you get a base 11 number.
// Unfortunately base 11 is a little awkward with big.Int, so just treat it
// as base 16.
func rank(l []big.Int) (r big.Int, err error) {
if len(l) == 0 {
return
}
s := make([]string, len(l))
for i, n := range l {
ns := n.String()
if ns[0] == '-' {
return r, fmt.Errorf("negative integers not mapped")
}
s[i] = "a" + ns
}
r.SetString(strings.Join(s, ""), 16)
return
}
 
// Split the base 16 representation at "a", recover the base 10 numbers.
func unrank(r big.Int) ([]big.Int, error) {
s16 := fmt.Sprintf("%x", &r)
switch {
case s16 == "0":
return nil, nil // empty list
case s16[0] != 'a':
return nil, fmt.Errorf("unrank not bijective")
}
s := strings.Split(s16[1:], "a")
l := make([]big.Int, len(s))
for i, s1 := range s {
if _, ok := l[i].SetString(s1, 10); !ok {
return nil, fmt.Errorf("unrank not bijective")
}
}
return l, nil
}
 
func main() {
// show empty list
var l []big.Int
r, _ := rank(l)
u, _ := unrank(r)
fmt.Println("Empty list:", l, &r, u)
 
// show random list
l = random()
r, _ = rank(l)
u, _ = unrank(r)
fmt.Println("\nList:")
for _, n := range l {
fmt.Println(" ", &n)
}
fmt.Println("Rank:")
fmt.Println(" ", &r)
fmt.Println("Unranked:")
for _, n := range u {
fmt.Println(" ", &n)
}
 
// show error with list containing negative
var n big.Int
n.SetInt64(-5)
_, err := rank([]big.Int{n})
fmt.Println("\nList element:", &n, err)
 
// show technique is not bijective
n.SetInt64(1)
_, err = unrank(n)
fmt.Println("Rank:", &n, err)
}
 
// returns 0 to 5 numbers in the range 1 to 2^100
func random() []big.Int {
r := rand.New(rand.NewSource(time.Now().Unix()))
l := make([]big.Int, r.Intn(6))
one := big.NewInt(1)
max := new(big.Int).Lsh(one, 100)
for i := range l {
l[i].Add(one, l[i].Rand(r, max))
}
return l
}
Output:
Empty list: [] 0 []

List:
   170245492534662309353778826165
   82227712638678862510272817700
Rank:
   17827272030291729487097780664374477811820701746650470453292650775464474368
Unranked:
   170245492534662309353778826165
   82227712638678862510272817700

List element: -5 negative integers not handled
Rank: 1 unrank not bijective

J

Explicit version

Implementation:

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
)

Example use:

   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

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 base 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.


Tacit versions

Base 11 encoding:

   rank  =. 11&#.@:}.@:>@:(,&:>/)@:(<@:(10&,)@:(10&#.^:_1)"0)@:x:
unrank=. 10&#.;._1@:(10&,)@:(11&#.^:_1)

Example use:

   rank 1 2 3 10 100 987654321 135792468107264516704251 7x
187573177082615698496949025806128189691804770100426
 
unrank 187573177082615698496949025806128189691804770100426x
1 2 3 10 100 987654321 135792468107264516704251 7

Prime factorization (Gödelian) encoding:

   rank=. */@:(^~ p:@:i.@:#)@:>:@:x:
unrank=. <:@:(#;.1@:~:@:q:)

Example use:

   rank 1 11 16 1 3 9 0 2 15 7 19 10
6857998574998940803374702726455974765530187550029640884386375715876970128518999225074067307280381624132537960815429687500
 
unrank 6857998574998940803374702726455974765530187550029640884386375715876970128518999225074067307280381624132537960815429687500x
1 11 16 1 3 9 0 2 15 7 19 10

Bijective

Using the method of the Python version (shifted):

   rank=. 1 -~ #.@:(1 , >@:(([ , 0 , ])&.>/)@:(<@:($&1)"0))@:x:
unrank=. #;._2@:((0 ,~ }.)@:(#.^:_1)@:(1&+))

Example use:

   >@:((] ; unrank ; rank@:unrank)&.>)@:i. 11
┌──┬───────┬──┐
000
├──┼───────┼──┤
10 01
├──┼───────┼──┤
212
├──┼───────┼──┤
30 0 03
├──┼───────┼──┤
40 14
├──┼───────┼──┤
51 05
├──┼───────┼──┤
626
├──┼───────┼──┤
70 0 0 07
├──┼───────┼──┤
80 0 18
├──┼───────┼──┤
90 1 09
├──┼───────┼──┤
100 210
└──┴───────┴──┘
 
(] ; rank ; unrank@:rank) 1 2 3 5 8
┌─────────┬────────┬─────────┐
1 2 3 5 8144012781 2 3 5 8
└─────────┴────────┴─────────┘

Java

Translation of Python via D

Works with: Java version 8
import java.math.BigInteger;
import static java.util.Arrays.stream;
import java.util.*;
import static java.util.stream.Collectors.*;
 
public class Test3 {
static BigInteger rank(int[] x) {
String s = stream(x).mapToObj(String::valueOf).collect(joining("F"));
return new BigInteger(s, 16);
}
 
static List<BigInteger> unrank(BigInteger n) {
BigInteger sixteen = BigInteger.valueOf(16);
String s = "";
while (!n.equals(BigInteger.ZERO)) {
s = "0123456789ABCDEF".charAt(n.mod(sixteen).intValue()) + s;
n = n.divide(sixteen);
}
return stream(s.split("F")).map(x -> new BigInteger(x)).collect(toList());
}
 
public static void main(String[] args) {
int[] s = {1, 2, 3, 10, 100, 987654321};
System.out.println(Arrays.toString(s));
System.out.println(rank(s));
System.out.println(unrank(rank(s)));
}
}
[1, 2, 3, 10, 100, 987654321]
37699814998383067155219233
[1, 2, 3, 10, 100, 987654321]

Kotlin

// version 1.1.2
 
import java.math.BigInteger
 
/* Separates each integer in the list with an 'a' then encodes in base 11. Empty list mapped to '-1' */
fun rank(li: List<Int>) = when (li.size) {
0 -> -BigInteger.ONE
else -> BigInteger(li.joinToString("a"), 11)
}
 
fun unrank(r: BigInteger) = when (r) {
-BigInteger.ONE -> emptyList<Int>()
else -> r.toString(11).split('a').map { if (it != "") it.toInt() else 0 }
}
 
 
/* Each integer n in the list mapped to '1' plus n '0's. Empty list mapped to '0' */
fun rank2(li:List<Int>): BigInteger {
if (li.isEmpty()) return BigInteger.ZERO
val sb = StringBuilder()
for (i in li) sb.append("1" + "0".repeat(i))
return BigInteger(sb.toString(), 2)
}
 
fun unrank2(r: BigInteger) = when (r) {
BigInteger.ZERO -> emptyList<Int>()
else -> r.toString(2).drop(1).split('1').map { it.length }
}
 
fun main(args: Array<String>) {
var li: List<Int>
var r: BigInteger
li = listOf(0, 1, 2, 3, 10, 100, 987654321)
println("Before ranking  : $li")
r = rank(li)
println("Rank = $r")
li = unrank(r)
println("After unranking  : $li")
 
println("\nAlternative approach (not suitable for large numbers)...\n")
li = li.dropLast(1)
println("Before ranking  : $li")
r = rank2(li)
println("Rank = $r")
li = unrank2(r)
println("After unranking  : $li")
 
println()
for (i in 0..10) {
val bi = BigInteger.valueOf(i.toLong())
li = unrank2(bi)
println("${"%2d".format(i)} -> ${li.toString().padEnd(9)} -> ${rank2(li)}")
}
}
Output:
Before ranking   : [0, 1, 2, 3, 10, 100, 987654321]
Rank = 828335141480036653618783
After unranking  : [0, 1, 2, 3, 10, 100, 987654321]

Alternative approach (not suitable for large numbers)...

Before ranking   : [0, 1, 2, 3, 10, 100]
Rank = 4364126777249122850009283661412696064
After unranking  : [0, 1, 2, 3, 10, 100]

 0 -> []        -> 0
 1 -> [0]       -> 1
 2 -> [1]       -> 2
 3 -> [0, 0]    -> 3
 4 -> [2]       -> 4
 5 -> [1, 0]    -> 5
 6 -> [0, 1]    -> 6
 7 -> [0, 0, 0] -> 7
 8 -> [3]       -> 8
 9 -> [2, 0]    -> 9
10 -> [1, 1]    -> 10

Perl 6

Here is a cheap solution using a base-11 encoding and string operations:

sub rank(*@n)      { :11(@n.join('A')) }
sub unrank(Int $n) { $n.base(11).split('A') }
 
say my @n = (^20).roll(12);
say my $n = rank(@n);
say unrank $n;
Output:
1 11 16 1 3 9 0 2 15 7 19 10
25155454474293912130094652799
1 11 16 1 3 9 0 2 15 7 19 10

Here is a bijective solution that does not use string operations.

multi infix:<rad> ()       { 0 }
multi infix:<rad> ($a) { $a }
multi infix:<rad> ($a, $b) { $a * $*RADIX + $b }
 
multi expand(Int $n is copy, 1) { $n }
multi expand(Int $n is copy, Int $*RADIX) {
my \RAD = $*RADIX;
 
my @reversed-digits = gather while $n > 0 {
take $n % RAD;
$n div= RAD;
}
 
eager for ^RAD {
[rad] reverse @reversed-digits[$_, * + RAD ... *]
}
}
 
multi compress(@n where @n == 1) { @n[0] }
multi compress(@n is copy) {
my \RAD = my $*RADIX = @n.elems;
 
[rad] reverse gather while @n.any > 0 {
(state $i = 0) %= RAD;
take @n[$i] % RAD;
@n[$i] div= RAD;
$i++;
}
}
 
sub rank(@n) { compress (compress(@n), @n - 1)}
sub unrank(Int $n) { my ($a, $b) = expand $n, 2; expand $a, $b + 1 }
 
my @list = (^10).roll((2..20).pick);
my $rank = rank @list;
say "[[email protected]] -> $rank -> [{unrank $rank}]";
 
for ^10 {
my @unrank = unrank $_;
say "$_ -> [[email protected]] -> {rank @unrank}";
}
Output:
[7 1 4 7 7 0 2 7 7 0 7 7] -> 20570633300796394530947471 -> [7 1 4 7 7 0 2 7 7 0 7 7]
0 -> [0] -> 0
1 -> [1] -> 1
2 -> [0 0] -> 2
3 -> [1 0] -> 3
4 -> [2] -> 4
5 -> [3] -> 5
6 -> [0 1] -> 6
7 -> [1 1] -> 7
8 -> [0 0 0] -> 8
9 -> [1 0 0] -> 9

Python

def rank(x): return int('a'.join(map(str, [1] + x)), 11)
 
def unrank(n):
s = ''
while n: s,n = "0123456789a"[n%11] + s, n//11
return map(int, s.split('a'))[1:]
 
l = [1, 2, 3, 10, 100, 987654321]
print l
n = rank(l)
print n
l = unrank(n)
print l
Output:
[0, 1, 2, 3, 10, 100, 987654321]
207672721333439869642567444
[0, 1, 2, 3, 10, 100, 987654321]

Bijection

Each number in the list is stored as a length of 1s, separated by 0s, and the resulting string is prefixed by '1', then taken as a binary number. Empty list is mapped to 0 as a special case. Don't use it on large numbers.

def unrank(n):
return map(len, bin(n)[3:].split("0")) if n else []
 
def rank(x):
return int('1' + '0'.join('1'*a for a in x), 2) if x else 0
 
for x in range(11):
print x, unrank(x), rank(unrank(x))
 
print
x = [1, 2, 3, 5, 8];
print x, rank(x), unrank(rank(x))
 
Output:
0 [] 0
1 [0] 1
2 [0, 0] 2
3 [1] 3
4 [0, 0, 0] 4
5 [0, 1] 5
6 [1, 0] 6
7 [2] 7
8 [0, 0, 0, 0] 8
9 [0, 0, 1] 9
10 [0, 1, 0] 10

[1, 2, 3, 5, 8] 14401279 [1, 2, 3, 5, 8]

Racket

Translation of: Tcl
(which gives credit to #D)
#lang racket/base
(require (only-in racket/string string-join string-split))
 
(define (integer->octal-string i)
(number->string i 8))
 
(define (octal-string->integer s)
(string->number s 8))
 
(define (rank is)
(string->number (string-join (map integer->octal-string is) "8")))
 
(define (unrank ranking)
(map octal-string->integer (string-split (number->string ranking 10) "8")))
 
(module+ test
(define loi '(1 2 3 10 100 987654321 135792468107264516704251 7))
(define rnk (rank loi))
(define urk (unrank rnk))
(displayln loi)
(displayln rnk)
(displayln urk))
Output:
(1 2 3 10 100 987654321 135792468107264516704251 7)
1828381281448726746426183460251416730347660304377387
(1 2 3 10 100 987654321 135792468107264516704251 7)

REXX

/*REXX program  assigns an integer for a  finite list  of  arbitrary positive integers. */
parse arg L /*obtain optional argument (int list).*/
if L='' then L=3 14 159 2653589793238 /*Not specified? Then use the default.*/
L=translate(space(L), ',', " ") /*use a commatized list of integers. */
numeric digits max(9, 2*length(L)) /*ensure enough dec. digits to handle L*/
 
say 'original list=' L /*display the original list of integers*/
N=rank(L); say ' map integer=' N /*generate and display the map integer.*/
O=unrank(N); say ' unrank=' O /*generate original integer and display*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
rank: return x2d( translate( space( arg(1) ), 'c', ",") )
unrank: return space( translate( d2x( arg(1) ), ',', "C") )
output   when using the default input:
original list= 3,14,159,2653589793238
  map integer= 72633563195984664801653304
       unrank= 3,14,159,2653589793238

Ruby

Translation of: Python
def rank(arr)
arr.join('a').to_i(11)
end
 
def unrank(n)
n.to_s(11).split('a').collect{|x| x.to_i}
end
 
l = [1, 2, 3, 10, 100, 987654321]
p l
n = rank(l)
p n
l = unrank(n)
p l
Output:
[1, 2, 3, 10, 100, 987654321]
14307647611639042485573
[1, 2, 3, 10, 100, 987654321]

Bijection

Translation of: Python
def unrank(n)
return [0] if n==1
n.to_s(2)[1..-1].split('0',-1).map(&:size)
end
 
def rank(x)
return 0 if x.empty?
('1' + x.map{ |a| '1'*a }.join('0')).to_i(2)
end
 
for x in 0..10
puts "%3d : %-18s: %d" % [x, a=unrank(x), rank(a)]
end
 
puts
x = [1, 2, 3, 5, 8]
puts "#{x} => #{rank(x)} => #{unrank(rank(x))}"
Output:
  0 : []                : 0
  1 : [0]               : 1
  2 : [0, 0]            : 2
  3 : [1]               : 3
  4 : [0, 0, 0]         : 4
  5 : [0, 1]            : 5
  6 : [1, 0]            : 6
  7 : [2]               : 7
  8 : [0, 0, 0, 0]      : 8
  9 : [0, 0, 1]         : 9
 10 : [0, 1, 0]         : 10

[1, 2, 3, 5, 8] => 14401279 => [1, 2, 3, 5, 8]

Sidef

Translation of: Ruby
func rank(Array arr) {
Number(arr.join('a'), 11)
}
 
func unrank(Number n) {
n.base(11).split('a').map { Num(_) }
}
 
var l = [1, 2, 3, 10, 100, 987654321]
say l
var n = rank(l)
say n
var l = unrank(n)
say l
Output:
[1, 2, 3, 10, 100, 987654321]
14307647611639042485573
[1, 2, 3, 10, 100, 987654321]

Bijection:

func unrank(Number n) {
n == 1 ? [0]
 : n.base(2).substr(1).split('0', -1).map{.len}
}
 
func rank(Array x) {
x.is_empty ? 0
 : Number('1' + x.map { '1' * _ }.join('0'), 2)
}
 
for x in (0..10) {
printf("%3d : %-18s: %d\n", x, unrank(x), rank(unrank(x)))
}
 
say ''
var x = [1, 2, 3, 5, 8]
say "#{x} => #{rank(x)} => #{unrank(rank(x))}"
Output:
  0 : []                : 0
  1 : [0]               : 1
  2 : [0, 0]            : 2
  3 : [1]               : 3
  4 : [0, 0, 0]         : 4
  5 : [0, 1]            : 5
  6 : [1, 0]            : 6
  7 : [2]               : 7
  8 : [0, 0, 0, 0]      : 8
  9 : [0, 0, 1]         : 9
 10 : [0, 1, 0]         : 10

[1, 2, 3, 5, 8] => 14401279 => [1, 2, 3, 5, 8]

Tcl

Works with: Tcl version 8.6

Inspired by the D solution.

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}
}

Demonstrating:

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

zkl

Using GMP, base 11 and sometimes strings to represent big ints.

var BN=Import("zklBigNum");
fcn rank(ns) { BN(ns.concat("A"),11) }
fcn unrank(bn) { bn.toString(11).split("a").apply("toInt") }
fcn unrankS(bn){ bn.toString(11).split("a") }
fcn rankz(ns,S=False){
ns.println();
rank(ns).println();
if(S) ns:rank(_):unrankS(_).println();
else ns:rank(_):unrank(_) .println();
}
rankz(T(1,2,3,10,100,987654321));
rankz(T(1,2,3,10,100,987654321,"135792468107264516704251",7),True);
Output:
L(1,2,3,10,100,987654321)
14307647611639042485573
L(1,2,3,10,100,987654321)
L(1,2,3,10,100,987654321,"135792468107264516704251",7)
187573177082615698496949025806128189691804770100426
L("1","2","3","10","100","987654321","135792468107264516704251","7")