Index finite lists of positive integers
It is known that the set of finite lists of positive integers is countable.
You are encouraged to solve this task according to the task description, using any language you may know.
This means that there exists a subset of natural integers which can be mapped to the set of finite lists of positive integers.
- Task
Implement such a mapping:
- write a function rank which assigns an integer to any finite, arbitrarily long list of arbitrary large positive integers.
- write a function unrank which is the rank 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.
There are many ways to do this. Feel free to choose any one you like.
- Extra credit
Make the rank function as a bijection and show unrank(n) for n varying from 0 to 10.
D
This solution isn't efficient.
<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]
FreeBASIC
Restricted to shortish lists with smallish integers, because the rank integers get really big really fast, and bloating the code with arbitrary precision arithmetic isn't illustrative. <lang FreeBASIC>type duple
A as ulongint B as ulongint
end type
function two_to_one( A as ulongint, B as ulongint ) as ulongint 'converts two numbers into one
dim as uinteger ret = A*A + B*B + 2*A*B - 3*A - B 'according to the table return 1 + ret/2 ' 1 2 3 4 5
end function ' -------------
' 1| 1 3 6 10 15
function one_to_two( R as ulongint ) as duple ' 2| 2 5 9 14 20
dim as uinteger t = int((-1+sqr(8*R-7))/2) ' 3| 4 8 13 19 26 dim as duple ret ' 4| 7 12 18 25 33 ret.A = (t*t+3*t+4)/2-R t = int((-1+sqr(8*R-7))/2) 'and the inverse of this ret.B = R-t*(t+1)/2 return ret
end function
function rank( N() as ulongint) as ulongint
dim as uinteger ret, num = ubound(N)+1
if num = 0 then return 0 'define a value of 0 for the empty list
if num = 1 then return two_to_one( N(0), 1 )
ret = two_to_one( N(0), N(1) )
for i as uinteger = 2 to num-1 'progressively encode the list by
ret = two_to_one( ret, N(i) ) 'applying 2to1 on the result of the
next i 'previous calculation with the next list element
return two_to_one(ret, num) 'store the length of the list as
end function 'the final component
sub unrank( R as ulongint, N() as ulongint )
dim as duple temp
if R = 0 then 'zero yields the empty list
redim N(-1)
return
end if
dim as ulongint num, Q(0 to 1)
temp = one_to_two( R )
num = temp.B 'first get the length of the encoded list
redim N(0 to num-1) as ulongint
if num = 1 then '(singleton handled as a special case)
N(0)=temp.A
return
end if
for i as integer = num-1 to 2 step -1 'get back the list elements one by one
temp = one_to_two( temp.A ) 'in the reverse order they were added
N(i) = temp.B
next i
temp = one_to_two( temp.A ) 'finally get the initial two list elements
N(0) = temp.A
N(1) = temp.B
end sub
sub show_list( L() as ulongint )
dim as integer num = ubound(L)
if num=-1 then
print "[]"
return
end if
print "[";
for i as integer = 0 to num-1
print str(L(i))+", ";
next i
print str(L(num))+"]"
end sub 'A few tests dim as duple temp redim as uinteger ex0(-1) 'empty list dim as ulongint R = rank(ex0()) R = rank(ex0()) print R, redim as ulongint X(0 to 1) unrank R, X() show_list(X())
dim as uinteger ex1(0 to 0) = {13} 'list with 1 element R = rank(ex1()) print R, unrank R, X() show_list(X())
dim as uinteger ex2(0 to 1) = {19, 361} 'list with 2 elements R = rank(ex2()) print R, redim as ulongint X(0 to 1) unrank R, X() show_list(X())
dim as uinteger ex6(0 to 5) = {1,2,1,2,3,1} 'list with 6 elements R = rank(ex6()) print R, unrank R, X() show_list(X())</lang>
- Output:
0 [] 79 [13] 2591460030 [19, 361] 9576882 [1, 2, 1, 2, 3, 1]
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. <lang go>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)
}</lang>
- 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. <lang go>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
}</lang>
- 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:
<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 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:
<lang j> rank =. 11&#.@:}.@:>@:(,&:>/)@:(<@:(10&,)@:(10&#.^:_1)"0)@:x:
unrank=. 10&#.;._1@:(10&,)@:(11&#.^:_1)</lang>
Example use:
<lang J> rank 1 2 3 10 100 987654321 135792468107264516704251 7x 187573177082615698496949025806128189691804770100426
unrank 187573177082615698496949025806128189691804770100426x
1 2 3 10 100 987654321 135792468107264516704251 7</lang>
Prime factorization (Gödelian) encoding:
<lang j> rank=. */@:(^~ p:@:i.@:#)@:>:@:x:
unrank=. <:@:(#;.1@:~:@:q:)</lang>
Example use:
<lang J> 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</lang>
Bijective
Using the method of the Python version (shifted):
<lang j> rank=. 1 -~ #.@:(1 , >@:(([ , 0 , ])&.>/)@:(<@:($&1)"0))@:x:
unrank=. #;._2@:((0 ,~ }.)@:(#.^:_1)@:(1&+))</lang>
Example use:
<lang J> >@:((] ; unrank ; rank@:unrank)&.>)@:i. 11 ┌──┬───────┬──┐ │0 │0 │0 │ ├──┼───────┼──┤ │1 │0 0 │1 │ ├──┼───────┼──┤ │2 │1 │2 │ ├──┼───────┼──┤ │3 │0 0 0 │3 │ ├──┼───────┼──┤ │4 │0 1 │4 │ ├──┼───────┼──┤ │5 │1 0 │5 │ ├──┼───────┼──┤ │6 │2 │6 │ ├──┼───────┼──┤ │7 │0 0 0 0│7 │ ├──┼───────┼──┤ │8 │0 0 1 │8 │ ├──┼───────┼──┤ │9 │0 1 0 │9 │ ├──┼───────┼──┤ │10│0 2 │10│ └──┴───────┴──┘
(] ; rank ; unrank@:rank) 1 2 3 5 8
┌─────────┬────────┬─────────┐ │1 2 3 5 8│14401278│1 2 3 5 8│ └─────────┴────────┴─────────┘</lang>
Java
<lang java>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)));
}
}</lang>
[1, 2, 3, 10, 100, 987654321] 37699814998383067155219233 [1, 2, 3, 10, 100, 987654321]
Julia
<lang julia>using LinearAlgebra LinearAlgebra.rank(x::Vector{<:Integer}) = parse(BigInt, "1a" * join(x, 'a'), base=11) function unrank(n::Integer)
s = ""
while !iszero(n)
ind = n % 11 + 1
n ÷= 11
s = "0123456789a"[ind:ind] * s
end
return parse.(Int, split(s, 'a'))[2:end]
end
v = [0, 1, 2, 3, 10, 100, 987654321] n = rank(v) v = unrank(n) println("# v = $v\n -> n = $n\n -> v = $v")</lang>
- Output:
# v = [0, 1, 2, 3, 10, 100, 987654321] -> n = 207672721333439869642567444 -> v = [0, 1, 2, 3, 10, 100, 987654321]
Kotlin
<lang scala>// 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)}")
}
}</lang>
- 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
Nim
<lang Nim>import strformat, strutils import bignum
func rank(list: openArray[uint]): Int =
result = newInt(0) for n in list: result = result shl (n + 1) result = result.setBit(n)
func unrank(n: Int): seq[uint] =
var m = n.clone var a = if m.isZero: 0u else: m.bitLen.uint while a > 0: m = m.clearBit(a - 1) let b = if m.isZero: 0u else: m.bitLen.uint result.add(a - b - 1) a = b
when isMainModule:
var b: Int
for i in 0..10:
b = newInt(i)
let u = b.unrank()
let r = u.rank()
echo &"{i:2d} {u:>9s} {r:>2s}"
b = newInt("12345678901234567890")
let u = b.unrank()
let r = u.rank()
echo &"\n{b}\n{u}\n{r}"</lang>
- 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
Perl
The base-11 approach requires bigint pragma for all but trivial lists. Using ntheory module for base conversions.
<lang perl>use bigint; use ntheory qw(fromdigits todigitstring); use feature 'say';
sub rank { join , fromdigits(join('a',@_), 11) } sub unrank { split 'a', todigitstring(@_[0], 11) }
say join ' ', @n = qw<12 11 0 7 9 15 15 5 7 13 5 5>; say $n = rank(@n); say join ' ', unrank $n;</lang>
- Output:
12 11 0 7 9 15 15 5 7 13 5 5 16588666500024842935939135419 12 11 0 7 9 15 15 5 7 13 5 5
Phix
base 11
<lang Phix>include mpfr.e
procedure rank(mpz r, sequence s)
for i=1 to length(s) do
s[i] = sprintf("%d",s[i])
end for
mpz_set_str(r,join(s,'a'),11)
end procedure
function unrank(mpz i)
sequence res = split(mpz_get_str(i,11),'a')
for j=1 to length(res) do
{{res[j]}} = scanf(res[j],"%d")
end for
return res
end function
sequence l = {1, 2, 3, 10, 100, 987654321} mpz n = mpz_init() rank(n,l) sequence u = unrank(n) ?{l,mpz_get_str(n),u}</lang>
- Output:
{{1,2,3,10,100,987654321},"14307647611639042485573",{1,2,3,10,100,987654321}}
bijective
<lang Phix>function unrank(atom n)
sequence res = sprintf("%0b",n)
if res="1" then return {0} end if
res = split(res[2..$],'0')
for i=1 to length(res) do res[i] = length(res[i]) end for
return res
end function
function rank(sequence x)
if x={} then return "0" end if
for i=1 to length(x) do
x[i] = repeat('1',x[i])
end for
atom Template:Res = scanf("0b1"&join(x,'0'),"%d")
return res
end function
for i=0 to 10 do
sequence a = unrank(i)
printf(1,"%3d : %-18v: %d\n",{i, a, rank(a)})
end for
sequence x = {1, 2, 3, 5, 8} printf(1,"%v => %d => %v\n",{x,rank(x),unrank(rank(x))})</lang>
- 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}
Python
<lang 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</lang>
- 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. <lang python>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)) </lang>
- 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
(which gives credit to #D)
<lang racket>#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))</lang>
- Output:
(1 2 3 10 100 987654321 135792468107264516704251 7) 1828381281448726746426183460251416730347660304377387 (1 2 3 10 100 987654321 135792468107264516704251 7)
Raku
(formerly Perl 6) Here is a cheap solution using a base-11 encoding and string operations: <lang perl6>sub rank(*@n) { :11(@n.join('A')) } sub unrank(Int $n) { $n.base(11).split('A') }
say my @n = (1..20).roll(12); say my $n = rank(@n); say unrank $n;</lang>
- Output:
1 11 16 1 3 9 0 2 15 7 19 10 25155454474293912130094652799 1 11 16 1 3 9 0 2 15 7 19 10
Here is a bijective solution that does not use string operations.
<lang perl6>multi infix:<rad> () { 0 } multi infix:<rad> ($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 $rank = rank @list; say "[$@list] -> $rank -> [{unrank $rank}]";
for ^10 {
my @unrank = unrank $_;
say "$_ -> [$@unrank] -> {rank @unrank}";
}</lang>
- 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
REXX
This REXX version can handle zeros as well as any sized (decimal) positive integers.
No checks are made that the numbers are non-negative integers or malformed integers. <lang rexx>/*REXX program assigns an integer for a finite list of arbitrary non-negative integers. */ parse arg $ /*obtain optional argument (int list).*/ if $= | $="," then $=3 14 159 265358979323846 /*Not specified? Then use the default.*/
/* [↑] kinda use decimal digits of pi.*/
$= translate( space($), ',', " ") /*use a commatized list of integers. */ numeric digits max(9, 2 * length($) ) /*ensure enough dec. digits to handle $*/
say 'original list=' $ /*display the original list of integers*/
N= rank($); 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") )</lang>
- output when using the default input:
original list= 3,14,159,265358979323846
map integer= 18594192178172074189223245894
unrank= 3,14,159,265358979323846
Ruby
<lang ruby>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</lang>
- Output:
[1, 2, 3, 10, 100, 987654321] 14307647611639042485573 [1, 2, 3, 10, 100, 987654321]
Bijection
<lang ruby>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))}"</lang>
- 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]
Scala
- Output:
Best seen in running your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
<lang Scala>object IndexFiniteList extends App {
val (defBase, s) = (10, Seq(1, 2, 3, 10, 100, 987654321))
def rank(x: Seq[Int], base: Int = defBase) = BigInt(x.map(Integer.toString(_, base)).mkString(base.toHexString), base + 1)
def unrank(n: BigInt, base: Int = defBase): List[BigInt] = n.toString(base + 1).split((base).toHexString).map(BigInt(_)).toList
val ranked = rank(s)
println(s.mkString("[", ", ", "]"))
println(ranked)
println(unrank(ranked).mkString("[", ", ", "]"))
}</lang>
Sidef
<lang 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</lang>
- Output:
[1, 2, 3, 10, 100, 987654321] 14307647611639042485573 [1, 2, 3, 10, 100, 987654321]
Bijection: <lang ruby>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))}"</lang>
- 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
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
Wren
<lang ecmascript>import "/big" for BigInt
// Separates each integer in the list with an 'a' then encodes in base 11. // Empty list mapped to '-1'. var rank = Fn.new { |li|
if (li.count == 0) return BigInt.minusOne
return BigInt.fromBaseString(li.join("a"), 11)
}
var unrank = Fn.new { |r|
if (r == BigInt.minusOne) return []
return r.toBaseString(11).split("a").map { |d| (d != "") ? Num.fromString(d) : 0 }.toList
}
// Each integer n in the list mapped to '1' plus n '0's. // Empty list mapped to '0' var rank2 = Fn.new { |li|
if (li.isEmpty) return BigInt.zero
var sb = ""
for (i in li) sb = sb + "1" + ("0" * i)
return BigInt.fromBaseString(sb, 2)
}
var unrank2 = Fn.new { |r|
if (r == BigInt.zero) return []
return r.toBaseString(2)[1..-1].split("1").map { |d| d.count }.toList
}
var li = [0, 1, 2, 3, 10, 100, 987654321] System.print("Before ranking : %(li)") var r = rank.call(li) System.print("Rank = %(r)") li = unrank.call(r) System.print("After unranking : %(li)") System.print("\nAlternative approach (not suitable for large numbers)...\n") li = li[0..-2] System.print("Before ranking : %(li)") r = rank2.call(li) System.print("Rank = %(r)") li = unrank2.call(r) System.print("After unranking : %(li)")</lang>
- 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]
zkl
Using GMP, base 11 and sometimes strings to represent big ints. <lang zkl>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") }</lang> <lang zkl>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);</lang>
- 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")