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Disarium numbers

From Rosetta Code
Task
Disarium numbers
You are encouraged to solve this task according to the task description, using any language you may know.

A Disarium number is an integer where the sum of each digit raised to the power of its position in the number, is equal to the number.


E.G.

135 is a Disarium number:

11 + 32 + 53 == 1 + 9 + 125 == 135

There are a finite number of Disarium numbers.


Task
  • Find and display the first 18 Disarium numbers.


Stretch
  • Find and display all 20 Disarium numbers.


See also



Arturo[edit]

disarium?: function [x][
j: 0
psum: sum map digits x 'dig [
j: j + 1
dig ^ j
]
return psum = x
]
 
cnt: 0
i: 0
while [cnt < 18][
if disarium? i [
print i
cnt: cnt + 1
]
i: i + 1
]
Output:
0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427

AWK[edit]

 
# syntax: GAWK -f DISARIUM_NUMBERS.AWK
BEGIN {
stop = 19
printf("The first %d Disarium numbers:\n",stop)
while (count < stop) {
if (is_disarium(n)) {
printf("%d ",n)
count++
}
n++
}
printf("\n")
exit(0)
}
function is_disarium(n, leng,sum,x) {
x = n
leng = length(n)
while (x != 0) {
sum += (x % 10) ^ leng
leng--
x = int(x/10)
}
return((sum == n) ? 1 : 0)
}
 
Output:
The first 19 Disarium numbers:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798

BASIC[edit]

BASIC256[edit]

function isDisarium(n)
digitos = length(string(n))
suma = 0
x = n
while x <> 0
suma += (x % 10) ^ digitos
digitos -= 1
x = x \ 10
end while
if suma = n then return True else return False
end function
 
limite = 19
cont = 0 : n = 0
print "The first"; limite; " Disarium numbers are:"
while cont < limite
if isDisarium(n) then
print n; " ";
cont += 1
endif
n += 1
end while
end
Output:
Igual que la entrada de FreeBASIC.

FreeBASIC[edit]

#define limite 19
 
Function isDisarium(n As Integer) As Boolean
Dim As Integer digitos = Len(Str(n))
Dim As Integer suma = 0, x = n
While x <> 0
suma += (x Mod 10) ^ digitos
digitos -= 1
x \= 10
Wend
Return Iif(suma = n, True, False)
End Function
 
Dim As Integer cont = 0, n = 0, i
Print "The first"; limite; " Disarium numbers are:"
Do While cont < limite
If isDisarium(n) Then
Print n; " ";
cont += 1
End If
n += 1
Loop
Sleep
Output:
Igual que la entrada de Python.

PureBasic[edit]

Procedure isDisarium(n.i)
digitos.i = Len(Str(n))
suma.i = 0
x.i = n
While x <> 0
r.i = (x % 10)
suma + Pow(r, digitos)
digitos - 1
x / 10
Wend
If suma = n
ProcedureReturn #True
Else
ProcedureReturn #False
EndIf
EndProcedure
 
OpenConsole()
limite.i = 19
cont.i = 0
n.i = 0
PrintN("The first" + Str(limite) + " Disarium numbers are:")
While cont < limite
If isDisarium(n)
Print(Str(n) + #TAB$)
cont + 1
EndIf
n + 1
Wend
Input()
CloseConsole()
Output:
Igual que la entrada de FreeBASIC.

C++[edit]

#include <vector>
#include <iostream>
#include <cmath>
#include <algorithm>
 
std::vector<int> decompose( int n ) {
std::vector<int> digits ;
while ( n != 0 ) {
digits.push_back( n % 10 ) ;
n /= 10 ;
}
std::reverse( digits.begin( ) , digits.end( ) ) ;
return digits ;
}
 
bool isDisarium( int n ) {
std::vector<int> digits( decompose( n ) ) ;
int exposum = 0 ;
for ( int i = 1 ; i < digits.size( ) + 1 ; i++ ) {
exposum += static_cast<int>( std::pow(
static_cast<double>(*(digits.begin( ) + i - 1 )) ,
static_cast<double>(i) )) ;
}
return exposum == n ;
}
 
int main( ) {
std::vector<int> disariums ;
int current = 0 ;
while ( disariums.size( ) != 18 ){
if ( isDisarium( current ) )
disariums.push_back( current ) ;
current++ ;
}
for ( int d : disariums )
std::cout << d << " " ;
std::cout << std::endl ;
return 0 ;
}
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427

Factor[edit]

Works with: Factor version 0.99 2021-06-02
USING: io kernel lists lists.lazy math.ranges math.text.utils
math.vectors prettyprint sequences ;
 
: disarium? ( n -- ? )
dup 1 digit-groups dup length 1 [a,b] v^ sum = ;
 
: disarium ( -- list ) 0 lfrom [ disarium? ] lfilter ;
 
19 disarium ltake [ pprint bl ] leach nl
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798 

Go[edit]

Translation of: Wren

A translation of Version 2.

Although Go has native unsigned 64 bit arithmetic, much quicker than I was expecting at a little under a minute.

package main
 
import (
"fmt"
"strconv"
)
 
const DMAX = 20 // maximum digits
const LIMIT = 20 // maximum number of disariums to find
 
func main() {
// Pre-calculated exponential and power serials
EXP := make([][]uint64, 1+DMAX)
POW := make([][]uint64, 1+DMAX)
 
EXP[0] = make([]uint64, 11)
EXP[1] = make([]uint64, 11)
POW[0] = make([]uint64, 11)
POW[1] = make([]uint64, 11)
for i := uint64(1); i <= 10; i++ {
EXP[1][i] = i
}
for i := uint64(1); i <= 9; i++ {
POW[1][i] = i
}
POW[1][10] = 9
 
for i := 2; i <= DMAX; i++ {
EXP[i] = make([]uint64, 11)
POW[i] = make([]uint64, 11)
}
for i := 1; i < DMAX; i++ {
for j := 0; j <= 9; j++ {
EXP[i+1][j] = EXP[i][j] * 10
POW[i+1][j] = POW[i][j] * uint64(j)
}
EXP[i+1][10] = EXP[i][10] * 10
POW[i+1][10] = POW[i][10] + POW[i+1][9]
}
 
// Digits of candidate and values of known low bits
DIGITS := make([]int, 1+DMAX) // Digits form
Exp := make([]uint64, 1+DMAX) // Number form
Pow := make([]uint64, 1+DMAX) // Powers form
 
var exp, pow, min, max uint64
start := 1
final := DMAX
count := 0
for digit := start; digit <= final; digit++ {
fmt.Println("# of digits:", digit)
level := 1
DIGITS[0] = 0
for {
// Check limits derived from already known low bit values
// to find the most possible candidates
for 0 < level && level < digit {
// Reset path to try next if checking in level is done
if DIGITS[level] > 9 {
DIGITS[level] = 0
level--
DIGITS[level]++
continue
}
 
// Update known low bit values
Exp[level] = Exp[level-1] + EXP[level][DIGITS[level]]
Pow[level] = Pow[level-1] + POW[digit+1-level][DIGITS[level]]
 
// Max possible value
pow = Pow[level] + POW[digit-level][10]
 
if pow < EXP[digit][1] { // Try next since upper limit is invalidly low
DIGITS[level]++
continue
}
 
max = pow % EXP[level][10]
pow -= max
if max < Exp[level] {
pow -= EXP[level][10]
}
max = pow + Exp[level]
 
if max < EXP[digit][1] { // Try next since upper limit is invalidly low
DIGITS[level]++
continue
}
 
// Min possible value
exp = Exp[level] + EXP[digit][1]
pow = Pow[level] + 1
 
if exp > max || max < pow { // Try next since upper limit is invalidly low
DIGITS[level]++
continue
}
 
if pow > exp {
min = pow % EXP[level][10]
pow -= min
if min > Exp[level] {
pow += EXP[level][10]
}
min = pow + Exp[level]
} else {
min = exp
}
 
// Check limits existence
if max < min {
DIGITS[level]++ // Try next number since current limits invalid
} else {
level++ // Go for further level checking since limits available
}
}
 
// All checking is done, escape from the main check loop
if level < 1 {
break
}
 
// Finally check last bit of the most possible candidates
// Update known low bit values
Exp[level] = Exp[level-1] + EXP[level][DIGITS[level]]
Pow[level] = Pow[level-1] + POW[digit+1-level][DIGITS[level]]
 
// Loop to check all last bits of candidates
for DIGITS[level] < 10 {
// Print out new Disarium number
if Exp[level] == Pow[level] {
s := ""
for i := DMAX; i > 0; i-- {
s += fmt.Sprintf("%d", DIGITS[i])
}
n, _ := strconv.ParseUint(s, 10, 64)
fmt.Println(n)
count++
if count == LIMIT {
fmt.Println("\nFound the first", LIMIT, "Disarium numbers.")
return
}
}
 
// Go to followed last bit candidate
DIGITS[level]++
Exp[level] += EXP[level][1]
Pow[level]++
}
 
// Reset to try next path
DIGITS[level] = 0
level--
DIGITS[level]++
}
fmt.Println()
}
}
Output:
# of digits: 1
0
1
2
3
4
5
6
7
8
9

# of digits: 2
89

# of digits: 3
135
175
518
598

# of digits: 4
1306
1676
2427

# of digits: 5

# of digits: 6

# of digits: 7
2646798

# of digits: 8

# of digits: 9

# of digits: 10

# of digits: 11

# of digits: 12

# of digits: 13

# of digits: 14

# of digits: 15

# of digits: 16

# of digits: 17

# of digits: 18

# of digits: 19

# of digits: 20
12157692622039623539

Found the first 20 Disarium numbers.

real	0m57.430s
user	0m57.420s
sys	0m0.105s

Haskell[edit]

module Disarium 
where
import Data.Char ( digitToInt)
 
isDisarium :: Int -> Bool
isDisarium n = (sum $ map (\(c , i ) -> (digitToInt c ) ^ i )
$ zip ( show n ) [1 , 2 ..]) == n
 
solution :: [Int]
solution = take 18 $ filter isDisarium [0, 1 ..]
 
Output:
[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427]

Julia[edit]

isdisarium(n) = sum(last(p)^first(p) for p in enumerate(reverse(digits(n)))) == n
 
function disariums(numberwanted)
n, ret = 0, Int[]
while length(ret) < numberwanted
isdisarium(n) && push!(ret, n)
n += 1
end
return ret
end
 
println(disariums(19))
@time disariums(19)
 
Output:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798]
  0.555962 seconds (5.29 M allocations: 562.335 MiB, 10.79% gc time)

Perl[edit]

use strict;
use warnings;
 
my ($n,@D) = (0, 0);
while (++$n) {
my($m,$sum);
map { $sum += $_ ** ++$m } split '', $n;
push @D, $n if $n == $sum;
last if 19 == @D;
}
print "@D\n";
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798

Phix[edit]

with javascript_semantics
constant limit = 19
integer count = 0, n = 0
printf(1,"The first 19 Disarium numbers are:\n")
while count<limit do
    atom dsum = 0
    string digits = sprintf("%d",n)
    for i=1 to length(digits) do
         dsum += power(digits[i]-'0',i)
    end for
    if dsum=n then
        printf(1," %d",n)
        count += 1
    end if
    n += 1
end while
Output:
The first 19 Disarium numbers are:
 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798

stretch[edit]

with javascript_semantics
-- translation of https://github.com/rgxgr/Disarium-Numbers/blob/master/Disarium.c
constant DMAX = iff(machine_bits()=64?20:7)

// Pre-calculated exponential & power serials
sequence exps = repeat(repeat(0,11),1+DMAX),
         pows = repeat(repeat(0,11),1+DMAX)
exps[1..2] = {{0,0,0,0,0,0,0,0,0,0,1},{0,1,2,3,4,5,6,7,8,9,10}}
pows[1..2] = {{0,0,0,0,0,0,0,0,0,0,0},{0,1,2,3,4,5,6,7,8,9, 9}}
for i=2 to DMAX do
    for j=1 to 10 do
        exps[i+1][j] = exps[i][j]*10
        pows[i+1][j] = pows[i][j]*(j-1)
    end for
    exps[i+1][11] = exps[i][11]*10
    pows[i+1][11] = pows[i][11] + pows[i+1][10]
end for

// Digits of candidate and values of known low bits
sequence digits = repeat(0,1+DMAX), // Digits form
         expl = repeat(0,1+DMAX),   // Number form
         powl = repeat(0,1+DMAX)    // Powers form

printf(1,"") -- (exclude console setup from timings [if pw.exe])
atom expn, powr, minn, maxx, t0 = time(), t1 = t0+1, count = 0
for digit=2 to DMAX+1 do
    printf(1,"Searching %d digits (started at %s):\n", {digit-1,elapsed(time()-t0)});
    integer level = 2
    digits[1] = 0
    while true do
        // Check limits derived from already known low bit values
        // to find the most possible candidates
        while 1<level and level<digit do
            // Reset path to try next if checking in level is done
            integer dl = digits[level]+1
            if dl>10 then
                digits[level] = 0;
                level -= 1
                digits[level] += 1
            else
                // Update known low bit values
                expl[level] = expl[level-1] + exps[level][dl]
                powl[level] = powl[level-1] + pows[digit-level+2][dl]

                // Max possible value
                powr = powl[level] + pows[digit-level+1][11]

                atom ed2 = exps[digit][2]
                if powr<ed2 then  // Try next since upper limit is invalidly low
                    digits[level] += 1
                else
                    atom el11 = exps[level][11],
                         el = expl[level]
                    maxx = remainder(powr,el11)
                    powr -= maxx
                    if maxx<el then
                        powr -= el11
                    end if
                    maxx = powr + el
                    if maxx<ed2 then  // Try next since upper limit is invalidly low
                        digits[level] += 1
                    else
                        // Min possible value
                        expn = el + ed2
                        powr = powl[level] + 1

                        if expn>maxx or maxx<powr then // Try next since upper limit is invalidly low
                            digits[level] += 1
                        else
                            if powr>expn then
                                minn = remainder(powr,el11)
                                powr -= minn
                                if minn>el then
                                    powr += el11
                                end if
                                minn = powr + el
                            else
                                minn = expn
                            end if

                            // Check limits existence
                            if maxx<minn then
                                digits[level] +=1   // Try next number since current limits invalid
                            else
                                level +=1   // Go for further level checking since limits available
                            end if
                        end if
                    end if
                end if
            end if
            if time()>t1 and platform()!=JS then
                progress("working:%v... (%s)",{digits,elapsed(time()-t0)})
                t1 = time()+1
            end if
        end while
      
        // All checking is done, escape from the main check loop
        if level<2 then exit end if

        // Final check last bit of the most possible candidates
        // Update known low bit values
        integer dlx = digits[level]+1
        expl[level] = expl[level-1] + exps[level][dlx];
        powl[level] = powl[level-1] + pows[digit+1-level][dlx];

        // Loop to check all last bit of candidates
        while digits[level]<10 do
            // Print out new disarium number
            if expl[level] == powl[level] then
                if platform()!=JS then progress("") end if
                integer ld = max(trim_tail(digits,0,true),2)
                printf(1,"%s\n",{reverse(join(apply(digits[2..ld],sprint),""))})
                count += 1
            end if

            // Go to followed last bit candidate
            digits[level] += 1
            expl[level] += exps[level][2]
            powl[level] += 1
        end while

        // Reset to try next path
        digits[level] = 0;
        level -= 1
        digits[level] += 1
    end while
    if platform()!=JS then progress("") end if
end for
printf(1,"%d disarium numbers found (%s)\n",{count,elapsed(time()-t0)})
Output:
Searching 1 digits (started at 0s):
0
1
2
3
4
5
6
7
8
9
Searching 2 digits (started at 0s):
89
Searching 3 digits (started at 0s):
135
175
518
598
Searching 4 digits (started at 0s):
1306
1676
2427
Searching 5 digits (started at 0.0s):
Searching 6 digits (started at 0.0s):
Searching 7 digits (started at 0.0s):
2646798
Searching 8 digits (started at 0.0s):
Searching 9 digits (started at 0.0s):
Searching 10 digits (started at 0.0s):
Searching 11 digits (started at 0.1s):
Searching 12 digits (started at 0.1s):
Searching 13 digits (started at 0.3s):
Searching 14 digits (started at 0.8s):
Searching 15 digits (started at 2.5s):
Searching 16 digits (started at 6.9s):
Searching 17 digits (started at 23.2s):
Searching 18 digits (started at 1 minute and 8s):
Searching 19 digits (started at 3 minutes and 35s):
Searching 20 digits (started at 10 minutes and 8s):
12157692622039623539
20 disarium numbers found (2 hours and 7s)

Takes about 48min to find the 20 digit number, then trundles away for over another hour. I think that technically it should also scan for 21 and 22 digit numbers to be absolutely sure there aren't any, but that certainly exceeds my patience.

Picat[edit]

Iterative approach[edit]

Translation of: Python
main =>
Limit = 19,
D = [],
N = 0,
printf("The first %d Disarium numbers are:\n",Limit),
while (D.len < Limit)
if disarium_number(N) then
D := D ++ [N]
end,
N := N + 1,
if N mod 10_000_000 == 0 then
println(test=N)
end
end,
println(D).
 
disarium_number(N) =>
Sum = 0,
Digits = N.to_string.len,
X = N,
while (X != 0, Sum <= N)
Sum := Sum + (X mod 10) ** Digits,
Digits := Digits - 1,
X := X div 10
end,
Sum == N.
Output:
The first 19 Disarium numbers are:
[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427,2646798]

2.905s

Constraint modelling[edit]

A faster approach is to use constraint modeling. It finds the first 19 Disarium numbers in 0.591s (vs 2.905s for the iterative approach).

Note that the domain of Picat's constraint variables is -2**56..2**56 (about 10**17) which means that this approach cannot be used to handle numbers of length 20.

The cp solver and sat solvers takes about the same time for finding the 7-digits number 2646798, but the sat solver is much faster for checking longer numbers; it took almost 9 minutes to prove that there are no Disarium numbers of length 8..17.

import sat.
% import cp.
 
main =>
D = [],
Limit = 19,
Base = 10,
foreach(Len in 1..20, D.len < Limit)
Nums = disarium_number_cp(Len,Base),
if Nums.len > 0 then
foreach(Num in Nums)
B = to_radix_string(Num,Base),
D := D ++ [B]
end
end
end,
printf("The first %d Disarius numbers in base %d:\n",D.len, Base),
println(D[1..Limit]),
nl.
 
% Find all Disarium of a certain length
disarium_number_cp(Len,Base) = findall(N,disarium_number_cp(Len,Base,N)).sort.
 
% Find a Disarium number of a certain length
disarium_number_cp(Len,Base,N) =>
X = new_list(Len),
X :: 0..Base-1,
N :: Base**(Len-1)-1..Base**Len-1,
N #= sum([X[I]**I : I in 1..Len]),
to_num(X,Base,N), % convert X <=> N
solve($[],X++[N]).
 
% Converts a number Num to/from a list of integer List given a base Base
to_num(List, Base, Num) =>
Len = length(List),
Num #= sum([List[I]*Base**(Len-I) : I in 1..Len]).
Output:
The first 19 Disarius numbers in base 10:
[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427,2646798]

0.591s

Finding the first 23 Disarium numbers in base 11 is easier:

The first 23 Disarius numbers in base 11:
[0,1,2,3,4,5,6,7,8,9,A,25,36,9A,105,438,488,609,85A,86A,2077,40509,43789]

0.015s

(And finding the first 36 Disarium numbers in base 36 is even easier: 0..Z.)


Python[edit]

#!/usr/bin/python
 
def isDisarium(n):
digitos = len(str(n))
suma = 0
x = n
while x != 0:
suma += (x % 10) ** digitos
digitos -= 1
x //= 10
if suma == n:
return True
else:
return False
 
if __name__ == '__main__':
limite = 19
cont = 0
n = 0
print("The first",limite,"Disarium numbers are:")
while cont < limite:
if isDisarium(n):
print(n, end = " ")
cont += 1
n += 1
Output:
The first 19 Disarium numbers are:
 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798

Raku[edit]

Not an efficient algorithm. First 18 in less than 1/4 second. 19th in around 45 seconds. Pretty much unusable for the 20th.

my $disarium = (^).hyper.map: { $_ if $_ == sum .polymod(10 xx *).reverse Z** 1..* };
 
put $disarium[^18];
put $disarium[18];
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427
2646798

Sidef[edit]

func is_disarium(n) {
n.digits.flip.sum_kv{|k,d| d**(k+1) } == n
}
 
say 18.by(is_disarium)
Output:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427]

Wren[edit]

Library: Wren-math

Version 1 (Brute force)[edit]

This version finds the first 19 Disarium numbers in 3.35 seconds though, clearly, finding the 20th is out of the question with this approach.

As a possible optimization, I tried caching all possible digit powers but there was no perceptible difference in running time for numbers up to 7 digits long.

import "./math" for Int
 
var limit = 19
var count = 0
var disarium = []
var n = 0
while (count < limit) {
var sum = 0
var digits = Int.digits(n)
for (i in 0...digits.count) sum = sum + digits[i].pow(i+1)
if (sum == n) {
disarium.add(n)
count = count + 1
}
n = n + 1
}
System.print("The first 19 Disarium numbers are:")
System.print(disarium)
Output:
The first 19 Disarium numbers are:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798]


Version 2 (Much faster)[edit]

This is a translation of the C code referred to in the Phix entry and finds the first 19 Disarium numbers in 0.012 seconds.

Efficient though this method is, unfortunately finding the 20th is still out of reasonable reach for Wren. If we let this run until 15 digit numbers have been examined (the most that 53 bit integer math can accurately manage), then the time taken rises to 19 seconds - roughly 3 times slower than Phix.

However, we need 64 bit integer arithmetic to get up to 20 digits and this requires the use of Wren-long which (as it's written entirely in Wren, not C) needs about 7 times longer (2 minutes 16 seconds) to even reach 15 digits. Using BigInt or GMP would be even slower.

So, if the Phix example requires 48 minutes to find the 20th number, it would probably take Wren the best part of a day to do the same which is far longer than I have patience for.

var DMAX  = 7  // maxmimum digits
var LIMIT = 19 // maximum number of Disariums to find
 
// Pre-calculated exponential and power serials
var EXP = List.filled(1 + DMAX, null)
var POW = List.filled(1 + DMAX, null)
EXP[0] = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
EXP[1] = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
POW[0] = List.filled(11, 0)
POW[1] = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9]
for (i in 2..DMAX) {
EXP[i] = List.filled(11, 0)
POW[i] = List.filled(11, 0)
}
for (i in 1...DMAX) {
for (j in 0..9) {
EXP[i+1][j] = EXP[i][j] * 10
POW[i+1][j] = POW[i][j] * j
}
EXP[i+1][10] = EXP[i][10] * 10
POW[i+1][10] = POW[i][10] + POW[i+1][9]
}
 
// Digits of candidate and values of known low bits
var DIGITS = List.filled(1 + DMAX, 0) // Digits form
var Exp = List.filled(1 + DMAX, 0) // Number form
var Pow = List.filled(1 + DMAX, 0) // Powers form
 
var exp
var pow
var min
var max
var start = 1
var final = DMAX
var count = 0
for (digit in start..final) {
System.print("# of digits: %(digit)")
var level = 1
DIGITS[0] = 0
while (true) {
// Check limits derived from already known low bit values
// to find the most possible candidates
while (0 < level && level < digit) {
// Reset path to try next if checking in level is done
if (DIGITS[level] > 9) {
DIGITS[level] = 0
level = level - 1
DIGITS[level] = DIGITS[level] + 1
continue
}
 
// Update known low bit values
Exp[level] = Exp[level - 1] + EXP[level][DIGITS[level]]
Pow[level] = Pow[level - 1] + POW[digit + 1 - level][DIGITS[level]]
 
// Max possible value
pow = Pow[level] + POW[digit - level][10]
 
if (pow < EXP[digit][1]) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}
 
max = pow % EXP[level][10]
pow = pow - max
if (max < Exp[level]) pow = pow - EXP[level][10]
max = pow + Exp[level]
 
if (max < EXP[digit][1]) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}
 
// Min possible value
exp = Exp[level] + EXP[digit][1]
pow = Pow[level] + 1
 
if (exp > max || max < pow) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}
 
if (pow > exp ) {
min = pow % EXP[level][10]
pow = pow - min
if (min > Exp[level]) {
pow = pow + EXP[level][10]
}
min = pow + Exp[level]
} else {
min = exp
}
 
// Check limits existence
if (max < min) {
DIGITS[level] = DIGITS[level] + 1 // Try next number since current limits invalid
} else {
level= level + 1 // Go for further level checking since limits available
}
}
 
// All checking is done, escape from the main check loop
if (level < 1) break
 
// Finally check last bit of the most possible candidates
// Update known low bit values
Exp[level] = Exp[level - 1] + EXP[level][DIGITS[level]]
Pow[level] = Pow[level - 1] + POW[digit + 1 - level][DIGITS[level]]
 
// Loop to check all last bits of candidates
while (DIGITS[level] < 10) {
// Print out new Disarium number
if (Exp[level] == Pow[level]) {
var s = ""
for (i in DMAX...0) s = s + DIGITS[i].toString
System.print(Num.fromString(s))
count = count + 1
if (count == LIMIT) {
System.print("\nFound the first %(LIMIT) Disarium numbers.")
return
}
}
 
// Go to followed last bit candidate
DIGITS[level] = DIGITS[level] + 1
Exp[level] = Exp[level] + EXP[level][1]
Pow[level] = Pow[level] + 1
}
 
// Reset to try next path
DIGITS[level] = 0
level = level - 1
DIGITS[level] = DIGITS[level] + 1
}
System.print()
}
Output:
# of digits: 1
0
1
2
3
4
5
6
7
8
9

# of digits: 2
89

# of digits: 3
135
175
518
598

# of digits: 4
1306
1676
2427

# of digits: 5

# of digits: 6

# of digits: 7
2646798

Found the first 19 Disarium numbers.

real	0m0.012s
user	0m0.008s
sys	0m0.004s


Version 3 (Embedded)[edit]

Library: Wren-i64

An initial outing for the above module which aims to speed up 64 bit arithmetic in Wren by wrapping the corresponding C99 fixed size types.

Early indications are that this is at least 4 times faster than Wren-long as it can search up to 15 digits in about 29 seconds which in turn is about 4 times slower than the Phix entry.

This suggested that the 20th Disarium number would be found in around 3.5 hours so I thought I'd have a go. However, after taking consistently 4 times longer than Phix to search each digit length up to 19, I was pleasantly surprised when the 20th number popped up after only 81 minutes!

I haven't bothered to search all 20 digits numbers up to the unsigned 64 limit as this would take far longer and, of course, be fruitless in any case.

import "./i64" for U64
 
var DMAX = 20 // maxmimum digits
var LIMIT = 20 // maximum number of disariums to find
 
// Pre-calculated exponential and power serials
var EXP = List.filled(1 + DMAX, null)
var POW = List.filled(1 + DMAX, null)
EXP[0] = List.filled(11, null)
EXP[1] = List.filled(11, null)
POW[0] = List.filled(11, null)
POW[1] = List.filled(11, null)
for (i in 0..9) EXP[0][i] = U64.zero
EXP[0][10] = U64.one
 
for (i in 0..10) EXP[1][i] = U64.from(i)
 
for (i in 0..10) POW[0][i] = U64.zero
 
for (i in 0..9) POW[1][i] = U64.from(i)
POW[1][10] = U64.from(9)
 
for (i in 2..DMAX) {
EXP[i] = List.filled(11, null)
POW[i] = List.filled(11, null)
for (j in 0..10) {
EXP[i][j] = U64.zero
POW[i][j] = U64.zero
}
}
for (i in 1...DMAX) {
for (j in 0..9) {
EXP[i+1][j] = EXP[i][j] * 10
POW[i+1][j] = POW[i][j] * j
}
EXP[i+1][10] = EXP[i][10] * 10
POW[i+1][10] = POW[i][10] + POW[i+1][9]
}
 
// Digits of candidate and values of known low bits
var DIGITS = List.filled(1 + DMAX, 0) // Digits form
var Exp = List.filled(1 + DMAX, null) // Number form
var Pow = List.filled(1 + DMAX, null) // Powers form
for (i in 0..DMAX) {
Exp[i] = U64.zero
Pow[i] = U64.zero
}
 
var exp = U64.new()
var pow = U64.new()
var min = U64.new()
var max = U64.new()
var start = 1
var final = DMAX
var count = 0
for (digit in start..final) {
System.print("# of digits: %(digit)")
var level = 1
DIGITS[0] = 0
while (true) {
// Check limits derived from already known low bit values
// to find the most possible candidates
while (0 < level && level < digit) {
// Reset path to try next if checking in level is done
if (DIGITS[level] > 9) {
DIGITS[level] = 0
level = level - 1
DIGITS[level] = DIGITS[level] + 1
continue
}
 
// Update known low bit values
Exp[level].add(Exp[level - 1], EXP[level][DIGITS[level]])
Pow[level].add(Pow[level - 1], POW[digit + 1 - level][DIGITS[level]])
 
// Max possible value
pow.add(Pow[level], POW[digit - level][10])
 
if (pow < EXP[digit][1]) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}
 
max.rem(pow, EXP[level][10])
pow.sub(max)
if (max < Exp[level]) pow.sub(EXP[level][10])
max.add(pow, Exp[level])
 
if (max < EXP[digit][1]) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}
 
// Min possible value
exp.add(Exp[level], EXP[digit][1])
pow.add(Pow[level], 1)
 
if (exp > max || max < pow) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}
 
if (pow > exp ) {
min.rem(pow, EXP[level][10])
pow.sub(min)
if (min > Exp[level]) {
pow.add(EXP[level][10])
}
min.add(pow, Exp[level])
} else {
min.set(exp)
}
 
// Check limits existence
if (max < min) {
DIGITS[level] = DIGITS[level] + 1 // Try next number since current limits invalid
} else {
level = level + 1 // Go for further level checking since limits available
}
}
 
// All checking is done, escape from the main check loop
if (level < 1) break
 
// Final check last bit of the most possible candidates
// Update known low bit values
Exp[level].add(Exp[level - 1], EXP[level][DIGITS[level]])
Pow[level].add(Pow[level - 1], POW[digit + 1 - level][DIGITS[level]])
 
// Loop to check all last bit of candidates
while (DIGITS[level] < 10) {
// Print out new disarium number
if (Exp[level] == Pow[level]) {
var s = ""
for (i in DMAX...0) s = s + DIGITS[i].toString
s = s.trimStart("0")
if (s == "") s = "0"
System.print(s)
count = count + 1
if (count == LIMIT) {
if (LIMIT < 20) {
System.print("\nFound the first %(LIMIT) Disarium numbers.")
} else {
System.print("\nFound all 20 Disarium numbers.")
}
return
}
}
 
// Go to followed last bit candidate
DIGITS[level] = DIGITS[level] + 1
Exp[level].add(Exp[level], EXP[level][1])
Pow[level].inc
}
 
// Reset to try next path
DIGITS[level] = 0
level = level - 1
DIGITS[level] = DIGITS[level] + 1
}
System.print()
}
Output:
# of digits: 1
0
1
2
3
4
5
6
7
8
9

# of digits: 2
89

# of digits: 3
135
175
518
598

# of digits: 4
1306
1676
2427

# of digits: 5

# of digits: 6

# of digits: 7
2646798

# of digits: 8

# of digits: 9

# of digits: 10

# of digits: 11

# of digits: 12

# of digits: 13

# of digits: 14

# of digits: 15

# of digits: 16

# of digits: 17

# of digits: 18

# of digits: 19

# of digits: 20
12157692622039623539

Found all 20 Disarium numbers.

real	81m16.365s
user	81m16.181s
sys	0m0.016s

XPL0[edit]

1.35 seconds on Pi4.

func Disarium(N);       \Return 'true' if N is a Disarium number
int N, N0, D(10), A(10), I, J, Sum;
[N0:= N;
for J:= 0 to 10-1 do A(J):= 1;
I:= 0;
repeat N:= N/10;
D(I):= rem(0);
I:= I+1;
for J:= 0 to I-1 do
A(J):= A(J) * D(J);
until N = 0;
Sum:= 0;
for J:= 0 to I-1 do
Sum:= Sum + A(J);
return Sum = N0;
];
 
int Cnt, N;
[Cnt:= 0; N:= 0;
loop [if Disarium(N) then
[IntOut(0, N); ChOut(0, ^ );
Cnt:= Cnt+1;
if Cnt >= 19 then quit;
];
N:= N+1;
];
]
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798