How to resolve the algorithm Disarium numbers step by step in the Wren programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Disarium numbers step by step in the Wren programming language
Table of Contents
Problem Statement
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.
135 is a Disarium number: 11 + 32 + 53 == 1 + 9 + 125 == 135 There are a finite number of Disarium numbers.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Disarium numbers step by step in the Wren programming language
Source code in the wren programming language
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)
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()
}
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()
}
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