Iterated digits squaring
You are encouraged to solve this task according to the task description, using any language you may know.
If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89:
15 -> 26 -> 40 -> 16 -> 37 -> 58 -> 89 7 -> 49 -> 97 -> 130 -> 10 -> 1
An example in Python:
<lang python>>>> step = lambda x: sum(int(d) ** 2 for d in str(x)) >>> iterate = lambda x: x if x in [1, 89] else iterate(step(x)) >>> [iterate(x) for x in xrange(1, 20)] [1, 89, 89, 89, 89, 89, 1, 89, 89, 1, 89, 89, 1, 89, 89, 89, 89, 89, 1]</lang>
- Task
- Count how many number chains for integers 1 <= n < 100_000_000 end with a value 89.
Or, for much less credit - (showing that your algorithm and/or language is slow):
- Count how many number chains for integers 1 <= n < 1_000_000 end with a value 89.
This problem derives from the Project Euler problem 92.
For a quick algorithm for this task see the talk page
- Cf
ALGOL 68
Brute-force with some caching. <lang algol68># count the how many numbers up to 100 000 000 have squared digit sums of 89 #
- compute a table of the sum of the squared digits of the numbers 00 to 99 #
[ 0 : 99 ]INT digit pair square sum; FOR d1 FROM 0 TO 9 DO
FOR d2 FROM 0 TO 9 DO
digit pair square sum[ ( d1 * 10 ) + d2 ] := ( d1 * d1 ) + ( d2 * d2 )
OD
OD;
- returns the sum of the squared digits of n #
PROC squared digit sum = ( INT n )INT:
BEGIN
INT result := 0;
INT rest := n;
WHILE rest /= 0 DO
INT digit pair = rest MOD 100;
result PLUSAB digit pair square sum[ digit pair ];
rest OVERAB 100
OD;
result
END # squared digit sum # ;
- for values up to 100 000 000, the largest squred digit sum will be that of 99 999 999 #
- i.e. 81 * 8 = 648, we will cache the values of the squared digit sums #
INT cache max = 81 * 8; [ 1 : cache max ]INT cache; FOR i TO cache max DO cache[ i ] := 0 OD;
INT count 89 := 0;
- fill in the cache #
FOR value FROM 2 TO cache max DO cache[ value ] := squared digit sum( value ) OD;
- we "know" that 89 and 1 are the terminal values #
cache[ 1 ] := 1; cache[ 89 ] := 89; FOR value FROM 2 TO cache max DO
INT sum := cache[ value ];
WHILE sum /= 1 AND sum /= 89 DO
sum := cache[ sum ]
OD;
cache[ value ] := sum
OD;
FOR value FROM 1 TO 100 000 000 DO
IF cache[ squared digit sum( value ) ] = 89 THEN count 89 +:= 1 FI
OD;
print( ( "Number of values whose squared digit sum is 89: ", whole( count 89, -10 ), newline ) )</lang>
- Output:
Number of values whose squared digit sum is 89: 85744333
BBC BASIC
Three versions timed on a 2.50GHz Intel Desktop. <lang bbcbasic> REM Version 1: Brute force
REM ---------------------------------------------------------
T%=TIME
N%=0
FOR I%=1 TO 100000000
J%=I%
REPEAT
K%=0:REPEAT K%+=(J%MOD10)^2:J%=J%DIV10:UNTIL J%=0
J%=K%
UNTIL J%=89 OR J%=1
IF J%>1 N%+=1
NEXT
PRINT "Version 1: ";N% " in ";(TIME-T%)/100 " seconds."
REM Version 2: Brute force + building lookup table
REM ---------------------------------------------------------
T%=TIME
DIM B% 9*9*8,H%(9)
N%=0
FOR I%=1 TO 100000000
J%=I%
H%=0
REPEAT
K%=0:REPEAT K%+=(J%MOD10)^2:J%=J%DIV10:UNTIL J%=0
H%(H%)=K%:H%+=1
J%=K%
IF B%?J%=1 EXIT REPEAT
UNTIL J%=89 OR J%=1
IF J%>1 N%+=1:WHILE H%>0:H%-=1:B%?H%(H%)=1:ENDWHILE
NEXT
PRINT "Version 2: ";N% " in ";(TIME-T%)/100 " seconds."
REM Version 3: Calc possible combinations (translation of C)
REM ---------------------------------------------------------
T%=TIME
DIM B%(9*9*8):B%(0)=1
FOR N%=1 TO 8
FOR I%=9*9*N% TO 1 STEP -1
FOR J%=1 TO 9
S%=J%*J%
IF S%>I% EXIT FOR
B%(I%)+=B%(I%-S%)
NEXT
NEXT
NEXT
N%=0
FOR I%=1 TO 9*9*8
J%=I%
REPEAT
K%=0:REPEAT K%+=(J%MOD10)^2:J%=J%DIV10:UNTIL J%=0
J%=K%
UNTIL J%=89 OR J%=1
IF J%>1 N%+=B%(I%)
NEXT
PRINT "Version 3: ";N% " in ";(TIME-T%)/100 " seconds."
END</lang>
- Output:
Version 1: 85744333 in 1447.08 seconds. Version 2: 85744333 in 718.04 seconds. Version 3: 85744333 in 0.02 seconds.
C
C99, tested with "gcc -std=c99". Record how many digit square sum combinations there are. This reduces numbers to , and the complexity is about . The 64 bit integer counter is good for up to , which takes practically no time to run. <lang c>#include <stdio.h>
typedef unsigned long long ull;
int is89(int x) { while (1) { int s = 0; do s += (x%10)*(x%10); while ((x /= 10));
if (s == 89) return 1; if (s == 1) return 0; x = s; } }
int main(void)
{
// array bounds is sort of random here, it's big enough for 64bit unsigned.
ull sums[32*81 + 1] = {1, 0};
for (int n = 1; ; n++) { for (int i = n*81; i; i--) { for (int j = 1; j < 10; j++) { int s = j*j; if (s > i) break; sums[i] += sums[i-s]; } }
ull count89 = 0; for (int i = 1; i < n*81 + 1; i++) { if (!is89(i)) continue;
if (sums[i] > ~0ULL - count89) { printf("counter overflow for 10^%d\n", n); return 0; } count89 += sums[i]; }
printf("1->10^%d: %llu\n", n, count89); }
return 0; }</lang>
- Output:
1->10^1: 7 1->10^2: 80 1->10^3: 857 1->10^4: 8558 1->10^5: 85623 1->10^6: 856929 1->10^7: 8581146 1->10^8: 85744333 1->10^9: 854325192 1->10^10: 8507390852 1->10^11: 84908800643 1->10^12: 850878696414 1->10^13: 8556721999130 1->10^14: 86229146720315 1->10^15: 869339034137667 1->10^16: 8754780882739336 1->10^17: 87975303595231975 1->10^18: 881773944919974509 1->10^19: 8816770037940618762 counter overflow for 10^20
Fast C implementation (<1 second my machine), which performs iterated digits squaring only once for each unique 8 digit combination. The cases 0 and 100,000,000 are ignored since they don't sum to 89: <lang c>
- include <stdio.h>
const int digits[] = { 0,1,2,3,4,5,6,7,8,9 };
// calculates factorial of a number int factorial(int n) {
return n == 0 ? 1 : n * factorial(n - 1);
}
// returns sum of squares of digits of n unsigned int sum_square_digits(unsigned int n) {
int i,num=n,sum=0;
// process digits one at a time until there are none left
while (num > 0) {
// peal off the last digit from the number
int digit=num % 10;
num=(num - digit)/10;
// add it's square to the sum
sum=sum+digit*digit;
}
return sum;
}
// builds all combinations digits 0-9 of length len // for each of these it will perform iterated digit squaring // and for those which result in 89 add to a counter which is // passed by pointer. long choose_sum_and_count_89(int * got, int n_chosen, int len, int at, int max_types, int *count89) {
int i;
long count = 0;
int digitcounts[10];
for (i=0; i < 10; i++) {
digitcounts[i]=0;
}
if (n_chosen == len) {
if (!got) return 1;
int sum=0;
for (i = 0; i < len; i++) {
int digit=digits[got[i]];
digitcounts[digit]++;
sum=sum + digit * digit;
}
if (sum == 0) {
return 1;
}
if ((sum != 1) && (sum != 89)) {
while ((sum != 1) && (sum != 89)) {
sum=sum_square_digits(sum);
}
}
if (sum == 89) {
int count_this_comb=factorial(len);
for (i=0; i<10; i++) {
count_this_comb/=factorial(digitcounts[i]);
}
(*count89)+=count_this_comb;
}
return 1;
}
for (i = at; i < max_types; i++) {
if (got) got[n_chosen] = i;
count += choose_sum_and_count_89(got, n_chosen + 1, len, i, max_types, count89);
}
return count;
}
int main(void) {
int chosen[10];
int count=0;
// build all unique 8 digit combinations which represent
// numbers 0-99,999,999 and count those
// whose iterated digit squaring sum to 89
// case 0, 100,000,000 are ignored since they don't sum to 89
choose_sum_and_count_89(chosen, 0, 8, 0, 10, &count);
printf("%d\n",count);
return 0;
} </lang>
- Output:
85744333
C++
Slow (~10 seconds on my machine) brute force C++ implementation: <lang cpp>
- include <iostream>
// returns sum of squares of digits of n unsigned int sum_square_digits(unsigned int n) {
int i,num=n,sum=0;
// process digits one at a time until there are none left
while (num > 0) {
// peal off the last digit from the number
int digit=num % 10;
num=(num - digit)/10;
// add it's square to the sum
sum+=digit*digit;
}
return sum;
} int main(void) {
unsigned int i=0,result=0, count=0;
for (i=1; i<=100000000; i++) {
// if not 1 or 89, start the iteration
if ((i != 1) || (i != 89)) {
result = sum_square_digits(i);
}
// otherwise we're done already
else {
result = i;
}
// while we haven't reached 1 or 89, keep iterating
while ((result != 1) && (result != 89)) {
result = sum_square_digits(result);
}
if (result == 89) {
count++;
}
}
std::cout << count << std::endl;
return 0;
} </lang>
- Output:
85744333
Ceylon
<lang ceylon>shared void run() {
function digitsSquaredSum(variable Integer n) { variable value total = 0; while(n > 0) { total += (n % 10) ^ 2; n /= 10; } return total; }
function lastSum(variable Integer n) { while(true) { n = digitsSquaredSum(n); if(n == 89 || n == 1) { return n; } } }
variable value eightyNines = 0; for(i in 1..100M - 1) { if(lastSum(i) == 89) { eightyNines++; } } print(eightyNines); }</lang>
Clojure
Direct Method
<lang lisp>(ns async-example.core
(:require [clojure.math.numeric-tower :as math]) (:use [criterium.core]) (:gen-class))
(defn sum-sqr [digits]
" Square sum of list of digits "
(let [digits-sqr (fn [n]
(apply + (map #(* % %) digits)))]
(digits-sqr digits)))
(defn get-digits [n]
" Converts a digit to a list of digits (e.g. 545 -> ((5) (4) (5)) (used for squaring digits) " (map #(Integer/valueOf (str %)) (String/valueOf n)))
(defn -isNot89 [x]
" Returns nil on 89 " (cond (= x 0) 0 (= x 89) nil (= x 1) 0 (< x 10) (recur (* x x)) :else (recur (sum-sqr (get-digits x)))))
- Cached version of isNot89 (i.e. remembers prevents inputs, and returns result by looking it up when input repeated)
(def isNot89 (memoize -isNot89))
(defn direct-method [ndigits]
" Simple approach of looping through all the numbers from 0 to 10^ndigits - 1 " (->> (math/expt 10 ndigits) (range 0) ; 0 to 10^ndigits (filter #(isNot89 (sum-sqr (get-digits %)))) ; filters out 89 (count) ; count non-89 (- (math/expt 10 ndigits)))) ; count 89 (10^ndigits - (count 89))
(time (println (direct-method 8)))
</lang>
- Output:
85744333 Time: 335 seconds
Using Combinations
<lang> (def DIGITS (range 0 10))
(defn -factorial [n]
(apply * (take n (iterate inc 1))))
- Cached version of factorial
(def factorial (memoize -factorial))
(defn -combinations [coll k]
" From http://rosettacode.org/wiki/Combinations_with_repetitions#Clojure " (when-let [[x & xs] coll] (if (= k 1) (map list coll) (concat (map (partial cons x) (-combinations coll (dec k))) (-combinations xs k)))))
- Cached version of combinations
(def combinations (memoize -combinations))
(defn comb [n r]
" count of n items select r " (/ (/ (factorial n) (factorial r)) (factorial (- n r))))
(defn count-digits [digit-list]
" count nunmber of occurences of digit in list "
(reduce (fn [m v] (update-in m [v] (fnil inc 0))) {} digit-list))
(defn count-patterns [c]
" Count of number of patterns with these digits "
(->>
c
(count-digits)
(reduce (fn [accum [k v]]
(* accum (factorial v)))
1)
(/ (factorial (count c)))))
(defn itertools-comb [ndigits]
(->>
ndigits
(combinations DIGITS)
(filter #(is89 (sum-sqr %))) ; items which are not 89 (i.e. 1 since lower count)
(reduce (fn [acc c]
(+ acc (count-patterns c)))
0)
(- (math/expt 10 ndigits))))
(println (itertools-comb 8))
- Time obtained using benchmark library (i.e. (bench (itertools-comb 8)) )
</lang>
{
- Output:
85744333
Time: 78 ms (i.e. using combinations was over 4,000 times faster
both tested on i7 CPU 920@2.67GHZ)
Common Lisp
<lang lisp> (defun square (number)
(expt number 2))
(defun list-digits (number)
"Return the `number' as a list of its digits."
(loop
:for (rest digit) := (multiple-value-list (truncate number 10))
:then (multiple-value-list (truncate rest 10))
:collect digit
:until (zerop rest)))
(defun next (number)
(loop :for digit :in (list-digits number) :sum (square digit)))
(defun chain-end (number)
"Return the ending number after summing the squaring of the digits of
`number'. Either 1 or 89."
(loop
:for next := (next number) :then (next next)
:until (or (eql next 1)
(eql next 89))
:finally (return next)))
(time
(loop
:with count := 0
:for candidate :from 1 :upto 100000000
:do (when (eql 89 (chain-end candidate))
(incf count))
:finally (return count)))
</lang>
- Output:
Evaluation took: 1128.773 seconds of real time 1126.231095 seconds of total run time (1117.296987 user, 8.934108 system) [ Run times consist of 56.419 seconds GC time, and 1069.813 seconds non-GC time. ] 99.77% CPU 2,815,545,509,836 processor cycles 580,663,356,272 bytes consed *
D
A simple memoizing partially-imperative brute-force solution: <lang d>import std.stdio, std.algorithm, std.range, std.functional;
uint step(uint x) pure nothrow @safe @nogc {
uint total = 0;
while (x) {
total += (x % 10) ^^ 2;
x /= 10;
}
return total;
}
uint iterate(in uint x) nothrow @safe {
return (x == 89 || x == 1) ? x : x.step.memoize!iterate;
}
void main() {
iota(1, 100_000_000).filter!(x => x.iterate == 89).count.writeln;
}</lang>
- Output:
85744333
The run-time is about 10 seconds compiled with ldc2.
A fast imperative brute-force solution: <lang d>void main() nothrow @nogc {
import core.stdc.stdio: printf;
enum uint magic = 89; enum uint limit = 100_000_000; uint[(9 ^^ 2) * 8 + 1] lookup = void;
uint[10] squares;
foreach (immutable i, ref x; squares)
x = i ^^ 2;
foreach (immutable uint i; 1 .. lookup.length) {
uint x = i;
while (x != magic && x != 1) {
uint total = 0;
while (x) {
total += squares[(x % 10)];
x /= 10;
}
x = total;
}
lookup[i] = x == magic; }
uint magicCount = 0;
foreach (immutable uint i; 1 .. limit) {
uint x = i;
uint total = 0;
while (x) {
total += squares[(x % 10)];
x /= 10;
}
magicCount += lookup[total]; }
printf("%u\n", magicCount);
}</lang> The output is the same. The run-time is less than 3 seconds compiled with ldc2.
A more efficient solution: <lang d>import core.stdc.stdio, std.algorithm, std.range;
enum factorial = (in uint n) pure nothrow @safe @nogc
=> reduce!q{a * b}(1u, iota(1u, n + 1));
uint iLog10(in uint x) pure nothrow @safe @nogc in {
assert(x > 0);
} body {
return (x >= 1_000_000_000) ? 9 :
(x >= 100_000_000) ? 8 :
(x >= 10_000_000) ? 7 :
(x >= 1_000_000) ? 6 :
(x >= 100_000) ? 5 :
(x >= 10_000) ? 4 :
(x >= 1_000) ? 3 :
(x >= 100) ? 2 :
(x >= 10) ? 1 : 0;
}
uint nextStep(uint x) pure nothrow @safe @nogc {
typeof(return) result = 0;
while (x > 0) {
result += (x % 10) ^^ 2;
x /= 10;
}
return result;
}
uint check(in uint[] number) pure nothrow @safe @nogc {
uint candidate = reduce!((tot, n) => tot * 10 + n)(0, number);
while (candidate != 89 && candidate != 1)
candidate = candidate.nextStep;
if (candidate == 89) {
uint[10] digitsCount;
foreach (immutable d; number)
digitsCount[d]++;
return reduce!((r, c) => r / c.factorial)
(number.length.factorial, digitsCount);
}
return 0;
}
void main() nothrow @nogc {
enum uint limit = 100_000_000; immutable uint cacheSize = limit.iLog10;
uint[cacheSize] number; uint result = 0; uint i = cacheSize - 1;
while (true) {
if (i == 0 && number[i] == 9)
break;
if (i == cacheSize - 1 && number[i] < 9) {
number[i]++;
result += number.check;
} else if (number[i] == 9) {
i--;
} else {
number[i]++;
number[i + 1 .. $] = number[i];
i = cacheSize - 1;
result += number.check;
}
}
printf("%u\n", result);
}</lang> The output is the same. The run-time is about 0.04 seconds or less. This third version was ported to D and improved from: mathblog.dk/project-euler-92-square-digits-number-chain/
A purely functional version, from the Haskell code. It includes two functions currently missing in Phobos used in the Haskell code.
<lang d>import std.stdio, std.typecons, std.traits, std.typetuple, std.range, std.algorithm;
auto divMod(T)(T x, T y) pure nothrow @safe @nogc {
return tuple(x / y, x % y);
}
auto expand(alias F, B)(B x) pure nothrow @safe @nogc if (isCallable!F &&
is(ParameterTypeTuple!F == TypeTuple!B)
&& __traits(isSame, TemplateOf!(ReturnType!F), Nullable)
&& isTuple!(TemplateArgsOf!(ReturnType!F)[0])
&& is(TemplateArgsOf!(TemplateArgsOf!(ReturnType!F)[0])[1] == B)) {
alias NAB = ReturnType!F; alias AB = TemplateArgsOf!NAB[0]; alias A = AB.Types[0];
struct Expand {
bool first;
NAB last;
@property bool empty() pure nothrow @safe @nogc {
if (first) {
first = false;
popFront;
}
return last.isNull;
}
@property A front() pure nothrow @safe @nogc {
if (first) {
first = false;
popFront;
}
return last.get[0];
}
void popFront() pure nothrow @safe @nogc { last = F(last.get[1]); }
}
return Expand(true, NAB(AB(A.init, x)));
}
//------------------------------------------------
uint step(uint x) pure nothrow @safe @nogc {
Nullable!(Tuple!(uint, uint)) f(uint n) pure nothrow @safe @nogc {
return (n == 0) ? typeof(return)() : typeof(return)(divMod(n, 10u).reverse);
}
return expand!f(x).map!(x => x ^^ 2).sum;
}
uint iter(uint x) pure nothrow @nogc {
return x.recurrence!((a, n) => step(a[n - 1])).filter!(y => y.among!(1, 89)).front;
}
void main() {
iota(1u, 100_000u).filter!(n => n.iter == 89).count.writeln;
}</lang> With a small back-porting (to run it with the Phobos of LDC2 2.065) it runs in about 15.5 seconds.
ERRE
<lang ERRE> PROGRAM ITERATION
BEGIN
PRINT(CHR$(12);) ! CLS
INPUT(N)
LOOP
N$=MID$(STR$(N),2)
S=0
FOR I=1 TO LEN(N$) DO
A=VAL(MID$(N$,I,1))
S=S+A*A
END FOR
IF S=89 OR S=1 THEN PRINT(S;) EXIT END IF
PRINT(S;)
N=S
END LOOP
PRINT
END PROGRAM </lang> This program verifies a number only. With a FOR..END FOR loop it's possible to verify a number range.
Factor
A brute-force approach with some optimizations. It uses the fact that the first digit-square-sum of any number < 100,000,000 is, at most, 648. These few chains are rapidly memoized as the results for all hundred-million numbers are calculated for the first time or looked up. <lang>USING: kernel math math.ranges math.text.utils memoize prettyprint sequences tools.time ; IN: rosetta-code.iterated-digits-squaring
- sum-digit-sq ( n -- m ) 1 digit-groups [ sq ] map-sum ;
MEMO: 1or89 ( n -- m )
[ dup [ 1 = ] [ 89 = ] bi or ] [ sum-digit-sq ] until ;
[
0 1
[
dup sum-digit-sq 1or89 89 = [ [ 1 + ] dip ] when
1 + dup 100,000,000 <
] loop
drop .
] time</lang>
- Output:
85744333 Running time: 55.76544594 seconds
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
' similar to C Language (first approach) ' timing for i3 @ 2.13 GHz
Function endsWith89(n As Integer) As Boolean
Dim As Integer digit, sum = 0
Do
While n > 0
digit = n Mod 10
sum += digit * digit
n \= 10
Wend
If sum = 89 Then Return True
If sum = 1 Then Return False
n = sum
sum = 0
Loop
End Function
Dim As Double start = timer Dim sums(0 To 8 * 81) As UInteger sums(0) = 1 sums(1) = 0 Dim s As Integer For n As Integer = 1 To 8
For i As Integer = n * 81 To 1 Step -1
For j As Integer = 1 To 9
s = j * j
If s > i Then Exit For
sums(i) += sums(i - s)
Next j
Next i
If n = 8 Then
Dim As UInteger count89 = 0
For i As Integer = 1 To n * 81
If Not endsWith89(i) Then Continue For
count89 += sums(i)
Next i
Print "There are";count89; " numbers from 1 to 100 million ending with 89"
End If
Next Print "Elapsed milliseconds ="; Int((timer - start) * 1000 + 0.5) Print Print "Press any key to quit" Sleep</lang>
- Output:
There are 85744333 numbers from 1 to 100 million ending with 89 Elapsed milliseconds = 2
Frink
<lang frink> total = 0 d = new dict var sum
for n = 1 to 100 million - 1 {
sum = n
do
{
if sum < 1000 and d@sum != undef
{
sum = d@sum
break
}
c = sum
sum = 0
for digit = integerDigits[c]
sum = sum + digit^2
} while (sum != 89) and (sum != 1)
if (n < 1000)
d@n = sum
if (sum == 89)
total = total + 1
}
println[total] </lang>
- Output:
85744333
Go
It's basic. Runs in about 30 seconds on an old laptop.
<lang go>package main
import ( "fmt" )
func main() { var d, n, o, u, u89 int64
for n = 1; n < 100000000; n++ { o = n for { u = 0 for { d = o%10 o = (o - d) / 10 u += d*d if o == 0 { break } } if u == 89 || u == 1 { if u == 89 { u89++ } break } o = u } } fmt.Println(u89) }</lang>
- Output:
85744333
Haskell
Basic solution that contains just a little more than the essence of this computation. This runs in less than eight minutes: <lang haskell>import Data.List (unfoldr) import Data.Tuple (swap)
step :: Int -> Int step = sum . map (^ 2) . unfoldr f where
f 0 = Nothing f n = Just . swap $ n `divMod` 10
iter :: Int -> Int iter = head . filter (`elem` [1, 89]) . iterate step
main = do
print $ length $ filter ((== 89) . iter) [1 .. 99999999]</lang>
- Output:
85744333
J
Here's an expression to turn a number into digits:
<lang J>digits=: 10&#.inv</lang>
And here's an expression to square them and find their sum: <lang J>sumdigsq=: +/"1@:*:@digits</lang>
But note that while the task description claims "you always end with either 1 or 89", that claim is somewhat arbitrary.
- But only somewhat the loop is 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89, so it only ends with 1 or one of the numbers in this loop. 42 is of course far more significant and the one I would choose!!--Nigel Galloway (talk) 10:12, 16 September 2014 (UTC)
<lang J> sumdigsq^:(i.16) 15 15 26 40 16 37 58 89 145 42 20 4 16 37 58 89 145</lang>
You could just as easily claim that you always end with either 1 or 4. So here's a routine which repeats the sum-square process until the sequence converges, or until it reaches the value 4:
<lang J>itdigsq4=:4 = sumdigsq^:(0=e.&4)^:_"0</lang>
But we do not actually need to iterate. The largest value after the first iteration would be:
<lang J> sumdigsq 99999999 648</lang>
So we could write a routine which works for the intended range, and stops after the first iteration: <lang J>itdigsq1=:1 = sumdigsq^:(0=e.&4)^:_"0 digsq1e8=:(I.itdigsq1 i.649) e.~ sumdigsq</lang>
In other words, if the result after the first iteration is any of the numbers in the range 0..648 which converges to 1, it's not a result which would converge to the other loop. This is considerably faster than trying to converge 1e8 sequences, and also evades having to pick an arbitrary stopping point for the sequence which loops for the bulk computation.
And this is sufficient to find our result. We don't want to compute the entire batch of values in one pass, however, so let's break this up into 100 batches of one million each:
<lang J> +/+/@:-.@digsq1e8"1(1+i.100 1e6) 85744333</lang>
Of course, there are faster ways of obtaining that result. The fastest is probably this: <lang J> 85744333 85744333</lang>
This might be thought of as representing the behavior of a highly optimized compiled program. We could abstract this further by using the previous expression at compile time, so we would not have to hard code it.
Julia
Brute force solution: <lang julia>function iterate(m::Integer)
while m != 1 && m != 89
s = 0
while m > 0 # compute sum of squares of digits
m, d = divrem(m, 10)
s += d ^ 2
end
m = s
end
return m
end itercount(k::Integer) = count(x -> iterate(x) == 89, 1:k)</lang>
More clever solution: <lang julia>using Combinatorics function itercountcombos(ndigits::Integer)
cnt = 0
f = factorial(ndigits)
# loop over all combinations of ndigits decimal digits:
for comb in combinations(1:(10+ndigits-1), ndigits)
s = 0
perms = 1
prevd = -1
rep = 1
for k = eachindex(comb) # sum digits ^ 2 and count permutations
d = comb[k] - k
s += d ^ 2
# accumulate number of permutations of repeated digits
if d == prevd
rep += 1
perms *= rep
else
prevd = d
rep = 1
end
end
if s > 0 && iterate(s) == 89
cnt += f ÷ perms # numbers we can get from digits
end
end
return cnt
end</lang>
Benchmarks <lang julia>@time itercount(100_000_000) @time itercountcombos(8) @time itercountcombos(17)</lang>
- Output:
8.866063 seconds (4.32 k allocations: 232.908 KiB) 0.053470 seconds (101.05 k allocations: 8.729 MiB) 1.588977 seconds (12.50 M allocations: 1.536 GiB, 16.94% gc time)
Java
<lang java>import java.util.stream.IntStream;
public class IteratedDigitsSquaring {
public static void main(String[] args) {
long r = IntStream.range(1, 100_000_000)
.parallel()
.filter(n -> calc(n) == 89)
.count();
System.out.println(r);
}
private static int calc(int n) {
while (n != 89 && n != 1) {
int total = 0;
while (n > 0) {
total += Math.pow(n % 10, 2);
n /= 10;
}
n = total;
}
return n;
}
}</lang>
85744333
jq
The algorithm presented here caches the results for 1 ... D*81 (where D is the relevant number of digits) and uses the combinatorial approach, but to keep things relatively brief, the factorials themselves are not cached.
Part 1: Foundations <lang jq>def factorial: reduce range(2;.+1) as $i (1; . * $i);
- Pick n items (with replacement) from the input array,
- but only consider distinct combinations:
def pick(n):
def pick(n; m): # pick n, from m onwards if n == 0 then [] elif m == length then empty elif n == 1 then (.[m:][] | [.]) else ([.[m]] + pick(n-1; m)), pick(n; m+1) end; pick(n;0) ;
- Given any array, produce an array of [item, count] pairs for each run.
def runs:
reduce .[] as $item
( [];
if . == [] then [ [ $item, 1] ]
else .[length-1] as $last
| if $last[0] == $item then (.[0:length-1] + [ [$item, $last[1] + 1] ] )
else . + $item, 1
end
end ) ;</lang>
Part 2: The Generic Task
Count how many number chains beginning with n (where 0 < n < 10^D) end with a value 89. <lang jq>def terminus:
# sum of the squared digits def ssdigits: tostring | explode | map(. - 48 | .*.) | add;
if . == 1 or . == 89 then . else ssdigits | terminus end;
- Count the number of integers i in [1... 10^D] with terminus equal to 89.
def task(D):
# The max sum of squares is D*81 so return an array that will instantly # reveal whether n|terminus is 89: def cache: reduce range(1; D*81+1) as $d ([false]; . + [$d|terminus == 89]);
# Compute n / (i1! * i2! * ... ) for the given combination, # which is assumed to be in order: def combinations(n): runs | map( .[1] | factorial) | reduce .[] as $i (n; ./$i);
cache as $cache
| (D|factorial) as $Dfactorial
| reduce ([range(0;10)] | pick(D)) as $digits
(0;
($digits | map(.*.) | add) as $ss
| if $cache[$ss] then . + ($digits|combinations($Dfactorial))
else .
end) ;</lang>
Part 3: D=8 <lang jq>task(8)</lang>
- Output:
<lang sh>$ jq -M -n -f Iterated_digits_squaring_using_pick.jq 85744333
- Using jq>1.4:
- user 0m2.595s
- sys 0m0.010s
- Using jq 1.4:
- user 0m3.942s
- sys 0m0.009s</lang>
Kotlin
<lang scala>// version 1.0.6
fun endsWith89(n: Int): Boolean {
var digit: Int
var sum = 0
var nn = n
while (true) {
while (nn > 0) {
digit = nn % 10
sum += digit * digit
nn /= 10
}
if (sum == 89) return true
if (sum == 1) return false
nn = sum
sum = 0
}
}
fun main(args: Array<String>) {
val sums = IntArray(8 * 81 + 1)
sums[0] = 1
sums[1] = 0
var s: Int
for (n in 1 .. 8)
for (i in n * 81 downTo 1)
for (j in 1 .. 9) {
s = j * j
if (s > i) break
sums[i] += sums[i - s]
}
var count89 = 0
for (i in 1 .. 8 * 81)
if (endsWith89(i)) count89 += sums[i]
println("There are $count89 numbers from 1 to 100 million ending with 89")
}</lang>
- Output:
There are 85744333 numbers from 1 to 100 million ending with 89
Lua
<lang lua>squares = {}
for i = 0, 9 do
for j = 0, 9 do
squares[i * 10 + j] = i * i + j * j
end
end
for i = 1, 99 do
for j = 0, 99 do
squares[i * 100 + j] = squares[i] + squares[j]
end
end
function sum_squares(n)
if n < 9999.5 then
return squares[n]
else
local m = math.floor(n / 10000)
return squares[n - 10000 * m] + sum_squares(m)
end
end
memory = {}
function calc_1_or_89(n)
local m = {}
n = memory[n] or n
while n ~= 1 and n ~= 89 do
n = memory[n] or sum_squares(n)
table.insert(m, n)
end
for _, i in pairs(m) do
memory[i] = n
end
return n
end
counter = 0
for i = 1, 100000000 do
if calc_1_or_89(i) == 89 then
counter = counter + 1
end
end
print(counter)</lang>
- Output:
85744333
Mathematica / Wolfram Language
<lang Mathematica>sumDigitsSquared[n_Integer] := Total[IntegerDigits[n]^2] stopValues = Join[{1}, NestList[sumDigitsSquared, 89, 7]]; iterate[n_Integer] :=
NestWhile[sumDigitsSquared, n, Intersection[stopValues, {#}] == {} &]
numberOfDigits = 8; maxSum = numberOfDigits 9^2; loopVariables =
ToExpression@Table["i" <> ToString[n], {n, numberOfDigits}];
iteratesToOne = Cases[Range@maxSum, _?(iterate[#] == 1 &)]; allIterators =
Flatten[{Reverse@#, 9}] & /@ Partition[loopVariables, 2, 1];
maxCombinations = numberOfDigits!;
ssd =
SparseArray[Table[n^2 -> numberOfDigits, {n, 9}], {maxSum}];
Do[
variables = loopVariables;; digitCount; iterators = allIterators;; digitCount - 1; Do[ssd += SparseArray[ Total[variables^2] -> maxCombinations/ Times @@ (Tally[PadRight[variables, numberOfDigits]][[All, 2]]!), {maxSum}], {i, 9}, Evaluate[Sequence @@ iterators]], {digitCount, 2, numberOfDigits}];
onesCount =
Total[Cases[
ArrayRules[ssd] /.
HoldPattern[{a_} -> b_] :> {a,
b}, {_?(MemberQ[iteratesToOne, #] &), _}]All, 2];
(10^numberOfDigits - 1) - onesCount</lang>
- Output:
85744333
Oberon-2
]]
} return $n
} for {set i 1} {$i <= 100000000} {incr i} {
incr count [expr {[ids $i] == 89}]
} puts $count</lang>
Intelligent Version
Conversion back and forth between numbers and strings is slow. Using math operations directly is much faster (around 4 times in informal testing). <lang tcl>proc ids n {
while {$n != 1 && $n != 89} {
for {set m 0} {$n} {set n [expr {$n / 10}]} { incr m [expr {($n%10)**2}] } set n $m
} return $n
} for {set i 1} {$i <= 100000000} {incr i} {
incr count [expr {[ids $i] == 89}]
} puts $count</lang>
Substantially More Intelligent Version
Using the observation that the maximum value after 1 step is obtained for 999999999, which is . Thus, running one step of the reduction and then using a lookup table (which we can construct quickly at the start of the run, and which has excellent performance) is much faster overall, approximately 3–4 times than the second version above (and over 12 times faster than the first version).
- Donald, you have 1 too many 9's the value after step 1 is 81*8 = 648. Not that that is the problem here, you can not afford to go around this loop 100 million times. Notice that IDS[21] == IDS[12], IDS[123] == IDS[132] == IDS[213} ... etc, etc. The Ruby version takes about a tenth of a second.--Nigel Galloway (talk) 12:47, 31 August 2014 (UTC)
<lang tcl># Basic implementation proc ids n {
while {$n != 1 && $n != 89} {
for {set m 0} {$n} {set n [expr {$n / 10}]} { incr m [expr {($n%10)**2}] } set n $m
} return $n
}
- Build the optimised version
set body {
# Microoptimisation to avoid an unnecessary alloc in the loop
for {set m 0} {$n} {set n [expr {"$n[unset n]" / 10}]} {
incr m [expr {($n%10)**2}]
}
} set map 0 for {set i 1} {$i <= 729} {incr i} {
lappend map [ids $i]
} proc ids2 n [append body "return \[lindex [list $map] \$m\]"]
- Put this in a lambda context for a little extra speed.
apply {{} {
set count 0
for {set i 1} {$i <= 100000000} {incr i} {
incr count [expr {[ids2 $i] == 89}]
} puts $count
}}</lang>
VBScript
<lang vb> start_time = Now cnt = 0 For i = 1 To 100000000 n = i sum = 0 Do Until n = 1 Or n = 89 For j = 1 To Len(n) sum = sum + (CLng(Mid(n,j,1))^2) Next n = sum sum = 0 Loop If n = 89 Then cnt = cnt + 1 End If Next end_time = Now
WScript.Echo "Elapse Time: " & DateDiff("s",start_time,end_time) &_ vbCrLf & "Count: " & cnt </lang>
- Output:
Elapse time is in seconds. Friends don't let friends do this in VBScript. :-)
Elapse Time: 2559 Count: 85744333
zkl
Using brute force is a never ending process so need to be clever, which takes under a second.
<lang zkl>fcn check(number){ // a list of digits: 13 is L(0,0,0,0,0,0,1,3)
candidate:=number.reduce(fcn(sum,n){ sum*10 + n },0); // digits to int
while(candidate != 89 and candidate != 1) // repeatedly sum squares of digits
{ candidate = candidate.split().reduce(fcn(sum,c){ sum + c*c },0); }
if(candidate == 89){ // count permutations of these digits, they all sum to 89
digitsCount:=List(0,0,0,0,0,0,0,0,0,0);
foreach d in (number){ digitsCount[d] += 1; }
return(digitsCount.reduce(fcn(r,c){ r/factorial(c) },cacheBang)); // cacheBang==number.len()!
}
0 // this number doesn't sum to 89 (ie sums to 1)
} fcn factorial(n) { (1).reduce(n,fcn(N,n){ N*n },1) }
limit:=0d100_000_000; cacheSize:=limit.toFloat().log10().ceil().toInt(); number:=(0).pump(cacheSize,List().write,0); // list of zeros result:=0; i:=cacheSize - 1; var cacheBang=factorial(cacheSize); //== number.len()!
while(True){ // create numbers s.t. no set of digits is repeated
if(i == 0 and number[i] == 9) break;
if(i == cacheSize - 1 and number[i] < 9){ number[i] += 1; result += check(number); }
else if(number[i] == 9) i -= 1;
else{
number[i] += 1;
foreach j in ([i + 1 .. cacheSize - 1]){ number[j] = number[i]; }
i = cacheSize - 1;
result += check(number);
}
} println(result);</lang>
- Output:
85744333
ZX Spectrum Basic
Very, very slow. Use a ZX Spectrum emulator and run with maximum speed option enabled. <lang zxbasic>10 LET n=0 20 FOR i=1 TO 1000 30 LET j=i 40 LET k=0 50 LET k=INT (k+FN m(j,10)^2) 60 LET j=INT (j/10) 70 IF j<>0 THEN GO TO 50 80 LET j=k 90 IF j=89 OR j=1 THEN GO TO 100 95 GO TO 40 100 IF j>1 THEN LET n=n+1 110 NEXT i 120 PRINT "Version 1: ";n 200 DEF FN m(a,b)=a-INT (a/b)*b: REM modulo</lang>
- Output:
Version 1: 857