First power of 2 that has leading decimal digits of 12: Difference between revisions
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p(123, 12345) = 3510491 |
p(123, 12345) = 3510491 |
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p(123, 678910) = 193060223</pre> |
p(123, 678910) = 193060223</pre> |
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=={{header|Racket}}== |
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{{trans|Pascal}} |
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First implementation is an untyped, naive revision of the Pascal algorithm. |
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But there is a lot of casting between the exact integers an floats. As well as runtime type checking. |
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Then there is the typed racket version, which is stricter about the use of types |
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(although, I'm not super happy at the algorithm counting using floating-point integers... but it's faster) |
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Compare and contrast... |
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===Untyped Racket=== |
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<lang racket>#lang racket |
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(define (fract-part f) |
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(- f (truncate f))) |
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(define ln10 (log 10)) |
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(define ln2/ln10 (/ (log 2) ln10)) |
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(define (inexact-p-test L) |
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(let ((digit-shift (let loop ((L (quotient L 10)) (shift 1)) |
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(if (zero? L) shift (loop (quotient L 10) (* 10 shift)))))) |
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(λ (p) (= L (truncate (* digit-shift (exp (* ln10 (fract-part (* p ln2/ln10)))))))))) |
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(define (p L n) |
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(let ((test? (inexact-p-test L))) |
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(let loop ((j 1) (n (sub1 n))) |
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(cond [(not (test? j)) (loop (add1 j) n)] |
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[(zero? n) j] |
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[else (loop (add1 j) (sub1 n))])))) |
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(module+ main |
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(define (report-p L n) |
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(time (printf "p(~a, ~a) = ~a~%" L n (p L n)))) |
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(report-p 12 1) |
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(report-p 12 2) |
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(report-p 123 45) |
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(report-p 123 12345) |
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(report-p 123 678910))</lang> |
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{{out}} |
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<pre>p(12, 1) = 7 |
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cpu time: 1 real time: 1 gc time: 0 |
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p(12, 2) = 80 |
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cpu time: 0 real time: 0 gc time: 0 |
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p(123, 45) = 12710 |
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cpu time: 5 real time: 5 gc time: 0 |
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p(123, 12345) = 3510491 |
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cpu time: 439 real time: 442 gc time: 23 |
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p(123, 678910) = 193060223 |
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cpu time: 27268 real time: 27354 gc time: 194</pre> |
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===Typed Racket=== |
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<lang racket></lang> |
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{{out}} |
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<pre></pre> |
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=={{header|Raku}}== |
=={{header|Raku}}== |
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Revision as of 11:44, 8 September 2021
You are encouraged to solve this task according to the task description, using any language you may know.
(This task is taken from a Project Euler problem.)
(All numbers herein are expressed in base ten.)
27 = 128 and 7 is
the first power of 2 whose leading decimal digits are 12.
The next power of 2 whose leading decimal digits
are 12 is 80,
280 = 1208925819614629174706176.
Define p(L,n) to be the nth-smallest
value of j such that the base ten representation
of 2j begins with the digits of L .
So p(12, 1) = 7 and
p(12, 2) = 80
You are also given that:
p(123, 45) = 12710
- Task
-
- find:
- p(12, 1)
- p(12, 2)
- p(123, 45)
- p(123, 12345)
- p(123, 678910)
- display the results here, on this page.
ALGOL 68
As wih the Go second sample, computes approximate integer values for the powers of 2.
Requires LONG INT to be at least 64 bits (as in Algol 68G).
<lang algol68># find values of p( L, n ) where p( L, n ) is the nth-smallest j such that #
- the decimal representation of 2^j starts with the digits of L #
BEGIN
# returns a string representation of n with commas #
PROC commatise = ( LONG INT n )STRING:
BEGIN
STRING result := "";
STRING unformatted = whole( n, 0 );
INT ch count := 0;
FOR c FROM UPB unformatted BY -1 TO LWB unformatted DO
IF ch count <= 2 THEN ch count +:= 1
ELSE ch count := 1; "," +=: result
FI;
unformatted[ c ] +=: result
OD;
result
END # commatise # ;
# returns p( prefix, occurance ) #
PROC p = ( INT prefix, INT occurance )LONG INT:
BEGIN
LONG INT quarter long max int = long max int OVER 4;
LONG INT p2 := 1;
INT count := 0;
INT power := 0;
WHILE count < occurance DO
power +:= 1;
p2 +:= p2;
LONG INT pre := p2;
WHILE pre > prefix DO
pre OVERAB 10
OD;
IF pre = prefix THEN
count +:= 1
FI;
IF p2 > quarter long max int THEN
p2 OVERAB 10 000
FI
OD;
power
END # p # ;
# prints p( prefix, occurance ) #
PROC print p = ( INT prefix, INT occurance )VOID:
print( ( "p(", whole( prefix, 0 ), ", ", whole( occurance, 0 ), ") = ", commatise( p( prefix, occurance ) ), newline ) );
# task test cases #
print p( 12, 1 );
print p( 12, 2 );
print p( 123, 45 );
print p( 123, 12345 );
print p( 123, 678910 )
END</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223
C
<lang c>#include <math.h>
- include <stdio.h>
int p(int l, int n) {
int test = 0;
double logv = log(2.0) / log(10.0);
int factor = 1;
int loop = l;
while (loop > 10) {
factor *= 10;
loop /= 10;
}
while (n > 0) {
int val;
test++;
val = (int)(factor * pow(10.0, fmod(test * logv, 1)));
if (val == l) {
n--;
}
}
return test;
}
void runTest(int l, int n) {
printf("p(%d, %d) = %d\n", l, n, p(l, n));
}
int main() {
runTest(12, 1); runTest(12, 2); runTest(123, 45); runTest(123, 12345); runTest(123, 678910);
return 0;
}</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12710 p(123, 12345) = 3510491 p(123, 678910) = 193060223
C++
<lang cpp>// a mini chrestomathy solution
- include <string>
- include <chrono>
- include <cmath>
- include <locale>
using namespace std; using namespace chrono;
// translated from java example unsigned int js(int l, int n) {
unsigned int res = 0, f = 1; double lf = log(2) / log(10), ip; for (int i = l; i > 10; i /= 10) f *= 10; while (n > 0) if ((int)(f * pow(10, modf(++res * lf, &ip))) == l) n--; return res;
}
// translated from go integer example (a.k.a. go translation of pascal alternative example) unsigned int gi(int ld, int n) {
string Ls = to_string(ld);
unsigned int res = 0, count = 0; unsigned long long f = 1;
for (int i = 1; i <= 18 - Ls.length(); i++) f *= 10;
const unsigned long long ten18 = 1e18; unsigned long long probe = 1;
do {
probe <<= 1; res++; if (probe >= ten18) {
do {
if (probe >= ten18) probe /= 10;
if (probe / f == ld) if (++count >= n) { count--; break; }
probe <<= 1; res++;
} while (1);
}
string ps = to_string(probe);
if (ps.substr(0, min(Ls.length(), ps.length())) == Ls) if (++count >= n) break;
} while (1);
return res;
}
// translated from pascal alternative example unsigned int pa(int ld, int n) {
const double L_float64 = pow(2, 64);
const unsigned long long Log10_2_64 = (unsigned long long)(L_float64 * log(2) / log(10));
double Log10Num; unsigned long long LmtUpper, LmtLower, Frac64;
int res = 0, dgts = 1, cnt;
for (int i = ld; i >= 10; i /= 10) dgts *= 10;
Log10Num = log((ld + 1.0) / dgts) / log(10);
// '316' was a limit
if (Log10Num >= 0.5) {
LmtUpper = (ld + 1.0) / dgts < 10.0 ? (unsigned long long)(Log10Num * (L_float64 * 0.5)) * 2 + (unsigned long long)(Log10Num * 2) : 0;
Log10Num = log((double)ld / dgts) / log(10);
LmtLower = (unsigned long long)(Log10Num * (L_float64 * 0.5)) * 2 + (unsigned long long)(Log10Num * 2);
} else {
LmtUpper = (unsigned long long)(Log10Num * L_float64);
LmtLower = (unsigned long long)(log((double)ld / dgts) / log(10) * L_float64);
}
cnt = 0; Frac64 = 0; if (LmtUpper != 0) {
do {
res++; Frac64 += Log10_2_64;
if ((Frac64 >= LmtLower) & (Frac64 < LmtUpper))
if (++cnt >= n) break;
} while (1);
} else { // '999..'
do {
res++; Frac64 += Log10_2_64;
if (Frac64 >= LmtLower) if (++cnt >= n) break;
} while (1);
};
return res;
}
int params[] = { 12, 1, 12, 2, 123, 45, 123, 12345, 123, 678910, 99, 1 };
void doOne(string name, unsigned int (*func)(int a, int b)) {
printf("%s version:\n", name.c_str());
auto start = steady_clock::now();
for (int i = 0; i < size(params); i += 2)
printf("p(%3d, %6d) = %'11u\n", params[i], params[i + 1], func(params[i], params[i + 1]));
printf("Took %f seconds\n\n", duration<double>(steady_clock::now() - start).count());
}
int main() {
setlocale(LC_ALL, "");
doOne("java simple", js);
doOne("go integer", gi);
doOne("pascal alternative", pa);
}</lang>
- Output:
Exeution times from Tio.run, language
java simple version: p( 12, 1) = 7 p( 12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 p( 99, 1) = 93 Took 11.350696 seconds go integer version: p( 12, 1) = 7 p( 12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 p( 99, 1) = 93 Took 1.787658 seconds pascal alternative version: p( 12, 1) = 7 p( 12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 p( 99, 1) = 93 Took 0.299815 seconds
Execution times on a core i7-7700 @ 3.6Ghz, g++, compiler flags: -O3 -std=c++17
java simple version: ... Took 15.060563 seconds go integer version: ... Took 1.268436 seconds pascal alternative version: ... Took 0.082407 seconds
C#
<lang cpp>// a mini chrestomathy solution
using System;
class Program {
// translated from java example
static long js(int l, int n) {
long res = 0, f = 1;
double lf = Math.Log10(2);
for (int i = l; i > 10; i /= 10) f *= 10;
while (n > 0)
if ((int)(f * Math.Pow(10, ++res * lf % 1)) == l) n--;
return res;
}
// translated from go integer example (a.k.a. go translation of pascal alternative example)
static long gi(int ld, int n) {
string Ls = ld.ToString();
long res = 0, count = 0, f = 1;
for (int i = 1; i <= 18 - Ls.Length; i++) f *= 10;
const long ten18 = (long)1e18; long probe = 1;
do {
probe <<= 1; res++; if (probe >= ten18)
do {
if (probe >= ten18) probe /= 10;
if (probe / f == ld)
if (++count >= n) { count--; break; }
probe <<= 1; res++;
} while (true);
string ps = probe.ToString();
if (ps.Substring(0, Math.Min(Ls.Length, ps.Length)) == Ls)
if (++count >= n) break;
} while (true);
return res;
}
// translated from pascal alternative example
static long pa(int ld, int n) {
double L_float64 = Math.Pow(2, 64);
ulong Log10_2_64 = (ulong)(L_float64 * Math.Log10(2));
double Log10Num; ulong LmtUpper, LmtLower, Frac64;
long res = 0, dgts = 1, cnt;
for (int i = ld; i >= 10; i /= 10) dgts *= 10;
Log10Num = Math.Log10((ld + 1.0) / dgts);
// '316' was a limit
if (Log10Num >= 0.5) {
LmtUpper = (ld + 1.0) / dgts < 10.0 ? (ulong)(Log10Num * (L_float64 * 0.5)) * 2 + (ulong)(Log10Num * 2) : 0;
Log10Num = Math.Log10((double)ld / dgts);
LmtLower = (ulong)(Log10Num * (L_float64 * 0.5)) * 2 + (ulong)(Log10Num * 2);
} else {
LmtUpper = (ulong)(Log10Num * L_float64);
LmtLower = (ulong)(Math.Log10((double)ld / dgts) * L_float64);
}
cnt = 0; Frac64 = 0; if (LmtUpper != 0)
do {
res++; Frac64 += Log10_2_64;
if ((Frac64 >= LmtLower) & (Frac64 < LmtUpper))
if (++cnt >= n) break;
} while (true);
else // '999..'
do {
res++; Frac64 += Log10_2_64;
if (Frac64 >= LmtLower) if (++cnt >= n) break;
} while (true);
return res;
}
static int[] values = new int[] { 12, 1, 12, 2, 123, 45, 123, 12345, 123, 678910, 99, 1 };
static void doOne(string name, Func<int, int, long> fun) {
Console.WriteLine("{0} version:", name);
var start = DateTime.Now;
for (int i = 0; i < values.Length; i += 2)
Console.WriteLine("p({0,3}, {1,6}) = {2,11:n0}", values[i], values[i + 1], fun(values[i], values[i + 1]));
Console.WriteLine("Took {0} seconds\n", DateTime.Now - start);
}
static void Main() {
doOne("java simple", js);
doOne("go integer", gi);
doOne("pascal alternative", pa);
}
}</lang>
- Output:
Results on the core i7-7700 @ 3.6Ghz.
java simple version: p( 12, 1) = 7 p( 12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 p( 99, 1) = 93 Took 00:00:09.5992122 seconds go integer version: p( 12, 1) = 7 p( 12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 p( 99, 1) = 93 Took 00:00:01.5335699 seconds pascal alternative version: p( 12, 1) = 7 p( 12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 p( 99, 1) = 93 Took 00:00:00.1391226 seconds
D
<lang d>import std.math; import std.stdio;
int p(int l, int n) {
int test = 0;
double logv = log(2.0) / log(10.0);
int factor = 1;
int loop = l;
while (loop > 10) {
factor *= 10;
loop /= 10;
}
while (n > 0) {
int val;
test++;
val = cast(int)(factor * pow(10.0, fmod(test * logv, 1)));
if (val == l) {
n--;
}
}
return test;
}
void runTest(int l, int n) {
writefln("p(%d, %d) = %d", l, n, p(l, n));
}
void main() {
runTest(12, 1); runTest(12, 2); runTest(123, 45); runTest(123, 12345); runTest(123, 678910);
}</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12710 p(123, 12345) = 3510491 p(123, 678910) = 193060223
Delphi
See Pascal.
F#
<lang fsharp> // First power of 2 that has specified leading decimal digits. Nigel Galloway: March 14th., 2021 let fG n g=let fN l=let l=10.0**(l-floor l) in l>=n && l<g in let f=log10 2.0 in seq{1..0x0FFFFFFF}|>Seq.filter(float>>(*)f>>fN) printfn "p(23,1)->%d" (int(Seq.item 0 (fG 2.3 2.4))) printfn "p(99,1)->%d" (int(Seq.item 0 (fG 9.9 10.0))) printfn "p(12,1)->%d" (int(Seq.item 0 (fG 1.2 1.3))) printfn "p(12,2)->%d" (int(Seq.item 1 (fG 1.2 1.3))) printfn "p(123,45)->%d" (int(Seq.item 44 (fG 1.23 1.24))) printfn "p(123,12345)->%d" (int(Seq.item 12344 (fG 1.23 1.24))) printfn "p(123,678910)->%d" (int(Seq.item 678909 (fG 1.23 1.24))) </lang>
- Output:
p(23,1)->61 p(99,1)->93 p(12,1)->7 p(12,2)->80 p(123,45)->12710 p(123,12345)->3510491 p(123,678910)->193060223 Real: 00:00:10.140
Factor
A translation of the first Pascal example:
<lang factor>USING: formatting fry generalizations kernel literals math math.functions math.parser sequences tools.time ;
CONSTANT: ld10 $[ 2 log 10 log / ]
- p ( L n -- m )
swap [ 0 0 ]
[ '[ over _ >= ] ]
[ [ log10 >integer 10^ ] keep ] tri*
'[
1 + dup ld10 * dup >integer - 10 log * e^ _ * truncate
_ number= [ [ 1 + ] dip ] when
] until nip ;
[
12 1 12 2 123 45 123 12345 123 678910 [ 2dup p "%d %d p = %d\n" printf ] 2 5 mnapply
] time</lang>
- Output:
12 1 p = 7 12 2 p = 80 123 45 p = 12710 123 12345 p = 3510491 123 678910 p = 193060223 Running time: 44.208249282 seconds
FreeBASIC
<lang freebasic>#define FAC 0.30102999566398119521373889472449302677
function p( L as uinteger, n as uinteger ) as uinteger
dim as uinteger count, j = 0
dim as double x, y
dim as string digits, LS = str(L)
while count < n
j+=1
x = FAC * j
if x < len(LS) then continue while
y = 10^(x-int(x))
y *= 10^len(LS)
digits = str(y)
if left(digits,len(LS)) = LS then count += 1
wend
return j
end function
print p(12, 1) print p(12, 2) print p(123, 45) print p(123, 12345) print p(123, 678910)</lang>
- Output:
7 80 12710 3510491 193060223
Go
<lang go>package main
import (
"fmt" "math" "time"
)
const ld10 = math.Ln2 / math.Ln10
func commatize(n uint64) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
func p(L, n uint64) uint64 {
i := L
digits := uint64(1)
for i >= 10 {
digits *= 10
i /= 10
}
count := uint64(0)
for i = 0; count < n; i++ {
e := math.Exp(math.Ln10 * math.Mod(float64(i)*ld10, 1))
if uint64(math.Trunc(e*float64(digits))) == L {
count++
}
}
return i - 1
}
func main() {
start := time.Now()
params := [][2]uint64{{12, 1}, {12, 2}, {123, 45}, {123, 12345}, {123, 678910}}
for _, param := range params {
fmt.Printf("p(%d, %d) = %s\n", param[0], param[1], commatize(p(param[0], param[1])))
}
fmt.Printf("\nTook %s\n", time.Since(start))
}</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 Took 38.015225244s
or, translating the alternative Pascal version as well, for good measure: <lang go>package main
import (
"fmt" "strconv" "time"
)
func p(L, n uint64) uint64 {
Ls := strconv.FormatUint(L, 10)
digits := uint64(1)
for d := 1; d <= 18-len(Ls); d++ {
digits *= 10
}
const ten18 uint64 = 1e18
var count, i, probe uint64 = 0, 0, 1
for {
probe += probe
i++
if probe >= ten18 {
for {
if probe >= ten18 {
probe /= 10
}
if probe/digits == L {
count++
if count >= n {
count--
break
}
}
probe += probe
i++
}
}
ps := strconv.FormatUint(probe, 10)
le := len(Ls)
if le > len(ps) {
le = len(ps)
}
if ps[0:le] == Ls {
count++
if count >= n {
break
}
}
}
return i
}
func commatize(n uint64) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
func main() {
start := time.Now()
params := [][2]uint64{{12, 1}, {12, 2}, {123, 45}, {123, 12345}, {123, 678910}}
for _, param := range params {
fmt.Printf("p(%d, %d) = %s\n", param[0], param[1], commatize(p(param[0], param[1])))
}
fmt.Printf("\nTook %s\n", time.Since(start))
}</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 Took 1.422321658s
Haskell
<lang haskell>import Control.Monad (guard) import Text.Printf (printf)
p :: Int -> Int -> Int p l n = calc !! pred n
where
digitCount = floor $ logBase 10 (fromIntegral l :: Float)
log10pwr = logBase 10 2
calc = do
raised <- [-1 ..]
let firstDigits = floor $ 10 ** (snd (properFraction $ log10pwr * realToFrac raised)
+ realToFrac digitCount)
guard (firstDigits == l)
[raised]
main :: IO () main = mapM_ (\(l, n) -> printf "p(%d, %d) = %d\n" l n (p l n))
[(12, 1), (12, 2), (123, 45), (123, 12345), (123, 678910)]</lang>
Which, desugaring a little from Control.Monad (guard) and the do syntax, could also be rewritten as: <lang haskell>import Text.Printf (printf)
p :: Int -> Int -> Int p l n = ([-1 ..] >>= f) !! pred n
where
digitCount =
floor $
logBase 10 (fromIntegral l :: Float)
log10pwr = logBase 10 2
f raised = [ds | l == ds]
where
ds =
floor $
10
** ( snd
( properFraction $
log10pwr * realToFrac raised
)
+ realToFrac digitCount
)
main :: IO () main =
mapM_
(\(l, n) -> printf "p(%d, %d) = %d\n" l n (p l n))
[ (12, 1),
(12, 2),
(123, 45),
(123, 12345),
(123, 678910)
]</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12710 p(123, 12345) = 3510491 p(123, 678910) = 193060223
J
Modeled on the python version. Depending on the part of speech defined, the variable names x, y, u, v, m, and n can have special meaning within explicit code. They are defined by the arguments. In the fonts I usually use, lower case "l" is troublesome as well. <lang J> p=: adverb define
el =. x en =. y pwr =. m el =. <. | el digitcount =. <. 10 ^. el log10pwr =. 10 ^. pwr 'raised found' =. _1 0 while. found < en do. raised =. >: raised firstdigits =. (<.!.0) 10^digitcount + 1 | log10pwr * raised found =. found + firstdigits = el end. raised
) </lang>
12 12 123 123 123 (,. ,. (2 p)&>) 1 2 45 12345 678910 12 1 7 12 2 80 123 45 12710 123 12345 3510491 123 678910 193060223
Java
<lang java> public class FirstPowerOfTwo {
public static void main(String[] args) {
runTest(12, 1);
runTest(12, 2);
runTest(123, 45);
runTest(123, 12345);
runTest(123, 678910);
}
private static void runTest(int l, int n) {
System.out.printf("p(%d, %d) = %,d%n", l, n, p(l, n));
}
public static int p(int l, int n) {
int test = 0;
double log = Math.log(2) / Math.log(10);
int factor = 1;
int loop = l;
while ( loop > 10 ) {
factor *= 10;
loop /= 10;
}
while ( n > 0) {
test++;
int val = (int) (factor * Math.pow(10, test * log % 1));
if ( val == l ) {
n--;
}
}
return test;
}
} </lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223
jq
This solution uses unbounded-precision integers represented as arrays of decimal digits beginning with the least significant digit, e.g. [1,2,3] represents 321.
To speed up the computation, a variable limit on the number of significant digits is used. Details can be seen in the implementation of p/2; for example, for p(123, 678910) the number of significant digits starts off at (10 + 30) = 40 and gradually tapers off to 10.
(Starting with (8+28) and tapering off to 8 is insufficient.) <lang jq>def normalize_base($base):
def n:
if length == 1 and .[0] < $base then .[0]
else .[0] % $base,
((.[0] / $base|floor) as $carry
|.[1:]
| .[0] += $carry
| n ) end; n;
def integers_as_arrays_times($n; $base):
map(. * $n) | [normalize_base($base)];
def p($L; $n):
# @assert($L > 0 and $n > 0)
($L|tostring|explode|reverse|map(. - 48)) as $digits # "0" is 48
| ($digits|length) as $ndigits
| (2*(2+($n|log|floor))) as $extra
| { m: $n,
i: 0,
keep: ( 2*(2+$ndigits) + $extra),
p: [1] # reverse-array representation of 1
}
| until(.m == 0;
.i += 1
| .p |= integers_as_arrays_times(2; 10)
| .p = .p[ -(.keep) :]
| if (.p[-$ndigits:]) == $digits
then
| .m += -1 | .keep -= ($extra/$n)
else . end )
| .i;
- The task
[[12, 1], [12, 2], [123, 45], [123, 12345], [123, 678910]][] | "With L = \(.[0]) and n = \(.[1]), p(L, n) = \( p(.[0]; .[1]))" </lang>
- Output:
As for Julia.
Julia
<lang julia>function p(L, n)
@assert(L > 0 && n > 0)
places, logof2, nfound = trunc(log(10, L)), log(10, 2), 0
for i in 1:typemax(Int)
if L == trunc(10^(((i * logof2) % 1) + places)) && (nfound += 1) == n
return i
end
end
end
for (L, n) in [(12, 1), (12, 2), (123, 45), (123, 12345), (123, 678910)]
println("With L = $L and n = $n, p(L, n) = ", p(L, n))
end
</lang>
- Output:
With L = 12 and n = 1, p(L, n) = 7 With L = 12 and n = 2, p(L, n) = 80 With L = 123 and n = 45, p(L, n) = 12710 With L = 123 and n = 12345, p(L, n) = 3510491 With L = 123 and n = 678910, p(L, n) = 193060223
Faster version
<lang julia>using Formatting, BenchmarkTools
function p(L, n)
@assert(L > 0 && n > 0)
Ls, ten18, nfound, i, probe = string(L), 10^18, 0, 0, 1
maxdigits = 10^(18 - ndigits(L))
while true
probe += probe
i += 1
if probe >= ten18
while true
(probe >= ten18) && (probe ÷= 10)
if probe ÷ maxdigits == L
if (nfound += 1) >= n
nfound -= 1
break
end
end
probe += probe
i += 1
end
end
ps = string(probe)
len = min(length(Ls), length(ps))
if ps[1:len] == Ls && (nfound += 1) >= n
break
end
end
return i
end
function testpLn(verbose)
for (L, n) in [(12, 1), (12, 2), (123, 45), (123, 12345), (123, 678910)]
i = p(L, n)
verbose && println("With L = $L and n = $n, p(L, n) = ", format(i, commas=true))
end
end
testpLn(true) @btime testpLn(false)
</lang>
- Output:
With L = 12 and n = 1, p(L, n) = 7 With L = 12 and n = 2, p(L, n) = 80 With L = 123 and n = 45, p(L, n) = 12,710 With L = 123 and n = 12345, p(L, n) = 3,510,491 With L = 123 and n = 678910, p(L, n) = 193,060,223 1.462 s (752 allocations: 32.19 KiB)
Kotlin
<lang scala>import kotlin.math.ln import kotlin.math.pow
fun main() {
runTest(12, 1) runTest(12, 2) runTest(123, 45) runTest(123, 12345) runTest(123, 678910)
}
private fun runTest(l: Int, n: Int) { // System.out.printf("p(%d, %d) = %,d%n", l, n, p(l, n))
println("p($l, $n) = %,d".format(p(l, n)))
}
fun p(l: Int, n: Int): Int {
var m = n
var test = 0
val log = ln(2.0) / ln(10.0)
var factor = 1
var loop = l
while (loop > 10) {
factor *= 10
loop /= 10
}
while (m > 0) {
test++
val value = (factor * 10.0.pow(test * log % 1)).toInt()
if (value == l) {
m--
}
}
return test
}</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223
Mathematica/Wolfram Language
Without compilation arbitrary precision will be used, and the algorithm will be slow, compiled is much faster: <lang Mathematica>f = Compile[{{ll, _Integer}, {n, _Integer}},
Module[{l, digitcount, log10power, raised, found, firstdigits, pwr = 2},
l = Abs[ll];
digitcount = Floor[Log[10, l]];
log10power = Log[10, pwr];
raised = -1;
found = 0;
While[found < n,
raised++;
firstdigits = Floor[10^(FractionalPart[log10power raised] + digitcount)];
If[firstdigits == l,
found += 1;
]
];
Return[raised]
]
];
f[12, 1] f[12, 2] f[123, 45] f[123, 12345] f[123, 678910]</lang>
- Output:
7 80 12710 3510491 193060223
Nim
This is a translation to Nim of the very efficient Pascal alternative algorithm, with minor adjustments. <lang Nim>import math, strformat
const
Lfloat64 = pow(2.0, 64) Log10_2_64 = int(Lfloat64 * log10(2.0))
- ---------------------------------------------------------------------------------------------------
func ordinal(n: int): string =
case n of 1: "1st" of 2: "2nd" of 3: "3rd" else: $n & "th"
- ---------------------------------------------------------------------------------------------------
proc findExp(number, countLimit: int) =
var i = number var digits = 1 while i >= 10: digits *= 10 i = i div 10
var lmtLower, lmtUpper: uint64
var log10num = log10((number + 1) / digits)
if log10num >= 0.5:
lmtUpper = if (number + 1) / digits < 10: uint(log10Num * (Lfloat64 * 0.5)) * 2 + uint(log10Num * 2)
else: 0
log10Num = log10(number / digits)
lmtLower = uint(log10Num * (Lfloat64 * 0.5)) * 2 + uint(log10Num * 2)
else:
lmtUpper = uint(log10Num * Lfloat64)
lmtLower = uint(log10(number / digits) * Lfloat64)
var count = 0
var frac64 = 0u64
var p = 0
if lmtUpper != 0:
while true:
inc p
inc frac64, Log10_2_64
if frac64 in lmtLower..lmtUpper:
inc count
if count >= countLimit:
break
else:
# Searching for "999...".
while true:
inc p
inc frac64, Log10_2_64
if frac64 >= lmtLower:
inc count
if count >= countLimit:
break
echo fmt"""The {ordinal(count)} occurrence of 2 raised to a power""" &
fmt""" whose product starts with "{number}" is {p}"""
- ———————————————————————————————————————————————————————————————————————————————————————————————————
findExp(12, 1) findExp(12, 2)
findExp(123, 45) findExp(123, 12345) findExp(123, 678910)</lang>
- Output:
Processor: Core I5 8250U
Compilation command: nim c -d:danger --passC:-flto leading12.nim
Time: about 110 ms
The 1st occurrence of 2 raised to a power whose product starts with "12" is 7 The 2nd occurrence of 2 raised to a power whose product starts with "12" is 80 The 45th occurrence of 2 raised to a power whose product starts with "123" is 12710 The 12345th occurrence of 2 raised to a power whose product starts with "123" is 3510491 The 678910th occurrence of 2 raised to a power whose product starts with "123" is 193060223
Pascal
First convert 2**i -> 10**x => x= ln(2)/ln(10) *i
The integer part of x is the position of the comma.Only the fraction of x leads to the digits.
0<= base ** frac(x) < base thats 1 digit before the comma
Only the first digits are needed.So I think, the accuracy is sufficient, because the results are the same :-)
<lang pascal>program Power2FirstDigits;
uses
sysutils, strUtils;
const {$IFDEF FPC}
{$MODE DELPHI}
ld10 :double = ln(2)/ln(10);// thats 1/log2(10)
{$ELSE}
ld10 = 0.30102999566398119521373889472449;
function Numb2USA(const S: string): string; var
i, NA: Integer;
begin
i := Length(S);
Result := S;
NA := 0;
while (i > 0) do
begin
if ((Length(Result) - i + 1 - NA) mod 3 = 0) and (i <> 1) then
begin
insert(',', Result, i);
inc(NA);
end;
Dec(i);
end;
end;
{$ENDIF}
function FindExp(CntLmt, Number: NativeUint): NativeUint; var
i, cnt, DgtShift: NativeUInt;
begin
//calc how many Digits needed i := Number; DgtShift := 1; while i >= 10 do begin DgtShift := DgtShift * 10; i := i div 10; end;
cnt := 0;
i := 0;
repeat
inc(i);
// x= i*ld10 -> 2^I = 10^x
// 10^frac(x) -> [0..10[ = exp(ln(10)*frac(i*lD10))
if Trunc(DgtShift * exp(ln(10) * frac(i * lD10))) = Number then
begin
inc(cnt);
if cnt >= CntLmt then
BREAK;
end;
until false;
write('The ', Numb2USA(IntToStr(cnt)), 'th occurrence of 2 raised to a power');
write(' whose product starts with "', Numb2USA(IntToStr(Number)));
writeln('" is ', Numb2USA(IntToStr(i)));
FindExp := i;
end;
begin
FindExp(1, 12); FindExp(2, 12);
FindExp(45, 123); FindExp(12345, 123); FindExp(678910, 123);
end.</lang>
- Output:
The 1th occurrence of 2 raised to a power whose product starts with "12" is 7 The 2th occurrence of 2 raised to a power whose product starts with "12" is 80 The 45th occurrence of 2 raised to a power whose product starts with "123" is 12,710 The 12,345th occurrence of 2 raised to a power whose product starts with "123" is 3,510,491 The 678,910th occurrence of 2 raised to a power whose product starts with "123" is 193,060,223 //64Bit real 0m43,031s //32Bit real 0m13,363s
alternative
Now only using the fractional part for maximum precision in Uint64
ignoring overflow so frac64 is [0..2**64-1] represent [0..1[
changed trunc(DgtShift*exp(ln(10)*frac(i*lD10))) = Number
No trunc Number/Digits <= .. < (Number+1)/Digits => no trunc
Logarithm (Ln(Number/Digits)/ln(10) <= frac(i*lD10) < ln((Number+1)/Digits)/ln(10) => no exp
<lang pascal>program Power2Digits; uses
sysutils,strUtils;
const
L_float64 = sqr(sqr(65536.0));//2**64 Log10_2_64 = TRUNC(L_float64*ln(2)/ln(10));
function FindExp(CntLmt,Number:NativeUint):NativeUint; var
Log10Num : extended; LmtUpper,LmtLower : UInt64; Frac64 : UInt64; i,dgts,cnt: NativeUInt;
begin
i := Number;
dgts := 1;
while i >= 10 do
Begin
dgts *= 10;
i := i div 10;
end;
//trunc is Int64 :-( so '316' was a limit
Log10Num :=ln((Number+1)/dgts)/ln(10);
IF Log10Num >= 0.5 then
Begin
IF (Number+1)/dgts < 10 then
Begin
LmtUpper := Trunc(Log10Num*(L_float64*0.5))*2;
LmtUpper += Trunc(Log10Num*2);
end
else
LmtUpper := 0;
Log10Num :=ln(Number/dgts)/ln(10);
LmtLower := Trunc(Log10Num*(L_float64*0.5))*2;
LmtLower += Trunc(Log10Num*2);
end
Else
Begin
LmtUpper := Trunc(Log10Num*L_float64);
LmtLower := Trunc(ln(Number/dgts)/ln(10)*L_float64);
end;
cnt := 0;
i := 0;
Frac64 := 0;
IF LmtUpper <> 0 then
Begin
repeat
inc(i);
inc(Frac64,Log10_2_64);
IF (Frac64>= LmtLower) AND (Frac64< LmtUpper) then
Begin
inc(cnt);
IF cnt>= CntLmt then
BREAK;
end;
until false
end
Else
//searching for '999..'
Begin
repeat
inc(i);
inc(Frac64,Log10_2_64);
IF (Frac64>= LmtLower) then
Begin
inc(cnt);
IF cnt>= CntLmt then
BREAK;
end;
until false
end;
write('The ',Numb2USA(IntToStr(cnt)),'th occurrence of 2 raised to a power');
write(' whose product starts with "',Numb2USA(IntToStr(number)));
writeln('" is ',Numb2USA(IntToStr(i)));
FindExp := i;
end;
Begin
FindExp(1,12); FindExp(2,12);
FindExp(45,223); FindExp(12345,123); FindExp(678910,123);
FindExp(1,99);
end.</lang>
- Output:
The 1th occurrence of 2 raised to a power whose product starts with "12" is 7 The 2th occurrence of 2 raised to a power whose product starts with "12" is 80 The 45th occurrence of 2 raised to a power whose product starts with "223" is 22,670 The 12,345th occurrence of 2 raised to a power whose product starts with "123" is 3,510,491 The 678,910th occurrence of 2 raised to a power whose product starts with "123" is 193,060,223 The 1th occurrence of 2 raised to a power whose product starts with "99" is 93 //64Bit real 0m0,138s //32Bit real 0m0,389s
Perl
<lang perl>use strict; use warnings; use feature 'say'; use feature 'state';
use POSIX qw(fmod); use Perl6::GatherTake;
use constant ln2ln10 => log(2) / log(10);
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub ordinal_digit {
my($d) = $_[0] =~ /(.)$/; $d eq '1' ? 'st' : $d eq '2' ? 'nd' : $d eq '3' ? 'rd' : 'th'
}
sub startswith12 {
my($nth) = @_;
state $i = 0;
state $n = 0;
while (1) {
next unless '1.2' eq substr(( 10 ** fmod(++$i * ln2ln10, 1) ), 0, 3);
return $i if ++$n eq $nth;
}
}
sub startswith123 {
my $pre = '1.23'; my ($this, $count) = (0, 0);
gather {
while (1) {
if ($this == 196) {
$this = 289;
$this = 485 unless $pre eq substr(( 10 ** fmod(($count+$this) * ln2ln10, 1) ), 0, 4);
} elsif ($this == 485) {
$this = 196;
$this = 485 unless $pre eq substr(( 10 ** fmod(($count+$this) * ln2ln10, 1) ), 0, 4);
} elsif ($this == 289) {
$this = 196
} elsif ($this == 90) {
$this = 289
} elsif ($this == 0) {
$this = 90;
}
take $count += $this;
}
}
}
my $start_123 = startswith123(); # lazy list
sub p {
my($prefix,$nth) = @_; $prefix eq '12' ? startswith12($nth) : $start_123->[$nth-1];
}
for ([12, 1], [12, 2], [123, 45], [123, 12345], [123, 678910]) {
my($prefix,$nth) = @$_;
printf "%-15s %9s power of two (2^n) that starts with %5s is at n = %s\n", "p($prefix, $nth):",
comma($nth) . ordinal_digit($nth), "'$prefix'", comma p($prefix, $nth);
}</lang>
- Output:
p(12, 1): 1st power of two (2^n) that starts with '12' is at n = 7 p(12, 2): 2nd power of two (2^n) that starts with '12' is at n = 80 p(123, 45): 45th power of two (2^n) that starts with '123' is at n = 12,710 p(123, 12345): 12,345th power of two (2^n) that starts with '123' is at n = 3,510,491 p(123, 678910): 678,910th power of two (2^n) that starts with '123' is at n = 193,060,223
Phix
with javascript_semantics function p(integer L, n) atom logof2 = log10(2) integer places = trunc(log10(L)), nfound = 0, i = 1 while true do atom a = i * logof2, b = trunc(power(10,a-trunc(a)+places)) if L == b then nfound += 1 if nfound == n then exit end if end if i += 1 end while return i end function constant tests = {{12, 1}, {12, 2}, {123, 45}, {123, 12345}, {123, 678910}} include ordinal.e include mpfr.e mpz z = mpz_init() atom t0 = time() for i=1 to length(tests)-(2*(platform()=JS)) do integer {L,n} = tests[i], pln = p(L,n) mpz_ui_pow_ui(z,2,pln) integer digits = mpz_sizeinbase(z,10) string st = iff(digits>2e6?sprintf("%,d digits",digits): shorten(mpz_get_str(z),"digits",5)) printf(1,"The %d%s power of 2 that starts with %d is %d [i.e. %s]\n",{n,ord(n),L,pln,st}) end for ?elapsed(time()-t0)
- Output:
The 1st power of 2 that starts with 12 is 7 [i.e. 128] The 2nd power of 2 that starts with 12 is 80 [i.e. 1208925819614629174706176] The 45th power of 2 that starts with 123 is 12710 [i.e. 12338...09024 (3,827 digits)] The 12345th power of 2 that starts with 123 is 3510491 [i.e. 12317...80448 (1,056,764 digits)] The 678910th power of 2 that starts with 123 is 193060223 [i.e. 58,116,919 digits]
(The last two tests are simply too much for pwa/p2js)
Python
Using logs, as seen first in the Pascal example.
<lang python>from math import log, modf, floor
def p(l, n, pwr=2):
l = int(abs(l))
digitcount = floor(log(l, 10))
log10pwr = log(pwr, 10)
raised, found = -1, 0
while found < n:
raised += 1
firstdigits = floor(10**(modf(log10pwr * raised)[0] + digitcount))
if firstdigits == l:
found += 1
return raised
if __name__ == '__main__':
for l, n in [(12, 1), (12, 2), (123, 45), (123, 12345), (123, 678910)]:
print(f"p({l}, {n}) =", p(l, n))</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12710 p(123, 12345) = 3510491 p(123, 678910) = 193060223
Racket
First implementation is an untyped, naive revision of the Pascal algorithm. But there is a lot of casting between the exact integers an floats. As well as runtime type checking.
Then there is the typed racket version, which is stricter about the use of types (although, I'm not super happy at the algorithm counting using floating-point integers... but it's faster)
Compare and contrast...
Untyped Racket
<lang racket>#lang racket (define (fract-part f)
(- f (truncate f)))
(define ln10 (log 10)) (define ln2/ln10 (/ (log 2) ln10))
(define (inexact-p-test L)
(let ((digit-shift (let loop ((L (quotient L 10)) (shift 1))
(if (zero? L) shift (loop (quotient L 10) (* 10 shift))))))
(λ (p) (= L (truncate (* digit-shift (exp (* ln10 (fract-part (* p ln2/ln10))))))))))
(define (p L n)
(let ((test? (inexact-p-test L)))
(let loop ((j 1) (n (sub1 n)))
(cond [(not (test? j)) (loop (add1 j) n)]
[(zero? n) j]
[else (loop (add1 j) (sub1 n))]))))
(module+ main
(define (report-p L n) (time (printf "p(~a, ~a) = ~a~%" L n (p L n)))) (report-p 12 1) (report-p 12 2) (report-p 123 45) (report-p 123 12345) (report-p 123 678910))</lang>
- Output:
p(12, 1) = 7 cpu time: 1 real time: 1 gc time: 0 p(12, 2) = 80 cpu time: 0 real time: 0 gc time: 0 p(123, 45) = 12710 cpu time: 5 real time: 5 gc time: 0 p(123, 12345) = 3510491 cpu time: 439 real time: 442 gc time: 23 p(123, 678910) = 193060223 cpu time: 27268 real time: 27354 gc time: 194
Typed Racket
<lang racket></lang>
- Output:
Raku
(formerly Perl 6)
Uses logs similar to Go and Pascal entries. Takes advantage of patterns in the powers to cut out a bunch of calculations.
<lang perl6>use Lingua::EN::Numbers;
constant $ln2ln10 = log(2) / log(10);
my @startswith12 = ^∞ .grep: { ( 10 ** ($_ * $ln2ln10 % 1) ).substr(0,3) eq '1.2' };
my @startswith123 = lazy gather loop {
state $pre = '1.23';
state $count = 0;
state $this = 0;
given $this {
when 196 {
$this = 289;
my \n = $count + $this;
$this = 485 unless ( 10 ** (n * $ln2ln10 % 1) ).substr(0,4) eq $pre;
}
when 485 {
$this = 196;
my \n = $count + $this;
$this = 485 unless ( 10 ** (n * $ln2ln10 % 1) ).substr(0,4) eq $pre;
}
when 289 { $this = 196 }
when 90 { $this = 289 }
when 0 { $this = 90 }
}
take $count += $this;
}
multi p ($prefix where *.chars == 2, $nth) { @startswith12[$nth-1] } multi p ($prefix where *.chars == 3, $nth) { @startswith123[$nth-1] }
- The Task
for < 12 1 12 2 123 45 123 12345 123 678910 > -> $prefix, $nth {
printf "%-15s %9s power of two (2^n) that starts with %5s is at n = %s\n", "p($prefix, $nth):",
comma($nth) ~ ordinal-digit($nth).substr(*-2), "'$prefix'", comma p($prefix, $nth);
}</lang>
- Output:
p(12, 1): 1st power of two (2^n) that starts with '12' is at n = 7 p(12, 2): 2nd power of two (2^n) that starts with '12' is at n = 80 p(123, 45): 45th power of two (2^n) that starts with '123' is at n = 12,710 p(123, 12345): 12,345th power of two (2^n) that starts with '123' is at n = 3,510,491 p(123, 678910): 678,910th power of two (2^n) that starts with '123' is at n = 193,060,223
REXX
<lang rexx>/*REXX program computes powers of two whose leading decimal digits are "12" (in base 10)*/ parse arg L n b . /*obtain optional arguments from the CL*/ if L== | L=="," then L= 12 /*Not specified? Then use the default.*/ if n== | n=="," then n= 1 /* " " " " " " */ if b== | b=="," then b= 2 /* " " " " " " */ LL= length(L) /*obtain the length of L for compares*/ fd= left(L, 1) /*obtain the first dec. digit of L.*/ fr= substr(L, 2) /* " " rest of dec. digits " " */ numeric digits max(20, LL+2) /*use an appropriate value of dec. digs*/ rest= LL - 1 /*the length of the rest of the digits.*/
- = 0 /*the number of occurrences of a result*/
x= 1 /*start with a product of unity (B**0).*/
do j=1 until #==n; x= x * b /*raise B to a whole bunch of powers.*/
parse var x _ 2 /*obtain the first decimal digit of X.*/
if _ \== fd then iterate /*check only the 1st digit at this time*/
if LL>1 then do /*check the rest of the digits, maybe. */
$= format(x, , , , 0) /*express X in exponential format. */
parse var $ '.' +1 f +(rest) /*obtain the rest of the digits. */
if f \== fr then iterate /*verify that X has the rest of digs.*/
end /* [↓] found an occurrence of an answer*/
#= # + 1 /*bump the number of occurrences so far*/
end /*j*/
say 'The ' th(n) ' occurrence of ' b ' raised to a power whose product starts with' ,
' "'L"'" ' is ' commas(j).
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: arg _; do c=length(_)-3 to 1 by -3; _= insert(',', _, c); end; return _ th: arg _; return _ || word('th st nd rd', 1 +_//10 * (_//100 % 10\==1) * (_//10 <4))</lang>
- output when using the inputs of: 12 1
The 1st occurrence of 2 raised to a power whose product starts with "12' is 7.
- output when using the inputs of: 12 2
The 2nd occurrence of 2 raised to a power whose product starts with "12' is 80.
- output when using the inputs of: 123 45
The 45th occurrence of 2 raised to a power whose product starts with "123' is 12,710.
- output when using the inputs of: 123 12345
The 12345th occurrence of 2 raised to a power whose product starts with "123' is 3,510,491.
- output when using the inputs of: 123 678910
The 678910th occurrence of 2 raised to a power whose product starts with "123' is 193,060,223.
Ruby
<lang ruby>def p(l, n)
test = 0
logv = Math.log(2.0) / Math.log(10.0)
factor = 1
loopv = l
while loopv > 10 do
factor = factor * 10
loopv = loopv / 10
end
while n > 0 do
test = test + 1
val = (factor * (10.0 ** ((test * logv).modulo(1.0)))).floor
if val == l then
n = n - 1
end
end
return test
end
def runTest(l, n)
print "P(%d, %d) = %d\n" % [l, n, p(l, n)]
end
runTest(12, 1) runTest(12, 2) runTest(123, 45) runTest(123, 12345) runTest(123, 678910)</lang>
- Output:
P(12, 1) = 7 P(12, 2) = 80 P(123, 45) = 12710 P(123, 12345) = 3510491 P(123, 678910) = 193060223
Rust
<lang rust>fn power_of_two(l: isize, n: isize) -> isize {
let mut test: isize = 0;
let log: f64 = 2.0_f64.ln() / 10.0_f64.ln();
let mut factor: isize = 1;
let mut looop = l;
let mut nn = n;
while looop > 10 {
factor *= 10;
looop /= 10;
}
while nn > 0 {
test = test + 1;
let val: isize = (factor as f64 * 10.0_f64.powf(test as f64 * log % 1.0)) as isize;
if val == l {
nn = nn - 1;
}
}
test
}
fn run_test(l: isize, n: isize) {
println!("p({}, {}) = {}", l, n, power_of_two(l, n));
}
fn main() {
run_test(12, 1); run_test(12, 2); run_test(123, 45); run_test(123, 12345); run_test(123, 678910);
} </lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12710 p(123, 12345) = 3510491 p(123, 678910) = 193060223
Scala
<lang scala>object FirstPowerOfTwo {
def p(l: Int, n: Int): Int = {
var n2 = n
var test = 0
val log = math.log(2) / math.log(10)
var factor = 1
var loop = l
while (loop > 10) {
factor *= 10
loop /= 10
}
while (n2 > 0) {
test += 1
val value = (factor * math.pow(10, test * log % 1)).asInstanceOf[Int]
if (value == l) {
n2 -= 1
}
}
test
}
def runTest(l: Int, n: Int): Unit = {
printf("p(%d, %d) = %,d%n", l, n, p(l, n))
}
def main(args: Array[String]): Unit = {
runTest(12, 1)
runTest(12, 2)
runTest(123, 45)
runTest(123, 12345)
runTest(123, 678910)
}
}</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223
Sidef
<lang ruby>func farey_approximations(r, callback) {
var (a1 = r.int, b1 = 1) var (a2 = a1+1, b2 = 1)
loop {
var a3 = a1+a2
var b3 = b1+b2
if (a3 < r*b3) {
(a1, b1) = (a3, b3)
}
else {
(a2, b2) = (a3, b3)
}
callback(a3 / b3) }
}
func p(L, nth) {
define ln2 = log(2) define ln5 = log(5) define ln10 = log(10)
var t = L.len-1
func isok(n) {
floor(exp(ln2*(n - floor((n*ln2)/ln10) + t) + ln5*(t - floor((n*ln2)/ln10)))) == L
}
var deltas = gather {
farey_approximations(ln2/ln10, {|r|
take(r.de) if (r.de.len == L.len)
break if (r.de.len > L.len)
})
}.sort.uniq
var c = 0 var k = (1..Inf -> first(isok))
loop {
return k if (++c == nth)
k += (deltas.first {|d| isok(k+d) } \\ die "error: #{k}")
}
}
var tests = [
[12, 1], [12, 2], [123, 45], [123, 12345], [123, 678910],
# extra [1234, 10000], [12345, 10000],
]
for a,b in (tests) {
say "p(#{a}, #{b}) = #{p(a,b)}"
}</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12710 p(123, 12345) = 3510491 p(123, 678910) = 193060223 p(1234, 10000) = 28417587 p(12345, 10000) = 284166722
Swift
Slower Version
<lang swift>let ld10 = log(2.0) / log(10.0)
func p(L: Int, n: Int) -> Int {
var l = L var digits = 1
while l >= 10 {
digits *= 10
l /= 10
}
var count = 0 var i = 0
while count < n {
let rhs = (Double(i) * ld10).truncatingRemainder(dividingBy: 1)
let e = exp(log(10.0) * rhs)
if Int(e * Double(digits)) == L {
count += 1
}
i += 1 }
return i - 1
}
let cases = [
(12, 1), (12, 2), (123, 45), (123, 12345), (123, 678910)
]
for (l, n) in cases {
print("p(\(l), \(n)) = \(p(L: l, n: n))")
}</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12710 p(123, 12345) = 3510491 p(123, 678910) = 193060223
Faster Version
<lang swift>import Foundation
func p2(L: Int, n: Int) -> Int {
let asString = String(L) var digits = 1
for _ in 1...18-asString.count {
digits *= 10
}
let ten18 = Int(1e18)
var count = 0, i = 0, probe = 1
while true {
probe += probe
i += 1
if probe >= ten18 {
while true {
if probe >= ten18 {
probe /= 10
}
if probe / digits == L {
count += 1
if count >= n {
count -= 1
break
}
}
probe += probe
i += 1
}
}
let probeString = String(probe) var len = asString.count
if asString.count > probeString.count {
len = probeString.count
}
if probeString.prefix(len) == asString {
count += 1
if count >= n {
break
}
}
}
return i
}
let cases = [
(12, 1), (12, 2), (123, 45), (123, 12345), (123, 678910)
]
for (l, n) in cases {
print("p(\(l), \(n)) = \(p2(L: l, n: n))")
}</lang>
- Output:
Same as before.
Visual Basic .NET
<lang vbnet>Module Module1
Function Func(ByVal l As Integer, ByVal n As Integer) As Long
Dim res As Long = 0, f As Long = 1
Dim lf As Double = Math.Log10(2)
Dim i As Integer = l
While i > 10
f *= 10
i /= 10
End While
While n > 0
res += 1
If CInt((f * Math.Pow(10, res * lf Mod 1))) = l Then
n -= 1
End If
End While
Return res End Function
Sub Main()
Dim values = {Tuple.Create(12, 1), Tuple.Create(12, 2), Tuple.Create(123, 45), Tuple.Create(123, 12345), Tuple.Create(123, 678910), Tuple.Create(99, 1)}
For Each pair In values
Console.WriteLine("p({0,3}, {1,6}) = {2,11:n0}", pair.Item1, pair.Item2, Func(pair.Item1, pair.Item2))
Next
End Sub
End Module</lang>
- Output:
p( 12, 1) = 60 p( 12, 2) = 70 p(123, 45) = 12,710 p(123, 12345) = 3,496,509 p(123, 678910) = 192,278,374 p( 99, 1) = 93
Wren
Just the first version which has turned out to be much quicker than the Go entry (around 37 seconds on the same machine). Frankly, I've no idea why. <lang ecmascript>import "/fmt" for Fmt import "/math" for Math
var ld10 = Math.ln2 / Math.ln10
var p = Fn.new { |L, n|
var i = L
var digits = 1
while (i >= 10) {
digits = digits * 10
i = (i/10).floor
}
var count = 0
i = 0
while (count < n) {
var e = Math.exp(Math.ln10 * (i * ld10).fraction)
if ((e * digits).truncate == L) count = count + 1
i = i + 1
}
return i - 1
}
var start = System.clock var params = [ [12, 1] , [12, 2], [123, 45], [123, 12345], [123, 678910] ] for (param in params) {
Fmt.print("p($d, $d) = $,d", param[0], param[1], p.call(param[0], param[1]))
}
System.print("\nTook %(System.clock - start) seconds.")</lang>
- Output:
p(12, 1) = 7 p(12, 2) = 80 p(123, 45) = 12,710 p(123, 12345) = 3,510,491 p(123, 678910) = 193,060,223 Took 27.851386 seconds.
zkl
Lots of float are slow so I've restricted the tests. <lang zkl>// float*int --> float and int*float --> int fcn p(L,nth){ // 2^j = <L><digits>
var [const] ln10=(10.0).log(), ld10=(2.0).log() / ln10;
digits := (10).pow(L.numDigits - 1);
foreach i in ([1..]){
z:=ld10*i;
if(L == ( ln10 * (z - z.toInt()) ).exp()*digits and (nth-=1) <= 0)
return(i);
}
}</lang>
GNU Multiple Precision Arithmetic Library
GMP is just used to give some context on the size of the numbers we are dealing with. <lang zkl>var [const] BI=Import("zklBigNum"); // libGMP tests:=T( T(12,1),T(12,2), T(123,45),T(123,12345), ); foreach L,nth in (tests){
n:=p(L,nth);
println("2^%-10,d is occurance %,d of 2^n == '%d<abc>' (%,d digits)"
.fmt(n,nth,L,BI(2).pow(n).len()));
}</lang>
- Output:
2^7 is occurance 1 of 2^n == '12<abc>' (3 digits) 2^80 is occurance 2 of 2^n == '12<abc>' (25 digits) 2^12,710 is occurance 45 of 2^n == '123<abc>' (3,827 digits) 2^3,510,491 is occurance 12,345 of 2^n == '123<abc>' (1,056,764 digits)