Code-Challenge: Non trivial powers Revisted

Question

Challenge

In this task you would be given an integer N you have to output all non-trivial power representation of the integer.

Non-trivial power means the base and exponent are > 1

The input N is given in a single line,the inputs are terminated by EOF.The number of inputs would not exceed 10000 values.

The challenge is to implement the fastest solution so that it can process a maximum of 10000 values as fast as possible.

Input

 1745041
 1760929
 1771561
 1852321
 1868689
 1874161
 1885129
 1907161
 1953125
 113795717542678488615120001
 113795717546049421004105281

Output

 1745041 = 1321^2 
 1760929 = 1327^2 
 1771561 = 11^6 121^3 1331^2 
 1852321 = 1361^2 
 1868689 = 1367^2 
 1874161 = 37^4 1369^2 
 1885129 = 1373^2 
 1907161 = 1381^2 
 1953125 = 5^9 125^3 
 113795717542678488615120001 = 10667507560001^2
 113795717546049421004105281 = 10667507560159^2

Constraints

• N is less than 30 digits long to test your favorite integer factorization algorithm ;-)
• Assume that for ever input N,at-least one non-trivial power representation is possible.
• You can use any language of your choice
• Fastest solution,with the most efficient time complexity wins!

Answer 1

C - No bigints

My solution use a 64 bit float and a 64 bit integer in combination to produce the result without using slow bignumber arithmetic. First a possible set of values that fit the high-order bits are produced using floats, then the low order is checked using integers. Due to 64 bit floats having 53 bit mantissa the method will break for numbers around 106 bit in length, but the task was only for up to 100 bit numbers.

#include "stdio.h"
#include "stdlib.h"
#include "time.h"
#include "math.h"
#define i64 unsigned long long
#define i32 unsigned int
#define f64 double

int main(){
    char inp[31];
    while(1==scanf("%30s",inp)){
        printf("%s =",inp);

        f64 upper=strtod(inp,NULL);
        //i64 lower=strtoull(inp,NULL,10);
        //i64 lower=atoll(inp);
        //Neither library function produce input mod 2^64 like one would expect, fantabulous!
        i64 lower=0;
        i32 c=0;
        while(inp[c]){
            lower*=10;
            lower+=inp[c]-48;
            c++;
        }
        f64 a,b;
        i64 candidate;
        f64 offvalue;
        f64 offfactor;
        i32 powleft;
        i64 product;
        i64 currentpow;
        for(a=2;a<50;a++){ //Loop for finding large bases (>3).
            b=1.0/a;
            b=pow(upper,b);
            candidate=(i64)(b+.5);
            if(candidate<4){
                break;
            }
            offvalue=b-(f64)candidate;
            offfactor=fabs(offvalue/b);
            if(offfactor<(f64)1e-14){
                product=1;
                powleft=(i32)a;
                currentpow=candidate;
                while(powleft){ //Integer exponentation loop.
                    if(powleft&1){
                        product*=currentpow;
                    }
                    currentpow*=currentpow;
                    powleft=powleft>>1;
                }
                if(product==lower){
                    printf(" %I64u^%.0lf",candidate,a);
                }
            }
        }
        for(candidate=3;candidate>1;candidate--){ //Loop for finding small bases (<4), 2 cycles of this saves 50 cycles of the previous loop.
            b=log(upper)/log(candidate);
            a=round(b);
            offfactor=fabs(a-b);
            if((offfactor<(f64)1e-14) && (a>1)){
                product=1;
                powleft=(i32)a;
                currentpow=candidate;
                while(powleft){ //Integer exponentation loop.
                    if(powleft&1){
                        product*=currentpow;
                    }
                    currentpow*=currentpow;
                    powleft=powleft>>1;
                }
                if(product==lower){
                    printf(" %I64u^%.0lf",candidate,a);
                }
            }
        }
        printf("\n");
        if(inp[0]==113){ //My keyboard lacks an EOF character, so q will have to do.
            return 0;
        }
    }
    return 0;
}

As for runtime, ideone produce a clean 0 http://www.ideone.com/HJl2f this should be a pretty superior method, but any attempt to verify this will drown in I/O.

Answer 2

Python, 207 chars

This takes O(log² N) bigint ops. O(log N) different exponents to try, and O(log N) binary search to find each base.

import sys
for N in map(int,sys.stdin.readlines()):
 p,s=2,''
 while 1<<p<=N:
  i,j=2,N
  while j>i+1:
   m=(i+j)/2
   if m**p>N:j=m
   else:i=m
  if i**p==N:s+=' %d^%d'%(i,p)
  p+=1
 if s:print N,'=',s[1:]

Answer 3

Ruby

Non-golfed, not-really-optimized solution. I hope I didn't mess up my math.

Roots are calculated first using core sqrt and cbrt methods, then using dumb approximation (binary search).

On my PC, 10000 long random inputs (not necessarily actually having non-trivial representations) take ~30 seconds (see the bench method).

def root start, exp, startapprox=nil
    if exp==1
        return start
    elsif exp==2
        return Math.sqrt(start).round
    elsif exp==3 and Math.respond_to? :cbrt # 1.9 only
        return Math.cbrt(start).round
    else
        lower=0
        upper=startapprox||Math.sqrt(start).ceil

        until upper-lower<=1
            res=(lower+upper)/2

            if res**exp<start
                lower=res+1
            else
                upper=res
            end
        end

        distlow=(start - lower**exp).abs
        distupp=(start - upper**exp).abs

        if distlow<distupp
            return lower
        else
            return upper
        end
    end
end

def powers number
    ret=[]
    skip=[]

    maxexp=(Math.log2 number).floor
    prev=nil

    2.upto(maxexp) do |exp|
        next if skip[exp]

        rt=root number, exp, prev
        if number==rt**exp
            ret<<"#{rt}^#{exp}"
        else
            (exp..maxexp).step(exp){|i| skip[i]=true} # if it's not 2nd power, it cant be 4th power etc
        end

        prev=rt
    end

    "#{number} = #{ret.reverse.join ' '}"
end

def bench n=10
    require 'benchmark'

    Array.new(n){Benchmark.realtime{powers rand 1e30}}.inject(:+)
end


# read input, output powers
if __FILE__==$0
    while (a=gets.strip).length>0
        puts powers a.to_i
    end
end