Problem 63: Powerful digit counts

(see projecteuler.net/problem=63)

The 5-digit number, 16807=7^5, is also a fifth power. Similarly, the 9-digit number, 134217728=8^9, is a ninth power.

How many n-digit positive integers exist which are also an nth power?

Algorithm

check(n) finds all such powers with n digits:
- first it generates the smallest n-digit number from and the largest n-digit number to
- e.g. for n=5 we have from=10000 and to=99999
- then all numbers 1^n to 9^n are computed, if they are between from and to we have a match

While experimenting I saw no number with more than 21 digits to fulfilled the problem's conditions.

A minor headache was that 21 digits don't fit into C++ unsigned long long anymore.
Therefore I switched to double which has a few rounding issues but they don't affect the original problem.

Alternative Approaches

It's easy to determine the number of digits using log10.
And as mentioned in the code, pow(a,b) returns a^b.
Both functions are available in the C++ standard library.

Modifications by HackerRank

Hackerrank wants you to print all powers with a certain number of digits - instead of finding the number of all such powers.
However, the rounding issues of double now come into play and the right-most digits are a bit off for large n.

Fortunately, Hackerrank wants us only to find the powers with at most 19 digits ... and a 64 bit unsigned long long is sufficient for this task.
That's why you find the #ifdef construct where the type of Number is defined.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

#include <iostream>
 
//#define ORIGINAL
 
// find all numbers where x^digits has digits digits (I'm loving that comment ...)
unsigned int check(unsigned int digits)
{
// numbe rof matches
unsigned int count = 0;
 
// unsigned long long isn't sufficient for the highest digits
// a double has some rounding issues but they don't affect to result
#ifdef ORIGINAL
typedef double Number;
#else
typedef unsigned long long Number;
#endif
 
// range of valid numbers
// from = 10^(digits-1)
// to = 10^digits - 1
Number to = 1;
for (unsigned int i = 1; i <= digits; i++)
to *= 10;
Number from = to / 10;
to--;
 
// try all single-digit base numbers
for (unsigned int base = 1; base <= 9; base++)
{
// compute power = base ^ digits
Number power = base;
for (unsigned int i = 1; i < digits && power <= to; i++)
power *= base;
// could use C++'s pow(), too
 
// right number of digits ?
if (power >= from && power <= to)
{
count++;
#ifndef ORIGINAL
std::cout << std::fixed << power << std::endl;
#endif
}
}
 
return count;
}
 
int main()
{
#ifdef ORIGINAL
// check all digits
unsigned int count = 0;
for (unsigned int digits = 1; digits <= 21; digits++) // I observed no results with more than 21 digits
count += check(digits);
std::cout << count << std::endl;
 
#else
 
// check only certain digits
unsigned int digits = 9;
std::cin >> digits;
check(digits);
#endif
 
return 0;
}

This solution contains 11 empty lines, 14 comments and 9 preprocessor commands.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):

This is equivalent to
echo 5 | ./63

Output:

(please click 'Go !')

(this interactive test is still under development, computations will be aborted after one second)

Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

Changelog

February 28, 2017 submitted solution
April 26, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler063

My code solves 10 out of 10 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

Links

projecteuler.net/thread=63 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-63-n-digit-nth-power/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p063.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem063.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler063.scala (written by Michael Bayne)

Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.
red problems are solved but exceed the time limit of one minute.

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The 206 solved problems had an average difficulty of 27.5% at Project Euler and
I scored 12,626 points (out of 14300 possible points, top rank was 20 out ouf ≈60000 in July 2017) at Hackerrank's Project Euler+.
Look at my progress and performance pages to get more details.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

more about me can be found on my homepage, especially in my coding blog.
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