<< problem 47 - Distinct primes factors Prime permutations - problem 49 >>

# Problem 48: Self powers

The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.

Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.

# Algorithm

This problem is pretty easy for languages with support for BigIntegers (such as Java and Python).
I don't want to use external libraries and thus have to write some code in C++ ...
welcome to the world of modular arithmetic !

There are two functions:
- mulmod is the same as (a*b)%modulo where a*b is allowed to exceed 64 bits (unsigned long long).
- powmod is the same as (base^exponent)%modulo where base^exponent is allowed to exceed 64 bits (unsigned long long).

mulmod has two paths:
- if both factors are small (fit in 32 bits) then there will be no overflow and the original formula (a * b) % modulo is executed, which is pretty fast
- else a and b are multiplied bitwise (similar to written multiplication), which is much slower

powmod employs the tricks of en.wikipedia.org/wiki/Exponentiation_by_squaring

## Alternative Approaches

GCC supports __int128 which allows to use the fast path of mulmod all the time.
See my toolbox for more alternatives.

# My code

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

       #include <iostream>

// return (a*b) % modulo
// note: NOT USED because mulmod() is faster (see below)
unsigned long long mulmodBitwise(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);

// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
result %= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}

// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
a  %= modulo;

// next bit
b >>= 1;
}

return result;
}

// return (a*b) % modulo
// very similar to mulmodBitWise, but multiple bits are processed at once (instead of just 1 bit per iteration)
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// count leading zero bits of modulo
unsigned long long m = modulo;
while ((m & 0x8000000000000000ULL) == 0)
{
m <<= 1;
}

// cover all bits of modulo

// blockwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
result += (b & mask) * a;
result %= modulo;

// next bits
a  %= modulo;
}
return result;
}

// return (base^exponent) % modulo
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = mulmod(result, base, modulo);

// even exponent ? a^b = (a*a)^(b/2)
base = mulmod(base, base, modulo);
exponent >>= 1;
}
return result;
}

int main()
{
// sum from 1^1 to x^x
unsigned int x;
std::cin >> x;

// keep the last 10 digits
const unsigned long long TenDigits = 10000000000ULL;

unsigned long long sum = 0;
// add all parts and don't forget the modulo ...
for (unsigned int i = 1; i <= x; i++)
sum += powmod(i, i, TenDigits);

std::cout << (sum % TenDigits) << std::endl;
return 0;
}


This solution contains 19 empty lines, 26 comments and 1 preprocessor command.

# Interactive test

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

Input data (separated by spaces or newlines):

This is equivalent to
echo 10 | ./48

Output:

Note: the original problem's input 1000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(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 26, 2017 submitted solution

# Hackerrank

My code solved 6 out of 6 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 never an option.

projecteuler.net/thread=48 - 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-48-last-ten-digits/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p048.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p048.mathematica (written by Nayuki)
C: github.com/eagletmt/project-euler-c/blob/master/40-49/problem48.c (written by eagletmt)
Javascript: github.com/dsernst/ProjectEuler/blob/master/48 Self powers.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler048.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 already 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.

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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
 << problem 47 - Distinct primes factors Prime permutations - problem 49 >>
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