Problem 122: Efficient exponentiation

(see projecteuler.net/problem=122)

The most naive way of computing n^15 requires fourteen multiplications:
n * n * ... * n = n^15

But using a "binary" method you can compute it in six multiplications:

n * n = n^2
n^2 * n^2 = n^4
n^4 * n^4 = n^8
n^8 * n^4 = n^12
n^12 * n^2 = n^14
n^14 * n = n^15

However it is yet possible to compute it in only five multiplications:

n * n = n^2
n^2 * n = n^3
n^3 * n^3 = n^6
n^6 * n^6 = n^12
n^12 * n^3 = n^15

We shall define m(k) to be the minimum number of multiplications to compute n^k; for example m(15) = 5.

For 1 <= k <= 200, find sum{m(k)}.

Algorithm

It took me quite some time to figure out a fast solution.
I realized pretty soon that I should generate a sequence which contains the exponents where a_i = a_k + a_l and i > k >= l.
A sequence for n^15 is { 1, 2, 3, 6, 12, 15 }. Using pen and paper I also found { 1, 2, 4, 5, 10, 15 }.
Therefore there is not one unique optimal solution for each exponent but there can be multiple.

First I tried an iterative depth-first search, where I append any combination of a_k + a_l to a list which starts with { 1 }.
The number of combinations was huge - too much for my little computer ...
Then I discovered that this problem is called "Addition chain" and there is substantial information available: en.wikipedia.org/wiki/Addition_chain
A list of chain lengths can be downloaded, too: wwwhomes.uni-bielefeld.de/achim/addition_chain.html

Especially the Brauer chain caught my eye: for n <= 2500 you find the correct solution by restricting k = i - 1.
In plain English: add every exponent to the biggest one only.

search is a recursive depth-first search, limited by maxDepth:
- it aborts immediately if search depth was exhausted, e.g. the current chain is longer than allowed
- it computes the sum of any element and the last element and returns true if it matches the exponent
- else it appends the sum and has to go deeper
The code runs about twice as fast when combining "high" numbers first: instead of a_0 + a_{last}, a_1 + a_{last}, ... a_{last} + a_{last} I start the other way around.

findChain slowly increases the search depth search returns a solution.

Modifications by HackerRank

Each solution has to be printed in a human-readable format.
Multiple test cases want you to compute the same formulas, therefore a small cache was added.

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>
#include <vector>
#include <map>
 
//#define ORIGINAL
 
// a single addition chain
typedef std::vector<unsigned int> Chain;
 
// iterative depth-first search of Brauer sequence
bool search(Chain& chain, unsigned int exponent, unsigned int maxDepth)
{
// too deep ?
if (chain.size() > maxDepth)
return false;
 
auto last = chain.back();
for (size_t i = 0; i < chain.size(); i++)
{
//auto sum = chain[i] + last;
auto sum = chain[chain.size() - 1 - i] + last; // try high exponents first => about twice as fast
if (sum == exponent)
return true;
 
chain.push_back(sum);
if (search(chain, exponent, maxDepth))
return true;
 
chain.pop_back();
}
 
return false;
}
 
// increase depth until a solution is found
Chain findChain(unsigned int exponent)
{
// cached ? (needed for Hackerrank only)
static std::map<unsigned int, Chain> cache;
auto lookup = cache.find(exponent);
if (lookup != cache.end())
return lookup->second;
 
// start iterative search
Chain chain;
unsigned int depth = 1;
while (true)
{
// reset chain
chain = { 1 };
// a start search
if (search(chain, exponent, depth))
break;
 
// failed, allow to go one step deeper
depth++;
}
 
cache[exponent] = chain;
return chain;
}
 
// print a single chain in Hackerrank format
void print(const Chain& chain)
{
// number of multiplications
std::cout << (chain.size() - 1) << std::endl;
// print each multiplication
for (size_t i = 1; i < chain.size(); i++)
{
// involved exponents
auto sum = chain[i];
auto add1 = chain[i - 1];
auto add2 = sum - add1;
 
std::cout << "n";
if (add1 > 1)
std::cout << "^" << add1;
std::cout << " * n";
if (add2 > 1)
std::cout << "^" << add2;
std::cout << " = n^" << sum << std::endl;
}
}
 
int main()
{
#ifdef ORIGINAL
 
unsigned int sum = 0;
// find all chains 2..200
for (unsigned int exponent = 2; exponent <= 200; exponent++)
{
auto chain = findChain(exponent);
// sum of all chains' lengths
sum += chain.size();
}
std::cout << sum << std::endl;
 
#else
 
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int exponent;
std::cin >> exponent;
 
// compute one chain (there might be different chains of the same length)
auto chain = findChain(exponent);
// append the exponent, which is not part of the chain yet
chain.push_back(exponent);
// and display
print(chain);
}
 
#endif
 
return 0;
}

This solution contains 20 empty lines, 20 comments and 6 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.

Number of test cases (1-5):

Input data (separated by spaces or newlines):
Note: Enter an exponent and one optimal solution (addition chain) will be displayed

This is equivalent to
echo "1 15" | ./122

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 0.11 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

May 15, 2017 submitted solution
May 15, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler122

My code solves 3 out of 3 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 40% (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.

Links

projecteuler.net/thread=122 - 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-122-efficient-exponentiation/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p122.java (written by Nayuki)
Scala: github.com/samskivert/euler-scala/blob/master/Euler122.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.

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The 163 solved problems had an average difficulty of 22.2% at Project Euler and I scored 11,907 points (out of 13200) at Hackerrank's Project Euler+.
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.
some names mentioned on this site may be trademarks of their respective owners.
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