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# 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 `print`

ed 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 `#ifdef`

s 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.

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

Output:

*(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.

*Please click on a problem's number to open my solution to that problem:*

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My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

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