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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)}.
My 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 theexponent
- else it appends the sum and has to go deeper
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.
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 toecho "1 15" | ./122
Output:
(this interactive test is still under development, computations will be aborted after one second)
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.
Benchmark
The correct solution to the original Project Euler problem was found in 0.11 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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 rarely 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:
C# 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)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
Heatmap
Please click on a problem's number to open my solution to that problem:
green | solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too | |
yellow | solutions score less than 100% at Hackerrank (but still solve the original problem easily) | |
gray | problems are already solved but I haven't published my solution yet | |
blue | solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much | |
orange | problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte | |
red | problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too | |
black | problems are solved but access to the solution is blocked for a few days until the next problem is published | |
the flashing problem is the one I solved most recently |
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I scored 13,486 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
Look at my progress and performance pages to get more details.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
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