<< problem 28 - Number spiral diagonals | Digit fifth powers - problem 30 >> |

# Problem 29: Distinct powers

(see projecteuler.net/problem=29)

Consider all integer combinations of a^b for 2 <= a <= 5 and 2 <= b <= 5:

2^2= 4, 2^3= 8, 2^4= 16, 2^5= 32

3^2= 9, 3^3= 27, 3^4= 81, 3^5= 243

4^2=16, 4^3= 64, 4^4=256, 4^5=1024

5^2=25, 5^3=125, 5^4=625, 5^5=3125

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated by ab for 2 <= a <= 100 and 2 <= b <= 100?

# My Algorithm

The core problem is: find all x^y where also an a^b exists with x>a and x <= limit (a "collision").

The only collision in the problem's example is x^y=4^2=16 and a^b=2^4=2^4=16.

My variable `repeated`

counts all numbers that appear repeatedly (such as 16).

The final result will be (limit-1)^2 - repeated (for the example we found that repeated=1 and therefore (5-1)^2 - 1 = 15).

The rest of this algorithm is all about an efficient way to find the value of repeated.

*Note:* we have limit=5 in the example but we need to solve for limit=100 (or limit=100000 in the modified Hackerrank problem).

Because of 4=2^2 we know that 4^y=2^{2y}: all powers with base 4 can be rewritten as 4^2=2^{2*2}=2^4, 4^3=2^{2*3}=2^6, 4^4=2^{2*4}=2^8 and 4^5=2^{2*5}=2^{10}.

Therefore when we convert all powers of 4 to powers of 2 we use those exponents: `2,3,4,5`

(for a=2) and `4,6,8,10`

(for x=4 → a=2^2).

Obviously `4`

is found in both sets. That's our collision.

The first step in my program is to find all exponents if we reduce all numbers x to their "parent" a^b:

1. we know for certain that all numbers from 2 to limit will be used: `2,3,4,5`

2. if x=a^2 is a valid base (that means x <= limit), then `2*2,2*3,2*4,2*5`

will be used for base a, too.

3. if x=a^3 is a valid base then `3*2,3*3,3*4,3*5`

will be used for base a, too.

4. if x=a^4 is a valid base then `4*2,4*3,4*4,4*5`

will be used for base a, too.

5. and so on ...

My container `minExponent`

records which exponents could be used. If there are multiple combinations, then the lowest is stored.

I hope an example makes things a bit clearer:

1. insert `1`

at positions `2,3,4,5`

2. insert `2`

at positions `6,8,10`

→ but not at position 4 because that one is already occupied

3. we can stop because there is no other base x=2^3 <= limit

`minExponent`

contains `2 → 1, 3 → 1, 4 → 1, 5 → 1, 6 → 2, 8 → 2, 10 → 2`

An `std::vector`

doesn't have gaps and starts at 0, that's why I there are some unused entries, too, filled with zero:

`minExponent = { 0, 0, 1, 1, 1, 1, 2, 0, 2, 0, 2 }`

The "highest" collision with base 2^b=x^y can be found for 2^b=64^y=2^{8y} (next power of 2 is 128, which is larger than 100).

For the modified Hackerrank problem, this "highest" collision with base 2^b=x^y can be found for x^y=65536^y=2^{16y} (limit is 100000 instead of 100).

That's why `MaxBasePower = 16`

.

My container `base`

is zero for every `base[x]`

where there no a^b=x exists. The data structure is incrementally updated and initially all entries are zero.

If we encounter such a `base[x]=0`

then we immediately fill `base[x*x]=x`

and `base[x*x*x]=x`

and `base[x*x*x*x]=x`

...

because `x*x`

, `x*x*x`

, ... will collide with `x`

.

In the example, for `x=2`

we'll find that `base[2]=0`

and hence fill `base[2*2]=2`

.

If `base[x] == 0`

then collisions with smaller x are impossible.

If `base[x] != 0`

then `x`

has a "parent" such that x=base[x]^exponent. Collisions are possible.

We scan through `minExponent`

and count all collisions (in `repeated`

):

Due to x^y = base[x]^{y*exponent} all elements `minExponent[y*exponent]`

have to be checked if that entry was

already used by some other `x`

.

## Alternative Approaches

The original problem can be solved by brute-force if your language supports big integers.

I haven't tried it but maybe you can store log{a^b} in a `std::set`

, too.

## Modifications by HackerRank

The higher limit of 100000 instead of 100 forced me to come up with my slightly more complex but much faster solution.

The result exceeds 32 bits (for limit close to 100000), that's why I introduced the `all`

variable as `unsigned long long`

.

# My code

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

#include <iostream>
#include <vector>
int main()
{
// I called it "limit" in the explanation
unsigned int maxExponent;
std::cin >> maxExponent;
// 2^17 > 10^5 (max. input)
const unsigned int MaxBasePower = 16;
// if a^b = base^(exponent*basePower)
// then b = (1..n) but exponent*basePower = (basePower, 2*basePower, 3*basePower, ..., n*basePower)
// store for each product exponent*basePower the smallest basePower
std::vector<unsigned int> minExponent((maxExponent+1)*MaxBasePower);
for (unsigned int i = 1; i <= MaxBasePower; i++)
for (unsigned int j = 1; j <= maxExponent; j++)
if (minExponent[i*j] == 0)
minExponent[i*j] = i;
// all "a" which can be composed as base^exponent, stored as [a] => [base]
std::vector<unsigned int> base(maxExponent + 1, 0);
// how often numbers were used multiple times (those are the collisions we are looking for)
unsigned int repeated = 0;
// analyze all bases
for (unsigned int x = 2; x <= maxExponent; x++) // maximum base is maxExponent, too
{
// is x = parent^exponent ?
auto parent = base[x];
if (parent == 0) // no
{
// find all future children where "x" is the parent
auto power = x * x;
// [x^2] = [x^3] = [x^4] = ... = x
while (power <= maxExponent)
{
base[power] = x;
power *= x;
}
// no x=a^b possible, "repeated" remains unchanged
continue;
}
// we have a parent, find exponent such that a = parent^exponent
unsigned int exponent = 0;
auto reduce = x;
while (reduce > 1)
{
reduce /= parent;
exponent++;
}
// analyze all pairs, and count all numbers we saw before (repeated++)
for (unsigned int y = 2; y <= maxExponent; y++)
{
// that exponent was already used ?
if (minExponent[y * exponent] < exponent)
repeated++;
}
}
// there are (maxExponent-1)^2 combinations, subtract all duplicates
unsigned long long all = maxExponent - 1;
auto result = all*all - repeated;
std::cout << result << std::endl;
return 0;
}

This solution contains 10 empty lines, 16 comments and 2 preprocessor commands.

# Interactive test

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

This is equivalent to`echo 5 | ./29`

Output:

*Note:* the original problem's input `100`

__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 24, 2017 submitted solution

April 7, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler029

My code solves **21** out of **21** 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=29 - **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-29-distinct-terms-sequence-ab/ (written by Kristian Edlund)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p029.hs (written by Nayuki)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p029.java (written by Nayuki)

Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p029.mathematica (written by Nayuki)

C: github.com/eagletmt/project-euler-c/blob/master/20-29/problem29.c (written by eagletmt)

Javascript: github.com/dsernst/ProjectEuler/blob/master/29 Distinct powers.js (written by David Ernst)

Scala: github.com/samskivert/euler-scala/blob/master/Euler029.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 or the memory limit of 256 MByte.

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

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I scored 12,983 points (out of 15100 possible points, top rank was 17 out of ≈60000 in August 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.

# 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.

<< problem 28 - Number spiral diagonals | Digit fifth powers - problem 30 >> |