<< problem 518 - Prime triples and geometric sequences Divisibility of factorials - problem 549 >>

# Problem 523: First Sort I

Consider the following algorithm for sorting a list:

1. Starting from the beginning of the list, check each pair of adjacent elements in turn.
2. If the elements are out of order:
2a. Move the smallest element of the pair at the beginning of the list.
2b. Restart the process from step 1.
3. If all pairs are in order, stop.

For example, the list { 4 1 3 2 } is sorted as follows:

4 1 3 2 (4 and 1 are out of order so move 1 to the front of the list)
1 4 3 2 (4 and 3 are out of order so move 3 to the front of the list)
3 1 4 2 (3 and 1 are out of order so move 1 to the front of the list)
1 3 4 2 (4 and 2 are out of order so move 2 to the front of the list)
2 1 3 4 (2 and 1 are out of order so move 1 to the front of the list)
1 2 3 4 (The list is now sorted)

Let F(L) be the number of times step 2a is executed to sort list L. For example, F({ 4 \space 1 \space 3 \space 2 }) = 5.

Let E(n) be the expected value of F(P) over all permutations P of the integers {1, 2, ..., n}.
You are given E(4) = 3.25 and E(10) = 115.725.

Find E(30). Give your answer rounded to two digits after the decimal point.

# My Algorithm

I solved this problem in a very empiric way:
- write code to solve the problem for small input values (see evaluate)
- stare at the output far too long
- finally see some inner structures after two days

The function evaluate can solve E(10) in a few seconds and will produce this output:

xE(x)#moves#permutations
1001
20.512
31.596
43.257824
56.25750120
611.428220720
720.421029005040
836.29146328040320
964.6223451120362880
10115.734199428803628800

These numbers didn't reveal anything at all - but then I computed the difference E(x) - E(x-1) (as fractions):

xE(x) - E(x-1)denominator = x
21/21/2
313/3
47/47/4
5315/5
631/631/6
7963/7
8127/8127/8
985/3255/9
10511/10511/10

The second columns already shows a pattern for E(x) - E(x - 1) when x is even:
the numerator is always one less than a power of two and the denominator is always x.

When enforcing the denominator to be x (third column) then the same pattern is appears for odd x as well and I finally have a formula:
E(x) - E(x - 1) = dfrac{2^{x-1} - 1}{x}

All I need is a simple for-loop iterating over all x <= 30.

## Note

I somehow prefer the way evaluate() finds the solutions: play around with data structures, STL algorithms, ... → let the computer do some hard work.
The formula used to determine E(30) is pretty simple (though a bit tricky to find) and could have been solved with a pencil and paper only.
That's not why I studied computer science and work as a software engineer ...

# 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 | ./523

Output:

Note: the original problem's input 30 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)

# My code

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

       #include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>

// execute all moves on all permutations of a given size
// → only needed to get an idea what the formula looks like, not needed anymore
double evaluate(unsigned int size)
{
// data = { 1,2,3,...,size }
std::vector<unsigned int> data, current;
for (unsigned int i = 1; i <= size; i++)
data.push_back(i);

// count moves and permutations
unsigned int moves = 0;
unsigned int permutations = 0;
do
{
// get current permutation
current = data;
permutations++;

// sort container
unsigned int pos = 1;
// compare data[pos - 1] and data[pos]
while (pos < size)
{
// pair in wrong order ?
if (current[pos] < current[pos - 1])
{
// rotate all elements from 0 to pos to the right once (last element becomes the first)
std::rotate(current.begin(), current.begin() + pos, current.begin() + pos + 1);
moves++;
// restart
pos = 1;
}
else
pos++;
}
} while (std::next_permutation(data.begin(), data.end()));

std::cout << "E(" << size << ")=" << moves / double(permutations) // result
<< " =" << moves << "/" << permutations << std::endl;   // and as a fraction
return moves / double(permutations);
}

int main()
{
unsigned int limit = 10;
std::cin >> limit;

std::cout << std::fixed << std::setprecision(2);

// solve for small input values
//for (unsigned int i = 1; i <= limit; i++)
//  evaluate(i);

double result = 0;
for (unsigned int i = 1; i <= limit; i++)
{
// 2^(i-1) - 1
auto numerator = (1 << (i - 1)) - 1; // 0,1,3,7,15,31,...
result += numerator / double(i);
}
std::cout << result << std::endl;
return 0;
}


This solution contains 8 empty lines, 15 comments and 4 preprocessor commands.

# 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

September 4, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 60% (out of 100%).

# 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

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The 270 solved problems (level 10) had an average difficulty of 31.3% at Project Euler and
I scored 13,386 points (out of 15600 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.

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