<< problem 218 - Perfect right-angled triangles Sphere Packing - problem 222 >>

# Problem 219: Skew-cost coding

Let A and B be bit strings (sequences of 0's and 1's).
If A is equal to the leftmost length(A) bits of B, then A is said to be a prefix of B.
For example, 00110 is a prefix of 001101001, but not of 00111 or 100110.

A prefix-free code of size n is a collection of n distinct bit strings such that no string is a prefix of any other.
For example, this is a prefix-free code of size 6:
0000, 0001, 001, 01, 10, 11

Now suppose that it costs one penny to transmit a '0' bit, but four pence to transmit a '1'.
Then the total cost of the prefix-free code shown above is 35 pence, which happens to be the cheapest possible for the skewed pricing scheme in question.
In short, we write Cost(6) = 35.

What is Cost(10^9) ?

# My Algorithm

The basic idea behind the construction of prefix-free codes is outlined on the Wikipedia page about Huffman codes (see en.wikipedia.org/wiki/Huffman_coding):

• add all codes to a priority-queue sorted by their weight
• pick the code from the queue's front and create two need codes: append a 0 and a 1 and insert those two codes in the priority queue
Initially there are two codes: 0 (weight 1) and 1 (weight 4).
The algorithm then has to find the remaining 10^9 - 2 codes and keep track of their cost.

Even though the correct result is found, this algorithm is pretty slow (queue needs 147 seconds).
When I looked at the lengths of the codes I saw that they are pretty short. That means that their cost is pretty low, too.
queue was repeatedly picking codes with the same cost from its storage. And adding children to the same two categories: plus 1 and plus 4 pence.

That's why I wrote a different approach called array:
Don't keep track of every single code - just count how many codes with a certain weight exists.
Initially there is one code with weight 1 and one code with weight 4.
Then the algorithm is as follows:
• pick all codes with the lowest weight from costs[x]
• append a zero and a one in order to create their children: costs[x + 1] += costs[x] and costs[x + 4] += costs[x]

## Note

The STL's priority_queue is a max-heap, that means that top() always returns the largest element.
However, I need the smallest element for my program and therefore needed to use std::greater for comparisons.
The peak memory usage of my first algorithm was about 1 GByte.

# 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 6 | ./219

Output:

Note: the original problem's input 1000000000 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. Or just jump to my GitHub repository.

       #include <iostream>
#include <queue>
#include <vector>
#include <functional>

// bit sequence
typedef unsigned char Cost;

// find result using a priority_queue
unsigned long long queue(unsigned int limit)
{
std::priority_queue<Cost, std::vector<Cost>, std::greater<Cost>> codes; // min-heap instead of default max-heap
// first two codes
codes.push(1);
codes.push(4);
unsigned long long totalCost = 5; // sum of the first two codes: 1+4

// until enough codes generated
unsigned int numCodes = 2;
while (numCodes < limit)
{
// pick the first
auto current = codes.top();
codes.pop();
// add two new children codes
codes.push(current + 1);
codes.push(current + 4);

// keep track of the cost
numCodes++;
totalCost += current + 1 + current + 4 - current; // same as current + 5
}

}

// find result using a bit-length counters
unsigned long long array(unsigned int limit)
{
std::vector<unsigned long long> costs(70, 0);
// initial single-bit codes: "0" => weight 1, "1" => weight 4
costs[1] = 1;
costs[4] = 1;
unsigned long long totalCost = 1 + 4;

auto current = 1;
// number of codes that I need to generate
auto remaining = limit - 2;
while (remaining > 0)
{
// all codes "used" of the current weight ? => look at higher weights
while (costs[current] == 0) // no gaps: "if" instead of "while" works as well
current++;

// try to process all codes of a certain weight at once
auto block = costs[current];
// except when I don't need all of them
if (block > remaining)
block = remaining;

remaining          -= block;
costs[current]     -= block;
costs[current + 1] += block;
costs[current + 4] += block;
// weight is block * (current + 1 + current + 4 - current)
totalCost          += block * (unsigned long long)(current + 5);
}

}

int main()
{
unsigned int limit = 1000000000;
std::cin >> limit;

// slow algorithm
//std::cout << queue(limit) << std::endl;

// fast algorithm
std::cout << array(limit) << std::endl;

return 0;
}


This solution contains 14 empty lines, 19 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 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

July 30, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 70% (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 black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 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.

 << problem 218 - Perfect right-angled triangles Sphere Packing - problem 222 >>
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