<< problem 215 - Crack-free Walls | Sphere Packing - problem 222 >> |

# Problem 219: Skew-cost coding

(see projecteuler.net/problem=219)

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.

# My code

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

#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
}
return totalCost;
}
// 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;
// start with the lowest weight (which is 1)
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;
// adjust counters
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);
}
return totalCost;
}
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.

# 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 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)*

# 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

July 30, 2017 submitted solution

July 30, 2017 added comments

# Difficulty

Project Euler ranks this problem at **70%** (out of 100%).

# Links

projecteuler.net/thread=219 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

# 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).

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red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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I scored 13,183 points (out of 15300 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 215 - Crack-free Walls | Sphere Packing - problem 222 >> |