<< problem 80 - Square root digital expansion Path sum: three ways - problem 82 >>

# Problem 81: Path sum: two ways

In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right,
by only moving to the right and down, is indicated in bold and is equal to 2427.

131 673 234 103 18
201 96 342 965 150
630 803 746 422 111
537 699 497 121 956
805 732 524 37 331

Find the minimal path sum, in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by only moving right and down.

# My Algorithm

That's a classic example for breadth-first search (see )
Initialization: create a priority-queue (descendingly ordered by the path sum / called weight in my program) and insert the upper left corner.

As long as we haven't reached the destination:

• pick the position with the lowest weight (=partial path sum)
• mark it as processed
• add its right and bottom neighbor to the priority queue, with their numbers added to the current weight
C++'s STL comes with a priority_queue. Unfortunately its top() returns the largest value but I need the smallest.
Therefore my struct Cell has operator<() implemented "the wrong way" on purpose.

## Note

Project Euler's file has all values stored in a CSV (comma-separated values) format.
The C++ code looks a bit ugly because C++ recognizes only whitespaces by default.

# 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.

Input data (separated by spaces or newlines):
Note: The first number is the width/height of the quadratic matrix, then follow all elements, separated by a space

This is equivalent to
echo "" | ./81

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, as well as the input data, too. Or just jump to my GitHub repository.

The code contains #ifdefs 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 <queue>
#include <vector>
#include <iostream>

// 2D matrix: unfortunately x and y are swapped, so we need to write matrix[y][x]
// instead of the more common matrix[x][y]
typedef std::vector<std::vector<unsigned int>> Matrix;

// use a priority queue to find the next cell to process
struct Cell
{
// position
unsigned int x, y;
// sum of shortest route so far
unsigned long long weight;

Cell(unsigned int x_, unsigned int y_, unsigned long long weight_)
: x(x_), y(y_), weight(weight_) {}

// std::priority_queue returns the LARGEST element, therefore I implement this function "the other way 'round"
bool operator<(const Cell& cell) const
{
return weight > cell.weight; // ">" is not a typo !
}
};

unsigned long long search(const Matrix& matrix)
{
// matrix height/width
const auto size = matrix.size();

std::vector<std::vector<bool>> processed(matrix.size());
for (auto& row : processed)
row.resize(matrix.size(), false);

// process cells in increasing distance from starting point
std::priority_queue<Cell> next;
// add starting point (upper left corner)
next.push(Cell(0, 0, matrix[0][0]));

while (!next.empty())
{
// get cell with the smallest weight
Cell cell = next.top();
// and remove it from the queue
next.pop();

// we have been here before ?
// must have been on a shorter route, hence discard current cell
if (processed[cell.y][cell.x])
continue;

processed[cell.y][cell.x] = true;

// finished ?
if (cell.x == size - 1 && cell.y == size - 1)
return cell.weight;

// one step right
if (cell.x + 1 < size)
next.push(Cell(cell.x + 1, cell.y, cell.weight + matrix[cell.y][cell.x + 1]));
// one step down
if (cell.y + 1 < size)
next.push(Cell(cell.x, cell.y + 1, cell.weight + matrix[cell.y + 1][cell.x]));
}

return -1; // failed
}

int main()
{
unsigned int size = 80;
//#define ORIGINAL
#ifndef ORIGINAL
std::cin >> size;
#endif

Matrix matrix(size);
for (auto& row : matrix)
{
row.resize(size);
for (auto& cell : row)
{
#ifdef ORIGINAL
// unfortunately, Project Euler used a CSV format which is a bit tricky to parse in C++
cell = 0;
// read until the number is complete or we run out of input
while (std::cin)
{
char c;
std::cin.get(c);
// number complete ?
if (c < '0' || c > '9')
break;

// add digit to current number
cell *= 10;
cell += c - '0';
}
#else
std::cin >> cell;
#endif
}
}

// go !
std::cout << search(matrix) << std::endl;

return 0;
}


This solution contains 18 empty lines, 24 comments and 8 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

March 12, 2017 submitted solution

# Hackerrank

My code solves 7 out of 7 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 10% (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.

# 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 80 - Square root digital expansion Path sum: three ways - problem 82 >>
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