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# Problem 15: Lattice paths

(see projecteuler.net/problem=15)

Starting in the top left corner of a 2x2 grid, and only being able to move to the right and down,

there are exactly 6 routes to the bottom right corner.

How many such routes are there through a 20x20 grid?

# Algorithm

Each grid consists of `(width+1)*(height+1)`

points. The grid in the problem statement has `(2+1)*(2+1)=9`

points.

I chose my coordinates to be [0][0] in the top left corner and [width][height] in the bottom right corner.

For each point in a grid, the number of routes from *this point* to the lower right corner

is the sum of all routes when going right plus all routes when going down:

`grid[x][y] = grid[x+1][y] + grid[x][y+1];`

There is a single route from the lower right corner to itself ("stay where you are"):

`grid[width][height] = 1;`

Now I perform a breadth-first search (en.wikipedia.org/wiki/Breadth-first_search) from the lower-right corner to the upper-left corner.

A queue named `next`

contains the coordinates of the next points to analyze - initially it holds the point above the lower-right corner

and the point to the left of the lower-right corner.

A loop processes each point in `next`

:

1. If this point was already processed, skip it

2. If it is possible to move right then get the number of routes stored in `grid[x+1][y]`

3. If it is possible to move down then get the number of routes stored in `grid[x][y+1]`

4. Write the sum of step 2 and 3 to `grid[x][y]`

, potentially avoid overflow (see Hackerrank modification)

5. If there is a left neighbor, enqueue it in `next`

6. If there is a neighbor above, enqueue it in `next`

Finally we have the result in `grid[0][0]`

.

## Alternative Approaches

I later found a posting on the internet that the result is

frac{(width + height)!}{width! height!}

but I'm not a math guy, I'm a programmer ...

however, the best explanation for that formula is as follows and unfortunately, it's about bits :-) :

If we take any route through the grid then a decision has to be made at each point: walk to the right or walk down.

That's similar to a binary number where each bit can be either 0 or 1. Let's say 0 means "right" and 1 means "down".

Then we have to make width + height decisions: exactly width times we have to go right and exactly height times we have to go down.

In our imaginary binary number, there are width + height bits with height 0s and width 1s.

All permutations of these bits are valid routes in the grid.

And as turns out, there are frac{(width + height)!}{width! height!} permutations, see en.wikipedia.org/wiki/Permutation

## Modifications by HackerRank

The result should be printed \mod (10^9 + 7)

# My code

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

The code contains `#ifdef`

s 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 <vector>
#include <deque>
#include <utility>
#include <iostream>
int main()
{
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int width, height;
std::cin >> width >> height;
// create a 2D array which contains the number of paths
// from the current lattice point to the lower-right corner
// there are (width + 1) * (height + 1) such points
// for the original problem, i.e. 21x21 numbers must be found
const unsigned long long Unknown = 0;
std::vector<std::vector<unsigned long long>> grid(width + 1);
for (auto& column : grid)
column.resize(height + 1, Unknown);
// one route if we are already at the goal
grid[width][height] = 1;
// enqueue the next unprocessed lattice points: left and upper neighbor of the lower-right corner
std::deque<std::pair<unsigned int, unsigned int>> next;
next.push_back(std::make_pair(width - 1, height));
next.push_back(std::make_pair(width, height - 1));
// as long as there are unprocessed points
while (!next.empty())
{
// get next point
auto current = next.front();
next.pop_front();
// I prefer names which are easier to read ...
auto x = current.first;
auto y = current.second;
// already solved ?
if (grid[x][y] != Unknown)
continue;
// sum of all path when going right plus when going down
unsigned long long routes = 0;
if (x < width) // can go right ?
routes += grid[x + 1][y];
if (y < height) // can go down ?
routes += grid[x][y + 1];
#define ORIGINAL
#ifndef ORIGINAL
routes %= 1000000007; // Hackerrank wants the result MOD 10^9 + 7
#endif
// solved number for current lattice point
grid[x][y] = routes;
// add left and upper neighbors for further processing
if (x > 0)
next.push_back(std::make_pair(x - 1, y));
if (y > 0)
next.push_back(std::make_pair(x, y - 1));
}
// we are done !
std::cout << grid[0][0] << std::endl;
}
return 0;
}

This solution contains 11 empty lines, 14 comments and 7 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 "1 2" | ./15`

Output:

*(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 23, 2017 submitted solution

March 31, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler015

My code solved **11** out of **11** 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 never an option.

# Links

projecteuler.net/thread=15 - **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-15/ (written by Kristian Edlund)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p015.hs (written by Nayuki)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p015.java (written by Nayuki)

Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p015.mathematica (written by Nayuki)

C: github.com/eagletmt/project-euler-c/blob/master/10-19/problem15.c (written by eagletmt)

Go: github.com/frrad/project-euler/blob/master/golang/Problem015.go (written by Frederick Robinson)

Javascript: github.com/dsernst/ProjectEuler/blob/master/15 Lattice paths.js (written by David Ernst)

Scala: github.com/samskivert/euler-scala/blob/master/Euler015.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 already 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.

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

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