<< problem 165 - Intersections | Number Rotations - problem 168 >> |

# Problem 166: Criss Cross

(see projecteuler.net/problem=166)

A 4x4 grid is filled with digits d, 0 <= d <= 9.

It can be seen that in the grid

6330

5043

0714

1245

the sum of each row and each column has the value 12. Moreover the sum of each diagonal is also 12.

In how many ways can you fill a 4x4 grid with the digits d, 0 <= d <= 9 so that each row, each column, and both diagonals have the same sum?

# My Algorithm

I brute-force the problem with a few performance tweaks.

Let's assign a variable to each cell of the grid:

abcd

efgh

ijkl

mnop

The first step is to find the current sum s (called `sum`

in my code):

s = a + b + c + d

The "fourth" variable of each row/column/diagonal depends on the three previous and s:

h = s - e - f - g (second row)

l = s - i - j - k (third row)

p = s - m - n - o (fourth row)

m = s - a - e - i (first column)

n = s - b - f - j (second column)

o = s - c - g - k (third column)

Now only 10 instead of 16 variables are unknown.

Actually, j isn't unknown - it can be computed, too:

j = s - d - g - m (anti-diagonal)

My program iterates over all possible values of the remaining 9 variables a, b, c, d, e, f, g, i and k.

Any combination where all variables are between `0`

and `maxDigit`

(which is 9 for the original problem) is valid - with one exception:

I have to check the diagonal, too (s = a + f + k + p), but I saw in my experiments that the fourth column doesn't need to be checked as well.

These are my performance tweaks:

1. If there is an even number of digits (`maxDigit = 9`

→ `0..9`

→ 10 digits), then a can be restricted to `0..(maxDigit - 1)/2`

)

For every solution found there is another "inverted" solution where each variable x' = maxDigit - x.

→ about twice as fast

2. The grid can be mirrored along the diagonal such that b <= e. For every b < e there is a second solution found by mirroring along the diagonal.

→ about twice as fast

3. All variables are unsigned: if a variable become negative it will be a huge positive number in C's two-complement representation.

The check `if (x < 0 || x > maxDigit)`

simplifies to `if (x > maxDigit)`

which is about 10% faster.

These three optimizations make my program about four times faster.

## Alternative Approaches

Nayuki's solution is based on f = b + c + 2d - e - i - k. Look at his code for an explanation - it makes the program about five times faster.

I didn't spend much time on a thorough mathematical analysis of the problem and didn't see that relationship.

## Modifications by HackerRank

The maximum digit can be changed via STDIN. The original problem has `maxDigit = 9`

.

# My code

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

#include <iostream>
int main()
{
// maximum digit
unsigned int maxDigit = 9;
std::cin >> maxDigit;
// restrict variable "a"
bool even = (maxDigit + 1) % 2 == 0;
auto maxA = even ? (maxDigit - 1) / 2 : maxDigit;
unsigned int result = 0;
for (unsigned int a = 0; a <= maxA; a++) // limit a to 0..4 => for every solution found there is a "inverted" version where x = 9 - x and x = a,b,c,d,...,p
for (unsigned int b = 0; b <= maxDigit; b++)
for (unsigned int c = 0; c <= maxDigit; c++)
for (unsigned int d = 0; d <= maxDigit; d++)
{
// all rows, columns, diagonals must share the same sum
auto sum = a + b + c + d;
for (unsigned int e = b; e <= maxDigit; e++) // !!! start at b instead of zero
for (unsigned int f = 0; f <= maxDigit; f++)
for (unsigned int g = 0; g <= maxDigit; g++)
{
// sum of second row must be identical to first row
auto h = sum - e - f - g;
if (h > maxDigit)
continue;
for (unsigned int i = 0; i <= maxDigit; i++)
{
// sum of first column must be identical to first row
auto m = sum - a - e - i;
if (m > maxDigit)
continue;
// sum of anti-diagonal must be identical to first row
auto j = sum - d - g - m;
if (j > maxDigit)
continue;
// sum of second column must be identical to first row
auto n = sum - b - f - j;
if (n > maxDigit)
continue;
for (unsigned int k = 0; k <= maxDigit; k++)
{
// sum of third column must be identical to first row
auto o = sum - c - g - k;
if (o > maxDigit)
continue;
// sum of third row must be identical to first row
auto l = sum - i - j - k;
if (l > maxDigit)
continue;
// sum of fourth row must be identical to first row
auto p = sum - m - n - o;
if (p > maxDigit)
continue;
// check diagonal, too
if (sum != a + f + k + p)
continue;
// yes, found a solution
result++;
// mirror grid along diagonal, too
if (b < e)
result++;
}
}
}
}
// add "inverted" solutions
if (even)
result *= 2;
std::cout << result << std::endl;
return 0;
}

This solution contains 14 empty lines, 14 comments and 1 preprocessor command.

# 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 | ./166`

Output:

*Note:* the original problem's input `9`

__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 0.17 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

June 14, 2017 submitted solution

June 14, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler166

My code solves **8** out of **8** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **50%** (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.

# Links

projecteuler.net/thread=166 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p166.java (written by Nayuki)

# 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 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.

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 165 - Intersections | Number Rotations - problem 168 >> |