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

# Problem 166: Criss Cross

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 = 90..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.
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.
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.

# 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):
Note: Enter the highest digit allowed (9 for the original problem)

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)

# 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++;
}
}
}
}

if (even)
result *= 2;

std::cout << result << std::endl;
return 0;
}


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

# Benchmark

The correct solution to the original Project Euler problem was found in 0.17 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

June 14, 2017 submitted solution

# Hackerrank

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.

# 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 165 - Intersections Number Rotations - problem 168 >>
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