<< problem 27 - Quadratic primes Distinct powers - problem 29 >>

# Problem 28: Number spiral diagonals

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is 101.

What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

# My Algorithm

My code contains two approaches: a basic brute-force version which easily solves the original problem
and a vastly different solution if the spiral is way bigger than 1001 (actually it can easily find the result for 10^18 spiral).

The spiral of the problem statement has size 5x5. We can split it into three circles/rings:
1. the inner ring (width=1) contains just 1
2. the middle ring (width=3) contains all numbers from 2 to 9
3. the outer ring (width=5) contains all numbers from 1 to 25

The "corners" of each ring are:
1. 1, sum is 1
2. 3, 5, 7 and 9, their sum is 24
3. 13, 17, 21 and 25, their sum is 76

The "largest" corner of each ring can be found in its upper-right corner (let's call it ur_{width}) and has an interesting pattern:
ur_1=1=1^2, ur_3=9=3^2, ur_5=25=5^2, and so on:
ur_i = i^2

All other corners can be derived from the upper-right corner (upper-left is ul_i, lower-left is ll_i, lower-right is lr_i):
ul_i = ur_i - (i - 1) = i^2 - (i - 1)
ll_i = ul_i - (i - 1) = i^2 - 2 * (i - 1)
lr_i = ll_i - (i - 1) = i^2 - 3 * (i - 1)

The sum of all corners of a ring is:
corners_i = ur_i + ul_i + ll_i + lr_i
= i^2 + i^2 - (i - 1) + i^2 - 2 * (i - 1) + i^2 - 3 * (i - 1)
= 4 * i^2 - 6i + 6

A for-loop running from 1, 3, 5, ... to 1001 (the maximum size of the spiral) computes the sum of all rings.
It shows the correct result pretty much instantly.

... but ...

Larger spirals (Hackerrank test cases ask for a width of up to 10^{18}-1) increase the computation time considerably.
Fortunately, there is a closed form to find the solution without any loops:

We saw that ur_i=i^2 where i is the width of the spiral (1, 3, 5, ...).
If we replace the width by the number x of the ring minus 1 (x=0 stands for the 1st ring, x=1 means 2nd ring, x=2 → 3rd ring, ...), then
i = 2x + 1 and x=floor{i/2}

Hence the upper-right corner of each ring is:
ur_x = 4x^2 + 4x + 1 (produces the series ur_0=1, ur_1=9, ur_2=25, ...)

The three other corners of each ring:
ul_x = ur_x - (i - 1) = ur_x - (2x + 1 - 1) = ur_x - 2x = 4x^2 + 4x + 1 - 2x
ll_x = ul_x - (i - 1) = 4x^2 + 4x + 1 - 4x = 4x^2 + 1
lr_x = ll_x - (i - 1) = 4x^2 + 4x + 1 - 6x = 4x^2 - 2x + 1

And their sum becomes:
corners_x = ur_x + ul_x + ll_x + lr_x = 16x^2 + 4x + 4

A spiral consists of \lfloor frac{width}{2} \rfloor rings, therefore the sums of the diagonals is the sum of all corners:
result = sum_{x=0..width/2}{16x^2 + 4x + 4}
=16 * sum_{x=0..width/2}{x^2} + 4 * sum_{x=0..width/2}{x} + sum_{x=0..width/2}{4}
=16 * sum{x^2} + 4 * sum{x} + sum{4}
(I removed the ranges from of the sums signs to make the formula shorter)

The series x^2 is called Square pyramidal numbers (see en.wikipedia.org/wiki/Square_pyramidal_number) and their sum is:

sum{x^2} = frac{x * (x + 1) * (2x + 1)}{6}

The series x is called Triangle numbers (see en.wikipedia.org/wiki/Triangular_number) and their sum is:

sum{x} = frac{x * (x + 1)}{2}

We can replace:
16 * sum{x^2} + 4 * sum{x} + sum{4}
=16x(x + 1)(2x + 1) / 6 + 4(x(x + 1) / 2) + 4x
=8x(x + 1)(2x + 1) / 3 + 2x(x + 1) + 4x

The only problem is that the sum formulas for Triangle numbers and Square pyramidal numbers are defined beginning at 1
whereas my original formula began at 0: our inner-most ring is missing. It is always 1 → let's add 1:
8x(x + 1)(2x + 1) / 3 + 2x(x + 1) + 4x + 1

## Modifications by HackerRank

Even though Hackerrank defines the maximum input to be 10^{18}, the output should be printed modulo 1000000007.
Languages that support big integer computations can easily evaluate this formula and are done.
But life is tough for C++ programmers ...

My C++ code needs a few more mathematical tricks to solve this problem. The rules for modular arithmetic are (see en.wikipedia.org/wiki/Modular_arithmetic):
(a + b) mod c = ((a mod c) + (b mod c)) mod c
(a * b) mod c = ((a mod c) * (b mod c)) mod c

In general, frac{a}{a} = a * a^{-1} = 1 which means that a^{-1} is the inverse of a

a^{-1} in modular arithmetic is defined as (see en.wikipedia.org/wiki/Modular_multiplicative_inverse):
if a mod b == 0 then frac{a}{b} mod c = ((a mod c) * inverse(b, c)) mod c

1000000007 is a prime number, therefore the inverse can be computed based on Fermat's little theorem (en.wikipedia.org/wiki/Fermat's_little_theorem):
a^{p-1} mod p = 1 (if a < p, in our case a=3 and p=10^9+7)
(a * a^{p-2}) mod p = 1a^{p-2} is the multicative inverse of a in p

I could precompute inverseModulo(3, 1000000007) (e.g. with this online tool: planetcalc.com/3311/)
but I decided to include the full source code to figure out the modular multiplicative inverse at runtime (inverseModulo and powerModulo).

My program splits the original formula into smaller parts:
8x(x + 1)(2x + 1) / 3 + 2x(x + 1) + 4x + 1
sharedTerm = (2*x * (x + 1)) % Modulo
sum1 = ((4 * sharedTerm * (2*x + 1) ) % Modulo) * inverseModulo(3, Modulo)
sum2 = sharedTerm + 4*x + 1
sum = (sum1 % Modulo + sum2 % Modulo) % Modulo

Note: the first half of sum1 must be a multiple of 3 (remember the condition a mod b == 0 when deriving the multiplicative inverse).
We know that the result as well as sum2 are integers. Therefore sum1 be an integer, too, and 8x(x + 1)(2x + 1) is always a multiple of 3.

I know, it's very awkward but gets the job done ... no other of my solutions for a Project Euler problem has so many words for so little code !

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 5" | ./28

Output:

Note: the original problem's input 1001 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++ and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       #include <iostream>

// return (base^exponent) % modulo
unsigned int powmod(unsigned long long base, unsigned int exponent, unsigned int modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// https://en.wikipedia.org/wiki/Exponentiation_by_squaring
// odd exponent ?
if (exponent % 2 == 1)
{
result = (result * base) % modulo;
exponent--;
}
else
{
base = (base * base) % modulo;
exponent /= 2;
}
}
return result;
}

// return modulo multiplicative inverse of a such that (a*inverse(a)) % p = 1
unsigned int inverseModulo(unsigned int a, unsigned int modulo)
{
// if p is prime, then https://en.wikipedia.org/wiki/Fermat%27s_little_theorem applies:
//      a^(p-1)  % p = 1
// (a * a^(p-2)) % p = 1
// that means a^(p-2) is the result
return powmod(a, modulo - 2, modulo);
// inspired by https://barkingbogart.wordpress.com/2013/02/21/cnk-mod-1000000007/
}

int main()
{
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned long long n;
std::cin >> n;

// sum along the diagonals, initially for width = 1
unsigned long long sum = 1;

// brute-force approach (good enough for original problem):
//for (int64_t width = 3; width <= n; width += 2)
//{
// all four corners
//sum     += width * width;                   // upper-right
//sum     += width * width -     (width - 1); // upper-left
//sum     += width * width - 2 * (width - 1); // lower-left
//sum     += width * width - 3 * (width - 1); // lower-right

// same as:
//  sum += 4*width*width - 6*width + 6;
//}

// direct computation:
// half side length
unsigned long long x = n / 2;
// the closed form is:
//sum  = 8 * x * (x + 1) * (2*x + 1) / 3   +   2 * x * (x + 1) + 4 * x + 1;

// apply Modulo whenever an overflow is possible
const unsigned int Modulo = 1000000007;

x %= Modulo;

// first part: 8 * x * (x + 1) * (2*x + 1) / 3
//           = 4 *  sharedTerm * (2*x + 1) / 3
unsigned long long sharedTerm = (2*x * (x + 1)) % Modulo;

// the division by 3 becomes a multiplication by its modulo multiplicative inverse
unsigned long long sum1 = ((4 * sharedTerm * (2*x + 1)) % Modulo) * inverseModulo(3, Modulo);

// second part: 2 * x * (x + 1)      + 4 * x + 1
//            =     secondTerm       + 4 * x + 1
unsigned long long sum2 = sharedTerm + 4*x + 1;

// sum = first part + second part
sum = (sum1 % Modulo + sum2 % Modulo) % Modulo;

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


This solution contains 14 empty lines, 32 comments and 1 preprocessor command.

# 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

February 23, 2017 submitted solution

# Hackerrank

My code solves 5 out of 5 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 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 27 - Quadratic primes Distinct powers - problem 29 >>
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