Author: Dmitry Brant

As long as I’m on a roll with my posts on number theory in C#, I thought I’d briefly discuss how to generate binomial coefficients, and then move on to Stirling numbers of the second kind, both of which are extremely useful in combinatorics and finite calculus.

Some prerequisites for calculating binomial coefficients include a standard factorial function, nothing special:

long factorial(long n)
{
    if (n == 0) return 1;
    long t = n;
    while(n-- > 2) t *= n;
    return t;
}

Note that I use longs whenever possible because, despite the performance hit from 64-bit operations, it’s worth it to be able to work with numbers that are double the magnitude of ints.

We will also need a function for calculating a falling power of a number, which is defined as: $$x^{\underline{n}} = x(x-1)(x-2)\ldots(x-(n-1))$$You’ll see why in a moment.

long fallingPower(long n, long p)
{
    long t = 1;
    for (long i = 0; i < p; i++) t *= n--;
    return t;
}

Notice that the above function handles the case $$n^{\underline{0}}$$ returning 1. However, it does not handle the case of p > n, which would give an incorrect result, but this is not necessary for our purposes.

Binomial Coefficients

Recall that the definition for the binomial coefficient is $${n \choose k} = \frac{n!}{k!(n-k)!}$$ However, using this exact formula to compute binomial coefficients is a bit naive. If we use falling powers (sometimes called falling factorials), the above formula easily reduces to: $$\frac{n^{\underline{k}}}{k!}$$

We can improve the algorithm a bit more by adding the condition:

$${n \choose k} =
\begin{cases}
\frac{n^{\underline{k}}}{k!} \quad \mbox{if } k \leq \lfloor n/2 \rfloor,\\
\frac{n^{\underline{n-k}}}{(n-k)!} \quad \mbox{if } k > \lfloor n/2 \rfloor.
\end{cases}
$$

Not only is this algorithm faster, but it can also handle larger coefficients than the original formula, since neither the falling power nor the factorial ever gets larger than n/2.
The code for this is straightforward:

long binomialCoeff(long n, long k)
{
    if ((k < 0) || (k > n)) return 0;
    k = (k > n / 2) ? n - k : k;
    return fallingPower(n, k) / factorial(k);
}

However, this is still not as optimal as it can be. The most optimal approach would be to accumulate the falling power while dividing by each factor of the factorial in place. This would minimize the chance of overflow errors, and allow for even larger coefficients to be calculated. The disadvantage of this algorithm is the necessary use of floating-point math:

long binomialCoeff(long n, long k) {
    if ((k < 0) || (k > n)) return 0;
    k = (k > n / 2) ? n - k : k;
    double a = 1;
    for (long i = 1; i <= k; i++) a = (a * (n-k+i)) / i;
    return a + 0.5;
}

Stirling Numbers

Stirling numbers (of the second kind) are useful for, among other things, enumerating the coefficients of the falling-power expansion of a regular power. For example, how would we express x³ in terms of falling powers of x? That is, how do we arrive at an equation of the form $$x^3 = ax^{\underline 3} + bx^{\underline 2} + cx^{\underline 1}$$ Well, we could just solve the equation directly, but this would get unwieldy for higher powers. A neat way of doing this involves the use of Stirling numbers of the second kind, $$\left\{\begin{matrix} n \\ k \end{matrix}\right\}$$ A useful theorem for computing these numbers is

$$ \left\{\begin{matrix} n \\ k \end{matrix}\right\}=\frac{1}{k!} \sum_{i=0}^{k}(-1)^i{k \choose i}(k-i)^n $$$

long stirling(long n, long k)
{
    long sum = 0, neg = 1;
    for (long i = 0; i <= k; i++)
    {
        sum += neg * binomialCoeff(k, i) * pow(k - i, n);
        neg = -neg;
    }
    sum /= factorial(k);
    return sum;
}

With the Stirling numbers in hand, we can now obtain the coefficients for falling power expansions:

$$x^m = \sum_{k=0}^{m}\left\{\begin{matrix} m \\ k \end{matrix}\right\}x^{\underline k}$$

I never thought that solving random math problems can be addictive, but Project Euler does exactly that. Not only does it make you flex your math muscles, it also challenges you to take your programming language of choice to its limits. So it’s a total win-win: you brush up on your math skills, and broaden your programming repertoire at the same time.

Once I discovered Project Euler, I couldn’t pull myself away from the computer until I solved as many problems as I could. As of this writing I’ve solved 187 of their 211 problems.

Problem 197 has to do with finding the n^th term of a particular recursively defined sequence: $$u_{n+1} = f(u_n)$$ with $$u_0 = -1, f(x) = \lfloor 2^{30.403243784-x^2}\rfloor \cdot 10^{-9}$$

Of course, as with most Project Euler problems, the value for n is set ridiculously high, presumably to eliminate the possibility of brute-forcing the problem (within the lifetime of the universe).

Fortunately, it takes little more than a superficial examination to see that this problem is actually quite simple in disguise. If we look at the first few terms of the sequence, we can already guess that this sequence appears to be “converging” to a function that oscillates between two values, approximately 0.681 and 1.029 (I won’t give precise numbers, since that would give away the solution).

This means that all we need to do is go far enough into the sequence that the deviation of the oscillations is less than the desired precision asked by the problem (10^-9). And it so happens that we don’t need to go out far at all. The sequence actually settles on its two oscillatory values as early as the 1000^th term (probably even earlier)! Therefore, the sum of the 1000^th and 1001^st term will be equivalent to the sum of the 10^12th and (10¹²+1)^st term, which is what the problem asks for!

The code to do this is elementary. I accomplished it with a mere 5 lines of C# code. Can you do better?

I thought I’d post a few code snippets in C# that have to do with basic number theory, since I use them in my programs from time to time. These snippets are by no means optimized, and the use of C# pretty much precludes their use in high-performance applications. Still, for relatively small arguments, these routines run surprisingly fast.

Prime numbers

To generate a list of prime numbers, we use the familiar Sieve of Eratosthenes. We can write one routine to generate the sieve:

bool[] GetPrimeSieve(long upTo)
{
    long sieveSize = upTo + 1;
    bool[] sieve = new bool[sieveSize];
    Array.Clear(sieve, 0, (int)sieveSize);
    sieve[0] = true;
    sieve[1] = true;
    long p, max = (long)Math.Sqrt(sieveSize) + 1;
    for (long i = 2; i <= max; i++)
    {
        if (sieve[i]) continue;
        p = i + i;
        while (p < sieveSize) { sieve[p] = true; p += i; }
    }
    return sieve;
}

The above function returns an array of booleans (the size of the given parameter), each of which is false if the array index is a prime number, or true if it's not a prime number. To get an actual list of prime numbers, we can use a function such as this:

long[] GetPrimesUpTo(long upTo)
{
    if (upTo < 2) return null;
    bool[] sieve = GetPrimeSieve(upTo);
    long[] primes = new long[upTo + 1];

    long index = 0;
    for (long i = 2; i <= upTo; i++) if (!sieve[i]) primes[index++] = i;

    Array.Resize(ref primes, (int)index);
    return primes;
}

The above function returns an actual list of primes less than or equal to the specified high limit. This means that we can easily compute the prime-counting function $$\pi(x)$$ for any integer x by counting the number of elements in the array returned by the function. We can write a similar function to generate a list of composites:

long[] GetCompositesUpTo(long upTo)
{
    if (upTo < 2) return null;
    bool[] sieve = GetPrimeSieve(upTo);
    long[] composites = new long[upTo + 1];

    long index = 0;
    for (long i = 2; i <= upTo; i++)
        if (sieve[i]) composites[index++] = i;

    Array.Resize(ref composites, (int)index);
    return composites;
}

As for determining if a single certain number is prime (without having to generate a giant sieve), we can simply use a function that attempts to factor the number. If the number happens to be divisible by an integer greater than 1 and less than or equal to its square root, then the number is not prime:

bool IsPrime(long n)
{
    long max = (long)Math.Sqrt(n);
    for (long i = 2; i <= max; i++)
        if (n % i == 0)
            return false;
    return true;
}

Greatest Common Divisor and Totient

To obtain the greatest common divisor (GCD) of two numbers, we use the usual Euclidean algorithm:

long gcd(long a, long b)
{
    long temp;
    while (b != 0)
    {
        temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

There is a slightly different binary algorithm for computing the GCD which is theoretically more efficient, but in practice (at least in C#) it's actually slightly less efficient than the algorithm above:

ulong gcdb(ulong a, ulong b)
{
    int shift;
    if (a == 0 || b == 0) return a | b;

    for (shift = 0; ((a | b) & 1) == 0; ++shift) { a >>= 1; b >>= 1; }
    while ((a & 1) == 0) a >>= 1;

    do {
        while ((b & 1) == 0) b >>= 1;

        if (a < b) {
            b -= a;
        } else {
            ulong temp = a - b;
            a = b;
            b = temp;
        }
        b >>= 1;
    } while (b != 0);
    return a << shift;
}

With the GCD readily available, determining whether two numbers are coprime is just a matter of telling whether or not their GCD is equal to 1. Also, calculating the LCM (least common multiple) of two numbers becomes trivial:

long lcm(long a, long b)
{
    long ret = 0, temp = gcd(a, b);
    if (temp != 0)
    {
        if (b > a) ret = (b / temp) * a;
        else ret = (a / temp) * b;
    }
    return ret;
}

Also using the GCD algorithm, it becomes easy to calculate Euler's totient function for a certain number, since the totient function is simply the number of integers less than or equal to n that are coprime to n:

long eulerTotient(long n)
{
    long sum = 0;
    for (long i = 1; i <= n; i++)
        if (gcd(i, n) == 1) sum++;
    return sum;
}

The above is a really naive algorithm. A much more efficient algorithm (one that is usually given in textbooks) is as follows:

int eulerTotient(int n)
{
    primes = GetPrimesUpTo(n+1);    //this can be precalculated beforehand
    int numPrimes = primes.Length;

    int totient = n;
    int currentNum = n, temp, p, prevP = 0;
    for (int i = 0; i < numPrimes; i++)
    {
        p = (int)primes[i];
        if (p > currentNum) break;
        temp = currentNum / p;
        if (temp * p == currentNum)
        {
            currentNum = temp;
            i--;
            if (prevP != p) { prevP = p; totient -= (totient / p); }
        }
    }
    return totient;
}