Multinomial Coefficients in Scala using Fold

Multinomial coefficients are often denoted as $$\left( n;~ k_1, \dots, k_m \right)$$ where $$n = \sum_{i=1}^m k_i$$. Another way to express a multinomial coefficient is using the $${\rm choose}$$ notation commonly employed in combinatorics for binomial coefficients. Given $$n \geq \sum_{i=1}^{m-1} k_i$$, we say that the following are equivalent:

$$\left( n;~ k_1, \dots, k_m \right) ~\mbox{where}~ k_m = n - \sum_{i=1}^{m-1} k_i \equiv {n \choose k_1, \dots, k_{m-1}}$$

For convenience, I'm going to use the first notation.

Theorem

I'm not sure if this has ever been stated before, but here's an observation that I made a couple months ago:

$$\left( n;~ k_1, \dots, k_m \right) = \left( n-k_m;~ k_1, \dots, k_{m-1} \right) \cdot \frac{1}{k_m !} \prod_{i=1}^{k_m} n-i+1$$

That is, there exists a linear, tail recursive definition for calculating multinomial coefficients.

Proof of Theorem

Without loss of generality, assume that $$n = \sum_{i=1}^{m} k_i$$. According to the equivalence noted above as well as the Multinomial Theorem, we have:

$$\left( n;~ k_1, \dots, k_m \right) = {n \choose k_1, \dots, k_m} = \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_m ! }$$

(1) If we pull out the term $$k_m$$, we get:

$$\left( n;~ k_1, \dots, k_m \right) = \frac{1}{k_m !} \cdot \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! }$$

(2) Now, let's find some $$\zeta$$ such that:

$$\frac{ (n-k_m)! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } \cdot \zeta = \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! }$$

(3) If we isolate $$\zeta$$ on the left-hand side, we observe the following:

$$\zeta = \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } \cdot \frac{ k_1 ! \cdot \cdots \cdot k_{m-1} ! }{ (n-k_m)! } = \frac{ n! }{ (n-k_m)! }$$

(4) We know that certain terms cancel in this fraction. It is easy to see that

$$\zeta = \frac{ n! }{ (n-k_m)! } = (n - k_m + 1) \cdot (n - k_m + 2) \cdot \cdots \cdot n = \prod_{i=1}^{k_m} n - k_m + i$$

(5) Since the product in $$\zeta$$ is from $$1$$ to $$k_m$$, we can be further simplify the expression to

$$\zeta = \prod_{i=1}^{k_m} n - k_m + i = \prod_{i=1}^{k_m} n-i+1$$

(6) Replacing $$\zeta$$ in (2) gives us the following:

$$\frac{ (n-k_m)! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } \cdot \prod_{i=1}^{k_m} n-i+1 = \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! }$$

(7) Now we use the result of (6) with (1) to get:

$$\left( n;~ k_1, \dots, k_m \right) = \frac{1}{k_m !} \cdot \frac{ (n-k_m)! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } \cdot \prod_{i=1}^{k_m} n-i+1$$

(8) Since the second fraction in (7) can be expressed as a multinomial, we have:

$$\left( n;~ k_1, \dots, k_m \right) = \left( n-k_m;~ k_1, \dots, k_{m-1} \right) \cdot \frac{1}{k_m !} \prod_{i=1}^{k_m} n-i+1$$

Therefore, the theorem is correct.

Some Notes

This result that we observed in (7) directly relates to the ratios between the coefficients on the face of a multidimensional Pascal triangle.

Corollary

It is easy to see that the following follows from the theorem:

$$\left( n;~ k_1, \dots, k_m \right) = \left( n-k_m;~ k_1, \dots, k_{m-1} \right) \cdot \prod_{i=1}^{k_m} \frac{n-i+1}{i}$$

Folding with scala

In Scala, the expression in the corollary can be expressed concisely using the foldLeft operator.This operator applies a binary operator to a start value and all elements of a list, going left to right. If we let our list be the numbers $$1, \dots, k_m$$, then we can foldLeft with the following binary operator:

$$\left(a,b\right) \Rightarrow \frac{a \left(k_m-b+1\right)}{b}$$

For more information on the general fold operator (with which you can pretty much do anything with), see "Introduction to Metamathematics" by Kleene (affiliate link).

The Code

Here is my recursive implementation of the recurrence relation mentioned in the corollary above:

I've omitted a an iterative implementation of the corollary that I've written for the sake of brevity. If anyone expresses some interest, I may include here or in a future blog post.

object Combinatorics {     /** Computes the multinomial coefficient (n; k_1, .., k_m)      *  where n = the sum of k_1, .., k_m.     *      *  This is a variadic convenience function that allows      *  someone to invoke <code>multi</code> without using an      *  array. Note, however, that the variadic parameter is      *  transformed into an array when this version of      *  <code>multi</code> is invoked.     *      *  @author Michael E. Cotterell <mepcotterell@gmail.com>     *  @param  k items to choose     */    def multi (k: Int*): Long = multi(k.toArray)     /** Computes the multinomial coefficient (n; k_1, .., k_m)      *  where n = the sum of k_1, .., k_m.      *      *  This implementation requires exactly n-many      *  multiplications and m-many recursive calls.     *       *  @see    http://michaelcotterell.com/2013/03/multinomial-coefficients/      *  @author Michael E. Cotterell <mepcotterell@gmail.com>     *  @param  k items to choose     */    def multi (k: Array[Int]): Long =     {        if (k.length == 1) 1L        else {            (1 to k.last).foldLeft(multi(k.slice(0, k.length - 1))){                (prev, i) => prev * (k.sum - i + 1) / i            }        } // if    } // multi     /** Computes the multinomial coefficient (n; k_1, .., k_m)      *  where n = the sum of k_1, .., k_m.     *      *  This implementation requires exactly n-many      *  multiplications and m-many recursive calls. Also, it is      *  experimentally slower than the <code>foldLeft</code>      *  implementation provided by the <code>multi</code>      *  function.     *       *  @see    http://michaelcotterell.com/2013/03/multinomial-coefficients/     *  @author Michael E. Cotterell <mepcotterell@gmail.com>     *  @param  k items to choose     */    def _multi (k: Array[Int]): Long =     {        if (k.length == 1) 1L        else {            var product = _multi(k.slice(0, k.length-1))            for(i <- 1 to k.last) product = product * (k.sum-i+1) / i                        product        } // if    } // _multi } // Combinatorics