In The Big Bang Theory, Sheldon Cooper explains an extension of the game Rock, Paper, Scissors by introducing two more possibilities, Lizard and Spock, so the game becomes Rock, Paper, Scissors, Lizard, Spock. Sam Kass and Karen Bryla invented the game before it became widely known via the television show.

The diagram below summarizes the rules:

Alternative rules

Imagine yourself in the position of Sam Kass and Karen Bryla designing the game. You first try adding one extra move, but it turns out that’s not possible.

The main property of Rock, Paper, Scissors is that no player has a winning strategy. That implies you can only add an even number of extra moves, keeping the total number of moves odd. That way, for every move one player makes, the other player has an equal number of winning and losing moves. Otherwise some moves would be better and others worse. So you can’t add just one move, say Lizard. You have to add two (or four, or six, …).

How many ways could you assign rules to an extension of Rock, Paper, Scissors adding two more moves?

Number the possible moves 1 through 5. We will make a table of which moves beat which, with +1 in the (i, k) position if move i beats move j and -1 if j beats i. There will be zeros on the diagonal since it’s a tie if both players make the same move.

Let’s start with the original Rock, Paper, Scissors. In order for this game to not have a winning strategy, the table must be filled in as below, with the only option being to set a = 1 or a = -1.

|---+----+----+----|
|   |  1 |  2 |  3 |
|---+----+----+----|
| 1 |  0 |  a | -a |
|---+----+----+----|
| 2 | -a |  0 |  a |
|---+----+----+----|
| 3 |  a | -a |  0 |
|---+----+----+----|

If 1, 2, and 3 correspond to Rock, Paper, and Scissors, then a = 1 according to the usual rules, but we’ll allow the possibility that the usual rules are reversed. (If you don’t like that, just set a = 1).

Next we fill in the rest of the moves. The table must be skew-symmetric, i.e. the (i, j) element must be the negative of the (j, i) element, because if (i, j) is a winning move then (j, i) is a losing move and vice versa. Also, the rows and columns must sum to zero. Together these requirements greatly reduce the number of possibilities.

|---+----+----+----+----+----|
|   |  1 |  2 |  3 |  4 |  5 |
|---+----+----+----+----+----|
| 1 |  0 |  a | -a |  b | -b |
|---+----+----+----+----+----|
| 2 | -a |  0 |  a |  c | -c |
|---+----+----+----+----+----|
| 3 |  a | -a |  0 |  d | -d |
|---+----+----+----+----+----|
| 4 | -b | -c | -d |  0 |  d |
|---+----+----+----+----+----|
| 5 |  b |  c |  d | -d |  0 |
|---+----+----+----+----+----|

The values of a, b, and c may each be chosen independently to be 1 or -1. If b + c = 0, then d can be chosen freely. Otherwise b and c have the same sign, and d must have the opposite sign. So all together there are 12 possibilities (6 if you insist a = 1). These are listed below.

|---+---+---+---|
| a | b | c | d |
|---+---+---+---|
| + | + | + | - |
| + | + | - | + |
| + | + | - | - |
| + | - | + | + |
| + | - | + | - |
| + | - | - | + |
| - | + | + | - |
| - | + | - | + |
| - | + | - | - |
| - | - | + | + |
| - | - | + | - |
| - | - | - | + |
|---+---+---+---|

The version of the rules used on The Big Bang Theory corresponds to the second row of the table above: a = b = d = 1 and c = -1.

Simpler solution

Here’s another way to count the number of possible designs. Suppose we start with tradition Rock, Paper, Scissors, corresponding to a = 1 in the notation above. Now let’s add the rules for Lizard. We can pick any two things from {Rock, Paper, Scissors, Spock} and say that Lizard beats them, and then the other two things must beat Lizard. There are 6 ways to choose 2 things from a set of 4.

Once we’ve decided the rules for Lizard, we have no choice regarding Spock. Spock’s rules must be the opposite of Lizard’s rules in order to balance everything out. If  we decide Lizard beats Rock, for example, then Rock must beat Spock so two things beat Rock and Rock beats two things.

If we’re willing to consider reversing the usual rules of Rock, Paper, Scissors, i.e. setting a = -1, then there are 6 more possibilities, for a total of 12.

Adding two more moves

By the way, we can see from the approach above how to add more moves. If we wanted to add Stephen Hawking and Velociraptor to our game, then we have 20 choices: we choose 3 things out of 6 for Hawking to beat, and the rest of the rules are determined by these choices. Velociraptor has to be the anti-Hawking. If we decide that Hawking beats Paper, Lizard, and Spock, then we’d get the rules in the diagram below.

Fair subsets

You might want to design the game so that for any subset of three moves you have a game with no winning strategy. Here’s an example why. If the subset (1 , 2, 4) is a fair game, then a = c. But if the subset (2, 3, 4) is a fair game, then a = −c. So one of the two games must have a winning strategy.

Graphing the rules

The first graph above was made with GraphVis using the code below.

digraph rock {
    node [fontname = "helvetica"]

    "Rock"     -> "Scissors" 
    "Rock"     -> "Lizard"
    "Paper"    -> "Rock"
    "Paper"    -> "Spock"
    "Scissors" -> "Paper"
    "Scissors" -> "Lizard"
    "Lizard"   -> "Spock"
    "Lizard"   -> "Paper"
    "Spock"    -> "Rock"
    "Spock"    -> "Scissors"

    layout = circo
}

Save the code to a file, say rock.gv, then run the command

dot -Tpng rock.gv > rock.png

to produce a PNG file.

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Inverting a power series

As a student, one of the things that seemed curious to me about power series was that a function might have a simple power series, but its inverse could be much more complicated. For example, the coefficients in the series for arctangent have a very simple pattern

but the coefficients in the series for tangent are mysterious.

There’s no obvious pattern to the coefficients. It is possible to write the sum in closed form, but this requires introducing the Bernoulli numbers which are only slightly less mysterious than the power series for tangent.

To a calculus student, this is bad news: a simple, familiar function has a complicated power series. But this is good news for combinatorics. Reading the equation from right to left, it says a complicated sequence has a simple generating function!

Computing the coefficients

The example above suggests that inverting a power series, i.e. starting with the power series for a function and computing the power series for its inverse, is going to be complicated. We introduce exponential Bell polynomials to encapsulate some of the complexity, analogous to introducing Bernoulli numbers above.

Assume the function we want to invert, f(x), satisfies f(0) = 0 and its derivative satisfies f‘(0) ≠ 0. Assume f and its inverse g have the series representations

and

Note that this isn’t the power series per se. The coefficients here are k! times the ordinary power series coefficients. (Compare with ordinary and exponential generating functions explained here.)

Then you can compute the series coefficients for g from the coefficients for f as follows. We have g₁ = 1/f₁ (things start out easy!) and for n ≥ 2,

where the B‘s are exponential Bell polynomials,

and

is the kth rising power of n. (More on rising and falling powers here.)

Example

Let’s do an example in Python. Suppose we want a series for the inverse of the gamma function near 2. The equations above assume we’re working in a neighborhood of 0 and that our function is 0 at 0. So we will invert the series for f(x) = Γ(x+2) − 1 and then adjust the result to find the inverse of Γ(x).

import numpy as np
from scipy.special import gamma
from sympy import factorial, bell 

def rising_power(a, k):
    return gamma(a+k)/gamma(a)

# Taylor series coefficients for gamma centered at 2
gamma_taylor_coeff = [
    1.00000000000000,
    0.42278433509846,
    0.41184033042643,
    0.08157691924708,
    0.07424901075351,
    -0.0002669820687,
    0.01115404571813,
    -0.0028526458211,
    0.00210393334069,
    -0.0009195738388,
    0.00049038845082,    
]
N = len(gamma_taylor_coeff)

f_exp_coeff = np.zeros(N)
for i in range(1, N):
    f_exp_coeff[i] = gamma_taylor_coeff[i]*factorial(i)

# Verify the theoretical requirements on f
assert( f_exp_coeff[0] == 0 )
assert( f_exp_coeff[1] != 0 )

f_hat = np.zeros(N-1)
for k in range(1, N-1):
    f_hat[k] = f_exp_coeff[k+1] / ((k+1)*f_exp_coeff[1])
    
g_exp_coeff = np.zeros(N)
g_exp_coeff[1] = 1/f_exp_coeff[1]

for n in range(2, N):
    s = 0
    for k in range(1, n):
        # Note that the last argument to bell must be a list,
        # and not a NumPy array.
        b = bell(n-1, k, list(f_hat[1:(n-k+1)]))
        s += (-1)**k * rising_power(n, k) * b
    g_exp_coeff[n] = s / f_exp_coeff[1]**n

def g(x):
    return sum(
        x**i * g_exp_coeff[i] / factorial(i) for i in range(1, N)
    )

def gamma_inverse(x):
    return 2. + g(x-1.)

# Verify you end up where you started when you do a round trip
for x in [1.9, 2.0, 2.1]:
    print( gamma_inverse(gamma(x)) )

Output:

1.90000003424331
2.00000000000000
2.09999984671755

Bell polynomials come up naturally in a variety of contexts: combinatorics, analysis, statistics, etc. Unfortunately, the variations on Bell polynomials are confusingly named.

Analogy with differential equations

There are Bell polynomials of one variable and Bell polynomials of several variables. The latter are sometimes called “partial” Bell polynomials. In differential equations, “ordinary” means univariate (ODEs) and “partial” means multivariate (PDEs). So if “partial” Bell polynomials are the multivariate form, you might assume that “ordinary” Bell polynomials are the univariate form. Unfortunately that’s not the case.

It seems that the “partial” in the context of Bell polynomials was taken from differential equations, but the term “ordinary” was not. Ordinary and partial are not opposites in the context of Bell polynomials. A Bell polynomial can be ordinary and partial!

Analogy with generating functions

“Ordinary” in this context is the opposite of “exponential,” not the opposite of “partial.” The analogy is with generating functions, not differential equations. The ordinary generating function of a sequence multiplies the nth term by xⁿ and sums. The exponential generating function of a sequence multiplies the nth term by xⁿ/n! and sums. For example, the ordinary generating function for the Fibonacci numbers is

while the exponential generating function is

where

The definitions of exponential and ordinary Bell polynomials are given below, but the difference between the two that I wish to point out is that the former divides the kth polynomial argument by k! while the latter does not. They also differ by a scaling factor. The exponential form of B_n,k has a factor of n! where the ordinary form has a factor of k!.

“Ordinary” as a technical term

Based on the colloquial meaning of “ordinary” you might assume that it is the default. And indeed that is the case with generating functions. Without further qualification, generating function means ordinary generating function. You’ll primarily see explicit references to ordinary generating functions in a context where it’s necessary to distinguish them from some other kind of generating function. Usually the word “ordinary” will be written in parentheses to emphasize that an otherwise implicit assumption is being made explicit. In short, “ordinary” and “customary” correspond in the context of generating functions.

But in the context of Bell polynomials, it seems that the exponential form is more customary. At least in some sources, an unqualified reference to Bell polynomials refers to the exponential variety. That’s the case in SymPy where the function bell() implements exponential Bell polynomials and there is no function to compute ordinary Bell polynomials.

Definitions

Now for the actual definitions. We can make our definitions more concise if we first define multi-index notation. A multi-index

is a set of n non-negative integers. The norm of a multi-index is defined by

and the factorial of a multi-index is defined by

If x = (x₁, x₂, …, x_n) is an n-tuple of real numbers then x raised to the power of a multi-index α is defined by

The ordinary Bell polynomial B_n,k is

where x is a (n – k +1)-tuple and α ranges over all multi-indices subject to two constraints: the norm of α is k and

The exponential Bell polynomials can then be defined in terms of the ordinary bell polynomials by

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