Skip to main content

Poisson Games and Sudden-Death Overtime

A Strange Recursive Relation, Automatic

Hofstadter mentions the following recursive relation in his great book "Gödel, Escher, Bach": \[
\begin{align}
g(0) &= 0;\\
g(n) &= n-g(g(n-1)).
\end{align}
\] I claim that \( g(n) = \left\lfloor \phi\cdot (n+1) \right\rfloor \), where \( \phi = \frac{-1+\sqrt{5}}{2}\), and I'll show this using a technique that makes proving many identities of this type nearly automatic.

Let \( \phi\cdot n = \left\lfloor \phi\cdot n \right\rfloor + e\), where \( 0 < e < 1\) as \( \phi \) is irrational, nor can \(e = 1-\phi\), and note that \(\phi\) satisfies \( {\phi}^2 + \phi - 1 = 0\). Some algebra gives \[
\begin{align}
n-\left\lfloor \left( \left\lfloor \phi\cdot n \right\rfloor + 1 \right) \cdot \phi \right\rfloor
&= n-\left\lfloor \left( n\cdot \phi - e + 1 \right) \cdot \phi \right\rfloor \\
&= n-\left\lfloor n\cdot {\phi}^2 - e\cdot \phi + \phi \right\rfloor \\
&= n-\left\lfloor n\cdot \left(1-\phi\right) - e\cdot \phi + \phi \right\rfloor \\
&= n-n-\left\lfloor -n\cdot \phi - e\cdot \phi + \phi \right\rfloor \\
&= -\left\lfloor -n\cdot \phi - e\cdot \phi + \phi \right\rfloor \\
&= -\left\lfloor -n \cdot \phi + e - e - e\cdot \phi + \phi \right\rfloor \\
&= \left\lfloor \phi\cdot n \right\rfloor -\left\lfloor - e - e\cdot \phi + \phi \right\rfloor.
\end{align}
\]
Now if \[
\begin{align}
0 < e < 1-\phi &\implies 0 < - e - e\cdot \phi + \phi < \phi;\\
1-\phi < e < 1 &\implies -1 < - e - e\cdot \phi + \phi < 0.
\end{align}
\]
This implies \[ n-\left\lfloor \left( \left\lfloor \phi\cdot n \right\rfloor + 1 \right) \cdot \phi \right\rfloor = \left\lfloor \phi\cdot (n+1) \right\rfloor .\] Since \( \left\lfloor \phi\cdot (0+1) \right\rfloor = 0\), we're done.

The point of the algebra was to move all terms involving \(n\) out, and then checking to see how the remaining term varied with \( e\). A simple idea, but very useful.

Comments

Post a Comment

Popular posts from this blog

A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written \( \Pr(A \mid B) \)
probability of B is written \( \Pr(B) \)

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: \[ \Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \]Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be \[ w=\frac{P^e}{P^e+1},\] where \(w\) is the win fraction expectation, \(P\) is runs/allowed (or similar) and \(e\) is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with \(g\) games is \[W = g\cdot w = g\cdot \frac{P^e}{P^e+1},\] so for one estimate we only need to compute the value of the partial derivative \(\frac{\partial W}{\partial P}\) at \(P=1\). Note that \[ W = g\left( 1-\frac{1}{P^e+1}\right), \] and so \[ \frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}\] and it follows \[ \frac{\partial W}{\partial P}(P=1) = …

Solving a Math Puzzle using Physics

The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds.


The first two parts require students to interpret the meaning of the components of the formula \(T(x) = 5 \sqrt{36+x^2} + 4(20-x) \), and the final "challenge" component involves finding the minimum of \( T(x) \) over \( 0 \leq x \leq 20 \). Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities.


One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},\] where \( \theta_1, \theta_2\) are the angles to the normal and \(v_1, v_2\) are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …