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### Touring Waldo; Overfitting Waldo; Scanning Waldo; Waldo, Waldo, Waldo

Randal Olson has written a nice article on finding Waldo - Here’s Waldo: Computing the optimal search strategy for finding Waldo. Randal presents a variety of machine learning methods to find very good search paths among the 68 known locations of Waldo. Of course, there's no need for an approximation; modern algorithms can optimize tiny problems like these exactly.

One approach would be to treat this as a traveling salesman problem with Euclidean distances as edge weights, but you'll need to add a dummy node that has edge weight 0 to every node. Once you have the optimal tour, delete the dummy node and you have your optimal Hamiltonian path.

I haven't coded in the dummy node yet, but here's the Waldo problem as a traveling salesman problem using TSPLIB format.

The Condorde software package optimizes this in a fraction of a second:

I'll be updating this article to graphically show you the results for the optimal Hamiltonian path. There are also many additional questions I'll address. Do we really want to use this as our search path? We're obviously overfitting. Do we want to assume Waldo will never appear in a place he hasn't appeared before? When searching for Waldo we see an entire little area, not a point, so a realistic approach would be to develop a scanning algorithm that covers the entire image and accounts for our viewing point and posterior Waldo density. We can also jump where we're looking at from point to point quickly while not searching for Waldo, but scans are much slower.

### Comments

1. Great/important stuff. Would love to see an efficient way to pull/clean CA's xls data!

### 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 …