### Solving the TopSpin Puzzle using GAP

TopSpin is an oval-track permutation puzzle that was made by Binary Arts; similar puzzles are made and sold by other manufacturers. Here's the Binary Arts TopSpin.

It's not a difficult puzzle to solve if you play around with it for a few hours and figure out how to generate various permutations. It's more interesting (and difficult) if you observe that the turntable has a distinguishable top and bottom. This suggests an interesting question - can you invert the turntable while keeping the numbers in the track in the same order?

The answer is, perhaps surprisingly, yes. Here's one way to find a sequence of operations that produces precisely this outcome.

GAP (Groups, Algorithms and Programming) is a freely available programming language that specializes in computational group theory, and it's perfect for solving permutation puzzles. Here's my GAP code for TopSpin. Label the top of the turntable with 21 and the bottom with 22. Flipping the turntable generates the permutation $(1,4)(2,3)(21,22);$ rotating the oval to the left generates the permutation $(2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,1).$ Denoting these two permutations as $$x$$ and $$y$$, we can now simply ask GAP to find a sequence of operations on the free group generated by $$x,y$$ that results in the pre-images $$(1,2)$$, flipping two adjacent numbers on the track, leaving everything else the same (including turntable parity); and $$(21,22)$$, flipping turntable parity, leaving everything else the same. This corresponds to the operation of flipping the turntable while keeping the order of the numbers in the track the same.

Running the code, we get these lovely results:
(1,2) = y*x^-1*y^-1*x^-1*y*x^-1*y*x^-1*y^-1*x^-1*y^-1*x^-1*y^2*x^-1*y^-1*x^-1*y^-1*x^-1*y^4*x^-1*y^-1*x^-1*y^2*x^-1*y^-2*x^-1*y*x^-1*y^2*x^-1*y^-3*x^-1*y^-3*x*y^-4*x*y*x*y^-1*x*y^4*x^-1*y^-5*x^-1*y^-1*x^-1*y^5*x^-1*y^-6*x^-1*y^2*x^-1*y^-1*x^-1*y^5*x*y*x*y*x*y*x*y*x*y^5*x*y*x*y^-1*x*y^-5*x^-1*y^-1*x^-1*y^-1*x^-1*y^-2*x*y*x*y^-1*x*y^4*x*y^-1*x*y^3*x^-1*y^-1*x^-1*y^6*x^-1*y^-1*x^-2*y^-4*x^-3*y^-3*x^-2*y^-1*x^-1*y*x^-1*y^2*x^-1*y^-2*x^-1*y^-1*x^-1*y*x*y*x^-1*y*x^-1*y^-1*x^-1*y^-2
(21,22) = y*x^-1*y^2*x*y^-1*x*y*x*y^-1*x*y^-2*x*y^2*x*y^-1*x*y*x*y*x*y^-2*x^-1*y*x^-1*y^-2*x^-1
Since there are sequences of operations that allow us to flip any two adjacent numbers or the parity of the turntable, it follows that all possible configurations are both solvable and achievable.

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### 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 …