## Among Us: Morse code puzzle

In the online game Among Us, players who visit the Comms room hear a fuzzy audio recording of a series of high-pitched beeps that sound like Morse code. I first heard the recording here, but this more recent video also plays it at around 5:00, followed by a good explanation of the problem with trying to decipher the code.

The following figure shows a spectrogram of the audio clip, with time on the x-axis, and each vertical slice showing the Fourier transform of a short (roughly 50 ms) sliding window of the signal centered at the corresponding time. We can clearly see the “dots” and “dashes” at around 1 kHz, with the corresponding translation overlaid in yellow.

Now that we have the Morse code extracted from the audio (which, for reference if you want to copy-paste and play with this problem, is “.-..--...-.---...-..-...“), we just need to decode it, right? The problem is that the dots and dashes are all uniformly spaced, without the required longer gaps between letters, let alone the still longer gaps that would be expected between words. Without knowing the intended locations of those gaps, the code is ambiguous: for example, the first dot could indicate the letter E, or the first dot and dash together could indicate an A, etc.

That turns out to be a big problem. The following figure shows the decoding trie for Morse code letters and digits; starting at the root, move to the left child vertex for each dot, or to the right child vertex for each dash. A red vertex indicates either an invalid code or other punctuation.

If we ignore the digits in the lowest level of the trie, we see that not only are Morse code letters ambiguous (i.e., not prefix-free), they are nearly “maximally ambiguous,” in the sense that the trie of letters is nearly complete. That is, for almost any prefix of four dots and dashes we may encounter, the gap indicating the end of the first letter could be after any of those first four symbols.

This would make a nice programming exercise for students, to show that this particular sequence of 24 symbols may be decoded into a sequence of letters in exactly 3,457,592 possible ways. Granted, most of these decodings result in nonsense, like AEABKGEAEAEEE. But a more interesting and challenging problem is to efficiently search for reasonable decodings, that is, messages consisting of actual (English?) words, perhaps additionally constrained by grammatical connections between words.

Of course, it’s also possible– probable?– that this audio clip is simply made up, a random sequence of dots and dashes meant to sound like “real” Morse code. And even if it’s not, we might not be able to tell the difference. Which is the interesting question that motivated this post: if we generate a completely random, and thus intentionally unintelligible, sequence of 24 dots and dashes, what is the probability that it still yields a “reasonable” possible decoding, for sufficiently large values of “reasonable”?

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## Counting Hotel Key Cards

Introduction

Suppose that you are the owner of a new hotel chain, and that you want to implement a mechanical key card locking system on all of the hotel room doors. Each key card will have a unique pattern of holes in it, so that when a card is inserted into the corresponding room’s door lock, a system of LEDs and detectors inside the lock will only recognize that unique pattern of holes as an indication to unlock the door.

(I have vague childhood memories of family vacations and my parents letting me use just such an exotic gadget to unlock our hotel room door.)

When you meet with a lock manufacturer, he shows you some examples of his innovative square key card design, with the “feature” that a key card may be safely inserted into the slot in a door lock in any of its eight possible orientations: any of the four edges of the square key card may be inserted first, with either side of the key card facing up. Each key card has a pattern of up to 36 holes aligned with a 6×6 grid of sensors in the lock that may “scan” the key card in any orientation.

The lock manufacturer agrees to provide locks and corresponding key cards for each room, with the following requirements:

1. A manufacturer-provided key card will only open its assigned manufacturer-provided lock and no other; and
2. A manufacturer-provided key card will open its assigned manufacturer-provided lock when inserted into the slot in any orientation.

How many distinct safely-locked rooms can the manufacturer support?

A simpler lock is a harder problem

The problem as stated above is a relatively straightforward application of Pólya counting, using the cycle index of the dihedral group of symmetries of the key card acting on (2-colorings of) the $n \times n$ grid of possible holes in the card. When $n$ is even, the cycle index (recently worked out in a similar problem here) is

$Z(D_4) = \frac{1}{8}(x_1^{n^2}+2x_4^{\frac{n^2}{4}}+3x_2^{\frac{n^2}{2}}+2x_1^n x_2^{\frac{n^2-n}{2}})$

Evaluating at $n=6, x_i=2$ yields a total of 8,590,557,312 distinct key cards– and corresponding hotel room door locks– that the manufacturer can provide.

However, these locks are expensive: the second requirement above means that each lock must contain not only the sensing hardware to scan the pattern of holes in a key card, but also the software to compare that detected pattern against the eight possibly distinct rotations and reflections of the pattern that unlocks the door. (For example, the key card on the left in the figure above “looks the same” to the sensor in any orientation; the key card in the middle, however, may present any of four distinct patterns of scanned holes; and the key card on the right “looks different” in each of its eight possible rotated or flipped orientations.)

Which leads to the problem that motivated this post: to reduce cost, let’s modify the second requirement above– but still retaining the first requirement– so that a manufacturer-provided key card will only open its assigned manufacturer-provided lock when inserted into the slot in a single correct orientation labeled on the key card. This way, the sensing hardware in the lock only needs to “look for” a single pattern of holes.

Now how many distinct key cards and corresponding room locks are possible?

Counting regular orbits

The idea is that, referring again to the figure above, key cards may only have patterns of holes like the example on the far right, without any rotation or reflection symmetries. In other words, given the (dihedral) group $G$ of symmetries acting on colorings of the set $X$ of possible key card hole positions, we are counting only regular orbits of this action– i.e., those orbits whose colorings are “fully asymmetric,” having a trivial stabilizer.

So how can we do this? My approach was to use inclusion-exclusion, counting those colorings fixed by none of the symmetries in $G-\{e\}$. To start, we represent each element of $G$ as a list of lists of elements of $X$, corresponding to the disjoint cycles in the permutation of $X$. For a given subset $S \subseteq G-\{e\}$ in the inclusion-exclusion summation, consider the equivalence relation on $X$ relating two key card hole positions if we can move one to the other by a sequence of symmetries in $S$. Then the desired number of $k$-colorings fixed by $S$ is $k^{m_S}$, where $m_S$ is the number of equivalence classes.

We can compute this equivalence relation using union-find to incrementally “merge” the sets of disjoint cycles in each permutation in $S$ (all of the code discussed here is available on GitHub):

def merge(s, p):
"""Merge union-find s with permutation p (as cycles)."""
def find(x):
while s[x] != x:
x = s[x]
return x
def union(x, y):
x = find(x)
s[find(y)] = x
return x
for cycle in p:
reduce(union, cycle)
for x in range(len(s)):
s[x] = find(x)
return s


It remains to compute the inclusion-exclusion alternating sum of these $(-1)^{|S|}k^{m_S}$ over all subsets $S \subseteq G-\{e\}$.

def cycle_index_term(s, k=2):
"""Convert union-find s to cycle index monomial at x[i]=k."""
#return prod(x[i]**j for i, j in Counter(Counter(s).values()).items())
return k ** sum(Counter(Counter(s).values()).values())

def asymmetric_colorings(group, k=2):
"""Number of k-colorings with no symmetries in the given group."""

# Group G acts on (colorings of) X = {0, 1, 2, ..., n-1}.
G = list(group)
n = sum(len(cycle) for cycle in G[0])

# Compute inclusion-exclusion sum over subsets of G-e.
G = [g for g in G if len(g) < n]
return sum((-1) ** len(subset) *
cycle_index_term(reduce(merge, subset, list(range(n))), k)
for subset in chain.from_iterable(combinations(G, r)
for r in range(len(G) + 1)))


Evaluating the result– and dividing by the size of each orbit $|G|=8$— yields 8,589,313,152 possible “fully asymmetric” key cards satisfying our requirements.

Questions

At first glance, this seems like a nice solution, with a concise implementation, that doesn’t require much detailed knowledge about the structure of the symmetry group involved in the action… but we get a bit lucky here. The time to compute the inclusion-exclusion summation is exponential in the order of the group, which just happens to be small in this case.

For a more complex example, imagine coloring each face of a fair die red or blue; how many of these colorings are “orientable,” so that if the die rests on a table and we pick it up, put it in a cup and shake it and roll the die to a random unknown orientation, we can inspect the face colors to unambiguously determine the die’s original resting orientation? We can use the same code above to answer this question for a cube or tetrahedron (0 ways) or octahedron (120/24=5 ways)… but the dodecahedron and icosahedron are beyond reach, with rotational symmetry groups of order 60.

Of course, in those particular cases, we can lean on the additional knowledge about the structure of the subgroup inclusion partial order to solve the problem with fewer than the $2^{60}$-ish operations required here. But is there a way to improve the efficiency of this algorithm in a way that is still generally applicable to arbitrary group actions?

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## Exploiting advantage from too few shuffles

Introduction

A few days ago a friend of mine referred me to an interesting podcast discussing card shuffling, framed as a friendly argument-turned-wager between a couple about how many times you should shuffle a deck of cards. A woman claims that the “rule” is that you riffle shuffle three times, then quit messing around and get to dealing. Her partner, on the other hand, feels like at least “four or more” riffle shuffles are needed for the cards to be sufficiently random.

A mathematician is brought into the discussion, who mentions the popular result that seven shuffles are needed… at least according to specific, but perhaps not necessarily practical, mathematical criteria for “randomness.” (There is some interesting preamble about the need to define exactly what is meant by “random,” which I was disappointed to hear defined as, “any card is equally likely to be in any position in the deck.” This isn’t really even close to good enough. For example, start with a brand new deck of cards in a known order, and simply cut the deck at a uniformly random position. Now each and every card is equally likely to be in any position in the deck, but the resulting arrangement of cards can hardly be called sufficiently random.)

A win for the man, right? But the woman’s side is vindicated in the end, by noting that even in casinos– where presumably this has been given a lot of thought– a standard poker deck is typically only shuffled three times. Several dealers are interviewed, each describing the process with the chant, “riffle, riffle, box, riffle, cut.”

The wash

A couple of observations occurred to me after listening to this discussion. First, it’s true that casino dealers don’t shuffle seven times… but they also don’t just shuffle three times. Particularly when presented with a brand new pack, before any riffle shuffling, they often start with a “wash,” consisting of spreading the cards haphazardly around the table, eventually collecting them back into a squared-up deck to begin the riffle-and-cut sequence.

Depending on how thorough it is, that initial wash alone is arguably sufficient to randomize the deck. If we think of a single riffle shuffle as applying a random selection of one of “only” $2^{52}$ possible permutations in a generating set, then the wash is roughly akin to making a single initial selection from a generating set of all 52! possible arrangements. If the wash is thorough enough that this selection is approximately uniform, then after that, any additional shuffling, riffle or otherwise, is just gravy.

When does it really matter?

The second observation is one made by a dealer interviewed in the podcast, who asks what I think is the critical practical question:

The real question is, what’s the goal of the shuffle? Is it to completely randomize the cards, or is it to make it so that it’s a fair game?

In other words, if we are going to argue that three, or any other number of shuffles, is not sufficient, then the burden is on us to show that this limited number of shuffles provides a practical advantage that we can actually exploit in whatever game we happen to be playing.

We have discussed some examples of this here before. For example, this wonderful card trick due to Charles Jordan involves finding a spectator’s secretly selected card, despite being buried in a thrice-shuffled deck. And even seven shuffles is insufficient to eliminate a huge advantage in the so-called New Age Solitaire wager.

But it’s an interesting question to consider whether there are “real” card games– not magic tricks or contrived wagers– where advantage may be gained by too few shuffles.

I struggled to think of such a practical example, and the following is the best I can come up with: let’s play a simplified version of the card game War (also discussed here recently). Start with a “brand new” deck of cards in the following order:

A “new deck order” of cards prior to shuffling for a simplified game of War.

Riffle shuffle the deck three times, and cut the deck. In fact, go ahead and cut the deck after each riffle shuffle. Then I will deal the cards into two equal piles of 26 cards, one for each of us. At each turn, we will simultaneously turn over the top card from our piles, and the higher card wins the “trick.” Let’s simplify the game by just playing through the deck one time, and instead of a “war” between cards of the same rank, let’s just discard the trick as a push. At the end of the game, whoever has taken the most tricks wins a dollar from the other player.

If three shuffles is really sufficient to make this a “fair” game, then the expected return for each player should be zero. Instead, I as the dealer will win over two out of three games, taking about 42 cents from you per game on average!

Of course, this is still contrived. Even the initial deck order above is cheating, since it isn’t the typical “new deck order” in most packs manufactured in the United States. And if we play the game repeatedly (with three shuffle-cuts in between), the advantage returns to near zero for reasonable methods of collecting the played cards back into the deck.

So, I wonder if there are better real, practical examples of this kind of exploitable advantage from too few shuffles? And can this advantage persist across multiple games, with the same too-few shuffles in between? It’s interesting to consider what types of games involve methods of collecting the played cards back into the deck to shuffle for the next round, that might retain some useful ordering; rummy-style games come to mind, for example, where we end up with “clumps” of cards of the same rank, or of consecutive ranks, etc.

## Giant Yahtzee

In the game of Yahtzee, players roll five standard dice, then repeatedly re-roll subsets of the dice, trying to obtain various scoring combinations, the most valuable of which is a “Yahtzee,” or five of a kind, i.e., all five dice showing the same value.

If we strip off the complexities of the multiple players, limited number of re-rolls, and various other scoring combinations (e.g., straights, full houses, etc.), there is a nice mathematical puzzle buried underneath:

Roll $n$ dice each with $d=6$ sides, and repeatedly re-roll any subset of the dice– you can “keep” any or none of your previous rolls, and you can re-roll dice you have previously kept– until all dice show the same value (e.g., all 1s, or all 2s, etc.). Using an optimal strategy, what is the (minimum) expected number of rolls required? In particular, can we solve this problem for “Giant Yahztee,” where we are playing with, say, $n=100$ dice?

Edit 2020-10-05: Following are my notes on this problem. Given that we (re)roll $r$ of the dice– setting aside the remaining $s=n-r$ already identical dice– let the random variable $X_r$ indicate the resulting new number of identical dice. The distribution of $X_r$ is given by

$P(X_r \leq t) = \frac{1}{d^r}[\frac{x^r}{r!}] \left(\sum\limits_{k=0}^t \frac{x^k}{k!}\right)^{d-1} \left(\sum\limits_{k=0}^{t-n+r} \frac{x^k}{k!}\right)$

$P(x_r = t) = P(X_r \leq t) - P(X_r \leq t-1)$

so that the transition matrix $P$ for the absorbing Markov chain with state space ${0, 1, 2, \ldots, n}$ indicating the current number of identical dice has entries

$P_{s,t} = P(X_{n-s}=t), 0 \leq s,t \leq n$

which we can use to compute the desired expected number of rolls. See the comments for a nice closed form solution for the cumulative distribution function for the number of rolls when $n=5$.

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## MATLAB’s colon operator and for loops

Introduction

The MATLAB colon operator is surprisingly complicated, given that its job seems pretty simple to describe: generate a vector of regularly-spaced values, with a specified starting point, step size, and endpoint. For example, to create the vector $x=(0, 0.1, 0.2, \ldots, 1.1, 1.2)$:

x = 0:0.1:1.2;


At least some complexity is understandable, since as in this example, the “intended” step size and/or the endpoints may not be represented exactly in double floating-point precision. But in MATLAB’s usual habit of trying to “helpfully” account for this, things get messier than they need to be. The motivation for this post is to describe two different behaviors of the colon operator: it behaves in one special way in for loops, and in a different way– well, everywhere else.

Creating vectors with colon syntax

First, the “everywhere else” case: as the documentation suggests,

The vector elements are roughly equal to [start, start+step, start+2*step, ...] … however, if [the step size] is not an integer, then floating point arithmetic plays a role in determining whether colon includes the endpoint in the vector.

That is, continuing the above example, note that ismember(1.2, x), despite the fact that 0+12*0.1 > 1.2. But the actual implementation is even more complex than just computing the “intended” endpoint. The output vector is effectively constructed in two halves, adding multiples of the step size to the starting point in the first half, and subtracting multiples of the step size from the (computed) endpoint in the second half.

So far, this seems reasonably well known, despite the broad strokes documentation. There is a good description of the details of how this works on Stack Overflow. Let’s not worry about those details here, though; instead, what seems less well known is that the same colon expression, such as in the example above, behaves differently when it appears in a for loop.

For loops with (and without) colon syntax

First, it’s worth noting that MATLAB for loops don’t have to use the colon operator at all. With not-quite-full-fledged iterator-ish semantics, you can iterate over the columns of an arbitrary array expression. For some examples:

for index = [2, 3, 5, 7]
disp(index); % 4 iterations
end

for index = x
disp(index); % 13 iterations
end


(Technically, iteration is over first-column “slices” of the possibly multi-dimensional array. This can cause some non-intuitive behavior. For example, how many iterations would you expect over ones(2,0,3)? What about ones(0,2,3)?)

But here is where things get weird. Consider the following example:

x = 0:0.1:1.2;
for index = 0:0.1:1.2
disp(find(x == index));
end


This loop only “finds” 7 of the 13 elements of the original vector above, which was created using exactly the same colon operator expression!

So what’s going on? First, while the colon operator documentation was perhaps merely incomplete, the for loop documentation is downright misleading, suggesting that the behavior is to “increment index by the step on each iteration.” That sounds to me like repeatedly adding the step size to the value at the previous iteration, which would be even worse in terms of error accumulation, and is fortunately not what’s happening here.

Instead, experiments suggest that what is happening is essentially the overly-simplified description in the colon operator (not the for loop) documentation: the statement for index = start:step:stop iterates over values of the form start+k*step — i.e., adding multiples of the step size to the starting point– with the added detail that the number of iterations (i.e., the stopping point) seems to be computed in the same way as the “normal” colon operator. That is, the documentation is also wrong in that it’s not as simple as incrementing “until index is greater than stop” (witness the example above, where the last value is allowed to slightly overshoot the given endpoint). I have been unable to find an example of a colon expression whose size is different depending on whether it’s in a for loop.

Conclusion

What I find most interesting about this is how hard MathWorks has to work– and is still working— to make this confusing. That is, the colon syntax in a for statement is a special case in the parser: there are necessarily extra lines of code to (1) detect the colon syntax in a for loop, and (2) do something different than they could have done by simply always evaluating whatever arbitrary array expression– colon or otherwise– is given to the right of the equals sign.

And this isn’t just old legacy behavior that no one is paying attention to anymore. Prior to R2019b, you could “trick” the parser into skipping the special case behavior in a for loop by wrapping the colon expression in redundant array brackets:

for index = [0:0.1:1.2]
disp(find(x == index)); % finds all 13 values
end


However, as of R2019b, this no longer “works;” short of using the explicit function notation colon(0,0.1,1.2), it now takes more sophisticated obfuscation on the order of [0:0.1:1.2, []] or similar nonsense to say, “No, really, use the colon version, not the for loop version.”

Posted in Uncategorized | 1 Comment

## Computing the angle between two vectors

Introduction

Given two vectors in three dimensions, what is the most accurate way to compute the angle between them? I have seen several different approaches to this problem recently in the wild, and although I knew some of them had potential issues, I wasn’t sure just how bad things might get in practice, nor which alternative was best as a replacement.

To make the setup more precise, let’s assume that we are given two non-zero input vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^3$, represented exactly by their double-precision coordinates, and we desire a function that returns a double-precision value that most closely approximates the angle $0 \leq \theta \leq \pi$ between the vectors, with all intermediate computation also done in double precision.

Kahan’s Mangled Angles

William Kahan discusses three formulas in the “Mangled Angles” section of the paper linked below. The first is the “usual” dot product formula:

$\theta = \cos^{-1}\frac{\mathbf{u}\cdot\mathbf{v}}{\left|\mathbf{u}\right|\left|\mathbf{v}\right|}$

with the following C++ implementation, which as Kahan points out requires clamping the double-precision dot product to the interval $[-1, 1]$ to avoid a NaN result for some vectors that are nearly parallel:

double angle(const Vector& u, const Vector& v)
{
return std::acos(std::min(1.0, std::max(-1.0,
dot(u, v) / (norm(u) * norm(v)))));
}


Kahan subsequently describes another formula using the cross product:

$\theta = \begin{cases}\sin^{-1}\frac{\left|\mathbf{u}\times\mathbf{v}\right|}{\left|\mathbf{u}\right|\left|\mathbf{v}\right|} & \text{if }\mathbf{u}\cdot\mathbf{v} \geq 0, \\ \pi-\sin^{-1}\frac{\left|\mathbf{u}\times\mathbf{v}\right|}{\left|\mathbf{u}\right|\left|\mathbf{v}\right|} & \text{if }\mathbf{u}\cdot\mathbf{v} < 0 \end{cases}$

with the following implementation:

double angle(const Vector& u, const Vector& v)
{
double angle = std::asin(std::min(1.0,
norm(cross(u, v)) / (norm(u) * norm(v))));
if (dot(u, v) < 0)
{
angle = 3.141592653589793 - angle;
}
return angle;
}


Interestingly, Kahan does not mention that this formula also requires clamping the asin argument to the interval $[-1, 1]$; following is an explicit example of inputs demonstrating the potential problem:

$\mathbf{u} = (-0.6171833037218851, -0.4342100935824679, 0.6561603190059907)$

$\mathbf{v} = (-0.32014601021553196, 0.9003703730169068, 0.29468580477598927)$

Finally, despite referring to the above formula as “the best known in three dimensions,” Kahan finishes with the following “better formula less well known than it deserves”:

$\theta = 2\tan^{-1}\frac{\left|{\left|\mathbf{v}\right|\mathbf{u} - \left|\mathbf{u}\right|\mathbf{v}}\right|}{\left|{\left|\mathbf{v}\right|\mathbf{u} + \left|\mathbf{u}\right|\mathbf{v}}\right|}$

with the following implementation:

double angle(const Vector& u, const Vector& v)
{
double nu = norm(u);
double nv = norm(v);
return 2 * std::atan2(norm(nv * u - nu * v), norm(nv * u + nu * v));
}


That’s a lot of square roots. I didn’t focus on performance here, but it would be an interesting follow-on analysis to compare the speed of each of these formulas.

Other approaches are possible; following is the formula that I thought was the most accurate, before reading Kahan’s paper:

$\theta = \tan^{-1}\frac{\left|\mathbf{u}\times\mathbf{v}\right|}{\mathbf{u}\cdot\mathbf{v}}$

double angle(const Vector& u, const Vector& v)
{
return std::atan2(norm(cross(u, v)), dot(u, v));
}


This has the added benefit of involving just a single square root. This is the formula that I used to compute the “true” angle between vectors to compare errors, using arbitrary-precision rational arithmetic to compute the square root (actually, a reciprocal square root, which is slightly easier) and arctangent. All of the source code is available on GitHub.

And finally, the approach that I saw most recently that motivated this post, using the Law of cosines:

$\theta = \cos^{-1}\frac{\left|\mathbf{u}\right|^2 + \left|\mathbf{v}\right|^2 - \left|\mathbf{u}-\mathbf{v}\right|^2}{2\left|\mathbf{u}\right|\left|\mathbf{v}\right|}$

double angle(const Vector& u, const Vector& v)
{
double u2 = u.x * u.x + u.y * u.y + u.z * u.z;
double v2 = v.x * v.x + v.y * v.y + v.z * v.z;
Vector d = u - v;
return std::acos(std::min(1.0, std::max(-1.0,
(u2 + v2 - (d.x * d.x + d.y * d.y + d.z * d.z)) /
(2 * std::sqrt(u2) * std::sqrt(v2)))));
}


Results

The relative accuracy of each formula depends on the magnitude of the angle between the input vectors. The following figure shows this comparison for angles near 0 (i.e., nearly parallel vectors), near $\pi/4$, near $\pi/2$ (i.e., nearly orthogonal vectors), and near $\pi$ (i.e., nearly “anti-parallel” vectors).

Absolute error between computed and exact rounded double-precision angle, vs. true/exact angle.

The x-axis indicates an offset from the “true” angle between the input vectors, computed to roughly 200-bit accuracy. The y-axis indicates the error in the double-precision output, compared against the true angle also rounded to double-precision. The points hugging the bottom of each figure are my poor man’s attempt at indicating zero error (note that these are on a log-log scale), i.e., the 64-bit double-precision output matched the corresponding value rounded from the 200-bit true angle. (In many ways this figure feels like a failure of visual display of quantitative information, and I’m not sure how best to improve it.)

So what’s the takeaway? If you don’t care about absolute errors smaller than a few dozen nanoradians, then it doesn’t really matter which formula you use. And if you do care about errors– and angles– smaller than that, then be sure that your inputs are accurate in the first place. For example, did you normalize your “real” input vectors to unit length first, and if so, how much error did you unintentionally incur as a result? We can construct very small angles between vectors if we restrict to “nice” two-dimensional inputs like $(1,0,0)$ and $(\cos \theta, \sin \theta, 0)$. But it’s an interesting exercise to see how difficult it is to construct vectors “in general position” (e.g., randomly rotated) with a prescribed small angle between them.

As expected, the two arccosine formulas behave poorly for nearly parallel/anti-parallel vectors, and as Kahan describes, the arcsine formula behaves poorly for nearly orthogonal vectors. The two arctangent formulas are the most consistently accurate, and when the one mentioned by Kahan is better, it’s typically much better.

Reference:

1. Kahan, W., How Futile are Mindless Assessments of Roundoff in Floating-Point Computation? [PDF]

## Ambiguous notation for logarithms

The motivation for this post is to respond to some questions about a recent video presentation titled, “Why you haven’t caught Covid-19 [sic],” presented by Anne Marie Knott, a professor in the Washington University St. Louis Olin Business School. The gist of the presentation is an argument against the “non-pharmaceutical interventions,” or stay-at-home orders, etc., in response to the current pandemic.

I am not interested in arguing about government policies, or even epidemiological models here. Frankly, this video is too easy a target. The error made in this video is a mathematical one– an error so simple, and yet so critical to the presenter’s argument, that it’s not worth bothering with the remainder of the presentation. Instead, I’d like to use this video as an excuse to rant about mathematical notation.

The problem starts at about 3:38 in the video, where the presenter attempts to analyze the COVID-19 outbreak on the aircraft carrier USS Theodore Roosevelt as a realization of the so-called “final size equation,” a model of the end-game, steady state extent of an epidemic in a closed system (since the sailors were isolated onboard the ship for a significant period of time). The final size equation is

$p = 1-e^{-R_0 p}$

where $p$ is the “final size” of the pandemic, or the fraction of the population that is eventually infected, and $R_0$ is the basic reproduction number, essentially the average number of additional people infected through contact with a person already infected, in the situation where everyone in the population is initially susceptible to infection.

As the presenter explains, there is a critical difference between a reproduction number less than one, resulting in “extinction” of the disease, and a value greater than one, resulting in an epidemic. Using the fact that 856 of the 4954 sailors onboard the Roosevelt eventually tested positive for COVID-19, corresponding to $p=856/4954$, we can estimate $R_0$ by solving for it in the final size equation, yielding

$R_0 = -\frac{\ln (1-p)}{p}$

It’s a simple exercise to verify that the resulting estimate of $R_0$ is about 1.1. It’s also a relatively simple exercise to verify that this estimation technique cannot possibly yield an estimate of $R_0$ that is less than one.

Despite this, the presenter manages– conveniently for her argument that the contagiousness of the virus is overblown– to compute a value of $R_0$ of about 0.48… by computing the base 10 logarithm $\log_{10}(1-p)$ instead of the natural logarithm $\ln(1-p)=\log_e(1-p)$ in the formula above.

It’s interesting to try to guess how the presenter managed to make this mistake. My guess is that she did this in an Excel spreadsheet; that is the only environment I know of where log(x) computes the base 10 logarithm. In any other programming environment I can think of, log(x) is the natural logarithm, and you have to work at it, so to speak, via log10(x), or log(x)/log(10), to compute the base 10 logarithm.

The mathematical notation situation is a bit of a mess as well. Sometimes I’m a mathematician, where $\ln x$ means the natural logarithm, and any other base is usually specified explicitly as $\log_b x$. But sometimes I am an engineer, where $\log x$ usually means base 10, but sometimes in a communications context it might mean base 2. Other times I am a computer scientist, where $\lg x$ is a common shorthand for base 2, and $\log x$ can mean pretty much anything, including “I don’t care about the base.”

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## Sliding rooks (and queens)

Introduction

Jacob Brazeal describes the following interesting puzzle in a recent MAA article (see reference below): starting with four rooks in the four corner squares of a chessboard, as shown in the figure below, move the rooks into the four center squares… where each single move is constrained to sliding a single rook, either horizontally along its rank or vertically along its file, as far as possible, “blocked” only by another rook or the edge of the board.

Starting configuration (left) and goal configuration (right) of sliding rooks puzzle.

Note that going in the other direction is easy– we can move the rooks from the center out to the corners in just 8 moves. But this problem is harder; it’s a nice programming exercise to determine the minimum number of moves required. The motivation for this post is to describe a slightly different approach to the problem than presented in the article, as well as a variant of the problem using queens instead of rooks that also has some interesting mathematical structure.

All of the code is available on GitHub.

We can view this problem as a directed graph, with a vertex $v$ for each possible state of the board, and a directed edge $v \to w$ if we can move a single rook in state $v$ to obtain state $w$. The goal is to find a minimum-length path from the starting vertex with the rooks at the corners to the goal vertex with the rooks in the center of the board.

It’s an interesting question whether there is a convenient admissible heuristic estimate of the number of moves required from a given board state, that would allow a more efficient informed search. I couldn’t come up with one; fortunately, simple breadth-first search turns out to be acceptably efficient for this problem:

from collections import deque

def bfs(neighbors, root):

Given a graph neighbors:V->V* and a root vertex, returns (p, d),
where p[v] is the predecessor of v on the path from the root, and
d[v] is the distance to v from the root.
"""
queue = deque([root])
parent = {root: None}
distance = {root: 0}
while queue:
vertex = queue.popleft()
for neighbor in neighbors(vertex):
if neighbor not in parent:
parent[neighbor] = vertex
distance[neighbor] = distance[vertex] + 1
queue.append(neighbor)
return (parent, distance)


It turns out that a minimum of 25 moves are required to solve the puzzle. That’s a lot– too many, really, so that this would probably not be very fun to explore by hand with an actual chess board (more on this shortly). And there are other configurations that are even more difficult to reach. The board that is “farthest” from the initial rooks-in-the-corners state is shown below, requiring 32 moves to reach:

The most difficult sliding rooks configuration, requiring 32 moves to reach.

Symmetry group action

How large is the directed graph that we need to explore? The referenced article describes a graph with ${64 \choose 4}$=635,376 vertices, one for each possible subset of four squares in which to place the rooks. This graph has some interesting structure, with one really large strongly connected component explored by the above search algorithm, containing 218,412– over one-third– of all possible board states. The remainder is made up of a large number of much smaller unreachable components: the next largest component contains just 278 vertices!

However, these numbers count configurations of rooks that are not usefully distinct. For example, the figure above shows just one of eight “different” vertices, all of which require 32 moves to reach from the initial vertex… but the other seven board states are merely rotations and/or mirror reflections of the board shown in the figure, and thus are reachable by correspondingly rotated and/or reflected versions of the same sequence of 32 moves.

In other words, let’s consider the dihedral group $D_4$ of symmetries of the board acting on the set of possible board states, and construct the (smaller) directed graph with a vertex for each orbit of that group action.

A standard trick for implementing this approach is to represent each orbit by one of its elements, chosen in some natural and consistent way; and a standard trick for making that choice is to impose some convenient total order on the set, and choose the least element of each orbit as its representative. In the case of this problem, as we encounter each board state v during the search, we “rename” it as min(orbit(v)), the lexicographically least tuple of rotated and/or reflected coordinates of the rook positions:

def orbit(pieces):
"""Orbit of dihedral group action on rooks on a chess board."""
for k in range(4):
yield pieces
yield tuple(sorted((n - 1 - x, y) for (x, y) in pieces))    # reflect
pieces = tuple(sorted((n - 1 - y, x) for (x, y) in pieces)) # rotate


This search space is almost– but not quite– eight times smaller. From the initial rooks-in-the-corners board state, we can reach 27,467 configurations unique up to rotations and reflections, out of a total of 79,920 possible configurations. We can compute the latter number without actually enumerating all possible board states: the cycle index of the dihedral group acting on the squares of an $n \times n$ board (assuming $n$ is even) is

$Z(D_4) = \frac{1}{8}(x_1^{n^2}+2x_4^{\frac{n^2}{4}}+3x_2^{\frac{n^2}{2}}+2x_1^n x_2^{\frac{n^2-n}{2}})$

and the number of possible board states with $r$ rooks is

$[y^r] \left. Z(D_4)\right|_{x_k=y^k+1}$

Finally, I think perhaps a more “fun” variant of this problem is to consider four queens in the corners, and try to move them to the four center squares as before, using the same “maximal” moves, but allowing diagonal moves as well as horizontal and vertical. This is more tractable to solve by hand, requiring only 12 moves to complete.

And the structure of the corresponding graph is also rather interesting: the large connected component is even larger, so that we can now reach 77,766 of the 79,920 possible configurations of four queens… but the remaining 2,154 configurations are all singleton components! That is, from any one of these 2,154 “lone” configurations, we can move into the large component with just a single move, and from there reach any of those 77,766 configurations… but we can’t get back, nor can we reach any of the other 2,153 lone unreachable configurations!

This was interesting enough that I wondered if it was true in general for other board sizes. It’s trivially true for 2×2 and 4×4 (since there are no unreachable board states), as well as 6×6, 8×8, and even 10×10… but unfortunately the pattern does not continue; the 12×12 board has larger-than-singleton connected components not reachable from the initial queens-in-the-corners state.

Reference:

1. Brazeal, J., Slides on a Chessboard, Math Horizons, 27(4) April 2020, p. 24-27 [link]
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## Seven riffle shuffles is not enough– except when it is

Introduction

How many times should you riffle shuffle a deck of cards? A commonly cited rule of thumb (see [1], as well as here, here, and here) is that seven riffle shuffles are sufficient to randomize a standard 52-card deck. The motivation for this post is to refine this in a couple of ways: first, even after seven riffle shuffles, enough order still remains in the deck that we can exploit it with a reasonably simple wager (see [2]). This seems to suggest that we need more than seven shuffles– and usually we do– but it is possible, at least in principle, to repeatedly riffle shuffle in such a way that (a) we can tell when it’s okay to stop, (b) sometimes after just seven or even as few as six shuffles, that (c) not just approximately but perfectly randomizes the deck.

Betting on seven shuffles

Alice and Bob are playing a game. They begin with a brand new deck of playing cards, with the cards in the standard “new deck order”:

New deck order. (Card images Copyright 2011,2019 Chris Aguilar, licensed under LGPL 3.0.)

The deck is shuffled, and cards are dealt one at a time from the top of the deck, placing each card dealt back on the bottom of the deck. As the cards are dealt, Alice is looking for the cards from the original top half of the deck: ace through king of hearts, followed by ace through king of clubs, in that order. Meanwhile, Bob is looking for the cards from the original bottom half of the deck, but in reverse order: ace through king of spades, followed by ace through king of diamonds.

When Alice’s first target card, the ace of hearts, is dealt, instead of returning it to the bottom of the deck, we remove it and set it aside in Alice’s pile. Then, once the two of hearts is dealt, we remove it and add it to Alice’s pile, etc. Similarly, when Bob’s first target card, the ace of spades, is dealt, we remove it and start Bob’s pile. The first player to complete his or her pile of 26 cards wins the game, receiving one dollar from the loser.

This should be a fair game, assuming that the deck is truly and thoroughly shuffled: Alice or Bob should each win with probability 1/2. However, starting from a new deck, even after riffle shuffling seven times, Alice wins over 80% of the time!

Probability of Alice winning the New Age Solitaire game, vs. number of initial riffle shuffles.

Reference [2] is accessible to undergraduates, and describes a beautiful formula for computing these probabilities of winning as a function of the number of initial riffle shuffles. But I think this game also makes a great simulation programming exercise, both to simulate the random riffle shuffles themselves, and to efficiently determine whether Alice or Bob wins given a particular shuffled arrangement of cards.

Riffle shuffles and inverse shuffles

The game described above suggests that seven shuffles is not enough to randomize the deck. So, how many more shuffles do we need?

First, let’s review the Gilbert-Shannon-Reeds model of a random riffle shuffle. As discussed recently here, we can represent a riffle shuffle of a deck of $n=52$ cards as a uniformly random string of $n$ bits. The number of zero bits indicates how many cards to cut from the top of the deck, and the positions of the zero bits indicate how those top cards are interleaved with the cards from the bottom part of the deck (represented by the one bits).

We can associate each such bit string encoding of a riffle shuffle with the corresponding permutation $\pi \in S_n$, indicating that the riffle shuffle moves the card initially in position $i$ to position $\pi(i)$. Repeated shuffling corresponds to composition, so that the effect of riffle shuffle $\pi_1$ followed by shuffle $\pi_2$ is $\pi_2 \pi_1$.

For reasons we will see shortly, it is also useful to consider the action of the inverse permutation $\pi^{-1}$ associated with a particular bit string encoding of a riffle shuffle. Imagine marking each card in the deck with the corresponding zero or one in the bit string; then “deinterleave,” sliding the “zero” cards out while preserving their relative order, and placing them on top of the remaining pack of “one” cards. The reader can verify that this “inverse shuffle” permutation $\pi^{-1}$ is indeed the inverse of the shuffle permutation $\pi$ corresponding to the same bit string encoding.

Stopping times

It turns out that inverse riffle shuffles are really handy, because of the following result (Lemma 9 in [1]). Suppose that we start with a new deck of cards, and repeatedly inverse shuffle the deck as follows:

1. Generate a random bit string, and mark each card with its corresponding 0 or 1 label.
2. Inverse shuffle the deck according to this bit string encoding; i.e., slide the 0 cards out and place them on top of the 1 cards.
3. Repeat steps 1 and 2… but in step 1, place each new randomly generated 0 or 1 label to the left of the previous labels on the card (from the previous inverse shuffles), effectively prepending a new most significant bit. Thus, during the $k$-th inverse shuffle, each card will have a $k$-bit integer label, and the execution of step 2 corresponds to a (stable) sorting of the cards by these integer labels.

If we continue inverse shuffling in this manner, stopping when we observe that the integer labels on the cards are all distinct, then the resulting (inverse) shuffled deck is fully randomized– that is, the resulting arrangement of cards in the deck is perfectly uniformly distributed. (More precisely, if the random variable $T$ is the minimum number of inverse shuffles required for all of the card labels to be distinct, then $T$ is a strong uniform stopping time.)

Shuffling backward in time

We’re not quite done yet. So far we have only described a method of inverse shuffling, and detecting when we can stop inverse shuffling, confident that the resulting arrangement of cards is perfectly randomized. How can we apply this to normal riffle shuffling?

The key observation (derived from Lemma 8 in [1]) is that the sequence of randomly generated “forward in time” riffle shuffles $\pi_1, \pi_2, \ldots, \pi_T$ results in a distribution of deck arrangements with the same distance from uniform as the sequence of corresponding inverse shuffles, in reverse order (i.e., executed “backward in time”).

Thus, as we riffle shuffle the deck, first with permutation $\pi_1$, then with $\pi_2$, etc., we can evaluate our stopping condition after $T$ shuffles by checking for distinct card labels resulting from the inverse shuffles $\pi^{-1}_T, \pi^{-1}_{T-1}, \ldots, \pi^{-1}_2, \pi^{-1}_1$, executed in that (reverse) order.

The following Python code implements this method, returning a list of individual riffle shuffles– as the corresponding bit strings– that when executed in order realizes a uniformly random permutation. Note that that reversed() is critical; without it, the resulting distribution of possible arrangements is observably non-uniform.

import numpy as np

def uniform_riffle_shuffles(n, rng=np.random.default_rng()):
"""Return list of encodings of riffle shuffles of deck of n cards."""
riffles = []
while True:

# Generate another riffle shuffle.
riffle = rng.integers(0, 2, n)
riffles.append(riffle)

# Perform all inverse shuffles in reverse order.
labels = np.zeros_like(np.arange(n))
for bit, riffle in enumerate(reversed(riffles)):
p = np.argsort(riffle, kind='stable')
labels = (labels + riffle * (2 ** bit))[p]

# Stop when card labels are all distinct.
if len(set(labels)) == n:
break
return riffles


Conclusion

To wrap up, how long does this perfectly random shuffling process take? This turns out to be an instance of the birthday problem: if the random variable $T$ indicates the number of shuffles required to randomize a deck with $n$ cards, then the cumulative distribution function $P(T \leq s)$ is the probability that $n$ card labels (think people), each of which is an $s$-bit integer (think possible birthdays), are all distinct:

$P(T \leq s) = \frac{(2^s)_n}{2^{sn}}$

where $(x)_n$ is the falling factorial. The following figure shows the resulting distribution, with the CDF in blue, the PDF in red, and the mean of approximately 11.7243 shuffles in black.

Distribution of number of riffle shuffles needed to perfectly randomize a 52-card deck.

References:

1. Aldous, D. and Diaconis, P., Shuffling Cards and Stopping Times, The American Mathematical Monthly, 93(5) 1986, p. 333-348 [PDF]
2. Zuylen, A. and Schalekamp, F., The Achilles’ Heel of the GSR Shuffle: A Note on New Age Solitaire, Probability in the Engineering and Informational Sciences, 18(3) July 2004, p. 315-328 [DOI]
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## History of COVID-19 cases

All of the raw data presented here was retrieved from this GitHub repository, which, as described there, is maintained by and thanks to the Johns Hopkins University Center for Systems Science and Engineering (JHU CSSE), with support by ESRI Living Atlas Team and the Johns Hopkins University Applied Physics Lab (JHU APL). I will reiterate their disclaimer that these data are strictly for educational purposes… and this post of mine is yet another possibly inaccurate perspective on data that is already derived from multiple and possibly conflicting sources. If you’re looking for medical guidance, look elsewhere.

The following figure shows the progression of COVID-19 over the last seven weeks or so, as measured by cumulative confirmed cases. I restricted attention to the eight countries currently having at least 2000 confirmed cases.

Cumulative confirmed cases vs. time, for each country currently having at least 2000 confirmed cases.

Although it’s interesting to try to interpret this view of past history, I think it’s difficult to use it to predict even the near future. Note how similar is the exponential growth (note the figure is on a logarithmic scale) for, well, pretty much everyone but China and South Korea, who appear to have taken the most drastic-but-apparently-successful measures to contain the virus. Comparing with Italy, which is now struggling with hospital capacity, we here in the United States appear to be on our way to very similar numbers of cases in a matter of 11 days or so, assuming recent growth continues.

Except that this is potentially misleading, for several reasons. On the pessimistic side, this figure only shows confirmed positive tests— the United States might already have (and in my opinion, almost certainly does have) many more people with the virus, given how little testing has been done so far.

On the other hand, the United States is a larger country than Italy, with roughly five times the population. The following figure attempts to account for this, showing the cumulative number of confirmed cases per million in population (population data obtained here).

Cumulative confirmed cases per million in population over time, for each country currently having at least 2000 confirmed cases.

Importantly, this has no effect on the “slope,” i.e., the exponential rate of growth of cases. It merely delays the same end result– this figure suggests that it might take two and a half weeks, instead of a week and a half… but we’re still headed where Italy is now.

I think an actual prediction of this sort is difficult to make confidently, though. Many interesting dials have been turned, even if only in the past few days. Human behavior has changed, with some significant steps taken on both large scales and small. Whether the eventual effects will be no more disastrous than waiting for the next truck to deliver more toilet paper, I’m not sure. The next two weeks or so will be interesting.

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