However, after a delay of several minutes, a flight attendant came on the PA and asked for two– specifically two– volunteers to give up their seat, in exchange for a flight later that afternoon. Two people immediately jumped up, left the airplane, and *then* we were ready to go… now with two empty seats.

The problem was *weight*: due to a changing forecast of bad weather, both in Baltimore and en route, we had taken on additional fuel at the last minute (e.g., to allow for diverting to a possibly now-more-distant alternate airport), resulting in the airplane exceeding its maximum takeoff weight. Something had to go, and apparently two passengers and their carry-on bags were a sufficient reduction in weight to allow us to take off.

What I found interesting about this episode was the relative *precision* of the change– 175 (or even 174) passengers bad, 173 passengers good– compared with the *uncertainty* in the total weight of the passengers, personal items, and carry-on bags remaining on board. That is, how does the airline know how much we weigh? Since Southwest does not ask individual passengers for their weight, let alone ask them to step on an actual scale prior to boarding, some method of estimation is required.

The FAA provides guidance on how to do this (see reference below): for large-cabin aircraft, the assumed average weight of an adult passenger, his or her clothing, personal items, and a carry-on bag is 190 pounds, with a standard deviation of 47 pounds. The figure below shows the resulting probability distribution of the *total* weight of all 175 passengers on the initially completely full flight:

It’s worth noting that the referenced Advisory Circular does provide a more detailed breakdown of assumed average passenger weight, to account for season of travel (5 more pounds of clothing in the winter), gender, children vs. adults, and “nonstandard weight groups” such as sports teams, etc. But for this summer flight, with a relatively even split of male and female passengers, the only simplifying assumption in the above figure is no kids.

The point is that this seems like a significant amount of uncertainty in the *actual* total weight of the airplane, for less than 400 pounds to be the difference between “Nope, we’re overweight” and “Okay, we’re safe to take off.”

**Reference:**

- Federal Aviation Administration Advisory Circular AC-120-27E, “Aircraft Weight and Balance Control,” 10 June 2005 [PDF]

You have once again been captured by mathematically-inclined pirates and threatened with walking the plank, unless you can win the following game: some number of black balls and white balls will be placed in an urn. The captain will randomly select and discard a ball from the urn, noting its color, then repeatedly draw and discard additional balls *as long as they are the same color*. The first drawn ball of a *different* color will be returned to the urn, and the whole process will be repeated. And repeated, and repeated, until the urn is empty. If the last ball drawn from the urn is black, you must walk the plank; if it is white, you will be set free.

You can choose any positive numbers and of black and white balls, respectively, to be placed in the urn at the start. How many of each should you start with to maximize your probability of survival?

]]>“Doctors say he’s got a 50/50 chance of living… though there’s only a 10% chance of that.”

I’ve lately had occasion to contemplate my own mortality. How long should I expect to live? The most recent life table published by the Centers for Disease Control (see the reference at the end of this post) indicates an expected lifespan of 76.5 years for a male. This is based on a model of age at death as a random variable with the probability density shown in the following figure.

The expected lifespan of 76.5 years is (using the red curve for males). In other words, if we observed a large number of hypothetical (male) infants born in the reference period 2014– and they continued to experience 2014 mortality rates throughout their lifetimes– then their ages at death would follow the above distribution, with an average of 76.5 years.

However, I have more information now: I have already survived roughly four decades of life. So it makes sense to ask, what is my *conditional* expected age at death, given that I have already survived to, say, age 40? In other words, what is ?

This value is 78.8 years; I can expect to live to a greater age now than I thought I would when I was first born. The following figure shows this conditional expected age at death , as well as the corresponding expected *additional* lifespan , as a function of current age .

For another example, suppose that I survive to age 70. Instead of expecting just another 6.5 years, my expected additional lifespan has jumped to 14.5 years.

Which brings us to the interesting observation motivating this post: suppose instead that I *die* at age 70. I will have missed out on an additional 14.5 years of life on average, compared to the rest of the septuagenarians around me. Put another way, at the moment of my death, I perceive that I am dying 14.5 years earlier than expected.

But this perceived “loss” *always* occurs, no matter when we die! (In terms of the above figure, the expected value is always positive.) We can average this effect over the entire population, and find that on average males die 12.2 years earlier than expected, and females die 10.8 years earlier than expected.

**Reference**:

- Arias, E., United States Life Tables 2014,
*National Vital Statistics Reports*,**66**(4) August 2017 [PDF]

Following are the probabilities for the United States 2014 period life table used in this post, derived from the NVSR data in the above reference, extended to maximum age 120 using the methodology described in the technical notes.

Age P(all) P(male) P(female) =========================================================== 0 0.005831 0.006325 0.005313 1 0.000367843 0.000391508 0.000343167 2 0.000246463 0.000276133 0.000216767 3 0.000182814 0.000206546 0.000157072 4 0.000156953 0.000183668 0.000129216 5 0.000141037 0.000160804 0.000120255 6 0.000125127 0.000142914 0.000106328 7 0.000112203 0.000128008 0.0000963806 8 0.000100276 0.000112117 0.0000884231 9 0.0000913317 0.0000992073 0.0000834481 10 0.0000883454 0.0000932456 0.0000824477 11 0.0000952854 0.000103156 0.0000874072 12 0.000119095 0.000137857 0.000101304 13 0.000164729 0.000203286 0.000124134 14 0.000227209 0.000294457 0.000155893 15 0.000293617 0.000391501 0.000190617 16 0.000362946 0.000491412 0.000227306 17 0.000442117 0.000609009 0.000265957 18 0.000529115 0.000743227 0.000302594 19 0.000616971 0.000881116 0.000338207 20 0.00070566 0.00101964 0.000373784 21 0.000786255 0.00114394 0.000408332 22 0.000847887 0.0012343 0.000438875 23 0.000886654 0.00128494 0.000465417 24 0.000909534 0.00130883 0.000490933 25 0.000929392 0.0013228 0.0005174 26 0.000952147 0.00134161 0.000545804 27 0.000977786 0.00136426 0.00057515 28 0.0010063 0.00139366 0.000605432 29 0.00103864 0.00142878 0.00063566 30 0.00107285 0.0014657 0.000668788 31 0.00110792 0.00150147 0.000705793 32 0.00114483 0.0015361 0.000745674 33 0.00118454 0.00156959 0.000792356 34 0.00122898 0.00160582 0.000845811 35 0.00128495 0.00165443 0.000908956 36 0.00135141 0.00171632 0.000981746 37 0.00142538 0.00178654 0.00106021 38 0.00150387 0.0018631 0.00114136 39 0.00158781 0.00194786 0.00122516 40 0.00168489 0.00204935 0.00131843 41 0.00179886 0.00217314 0.00142304 42 0.00192761 0.00231898 0.00153401 43 0.00207581 0.00249327 0.00165615 44 0.00224899 0.00270038 0.00179419 45 0.00243725 0.00292846 0.00194116 46 0.00265083 0.0031884 0.00210855 47 0.0029103 0.00350311 0.00231349 48 0.00321588 0.00387243 0.0025555 49 0.0035468 0.00427362 0.00281771 50 0.00387592 0.00467212 0.00307957 51 0.00420287 0.00507013 0.00333606 52 0.00454693 0.00549737 0.00359937 53 0.00492128 0.00597165 0.0038758 54 0.00532664 0.00649003 0.00417147 55 0.00575619 0.00703581 0.00448667 56 0.00619215 0.00758275 0.00481225 57 0.00662626 0.00813029 0.00513737 58 0.00705499 0.00866988 0.00545972 59 0.00748745 0.00921066 0.00578812 60 0.00794918 0.00978879 0.0061402 61 0.0084469 0.010399 0.00653116 62 0.0089597 0.010994 0.00696556 63 0.00947691 0.0115436 0.00744933 64 0.0100035 0.0120603 0.00797895 65 0.0105466 0.0125691 0.00854801 66 0.0111425 0.0131347 0.00916566 67 0.0118165 0.0137895 0.00985298 68 0.0126025 0.0145881 0.010625 69 0.0135386 0.0155721 0.0115183 70 0.014622 0.016711 0.0125556 71 0.0157853 0.0179169 0.0136863 72 0.0169733 0.0191484 0.0148415 73 0.0181664 0.0203416 0.0160437 74 0.0193544 0.0214907 0.0172763 75 0.0205581 0.0226235 0.0185559 76 0.0219039 0.023887 0.0199909 77 0.0233782 0.0252875 0.0215559 78 0.0249405 0.0266573 0.0233399 79 0.0266659 0.0281283 0.0253501 80 0.0283006 0.0295587 0.0272207 81 0.0298041 0.0307938 0.0290306 82 0.0311707 0.0318902 0.0307088 83 0.0326118 0.0329808 0.0325375 84 0.0338734 0.0336728 0.0344093 85 0.0348103 0.0342521 0.0357896 86 0.0356915 0.0345144 0.0373244 87 0.036144 0.0342741 0.0384714 88 0.0361034 0.0334914 0.0391388 89 0.0355212 0.0321521 0.0392438 90 0.0343716 0.0302738 0.0387212 91 0.0326583 0.0279093 0.0375332 92 0.0304192 0.0251463 0.0356786 93 0.0277276 0.0221028 0.0331989 94 0.02469 0.0189177 0.030181 95 0.0214386 0.0157381 0.0267542 96 0.0181203 0.0127037 0.0230805 97 0.0148823 0.00993297 0.0193396 98 0.011857 0.00751144 0.0157101 99 0.00914934 0.00548606 0.0123497 100 0.00682791 0.0038652 0.00937893 101 0.0049216 0.00262443 0.0068711 102 0.00342273 0.00171608 0.00484973 103 0.00229461 0.00108015 0.00329442 104 0.00148199 0.000654343 0.00215221 105 0.000921789 0.000381551 0.00135156 106 0.000552114 0.000214241 0.000815772 107 0.00031851 0.000115918 0.00047333 108 0.000177059 0.0000604924 0.000264139 109 0.0000949139 0.0000304833 0.000141878 110 0.0000491106 0.0000148533 0.0000734294 111 0.0000245565 0.00000700891 0.0000366663 112 0.0000118822 0.00000320819 0.0000176914 113 0.00000557214 0.00000142697 0.00000826206 114 0.00000253659 0.000000617876 0.00000374136 115 0.00000112287 0.000000260924 0.00000164594 116 0.0000004842 0.00000010766 0.00000070483 117 0.000000203758 0.0000000434809 0.000000294373 118 0.0000000838243 0.0000000172192 0.000000120143 119 0.0000000337717 0.0000000066978 0.0000000480077 120 0.0000000216956 0.00000000409582 0.0000000304297]]>

The title question came up recently, which I think makes for an interesting combinatorics exercise. The idea is that, given the extent of “borrowing” of past musical ideas by later artists, are we in danger of running out of new music?

We can turn this into a combinatorics problem by focusing solely on *pitch *and* rhythm*: in a single bar of music, how many possible melodies are there, consisting of a sequence of notes and rests of varying pitch and duration? The point is that this number is *finite*; it may be astronomical, but *how* astronomical?

This is not a new question, and there are plenty of answers out there which make various simplifying assumptions. For example, Vsauce has a video, “Will We Ever Run Out of New Music?,” which in turn refers to a write-up, “How many melodies are there in the universe?,” describing a calculation based on a recurrence relation that effectively requires cutting segments of a bar exactly in halves– undercounting in a way that I suspect was not intentional, implicitly prohibiting even relatively simple melodies like “I’ll Be Home for Christmas.”

But the most common simplifying assumption seems to be a lack of treatment of *rests*— that is, only counting melodies consisting of a sequence of *notes*. Rests are an interesting wrinkle that complicates the counting problem: for example, a half note is different from two consecutive quarter notes of the same pitch, but a half rest *sounds the same* as two consecutive quarter rests. The objective of this post is to add this “expressive power” to the calculation of possible melodies.

**Solution**

Consider a single bar in 4/4 time, consisting of a sequence of whole, half, quarter, eighth, and sixteenth notes and/or rests, with notes chosen from 13 possible pitches, allowing melodies within an octave of the 12-pitch chromatic scale, but also allowing an octave jump (e.g., “Take Me Out to the Ball Game,” “Over the Rainbow,” etc.).

We can encode the choice of a single note of possible pitches with the following generating function, weighted by duration:

and all possible rests with

Then the generating function for the number of possible melodies is

Intuitively, a melody consists of zero or one rest, followed by a sequence of zero or more sub-sequences, each consisting of a note followed by zero or one rest. The coefficient is the number of one-bar melodies given the above constraints… but this counts two melodies as distinct even if they only differ in relative pitch. The number of possible melodies consisting of sequences of *intervals* confined to at most an octave jump is

where the +1 accounts for the single “silent melody” of a whole rest. The result is 3,674,912,999,046,911,152, or about 3.7 billion billion possible melodies.

**Results**

The machinery described above may be easily extended to consider different sets of assumptions: different time signatures, longer or shorter lists of possible notes to choose from, dotted notes and rests, triplets (e.g., the *Star Wars* theme), etc. The figure below shows the number of possible one-bar melodies for a variety of such assumptions.

As might be expected, dotted notes and/or rests do not affect the “space” of possible melodies nearly as much as note *duration*: halve the shortest allowable note value, and you very roughly double the number of “bits” in the representation of a melody. If we extend our expressive power to allow 32nd (possibly dotted) notes and rests, then there are 6,150,996,564,625,709,162,647,180,518,925,064,281,006 possible melodies.

Of course, these calculations only address the question of how many melodies are *possible*— not how many of such melodies are actually appealing to our human ears.

Every year, there are upsets and wild outcomes during the NCAA men’s basketball tournament. But this year felt, well, *wilder*. For example, for the first time in 136 games over the 34 years of the tournament’s current 64-team format, a #16 seed (UMBC) beat a #1 seed (Virginia) in the first round. (I refuse to acknowledge the abomination of the 4 “play in” games in the zero-th round.) And I am a Kansas State fan, who watched my #9 seed Wildcats beat *Kentucky*, a team that went to the Final Four in 4 of the last 8 years.

So I wondered whether this was indeed the “wildest” tournament ever… and it turns out that it was, by several reasonable metrics.

**Modeling game probabilities**

To compare the tournaments in different years, we assume that the probability of outcome of any particular game depends only on the *seeds* of the opposing teams. Schwertman et. al. (see reference below) suggest a reasonable model of the form

where is some measure of the “strength” of seed (ranging from 1 to 16), and the scale factor calibrates the range of resulting probabilities, selected here so that the most extreme value matches the current maximum likelihood estimate based on the 136 observations over the past 34 years.

One simple strength function is the linear , although this would suggest, for example, that #1 vs. #5 and #11 vs. #15 are essentially identical match-ups. A better fit is

where is the quantile of the normal distribution, and is the number of teams in all of Division I. The idea is that team strength is normally distributed, and the tournament invites the 64 teams in the upper tail of the distribution, as shown in the figure below.

**Probability of a perfect bracket**

Armed with these candidate models, I looked at all of the tournaments since 1985, the first year of the current 64-team format. I have provided summary data sets before (a search of this blog for “NCAA” will yield several posts on this subject), but this analysis required more raw data, all of which is now available at the usual location here.

For each year of the tournament, we can ask what is the probability of picking a perfect bracket in that year, correctly identifying the winners of all 63 games? Actually, there are three reasonable variants of this question:

- If we flip a coin to pick each game, what is the probability of picking every game correctly?
- If we pick a “chalk” bracket, always picking the favored higher-seeded (i.e., lower-numbered) team to win each game, what is the probability of picking every game correctly?
- If we managed to pick the perfect bracket for a given year, what is the prior probability of that particular outcome?

The answer to the first question is the 1 in , or the “1 in 9.2 quintillion” that appears in popular press. And this is always exactly correct, no matter how individual teams actually match up in any given year, as long as we are flipping a coin to guess the outcome of each game. But this isn’t very realistic, since seed match-ups *do* matter; a #1 seed will beat a #16 seed… well, *almost* all of the time.

So the second question is more interesting, but also more complicated, since it *does* depend on our model of how different seeds match up… but it doesn’t depend on which year of the tournament we’re talking about, at least as long as we always use the same model. Using the strength models described above, a chalk bracket has a probability of around 1 in 100 billion of being correct (1 in 186 billion for the linear strength model, or 1 in 90 billion for the normal strength model).

The third question is the motivation for this post: the probability of a given year’s actual outcome will generally lie somewhere between the other two “extremes.” How has this probability varied over the years, and was 2018 really an outlier? The results are shown in the figure below.

The constant black line at the bottom is the 1 in 9.2 quintillion coin flip. The constant red and blue lines at the top are the probabilities of a chalk bracket, assuming the linear or normal strength models, respectively.

And in between are the actual outcomes of each tournament. (*Aside*: I tried a bar chart for this, but I think the line plot more clearly shows the comparison of the two models, as well as both the maximum *and* minimum behavior that we’re interested in here.) This year’s 2018 tournament was indeed the most unlikely, so to speak, although it has close competition, all in this decade. At the other extreme, 2007 was the *most* likely bracket.

**Reference:**

- Schwertman, N., McCready, T., and Howard, L., Probability Models for the NCAA Regional Basketball Tournaments,
*The American Statistician*,**45**(1) February 1991, p. 35-38 [JSTOR]

- Paul says, “Neither Steve nor Ted was in on it.”
- Quinn says, “Ray wasn’t in on it, but Paul was.”
- Ray says, “If Ted was in on it, then so was Steve.”
- Steve says, “Paul wasn’t in on it, but Quinn was.”
- Ted says, “Quinn wasn’t in on it, but Paul was.”

You do not know which, nor even how many, of the five suspects were involved in the crime. However, you do know that every guilty suspect is lying, and every innocent suspect is telling the truth. Which suspect or suspects committed the crime?

I think puzzles similar to this one make good, fun homework problems in a discrete mathematics course introducing propositional logic. However, this particular puzzle is a bit more complex than the usual “Which one of three suspects is guilty?” type, not just because there are more suspects, but also because we don’t know *how many* suspects are guilty.

That added complexity is motivated by trying to transform this typically pencil-and-paper mathematical logic problem into a potentially nice computer science programming exercise: consider writing a program to *automate* solving this problem… or even better, writing a program to *generate new random instances* of problems like this one, while ensuring that the resulting puzzle has some reasonably “nice” properties. For example, the solution should be unique; but the puzzle should also be “interesting,” in that we should need *all* of the suspects’ statements to deduce who is guilty (that is, any proper subset of the statements should imply at least two distinct possible solutions).

This is a follow-up to one of last month’s posts that contained some images of randomly-generated “lozenge tilings.” The focus here is not on the tilings themselves, but on the *perfectly random* sampling technique due to Propp and Wilson known as “coupling from the past” (see references below) that was used to generate them. As is often the case here, I don’t have any new mathematics to contribute; the objective of this post is just to go beyond the pseudo-code descriptions of the technique in most of the literature, and provide working Python code with a couple of different example applications.

The basic problem is this: suppose that we want to sample from some probability distribution over a finite discrete space , and that we have an ergodic Markov chain whose steady state distribution is . An *approximate* sampling procedure is to pick an arbitrary initial state, then just run the chain for some large number of iterations, with the final state as your sample. This idea has been discussed here before, in the context of card shuffling, the board game Monopoly, as well as the lozenge tilings from last month.

For example, consider shuffling a deck of cards using the following iterative procedure: select a random *adjacent* pair of cards, and flip a coin to decide whether to put them in ascending or descending order (assuming any convenient total ordering, e.g. first by rank, then by suit alphabetically). Repeat for some “large” number of iterations.

There are two problems with this approach:

- It’s approximate; the longer we iterate the chain, the closer we get to the steady state distribution… but (usually) no matter when we stop, the distribution of the resulting state is never exactly .
- How many iterations are enough? For many Markov chains, the mixing time may be difficult to analyze or even estimate.

**Coupling from the past**

Coupling from the past can be used to address both of these problems. For a surprisingly large class of applications, it is possible to sample *exactly* from the stationary distribution of a chain, by iterating multiple realizations of the chain until they are “coupled,” without needing to know ahead of time when to stop.

First, suppose that we can express the random state transition behavior of the chain using a fixed, deterministic function , so that for a random variable uniformly distributed on the unit interval,

In other words, given a current state and random draw , the next state is .

Now consider a single infinite sequence of random draws . We will use this same *single* source of randomness to iterate *multiple* realizations of the chain, one for each of the possible initial states… but starting *in the past*, and running forward to time zero. More precisely, let’s focus on two particular chains with different initial states and at time , so that

(Notice how the more random draws we generate, the farther *back* in time they are used. More on this shortly.)

A key observation is that if at any point, then the chains are “coupled:” since they experience the same sequence of randomness influencing their behavior, they will continue to move in lockstep for all subsequent iterations as well, up to . Furthermore, if the states at time zero are *all* the same for *all* possible initial states run forward from time , then the distribution of that final state is the stationary distribution of the chain.

But what if all initial states *don’t* end at the same final state? This is where the single source of randomness is key: we can simply look farther back in the past, say time steps instead of , by extending our sequence of random draws… as long as we *re-use the existing random draws* to make the same “updates” to the states at later times (i.e., closer to the end time zero).

**Monotone coupling**

So all we need to do to perfectly sample from the stationary distribution is

- Choose a sufficiently large to look far enough into the past.
- Run realizations of the chain, one for each initial state, all starting at time and running up to time zero.
- As long as the final state is the same for all initial states, output as the sample. Otherwise, look farther back in the past (), generate additional random draws accordingly, and go back to step 2.

This doesn’t seem very helpful, since step 2 requires enumerating elements of the typically enormous state space. Fortunately, in many cases, including both lozenge tiling and card shuffling, it is possible to simplify the procedure by imposing a partial order on the state space– with minimum and maximum elements– that is preserved by the update function . That is, suppose that for all states and random draws , if , then .

Then instead of running the chain for *all* possible initial states, we can just run *two* chains, starting with the minimum and maximum elements in the partial order. If those two chains are coupled by time zero, then they will also “squeeze” any other initial state into the same coupling.

Following is the resulting Python implementation:

import random def monotone_cftp(mc): """Return stationary distribution sample using monotone CFTP.""" updates = [mc.random_update()] while True: s0, s1 = mc.min_max() for u in updates: s0.update(u) s1.update(u) if s0 == s1: break updates = [mc.random_update() for k in range(len(updates) + 1)] + updates return s0

Sort of like the Fisher-Yates shuffle, this is one of those algorithms that is potentially easy to implement incorrectly. In particular, note that the list of “updates,” which is traversed *in order* during iteration of the two chains, is *prepended* with additional blocks of random draws as needed to start farther back in the past. The figure below shows the resulting behavior, mapping each output of the random number generator to the time at which it is used to update the chains.

Coming back once again to the card shuffling example, the code below uses coupling from the past with random adjacent transpositions to generate an exactly uniform random permutation . (A similar example generating random lozenge tilings may be found at the usual location here.)

class Shuffle: def __init__(self, n, descending=False): self.p = list(range(n)) if descending: self.p.reverse() def min_max(self): n = len(self.p) return Shuffle(n, False), Shuffle(n, True) def __eq__(self, other): return self.p == other.p def random_update(self): return (random.randint(0, len(self.p) - 2), random.randint(0, 1)) def update(self, u): k, c = u if (c == 1) != (self.p[k] < self.p[k + 1]): self.p[k], self.p[k + 1] = self.p[k + 1], self.p[k] print(monotone_cftp(Shuffle(52)).p)

The minimum and maximum elements of the partial order are the identity and reversal permutations, as might be expected; but it’s an interesting puzzle to determine the actual partial order that these random adjacent transpositions preserve. Consider, for example, a similar process where we select a random adjacent pair of cards, then flip a coin to decide *whether to swap them or not* (vs. whether to put them in ascending or descending order). This process has the same uniform stationary distribution, but won’t work when applied to monotone coupling from the past.

**Re-generating vs. storing random draws**

One final note: although the above implementation is relatively easy to read, it may be prohibitively expensive to store all of the random draws as they are generated. The slightly more complex implementation below is identical in output behavior, but is more space-efficient by only storing “markers” in the random stream, re-generating the draws themselves as needed when looking farther back into the past.

import random def monotone_cftp(mc): """Return stationary distribution sample using monotone CFTP.""" updates = [(random.getstate(), 1)] while True: s0, s1 = mc.min_max() rng_next = None for rng, steps in updates: random.setstate(rng) for t in range(steps): u = mc.random_update() s0.update(u) s1.update(u) if rng_next is None: rng_next = random.getstate() if s0 == s1: break updates.insert(0, (rng_next, 2 ** len(updates))) random.setstate(rng_next) return s0

**References:**

- Propp, J. and Wilson, D., Exact Sampling with Coupled Markov Chains and Applications to Statistical Mechanics,
*Random Structures and Algorithms*,**9**1996, p. 223-252 [PDF] - Wilson, D., Mixing times of lozenge tiling and card shuffling Markov chains,
*Annals of Applied Probability*,**14**(1) 2004, p. 274-325 [PDF]

I recently saw several different checks, all from the same bank, each with a different dollar amount printed in a fixed-width font, similar to the shortened example below:

`**PAY EXACTLY**twelve thousand thirty-four and 56/100****************`

Counting asterisks confirmed that the amount fields on each check were all padded to exactly the same length… which made me wonder, what is the largest check that the bank could print, using English words for the dollar amount?

Actually, that’s not quite the question I had in mind. There are very short names for very large numbers, but it’s not very helpful to say that a bank can cut a check for, say, “one googol” dollars, when most *smaller* amounts would not fit in the same length of field.

So to make the problem less linguistic and more mathematical, let’s instead ask the following more precise question: what is the largest integer such that *every* dollar amount from zero to can be printed in words using at most characters?

**Converting integers into words**

For short fields, such as in the case of the bank checks, brute force is sufficient; we just need a function to convert integers into words. This is a pretty common programming problem, with one Python implementation shown below, where `integer_name(n)`

works for any :

units = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen'] tens = ['zero', 'ten', 'twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety'] powers = ['zero', 'thousand', 'million', 'billion', 'trillion', 'quadrillion', 'quintillion', 'sextillion', 'septillion', 'octillion', 'nonillion', 'decillion'] hundred = 'hundred' minus = 'minus' comma = ',' and_ = ' and' space = ' ' hyphen = '-' empty = '' def small_integer_name(n, use_and=False): s = empty if n >= 100: q, n = divmod(n, 100) s = units[q] + space + hundred + ( (and_ if use_and else empty) + space if n > 0 else empty) if n >= 20: q, n = divmod(n, 10) s += tens[q] + (hyphen if n > 0 else empty) return (s + units[n] if n > 0 else s) def integer_name(n, use_comma=False, use_and=False, power=0): if n < 0: return minus + space + integer_name(-n, use_comma, use_and) elif n == 0: return units[0] s = empty if n >= 1000: q, n = divmod(n, 1000) s = integer_name(q, use_comma, use_and, power + 1) + ( (comma if use_comma else empty) + space if n > 0 else empty) return (s + small_integer_name(n, use_and) + (space + powers[power] if power > 0 else empty) if n > 0 else s)

There are a couple of things to note. First, although typical American style does not include commas separating thousands, or the word *and* between hundreds and tens (e.g., “one hundred and twenty-three”), it’s nice to have the option here, since it affects the calculation of .

Also, “everything” is a variable, including the `hyphen`

, `comma`

, `space`

, even the `empty`

string. The addition operator is overloaded for string concatenation, but it’s easy to replace the string constants with, say, integer word counts, or syllable counts, or whatever, without changing the code. For example, we can automatically generate lousy haiku:

One hundred twenty-

one thousand seven hundred

seventy-seven

Coming back to the bank checks, it turns out that , so that the bank can print a check for any amount up to $1,113,322.99. Which doesn’t seem terribly large– I wonder if they fall back to using numerals if the amount doesn’t fit using words?

**Twitter scale**

This isn’t a new problem. A couple of years ago, FiveThirtyEight.com’s Riddler posed a problem involving the Twitter account @CountVonCount, of Sesame Street fame, essentially asking for … and then again asking for just last month, in response to the recent increase in maximum length of a tweet from 140 to 280 characters (in each case, less one character to account for an exclamation point).

These field widths are large enough that a brute force approach is no longer feasible. It’s a nice problem to implement a function to compute efficiently and exactly, with results as shown in the figure below.

All of the source code is available at the usual location here.

]]>In the figure below, a regular hexagon with side length 12 is tiled with “lozenges,” sort of like triangular dominoes, each consisting of a pair of unit-length equilateral triangles joined at a common side.

Each lozenge is in one of three possible orientations. Although the orientations appear random, if we count carefully, we find that there are exactly the same number of lozenges in each of the three orientations. In fact, no matter how we tile the hexagon, it is always the case that the number of lozenges (144 in this case) of each “type” is fixed. Can you prove this?

**Proof without words**

Although the (proof of the) Pythagorean theorem seems to be the most common example of a proof without words— usually a picture or diagram that doesn’t require any words to explain– this problem, with its “proof” below, is probably my favorite.

One reason I like this problem, particularly as motivation for student discussion, is that it is somewhat controversial. Is the above “proof without words” really a satisfactory proof?

**Theorem without words**

Moving away from random tilings for the moment, the original motivation for this post wasn’t actually a *proof* without words, but a *theorem* without words. I recently learned an interesting result I had not seen before, involving no more than high school-level physics, that I thought could be presented as an animation without any accompanying explanation:

Unfortunately, I’m not sure this quite succeeds as a “theorem without words.” As with the tiling proof, there is more going on here than the above animation arguably conveys. In particular, one of the most interesting things about this problem– that the animation *doesn’t* really show– is that the *shape* (that is, the eccentricity) of the ellipse of apogees is invariant: it does not depend on the speed of the projectile, or gravity, or any relationship between the two.

**More on random lozenge tilings**

Finally, some source code: although there are plenty of papers and web sites with images of random tilings like those above, I wanted to make my own images, and working code is harder to find. See here for my Python implementation for generating a random lozenge tiling of a hexagon of a given size. For example, the following figure shows a random tiling of a hexagon with side length 64:

My implementation is quick and dirty in the sense that it simply iterates the Markov chain of single-step up-or-down moves in the corresponding family of non-intersecting lattice paths, essentially shuffling long enough for the resulting random tiling to be *approximately* uniform. (Note that the tilings in the above images are more random than they might appear, despite the “frozen” regions near the corners.) It would be interesting to extend this implementation to use coupling from the past, to generate an *exactly* uniform random tiling.

**References:**

This post was motivated by a recent attempt to transform a photograph into a large digital print, in what I hoped would be a mathematically interesting way. The idea was pretty simple: convert the (originally color) image into a black-on-white line drawing– a white background, with a single, continuous, convoluted black curve, one pixel wide.

This isn’t a new idea. One example of how to do this is to convert the image into a solution of an instance of the traveling salesman problem, with more “cities” clustered in darker regions of the source image. But I wanted to do something slightly different, with more explicitly visible structure… which doesn’t necessarily translate to more visual appeal: draw a Hilbert space-filling curve, but vary the *order* of the curve (roughly, the depth of recursion) locally according to the gray level of the corresponding pixels of the source image.

After some experimenting, I settled on the transformation described in the figure below. Each pixel of the source image is “inflated” to an 8-by-8 block of pixels in the output, with a black pixel (lower left) represented by a second-order Hilbert curve, and a white pixel (upper left) by just a line segment directly connecting the endpoints of the block, with two additional gray levels in between, each connecting progressively more/fewer points along the curve.

**Example**

The figure below shows an example of creating an image for input to the algorithm. There are two challenges to consider:

The 8-fold inflation of each pixel means that if we want the final output to be 1024-by-1024, then the input image must be 128-by-128, as shown here. (For my project, I could afford a 512-by-512 input, with larger than single-pixel steps between points on the curve, yielding output suitable for a 36-inch print.)*Size:*The color in the input image must be quantized to just four gray levels, using your favorite photo editor. (I did this in Mathematica.)**Grayscale:**

The figure below shows the resulting output. You may have to zoom in to see the details of the curve, especially in the black regions.

**Source code**

Following is the Python source that does the transformation. It uses the Hilbert curve encoding module from my allRGB experiment, and I used Pygame for the image formatting so that I could watch the output as it was created.

import hilbert import pygame BLACK = (0, 0, 0, 255) DARK_GRAY = (85, 85, 85, 255) LIGHT_GRAY = (160, 160, 160, 255) WHITE = (255, 255, 255, 255) STEPS = {BLACK: [1] * 15, DARK_GRAY: [1, 1, 1, 1, 4, 1, 1, 5], LIGHT_GRAY: [4, 7, 4], WHITE: [15]} class Halftone: def __init__(self, image, step): self.h = hilbert.Hilbert(2) self.index = -1 self.pos = (0, 0) width, height = image.get_size() self.target = pygame.Surface((step * 4 * width, step * 4 * height)) self.target.fill(WHITE) for pixel in range(width * height): self.move(1, step) for n in STEPS[tuple(image.get_at([w // 4 for w in self.pos]))]: self.move(n, step) def move(self, n, step): self.index = self.index + n next_pos = self.h.encode(self.index) pygame.draw.line(self.target, BLACK, [step * w for w in self.pos], [step * w for w in next_pos]) self.pos = next_pos if __name__ == '__main__': import sys step = int(sys.argv[1]) for filename in sys.argv[2:]: pygame.image.save(Halftone(pygame.image.load(filename), step).target, filename + '.ht.png')]]>