## Uncertainty in trading passengers for fuel

I had an interesting experience recently while preparing for a flight from Los Angeles to Baltimore. It was a completely full flight– initially, at least– with myself and 174 other passengers who had already boarded the Southwest 737-800, seemingly ready to push back and get on our way.

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:

Distribution of total weight of 175 passengers on a Southwest Boeing 737-800.

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]
Posted in Uncategorized | 4 Comments

## An urn puzzle

It has been a while again since I last posted a puzzle, so…

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 $b$ and $w$ 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?

Posted in Uncategorized | 2 Comments

## On average, we die a decade earlier than expected

“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 $X$ with the probability density shown in the following figure.

Probability distributions of age at death based on the United States period life table for 2014.

The expected lifespan of 76.5 years is $E[X]$ (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 $E[X | X \geq 40]$?

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 $E[X | X \geq x]$, as well as the corresponding expected additional lifespan $E[X-x | X \geq x]$, as a function of current age $x$.

Conditional expected age at death and expected additional lifespan, vs. 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 $E[X-x | X \geq x]$ 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:

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

Following are the probabilities $P(\left\lfloor{X}\right\rfloor = x)$ 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

Posted in Uncategorized | 1 Comment

## How many melodies are there?

Introduction

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.).

Twelve notes of the chromatic scale, and a thirteenth allowing melodies with an octave jump.

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

$g_n(x) = n(x+x^2+x^4+x^8+x^{16})$

and all possible rests with

$h(x) = \frac{x}{1-x}$

Then the generating function for the number of possible melodies is

$f_n(x) = \frac{1+h(x)}{1-g_n(x)(1+h(x))}$

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 $[x^{16}]f_{13}(x)$ 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

$[x^{16}]f_{13}(x) - [x^{16}]f_{12}(x) + 1$

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.

## Picking a perfect NCAA bracket: 2018 was the most unlikely tournament so far

Introduction

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

$p_{i,j} = 1-p_{j,i} = \frac{1}{2}+k(s_i-s_j)$

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

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

$s_i = \Phi^{-1}(1-\frac{4i}{n})$

where $\Phi$ is the quantile of the normal distribution, and $n=351$ 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.

Normally-distributed strength s_i for each seed.

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, as well as on GitHub.

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:

1. If we flip a coin to pick each game, what is the probability of picking every game correctly?
2. 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?
3. 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 $2^{63}$, 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.

Probability of a perfect bracket, 1985-2018.

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:

1. Schwertman, N., McCready, T., and Howard, L., Probability Models for the NCAA Regional Basketball Tournaments, The American Statistician45(1) February 1991, p. 35-38 [JSTOR]
Posted in Uncategorized | 2 Comments

## Whodunit logic puzzle

You are a detective investigating a robbery, with five suspects having made the following statements:

• 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).

Posted in Uncategorized | 3 Comments

## Coupling from the past

Introduction

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 $\pi$ over a finite discrete space $S$, and that we have an ergodic Markov chain whose steady state distribution is $\pi$. 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:

1. 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 $\pi$.
2. 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 $f:S \times [0,1) \to S$, so that for a random variable $U$ uniformly distributed on the unit interval,

$\forall i,j\ P(f(s_i, U)=s_j) = p_{i,j}$

In other words, given a current state $X_t$ and random draw $U_t$, the next state is $X_{t+1}=f(X_t, U_t)$.

Now consider a single infinite sequence of random draws $(u_1, u_2, u_3, \ldots)$. We will use this same single source of randomness to iterate multiple realizations of the chain, one for each of the $|S|$ 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 $s_i$ and $s_j$ at time $-n$, so that

$X_{-n}=s_i, Y_{-n}=s_j$

$X_{-(n-1)}=f(X_{-n}, u_n), Y_{-(n-1)}=f(Y_{-n}, u_n)$

$\cdots$

$X_{-1}=f(X_{-2}, u_2), Y_{-1}=f(Y_{-2}, u_2)$

$X_0=f(X_{-1}, u_1), Y_0=f(Y_{-1}, u_1)$

(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 $X_t=Y_t$ 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 $X_0=Y_0$. Furthermore, if the states at time zero are all the same for all possible initial states run forward from time $-n$, then the distribution of that final state $X_0$ 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 $2n$ time steps instead of $n$, 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

1. Choose a sufficiently large $n$ to look far enough into the past.
2. Run $|S|$ realizations of the chain, one for each initial state, all starting at time $-n$ and running up to time zero.
3. As long as the final state $X_0$ is the same for all initial states, output $X_0$ as the sample. Otherwise, look farther back in the past ($n \leftarrow 2n$), 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 $\preceq$ on the state space– with minimum and maximum elements– that is preserved by the update function $f$. That is, suppose that for all states $s_i,s_j \in S$ and random draws $u$, if $s_i \preceq s_j$, then $f(s_i, u) \preceq f(s_j,u)$.

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."""
while True:
s0, s1 = mc.min_max()
s0.update(u)
s1.update(u)
if s0 == s1:
break
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.

Random draws in the order they are generated, vs. the order in which they are used in state transitions.

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 $p$. (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."""
while True:
s0, s1 = mc.min_max()
rng_next = None
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