The distribution of “roll and keep” dice

Introduction

Some role-playing games involve a “roll and keep” dice mechanic, where you roll some number of dice, but only “keep” a specified number of them with the largest values, where the result of the roll is the sum of the kept dice. For example, rolling five dice and keeping the largest three (sometimes denoted 5d6k3) could yield a score between 3 and 18, inclusive. What is the probability of each possible score?

The motivation for this post is (1) to capture my notes on the general solution to this problem, and (2) to describe some potential optimizations in implementation to handle very large instances of the problem, such as this Hacker Rank version of the problem, where we may be rolling as many as 10,000 dice.

The solution

To start, let’s simplify the problem by just rolling and keeping all of the dice without discarding any, since the order statistics are what make this problem complicated. Given n dice each with d sides, the number r_{n,d}(s) of equally likely ways to roll a score s is given by

r_{0,0}(s) = 1

r_{n,d}(0) = 0

r_{n,d}(s) = \sum\limits_{k=0}^{\lfloor\frac{s-n}{d}\rfloor} (-1)^k {n \choose k} {{s-k d-1} \choose {n-1}}

so that the probability of rolling a score s is r_{n,d}(s)/d^n.

If we now consider keeping only m \leq n of the dice, then we can group the r_{n,d,m}(s) desired outcomes with score s by:

  • the smallest value a among the m largest dice that we keep,
  • the number b of kept dice that are strictly greater than a, and
  • the number c of discarded dice that are strictly less than a.

This yields the following summation:

r_{n,d,m}(s) = \sum\limits_{a=1}^d \sum\limits_{b=0}^{m-1} \sum\limits_{c=0}^{n-m} {n \choose {b,c}} (a-1)^c r_{b,d-a}(s-a m)

Implementation details

At this point, we’re done… except that if the number n of dice rolled is large, then a straightforward implementation of this nested summation will be pretty slow, involving calculation of a large number of very large multinomial coefficients.

My first thought was to speed up calculation of those very large coefficients via their prime factorization using Legendre’s formula. The paper by Goetgheluck linked at the end of this post describes a more optimized approach, but the following simple Python implementation is already significantly faster than the usual iterative algorithm for sufficiently large inputs (the implementation of primes(n) is left as an exercise for the reader, or a future post):

def binomial(n, k):
    """Large binomial coefficient n choose k."""
    if k < 0 or k > n:
        return 0
    c = 1
    for p in primes(n):
        c = c * p ** (v_fact(n, p) - v_fact(k, p) - v_fact(n - k, p))
    return c

def v_fact(n, p):
    """Largest power of prime p dividing n factorial."""
    v = 0
    while n > 0:
        n = n // p
        v = v + n
    return v

(It’s worth noting that the Hacker Rank problem actually only asks for the number of ways to roll a given total modulo a prime, suggesting that a similar but even more effective optimization using Lucas’s theorem is probably intended.)

But we can do much better than this, by observing that in the “usual” iterative algorithm, all of the intermediate products inside the loop are also binomial coefficients… and we need all of them for this problem. That is, we can rewrite the summation as

r_{n,d,m}(s) = \sum\limits_{a=1}^d \sum\limits_{b=0}^{m-1} {n \choose b} r_{b,d-a}(s-a m) \sum\limits_{c=0}^{n-m} {n-b \choose c} (a-1)^c

and then “embed” the usual iterative calculation of the binomial coefficients inside the nested summations. The following implementation eliminates all but a single explicit calculation of a binomial coefficient:

def rolls_keep(s, n, d, m):
    """Ways to roll sum s with n d-sided dice keeping m largest."""
    result = 0
    for a in range(1, d + 1):
        choose_b = 1
        for b in range(0, m):
            sum_c = 0
            choose_c = 1
            pow_c = 1
            for c in range(0, n - m + 1):
                sum_c = sum_c + choose_c * pow_c
                choose_c = choose_c * (n - b - c) // (c + 1)
                pow_c = pow_c * (a - 1)
            result = result + choose_b * rolls(s - a * m, b, d - a) * sum_c
            choose_b = choose_b * (n - b) // (b + 1)
    return result

def rolls(s, n, d):
    """Ways to roll sum s with n d-sided dice."""
    if s == 0 and n == 0:
        return 1
    if d == 0:
        return 0
    result = 0
    choose_k = 1
    for k in range(0, (s - n) // d + 1):
        result = result + (-1) ** k * choose_k * binomial(s - k * d - 1, n - 1)
        choose_k = choose_k * (n - k) // (k + 1)
    return result

Reference:

  1. Goetgheluck, P., Computing Binomial Coefficients, The American Mathematical Monthly, 94(4) April 1987, p. 360-365 [JSTOR]
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Snap card game probability

Introduction

In the children’s card game Snap, a deck of cards is shuffled and divided as evenly as possible among two or more players. In alternating turns, each player deals a card from her stack onto a face-up pile in front of her. If at any time, the top cards of any two players’ face-up piles have the same rank, the first player to shout “Snap!” takes both face-up piles and adds them to the bottom of her remaining stack. The objective is to accumulate all of the cards.

It is possible for the players to deal through their entire stacks without a single snap, i.e., at no time do the top cards of two piles have the same rank. Let us call such a game boring (a term that may be suggested by, say, your niece to whom you are introducing the game). What is the probability of a boring game of Snap?

One of the reasons I love combinatorics is that it is so easy to ask a hard question. That is, we can pose a series of very similar-sounding problems, all of which are easy to state and understand, but whose minor differences result in major changes in complexity of their solutions. The motivation for this post is to describe a simple children’s card game as an example of this phenomenon.

Frustration Solitaire

Before tackling the actual problem described above, let’s consider a slightly modified version of the two-player game that is easier to analyze:

  1. Instead of dividing a single shuffled deck in half, start with two separate complete shuffled decks, one for each player.
  2. Instead of alternating turns dealing a card from each player’s stack, at each turn both players simultaneously deal a card face-up from their remainder.

Dealing through the entirety of both decks without a snap is effectively equivalent to winning a game of Frustration Solitaire, where a single player deals cards from a single shuffled deck, saying ranks “Ace, two, three, …, king” repeatedly, losing the game if the dealt card ever matches the called rank.

Even within this already-modified context, there are three slight variations that range from very simple to– I think– intractable:

  1. If a snap requires that both cards match in rank and suit, then we are counting derangements, and the probability of a boring game is approximately 1/e, or about 0.367879.
  2. If a snap requires that both cards match in rank only, as in Frustration Solitaire, then we have discussed this problem before here, in the context of a Secret Santa drawing among 13 families each with 4 family members. In this case, the probability of a boring game is approximately 0.0162327.
  3. If a snap requires that both cards match in rank or in suit… well, although this is still effectively a problem of counting permutations with restricted positions, I think this problem is much harder in practice, since the board of restricted positions can’t be nicely decomposed into sub-boards with no rows or columns in common.

War

Let’s consider one more modified version of the game, this time actually rolling back one of the changes above: let’s return to playing with a single shuffled deck divided evenly among the two players, but retain the change where the players deal simultaneously from their stacks.

This version of the game has a lot of structure in common with another children’s card game, War. In this case, a boring game of Snap corresponds to War with no war– that is, dealing through both deck halves without any pair of cards matching in rank.

This is a relatively straightforward inclusion-exclusion problem; with a deck with r=13 ranks and s=4 suits, the probability of a boring game is

\frac{1}{(r s)!} \sum\limits_{j=0}^n (-1)^j {n \choose j} j!(r s-2j)! [x^j]g(x)^r

where

n = \lfloor\frac{r s}{2}\rfloor

g(x) = \sum\limits_{k=0}^{s/2} \frac{s!}{k!(s-2k)!} x^k

which for a standard 52-card deck yields a probability of a boring game of about 0.210214.

Counting “words” with prohibited subwords

Coming back finally to the original rules of Snap, counting boring games is complicated by the fact that the players alternate turning individual cards face-up, rather than simultaneously revealing a pair of cards at a time, so that a snap may “start” with either the first or the second player’s deal. Intuitively, consider skipping the initial separation of the deck into halves, and simply deal cards one at a time from the single shuffled deck; a boring game is one in which no two consecutively dealt cards are the same rank.

There is a wonderful paper by Jair Taylor (referenced below) describing a very general but more sophisticated technique for counting arrangements with these types of restrictions. Applying this technique to Snap, the probability of a boring game using a deck with r ranks and s suits is

\frac{(s!)^r}{(r s)!} \int_{t=0}^{\infty} e^{-t} g_s(t)^r dt

where

g_k(t) = (-1)^k \sum\limits_{i=0}^k (-1)^i {k-1 \choose k-i} \frac{t^i}{i!}

yielding a probability of approximately 0.0454763, or about once every 22 shuffles, that a game of Snap will be boring.

Once again, however, it’s very easy to make this problem much harder: if we allow a snap to involve adjacent cards matching in rank or in suit, what is the resulting probability of a boring game? What if there are more than two players?

Reference:

  1. Taylor, J., Counting words with Laguerre series [arXiv]
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Poker puzzle

A variant of this problem came up in discussion recently, that I think lends itself to attack via either mathematical “pencil and paper” or “write a program” analysis:

What is the median poker hand? More precisely, among all {52 \choose 5}=2598960 possible five-card poker hands, what hand ranks higher than exactly half of them?

Of course, the solution is possibly not quite unique, since the ranking of hands is a weak order: suits never decide the relative ranking of hands. For example, there are four possible royal flushes, and no one royal flush beats another. So, as a follow-up combinatorics problem: how many “distinct” hand ranks are there? That is, how many equivalence classes are there in the incomparability relation “hand x ties with hand y” on the set of possible hands?

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Calling Java from MATLAB with pass-by-reference arrays

MATLAB has a pretty intimate connection with Java, supporting the creation and manipulation of native Java objects directly in MATLAB source code. For example:

x = java.lang.Double(pi);
disp(x.isInfinite());

Contrast this with C++, where it is typically necessary to write an additional small amount of “wrapper” C code using the MEX API to bridge the gap between languages.

For the most part, conversions between MATLAB and Java data types are automatic, as expected, and as desired… with one exception: passing a primitive array by reference. For example, suppose that we want to call a static method in a third-party Java library that converts an input 6-state vector (as a primitive array of 6 doubles) from one coordinate frame to another, by populating a given “output” array of 6 doubles:

public static int convert(double[] x_in, double[] x_out) { ... }

As far as I can tell, it is impossible to use a method like this directly from MATLAB. The problem is that we can never construct the necessary output buffer x_out, and hold onto it as a reference that could be passed to convert. MATLAB insists on automatically converting any such reference to a Java array of primitive type into its “value” as a MATLAB array of the corresponding native type, leaving the reference to the actual “buffer” behind in the Java address space.

Interestingly, MATLAB’s built-in function javaArray seems like it might have been intended for situations like this. But the arrays created by this built-in function are not of the necessary primitive data types, but of their “boxed” counterparts, which seems almost entirely useless to me, since I don’t know of any Java libraries whose interface involves arrays of boxed primitive types.

So, it seems that we need at least some additional wrapper Java code to make methods like this usefully accessible from MATLAB. The objective of this post is to provide an implementation that I think fills this gap in a reasonably general way, by using Java’s reflection mechanism to allow creating primitive arrays and passing them by reference to any Java method directly from MATLAB.

(Aside: I suppose I get some masochistic enjoyment from repeatedly using the phrase “pass by reference” here, hoping for pedantic complaints arguing that “Java is pass by value, not pass by reference.”)

The code is available on GitHub, as well as the old location here. Following is a simple example showing how it works:

s = java.lang.String('foo');
dst = matlab.JavaArray('java.lang.Character', 3);
callJava('getChars', s, int32(0), int32(3), dst, int32(0));
dst.get()

The idea is similar to the libpointer and calllib interface to C shared libraries: a matlab.JavaArray acts like a libpointer, and callJava acts like calllib, automatically “unwrapping” any matlab.JavaArray arguments into their underlying primitive array references.

Finally, a disclaimer: the method dispatch is perhaps overly simple, keying only on the desired method name and provided number of arguments. If your Java class has multiple methods with the same name, and the same number of parameters– regardless of their type, or of the return type– the first method matching these two criteria will be called.

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How to tell time using GPS, and the recent Collins glitch

Introduction

Last week on Sunday 9 June, hundreds of airplanes experienced a failure mode with some variants of the Collins Aerospace GPS-4000S sensor, grounding many business jets and even causing some delays among regional airlines.

Following is the initial description of the problem from Collins, as I first saw it quoted in Aviation International News:

“The root cause is a software design error that misinterprets GPS time updates. A ‘leap second’ event occurs once every 2.5 years within the U.S. Government GPS satellite almanac update. Our GPS-4000S (P/N 822-2189-100) and GLU-2100 (P/N 822-2532-100) software’s timing calculations have reacted to this leap second by not tracking satellites upon power-up and subsequently failing. A regularly scheduled almanac update with this ‘leap second’ was distributed by the U.S. government on 0:00 GMT Sunday, June 9, 2019, and the failures began to occur after this event.”

This sounded very suspicious to me… but suspicious in a mathematically interesting way, hence the motivation for this post.

The suggestion, re-interpreted and re-propagated in subsequent aviation news articles and comment threads, seems to be that the gummint introduced a “leap second event” in the recent GPS almanac update, and the Collins receivers weren’t expecting it. The “almanac” is a periodic, low-throughput message broadcast by the satellites in the GPS constellation, that newly powered-on receivers can use to get an initial “rough idea” of where they are and what time it is.

It is true that one of the fields in the almanac is a count of the number of leap seconds added to UTC time since the GPS epoch, that is, since the GPS “clock started ticking” on 6 January 1980 at 00:00:00 UTC. So what’s a leap second? Briefly, the earth’s rotation is slowing down, and so to keep our UTC clocks reasonably consistent with another astronomical time reference known as UT1, we periodically “delay” the advance of UTC time by introducing a leap second, which is an extra 61st second of the last minute of a month, historically either at the end of June or the end of December.

There have been 18 leap seconds added to UTC since 1980… but leap seconds are scheduled months in advance, and it has already been announced that there will not be a leap second at the end of this month, and there certainly wasn’t a leap second added this past June 9th.

So what really happened last week? The remainder of this post is purely speculation; I have no affiliation with Collins Aerospace nor any of its competitors, so I don’t have any knowledge of the actual software whose design was reported to be in error. This is just guesswork from a mathematician with an interest in aviation.

GPS week number roll-over

The Global Positioning System has an interesting problem: it’s hard to figure out what time it is. That is, if your GPS receiver wants to learn not just its location, but also the current time, without relying on or comparing with any external clock, then that is somewhere between impossible and difficult, depending on how fancy we want to get. The only “timestamp” that is broadcast by the satellites is a week number– indicating the integer number of weeks elapsed since the GPS epoch on 6 January 1980– and a number of seconds elapsed within the current week.

The problem is that the message field for the week number is only 10 bits long, meaning that we can only encode week 0 through week 1023. After that, on “week 1024,” the odometer rolls over, so to speak, back to indicating week 0 again.

This has already happened twice: the first roll-over was between 21 and 22 August 1999, and the second was just a couple of months ago, between 6 and 7 April 2019. An old GPS receiver whose software had not been updated to account for these roll-overs might show the time transition from 21 August 1999 “back” to 6 January 1980, for example.

It’s worth noting that those roll-overs didn’t actually occur exactly at midnight… or at least, not at midnight UTC. The GPS “clock” does not include leap seconds, but instead ticks happily along, always exactly 60x60x24x7=604,800 seconds per week. So, for example, GPS time is currently “ahead” of UTC time by 18 seconds, corresponding to the 18 leap seconds that have contributed to the “slowing” of the advance of the UTC clock. The GPS week number most recently rolled over on 6 April, not at midnight, but at 23:59:42 UTC.

Using the leap second count

It turns out that we could modify our GPS receiver software to extend its ability to tell time beyond a single 1024-week cycle, by using the leap second count field together with the rolling week number. The idea is that by predicting the addition of more leap seconds at reasonably regular intervals in the future, we can use the week number to determine the time within a 1024-week cycle, and use the leap second count to determine which 1024-week cycle we are in.

This is not a new idea; there is a good description of the approach here, in the form of a patent application by Trimble, a popular manufacturer of GPS receivers. Given the current week number 0 \leq W < 1024 and the leap second count L in the almanac, the suggested formula for the “absolute” number of weeks W' since GPS epoch is given by

W' = t + ((W-t+512)\mod 1024) - 512

t = \lfloor 84.56 + 70.535L \rfloor

where the intermediate value t is essentially an estimate of the week number obtained by a linear fit against the historical rate of introduction of leap seconds.

This formula was proposed in 1996, and it would indeed have worked well past the 1999 roll-over… but although it was predicted to “provide a solution to the GPS rollover problem for about 173 years,” unfortunately it would really only have lasted for about 12 years, first yielding an incorrect time in week 1654 on 18 September 2011.

The problem is shown in the figure below, indicating the number of leap seconds that have been added over time since the GPS epoch in 1980, with the red bars indicating the “week zero” epoch and roll-overs up to this point:

Leap seconds added to UTC time since GPS epoch, with GPS epoch and 1024-week roll-overs shown in red.

Right after the first roll-over in 1999, the reasonably regular introduction of leap seconds stopped, and even once they started to become “regular” again, they were regular at a lesser rate (although still more frequently than the “2.5 years” suggested by the Collins report).

Conclusion

Could something like this be the cause of this past week’s sensor failures? It’s certainly possible: it’s a relatively simple programming exercise to search for different linear fit coefficients in the above formula– a variant of which might have been used on these Collins receivers– that

  1. Yield the correct absolute week number for a longer time than the above formula, including continuing past the second roll-over this past April; but
  2. Yield the incorrect absolute week number, for the first time, on 9 June (i.e., the start of week 2057).

Such coefficients aren’t hard to find; for example, the reader can verify that the following satisfies the above criteria:

t = \lfloor -291 + 102L \rfloor

which corresponds to an estimate of one additional leap second approximately every 2 years.

Edit: See Steve Allen’s comment (and my reply) for a description of what I think is probably a more likely root cause of the problem– still related to interpreting the combination of week number roll-over and leap second occurrences, but with a slightly different failure mode.

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A potential exploit of a Mountain Dew promotion

Introduction

This past Monday marked the start of a 10-week promotion where you can buy bottles of Mountain Dew, each with a label showing one of the 50 United States. Collect all 50 state labels, and win $100 (in the form of a prepaid Visa card).

How many bottles of Mountain Dew should you expect to have to buy to win one of these $100 gift cards? The motivation for this post is to discuss this promotion as an instance of the classic coupon collector’s problem… but with a couple of interesting wrinkles, one of which appears to make this promotion vulnerable to exploitation by a determined participant, with an opportunity to net a significant positive return with almost no risk.

Group drawings: buying six-packs

The first wrinkle is to suppose that we plan to buy our bottles of Mountain Dew in six-packs, instead of one bottle at a time. Buying in bulk reduces the price per bottle: an individual 16.9-ounce bottle is easily a dollar or more, while I can find six-packs in my neighborhood for about $3.48, or 58 cents per bottle. I will assume this price for the remainder of this discussion.

So how many six-packs should we expect to have to buy to collect a complete set of 50 state labels? More generally, let m=6 be the number of bottles in each pack, where each bottle has a label chosen independently and uniformly from s=50 possible states, and let X_1 be the random variable indicating the number of packs we must buy until we first collect at least one of every state label. What is the distribution of X_1?

Stadje refers to this as the coupon collector’s problem “with group drawings” (see references at the end of this post). However, he addresses a slightly different variant, where the labels in a six-pack must be distinct, that is, sampled without replacement from the 50 possible labels (similar to the NFL drug testing procedure discussed in a recent post). In our case, each individual bottle’s label is an independent sample, so that a six-pack may contain duplicates. Inspection of a few six-packs at my local grocery store verified that this is indeed the more appropriate model for this problem.

The probability that we have collected all state labels after buying n packs is given by

P(X_1 \leq n) = \sum\limits_{k=0}^s (-1)^k {s \choose k} (1-\frac{k}{s})^{mn}

It’s a nice exercise to show that, from this, we can derive the expected number of packs as

E[X_1] = -\sum\limits_{k=1}^s (-1)^k {s \choose k} \frac{1}{1-(1-k/s)^m}

which evaluated for m=6 and s=50 yields an average of about 37.91 six-packs, or about $131.92. So if we buy six-packs until we collect all 50 state labels, thus winning $100, we should expect to lose about $31.92 on average as a result of our venture, with a probability of only about 0.16 that we manage to net any positive return. Even worse, there is no upper bound on how much we might lose in that remaining 84% case. We will fix this shortly.

The Double Dixie cup problem: decreasing risk

The much more interesting wrinkle is that the terms of the promotion allow a single participant to win more than once: up to five times during the 10-week period, collecting a total of $500 worth of gift cards. This helps us a lot, because although we expect to need to buy about 38 six-packs to collect the first set of 50 state labels, we typically need less than half as many additional packs to collect the second and subsequent complete sets of labels.

Newman refers to this as the “double Dixie cup problem” (see references below). Let’s extend the notation described above, and define the random variable X_d to indicate the number of packs that we must buy to collect d complete sets of state labels. Then the cumulative distribution for X_d is given by

P(X_d \leq n) = \frac{1}{s^{mn}} [\frac{x^{mn}}{{mn}!}][(e^x - \sum\limits_{k=0}^{d-1} \frac{x^k}{k!})^s]

Now, let’s suppose that we start buying six-packs, trying to collect d=5 complete sets of bottle labels, thus winning $500. At $3.48 per six-pack, we will lose money only if we have to buy more than 143 six-packs.

Evaluating 1-P(X_5 \leq 143) yields a probability of less than 0.008 that we will lose money in our search for five complete sets of labels. We can bound the possible amount lost by stopping at 144 six-packs, no matter what, and collecting $100 gift cards for however many complete sets of bottle labels we have managed to collect at that point. The figure below shows the resulting distribution of possible net returns, with the expected return of $167.53 shown in red.

Probability distribution of net return from strategy of buying six-packs until collecting 5 complete sets, or 144 packs, whichever comes first.

Open questions

A 99.2679% chance of making money, with an expected return of $167.53, sounds like a pretty good wager. But even this may not be the best we could possibly do. Although there is certainly no advantage to buying any more than 144 six-packs in total, it may be advantageous to stop before that point, even without having collected all five complete sets of bottle labels, if the expected additional gain from continuing to buy packs is negative. Such an optimal stopping strategy would likely be very complex, but it would be interesting to investigate how much the above approach could be improved.

Finally, I have already sunk about $300 into a bunch of Skittles that I didn’t eat; I have no plans to invest in another experiment buying gallons of Mountain Dew that I won’t drink. But there is a good reason why I’m more reluctant in this case: all of the above analysis assumes that each of the 50 states are equally likely to appear on any bottle. Perhaps they truly are uniformly distributed– the promotion is intended to “celebrate what makes every state epic,” after all, so any state that is more scarce than it should be would feel justifiably slighted.

We made the same uniformity assumption in the recent Skittles experiment, which turned out to be mostly correct. But in that case, any departure from uniformity would only have helped us, making it more likely to find an identical pair of packs. In this case, however, a non-uniform distribution of state labels makes it harder to collect complete sets.

References:

  1. Newman, Donald J., The Double Dixie Cup Problem, The American Mathematical Monthly, 67(1) January 1960, p. 58-61 [JSTOR]
  2. Stadje, Wolfgang, The Collector’s Problem with Group Drawings, Advances in Applied Probability, 22(4) December 1990, p. 866-882 [JSTOR]
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Tracking heap memory use in Windows

Introduction

I recently encountered a problem with a C++ program that was allocating more memory than it should. It wasn’t leaking memory– that is, the program was well-behaved in the sense of eventually releasing every byte of memory that it allocated. But it allocated a lot, enough so that running multiple instances of the program simultaneously on cluster nodes would exceed physical memory capacity.

It should be relatively easy to troubleshoot memory issues like this. However, let’s suppose that our application runs on Windows, compiled in Visual Studio, and for mostly stupid reasons, the built-in Diagnostic Tools are unavailable, and we have neither administrative privilege nor support to install any of the other various existing tools for doing this sort of thing. How can we reinvent this wheel with a minimal amount of effort?

This turned out to be an interesting exercise, with a relatively simple implementation that solved the problem at hand, but also raised some questions that I don’t have good answers for.

Implementation

The result is available here, as well as on GitHub. It’s pretty easy to use: there is no need to modify application source code at all, just compile and link with mem_log.cpp in a debug build. There is only one function exposed in the corresponding header: mem::print_log(std::ostream&), which you can call yourself if desired, but it will be called automatically at program termination, printing a log of heap memory usage to stdout. For example, the following program:

int main()
{
    for (int i = 10; i > 0; --i)
    {
        int *a = new int[i];
        delete[] a;
    }
}

yields the following output:

                    Heap            Freed      Max. Alloc.
==========================================================
Caller: c:\users\username\test_mem\test_mem.cpp(5):main
Blocks:                0               10                1
Bytes:                 0              220               40

The first column indicates the current (and in the usual case, final) state of the heap, as the number of allocated blocks (i.e., calls to operator new) and corresponding total number of bytes. Non-zero values here suggest a memory leak… although more on this shortly.

The second column indicates the number of previously allocated blocks and bytes released by eventual calls to operator delete. Finally, the third column is the one that helped solve this problem, indicating the maximum number of bytes (and corresponding blocks) allocated at any one time.

There is one tricky note: in general there will be multiple records in the log, one for each line of application source code that calls operator new. Such a call may not be explicit as in the above example, but instead perhaps an indirect result of, say, std::vector::push_back, for example. We want to identify this call with the application’s code, not the standard library’s. We do this by walking the stack trace, starting from the globally replaced implementation of operator new, inspecting each corresponding source code filename until the first one is found containing the case-sensitive prefix specified by the preprocessor definition MEM_LOG_PATH (whose default is c:\users). If your application source code lives somewhere else, define MEM_LOG_PATH accordingly.

Limitations and questions

Although this code helped to solve this particular problem, it is still of limited use:

  1. It only works on Windows. I have indicated in source code comments where the Windows-specific stuff is, but have made no effort to implement the corresponding Linux calls to backtrace, etc.
  2. It is not thread-safe.
  3. It is not a leak detector. I only replaced operator new and delete, which is all that we really need to track memory usage in a well-behaved program that doesn’t care about alignment. We would also need to replace the array forms new[] and delete[] (which otherwise call our non-array implementations) to properly check for common undefined-behavior errors like freeing an array allocated via new[] with the non-array delete.
  4. Along the same lines as (3), I didn’t bother with the arguably much harder problem of tracking application memory usage via std::malloc and std::free.

Finally, the special, automagic global replacement behavior of implementing our own operator new and delete presents an interesting puzzle that I’m not sure how to solve, or whether it is even solvable. First, consider that we want to track every memory allocation and release, no matter how early (or late) it occurs during program execution. Since that tracking involves manipulation of our own internal data structure, we need to ensure that our internal object gets fully constructed– and stays constructed– by the time we need it to record any such unpredictably timed allocation.

One easy way to do this is with the construct on first use idiom: we instantiate our internal object as a function static variable, allocated on the heap the first time we need it, and never released. This intentionally “leaks” our object, but (1) so what? And (2) we have to do this, since there is no guarantee– as far as I can tell– that any particular non-trivial static object’s lifetime would persist beyond the last call to the allocation operators we are trying to track.

This works, in the sense that we can confidently track all heap allocations and releases, no matter when they occur… but now consider the final automatic call to mem::print_log(std::cout), which is realized in the destructor of a “dummy” static object whose only purpose is to try to be the last thing that happens in the program. But because of the static initialization order fiasco, there is still a possibility that we might “miss” a subsequent heap allocation or release during some later static object’s destruction. (That is, although we will track it, we won’t see it in the final printed log.) Is there any way to fix this? I think the answer is no, but I’m not sure.

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