# Searching for RH Counterexamples — Exploring Data

We’re ironically searching for counterexamples to the Riemann Hypothesis.

In the last article we added a menagerie of “production readiness” features like continuous integration tooling (automating test running and static analysis), alerting, and a simple deployment automation. Then I let it loose on AWS, got extremely busy with buying a house, forgot about this program for a few weeks (no alerts means it worked flawlessly!), and then saw my AWS bill.

So I copied the database off AWS using pg_dump (piped to gzip), terminated the instances, and inspected the results. A copy of the database is here. You may need git-lfs to clone it. If I wanted to start it back up again, I could spin them back up, and use gunzip | psql to restore the database, and it would start back up from where it left off. A nice benefit of all the software engineering work done thus far.

This article will summarize some of the data, show plots, and try out some exploratory data analysis techniques.

## Summary

We stopped the search mid-way through the set of numbers with 136 prime divisors.

The largest number processed was

1255923956750926940807079376257388805204
00410625719434151527143279285143764977392
49474111379646103102793414829651500824447
17178682617437033476033026987651806835743
3694669721205424205654368862231754214894
07691711699791787732382878164959602478352
11435434547040000

Which in factored form is the product of these terms

  2^8   3^7   5^4   7^4  11^3  13^3  17^2  19^2  23^2  29^2
31^2  37^2  41^2  43^1  47^1  53^1  59^1  61^1  67^1  71^1
73^1  79^1  83^1  89^1  97^1 101^1 103^1 107^1 109^1 113^1
127^1 131^1 137^1 139^1 149^1 151^1 157^1 163^1 167^1 173^1
179^1 181^1 191^1 193^1 197^1 199^1 211^1 223^1 227^1 229^1
233^1 239^1 241^1 251^1 257^1 263^1 269^1 271^1 277^1 281^1
283^1 293^1 307^1 311^1 313^1 317^1 331^1 337^1 347^1 349^1
353^1 359^1 367^1 373^1 379^1 383^1 389^1 397^1 401^1 409^1
419^1 421^1 431^1 433^1 439^1 443^1 449^1 457^1 461^1 463^1
467^1 479^1 487^1 491^1 499^1 503^1 509^1 521^1 523^1 541^1
547^1 557^1 563^1 569^1 571^1 577^1


The best witness—the number with the largest witness value—was

38824169178385494306912668979787078930475
9208283469279319659854547822438432284497
11994812030251439907246255647505123032869
03750131483244222351596015366602420554736
87070007801035106854341150889235475446938
52188272225341139870856016797627204990720000

which has witness value 1.7707954880001586, which is still significantly smaller than the needed 1.782 to disprove RH.

The factored form of the best witness is

 2^11   3^7   5^4   7^3  11^3  13^2  17^2  19^2  23^2  29^2
31^2  37^1  41^1  43^1  47^1  53^1  59^1  61^1  67^1  71^1
73^1  79^1  83^1  89^1  97^1 101^1 103^1 107^1 109^1 113^1
127^1 131^1 137^1 139^1 149^1 151^1 157^1 163^1 167^1 173^1
179^1 181^1 191^1 193^1 197^1 199^1 211^1 223^1 227^1 229^1
233^1 239^1 241^1 251^1 257^1 263^1 269^1 271^1 277^1 281^1
283^1 293^1 307^1 311^1 313^1 317^1 331^1 337^1 347^1 349^1
353^1 359^1 367^1 373^1 379^1 383^1 389^1 397^1 401^1 409^1
419^1 421^1 431^1 433^1 439^1 443^1 449^1 457^1 461^1 463^1
467^1 479^1 487^1 491^1 499^1 503^1 509^1 521^1 523^1 541^1
547^1 557^1 563^1


The average search block took 4m15s to compute, while the max took 7m7s and the min took 36s.

The search ran for about 55 days (hiccups included), starting at 2021-03-05 05:47:53 and stopping at 2021-04-28 15:06:25. The total AWS bill—including development, and periods where the application was broken but the instances still running, and including instances I wasn’t using but forgot to turn off—was $380.25. When the application was running at its peak, the bill worked out to about$100/month, though I think I could get it much lower by deploying fewer instances, after we made the performance optimizations that reduced the need for resource-heavy instances. There is also the possibility of using something that integrates more tightly with AWS, such as serverless jobs for the cleanup, generate, and process worker jobs.

## Plots

When in doubt, plot it out. I started by writing an export function to get the data into a simpler CSV, which for each $n$ only stored $\log(n)$ and the witness value.

I did this once for the final computation results. I’ll call this the “small” database because it only contains the largest witness value in each block. I did it again for an earlier version of the database before we introduced optimizations (I’ll call this the “large” database), which had all witness values for all superabundant numbers processed up to 80 prime factors.. The small database was only a few dozen megabytes in size, and the large database was ~40 GiB, so I had to use postgres cursors to avoid loading the large database into memory. Moreover, then generated CSV was about 8 GiB in size, and so it required a few extra steps to sort it, and get it into a format that could be plotted in a reasonable amount of time.

First, using GNU sort to sort the file by the first column, $\log(n)$

sort -t , -n -k 1 divisor_sums.csv -o divisor_sums_sorted.csv


Then, I needed to do some simple operations on massive CSV files, including computing a cumulative max, and filtering down to a subset of rows that are sufficient for plotting. After trying to use pandas and vaex, I realized that the old awk command line tool would be great at this job. So I wrote a simple awk script to process the data, and compute data used for the cumulative max witness value plots below.

Then finally we can use vaex to create two plots. The first is a heatmap of witness value counts. The second is a plot of the cumulative max witness value. For the large database:

And for the small database

Note, the two ridges disagree slightly (the large database shows a longer flat line than the small database for the same range), because of the way that the superabundant enumeration doesn’t go in increasing order of $n$. So larger witness values in the range 400-500 are found later.

## Estimating the max witness value growth rate

The next obvious question is whether we can fit the curves above to provide an estimate of how far we might have to look to find the first witness value that exceeds the desired 1.782 threshold. Of course, this will obviously depend on the appropriateness of the underlying model.

A simple first guess would be split between two options: the real data is asymptotic like $a + b / x$ approaching some number less than 1.782 (and hence this approach cannot disprove RH), or the real data grows slowly (perhaps doubly-logarithmic) like $a + b \log \log x$, but eventually surpasses 1.782 (and RH is false). For both cases, we can use scipy’s curve fitting routine as in this pull request.

For the large database (roughly using log n < 400 since that’s when the curve flatlines due to the enumeration order), we get a reciprocal fit of

$\displaystyle f(x) \approx 1.77579122 – 2.72527824 / x$

and a logarithmic fit of

$\displaystyle f(x) \approx 1.65074314 + 0.06642373 \log(\log(x))$

The estimated asymptote is around 1.7757 in the first case, and the second case estimates we’d find an RH counterexample at around $log(n) = 1359$.

For the small database of only sufficiently large witness values, this time going up to about $log(n) \approx 575$, the asymptotic approximation is

$\displaystyle 1.77481154 -2.31226382 / x$

And the logarithmic approximation is

$\displaystyle 1.70825262 + 0.03390312 \log(\log(x))$

Now the asymptote is slightly lower, at 1.7748, and the logarithmic model approximates the counterexample can be found at approximately $\log(n) = 6663$.

Both of the logarithmic approximations suggest that if we want to find an RH counterexample, we would need to look at numbers with thousands of prime factors. The first estimate puts a counterexample at about $2^{1960}$, the second at $2^{9612}$, so let’s say between 1k and 10k prime factors.

Luckily, we can actually jump forward in the superabundant enumeration to exactly the set of candidates with $m$ prime factors. So it might make sense to jump ahead to, say, 5k prime factors and search in that region. However, the size of a level set of the superabundant enumeration still grows exponentially in $m$. Perhaps we should (heuristically) narrow down the search space by looking for patterns in the distribution of prime factors for the best witness values we’ve found thus far. Perhaps the values of $n$ with the best witness values tend to have a certain concentration of prime factors.

## Exploring prime factorizations

At first, my thought was to take the largest witness values, look at their prime factorizations, and try to see a pattern when compared to smaller witness values. However, other than the obvious fact that the larger witness values correspond to larger numbers (more and larger prime factors), I didn’t see an obvious pattern from squinting at plots.

To go in a completely different direction, I wanted to try out the UMAP software package, a very nice and mathematically sophisticated for high dimensional data visualization. It’s properly termed a dimensionality reduction technique, meaning it takes as input a high-dimensional set of data, and produces as output a low-dimensional embedding of that data that tries to maintain the same shape as the input, where “shape” is in the sense of a certain Riemannian metric inferred from the high dimensional data. If there is structure among the prime factorizations, then UMAP should plot a pretty picture, and perhaps that will suggest some clearer approach.

To apply this to the RH witness value dataset, we can take each pair $(n, \sigma(n)/(n \log \log n))$, and associate that with a new (high dimensional) data point corresponding to the witness value paired with the number’s prime factorization

$\displaystyle (\sigma(n)/(n \log \log n), k_1, k_2, \dots, k_d)$,

where $n = 2^{k_1} 3^{k_2} 5^{k_3} \dots p_d^{k_d}$, with zero-exponents included so that all points have the same dimension. This pull request adds the ability to factorize and export the witness values to a CSV file as specified, and this pull request adds the CSV data (using git-lfs), along with the script to run UMAP, the resulting plots shown below, and the saved embeddings as .npy files (numpy arrays).

When we do nothing special to the data and run it through UMAP we see this plot.

It looks cool, but if you stare at it for long enough (and if you zoom in when you generate the plot yourself in matplotlib) you can convince yourself that it’s not finding much useful structure. The red dots dominate (lower witness values) and the blue dots are kind of spread haphazardly throughout the red regions. The “ridges” along the chart are probably due to how the superabundant enumeration skips lots of numbers, and that’s why it thins out on one end: the thinning out corresponds to fewer numbers processed that are that large since the enumeration is not uniform.

It also seemed like there is too much data. The plot above has some 80k points on it. After filtering down to just those points whose witness values are bigger than 1.769, we get a more manageable plot.

This is a bit more reasonable. You can see a stripe of blue dots going through the middle of the plot.

Before figuring out how that blue ridge might relate to the prime factor patterns, let’s take this a few steps further. Typically in machine learning contexts, it helps to normalize your data, i.e., to transform each input dimension into a standard Z-score with respect to the set of values seen in that dimension, subtracting the mean and dividing by the standard deviation. Since the witness values are so close to each other, they’re a good candidate for such normalization. Here’s what UMAP plots when we normalize the witness value column only.

Now this is a bit more interesting! Here the colormap on the right is in units of standard deviation of witness values. You can see a definite bluest region, and it appears that the data is organized into long brushstrokes, where the witness values increase as you move from one end of the stroke to the other. At worst, this suggests that the dataset has structure that a learning algorithm could discover.

Going even one step further, what if we normalize all the columns? Well, it’s not as interesting.

If you zoom in, you can see that the same sort of “brushstroke” idea is occurring here too, with blue on one end and red on the other. It’s just harder to see.

We would like to study the prettiest picture and see if we can determine what pattern of prime numbers the blue region has, if any. The embedding files are stored on github, and I put up (one version of) the UMAP visualization as an interactive plot via this pull request.

I’ve been sitting on this draft for a while, and while this article didn’t make a ton of headway, the pictures will have to do while I’m still dealing with my new home purchase.

Some ideas for next steps:

• Squint harder at the distributions of primes for the largest witness values in comparison to the rest.
• See if a machine learning algorithm can regress witness values based on their prime factorizations (and any other useful features I can derive). Study the resulting hypothesis to determine which features are the most important. Use that to refine the search strategy.
• Try searching randomly in the superabundant enumeration around 1k and 10k prime factors, and see if the best witness values found there match the log-log regression.
• Since witness values above a given threshold seem to be quite common, and because the UMAP visualization shows some possible “locality” structure for larger witness values, it suggests if there is a counterexample to RH then there are probably many. So a local search method (e.g., local neighborhood search/discrete gradient ascent with random restarts) might allow us to get a better sense for whether we are on the right track.

Until next time!

# Formulating the Support Vector Machine Optimization Problem

## The hypothesis and the setup

This blog post has an interactive demo (mostly used toward the end of the post). The source for this demo is available in a Github repository.

Last time we saw how the inner product of two vectors gives rise to a decision rule: if $w$ is the normal to a line (or hyperplane) $L$, the sign of the inner product $\langle x, w \rangle$ tells you whether $x$ is on the same side of $L$ as $w$.

Let’s translate this to the parlance of machine-learning. Let $x \in \mathbb{R}^n$ be a training data point, and $y \in \{ 1, -1 \}$ is its label (green and red, in the images in this post). Suppose you want to find a hyperplane which separates all the points with -1 labels from those with +1 labels (assume for the moment that this is possible). For this and all examples in this post, we’ll use data in two dimensions, but the math will apply to any dimension.

Some data labeled red and green, which is separable by a hyperplane (line).

The hypothesis we’re proposing to separate these points is a hyperplane, i.e. a linear subspace that splits all of $\mathbb{R}^n$ into two halves. The data that represents this hyperplane is a single vector $w$, the normal to the hyperplane, so that the hyperplane is defined by the solutions to the equation $\langle x, w \rangle = 0$.

As we saw last time, $w$ encodes the following rule for deciding if a new point $z$ has a positive or negative label.

$\displaystyle h_w(z) = \textup{sign}(\langle w, x \rangle)$

You’ll notice that this formula only works for the normals $w$ of hyperplanes that pass through the origin, and generally we want to work with data that can be shifted elsewhere. We can resolve this by either adding a fixed term $b \in \mathbb{R}$—often called a bias because statisticians came up with it—so that the shifted hyperplane is the set of solutions to $\langle x, w \rangle + b = 0$. The shifted decision rule is:

$\displaystyle h_w(z) = \textup{sign}(\langle w, x \rangle + b)$

Now the hypothesis is the pair of vector-and-scalar $w, b$.

The key intuitive idea behind the formulation of the SVM problem is that there are many possible separating hyperplanes for a given set of labeled training data. For example, here is a gif showing infinitely many choices.

The question is: how can we find the separating hyperplane that not only separates the training data, but generalizes as well as possible to new data? The assumption of the SVM is that a hyperplane which separates the points, but is also as far away from any training point as possible, will generalize best.

While contrived, it’s easy to see that the separating hyperplane is as far as possible from any training point.

More specifically, fix a labeled dataset of points $(x_i, y_i)$, or more precisely:

$\displaystyle D = \{ (x_i, y_i) \mid i = 1, \dots, m, x_i \in \mathbb{R}^{n}, y_i \in \{1, -1\} \}$

And a hypothesis defined by the normal $w \in \mathbb{R}^{n}$ and a shift $b \in \mathbb{R}$. Let’s also suppose that $(w,b)$ defines a hyperplane that correctly separates all the training data into the two labeled classes, and we just want to measure its quality. That measure of quality is the length of its margin.

Definition: The geometric margin of a hyperplane $w$ with respect to a dataset $D$ is the shortest distance from a training point $x_i$ to the hyperplane defined by $w$.

The best hyperplane has the largest possible margin.

This margin can even be computed quite easily using our work from last post. The distance from $x$ to the hyperplane defined by $w$ is the same as the length of the projection of $x$ onto $w$. And this is just computed by an inner product.

If the tip of the $x$ arrow is the point in question, then $a$ is the dot product, and $b$ the distance from $x$ to the hyperplane $L$ defined by $w$.

## A naive optimization objective

If we wanted to, we could stop now and define an optimization problem that would be very hard to solve. It would look like this:

\displaystyle \begin{aligned} & \max_{w} \min_{x_i} \left | \left \langle x_i, \frac{w}{\|w\|} \right \rangle + b \right | & \\ \textup{subject to \ \ } & \textup{sign}(\langle x_i, w \rangle + b) = \textup{sign}(y_i) & \textup{ for every } i = 1, \dots, m \end{aligned}

The formulation is hard. The reason is it’s horrifyingly nonlinear. In more detail:

1. The constraints are nonlinear due to the sign comparisons.
2. There’s a min and a max! A priori, we have to do this because we don’t know which point is going to be the closest to the hyperplane.
3. The objective is nonlinear in two ways: the absolute value and the projection requires you to take a norm and divide.

The rest of this post (and indeed, a lot of the work in grokking SVMs) is dedicated to converting this optimization problem to one in which the constraints are all linear inequalities and the objective is a single, quadratic polynomial we want to minimize or maximize.

Along the way, we’ll notice some neat features of the SVM.

## Trick 1: linearizing the constraints

To solve the first problem, we can use a trick. We want to know whether $\textup{sign}(\langle x_i, w \rangle + b) = \textup{sign}(y_i)$ for a labeled training point $(x_i, y_i)$. The trick is to multiply them together. If their signs agree, then their product will be positive, otherwise it will be negative.

So each constraint becomes:

$\displaystyle (\langle x_i, w \rangle + b) \cdot y_i \geq 0$

This is still linear because $y_i$ is a constant (input) to the optimization problem. The variables are the coefficients of $w$.

The left hand side of this inequality is often called the functional margin of a training point, since, as we will see, it still works to classify $x_i$, even if $w$ is scaled so that it is no longer a unit vector. Indeed, the sign of the inner product is independent of how $w$ is scaled.

## Trick 1.5: the optimal solution is midway between classes

This small trick is to notice that if $w$ is the supposed optimal separating hyperplane, i.e. its margin is maximized, then it must necessarily be exactly halfway in between the closest points in the positive and negative classes.

In other words, if $x_+$ and $x_-$ are the closest points in the positive and negative classes, respectively, then $\langle x_{+}, w \rangle + b = -(\langle x_{-}, w \rangle + b)$. If this were not the case, then you could adjust the bias, shifting the decision boundary along $w$ until it they are exactly equal, and you will have increased the margin. The closest point, say $x_+$ will have gotten farther away, and the closest point in the opposite class, $x_-$ will have gotten closer, but will not be closer than $x_+$.

## Trick 2: getting rid of the max + min

Resolving this problem essentially uses the fact that the hypothesis, which comes in the form of the normal vector $w$, has a degree of freedom in its length. To explain the details of this trick, we’ll set $b=0$ which simplifies the intuition.

Indeed, in the animation below, I can increase or decrease the length of $w$ without changing the decision boundary.

I have to keep my hand very steady (because I was too lazy to program it so that it only increases/decreases in length), but you can see the point. The line is perpendicular to the normal vector, and it doesn’t depend on the length.

Let’s combine this with tricks 1 and 1.5. If we increase the length of $w$, that means the absolute values of the dot products $\langle x_i, w \rangle$ used in the constraints will all increase by the same amount (without changing their sign). Indeed, for any vector $a$ we have $\langle a, w \rangle = \|w \| \cdot \langle a, w / \| w \| \rangle$.

In this world, the inner product measurement of distance from a point to the hyperplane is no longer faithful. The true distance is $\langle a, w / \| w \| \rangle$, but the distance measured by $\langle a, w \rangle$ is measured in units of $1 / \| w \|$.

In this example, the two numbers next to the green dot represent the true distance of the point from the hyperplane, and the dot product of the point with the normal (respectively). The dashed lines are the solutions to <x, w> = 1. The magnitude of w is 2.2, the inverse of that is 0.46, and indeed 2.2 = 4.8 * 0.46 (we’ve rounded the numbers).

Now suppose we had the optimal hyperplane and its normal $w$. No matter how near (or far) the nearest positively labeled training point $x$ is, we could scale the length of $w$ to force $\langle x, w \rangle = 1$. This is the core of the trick. One consequence is that the actual distance from $x$ to the hyperplane is $\frac{1}{\| w \|} = \langle x, w / \| w \| \rangle$.

The same as above, but with the roles reversed. We’re forcing the inner product of the point with w to be 1. The true distance is unchanged.

In particular, if we force the closest point to have inner product 1, then all other points will have inner product at least 1. This has two consequences. First, our constraints change to $\langle x_i, w \rangle \cdot y_i \geq 1$ instead of $\geq 0$. Second, we no longer need to ask which point is closest to the candidate hyperplane! Because after all, we never cared which point it was, just how far away that closest point was. And now we know that it’s exactly $1 / \| w \|$ away. Indeed, if the optimal points weren’t at that distance, then that means the closest point doesn’t exactly meet the constraint, i.e. that $\langle x, w \rangle > 1$ for every training point $x$. We could then scale $w$ shorter until $\langle x, w \rangle = 1$, hence increasing the margin $1 / \| w \|$.

In other words, the coup de grâce, provided all the constraints are satisfied, the optimization objective is just to maximize $1 / \| w \|$, a.k.a. to minimize $\| w \|$.

This intuition is clear from the following demonstration, which you can try for yourself. In it I have a bunch of positively and negatively labeled points, and the line in the center is the candidate hyperplane with normal $w$ that you can drag around. Each training point has two numbers next to it. The first is the true distance from that point to the candidate hyperplane; the second is the inner product with $w$. The two blue dashed lines are the solutions to $\langle x, w \rangle = \pm 1$. To solve the SVM by hand, you have to ensure the second number is at least 1 for all green points, at most -1 for all red points, and then you have to make $w$ as short as possible. As we’ve discussed, shrinking $w$ moves the blue lines farther away from the separator, but in order to satisfy the constraints the blue lines can’t go further than any training point. Indeed, the optimum will have those blue lines touching a training point on each side.

I bet you enjoyed watching me struggle to solve it. And while it’s probably not the optimal solution, the idea should be clear.

The final note is that, since we are now minimizing $\| w \|$, a formula which includes a square root, we may as well minimize its square $\| w \|^2 = \sum_j w_j^2$. We will also multiply the objective by $1/2$, because when we eventually analyze this problem we will take a derivative, and the square in the exponent and the $1/2$ will cancel.

## The final form of the problem

Our optimization problem is now the following (including the bias again):

\displaystyle \begin{aligned} & \min_{w} \frac{1}{2} \| w \|^2 & \\ \textup{subject to \ \ } & (\langle x_i, w \rangle + b) \cdot y_i \geq 1 & \textup{ for every } i = 1, \dots, m \end{aligned}

This is much simpler to analyze. The constraints are all linear inequalities (which, because of linear programming, we know are tractable to optimize). The objective to minimize, however, is a convex quadratic function of the input variables—a sum of squares of the inputs.

Such problems are generally called quadratic programming problems (or QPs, for short). There are general methods to find solutions! However, they often suffer from numerical stability issues and have less-than-satisfactory runtime. Luckily, the form in which we’ve expressed the support vector machine problem is specific enough that we can analyze it directly, and find a way to solve it without appealing to general-purpose numerical solvers.

We will tackle this problem in a future post (planned for two posts sequel to this one). Before we close, let’s just make a few more observations about the solution to the optimization problem.

## Support Vectors

In Trick 1.5 we saw that the optimal separating hyperplane has to be exactly halfway between the two closest points of opposite classes. Moreover, we noticed that, provided we’ve scaled $\| w \|$ properly, these closest points (there may be multiple for positive and negative labels) have to be exactly “distance” 1 away from the separating hyperplane.

Another way to phrase this without putting “distance” in scare quotes is to say that, if $w$ is the normal vector of the optimal separating hyperplane, the closest points lie on the two lines $\langle x_i, w \rangle + b = \pm 1$.

Now that we have some intuition for the formulation of this problem, it isn’t a stretch to realize the following. While a dataset may include many points from either class on these two lines $\langle x_i, w \rangle = \pm 1$, the optimal hyperplane itself does not depend on any of the other points except these closest points.

This fact is enough to give these closest points a special name: the support vectors.

We’ll actually prove that support vectors “are all you need” with full rigor and detail next time, when we cast the optimization problem in this post into the “dual” setting. To avoid vague names, the formulation described in this post called the “primal” problem. The dual problem is derived from the primal problem, with special variables and constraints chosen based on the primal variables and constraints. Next time we’ll describe in brief detail what the dual does and why it’s important, but we won’t have nearly enough time to give a full understanding of duality in optimization (such a treatment would fill a book).

When we compute the dual of the SVM problem, we will see explicitly that the hyperplane can be written as a linear combination of the support vectors. As such, once you’ve found the optimal hyperplane, you can compress the training set into just the support vectors, and reproducing the same optimal solution becomes much, much faster. You can also use the support vectors to augment the SVM to incorporate streaming data (throw out all non-support vectors after every retraining).

Eventually, when we get to implementing the SVM from scratch, we’ll see all this in action.

Until then!

# Holidays and Homicide

## A Study In Data

Just before midnight on Thanksgiving, there was a murder by gunshot about four blocks from my home. Luckily I was in bed by then, but all of the commotion over the incident got me thinking: is murder disproportionately more common on Thanksgiving? What about Christmas, Valentine’s Day, or Saint Patrick’s Day?

Of course, with the right data set these are the kinds of questions one can answer! After an arduous Google search for agreeable data sets, I came across this project called the History of Violence Database (perhaps the most ominous database name ever!). Unfortunately most of the work in progress there puts the emphasis on history, cataloging homicide incidents only earlier than the 1920’s.

But one page I came across contained a complete list of the dates of each known homicide in San Francisco from 1849-2003. What’s more, it is available in a simple, comma-delimited format. (To all data compilers everywhere: this is how data should be transferred! Don’t only provide it in Excel, or SPSS, or whatever proprietary software you might use. Text files are universally accessible.)

With a little sed preprocessing and some Mathematica magic, I whipped together this chart of homicide counts by day of the year (click on the image to get a larger view):

Homicides in San Francsico 1849 – 2003, organized by day of the year.

Here the red grid lines mark the highest ten homicide counts, the green grid lines mark the lowest ten, and the blue lines mark specific holidays. Some of the blue and red lines cover the same date, and so in that case the red line is the one displayed. Finally, the horizontal yellow line represents the median homicide count of 19, so one can compare individual dates against the median.

Now it would be a terrible thing to “conclude” something about general human behavior from this particular data set. But I’m going to do it anyway because it’s fun, and it lets me make fascinating and controversial observations. Plus, I need interesting tidbits of information to drop at parties with math and statistics students.

Here are my observations:

• There is a correlation between some holidays and homicide.
• New Year’s Day is by far the most violent day of the year, followed by Christmas Day. On the other hand, Christmas Eve is only slightly above average.
• The safest days of the year is January 5th. Having no other special recognition, it should be deemed National Peace Day, or at least National Too Pooped from New Year’s to Commit Murder Day.
• New Year’s Day (likely, Morning) is not dangerous because of excessive alcohol consumption alone. If it were, Saint Patrick’s Day would surely be similar. Although to be fair, one should compare this with the same statistic for Boston and Chicago.
• None of the following holidays are significantly more dangerous than the average day: Groundhog Day, Valentine’s Day, St. Patrick’s Day, April Fool’s Day, Cinco de Mayo, Halloween, Veteran’s Day, and New Year’s Eve (before midnight).
• On the other hand, the days following July 4th are quite vicious, even though July 3rd is very quiet.
• The days near November 24th are particularly violent because Thanksgiving (and Black Friday?) often fall on those days. Family gatherings are clearly high-risk events.
• Last-minute Christmas shopping (Dec. 13th, 15th) obviously brings out the rage in everyone.
• March and October are the docile months, while January, June, July, and December are the worst. Murder happens more often in the usual vacation months than at other times of the year.

Of course, some of the above points are meant to be facetious, but they bring up some interesting questions. Why does homicide show up more often during certain months of the year? It’s well known that most murder victims are personal acquaintances of the assailant (specifically, friends and family members); does this mean that family time rejuvenates or fuels strife? Are people more stressed during these times of year? Are June and July more violent simply because heat aggravates people, or at least gets them to interact with others more?

Unfortunately these aren’t the kinds of questions that lend themselves to mathematical proof, and as such I can’t give a definitive answer. But feel free to voice your own opinion in the comments!

For your viewing pleasure, here are the homicide counts on each day for the top and bottom 67 days, respectively, in ascending order:

highest:
{{"04/05", 24}, {"04/19", 24}, {"05/15", 24}, {"05/24", 24},
{"05/25", 24}, {"06/28", 24}, {"08/05", 24}, {"08/17", 24},
{"09/01", 24}, {"09/03", 24}, {"09/19", 24}, {"09/20", 24},
{"10/26", 24}, {"11/02", 24}, {"11/23", 24}, {"02/01", 25},
{"02/08", 25}, {"02/11", 25}, {"02/15", 25}, {"04/21", 25},
{"05/07", 25}, {"06/04", 25}, {"06/07", 25}, {"07/24", 25},
{"08/01", 25}, {"08/25", 25}, {"09/07", 25}, {"10/21", 25},
{"10/23", 25}, {"11/03", 25}, {"11/10", 25}, {"11/27", 25},
{"01/06", 26}, {"01/18", 26}, {"03/01", 26}, {"03/24", 26},
{"05/30", 26}, {"07/04", 26}, {"07/29", 26}, {"08/09", 26},
{"09/21", 26}, {"11/09", 26}, {"11/25", 26}, {"12/06", 26},
{"05/02", 27}, {"09/06", 27}, {"09/24", 27}, {"10/11", 27},
{"11/08", 27}, {"12/12", 27}, {"12/20", 27}, {"07/18", 28},
{"09/04", 28}, {"12/02", 28}, {"05/18", 29}, {"07/01", 29},
{"07/07", 29}, {"10/27", 29}, {"11/24", 29}, {"01/28", 30},
{"01/24", 31}, {"07/05", 31}, {"08/02", 31}, {"12/15", 31},
{"12/13", 32}, {"12/25", 36}, {"01/01", 43}}
lowest:
{{"01/05", 6}, {"02/29", 6}, {"06/20", 7}, {"07/03", 7},
{"04/15", 8}, {"02/03", 9}, {"02/10", 9}, {"09/16", 9},
{"03/04", 10}, {"05/16", 10}, {"09/29", 10}, {"10/12", 10},
{"12/16", 10}, {"02/04", 11}, {"05/17", 11}, {"05/23", 11},
{"06/06", 11}, {"06/12", 11}, {"07/11", 11}, {"09/22", 11},
{"11/12", 11}, {"11/16", 11}, {"12/19", 11}, {"03/13", 12},
{"03/20", 12}, {"05/21", 12}, {"05/29", 12}, {"07/14", 12},
{"08/13", 12}, {"09/05", 12}, {"10/28", 12}, {"11/22", 12},
{"01/19", 13}, {"02/06", 13}, {"02/09", 13}, {"04/22", 13},
{"04/28", 13}, {"05/04", 13}, {"05/11", 13}, {"08/27", 13},
{"09/15", 13}, {"10/03", 13}, {"10/13", 13}, {"12/03", 13},
{"01/10", 14}, {"01/26", 14}, {"01/31", 14}, {"02/13", 14},
{"02/18", 14}, {"02/23", 14}, {"03/14", 14}, {"03/15", 14},
{"03/26", 14}, {"03/31", 14}, {"04/11", 14}, {"05/12", 14},
{"05/28", 14}, {"06/19", 14}, {"06/24", 14}, {"07/20", 14},
{"07/31", 14}, {"08/10", 14}, {"10/22", 14}, {"12/22", 14},
{"01/07", 15}, {"01/14", 15}, {"01/16", 15}}

Unfortunately many holidays are defined year by year. Thanksgiving, Easter, Mother’s Day, Father’s Day, MLK Jr. Day, Memorial Day, and the Chinese New Year all can’t be analyzed with this chart because they fall on different days each year. It would not be very difficult to organize this data set according to the rules of those special holidays. We leave it as an exercise to the reader to do so. Remember that the data and Mathematica notebook used in this post are available from this blog’s Github page.

And of course, it would be quite amazing to find a data set which provides the dates of (even the last fifty years of) homicide history in the entire US, and not just one city. If any readers know of such a data set, please let me know.

Until next time!