Category Archives: Expository

How do we decide how many representatives there are for each state?

The US House of Representatives has 435 voting members (and 6 non-voting members: one each from Washington DC, Puerto Rico, American Samoa, Guam, the Northern Mariana Islands, and the US Virgin Islands). Roughly speaking, the higher the population of a state is, the more representatives it should have.

But what does this really mean?

If we looked at the US Constitution to make this clear, we would find little help. The third clause of Article I, Section II of the Constitution says

Representatives and direct Taxes shall be apportioned among the several States which may be included within this Union, according to their respective Numbers … The number of Representatives shall not exceed one for every thirty thousand, but each state shall have at least one Representative.

This doesn’t give clarity.1 In fact, uncertainty surrounding proper apportionment of representatives led to the first presidential veto.

The Apportionment Act of 1792

According to the 1790 Census, there were 3199415 free people and 694280 slaves in the United States.2

When Congress sat to decide on apportionment in 1792, they initially computed the total (weighted) population of the United States to be 3199415 + (3/5)⋅694280 ≈ 3615923. They noted that the Constitution says there should be no more than 1 representative for every 30000, so they divided the total population by 30000 and rounded down, getting 3615983/30000 ≈ 120.5.

Thus there were to be 120 representatives. If one takes each state and divides their populations by 30000, one sees that the states should get the following numbers of representatives3

State          ideal    rounded_down
Vermont        2.851    2
NewHampshire   4.727    4
Maine          3.218    3
Massachusetts  12.62    12
RhodeIsland    2.281    2
Connecticut    7.894    7
NewYork        11.05    11
NewJersey      5.985    5
Pennsylvania   14.42    14
Delaware       1.851    1
Maryland       9.283    9
Virginia       21.01    21
Kentucky       2.290    2
NorthCarolina  11.78    11
SouthCarolina  6.874    6
Georgia        2.361    2

But here is a problem: the total number of rounded down representatives is only 112. So there are 8 more representatives to give out. How did they decide which to assign these representatives to? They chose the 8 states with the largest fractional “ideal” parts:

  1. New Jersey (0.985)
  2. Connecticut (0.894)
  3. South Carolina (0.874)
  4. Vermont (0.851)
  5. Delaware (0.851)
  6. Massachusetts+Maine (0.838)
  7. North Carolina (0.78)
  8. New Hampshire (0.727)

(Maine was part of Massachuestts at the time, which is why I combine their fractional parts). Thus the original proposed apportionment gave each of these states one additional representative. Is this a reasonable conclusion?

Perhaps. But these 8 states each ended up having more than 1 representative for each 30000. Was this limit in the Constitution meant country-wide (so that 120 across the country is a fine number) or state-by-state (so that, for instance, Delaware, which had 59000 total population, should not be allowed to have more than 1 representative)?

There is the other problem that New Jersey, Connecticut, Vermont, New Hampshire, and Massachusetts were undoubtedly Northern states. Thus Southern representatives asked, Is it not unfair that the fractional apportionment favours the North?4

Regardless of the exact reasoning, the Secretary of State Thomas Jefferson and Attorney General Edmond Randalph (both from Virginia) urged President Washington to veto the bill, and he did. This was the first use of the Presidential veto.

Afterwards, Congress got together and decided on starting with 33000 people per representative and ignoring fractional parts entirely. The exact method became known as the Jefferson Method of Apportionment, and was used in the US until 1830. The subtle part of the method involves deciding on the number 33000. In the US, the exact number of representatives sometimes changed from election to election. This number is closely related to the population-per-representative, but these were often chosen through political maneuvering as opposed to exact decision.

As an aside, it’s interesting to note that this method of apportionment is widely used in the rest of the world, even though it was abandoned in the US.5 In fact, it is still used in Albania, Angola, Argentina, Armenia, Aruba, Austria, Belgium, Bolivia, Brazil, Bulgaria, Burundi, Cambodia, Cape Verde, Chile, Colombia, Croatia, the Czech Republic, Denmark, the Dominican Republic, East Timor, Ecuador, El Salvador, Estonia, Fiji, Finland, Guatemala, Hungary, Iceland, Israel, Japan, Kosovo, Luxembourg, Macedonia, Moldova, Monaco, Montenegro, Mozambique, Netherlands, Nicaragua, Northern Ireland, Paraguay, Peru, Poland, Portugal, Romania, San Marino, Scotland, Serbia, Slovenia, Spain, Switzerland, Turkey, Uruguay, Venezuela and Wales — as well as in many countries for election to the European Parliament.

Apportionment Act of 1792

Measuring the fairness of an apportionment method

At the core of different ideas for apportionment is fairness. How can we decide if an apportionment fair?

We’ll consider this question in the context of the post-1911 United States — after the number of seats in the House of Representatives was established. This number was set at 433, but with the proviso that anticipated new states Arizona and New Mexico would each come with an additional seat.6

So given that there are 435 seats to apportion, how might we decide if an apportionment is fair? Fundamentally, this should relate to the number of people each representative actually represents.

For example, in the 1792 apportionment, the single Delawaran representative was there to represent all 55000 of its population, while each of the two Rhode Island representatives corresponded to 34000 Rhode Islanders. Within the House of Representatives, it was as though the voice of each Delawaran only counted 61 percent as much as the voice of each Rhode Islander7

The number of people each representative actually represent is at the core of the notion of fairness — but even then, it’s not obvious.

Suppose we enumerate the states, so that Si refers to state i. We’ll also denote by Pi the population of state i, and we’ll let Ri denote the number of representatives allotted to state i.

In the ideal scenario, every representative would represent the exact same number of people. That is, we would have
$$\text{pop. per rep. in state i}
= \frac{P_i}{R_i}
= \frac{P_j}{R_j}
= \text{pop. per rep. in state j}$$

for every pair of states i and j. But this won’t ever happen in practice.

Generally, we should expect $\frac{P_i}{R_i} \neq \frac{P_j}{R_j}$ for every pair of distinct states. If
$$
\frac{P_i}{R_i} > \frac{P_j}{R_j}, \tag{1}
$$

then we can say that each representative in state i represents more people, and thus those people have a diluted vote.

Amounts of Inequality

There are lots of pairs of states. How do we actually measure these inequalities? This would make an excellent question in a statistics class (illustrating how one can answer the same question in different, equally reasonable ways) or even a civics class.

A few natural ideas emerge:

  • We might try to minimize the differences of constituency size: $\left \lvert \frac{P_i}{R_i} – \frac{P_j}{R_j} \right \rvert$.
  • We might try to minimize the differences in per capita representation: $\left \lvert \frac{R_i}{P_i} – \frac{R_j}{P_j} \right \rvert$.
  • We might take overall size into account, and try to minimize both the relative constituency size and relative difference in per capita representation.

This last one needs a bit of explanation. Define the relative difference between two numbers x and y to be
$$
\frac{\lvert x – y \rvert}{\min(x, y)}.
$$

Suppose that for a pair of states, we have that $(1)$ holds, i.e. that representatives in state j have smaller constituencies than in state i (and therefore people in state j have more powerful votes). Then the relative difference in constituency size is
$$
\frac{P_i/R_i – P_j/R_j}{P_j/R_j} = \frac{P_i/R_i}{P_j/R_j} – 1.
$$

The relative difference in per capita representation is
$$
\frac{R_j/P_j – R_i/P_i}{R_i/P_i} = \frac{R_j/P_j}{R_i/P_i} – 1 =
\frac{P_i/R_i}{P_j/R_j} – 1.
$$

Thus these are the same! By accounting for differences in size by taking relative proportions, we see that minimizing relative difference in constituency size and minimizing relative difference in per capita representation are actually the same.

All three of these measures seem reasonable at first inspection. Unfortunately, they all give different apportionments (and all are different from Jefferson’s scheme — though to be fair, Jefferson’s scheme doesn’t seek to minimize inequality and there is no reason to think it should behave the same).

Each of these ideas leads to a different apportionment scheme, and in fact each has a name.

  • Minimizing differences in constituency size is the Dean method.
  • Minimizing differences in per capita representation is the Webster method.
  • Minimizing relative differences between both constituency size and per capita representation is the Hill (or sometimes Huntington-Hill) method.

Further, each of these schemes has been used at some time in US history. Webster’s method was used immediately after the 1840 census, but for the 1850 census the original Alexander Hamilton scheme (the scheme vetoed by Washington in 1792) was used. In fact, the Apportionment Act of 1850 set the Hamilton method as the primary method, and this was nominally used until 1900.8 The Webster method was used again immediately after the 1910 census. Due to claims of incomplete and inaccurate census counts, no apportionment occurred based on the 1920 census.9

In 1929 an automatic apportionment act was passed.10 In it, up to three different apportionment schemes would be provided to Congress after each census, based on a total of 435 seats:

  1. The apportionment that would come from whatever scheme was most recently used. (In 1930, this would be the Webster method).
  2. The apportionment that would come from the Webster method.
  3. The apportionment that would come from the newly introduced Hill method.

If one reads congressional discussion from the time, then it will be good to note that Webster’s method is sometimes called the method of major fractions and Hill’s method is sometimes called the method of equal proportions. Further, in a letter written by Bliss, Brown, Eisenhart, and Pearl of the National Academy of Sciences, Hill’s method was declared to be the recommendation of the Academy.11 From 1930 on, Hill’s method has been used.

Why use the Hill method?

The Hamilton method led to a few paradoxes and highly counterintuitive behavior that many representatives found disagreeable. In 1880, a paradox now called the Alabama paradox was noted. When deciding on the number of representatives that should be in the House, it was noted that if the House had 299 members, Alabama would have 8 representatives. But if the House had 300 members, Alabama would have 7 representatives — that is, making one more seat available led to Alabama receiving one fewer seat.

The problem is the fluctuating relationships between the many fractional parts of the ideal number of representatives per state (similar to those tallied in the table in the section The Apportionment Act of 1792).

Another paradox was discovered in 1900, known as the Population paradox. This is a scenario in which a state with a large population and rapid growth can lose a seat to a state with a small population and smaller population growth. In 1900, Virginia lost a seat to Maine, even though Virginia’s population was larger and growing much more rapidly.

In particular, in 1900, Virginia had 1854184 people and Maine had 694466 people, so Virginia had 2.67 times the population as Maine. In 1901, Virginia had 1873951 people and Maine had 699114 people, so Virginia had 2.68 times the number of people. And yet Hamilton apportionment would have given 10 seats to Virginia and 3 to Maine in 1900, but 9 to Virginia and 4 to Maine in 1901.

Central to this paradox is that even though Virginia was growing faster than Maine the rest of the nation was growing fast still, and proportionally Virginia lost more because it was a larger state. But it’s still paradoxical for a state to lose a representative to a second state that is both smaller in population and is growing less rapidly each census.12

The Hill method can be shown to not suffer from either the Alabama paradox or the Population paradox. That it doesn’t suffer from these paradoxical behaviours and that it seeks to minimize a meaningful measure of inequality led to its adoption in the US.13

Understanding the modern Hill method in practice

Since 1930, the US has used the Hill method to apportion seats for the House of Representatives. But as described above, it may be hard to understand how to actually apply the Hill method. Recall that Pi is the population of state i, and Ri is the number of representatives allocated to state i. The Hill method seeks to minimize
$$
\frac{P_i/R_i – P_j/R_j}{P_j/R_j} = \frac{P_i/R_i}{P_j/R_j} – 1
$$

whenever Pi/Ri > Pj/Rj. Stated differently, the Hill method seeks to guarantee the smallest relative differences in constituency size.

We can work out a different way of understanding this apportionment that is easier to implement in practice.

Suppose that we have allocated all of the representatives to each state and state j has Rj representatives, and suppose that this allocation successfully minimizes relative differences in constituency size. Take two different states i and j with Pi/Ri > Pj/Rj. (If this isn’t possible then the allocation is perfect).

We can ask if it would be a good idea to move one representative from state j to state i, since state j‘s constituency sizes are smaller. This can be thought of as working with Ri′=Ri + 1 and Rj′=Rj − 1. If this transfer lessens the inequality then it should be made — but since we are supposing that the allocation successfully minimizes relative difference in constituency size, we must have that the inequality is at least as large. This necessarily means that Pj/Rj′>Pi/Ri (since otherwise the relative difference is strictly smaller) and
$$
\frac{P_jR_i’}{P_iR_j’} – 1 \geq \frac{P_iR_j}{P_jR_i} – 1
$$

(since the relative difference must be at least as large). This is equivalent to
$$
\frac{P_j(R_i+1)}{P_i(R_j-1)} \geq \frac{P_iR_j}{P_jR_i}
\iff
\frac{P_j^2}{(R_j-1)R_j} \geq \frac{P_i^2}{R_i(R_i+1)}.
$$

As every variable is positive, we can rewrite this as
$$
\frac{P_j}{\sqrt{(R_j – 1)R_j}} \geq \frac{P_i}{\sqrt{R_i(R_i+1)}}. \tag{2}
$$

We’ve shown that $(2)$ must hold whenever Pi/Ri > Pj/Rj in a system that minimizes relative difference in constituency size. But in fact it must hold for all pairs of states i and j.

Clearly it holds if i = j as the denominator on the left is strictly smaller.

If we are in the case when Pj/Rj > Pi/Ri, then we necessarily have the chain Pj/(Rj − 1)>Pj/Rj > Pi/Ri > Pi/(Ri + 1). Multiplying the inner and outer inequalities shows that $(2)$ holds trivially in this case.

This inequality shows that the greatest obstruction to being perfectly apportioned as per Hill’s method is the largest fraction
$$ \frac{R_i}{\sqrt{P_i(P_i+1)}} $$
being too large. (Some call this term the Hill rank-index).

An iterative Hill apportionment

This observation leads to the following iterative construction of a Hill apportionment. Initially, assign every state 1 representative (since by the Constitution, each state gets at least one representative). Then, given an apportionment for n seats, we can get an apportionment for n + 1 seats by assigning the additional seat the any state i which maximizes the Hill rank-index $R_i/\sqrt{P_i(P_i+1)}$.

Further, it can be shown that there is a unique apportionment in Hill’s method (except for ties in the Hill rank-index, which are exceedingly rare in practice). Thus the apportionment is unique.

This is very quickly and easily implemented in code. In a later note, I will share the code I used to compute the various data for this note, as well as an implementation of Hill apportionment.

Additional notes: Consequences of the 1870 and 1990 Apportionments

The 1870 Apportionment

Officially, Dean’s method of apportionment has never been used. But it was perhaps used in 1870 without being described. Officially, Hamilton’s method was in place and the size of the House was agreed to be 292. But the actual apportionment that occurred agreed with Dean’s method, not Hamilton’s method. Specifically, New York and Illinois were each given one fewer seat than Hamilton’s method would have given, while New Hampshire and Florida were given one additional seat each.

There are many circumstances surrounding the 1870 census and apportionment that make this a particularly convoluted time. Firstly, the US had just experienced its Civil War, where millions of people died and millions others moved or were displaced. Animosity and reconstruction were both in full swing. Secondly, the US passed the 14th amendment in 1868, so that suddenly the populations of Southern states grew as former slaves were finally allowed to be counted fully.

One might think that having two pairs of states swap a representative would be mostly inconsequential. But this difference — using Dean’s method instead of the agreed on Hamilton method, changed the result of the 1876 Presidential election. In this election, Samuel Tilden won New York while Rutherford B. Hayes won Illinois, New Hampshire, and Florida. As a result, Tilden received one fewer electoral vote and Hayes received one additional electoral vote — and the total electoral voting in the end had Hayes win with 185 votes to Tilden’s 184.

There is still one further mitigating factor, however, that causes this to be yet more convoluted. The 1876 election is perhaps the most disputed presidential election. In Florida, Louisiana, and South Carolina, each party reported that its candidate had won the state. Legitimacy was in question, and it’s widely believed that a deal was struck between the Democratic and Republican parties (see wikipedia and 270 to win). As a result of this deal, the Republican candidate Rutherford B. Hayes would gain all disputed votes and remove federal troops (which had been propping up reconstructive efforts) from the South. This marked the end of the “Reconstruction” period, and allowed the rise of the Democratic Redeemers (and their rampant black voter disenfranchisement) in the South.

The 1990 Apportionment

Similar in consequence though not in controversy, the apportionment of 1990 influenced the results of the 2000 presidential election between George W. Bush and Al Gore (as the 2000 census is not complete before the election takes place, so the election occurs with the 1990 electoral college sizes). The modern Hill apportionment method was used, as it has been since 1930. But interestingly, if the originally proposed Hamilton method of 1792 was used, the electoral college would have been tied at 26914. If Jefferson’s method had been used, then Gore would have won with 271 votes to Bush’s 266.

These decisions have far-reaching consequences!

Sources

  1. Balinski, Michel L., and H. Peyton Young. Fair representation: meeting the ideal of one man, one vote. Brookings Institution Press, 2010.
  2. Balinski, Michel L., and H. Peyton Young. “The quota method of apportionment.” The American Mathematical Monthly 82.7 (1975): 701-730.
  3. Bliss, G. A., Brown, E. W., Eisenhart, L. P., & Pearl, R. (1929). Report to the President of the National Academy of Sciences. February, 9, 1015-1047.
  4. Crocker, R. House of Representatives Apportionment Formula: An Analysis of Proposals for Change and Their Impact on States. DIANE Publishing, 2011.
  5. Huntington, The Apportionment of Representatives in Congress, Transactions of the American Mathematical Society 30 (1928), 85–110.
  6. Peskin, Allan. “Was there a Compromise of 1877.” The Journal of American History 60.1 (1973): 63-75.
  7. US Census Results
  8. US Constitution
  9. US Congressional Record, as collected at https://memory.loc.gov/ammem/amlaw/lwaclink.html
  10. George Washington’s collected papers, as archived at https://web.archive.org/web/20090124222206/http://gwpapers.virginia.edu/documents/presidential/veto.html
  11. Wikipedia on the Compromise of 1877, at https://en.wikipedia.org/wiki/Compromise_of_1877
  12. Wikipedia on Arthur Vandenberg, at https://en.wikipedia.org/wiki/Arthur_Vandenberg
Posted in Data, Expository, Mathematics, Politics, Story | Tagged , , | Leave a comment

Talk: Finding Congruent Numbers, Arithmetic Progressions of Squares, and Triangles

Here are some notes for my talk Finding Congruent Numbers, Arithmetic Progressions of Squares, and Triangles (an invitation to analytic number theory), which I’m giving on Tuesday 26 February at Macalester College.

The slides for my talk are available here.

The overarching idea of the talk is to explore the deep relationship between

  1. right triangles with rational side lengths and area $n$,
  2. three-term arithmetic progressions of squares with common difference $n$, and
  3. rational points on the elliptic curve $Y^2 = X^3 – n^2 X$.

If one of these exist, then all three exist, and in fact there are one-to-one correspondences between each of them. Such an $n$ is called a congruent number.

By understanding this relationship, we also describe the ideas and results in the paper A Shifted Sum for the Congruent Number Problem, which I wrote jointly with Tom Hulse, Chan Ieong Kuan, and Alex Walker.

Towards the end of the talk, I say that in practice, the best way to decide if a (reasonably sized) number is congruent is through elliptic curves. Given a computer, we can investigate whether the number $n$ is congruent through a computer algebra system like sage.1

For the rest of this note, I’ll describe how one can use sage to determine whether a number is congruent, and how to use sage to add points on elliptic curves to generate more triangles corresponding to a particular congruent number.

Firstly, one needs access to sage. It’s free to install, but it’s quite large. The easiest way to begin using sage immediately is to use cocalc.com,  a free interface to sage (and other tools) that was created by William Stein, who also created sage.

In a sage session, we can create an elliptic curve through


> E6 = EllipticCurve([-36, 0])
> E6
Elliptic Curve defined by y^2 = x^3 - 36*x over Rational Field

More generally, to create the curve corresponding to whether or not $n$ is congruent, you can use


> n = 6   # (or anything you want)
> E = EllipticCurve([-n**2, 0])

We can ask sage whether our curve has many rational points by asking it to (try to) compute the rank.


> E6.rank()
1

If the rank is at least $1$, then there are infinitely many rational points on the curve and $n$ is a congruent number. If the rank is $0$, then $n$ is not congruent.2

For the curve $Y^2 = X^3 – 36 X$ corresponding to whether $6$ is congruent, sage returns that the rank is $1$. We can ask sage to try to find a rational point on the elliptic curve through


> E6.point_search(10)
[(-3 : 9 : 1)]

The 10 in this code is a limit on the complexity of the point. The precise definition isn’t important — using $10$ is a reasonable limit for us.

We see that this output something. When sage examines the elliptic curve, it uses the equation $Y^2 Z = X^3 – 36 X Z^2$ — it turns out that in many cases, it’s easier to perform computations when every term is a polynomial of the same degree. The coordinates it’s giving us are of the form $(X : Y : Z)$, which looks a bit odd. We can ask sage to return just the XY coordinates as well.


> Pt = E6.point_search(10)[0]  # The [0] means to return the first element of the list
> Pt.xy()
(-3, 9)

In my talk, I describe a correspondence between points on elliptic curves and rational right triangles. In the talk, it arises as the choice of coordinates. But what matters for us right now is that the correspondence taking a point $(x, y)$ on an elliptic curve to a triangle $(a, b, c)$ is given by
$$(x, y) \mapsto \Big( \frac{n^2-x^2}{y}, \frac{-2 \cdot x \cdot y}{y}, \frac{n^2 + x^2}{y} \Big).$$

We can write a sage function to perform this map for us, through


> def pt_to_triangle(P):
    x, y = P.xy()
    return (36 - x**2)/y, (-2*x*6/y), (36+x**2)/y

> pt_to_triangle(Pt)
(3, 4, 5)

This returns the $(3, 4, 5)$ triangle!

Of course, we knew this triangle the whole time. But we can use sage to get more points. A very cool fact is that rational points on elliptic curves form a group under a sort of addition — we can add points on elliptic curves together and get more rational points. Sage is very happy to perform this addition for us, and then to see what triangle results.


> Pt2 = Pt + Pt
> Pt2.xy()
(25/4, -35/8)
> pt_to_triangle(Pt2)
(7/10, 120/7, -1201/70)

Another rational triangle with area $6$ is the $(7/10, 120/7, 1201/70)$ triangle. (You might notice that sage returned a negative hypotenuse, but it’s the absolute values that matter for the area). After scaling this to an integer triangle, we get the integer right triangle $(49, 1200, 1201)$ (and we can check that the squarefree part of the area is $6$).

Let’s do one more.


> Pt3 = Pt + Pt + Pt
> Pt3.xy()
(-1587/1369, -321057/50653)
> pt_to_triangle(Pt3)
(-4653/851, -3404/1551, -7776485/1319901)

That’s a complicated triangle! It may be fun to experiment some more — the triangles rapidly become very, very complicated. In fact, it was very important to the main result of our paper that these triangles become so complicated so quickly!

Posted in Expository, Math.NT, Mathematics, Programming, sage, sagemath, sagemath | Leave a comment

Update to Second Moments in the Generalized Gauss Circle Problem

Last year, my coauthors Tom Hulse, Chan Ieong Kuan, and Alex Walker posted a paper to the arXiv called “Second Moments in the Generalized Gauss Circle Problem”. I’ve briefly described its contents before.

This paper has been accepted and will appear in Forum of Mathematics: Sigma.More randomized squares art

This is the first time I’ve submitted to the Forum of Mathematics, and I must say that this has been a very good journal experience. One interesting aspect about FoM: Sigma is that they are immediate (gold) open access, and they don’t release in issues. Instead, articles become available (for free) from them once the submission process is done. I was reviewing a publication-proof of the paper yesterday, and they appear to be very quick with regards to editing. Perhaps the paper will appear before the end of the year.

An updated version (the version from before the handling of proofs at the journal, so there will be a number of mostly aesthetic differences with the published version) of the paper will appear on the arXiv on Monday 10 December.1

A new appendix has appeared

There is one major addition to the paper that didn’t appear in the original preprint. At one of the referee’s suggestions, Chan and I wrote an appendix. The major content of this appendix concerns a technical detail about Rankin-Selberg convolutions.

If $f$ and $g$ are weight $k$ cusp forms on $\mathrm{SL}(2, \mathbb{Z})$ with expansions $$ f(z) = \sum_ {n \geq 1} a(n) e(nz), \quad g(z) = \sum_ {n \geq 1} b(n) e(nz), $$ then one can use a (real analytic) Eisenstein series $$ E(s, z) = \sum_ {\gamma \in \mathrm{SL}(2, \mathbb{Z})_ \infty \backslash \mathrm{SL}(2, \mathbb{Q})} \mathrm{Im}(\gamma z)^s $$ to recognize the Rankin-Selberg $L$-function \begin{equation}\label{RS} L(s, f \otimes g) := \zeta(s) \sum_ {n \geq 1} \frac{a(n)b(n)}{n^{s + k – 1}} = h(s) \langle f g y^k, E(s, z) \rangle, \end{equation} where $h(s)$ is an easily-understandable function of $s$ and where $\langle \cdot, \cdot \rangle$ denotes the Petersson inner product.

When $f$ and $g$ are not cusp forms, or when $f$ and $g$ are modular with respect to a congruence subgroup of $\mathrm{SL}(2, \mathbb{Z})$, then there are adjustments that must be made to the typical construction of $L(s, f \otimes g)$.

When $f$ and $g$ are not cusp forms, then Zagier2 provided a way to recognize $L(s, f \otimes g)$ when $f$ and $g$ are modular on the full modular group $\mathrm{SL}(2, \mathbb{Z})$. And under certain conditions that he describes, he shows that one can still recognize $L(s, f \otimes g)$ as an inner product with an Eisenstein series as in \eqref{RS}.

In principle, his method of proof would apply for non-cuspidal forms defined on congruence subgroups, but in practice this becomes too annoying and bogged down with details to work with. Fortunately, in 2000, Gupta3 gave a different construction of $L(s, f \otimes g)$ that generalizes more readily to non-cuspidal forms on congruence subgroups. His construction is very convenient, and it shows that $L(s, f \otimes g)$ has all of the properties expected of it.

However Gupta does not show that there are certain conditions under which one can recognize $L(s, f \otimes g)$ as an inner product against an Eisenstein series.4 For this paper, we need to deal very explicitly and concretely with $L(s, \theta^2 \otimes \overline{\theta^2})$, which is formed from the modular form $\theta^2$, non-cuspidal on a congruence subgroup.

The Appendix to the paper can be thought of as an extension of Gupta’s paper: it uses Gupta’s ideas and techniques to prove a result analogous to \eqref{RS}. We then use this to get the explicit understanding necessary to tackle the Gauss Sphere problem.

There is more to this story. I’ll return to it in a later note.

Other submission details for FoM: Sigma

I should say that there are many other revisions between the original preprint and the final one. These are mainly due to the extraordinary efforts of two Referees. One Referee was kind enough to give us approximately 10 pages of itemized suggestions and comments.

When I first opened these comments, I was a bit afraid. Having so many comments was daunting. But this Referee really took his or her time to point us in the right direction, and the resulting paper is vastly improved (and in many cases shortened, although the appendix has hidden the simplified arguments cut in length).

More broadly, the Referee acted as a sort of mentor with respect to my technical writing. I have a lot of opinions on technical writing,5 but this process changed and helped sharpen my ideas concerning good technical math writing.

I sometimes hear lots of negative aspects about peer review, but this particular pair of Referees turned the publication process into an opportunity to learn about good mathematical exposition — I didn’t expect this.

I was also surprised by the infrastructure that existed at the University of Warwick for handling a gold open access submission. As part of their open access funding, Forum of Math: Sigma has an author-pays model. Or rather, the author’s institution pays. It took essentially no time at all for Warwick to arrange the payment (about 500 pounds).

This is a not-inconsequential amount of money, but it is much less than the 1500 dollars that PLoS One uses. The comparison with PLoS One is perhaps apt. PLoS is older, and perhaps paved the way for modern gold open access journals like FoM. PLoS was started by group of established biologists and chemists, including a Nobel prize winner; FoM was started by a group of established mathematicians, including multiple Fields medalists.6

I will certainly consider Forum of Mathematics in the future.

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Notes from a Talk at Building Bridges 4

On 18 July 2018 I gave a talk at the 4th Building Bridges Automorphic Forms Workshop, which is hosted at the Renyi Institute in Budapest, Hungary this year. In this talk, I spoke about counting points on hyperboloids, with a certain focus on counting points on the three dimensional hyperboloid

$$\begin{equation} X^2 + Y^2 = Z^2 + h \end{equation}$$

for any fixed integer $h$.

I gave a similar talk at the 32nd Automorphic Forms Workshop in Tufts in March. I don’t say this during my talk, but a big reason for giving these talks is to continue to inspire me to finish the corresponding paper. (There are still a couple of rough edges that need some attention).

The methodology for the result relies on the spectral expansion of half-integral weight modular forms. This is unfriendly to those unfamiliar with the subject, and particularly mysterious to students. But there is a nice connection to a topic discussed by Arpad Toth during the previous week’s associated summer school.

Arpad sketched a proof of the spectral decomposition of holomorphic modular cusp forms on $\Gamma = \mathrm{SL}(2, \mathbb{Z})$. He showed that
$$\begin{equation} L^2(\Gamma \backslash \mathcal{H}) = \textrm{cuspidal} \oplus \textrm{Eisenstein}, \tag{1}
\end{equation}$$
where the cuspidal contribution comes from Maass forms and the Eisenstein contribution comes from line integrals against Eisenstein series.

The typical Eisenstein series $$\begin{equation} E(z, s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \textrm{Im}(\gamma z)^s \end{equation}$$ only converges for $\mathrm{Re}(s) > 1$, and the initial decomposition in $(1)$ implicitly has $s$ in this range.

To write down the integrals appearing in the Eisenstein spectrum explicitly, one normally shifts the line of integration to $1/2$. As Arpad explained, classically this produces a pole at $s = 1$ (which is the constant function).

In half-integral weight, the Eisenstein series has a pole at $s = 3/4$, with the standard theta function

$$\begin{equation} \theta(z) = \sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 z} \end{equation}$$

as the residue. (More precisely, it’s a constant times $y^{1/4} \theta(z)$, or a related theta function for $\Gamma_0(N)$). I refer to this portion of the spectrum as the residual spectrum, since it comes from often-forgotten residues of Eisenstein series. Thus the spectral decomposition for half-integral weight objects is a bit more complicated than the normal case.

When giving talks involving half-integral weight spectral expansions to audiences including non-experts, I usually omit description of this. But for those who attended the summer school, it’s possible to at least recognize where these additional terms come from.

The slides for this talk are available here.

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Paper Announcement: A Shifted Sum for the Congruent Number Problem

Tom Hulse, Chan Ieong Kuan, Alex Walker, and I have just uploaded a new paper to the arXiv titled A Shifted Sum for the Congruent Number Problem. In this charming, short paper, we investigate a particular sum of terms which are products of square-indicator functions and show that its asymptotics are deeply connected to congruent numbers. This note serves to describe and provide additional context for these results. (This note is also available as a pdf).

Congruent Numbers

We consider some triangles. There are many right triangles, such as the triangle with sides $(3, 4, 5)$ or the triangle with sides $(1, 1, \sqrt{2})$. We call a right triangle rational when all its side lengths are rational numbers. For illustration, $(3, 4, 5)$ is rational, while $(1, 1, \sqrt{2})$ is not. $\DeclareMathOperator{\sqfree}{sqfree}$

There is mythology surrounding rational right triangles. According to legend, the ancient Greeks, led both philosophcally and mathematically by Pythagoras (who was the first person to call himself a philosopher and essentially the first to begin to distill and codify mathematics), believed all numbers and quantities were ratios of integers (rational). When a disciple of Pythagoras named Hippasus found that the side lengths of the right triangle $(1, 1, \sqrt{2})$ were not rational multiples of each other, the other followers of Pythagoras killed him by casting him overboard while at sea for having produced an element which contradicted the gods. (It with some irony that we now attribute this as a simple consequence of the Pythagorean Theorem).

This mythology is uncertain, but what is certain is that even the ancient Greeks were interested in studying rational right triangles, and they began to investigate what we now call the Congruent Number Problem. By the year 972 the CNP appears in Arabic manuscripts in (essentially) its modern formulation. The Congruent Number Problem (CNP) may be the oldest unresolved math problem.

We call a positive rational number $t$ congruent if there is a rational right triangle with area $t$. The triangle $(3,4,5)$ shows that $6 = 3 \cdot 4 / 2$ is congruent. The CNP is to describe all congruent numbers. Alternately, the CNP asks whether there is an algorithm to show definitively whether or not $t$ is a congruent number for any $t$.

We can reduce the problem to a statement about integers. If the rational number $t = p/q$ is the area of a triangle with legs $a$ and $b$, then the triangle $aq$ and $bq$ has area $tq^2 = pq$. It follows that to every rational number there is an associated squarefree integer for which either both are congruent or neither are congruent. Further, if $t$ is congruent, then $ty^2$ and $t/y^2$ are congruent for any integer $y$.

We may also restrict to integer-sided triangles if we allow ourselves to look for those triangles with squarefree area $t$. That is, if $t$ is the area of a triangle with rational sides $a/A$ and $b/B$, then $tA^2 B^2$ is the area of the triangle with integer sides $aB$ and $bA$.

It is in this form that we consider the CNP today.

Congruent Number Problem

Given a squarefree integer $t$, does there exist a triangle with integer side lengths such that the squarefree part of the area of the triangle is $t$?

We will write this description a lot, so for a triangle $T$ we introduce the notation
\begin{equation}
\sqfree(T) = \text{The squarefree part of the area of } T.
\end{equation}
For example, the area of the triangle $T = (6, 8, 10)$ is $24 = 6 \cdot 2^2$, and so $\sqfree(T) = 6$. We should expect this, as $T$ is exactly a doubled-in-size $(3,4,5)$ triangle, which also corresponds to the congruent number $6$. Note that this allows us to only consider primitive right triangles.

Main Result

Let $\tau(n)$ denote the square-indicator function. That is, $\tau(n)$ is $1$ if $n$ is a square, and is $0$ otherwise. Then the main result of the paper is that the sum
\begin{equation}
S_t(X) := \sum_{m = 1}^X \sum_{n = 1}^X \tau(m-n)\tau(m)\tau(nt)\tau(m+n)
\end{equation}
is related to congruent numbers through the asymptotic
\begin{equation}
S_t(X) = C_t \sqrt X + O_t\Big( \log^{r/2} X\Big),
\end{equation}
where
\begin{equation}
C_t = \sum_{h_i \in \mathcal{H}(t)} \frac{1}{h_i}.
\end{equation}
Each $h_i$ is a hypotenuse of a primitive integer right triangle $T$ with $\sqfree(T) = t$. Each hypotnesue will occur in a pair of similar triangles $(a,b, h_i)$ and $(b, a, h_i)$; $\mathcal{H}(t)$ is the family of these triangles, choosing only one triangle from each similar pair. The exponent $r$ in the error term is the rank of the elliptic curve
\begin{equation}
E_t(\mathbb{Q}): y^2 = x^3 – t^2 x.
\end{equation}

What this says is that $S_t(X)$ will have a main term if and only if $t$ is a congruent number, so that computing $S_t(X)$ for sufficiently large $X$ will show whether $t$ is congruent. (In fact, it’s easy to show that $S_t(X) \neq 0$ if and only if $t$ is congruent, so the added value here is the nature of the asymptotic).

We should be careful to note that this does not solve the CNP, since the error term depends in an inexplicit way on the desired number $t$. What this really means is that we do not have a good way of recognizing when the first nonzero term should occur in the double sum. We can only guarantee that for any $t$, understanding $S_t(X)$ for sufficiently large $X$ will allow one to understand whether $t$ is congruent or not.

Intuition and Methodology

There are four primary components to this result:

  1. There is a bijection between primitive integer right triangles $T$ with
    $\sqfree(T) = t$ and arithmetic progressions of squares $m^2 – tn^2, m^2,
    m^2 + tn^2$ (where each term is itself a square).
  2. There is a bijection between primitive integer right triangles $T$ with
    $\sqfree(T) = t$ and points on the elliptic curve $E_t(\mathbb{Q}): y^2 = x^3
    – t x$ with $y \neq 0 $.
  3. If the triangle $T$ corresponds to a point $P$ on the curve $E_t$, then
    the size of the hypotenuse of $T$ can be bounded below by $H(P)$, the
    (naive) height of the point on the elliptic curve.
  4. Néron (and perhaps Mordell, but I’m not quite fluent in the initial
    history of the theory of elliptic curves) proved strong (upper) bounds on
    the number of points on an elliptic curve up to a given height. (In fact,
    they proved asymptotics which are much stronger than we use).

In this paper, we use $(1)$ to relate triangles $T$ to the sum $S_t(X)$ and we use $(2)$ to relate these triangles to points on the elliptic curve. Tracking the exact nature of the hypotenuses through these bijections allows us to relate the sum to certain points on elliptic curves. In order to facilitate the tracking of these hypotenuses, we phrase these bijections in slightly different ways than have appeared in the literature. By $(3)$ and $(4)$, we can bound the number and size of the hypotenuses which appear in terms of numbers of points on the elliptic curve up to a certain height. Intuitively this is why the higher the rank of the elliptic curve (corresponding roughly to the existence of many more points on the curve), the worse the error term in our asymptotic.

I would further conjecture that the error term in our asymptotic is essentially best-possible, even though we have thrown away some information in our proof.

Additional Context

We are not the first to note either the bijection between triangles $T$ and arithmetic progressions of squares or between triangles $T$ and points on a particular elliptic curve. The first is surely an ancient observation, but I don’t know who first considered the relation to elliptic curves. But it’s certain that this was a fundamental aspect in Tunnell’s famous work A Classical Diophantine Problem and Modular Forms of Weight 3/2 from 1983, where he used the properties of the elliptic curve $E_t$ to relate the CNP to the Birch and Swinnerton-Dyer Conjecture.

One statement following from the Birch and Swinnerton-Dyer conjecture (BSD) is that if an elliptic curve $E$ has rank $r$, then the $L$-function $L(s, E)$ has a zero of order $r$ at $1$. The relation between lots of points on the curve and the existence of a zero is intuitive from the approximate relation that
\begin{equation}
L(1, E) \approx \lim_{X} \prod_{p \leq X} \frac{p}{\#E(\mathbb{F}_p)},
\end{equation}
so if $E$ has lots and lots of points then we should expect the multiplicands to be very small.

On the other hand, the elliptic curve $E_t: y^2 = x^3 – t^2 x$ has the interesting property that any point with $y \neq 0$ generates a free group of points on the curve. From the bijections alluded to above, a primitive right integer triangle $T$ with $\sqfree(T) = t$ corresponds to a point on $E_t$ with $y \neq 0$, and thus guarantees that there are lots of points on the curve. Tunnell showed that what I described as “lots of points” is actually enough points that $L(1, E)$ must be zero (assuming the relation between the rank of the curve and the value of $L(1, E)$ from BSD).

Tunnell proved that if BSD is true, then $L(1, E) = 0$ if and only if $n$ is a congruent number.

Yet for any elliptic curve we know how to compute $L(1, E)$ to guaranteed accuracy (for instance by using Dokchitser’s algorithm). Thus a corollary of Tunnell’s theorem is that BSD implies that there is an algorithm which can be used to determine definitively whether or not any particular integer $t$ is congruent.

This is the state of the art on the congruent number problem. Unfortunately, BSD (or even the somewhat weaker between BSD and mere nonzero rank of elliptic curves as is necessary for Tunnell’s result for the CNP) is quite far from being proven.

In this context, the main result of this paper is not as effective at actually determining whether a number is congruent or not. But it does have the benefit of not relying on any unknown conjecture.

And there is some potential follow-up questions. The sum $S_t(X)$ appears as an integral transform of the multiple Dirichlet series
\begin{equation}
\sum_{m,n} \frac{\tau(m-n)\tau(m)\tau(nt)\tau(m+n)}{m^s n^w}
\approx
\sum_{m,n} \frac{r_1(m-n)r_1(m)r_1(nt)r_1(m+n)}{m^s n^w},
\end{equation}
where $r_1(n)$ is $1$ if $n = 0$ or $2$ if $n$ is a positive square, and $0$ otherwise. Then $r_1(n)$ appears as the Fourier coefficients of the half-integral weight standard theta function
\begin{equation}
\theta(z)
= \sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 z}
= \sum_{n \geq 0} r_1(n) e^{2 \pi i n z},
\end{equation}
and $S_t(X)$ is a shifted convolution sum coming from some products of modular forms related to $\theta(z)$.

It may be possible to gain further understanding of the behavior of $S_t(X)$ (and therefore the congruent number problem) by studying the shifted convolution as coming from theta functions.

I would guess that there is a deep relation to Tunnell’s analysis in his 1983 paper, as in some sense he constructs appropriate products of three theta functions and uses them centrally in his proof. But I do not understand this relationship well enough yet to know whether it is possible to deepen our understanding of the CNP, BSD, or Tunnell’s proof. That is something to explore in the future.

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Notes from a talk at Tufts, Automorphic Forms Workshop

On 19 March I gave a talk at the 32nd Automorphic Forms Workshop, which ishosted by Tufts this year. The content of the talk concerned counting points on hyperboloids, and inparticular counting points on the three dimensional hyperboloid

$$\begin{equation}
X^2 + Y^2 = Z^2 + h
\end{equation}$$

for any fixed integer $h$. But thematically, I wanted to give another concrete example of using modularforms to compute some sort of arithmetic data, and to mention how the perhapsapparently unrelated topic of spectral theory appears even in such an arithmeticapplication.

Somehow, starting from counting points on $X^2 + Y^2 = Z^2 + h$ (which appearssimple enough on its own that I could probably put this in front of anelementary number theory class and they would feel comfortable experimentingaway on the topic), one gets to very scary-looking expressions like

$$\begin{equation}
\sum_{t_j}
\langle P_h^k, \mu_j \rangle
\langle \theta^2 \overline{\theta} y^{3/4}, \mu_j \rangle +
\sum_{\mathfrak{a}}\int_{(1/2)}
\langle P_h^k, E_h^k(\cdot, u) \rangle
\langle \theta^2 \overline{\theta} y^{3/4}, E_h^k(\cdot, u) \rangle du,
\end{equation}$$

which is full of lots of non-obvious symbols and is generically intimidating.

Part of the theme of this talk is to give a very direct idea of how one gets tothe very complicated spectral expansion from the original lattice-countingproblem. Stated differently, perhaps part of the theme is to describe a simple-lookingnail and a scary-looking hammer, and show that the hammer actually works quitewell in this case.

The slides for this talk are available here.

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Segregation, Gerrymandering, and Schelling’s Model

[This note is more about modeling some of the mathematics behind political events than politics themselves. And there are pretty pictures.]

Gerrymandering has become a recurring topic in the news. The Supreme Court of the US, as well as more state courts and supreme courts, is hearing multiple cases on partisan gerrymandering (all beginning with a case in Wisconsin).

Intuitively, it is clear that gerrymandering is bad. It allows politicians to choose their voters, instead of the other way around. And it allows the majority party to quash minority voices.

But how can one identify a gerrymandered map? To quote Justice Kennedy in his Concurrence the 2004 Supreme Court case Vieth v. Jubelirer:

When presented with a claim of injury from partisan gerrymandering, courts confront two obstacles. First is the lack of comprehensive and neutral principles for drawing electoral boundaries. No substantive definition of fairness in districting seems to command general assent. Second is the absence of rules to limit and confine judicial intervention. With uncertain limits, intervening courts–even when proceeding with best intentions–would risk assuming political, not legal, responsibility for a process that often produces ill will and distrust.

Later, he adds to the first obstacle, saying:

The object of districting is to establish “fair and effective representation for all citizens.” Reynolds v. Sims, 377 U.S. 533, 565—568 (1964). At first it might seem that courts could determine, by the exercise of their own judgment, whether political classifications are related to this object or instead burden representational rights. The lack, however, of any agreed upon model of fair and effective representation makes this analysis difficult to pursue.

From Justice Kennedy’s Concurrence emerges a theme — a “workable standard” of identifying gerrymandering would open up the possibility of limiting partisan gerrymandering through the courts. Indeed, at the core of the Wisconsin gerrymandering case is a proposed “workable standard”, based around the efficiency gap.

 

Thomas Schelling and Segregation

In 1971, American economist Thomas Schelling (who later won the Nobel Prize in Economics in 2005) published Dynamic Models of Segregation (Journal of Mathematical Sociology, 1971, Vol 1, pp 143–186). He sought to understand why racial segregation in the United States seems so difficult to combat.

He introduced a simple model of segregation suggesting that even if each individual person doesn’t mind living with others of a different race, they might still choose to segregate themselves through mild preferences. As each individual makes these choices, overall segregation increases.

I write this post because I wondered what happens if we adapt Schelling’s model to instead model a state and its district voting map. In place of racial segregation, I consider political segregation. Supposing the district voting map does not change, I wondered how the efficiency gap will change over time as people further segregate themselves.

It seemed intuitive to me that political segregation (where people who had the same political beliefs stayed largely together and separated from those with different political beliefs) might correspond to more egregious cases of gerrymandering. But to my surprise, I was (mostly) wrong.

Let’s set up and see the model.

(more…)

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Advent of Code: Day 4

This is a very short post in my collection working through this year’s Advent of Code challenges. Unlike the previous ones, this has no mathematical comments, as it was a very short exercise. This notebook is available in its original format on my github.

Day 4: High Entropy Passphrases

Given a list of strings, determine how many strings have no duplicate words.

This is a classic problem, and it’s particularly easy to solve this in python. Some might use collections.Counter, but I think it’s more straightforward to use sets.

The key idea is that the set of words in a sentence will not include duplicates. So if taking the set of a sentence reduces its length, then there was a duplicate word.

In [1]:
with open("input.txt", "r") as f:
    lines = f.readlines()
    
def count_lines_with_unique_words(lines):
    num_pass = 0
    for line in lines:
        s = line.split()
        if len(s) == len(set(s)):
            num_pass += 1
    return num_pass

count_lines_with_unique_words(lines)
Out[1]:
455

I think this is the first day where I would have had a shot at the leaderboard if I’d been gunning for it.

Part 2

Let’s add in another constraint. Determine how many strings have no duplicate words, even after anagramming. Thus the string

abc bac

is not valid, since the second word is an anagram of the first. There are many ways to tackle this as well, but I will handle anagrams by sorting the letters in each word first, and then running the bit from part 1 to identify repeated words.

In [2]:
with open("input.txt", "r") as f:
    lines = f.readlines()
    
sorted_lines = []
for line in lines:
    sorted_line = ' '.join([''.join(l) for l in map(sorted, line.split())])
    sorted_lines.append(sorted_line)

sorted_lines[:2]
    
Out[2]:
['bddjjow acimrv bcjjm anr flmmos fiosv',
 'bcmnoxy dfinyzz dgmp dfgioy hinrrv eeklpuu adgpw kqv']
In [3]:
count_lines_with_unique_words(sorted_lines)
Out[3]:
186
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Advent of Code: Day 3

This is the third notebook in my posts on the Advent of Code challenges. The notebook in its original format can be found on my github.

Day 3: Spiral Memory

Numbers are arranged in a spiral

17  16  15  14  13
18   5   4   3  12
19   6   1   2  11
20   7   8   9  10
21  22  23---> ...

Given an integer n, what is its Manhattan Distance from the center (1) of the spiral? For instance, the distance of 3 is $2 = 1 + 1$, since it’s one space to the right and one space up from the center.

Here’s my idea. The bottom right corner of the $k$th layer is the integer $(2k+1)^2$, since that’s how many integers are contained within that square. The other three corners in that layer are $(2k+1)^2 – 2k, (2k+1)^2 – 4k$, and $(2k+1)^2 – 6k$. Finally, the closest spot on the $k$th layer to the origin is at distance $k$: these are the four “axis locations” halfway between the corners, at $(2k+1)^2 – k, (2k+1)^2 – 3k, (2k+1)^2 – 5k$, and $(2k+1)^2 – 7k$.

For instance when $k = 1$, the bottom right is $(2 + 1)^2 = 9$, and the four “axis locations” are $9 – 1, 9 – 3, 9-5$, and $9-7$. The “axis locations” are $k$ away, and the corners are $2k$ away.

So I will first find which layer the number is on. Then I’ll figure out which side it’s on, and then how far away it is from the nearest “axis location” or “corner”.

My given number happens to be 289326.

In [1]:
import math

def find_lowest_larger_odd_square(n):
    upper = math.ceil(n**.5)
    if upper %2 == 0:
        upper += 1
    return upper
In [2]:
assert find_lowest_larger_odd_square(39) == 7
assert find_lowest_larger_odd_square(26) == 7
assert find_lowest_larger_odd_square(25) == 5
In [3]:
find_lowest_larger_odd_square(289326)
Out[3]:
539
In [4]:
539**2 - 289326
Out[4]:
1195

It happens to be that our integer is very close to an odd square.
The square is $539^2$, and the distance to that square is $538$ from the center.

Note that $539 = 2(269) + 1$, so this is the $269$th layer of the square.
The previous corner to $539^2$ is $539^2 – 538$, and the previous corner to that is $539^2 – 2\cdot538 = 539^2 – 1076$.
This is the nearest corner.
How far away from the square is this corner?

(more…)

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Advent of Code: Day 2

This is the second notebook in my posts on the Advent of Code challenges. This notebook in its original format can be found on my github.

Day 2: Corruption Checksum, part I

You are given a table of integers. Find the difference between the maximum and minimum of each row, and add these differences together.

There is not a lot to say about this challenge. The plan is to read the file linewise, compute the difference on each line, and sum them up.

In [1]:
with open("input.txt", "r") as f:
    lines = f.readlines()
lines[0]
Out[1]:
'5048\t177\t5280\t5058\t4504\t3805\t5735\t220\t4362\t1809\t1521\t230\t772\t1088\t178\t1794\n'
In [2]:
l = lines[0]
l = l.split()
l
Out[2]:
['5048',
 '177',
 '5280',
 '5058',
 '4504',
 '3805',
 '5735',
 '220',
 '4362',
 '1809',
 '1521',
 '230',
 '772',
 '1088',
 '178',
 '1794']
In [3]:
def max_minus_min(line):
    '''Compute the difference between the largest and smallest integer in a line'''
    line = list(map(int, line.split()))
    return max(line) - min(line)

def sum_differences(lines):
    '''Sum the value of `max_minus_min` for each line in `lines`'''
    return sum(max_minus_min(line) for line in lines)
In [4]:
testcase = ['5 1 9 5','7 5 3', '2 4 6 8']
assert sum_differences(testcase) == 18
In [5]:
sum_differences(lines)
Out[5]:
58975

Mathematical Interlude

In line with the first day’s challenge, I’m inclined to ask what we should “expect.” But what we should expect is not well-defined in this case. Let us rephrase the problem in a randomized sense.

Suppose we are given a table, $n$ lines long, where each line consists of $m$ elements, that are each uniformly randomly chosen integers from $1$ to $10$. We might ask what is the expected value of this operation, of summing the differences between the maxima and minima of each row, on this table. What should we expect?

As each line is independent of the others, we are really asking what is the expected value across a single row. So given $m$ integers uniformly randomly chosen from $1$ to $10$, what is the expected value of the maximum, and what is the expected value of the minimum?

 

Expected Minimum

Let’s begin with the minimum. The minimum is $1$ unless all the integers are greater than $2$. This has probability
$$ 1 – \left( \frac{9}{10} \right)^m = \frac{10^m – 9^m}{10^m}$$
of occurring. We rewrite it as the version on the right for reasons that will soon be clear.
The minimum is $2$ if all the integers are at least $2$ (which can occur in $9$ different ways for each integer), but not all the integers are at least $3$ (each integer has $8$ different ways of being at least $3$). Thus this has probability
$$ \frac{9^m – 8^m}{10^m}.$$
Continuing to do one more for posterity, the minimum is $3$ if all the integers are at least $3$ (each integer has $8$ different ways of being at least $3$), but not all integers are at least $4$ (each integer has $7$ different ways of being at least $4$). Thus this has probability

$$ \frac{8^m – 7^m}{10^m}.$$

And so on.

Recall that the expected value of a random variable is

$$ E[X] = \sum x_i P(X = x_i),$$

so the expected value of the minimum is

$$ \frac{1}{10^m} \big( 1(10^m – 9^m) + 2(9^m – 8^m) + 3(8^m – 7^m) + \cdots + 9(2^m – 1^m) + 10(1^m – 0^m)\big).$$

This simplifies nicely to

$$ \sum_ {k = 1}^{10} \frac{k^m}{10^m}. $$

Expected Maximum

The same style of thinking shows that the expected value of the maximum is

$$ \frac{1}{10^m} \big( 10(10^m – 9^m) + 9(9^m – 8^m) + 8(8^m – 7^m) + \cdots + 2(2^m – 1^m) + 1(1^m – 0^m)\big).$$

This simplifies to

$$ \frac{1}{10^m} \big( 10 \cdot 10^m – 9^m – 8^m – \cdots – 2^m – 1^m \big) = 10 – \sum_ {k = 1}^{9} \frac{k^m}{10^m}.$$

Expected Difference

Subtracting, we find that the expected difference is

$$ 9 – 2\sum_ {k=1}^{9} \frac{k^m}{10^m}. $$

From this we can compute this for each list-length $m$. It is good to note that as $m \to \infty$, the expected value is $9$. Does this make sense? Yes, as when there are lots of values we should expect one to be a $10$ and one to be a $1$. It’s also pretty straightforward to see how to extend this to values of integers from $1$ to $N$.

Looking at the data, it does not appear that the integers were randomly chosen. Instead, there are very many relatively small integers and some relatively large integers. So we shouldn’t expect this toy analysis to accurately model this problem — the distribution is definitely not uniform random.
But we can try it out anyway.

(more…)

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