A trailer for p-adic analysis, first half: USA TST 2003

I think this post is more than two years late in coming, but anywhow…

This post introduces the -adic integers , and the -adic numbers . The one-sentence description is that these are “integers/rationals carrying full mod information” (and only that information).

The first four sections will cover the founding definitions culminating in a short solution to a USA TST problem.

In this whole post, is always a prime. Much of this is based off of Chapter 6B from Straight from the Book.

1. Motivation

Before really telling you what and are, let me tell you what you might expect them to do.

In elementary/olympiad number theory, we’re already well-familiar with the following two ideas:

• Taking modulo a prime or prime , and
• Looking at the exponent .

Let me expand on the first point. Suppose we have some Diophantine equation. In olympiad contexts, one can take an equation modulo to gain something else to work with. Unfortunately, taking modulo loses some information: (the reduction is far from injective).

If we want finer control, we could consider instead taking modulo , rather than taking modulo . This can also give some new information (cubes modulo , anyone?), but it has the disadvantage that isn’t a field, so we lose a lot of the nice algebraic properties that we got if we take modulo .

One of the goals of -adic numbers is that we can get around these two issues I described. The -adic numbers we introduce is going to have the following properties:

1. You can “take modulo for all at once”.

   In olympiad contexts, we are used to picking a particular modulus and then seeing what happens if we take that modulus. But with -adic numbers, we won’t have to make that choice. An equation of -adic numbers carries enough information to take modulo .

2. The numbers form a field, the nicest possible algebraic structure: makes sense. Contrast this with , which is not even an integral domain.
3. It doesn’t lose as much information as taking modulo does: rather than the surjective we have an injective map .
4. Despite this, you “ignore” some “irrelevant” data. Just like taking modulo , you want to zoom-in on a particular type of algebraic information, and this means necessarily losing sight of other things.

   Let me draw an analogy. We know that the equation

   

   has no integer solutions, because, well, squares are nonnegative. But you will find that this equation has solutions modulo any prime , because once you take modulo you stop being able to talk about numbers being nonnegative. The same thing will happen if we work in -adics: the above equation has a solution in for every prime .

So, you can think of -adic numbers as the right tool to use if you only really care about modulo information, but normal isn’t quite powerful enough.

To be more concrete, I’ll give a poster example now:

Example 1 (USA TST 2002/2)

For a prime , show the value of

does not depend on .

Here is a problem where we clearly only care about -type information. Yet it’s a nontrivial challenge to do the necessary manipulations mod (try it!). The basic issue is that there is no good way to deal with the denominators modulo (in part is not even an integral domain).

However, with -adic analysis we’re going to be able to overcome these limitations and give a “straightforward” proof by using the identity

Such an identity makes no sense over or for converge reasons, but it will work fine over the , which is all we need.

2. Algebraic perspective

We now construct and . I promised earlier that a -adic integer will let you look at “all residues modulo ” at once. This definition will formalize this.

2.1. Definition of

Definition 2 (Introducing )

A -adic integer is a sequence

of residues modulo for each integer , satisfying the compatibility relations for .

The set of -adic integers forms a ring under component-wise addition and multiplication.

Example 3 (Some -adic integers)

Let . Every usual integer generates a (compatible) sequence of residues modulo for each , so we can view each ordinary integer as -adic one:

On the other hand, there are sequences of residues which do not correspond to any usual integer despite satisfying compatibility relations, such as

which can be thought of as .

In this way we get an injective map

which is not surjective. So there are more -adic integers than usual integers.

(Remark for experts: those of you familiar with category theory might recognize that this definition can be written concisely as

where the inverse limit is taken across .)

Exercise 4

Check that is an integral domain.

2.2. Base expansion

Here is another way to think about -adic integers using “base ”. As in the example earlier, every usual integer can be written in base , for example

More generally, given any , we can write down a “base ” expansion in the sense that there are exactly choices of given . Continuing the example earlier, we would write

and in general we can write

where , such that the equation holds modulo for each . Note the expansion is infinite to the left, which is different from what you’re used to.

(Amusingly, negative integers also have infinite base expansions: , corresponding to .)

Thus you may often hear the advertisement that a -adic integer is an “possibly infinite base expansion”. This is correct, but later on we’ll be thinking of in a more and more “analytic” way, and so I prefer to think of this as a “Taylor series with base ”. Indeed, much of your intuition from generating functions (where is a field) will carry over to .

2.3. Constructing

Here is one way in which your intuition from generating functions carries over:

Proposition 5 (Non-multiples of are all invertible)

The number is invertible if and only if . In symbols,

Contrast this with the corresponding statement for : a generating function is invertible iff .

Proof: If then , so clearly not invertible. Otherwise, for all , so we can take an inverse modulo , with . As the are themselves compatible, the element is an inverse.

Example 6 (We have )

We claim the earlier example is actually

Indeed, multiplying it by gives

(Compare this with the “geometric series” . We’ll actually be able to formalize this later, but not yet.)

Remark 7 ( is an integer for )

The earlier proposition implies that (among other things); your intuition about what is an “integer” is different here! In olympiad terms, we already knew made sense, which is why calling an “integer” in the -adics is correct, even though it doesn’t correspond to any element of .

Fun (but trickier) exercise: rational numbers correspond exactly to eventually periodic base expansions.

With this observation, here is now the definition of .

Definition 8 (Introducing )

Since is an integral domain, we let denote its field of fractions. These are the -adic numbers.

Continuing our generating functions analogy:

This means is “Laurent series with base ”, and in particular according to the earlier proposition we deduce:

Proposition 9 ( looks like formal Laurent series)

Every nonzero element of is uniquely of the form

Thus, continuing our base analogy, elements of are in bijection with “Laurent series”

for . So the base representations of elements of can be thought of as the same as usual, but extending infinitely far to the left (rather than to the right).

(Fair warning: the field has characteristic zero, not .)

Remark 10 (Warning on fraction field)

This result implies that you shouldn’t think about elements of as (for ) in practice, even though this is the official definition (and what you’d expect from the name ). The only denominators you need are powers of .

To keep pushing the formal Laurent series analogy, is usually not thought of as quotient of generating functions but rather as “formal series with some negative exponents”. You should apply the same intuition on .

(At this point I want to make a remark about the fact , connecting it to the wish-list of properties I had before. In elementary number theory you can take equations modulo , but if you do the quantity doesn’t make sense unless you know . You can’t fix this by just taking modulo since then you need to get , ad infinitum. You can work around issues like this, but the nice feature of and is that you have modulo information for “all at once”: the information of packages all the modulo information simultaneously. So you can divide by with no repercussions.)

3. Analytic perspective

3.1. Definition

Up until now we’ve been thinking about things mostly algebraically, but moving forward it will be helpful to start using the language of analysis. Usually, two real numbers are considered “close” if they are close on the number of line, but for -adic purposes we only care about modulo information. So, we’ll instead think of two elements of or as “close” if they differ by a large multiple of .

For this we’ll borrow the familiar from elementary number theory.

Definition 11 (-adic valuation and absolute value)

We define the -adic valuation in the following two equivalent ways:

• For we let be the largest such that (or if ). Then extend to all of by .
• Each can be written uniquely as for , . We let .

By convention we set . Finally, define the -adic absolute value by

In particular .

This fulfills the promise that and are close if they look the same modulo for large ; in that case is large and accordingly is small.

3.2. Ultrametric space

In this way, and becomes a metric space with metric given by .

Exercise 12

Suppose is continuous and for every . Prove that .

In fact, these spaces satisfy a stronger form of the triangle inequality than you are used to from .

Proposition 13 ( is an ultrametric)

For any , we have the strong triangle inequality

Equality holds if (but not only if) .

However, is more than just a metric space: it is a field, with its own addition and multiplication. This means we can do analysis just like in or : basically, any notion such as “continuous function”, “convergent series”, et cetera has a -adic analog. In particular, we can define what it means for an infinite sum to converge:

Definition 14 (Convergence notions)

Here are some examples of -adic analogs of “real-world” notions.

• A sequence , \dots converges to a limit if .
• The infinite series converges if the sequence of partial sums , , \dots, converges to some limit.
• \dots et cetera \dots

With this definition in place, the “base ” discussion we had earlier is now true in the analytic sense: if then

Indeed, the th partial sum is divisible by , hence the partial sums approach as .

While the definitions are all the same, there are some changes in properties that should be true. For example, in convergence of partial sums is simpler:

Proposition 15 ( iff convergence of series)

A series in converges to some limit if and only if .

Contrast this with in . You can think of this as a consequence of strong triangle inequality. Proof: By multiplying by a large enough power of , we may assume . (This isn’t actually necessary, but makes the notation nicer.)

Observe that must eventually stabilize, since for large enough we have . So let be the eventual residue modulo of for large . In the same way let be the eventual residue modulo , and so on. Then one can check we approach the limit .

Here’s a couple exercises to get you used to thinking of and as metric spaces.

Exercise 16 ( is compact)

Show that is not compact, but is. (For the latter, I recommend using sequential continuity.)

Exercise 17 (Totally disconnected)

Show that both and are totally disconnected: there are no connected sets other than the empty set and singleton sets.

3.3. More fun with geometric series

While we’re at it, let’s finally state the -adic analog of the geometric series formula.

Proposition 18 (Geometric series)

Let with . Then

Proof: Note that the partial sums satisfy , and as since .

So, is really a correct convergence in . And so on.

If you buy the analogy that is generating functions with base , then all the olympiad generating functions you might be used to have -adic analogs. For example, you can prove more generally that:

Theorem 19 (Generalized binomial theorem)

If and , then for any we have the series convergence

(I haven’t defined , but it has the properties you expect.) The proof is as in the real case; even the theorem statement is the same except for the change for the extra subscript of . I won’t elaborate too much on this now, since -adic exponentiation will be described in much more detail in the next post.

3.4. Completeness

Note that the definition of could have been given for as well; we didn’t need to introduce it (after all, we have in olympiads already). The big important theorem I must state now is:

Theorem 20 ( is complete)

The space is the completion of with respect to .

This is the definition of you’ll see more frequently; one then defines in terms of (rather than vice-versa) according to

(Remark for experts: is a field with a non-Arcihmedian valuation; then is its valuation ring.)

Let me justify why this definition is philosophically nice.

Suppose you are a numerical analyst and you want to estimate the value of the sum

to within . The sum consists entirely of rational numbers, so the problem statement would be fair game for ancient Greece. But it turns out that in order to get a good estimate, it really helps if you know about the real numbers: because then you can construct the infinite series , and deduce that , up to some small error term from the terms past , which can be bounded.

Of course, in order to have access to enough theory to prove that , you need to have the real numbers; it’s impossible to do serious analysis in the non-complete space , where e.g. the sequence , , , , \dots is considered “not convergent” because . Instead, all analysis is done in the completion of , namely .

Now suppose you are an olympiad contestant and want to estimate the sum

to within mod (i.e. to within in ). Even though is a rational number, it still helps to be able to do analysis with infinite sums, and then bound the error term (i.e. take mod ). But the space is not complete with respect to either, and thus it makes sense to work in the completion of with respect to . This is exactly .

4. Solving USA TST 2002/2

Let’s finally solve Example~1, which asks to compute

Armed with the generalized binomial theorem, this becomes straightforward.

Using the elementary facts that and , this solves the problem.

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A story of block-ascending permutations

I recently had a combinatorics paper appear in the EJC. In this post I want to brag a bit by telling the “story” of this paper: what motivated it, how I found the conjecture that I originally did, and the process that eventually led me to the proof, and so on.

This work was part of the Duluth REU 2017, and I thank Joe Gallian for suggesting the problem.

1. Background

Let me begin by formulating the problem as it was given to me. First, here is the definition and notation for a “block-ascending” permutation.

Definition 1

For nonnegative integers , …, an -ascending permutation is a permutation on whose descent set is contained in . In other words the permutation ascends in blocks of length , , …, , and thus has the form

for which for all .

It turns out that block-ascending permutations which also avoid an increasing subsequence of certain length have nice enumerative properties. To this end, we define the following notation.

Definition 2

Let denote the set of -ascending permutations which avoid the pattern .

(The reason for using will be explained later.) In particular, if .

Example 3

Here is a picture of a permutation in (but not in , since one can see an increasing length subsequence shaded). We would denote it .

Now on to the results. A 2011 paper by Joel Brewster Lewis (JBL) proved (among other things) the following result:

Theorem 4 (Lewis 2011)

The sets and are in bijection with Young tableau of shape .

Remark 5

When , this implies , which is the set of -avoiding permutations of length , is in bijection with the Catalan numbers; so is which is the set of -avoiding zig-zag permutations.

Just before the Duluth REU in 2017, Mei and Wang proved that in fact, in Lewis’ result one may freely mix and ‘s. To simplify notation,

Definition 6

Let . Then denotes where

Theorem 7 (Mei, Wang 2017)

The sets are also in bijection with Young tableau of shape .

The proof uses the RSK correspondence, but the authors posed at the end of the paper the following open problem:

Problem

Find a direct bijection between the sets above, not involving the RSK correspondence.

This was the first problem that I was asked to work on. (I remember I received the problem on Sunday morning; this actually matters a bit for the narrative later.)

At this point I should pause to mention that this notation is my own invention, and did not exist when I originally started working on the problem. Indeed, all the results are restricted to the case where for each , and so it was unnecessary to think about other possibilities for : Mei and Wang’s paper use the notation . So while I’ll continue to use the notation in the blog post for readability, it will make some of the steps more obvious than they actually were.

2. Setting out

Mei and Wang’s paper originally suggested that rather than finding a bijection for any and , it would suffice to biject

and then compose two such bijections. I didn’t see why this should be much easier, but it didn’t seem to hurt either.

As an example, they show how to do this bijection with and . Indeed, suppose . Then is an increasing sequence of length right at the start of . So had better be the largest element in the permutation: otherwise later in the biggest element would complete an ascending permutation of length already! So removing gives a bijection between .

But if you look carefully, this proof does essentially nothing with the later blocks. The exact same proof gives:

Proposition 8

Suppose . Then there is a bijection

by deleting the st element of the permutation (which must be largest one).

Once I found this proposition I rejected the initial suggestion of specializing . The “easy case” I had found told me that I could take a set and delete the single element from it. So empirically, my intuition from this toy example told me that it would be easier to find bijections whee and were only “a little different”, and hope that the resulting bijection only changed things a little bit (in the same way that in the toy example, all the bijection did was delete one element). So I shifted to trying to find small changes of this form.

3. The fork in the road

3.1. Wishful thinking

I had a lucky break of wishful thinking here. In the notation with , I had found that one could replace with either or freely. (But this proof relied heavily on the fact the block really being on the far left.) So what other changes might I be able to make?

There were two immediate possibilities that came to my mind.

• Deletion: We already showed could be changed from to for any . If we can do a similar deletion with for any , not just , then we would be done.
• Swapping: If we can show that two adjacent ‘s could be swapped, that would be sufficient as well. (It’s also possible to swap non-adjacent ‘s, but that would cause more disruption for no extra benefit.)

Now, I had two paths that both seemed plausible to chase after. How was I supposed to know which one to pick? (Of course, it’s possible neither work, but you have to start somewhere.)

Well, maybe the correct thing to do would have to just try both. But it was Sunday afternoon by the time I got to this point. Granted, it was summer already, but I knew that come Monday I would have doctor appointments and other trivial errands to distract me, so I decided I should pick one of them and throw the rest of the day into it. But that meant I had to pick one.

(I confess that I actually already had a prior guess: the deletion approach seemed less likely to work than the swapping approach. In the deletion approach, if is somewhere in the middle of the permutation, it seemed like deleting an element could cause a lot of disruption. But the swapping approach preserved the total number of elements involved, and so seemed more likely that I could preserve structure. But really I was just grasping at straws.)

3.2. Enter C++

Yeah, I cheated. Sorry.

Those of you that know anything about my style of math know that I am an algebraist by nature — sort of. It’s more accurate to say that I depend on having concrete examples to function. True, I can’t do complexity theory for my life, but I also haven’t been able to get the hang of algebraic geometry, despite having tried to learn it three or four times by now. But enumerative combinatorics? OH LOOK EXAMPLES.

Here’s the plan: let . Then using a C++ computer program:

• Enumerate all the permutations in .
• Enumerate all the permutations in .
• Enumerate all the permutations in .

If the deletion approach is right, then I would hope and look pretty similar. On the flip side, if the swapping approach is right, then and should look close to each other instead.

It’s moments like this where my style of math really shines. I don’t have to make decisions like the above off gut-feeling: do the “data science” instead.

3.3. A twist of fate

Except this isn’t actually what I did, since there was one problem. Computing the longest increasing subsequence of a length permutation takes time, and there are or so permutations. But when , we have , which is a pretty big number. Unfortunately, my computer is not really that fast, and I didn’t really have the patience to implement the “correct” algorithms to bring the runtime down.

The solution? Use instead.

In a deep irony that I didn’t realize at the time, it was this moment when I introduced the notation, and for the first time allowed the to not be in . My reasoning was that since I was only doing this for heuristic reasons, I could instead work with and probably not change much about the structure of the problem, while replacing , which would run times faster. This was okay since all I wanted to do was see how much changing the “middle” would disrupt the structure.

And so the new plan was:

• Enumerate all the permutations in .
• Enumerate all the permutations in .
• Enumerate all the permutations in .

I admit I never actually ran the enumeration with , because the route with and turned out to be even more promising than I expected. When I compared the empirical data for the sets and , I found that the number of permutations with any particular triple were equal. In other words, the outer blocks were preserved: the bijection

does not tamper with the outside blocks of length and .

This meant I was ready to make the following conjecture. Suppose , . There is a bijection

which only involves rearranging the elements of the th and st blocks.

4. Rooting out the bijection

At this point I was in a quite good position. I had pinned down the problem to a finding a particular bijection that I was confident had to exist, since it was showing up to the empirical detail.

Let’s call this mythical bijection . How could I figure out what it was?

4.1. Hunch: preserves order-isomorphism

Let me quickly introduce a definition.

Definition 9

We say two words and are order-isomorphic if if and only . Then order-isomorphism gives equivalence classes, and there is a canonical representative where the letters are ; this is called a reduced word.

Example 10

The words , and are order-isomorphic; the last is reduced.

Now I guessed one more property of : this should order-isomorphism.

What do I mean by this? Suppose in one context changed to ; then we would expect that in another situation we should have changing to . Indeed, we expect (empirically) to not touch surrounding outside blocks, and so it would be very strange if behaved differently due to far-away numbers it wasn’t even touching.

So actually I’ll just write

for this example, reducing the words in question.

4.2. Keep cheating

With this hunch it’s possible to cheat with C++ again. Here’s how.

Let’s for concreteness suppose and the particular sets

Well, it turns out if you look at the data:

• The only element of which starts with and ends with is .
• The only element of which starts with and ends with is .

So that means that is changed to . Thus the empirical data shows that

In general, it might not be that clear cut. For example, if we look at the permutations starting with and , there is more than one.

• and are both in .
• and are both in in .

Thus

but we can’t tell which one goes to which (although you might be able to guess).

Fortunately, there is lots of data. This example narrowed down to two values, but if you look at other places you might have different data on . Since we think is behaving the same “globally”, we can piece together different pieces of data to get narrower sets. Even better, is a bijection, so once we match either of or , we’ve matched the other.

You know what this sounds like? Perfect matchings.

So here’s the experimental procedure.

• Enumerate all permutations in and .
• Take each possible tuple , and look at the permutations that start and end with those particular four elements. Record the reductions of and for all these permutations. We call these input words and output words, respectively. Each output word is a “candidate” of for a input word.
• For each input word that appeared, take the intersection of all output words that appeared. This gives a bipartite graph , with input words being matched to their candidates.
• Find perfect matchings of the graph.

And with any luck that would tell us what is.

4.3. Results

Luckily, the bipartite graph is quite sparse, and there was only one perfect matching.

246|1357 => 2467|135
247|1356 => 2457|136
256|1347 => 2567|134
257|1346 => 2357|146
267|1345 => 2367|145
346|1257 => 3467|125
347|1256 => 3457|126
356|1247 => 3567|124
357|1246 => 1357|246
367|1245 => 1367|245
456|1237 => 4567|123
457|1236 => 1457|236
467|1235 => 1467|235
567|1234 => 1567|234

If you look at the data, well, there are some clear patterns. Exactly one number is “moving” over from the right half, each time. Also, if is on the right half, then it always moves over.

Anyways, if you stare at this for an hour, you can actually figure out the exact rule:

Claim 11

Given an input , move if is the largest index for which , or if no such index exists.

And indeed, once I have this bijection, it takes maybe only another hour of thinking to verify that this bijection works as advertised, thus solving the original problem.

Rather than writing up what I had found, I celebrated that Sunday evening by playing Wesnoth for 2.5 hours.

5. Generalization

5.1. Surprise

On Monday morning I was mindlessly feeding inputs to the program I had worked on earlier and finally noticed that in fact and also had the same cardinality. Huh.

It seemed too good to be true, but I played around some more, and sure enough, the cardinality of seemed to only depend on the order of the ‘s. And so at last I stumbled upon the final form the conjecture, realizing that all along the assumption that I had been working with was a red herring, and that the bijection was really true in much vaster generality. There is a bijection

which only involves rearranging the elements of the th and st blocks.

It also meant I had more work to do, and so I was now glad that I hadn’t written up my work from yesterday night.

5.2. More data science

I re-ran the experiment I had done before, now with . (This was interesting, because the elements in question could now have either longest increasing subsequence of length , or instead of length .)

The data I obtained was:

246|13578 => 24678|135
247|13568 => 24578|136
248|13567 => 24568|137
256|13478 => 25678|134
257|13468 => 23578|146
258|13467 => 23568|147
267|13458 => 23678|145
268|13457 => 23468|157
278|13456 => 23478|156
346|12578 => 34678|125
347|12568 => 34578|126
348|12567 => 34568|127
356|12478 => 35678|124
357|12468 => 13578|246
358|12467 => 13568|247
367|12458 => 13678|245
368|12457 => 13468|257
378|12456 => 13478|256
456|12378 => 45678|123
457|12368 => 14578|236
458|12367 => 14568|237
467|12358 => 14678|235
468|12357 => 12468|357
478|12356 => 12478|356
567|12348 => 15678|234
568|12347 => 12568|347
578|12346 => 12578|346
678|12345 => 12678|345

Okay, so it looks like:

• exactly two numbers are moving each time, and
• the length of the longest run is preserved.

Eventually, I was able to work out the details, but they’re more involved than I want to reproduce here. But the idea is that you can move elements “one at a time”: something like

while preserving the length of increasing subsequences at each step.

So, together with the easy observation from the beginning, this not only resolves the original problem, but also gives an elegant generalization. I had now proved:

Theorem 12

For any , …, , the cardinality

does not depend on the order of the ‘s.

6. Discovered vs invented

Whenever I look back on this, I can’t help thinking just how incredibly lucky I got on this project.

There’s this perpetual debate about whether mathematics is discovered or invented. I think it’s results like this which make the case for “discovered”. I did not really construct the bijection myself: it was “already there” and I found it by examining the data. In another world where did not exist, all the creativity in the world wouldn’t have changed anything.

So anyways, that’s the behind-the-scenes tour of my favorite combinatorics paper.

DNSCrypt Setup with PDNSD

Here are notes for setting up DNSCrypt on Arch Linux, using pdnsd as a DNS cache, assuming the use of NetworkManager. I needed it one day since the network I was using blocked traffic to external DNS servers (parental controls), and the DNS server provided had an outdated entry for hmmt.co. (My dad then pointed out to me I could have just hard-coded the necessary IP address in /etc/hosts, oops.)

For the whole process, useful commands to test with are:

• nslookup hmmt.co will tell you the IP used and the server from which it came.
• dig http://www.hmmt.co gives much more detailed information to this effect. (From bind-tools.)
• dig @127.0.0.1 http://www.hmmt.co lets you query a specific DNS server (in this case 127.0.0.1).
• drill @127.0.0.1 http://www.hmmt.co behaves similarly.

First, pacman -S pdnsd dnscrypt-proxy (with sudo ostensibly, but I’ll leave that out here and henceforth).

Run systemctl edit dnscrypt-proxy.socket and fill in override.conf with

[Socket]
ListenStream=
ListenDatagram=
ListenStream=127.0.0.1:40
ListenDatagram=127.0.0.1:40

Optionally, one can also specify which server which DNS serve to use with systemctl edit dnscrypt-proxy.service. For example for cs-uswest I write

[Service]
ExecStart=
ExecStart=/usr/bin/dnscrypt-proxy \
      -R cs-uswest

The empty ExecStart= is necessary, since otherwise systemctl will complain about multiple ExecStart commands.

This thus configures dnscrypt-proxy to listen on 127.0.0.1, port 40.

Now we configure pdnsd to listen on port 53 (default) for cache, and relay cache misses to dnscrypt-proxy. This is accomplished by using the following for /etc/pdnsd.conf:

global {
    perm_cache = 1024;
    cache_dir = "/var/cache/pdnsd";
    run_as = "pdnsd";
    server_ip = 127.0.0.1;
    status_ctl = on;
    query_method = udp_tcp;
    min_ttl = 15m;       # Retain cached entries at least 15 minutes.
    max_ttl = 1w;        # One week.
    timeout = 10;        # Global timeout option (10 seconds).
    neg_domain_pol = on;
    udpbufsize = 1024;   # Upper limit on the size of UDP messages.
}

server {
    label = "dnscrypt-proxy";
    ip = 127.0.0.1;
    port = 40;
    timeout = 4;
    proxy_only = on;
}

source {
    owner = localhost;
    file = "/etc/hosts";
}

Now it remains to change the DNS server from whatever default is used into 127.0.0.1. For NetworkManager users, it is necessary to edit /etc/NetworkManager/NetworkManager.conf to prevent it from overriding this file:

[main]
...
dns=none

This will cause resolv.conf to be written as an empty file by NetworkManager: in this case, the default 127.0.0.1 is used as the nameserver, which is what we want.

Needless to say, one finishes with

systemctl enable dnscrypt-proxy
systemctl start dnscrypt-proxy
systemctl enable pdnsd
systemctl start pdnsd

Shifting PDF’s using gs

Some time ago I was reading the 18.785 analytic NT notes
to try and figure out some sections of Davenport that I couldn’t understand.
These notes looked nice enough that I decided I should probably print them out,
But much to my annoyance, I found that almost all the top margins were too tiny, and the bottom margins too big.
(I say “almost all” since the lectures 19 and 24 (Bombieri proof and elliptic curves) were totally fine, for inexplicable reasons).

Thus, instead of reading Davenport like I told myself to, I ended up learning enough GhostScript flags to write the following short script,
which I’m going to share today so that other people can find better things to do with their time.

    #!/bin/bash
    for file in $@
    do
        echo "Shifting $file ..."
        gs \
            -sDEVICE=pdfwrite \
            -o shifted-$file \
            -dPDFSETTINGS=/prepress \
            -c "<</PageOffset [0 -36]>>; setpagedevice" \
            -f $file
    done
    

The arguments 0 and -36 indicate to not change the left/right margins, but to shift the content vertically downwards by 36pt (a half inch).
Of course, they can and should be adjusted depending on specific task.
Invocation is the standard ./script-name.sh *.pdf (or whatever).

(Aside: ironically, this decreased the file sizes of the affected PDF’s.)

Git Aliases

For Git users:

I’ve recently discovered the joy that is git aliases, courtesy of this blog post. To return to the favor, I thought I’d share the ones that I came up with.

For those of you that don’t already know, Git allows you to make aliases — shortcuts for commands. Specifically, if you add the following lines to your .gitconfig:

[alias]
    cm = commit
    co = checkout
    br = branch

Then running git cm will expand as git commit, and git co master is git checkout master, and so on. You can see how this might make you happy because it could save a few keystrokes. But I think it’s more useful than that — let me share what I did.

The first thing I did was add

pu = pull origin
psh = push origin

and permanently save myself the frustration of forgetting to type origin. Not bad. Even more helpful was the command

undo = reset --soft HEAD~1

Thus if I make a commit and then decide I want to undo it, rather than having to remember (or Google) what the correct incantations were, I just have to type git undo. It’s really an undo button!

Now for the fun part — some of Git’s useful commands are pretty verbose and take up lots of space. For example, here’s what git status looks like:

Kind of verbose if you ask me, and by now I know what “git pull” does. Fortunately, it turns out that there are some options you can run to make this look nicer. All you have to do is say git status -s -b, or in the context of this post, set the alias

ss = status -s -b

Then you get

which is much cooler.

Similarly, git log takes up a lot of space. I have the following format, which I’ve edited from the above blog post to suit my own tastes.

ls = log -n 16 --pretty=format:"%C(yellow)%h\\ %C(cyan)[%cn]\\ %C(reset)%s\\ %C(red)%d" --decorate 
ll = log -n 6 --pretty=format:"%C(yellow)%h\\ %C(cyan)[%cn]\\ %C(reset)%s\\ %C(red)%ad" --decorate --date=short --stat

These give in my opinion the much more readable format

If you’re on a branch that does merges, you might also have fun with

tree = log -n 16 --pretty=format:"%C(yellow)%h\\ %C(cyan)[%cn]\\ %C(reset)%s\\ %C(red)%d" --decorate --graph

which will put these into a graphical tree for your viewing pleasure.

And finally a few more that I find nice, some again taken directly from the link above:

fail = commit --amend # to avoid stupid "oops typo" commits
rb = rebase
rbc = rebase --continue
bis = bisect
dc = checkout --
assume = update-index --assume-unchanged
unassume = update-index --no-assume-unchanged
assumed = "!git ls-files -v | grep ^h | cut -c 3-"

(Here “dc” is short for “discard”, since git dc file discards the changes to that file.) And that’s just the beginning of what you can do!

Pre-emptive answer: I’m also using git-completion (for tab-completing in git) and git-prompt with the line

export PS1='\[33[0;32m\]${debian_chroot:+($debian_chroot)}\u@\h \[33[0;33m\]\w$(__git_ps1 " \[33[1;31m\]#%s")\n\[33[0m\]\$ '

in my bashrc. That’s where the branch indicators are coming from. The terminal is XFCE4.