Yeah, I like subleq.

• compiler is extremely fast, faster even than tinycc
• strongly statically typed: all values are ints. Since it’s all of them, you don’t even need to write it!
• memory safe: the entire (virtual) address space is guaranteed to be accessible at all times so there’s no way to leak any of it (can’t release it anyway) or to segfault (it’s all accessible).

Subleq is the obvious winner in my mind.

There is no official definition, in part because there isn’t any formal way to define the term that satisfies our intuition.

Most treatments will handle “transpiling” as a special case of “compiling” and some will even handle decompilation as a special case where the object language is higher level than the source. Of course, even defining “higher level” can be quite hard.

Plenty of languages “compile to C” and I see no issue with saying something “compiles to js,” especially given that js mostly lacks features of purescript rather than the other way around.

“Monadic type” has something like three meanings depending on context, and it’s not clear which one you mean. One of them is common in math, but not so common in programming, so probably not that. But neither “parametric types with a single argument” nor “types that encode a category-theoretic monad” have the property you say, as far as I know.

I imagine you’re probably referring to the latter, since the optional monad exists. That’s very different from returning null. The inhabitants of Integer in Java, for example, are the boxed machine ints and null. The inhabitants of Optional[Integer] (it won’t let me use angle brackets here) are Optional.of(i) for each machine int i, Optional.empty(), and null.

Optional.empty() is not null and should not be called a “Null object.” It’s also not of type Integer, so you’re not even allowed to return it unless the function type explicitly says so. Writing such function types is pretty uncommon to do in java programs but it’s more normal in kotlin. In languages like Haskell, which don’t have null at all, this is idiomatic.

I’ve only ever seen “one-time” in cryptography to refer to One-Time Pads (OTP). They are literally uncrackable (because every possible plaintext could be encoded by every possible ciphertext) but they achieve that by using a shared private key. The cipher becomes attackable if the key is re-used, hence the “one-time.”

But that key has to be exchanged somehow, and that exchange can be attacked instead. Key exchange algorithms can’t necessarily transfer every possible OTP which means eavesdropping on the exchange would make an OTP attackable. So the best option we know of that doesn’t require secret meetings to share OTPs* really is to use RSA encryption. Once we have efficient quantum-resistant schemes, they’ll be the best option we know.

* and let’s be honest, secret meetings can be eavesdropped on as well.

Bril is the only compiler IL I know of that is specifically designed for education.

R. Kent Dybvig’s compilers course has had approximately 15 “intermediate” representations designed for his course since at least 2004 – a consequence of teaching the course using the nanopass compiler framework for scheme. You could broadly divide these into “representations that are restrictions of scheme,” and “representations that are increasingly-annotated versions of UIL” where UIL is the underlying intermediate representation. As far as I know, UIL was also designed for this course.

How does this compare with GumTree? It’s weird that the page doesn’t even mention existing state-of-the-art tools for this task.

edit: I’ve compared GumTree and difftastic myself while working on a project this past week. Difftastic is harder to use programatically (the JSON format is unstable and leaves something to be desired) but other than that it’s miles and miles better.

Only the number of shuffles is linear. Shuffling an array and marking/deleting correctly-placed elements still take linear time even with a “placement oracle.” It’s at best O(n^2) so the algorithm still wouldn’t be a good sorting algorithm.

It’s like doing selection sort, except we’re selecting a random set of elements (from that poisson distribution) instead of the smallest one.

I’m not sure the median is what you want. The worst case behavior is unbounded. There is no guarantee that such an algorithm ever actually terminates, and in fact (with very low probability) it may not! But that means there is no well-defined median; we can’t enumerate the space.

So let’s instead ask about the average, which is meaningful, as the increasingly high iteration-count datapoints are also decreasingly likely, in a way that we can compute without trying to enumerate all possible sequences of shuffles.

Consider the problem like this: at every iteration, remove the elements that are in the correct positions and continue sorting a shorter list. As long as we keep getting shuffles where nothing is in the correct position, we can go forever. Such shuffles are called derangements, and the probability of getting one is 1/e. That is, the number of derangements of n items is the nearest integer to n!/e, so the probability of a derangement would be 1/n! * [n!/e]. This number converges to 1/e incredibly quickly as n grows - unsurprisingly, the number of correct digits is on the order of the factorial of n.

We’re now interested in partial derangements D_{n,k}; the number of permutations of n elements which have k fixed points. D_{n,0} is the number of derangements, as established that is [n!/e]. Suppose k isn’t 0. Then we can pick k points to be correctly sorted, and multiply by the number of derangements of the others, for a total of nCk * [(n-k)!/e]. Note that [1/e] is 0, indeed, it’s not possible for exactly one element to be out of place.

So what’s the probability of a particular partial derangement? Well now we’re asking for D_{n,k}/n!. That would be nCk/n! * [(n-k)!/e]. Let’s drop the nearest integer bit and call it an approximation, then (nCk * (n-k)!)/(n! * e) = 1/(k!*e). Look familiar? That’s a Poisson distribution with λ = 1!

But if we have a Poisson distribution with λ = 1, then that means that on average we expect one new sorted element per shuffle, and hence we expect to take n shuffles. I’ll admit, I was not expecting that when I started working this out. I wrote a quick program to average some trials as a sanity check and it seems to hold.