Burrows–Wheeler Transform (opens in new tab)

It's frustrating that your link to an "explanation" is an intentionally buggy implementation of the algorithm :(

I understand the point of the debugging game, but in this context, a link to the solutions page would be more helpful....

> Wikipedia gives an impractical, inefficient account of the algorithm. It's almost comically bad.

I encourage you to edit and fix it.

I love your site! Hope you don't mind that I sent you an email.

It's a fantastic example of minimal yet fully functional!

General-purpose compression algorithms tend to be conceptually simple. They must be fast, so they can't try many different approaches to see what works best. And because they are general-purpose, they can't assume much about the data.

The typical data compression algorithm starts with a combinatorial transformation of the data. Then it creates a statistical model of the transformed data and encodes it according to the model. There are two successful families of combinatorial transformations: Lempel–Ziv and Burrows–Wheeler. Neither of them is superior to the other, but the compression performance depends on the properties of the data.

Lempel–Ziv parsing replaces copies of repeated substrings with references to an earlier copy. It works best when there are long repeated substrings in the data. The modeling/encoding parts of an LZ compressor are usually pretty simple.

Burrows-Wheeler transform reorders the symbols according to the context they occur in. It's often but not always followed by run-length encoding. Because the symbols are reordered by context, BWT-based compressors work best when there are many copies of each repeated substring.

Unlike the LZ, the BWT does not compress the data on its own. It relies entirely on statistical compression. Because the symbols are ordered by context, the BWT is an order-n model for every n simultaneously. Hence you get something similar to PPM by simply partitioning the BWT into blocks and encoding each block independently with an order-0 encoder.

Not necessarily. After the transform, there are more runs of characters, but it could be that this comes at the expense of recognizable patters (like common words) in the input. It likely helps simple compression algorithms more than sophisticated ones.

Same here. m2^2 ;) toying around with these kind of indices to detect / find repetitions for years already.

Btw: what's the best source for these kind of one of a kind, contra-intuitive algos and datastructures? Be it online, be it books...

BWT can't be the only one, right?

If you randomize a list of numbers there will always be cycles like described (similar to how there will always be cycles if you would draw lots for Secret Santa and then reveal who drew who), and these cycles have a certain length. It might be one cycle of length 100, but more commonly there will be lots of shorter cycles. If you start with your own number, you are guaranteed to be in a cycle that will return back to you, you just don't know how long that cycle will be (could be 1, could be 100, more likely something between). And apparently you can calculate that in ~31% of cases all cycles will have a length of 50 or less. This is just a neat way to exploit a kind of structure that there always will be within the randomness.

The key insight is that this strategy causes there to be a strong correlation between the success of different prisoners.

As an example, let's say prisoner #10 opens his box and finds #21, then finds #5 in that box, then #84, then #51, then finally succeeds and finds his number 10 in box #51. These boxes form a cycle 10-21-5-84-51 (and then back to 10). Anyone who opens any of these boxes will eventually see the same set numbers that #10 did, so that means that we know prisoners #5, #10, #21, #51 and #84 will all succeed in finding their number by starting at their own box.

Compare that to the situation where they just randomly look in boxes and each independently have only a 50% chance to succeed - then the odds that those 5 prisoners all succeed would be (1/2)^5 = 1/32, or only about 3%. In every cycle containing n prisoners, instead of their success rate being (1/2)^n, now they all succeed together or all fail together.

Now that the success of every prisoner in the same cycle is perfectly correlated with each other, the prisoners' overall chance of success depends only on whether the random permutation created by the warden has any cycles >50 in length. If so, then everyone in that cycle will fail. If not, then all the cycles across the 100 boxes are short enough that all 100 prisoners will succeed.

It doesn't affect the chance of any one prisoner finding their number - that's still 50%. But the joint strategy massively reduces the independence of the individual prisoners' success, and hence the distribution of the number of successful prisoners.

Rather than a normal distribution (sum of independent outcomes) with a big peak around 50% of the prisoners finding their number and a vanishingly small chance of them all (or none) finding the right number, you end up with a much more complex pattern - there's an approximately uniform distribution in the 0-50 successes range with ~70% of the overall outcome, and a huge peak at the 100 successes point with ~30%.

https://www.r-bloggers.com/2014/08/update-100-prisoners-100-... has a bunch of distribution plots showing this.

No. That rules means you are definitely on the right track.

Think of it like having to search for someone in a forest with only enough time to search half the forest.

Would you want to search random spots or be dropped randomly into the trail to the person?

You might yet run out of time, but on average it'll be much more efficient. Surprisingly so in the 100 boxes context.

Veritasium has an excellent video on it but I'm on mobile and can't dig out a link. I'd recommend it though.

Hm, it seems obvious that Guido and Python have more impact on the industry (and I'm aware of Burrows' antipathy toward Python, which was well known at Google. As far as I remember it was due to debugging a syntax error caused by indentation in SWIG/Python)

JPL and Google were early Python users. Google's crawler and the home page started out in Python (trivia: the ancient asyncore stdlib module and its author played some role, as far as I remember)

Pretty much all AI and self driving cars use Python, BitTorrent was written in Python, Python is embedded in GDB, embedded in huge commercial applications like Maya for visual effects, etc. I think Ethereum had an early Python implementation

Python's design achieved Guy Steele's "growing a language" vision in his famous talk, i.e. using operator overloading to allow scientists to create their own lingua franca -- https://news.ycombinator.com/item?id=29171519 (i.e. the talk was about adding operator overloading to Java, as well "generics" and value types)

Mike Burrows is a great (even legendary) programmer and computer scientist, but if you're talking about "impact on the industry", there are levels :) I have to say I'm a big Alta Vista fan though

And a really nice person to work with

Not only that, the suffix array [1] is using counting sort O(n) instead of a comparison based O(n log n) sort. That is because they assume integer alphabets. For general alphabets they can only achieve O(n log n).

[1] https://arxiv.org/abs/1610.08305

Can you provide examples of other BWT implementations, and instructions and perhaps some of the weaknesses of bzip2? Genuinely interested.

bzip2 beats xz for 1gb of zeros as well.

Isn't bzip notoriously slow though?