

let p x = x ++ show x in putStrLn $ p"let p x = x ++ show x in putStrLn $ p"







main(a){printf(a="main(a){printf(a=%c%s%c,34,a,34);}",34,a,34);}





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[P](Q 0 (Q 0 )) = [Q 0 ](Q 0 ) (by definition of Q 0 )

= [Q 0 (Q 0 )] (by definition of P(x))



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I'm guessing that anyone reading this is already familiar with the idea of programs that output themselves. If you're not there's a great list of such programs here . But what I was surprised to discover at the weekend was that there is in fact a bit of general theory about this. In particular, what do we need in a computer language to guarantee we can write a self-replicator?Consider the following in Haskell:Evaluate this expression in an interactive Haskell session and it prints itself out. But there's a nice little cheat that made this easy: the Haskell 'show' function conveniently wraps a string in quotation marks. So we simply have two copies of once piece of code: one without quotes followed by one in quotes. In C, on the other hand, there is a bit of a gotcha. You need to explicitly write code to print those extra quotation marks. And of course, just like in Haskell, this code needs to appear twice, once out of quotes and once in. But the version in quotes needs the quotation marks to be 'escaped' using backslash so it's notactually the same as the first version. And that means we can't use exactly the same method as with Haskell. The standard workaround is not to represent the quotation marks directly in the strings, but instead to use the ASCII code for this character and use C's convenient %c mechanism to print at. For example:Again we were lucky, C provides this great %c mechanism. What do you need in a language to be sure you can write a self-replicator?It turns out there is a very general approach to writing self-replicators that's described in Vicious Circles. What follows is essentially from there except that I've simplified the proofs by reducing generality.We'll use capital letters to represent programs. Typically these mean 'inert' strings of characters. I'll use square brackets to indicate the function that the program evaluates. So if P is a program to compute the mathematical function p, we write [P](x) = p(x). P is a program and [P] is a function. We'll consider both programs that take arguments like the P I just mentioned, and also programs, R, that take no arguments, so [R] is simply the output or return value of the program R.Now we come to an important operation. We've defined [P](x) to be the result of running P with input x. Now we define P(x) to be the program P modified so that it no longer takes an argument or input but instead substitutes the 'hard-coded' value of x instead. In other words [P(x)] = [P](x). P(x) is, of course, another program. There are also many ways of implementing P(x). We could simply evaluate [P](x) and write a program to simply print this out or return it. On the other hand, we could do the absolute minimum and write a new piece of code that simply calls P and supplies it with a hard-coded argument. Whatever we choose is irrelevant to the following discussion. So here's the demand that we make of our programming language: that it's powerful enough for us to write a program that can compute P(x) from inputs P and x. This might not be a trivial program to write, but it's not conceptually hard either. It doesn't have gotchas like the quotation mark issue above. Typically we can compute P(x) by some kind of textual substitution on P.With that assumption in mind, here's a theorem: any program P that takes one argument or input has a fixed point, X, in the sense that running P with input X gives the same result as just running X. Given an input X, P acts just like an interpreter for the programming language as it outputs the same thing as aninterpreter would given input X.So here's a proof:Define the function f(Q) = [P](Q(Q)). We've assumed that we can write a program that computes P(x) from P and x so we know we can write a program to compute Q(Q) for any Q. We can then feed this as an input to [P]. So f is obviously computable by some program which we call Q. So [Q](Q) = [P](Q(Q)).Now the fun starts:In other words Q(Q) is our fixed point.So now take P to compute the identity function. Then [Q(Q)] = [P](Q(Q)) = Q(Q). So Q(Q) outputs itself when run! What's more, this also tells us how to do other fun stuff like write a program to print itself out backwards. And it tells us how to do this in any reasonably powerful programming language. We don't need to worry about having to work around problems like 'escaping' quotation marks - we can always find a way to replicate the escape mechanism too.So does it work in practice? Well it does for Haskell - I derived the Haskell fragment above by applying this theorem directly, and then simplifying a bit. For C++, however, it might give you a piece of code that is longer than you want. In fact, you can go one step further and write a program that automatically generates a self-replicator. Check out Samuel Moelius 's kpp. It is a preprocessor that converts an ordinary C++ program into one that can access its own source code by including the code to generate its own source within it.Another example of an application of these methods is Futamura's theorem which states that there exists a program that can take as input an interpreter for a language and output a compiler. I personally think this is a little bogus.