Deriving a Quine in a Lisp

栏目: IT技术 · 发布时间: 6年前

内容简介:As with myprevious post, this post is another excerpt that will be included in my final Master’s thesis, but I decided it is interesting enough to post it on its own.We start with a definition ofDefinition 1: For an expression

As with myprevious post, this post is another excerpt that will be included in my final Master’s thesis, but I decided it is interesting enough to post it on its own.

We start with a definition of diagonalization (or quotation), as discussed in The Gödelian Puzzle Book :

Definition 1: For an expression Deriving a Quine in a Lisp in which a variable occurs, we say that its diagonalization Deriving a Quine in a Lisp is the substitution of the variable with the quoted expression Deriving a Quine in a Lisp .

This definition allows us to represent self-referential expressions.

For example, let Deriving a Quine in a Lisp . Then, the diagonalization of it will be Deriving a Quine in a Lisp . In Lisp code:

> (define p (lambda (x) (list 'Boro 'is 'reading x)))
> (p 'a-book)
'(Boro is reading a-book)
> (define d (lambda (p x) (p (list 'quote (p x)))))
> (d p 'a-book)
'(Boro is reading '(Boro is reading a-book))

As another example, let Deriving a Quine in a Lisp stand for “Boro is reading the diagonalization of x”, and let Deriving a Quine in a Lisp stand for `Boro is reading the diagonalization of “Boro is reading the diagonalization of x”`. That is, Deriving a Quine in a Lisp . But, the diagonalization of Deriving a Quine in a Lisp is simply Deriving a Quine in a Lisp , i.e. Deriving a Quine in a Lisp . So, Deriving a Quine in a Lisp refers to itself. In Lisp:

> (define q (lambda (x) (list 'Boro 'is 'reading (d identity x))))
> (define r (lambda (x) (d q x)))
> (equal? (d q 'test) (r 'test))
#t

Based on these definitions, we will now show how to derive a Quine , which is a program that when evaluated returns its source code as an output – a metaprogram.

Note how we used (d identity x) earlier, i.e. the diagonalization of x. This simply evaluates to (list 'quote x) .

Now, let’s consider the expression (list x (list 'quote x)) which will return a list with two members: x and its diagonalization:

> (define quine-1 (lambda (x) (list x (list 'quote x))))

What if we apply it to itself?

> (quine-1 quine-1)
'(#<procedure:quine-1> '#<procedure:quine-1>)

Nothing useful. How about applying its quoted version?

> (quine-1 'quine-1)
'(quine-1 'quine-1)

This is exactly why we picked the expression that contains a list of x and its diagonalization. We wanted the evaluation of that expression to return the same expression.

It looks like we are on the right track. Finally, the Quine code that will reproduce itself is just taking the lambda and applying it to its quoted version:

((lambda (x) (list x (list 'quote x)))
  '(lambda (x) (list x (list 'quote x))))

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