Text-based interpreter: recursive, fixed-point, fix-expression, letrec-expression

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= = Below is a meaningless passage.

I am a person who enjoys the pleasure of learning knowledge and enjoys sharing it. I think most of the knowledge can be learned as long as it takes some time. A few years ago, I was hooked on differential geometry. I tell every friend that it's fun to take a little time and energy to learn. They replied, alas, there is no time to know where to go. Later on, I was fascinated by quantum mechanics. I say to every friend that it's worth learning, just take a little time and energy. They answered, alas, too much trouble to learn. Later, I was hooked on the language of programming. I say to every friend that you can master this if you squeeze a little time. They replied, "I don't know what I've been doing. Now, they're all married, and I'm still single.

= = Then, the body begins

No move.

Alas, it is also a mathematical concept, it is not a practical meaning of the section. But as a recursive-related common sense, it has to be mentioned. In other words, as a common sense, I am probably a distortion of values, the importance of ranking upside down. It is said that many research program languages, the study of partial differential equations of the people are like this.

Then, a--er--something $x$ is called the fixed point of the function $f$ if it satisfies the following conditions: \[(f; x) = x \]

What is the relationship between fixed points and recursion? $ ({Y} \ f) $ is not only a recursive function on the $f$ of an auxiliary function, but also a fixed point for a $f$. The proofs are as follows: \begin{eqnarray*} ({Y} \ f) &=& (\lambda X. (\lambda v. (v; v) \ \lambda F. (x \; (f; f)) \; f) \ &=& (\lambda x. (\lambda v. (v; v) \ \lambda p. (x \; (p; p)) \; f) \ &=& (\lambda v. (v; v) \ \lambda P. (f; (p; p)) \ &=& (\lambda P. (f;; (p; p)) \; \lambda P. (f; (p; p)) \ &=& (f \; (\lambda P. (f;; (p; p)) \; \lambda P. (f; (p; p))) \ &=& (f \; (\lambda v. (v; v) \ \lambda p. (f \; (p; p))) \ &=& (f \; (\lambda X. (\lambda v. (v; v) \ \lambda p. (x \; (p; p)) \; f) \ &=& (f \; ({y} \ f) \end{eqnarray*} The recursive function must be a fixed point of a function $f$ because the recursive function can certainly be written as a form of $ ({y} \ f) $ (see the previous mkdouble construction method).

Add Fix expression

It is always too troublesome to construct a recursive function with Y-combination. So we need an expression that can directly construct a recursive function. Let's look at what elements this expression requires. First of all, recursive functions to call themselves, need a point to their own variables, recorded as $x_1$. Second, the recursive function has a parameter, which is recorded as $x_2$. Finally, the function body of the recursive function is recorded as $m$. To construct a recursive function is to look for a fixed point, so this expression is called the fix expression (fixed point). Fix expression Long this sample: \[({fix} \; X_1 \; X_2 \; M) \] The easiest way to add a fix expression is to define it as a macro: \[({fix} \; X_1 \; X_2 \; M) = ({Y} \ \lambda X_1.\lambda x_2.m) \] But doing so reintroduced the Y-combo. This time in another way, not to define macros, but to add the fix expression syntax: \begin{eqnarray*} M, N, L &=& ... \ &|& ({fix} \; X_1 \; X_2 \; M) \end{eqnarray*}

Now examine the evaluation process for the fix expression. The result of the fix expression is a function, and the argument is $x_2$: \[eval ({fix} \; X_1 \; X_2 \; M) = \lambda x_2.? \] The function of the most suitable place at the question mark is $m$, but $m$ contains the free variable $x_1$, before using $m$ to get rid of $x_1$ first. $X _1$ represents the recursive function itself, which is $ ({fix} \; X_1 \; X_2 \; M) $. To remove the free variable $x_1$ in $m$, replace $x_1$ with a recursive function $ ({fix} \; X_1 \; X_2 \; M) $. In summary, the evaluation procedure for the fix expression is: \[eval ({fix} \; X_1 \; X_2 \; M) = \lambda x_2.m[x_1 \leftarrow ({fix} \; X_1 \; X_2 \; M)] \] Code:

Added a new syntax to add the appropriate replacement process (this replacement process is similar to the $\LAMBDA x.m$ replacement process, but a bit troublesome): \begin{eqnarray*} ({fix} \; X_1 \; X_2 \; M) [x_1 \leftarrow N] &=& ({fix} \; X_1 \; X_2 \; M) \ ({fix} \; X_1 \; X_2 \; M) [x_2 \leftarrow N] &=& ({fix} \; X_1 \; X_2 \; M) \ ({fix} \; X_1 \; X_2 \; M) [X_3 \leftarrow N] &=& ({fix} \; X_4 \; X_5 \; M[x_2 \leftarrow x_5][x_1 \leftarrow x_4][x_3 \leftarrow N] \ & where &x_3 \neq x_1, x_3 \neq x_2, \ &&X_4 \ Notin FV (n), X_4 \notin FV (m) \backslash\{x_1\}, \ &&x_5 \notin FV (n), X_5 \notin FV (m) \backslash\{x_2\} \end{eqna rray*} code:

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