I am trying to test some of the lambda calculus functions that I wrote using Racket but not having much luck with the testcases. For example given a definition
; successor function
(define my_succ (λ (one)
(λ (two)
(λ (three)
(two ((one two) three))))))
I am trying to apply it to 1 2 3, expecting the successor of 2 to be 3 by doing
(((my_succ 1) 2) 3)
logic being that since my_succ is a function that takes one arg and passes it to another function that takes one arg which passes it to the third function that takes one arg. But I get
application: not a procedure;
expected a procedure that can be applied to arguments
given: 1
arguments.:
I tried Googling and found a lot of code for the rules, but no examples of application of these rules. How should I call the above successor function in order to test it?
You are mixing two completely different things: lambda terms and functions in Racket.
In Racket you can have anonymous functions, that can be written in the λ notation (like (λ(x) (+ x 1)) that returns the successor of an integer, so that ((λ(x) (+ x 1)) 1) returns 2),
in Pure Lambda Calculus you have only lambda terms, that are written with in a similar notation, and that can be interpreted as functions.
In the second domain, you do not have natural numbers like 0, 1, 2, ..., but you have only lambda terms, and represent numbers as such. For instance, if you use the so-called Church numerals, you represent (in technical term encode) the number 0 with the lambda term λf.λx.x, 1 with λf.λx.f x, 2 with λf.λx.f (f x) and so on.
So, the function successor (for numbers represented with this encoding) correspond to a term which, in Racket notation, is the function that you have written, but you cannot apply it to numbers like 0, 1, etc., but only to other lambda expressions, that is you could write something like this:
(define zero (λ(f) (λ (x) x))) ; this correspond to λf.λx.x
(successor zero)
The result in Racket is a procedure (it will be printed as: #<procedure>), but if you try to test that your result is correct, comparing it with the functional encoding of 1, you will find something strange. In fact:
(equal? (successor zero) (λ(f) (λ(x) (f x))))
produces #f, since if you compare two procedures in Racket you obtain always false (e.g. (equal? (λ(x)x) (λ(x)x)) produces #f), unless you compare the “identical” (in the sense of “same memory cell”) value ((equal? zero zero) gives #t). This is due to the fact that, for comparing correctly two functions, you should compare infinite sets of couples (input, output)!
Another possibility would be representing lambda terms as some kind of structure in Racket, so you can represent Church numerals, as well as "normal" lambda terms, and define a function apply (or better reduce) the perform lambda-reduction.
You are trying to apply currying.
(define my_succ
(lambda(x)(
lambda(y)(
lambda(z)(
(f x y z)))))
(define (add x y z)
(+ x y z))
((( (my_succ add)1)2)3)
Implementation in DR Racket:
Related
I am reading a Gentle Introduction to Symbolic Computation and it asks this question. Basically, the previous content deals with making up bigger functions with small ones. (Like 2- will be made of two 1- (decrement operators for lisp))
So one of the questions is what are the two different ways to define a function HALF which returns one half of its input. I have been able to come up with the obvious one (dividing number by 2) but then get stuck. I was thinking of subtracting HALF of the number from itself to get half but then the first half also has to be calculated...(I don't think the author intended to introduce recursion so soon in the book, so I am most probably wrong).
So my question is what is the other way? And are there only two ways?
EDIT : Example HALF(5) gives 2.5
P.S - the book deals with teaching LISP of which I know nothing about but apparently has a specific bent towards using smaller blocks to build bigger ones, so please try to answer using that approach.
P.P.S - I found this so far, but it is on a completely different topic - How to define that float is half of the number?
Pdf of book available here - http://www.cs.cmu.edu/~dst/LispBook/book.pdf (ctrl+f "two different ways")
It's seems to be you are describing peano arithmetic. In practice it works the same way as doing computation with fluids using cups and buckets.
You add by taking cups from the source(s) to a target bucket until the source(s) is empty. Multiplication and division is just advanced adding and substraction. To halve you take from source to two buckets in alterations until the source is empty. Of course this will either do ceil or floor depending on what bucket you choose to use as answer.
(defun halve (x)
;; make an auxillary procedure to do the job
(labels ((loop (x even acc)
(if (zerop x)
(if even (+ acc 0.5) acc)
(loop (- x 1) (not even) (if even (+ acc 1) acc)))))
;; use the auxillary procedure
(loop x nil 0)))
Originally i provided a Scheme version (since you just tagged lisp)
(define (halve x)
(let loop ((x x) (even #f) (acc 0))
(if (zero? x)
(if even (+ acc 0.5) acc)
(loop (- x 1) (not even) (if even (+ acc 1) acc)))))
Edit: Okay, lets see if I can describe this step by step. I'll break the function into multiple lines also.
(defun half (n)
;Takes integer n, returns half of n
(+
(ash n -1) ;Line A
(if (= (mod n 2) 1) .5 0))) ;Line B
So this whole function is an addition problem. It is simply adding two numbers, but to calculate the values of those two numbers requires additional function calls within the "+" function.
Line A: This performs a bit-shift on n. The -1 tells the function to shift n to the right one bit. To explain this we'll have to look at bit strings.
Suppose we have the number 8, represented in binary. Then we shift it one to the right.
1000| --> 100|0
The vertical bar is the end of the number. When we shift one to the right, the rightmost bit pops off and is not part of the number, leaving us with 100. This is the binary for 4.
We get the same value, however if we perform the shift on nine:
1001| --> 100|1
Once, again we get the value 4. We can see from this example that bit-shifting truncates the value and we need some way to account for the lost .5 on odd numbers, which is where Line B comes in.
Line B: First this line tests to see if n is even or odd. It does this by using the modulus operation, which returns the remainder of a division problem. In our case, the function call is (mod n 2), which returns the remainder of n divided by 2. If n is even, this will return 0, if it is odd, it will return 1.
Something that might be tripping you up is the lisp "=" function. It takes a conditional as its first parameter. The next parameter is the value the "=" function returns if the conditional is true, and the final parameter is what to return if the conditional is false.
So, in this case, we test to see if (mod n 2) is equal to one, which means we are testing to see if n is odd. If it is odd, we add .5 to our value from Line A, if it is not odd, we add nothing (0) to our value from Line A.
Using DrRacket, on both linux and Mac OS, the following code gives this error
*: expects type <number> as 1st argument, given #<undefined>
but if I uncomment the (newline) at the beginning of the procedure definition, it works fine, yielding the expected value, 9.
#lang r5rs
(define (quadr x y)
;(newline)
(define xx (* x x))
(define yy (* y y))
(define xxyy (* xx yy))
(+ xx yy xxyy))
(display (quadr 1 2))
(newline)
Is this a bug in Racket's scheme interpreter, or is the language specified so that the nested invocations of (define ...) may happen out of order? If the latter is the case, where can I find the relevant bit of the language specification?
As an aside, I am very aware of the "let" construct and know that this is not the recommended way to define such a procedure. Still, I didn't expect this error.
Here's the relevant link to the R5RS specification explaining the behavior of internal definitions. Notice that in section §5.2.2 is stated that:
... it must be possible to evaluate each <expression> of every internal definition in a <body> without assigning or referring to the value of any <variable> being defined.
In other words, you can't count on a correct behavior if you define values that depend on previously defined values within the same internal definition. Use let* for this:
(define (quadr x y)
(let* ((xx (* x x))
(yy (* y y))
(xxyy (* xx yy)))
(+ xx yy xxyy)))
Or a bit more verbose, using a couple of nested lets:
(define (quadr x y)
(let ((xx (* x x))
(yy (* y y)))
(let ((xxyy (* xx yy)))
(+ xx yy xxyy))))
It's very peculiar that inserting a (newline) causes the variable definition to work for this example, but when you're dealing with undefined behavior, anything can happen. As a side note, if I use #lang racket instead of #lang r5rs, the original code works for me without the extra (newline).
While coding a predicate that tests whether or not a number is divisible by all integers across a certain range, I was wondering if it is possible to make rules about input, maybe through the "declare" symbol?
Code:
(defun integer-divisiblep (n m i)
(declare (integer n m i))
(do ((x m (- x 1)))
((< x n) (return t))
(when (not (integerp (/ i x)))
(return nil))))
In this case I may like to specify that the input value "n" must be smaller than "m". Is there anyway to do this with an inbuilt function? I can't seem to find what I want with the declaration-identifiers on the Hyperspec.
Also, I'm using SBCL, if that makes a difference.
Common Lisp does not provide static type checks for argument types. Some Common Lisp compilers do it as an extension, most notably CMUCL and SBCL. These static type checks use the typical declarations of variable types provided by DECLARE. You need to see the syntax of the various types to see what can be declared.
Dynamic checks at runtime are best done with CHECK-TYPE and ASSERT.
In this case I may like to specify that the input value "n" must be smaller than "m"
This is something like:
(assert (and (numberp m) (numberp n) (< n m)) (m n))
The list (m n) at the end is a list of variables which can be interactively set by the user, if the assert is violated. After entering a different value, the assertion will be checked again, until the assertion is satisfied.
I'm trying to write a function that can calculate GPA. Now I can do limited calculation(only 3 ),but I stuck on how to calculate more , without using loop or recursion (that's the requirement of subject) how to expend nth function? like: (nth n) ,if so ,is that mean i need to write a lambda expression? As an newbie, I maynot describe the question clearly, really need some help..
Glist is grade points Clist is credit hours.
GPA=( gradepoint *credithour + gradepoint *credithour) / ( the sum of credithour) like: (3*1+3*2+4*1)/(1+2+1)
here is my code:
(defun gpa (Glist Clist)
(format t "~3,2f~%"
(/
(+(nth 0 (mapcar #' * Glist Clist))
(nth 1 (mapcar #' * Glist Clist))
(nth 2 (mapcar #' * Glist Clist)))
(+ (nth 0 Clist)
(nth 1 Clist)
(nth 2 Clist))
);end "/"
);end "format"
(values) );end
EDIT
This seems like a good opportunity to emphasize some common (little c) Lisp ideas, so I fleshed out my answer to illustrate.
As mentioned in another answer, you could use a sum function that operates on lists (of numbers):
(defun sum (nums)
(reduce #'+ nums))
The dot product is the multiplicative sum of two (equal-length) vectors:
(defun dot-product (x y)
(sum (mapcar #'* x y)))
The function gpa is a simple combination of the two:
(defun gpa (grades credits)
(/ (dot-product grades credits) (sum credits)))
The example from the question results in the answer we expect (minus being formatted as a float):
(gpa '(3 3 4) '(1 2 1))
> 13/4
There are a few things worth mentioning from this example:
You should learn about map, reduce, and their variants and relatives. These functions are very important to Lisp and are very useful for operating on lists. map* functions generally map sequences to a sequence, and reduce usually transforms a sequence into to a single value (you can however use forms like (reduce #'cons '(1 2 3))).
This is a good example of the "bottom-up" approach to programming; by programming simple functions like sum that are often useful, you make it easy to write dot-product on top of it. Now the gpa function is a simple, readable function built on top of the other two. These are all one-liners, and all are easily readable to anyone who has a basic knowledge of CL. This is in contrast to the methodology usually applied to OOP.
There is no repetition of code. Sure, sum is used more than once, but only where it makes sense. You can do very little more to abstract the notion of a sum of the elements of a list. It's more natural in Scheme to write functions with functions, and that's a whole different topic. This is a simple example, but no two functions are doing the same thing.
If you're using nth to traverse a list, you're doing it wrong. In this case, you might want to write a summing function:
(defun sum (items)
(reduce #'+ items))
I never really thought about this until I was explaining some clojure code to a coworker who wasn't familiar with clojure. I was explaining let to him when he asked why you use a vector to declare the bindings rather than a list. I didn't really have an answer for him. But the language does restrict you from using lists:
=> (let (x 1) x)
java.lang.IllegalArgumentException: let requires a vector for its binding (NO_SOURCE_FILE:0)
Why exactly is this?
Mostly readability, I imagine. Whenever bindings are needed in Clojure, a vector is pretty consistently used. A lot of people agree that vectors for bindings make things flow better, and make it easier to discern what the bindings are and what the running code is.
Just for fun:
user=> (defmacro list-let [bindings & body] `(let ~(vec bindings) ~#body))
#'user/list-let
user=> (macroexpand-1 '(list-let (x 0) (println x)))
(clojure.core/let [x 0] (println x))
user=> (list-let (x 0 y 1) (println x y))
0 1
nil
This is an idiom from Scheme. In many Scheme implementations, square brackets can be used interchangeably with round parentheses in list literals. In those Scheme implementations, square brackets are often used to distinguish parameter lists, argument lists and bindings from S-expressions or data lists.
In Clojure, parentheses and brackets mean different things, but they are used the same way in binding declarations.
Clojure tries very hard to be consistent. There is no technical reason with a list form could not have been used in let, fn, with-open, etc... In fact, you can create your own my-let easily enough that uses one instead. However, aside from standing out visibly, the vector is used consistently across forms to mean "here are some bindings". You should strive to uphold that ideal in your own code.
my guess is that it's a convention
fn used it, defn used it, loop uses.
it seems that it's for everything that resembles a block of code that has some parameters; more specific, the square brackets are for marking those parameters
other forms for blocks of code don't use it, like if or do. they don't have any parameters
Another way to think about this is that let is simply derived from lambda. These two expressions are equivalent:
((fn [y] (+ y 42)) 10)
(let [y 10] (+ 42 y))
So as an academic or instructional point, you could even write your own very rudimentary version of let that took a list as well as a vector:
(defmacro my-let [x body]
(list (list `fn[(first x)]
`~body)
(last x)))
(my-let (z 42) (* z z))
although there would be no practical reason to do this.