Is there any difference between free variable and free occurrence of a variable in context of lambda calculus? If yes, then please explain with an example or two.
Actually I was going through the conversion rules for lambda expression where I came across the following line:
In stating the conversion rules the notation E[E'/V] is used to mean the result of substituting E' for each free occurrence of V in E
Let's take the term T:
t\q\p\ (t x (x\ q x) (p q x)
(where x\ t means lambda x.t - this is Lambda-Prolog notation)
There is one free variable: x, and four bound variables, one of which is also named x. But the two "x" are not the same variable (in the sense that the term can be alpha renamed to t\q\p\ (t x (y\ q y) (p q x)
but not, for example, to: t\q\p\ (t x (y\ q y) (p q y)
In the term T above, there are two free occurrences of the variable x and one bound occurrence of another variable, also named x.
Now, if your question is "can there be, in the same term, both free occurrences and bound occurrences of the same variable, be it bound or free?", I don't think so.
Related
Revising for a course on automated reasoning and I don't quite understand how to answer this question:
Show how the notion of pairs (x, y) can be defined in higher-order logic using a lambda abstraction. Define a function π1 that returns the first element of such a pair. Finally, show that π1(x, y) = x.
I've found similar questions on stackoverflow, but they're all to do with scheme, which I've never used. An explanation in English/relevant mathematical notation would be appreciated
Here you go
PAIR := λx. λy. λp. p x y
π1 := λp. p (λx. λy. x)
π2 := λp. p (λx. λy. y)
π1 (PAIR a b) => a
π2 (PAIR a b) => b
Check the wiki entry on Church encoding for some good examples, too
The main topic of this question is to understand how data can be represented as functions. When you're working with other paradigms , the normal way of thinking is "data = something that's stored in a variable" (could be an array, object, whatever structure you want).
But when we're in functional programming, we can also represent data as functions.
So let's say you want a function pair(x,y)
This is "pseudo" lisp language:
(function pair x y =
lambda(pick)
if pick = 1 return x
else return y )
That example, is showing a function that returns a lambda function which expects a parameter.
(function pi this-is-pair = this-is-pair 1)
this-is-pair should be constructed with a pair function, therefore, the parameter is a function which expects other parameter (pick).
And now, you can test what you need
(pi (pair x y ) ) should return x
I would highly recommend you to see this video about compound data. Most of the examples are made on lisp, but it's great to understand a concept like that.
Pairs or tuples describes Products Domain, is the union of all elements of the set A and all elements of the set B:
A × B = { (a, b) | a ∈ A, b ∈ B }
Here, A and B are diferent types, so if you for example are in a laguage program like C, Java, you can have pair like (String, Integer), (Char, Boolean), (Double, Double)
Now, the function π1, is just a function that takes a pair and returns the first element, this function is called in usually first, and that's how it looks like π1(x, y) = x, on the other hand you have second, doing the same thing but returning the second element:
fst(a, b) = a
snd(a, b) = b
When I studied the signature "Characteristics of the programming languages" in college our professor recommended this book, see the chapter Product Domain to understand well all this concepts.
Define a function which, given a list L, an object x, and a positive
integer k, returns the position of the k-th occurrence of x in L if x
appears at least k times in L otherwise 0. For example, if L is
[#"a", #"b", #"c", #"b"], x is #"b", and k is 2, then the function
returns 4.
For this problem, I can't use any auxiliary functions and can't use the length function. Any ideas on how I would go about solving it?
Assume find is the function you have to implement. Two things are immediate from the description:
find returns an integer
find takes at least three arguments (L, x, and k)
Furthermore, functions on lists typically distinguish between two cases: either the given list is empty ([]) or it contains at least one element (y::ys). Thus we can use the following skeleton for find
fun find [] x k = ...
| find (y::ys) x k = ...
You should start by thinking about what to do in either case. If L is empty there are definitely no occurrences of x in it, thus the result is 0. Otherwise, L consists of a head-element y and the remaining list ys. If x is equal to y, we just found an occurrence of x, otherwise we have to continue to search in ys. According to this, the skeleton can be refined into
fun find [] x k = 0
| find (y::ys) x k =
if x = y then ...
else find ys x k
At this point the only remaining part is when L is not empty and x = y. But since we are interested in the k-th occurrence of x, we have to check whether we found it or not. This could be done by decreasing k each time we find an occurrence of x and finally, when we find an occurrence and k is 1 at the same time, we found the desired k-th occurrence of x.
fun find [] x k = 0
| find (y::ys) x k =
if x = y andalso k > 1 then find ys x (k - 1)
else if x = y then ...
else find ys x k
The remaining case is that we found the k-th occurrence of x. So we should return the index i of this occurrence. But where should we get i from? In some way we have to compute it inside our function. How to do this, is for you to find out.
I am a complete beginner to J. My first attempt at writing a function is something like the totient function. It takes an array, assumed to be i.num for some number, num. And then computes the number of elements coprime to the tally of the array.
If I do it manually like this:
numbers =: i.7
#(1=( #numbers)+./numbers)#numbers
it works. The result is 6.
So I want to turn this into a general function.
Tot =: monad :'(1=( #x)+./x)#x'
Tot i.11
This fails and I get a value error for x. I don't know why.
The monad's variable should be y, not x:
Tot =: monad :'(1=( #y)+./y)#y'
x is the left argument of a dyad.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Functional programming and non-functional programming
I'm afraid Wikipedia did not bring me any further.
Many thanks
PS: This past thread is also very good, however I am happy I asked this question again as the new answers were great - thanks
Functional programming and non-functional programming
First learn what a Turing machine is (from wikipedia).
A Turing machine is a device that manipulates symbols on a strip of tape according to a table of rules. Despite its simplicity, a Turing machine can be adapted to simulate the logic of any computer algorithm, and is particularly useful in explaining the functions of a CPU inside a computer.
This is about Lamda calculus. (from wikipedia).
In mathematical logic and computer science, the lambda calculus, also written as the λ-calculus, is a formal system for studying computable recursive functions, a la computability theory, and related phenomena such as variable binding and substitution.
The functional programming languages use, as their fundamental model of computation, the lambda calculus, while all the other programming languages use the Turing machine as their fundamental model of computation. (Well, technically, I should say functional programming languages vr.s imperative programming languages- as languages in other paradigms use other models. For example, SQL uses the relational model, Prolog uses a logic model, and so on. However, pretty much all the languages people actually think about when discussing programming languages are either functional or imperative, so I’ll stick with the easy generality.)
What do I mean by “fundamental model of computation”? Well, all languages can be thought of in two layers: one, some core Turing-complete language, and then layers of either abstractions or syntactic sugar (depending upon whether you like them or not) which are defined in terms of the base Turing-complete language. The core language for imperative languages is then a variant of the classic Turing machine model of computation one might call “the C language”. In this language, memory is an array of bytes that can be read from and written to, and you have one or more CPUs which read memory, perform simple arithmetic, branch on conditions, and so on. That’s what I mean by the fundamental model of computation of these languages is the Turing Machine.
The fundamental model of computation for functional languages is the Lambda Calculus, and this shows up in two different ways. First, one thing that many functional languages do is to write their specifications explicitly in terms of a translation to the lambda calculus to specify the behavior of a program written in the language (this is known as “denotational semantics”). And second, almost all functional programming languages implement their compilers to use an explicit lambda-calculus-like intermediate language- Haskell has Core, Lisp and Scheme have their “desugared” representation (after all macros have been applied), Ocaml (Objective Categorical Abstract Machine Language) has it’s lispish intermediate representation, and so on.
So what is this lambda calculus I’ve been going on about? Well, the basic idea is that, to do any computation, you only need two things. The first thing you need is function abstraction- the definition of an unnamed, single-argument, function. Alonzo Church, who first defined the Lambda calculus used the rather obscure notation to define a function as the greek letter lambda, followed by the one-character name of the argument to the function, followed by a period, followed by the expression which was the body of the function. So the identity function, which given any value, simply returns that value, would look like “λx.x” I’m going to use a slight more human-readable approach- I’m going to replace the λ character with the word “fun”, the period with “->”, and allow white space and allow multi-character names. So I might write the identity function as “fun x -> x”, or even “fun whatever -> whatever”. The change in notation doesn’t change the fundamental nature. Note that this is the source of the name “lambda expression” in languages like Haskell and Lisp- expressions that introduce unnamed local functions.
The only other thing you can do in the Lambda Calculus is to call functions. You call a function by applying an argument to it. I’m going to follow the standard convention that application is just the two names in a row- so f x is applying the value x to the function named f. We can replace f with some other expression, including a Lambda expression, if we want- and we can When you apply an argument to an expression, you replace the application with the body of the function, with all the occurrences of the argument name replaced with whatever value was applied. So the expression (fun x -> x x) y becomes y y.
The theoreticians went to great lengths to precisely define what they mean by “replacing all occurrences of the variable with the the value applied”, and can go on at great lengths about how precisely this works (throwing around terms like “alpha renaming”), but in the end things work exactly like you expect them to. The expression (fun x -> x x) (x y) becomes (x y) (x y)- there is no confusion between the argument x within the anonymous function, and the x in the value being applied. This works even in multiple levels- the expression (fun x -> (fun x -> x x)) (x x)) (x y) becomes first (fun x -> x x) ((x y) (x y)) and then ((x y) (x y)) ((x y) (x y)). The x in the innermost function (“(fun x -> x x)”) is a different x than the other x’s.
It is perfectly valid to think of function application as a string manipulation. If I have a (fun x -> some expression), and I apply some value to it, then the result is just some expression with all the x’s textually replaced with the “some value” (except for those which are shadowed by another argument).
As an aside, I will add parenthesis where needed to disambiguate things, and also elide them where not needed. The only difference they make is grouping, they have no other meaning.
So that’s all there is too it to the Lambda calculus. No, really, that’s all- just anonymous function abstraction, and function application. I can see you’re doubtful about this, so let me address some of your concerns.
First, I specified that a function only took one argument- how do you have a function that takes two, or more, arguments? Easy- you have a function that takes one argument, and returns a function that takes the second argument. For example, function composition could be defined as fun f -> (fun g -> (fun x -> f (g x))) – read that as a function that takes an argument f, and returns a function that takes an argument g and return a function that takes an argument x and return f (g x).
So how do we represent integers, using only functions and applications? Easily (if not obviously)- the number one, for instance, is a function fun s -> fun z -> s z – given a “successor” function s and a “zero” z, one is then the successor to zero. Two is fun s -> fun z -> s s z, the successor to the successor to zero, three is fun s -> fun z -> s s s z, and so on.
To add two numbers, say x and y, is again simple, if subtle. The addition function is just fun x -> fun y -> fun s -> fun z -> x s (y s z). This looks odd, so let me run you through an example to show that it does, in fact work- let’s add the numbers 3 and 2. Now, three is just (fun s -> fun z -> s s s z) and two is just (fun s -> fun z -> s s z), so then we get (each step applying one argument to one function, in no particular order):
(fun x -> fun y -> fun s -> fun z -> x s (y s z)) (fun s -> fun z -> s s s z) (fun s -> fun z -> s s z)
(fun y -> fun s -> fun z -> (fun s -> fun z -> s s s z) s (y s z)) (fun s -> fun z -> s s z)
(fun y -> fun s -> fun z -> (fun z -> s s s z) (y s z)) (fun s -> fun z -> s s z)
(fun y -> fun s -> fun z -> s s s (y s z)) (fun s -> fun z -> s s z)
(fun s -> fun z -> s s s ((fun s -> fun z -> s s z) s z))
(fun s -> fun z -> s s s (fun z -> s s z) z)
(fun s -> fun z -> s s s s s z)
And at the end we get the unsurprising answer of the successor to the successor to the successor to successor to the successor to zero, known more colloquially as five. Addition works by replacing the zero (or where we start counting) of the x value with the y value- to define multiplication, we instead diddle with the concept of “successor”:
(fun x -> fun y -> fun s -> fun z -> x (y s) z)
I’ll leave it to you to verify that the above code does
Wikipedia says
Imperative programs tend to emphasize the series of steps taken by a program in carrying out an action, while functional programs tend to emphasize the composition and arrangement of functions, often without specifying explicit steps. A simple example illustrates this with two solutions to the same programming goal (calculating Fibonacci numbers). The imperative example is in C++.
// Fibonacci numbers, imperative style
int fibonacci(int iterations)
{
int first = 0, second = 1; // seed values
for (int i = 0; i < iterations; ++i) {
int sum = first + second;
first = second;
second = sum;
}
return first;
}
std::cout << fibonacci(10) << "\n";
A functional version (in Haskell) has a different feel to it:
-- Fibonacci numbers, functional style
-- describe an infinite list based on the recurrence relation for Fibonacci numbers
fibRecurrence first second = first : fibRecurrence second (first + second)
-- describe fibonacci list as fibRecurrence with initial values 0 and 1
fibonacci = fibRecurrence 0 1
-- describe action to print the 10th element of the fibonacci list
main = print (fibonacci !! 10)
See this PDF also
(A) Functional programming describes solutions mechanically. You define a machine that is constantly outputting correctly, e.g. with Caml:
let rec factorial = function
| 0 -> 1
| n -> n * factorial(n - 1);;
(B) Procedural programming describes solutions temporally. You describe a series of steps to transform a given input into the correct output, e.g. with Java:
int factorial(int n) {
int result = 1;
while (n > 0) {
result *= n--;
}
return result;
}
A Functional Programming Language wants you to do (A) all the time. To me, the greatest tangible uniqueness of purely functional programming is statelessness: you never declare variables independently of what they are used for.
how can I accomplish this:
Give a tail-recursive definition for each of the following predicates.
power(X,Y,Z): XY=Z.
gcd(X,Y,Z): The greatest common divisor of X and Y is Z.
sum(L,Sum): Sum is the sum of the elements in L.
so far I have done this but not sure if that's correct
power(_,0,1) :- !.
power(X,Y,Z) :- Y1 is Y - 1,power(X,Y1,Z1),Z is X * Z1.
sum(void,0).
sum(t(V,L,R),S) :- sum(L,S1),sum(R,S2), S is V + S1 + S2.
These are not tail recursive. You can write tail recursive variants by using an accumulator, see this answer.
Your sum is over a tree, which is unusual, normally one would use a list. In Prolog [] is the empty list and [X|R] is the pattern for a nonempty list with the head X and the tail R.