I am trying to write a rule that can return the sum of the product of each element from two lists (same length).
Here is what I have right now:
sum(0, _, []).
sum(Result, [H1|T1], [H2|T2]) :-
sum(Remaining,T1, T2),
Remaining is Result - (H1*H2).
It won't work when one of the list is not instantiated. What changes I need to make in order to make the following possible?
sum([1,2],X,3).
X = [3,0].
Thanks.
What you are calculating is commonly referred to as a dot product (also known as scalar product or inner product).
You write you are not allowed to use libraries. That surely refers to external libraries---not to the standard library that is part of SWI Prolog, right?
The following predicate list_list_dotProduct/3 roughly corresponds to the code you implemented. It uses finite domain constraints (#>=)/2 and (#=)/2 to allow for non-unidirectional integer arithmetic:
:- use_module(library(clpfd)).
list_list_dotProduct([],[],0).
list_list_dotProduct([X|Xs],[Y|Ys],Sum) :-
X #>= 0,
Y #>= 0,
Sum #= X*Y + Sum0,
list_list_dotProduct(Xs,Ys,Sum0).
Consider the following query:
?- list_list_dotProduct([1,2],Xs,3), label(Xs).
Xs = [1, 1] ;
Xs = [3, 0].
As an added bonus, here's an alternative implementation that is based on the predefined predicates same_length/2, ins/2, and scalar_product/4:
list_list_dotProduct(Xs,Ys,Prod) :-
same_length(Xs,Ys),
Xs ins 0..sup,
Ys ins 0..sup,
scalar_product(Xs,Ys,#=,Prod).
Related
I'm supposed to write a predicate that does some math stuff. But I don't know how to pass numbers or return numbers.
Maybe you can give me an example?
Let's say a predicate divide/2 that takes two numbers a and b and returns a/b.
Yes, you pass numbers in in some arguments, and you get the result back in some other argument(s) (usually last). For example
divide( N, D, R) :-
R is N / D.
Trying:
112 ?- divide(100,5,X).
X = 20.
113 ?- divide(100,7,X).
X = 14.285714285714286.
Now, this predicate is divide/3, because it has three arguments: two for inputs and one for the output "information flow".
This is a simplified, restricted version of what a Prolog predicate can do. Which is, to not be that uni-directional.
I guess "return" is a vague term. Expression languages have expressions e-value-ated so a function's last expression's value becomes that function's "return" value; Prolog does not do that. But command-oriented languages return values by putting them into some special register. That's not much different conceptually from Prolog putting some value into some logvar.
Of course unification is more complex, and more versatile. But still, functions are relations too. Predicates "return" values by successfully unifying their arguments with them, or fail to do so, as shown in the other answer.
Prolog is all about unifying variables. Predicates don't return values, they just succeed or fail.
Typically when a predicate is expected to produce values based on some of the arguments then the left-most arguments are inputs and the right-most are the outputs. However, many predicates work with allowing any argument to be an input and any to be a output.
Here's an example for multiply showing how it is used to perform divide.
multiply(X,Y,Z) :- number(X),number(Y),Z is X * Y.
multiply(X,Y,Z) :- number(X),number(Z),X \= 0,Y is Z / X.
multiply(X,Y,Z) :- number(Y),number(Z),Y \= 0,X is Z / Y.
Now I can query it like this:
?- multiply(5,9,X).
X = 45 .
But I can easily do divide:
?- multiply(5,X,9).
X = 1.8 .
It even fails if I try to do a division by 0:
?- multiply(X,0,9).
false.
Here's another approach. So let's say you have a list [22,24,34,66] and you want to divide each answer by the number 2. First we have the base predicate where if the list is empty and the number is zero so cut. Cut means to come out of the program or just stop don't go to the further predicates. The next predicate checks each Head of the list and divides it by the number A, meaning (2). And then we simply print the Answer. In order for it to go through each element of the list we send back the Tail [24,34,66] to redo the steps. So for the next step 24 becomes the Head and the remaining digits [34,66] become the Tail.
divideList([],0,0):-!.
divideList([H|T],A,Answer):-
Answer is H//A,
writeln(Answer),
divideList(T,A,_).
?- divideList([22,24,34,66],2,L).
OUTPUT:
11
12
17
33
Another simpler approach:
divideList([],_,[]).
divideList([H|T],A,[H1|L]):-
H1 is H//A,!,
divideList(T,A,L).
?-divideList([22,4,56,38],2,Answer).
Answer = [11, 2, 28, 19]
I have a relation R :: w => w => bool that is both transitive an irreflexive.
I have the axiom Ax1: "finite {x::w. True}". Therefore, for each x there is always a longest sequence of wn R ... R w2 R w1 R x.
I need a function F:: w => nat, that -for a given x - gives back the "lenght" of this sequence (or 0 if there is no y such that xRy). How would I go about building one in isabelle.
Also: Is Ax1 a good way to axiomatize the "finiteness of type w" or is there a better one?
First of all, a more idiomatic way of writing {x::w. True} is UNIV :: w set. I suggest writing finite (UNIV :: w set), or possibly using the finite type class, although that might make your theorem more difficult to apply because you need a finite instance for your type. I think it's not really necessary or helpful for your use case.
I then suggest the following approach:
Define an inductive predicate (using inductive) on lists of type w list stating that the first element is x and for each two successive list elements y and z, R y z holds, i.e. the list is an ascending chain w.r.t. R.
Show that any list that is such a chain must have distinct elements (cf. distinct :: 'a list ⇒ bool).
Show that there are finitely many distinct lists over a finite set.
Use the Max operator to find the biggest n such that there exists a list of length n that is an ascending chain w.r.t. R. That this works should be easy since there is at least one such chain, and you've already shown that there are only finitely many chains.
I'm currently learning programming language concepts and pragmatics, hence I feel like I need help in differentiating two subbranches of declarative language family.
Consider the following code snippets which are written in Scheme and Prolog, respectively:
;Scheme
(define gcd
(lambda (a b)
(cond ((= a b) a)
((> a b) (gcd (- a b) b))
(else (gcd (- b a) a)))))
%Prolog
gcd(A, B, G) :- A = B, G = A.
gcd(A, B, G) :- A > B, C is A-B, gcd(C, B, G).
gcd(A, B, G) :- B > A, C is B-A, gcd(C, A, G).
The thing that I didn't understand is:
How do these two different programming languages behave
differently?
Where do we make the difference so that they are categorized either
Functional or Logic-based programming language?
As far as I'm concerned, they do exactly the same thing, calling recursive functions until it terminates.
Since you are using very low-level predicates in your logic programming version, you cannot easily see the increased generality that logic programming gives you over functional programming.
Consider this slightly edited version of your code, which uses CLP(FD) constraints for declarative integer arithmetic instead of the low-level arithmetic you are currently using:
gcd(A, A, A).
gcd(A, B, G) :- A #> B, C #= A - B, gcd(C, B, G).
gcd(A, B, G) :- B #> A, C #= B - A, gcd(C, A, G).
Importantly, we can use this as a true relation, which makes sense in all directions.
For example, we can ask:
Are there two integers X and Y such that their GCD is 3?
That is, we can use this relation in the other direction too! Not only can we, given two integers, compute their GCD. No! We can also ask, using the same program:
?- gcd(X, Y, 3).
X = Y, Y = 3 ;
X = 6,
Y = 3 ;
X = 9,
Y = 3 ;
X = 12,
Y = 3 ;
etc.
We can also post even more general queries and still obtain answers:
?- gcd(X, Y, Z).
X = Y, Y = Z ;
Y = Z,
Z#=>X+ -1,
2*Z#=X ;
Y = Z,
_1712+Z#=X,
Z#=>X+ -1,
Z#=>_1712+ -1,
2*Z#=_1712 ;
etc.
That's a true relation, which is more general than a function of two arguments!
See clpfd for more information.
The GCD example only lightly touches on the differences between logic programming and functional programming as they are much closer to each other than to imperative programming. I will concentrate on Prolog and OCaml, but I believe it is quite representative.
Logical Variables and Unification:
Prolog allows to express partial datastructures e.g. in the term node(24,Left,Right) we don't need to specify what Left and Right stand for, they might be any term. A functional language might insert a lazy function or a thunk which is evaluated later on, but at the creation of the term, we need to know what to insert.
Logical variables can also be unified (i.e. made equal). A search function in OCaml might look like:
let rec find v = function
| [] -> false
| x::_ when v = x -> true
| _::xs (* otherwise *) -> find v xs
While the Prolog implementation can use unification instead of v=x:
member_of(X,[X|_]).
member_of(X,[_|Xs]) :-
member_of(X,Xs).
For the sake of simplicity, the Prolog version has some drawbacks (see below in backtracking).
Backtracking:
Prolog's strength lies in successively instantiating variables which can be easily undone. If you try the above program with variables, Prolog will return you all possible values for them:
?- member_of(X,[1,2,3,1]).
X = 1 ;
X = 2 ;
X = 3 ;
X = 1 ;
false.
This is particularly handy when you need to explore search trees but it comes at a price. If we did not specify the size of the list, we will successively create all lists fulfilling our property - in this case infinitely many:
?- member_of(X,Xs).
Xs = [X|_3836] ;
Xs = [_3834, X|_3842] ;
Xs = [_3834, _3840, X|_3848] ;
Xs = [_3834, _3840, _3846, X|_3854] ;
Xs = [_3834, _3840, _3846, _3852, X|_3860] ;
Xs = [_3834, _3840, _3846, _3852, _3858, X|_3866] ;
Xs = [_3834, _3840, _3846, _3852, _3858, _3864, X|_3872]
[etc etc etc]
This means that you need to be more careful using Prolog, because termination is harder to control. In particular, the old-style ways (the cut operator !) to do that are pretty hard to use correctly and there's still some discussion about the merits of recent approaches (deferring goals (with e.g. dif), constraint arithmetic or a reified if). In a functional programming language, backtracking is usually implemented by using a stack or a backtracking state monad.
Invertible Programs:
Perhaps one more appetizer for using Prolog: functional programming has a direction of evaluation. We can use the find function only to check if some v is a member of a list, but we can not ask which lists fulfill this. In Prolog, this is possible:
?- Xs = [A,B,C], member_of(1,Xs).
Xs = [1, B, C],
A = 1 ;
Xs = [A, 1, C],
B = 1 ;
Xs = [A, B, 1],
C = 1 ;
false.
These are exactly the lists with three elements which contain (at least) one element 1. Unfortunately the standard arithmetic predicates are not invertible and together with the fact that the GCD of two numbers is always unique is the reason why you could not find too much of a difference between functional and logic programming.
To summarize: logic programming has variables which allow for easier pattern matching, invertibility and exploring multiple solutions of the search tree. This comes at the cost of complicated flow control. Depending on the problem it is easier to have a backtracking execution which is sometimes restricted or to add backtracking to a functional language.
The difference is not very clear from one example. Programming language are categorized to logic,functional,... based on some characteristics that they support and as a result they are designed in order to be more easy for programmers in each field (logic,functional...). As an example imperative programming languages (like c) are very different from object oriented (like java,C++) and here the differences are more obvious.
More specifically, in your question the Prolog programming language has adopted he philosophy of logic programming and this is obvious for someone who knows a little bit about mathematical logic. Prolog has predicates (rather than functions-basically almost the same) which return true or false based on the "world" we have defined which is for example what facts and clauses do we have already defined, what mathematical facts are defined and more....All these things are inherited by mathematical logic (propositional and first order logic). So we could say that Prolog is used as a model to logic which makes logical problems (like games,puzzles...) more easy to solve. Moreover Prolog has some features that general-purpose languages have. For example you could write a program in your example to calculate gcd:
gcd(A, B, G) :- A = B, G = A.
gcd(A, B, G) :- A > B, C is A-B, gcd(C, B, G).
gcd(A, B, G) :- B > A, C is B-A, gcd(C, A, G).
In your program you use a predicate gcd in returns TRUE if G unifies with GCD of A,B, and you use multiple clauses to match all cases. When you query gcd(2,5,1). will return True (NOTE that in other languages like shceme you can't give the result as parameter), while if you query gcd(2,5,G). it unifies G with gcd of A,B and returns 1, it is like asking Prolog what should be G in order gcd(2,5,G). be true. So you can understand that it is all about when the predicate succeeds and for that reason you can have more than one solutions, while in functional programming languages you can't.
Functional languages are based in functions so always return the SAME
TYPE of result. This doesn't stand always in Prolog you could have a predicate predicate_example(Number,List). and query predicate_example(5,List). which returns List=... (a list) and also query
predicate_example(Number,[1,2,3]). and return N=... (a number).
The result should be unique, In mathematics, a function is a relation
between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output
Should be clear what parameter is the variable that will be returned
for example gcd function is of type : N * N -> R so gets A,B parameters which belong to N (natural numbers) and returns gcd. But prolog (with some changes in your program) could return the parameter A,so querying gcd(A,5,1). would give all possible A such that predicate gcd succeeds,A=1,2,3,4,5 .
Prolog in order to find gcd tries every possible way with choice
points so in every step it will try all of you three clauses and will
find every possible solutions. Functional programming languages on
the other hand, like functions should have well unique defined steps
to find the solution.
So you can understand that the difference between Functional and logic languages may not be always visible but they are based on different philosophy-way of thinking.
Imagine how hard would be to solve tic-tac-toe or N queens problem or man-goat-wolf-cabbage problem in Scheme.
This is supposed to calculate the sum of two lists. The lists can be of different size.
sum([],[],[]).
sum(A,[],A).
sum([],B,B).
sum([A|Int1],[B|Int2],[C|Int3]) :-
(
C =:= A + B
;
((C =:= A), B = [])
;
((C =:= B), A = [])
),
sum(Int1,Int2,Int3).
It seems to work correctly, except when trying to find the sum of two lists. Then it gives the following error:
ERROR: =:=/2: Arguments are not sufficiently instantiated
I don't see why. There's a recursive and a basis step, what exactly is not yet instantiated and how do I fix it?
[1] While your disjunctions in the last clause are -- to some extent -- conceptually correct, Prolog considers these disjunctions in sequence. So it first considers C =:= A + B. But either A or B can be the empty list! This is what causes the error you reported, since the empty list is not allowed to occur in a numeric operation.
[2] You need to use C is A + b (assignment) i.o. C =:= A + B (numeric equivalence).
[3] If you say [A|Int1] and then A = [], then this means that [A|Int1] is not (only) a list of integers (as you claim it is) but (also) a list of lists! You probably intend to check whether the first or the second list is empty, not whether either contains the empty list.
Staying close to your original program, I would suggest to reorder and change things in the following way:
sumOf([], [], []):- !.
sumOf([], [B|Bs], [C|Cs]):- !,
C is B,
sumOf([], Bs, Cs).
sumOf([A|As], [], [C|Cs]):- !,
C is A,
sumOf(As, [], Cs).
sumOf([A|As], [B|Bs], [C|Cs]):-
C is A + B,
sumOf(As, Bs, Cs).
For example:
?- sumOf([1,2,3], [1,-90], X).
X = [2, -88, 3]
Notice my use of the cut (symbol !) in the above. This makes sure that the same answer is not given multiple times or -- more technically -- that no choicepoints are kept (and is called determinism).
You should read a tutorial or a book. Anyway, this is how you add two things to each other:
Result is A + B
This is how you could add all elements of one list:
sum([], 0). % because the sum of nothing is zero
sum([X|Xs], Sum) :-
sum(Xs, Sum0),
Sum is X + Sum0.
And this is how you could add the sums of a list of lists:
sums([], 0).
sums([L|Ls], Sums) :-
sums(Ls, Sums0),
sum(L, S),
Sums is Sums0 + S.
I am currently working on a Mathematica project to calculate Riemann's sums and put them in a table. I am having trouble printing the row numbers (intervals). (The row numbers are also parameters to the secondary functions). I don't know of any way to just access the index of the iterator in a Mathematica Table, so I am trying to compute them using the function parameters.
Here is an example of what I'd like to print, for the integral of x^2 over the range {0, 1}, with 10 subdivisions.
tableRiemannSums[#^2 &, {0, 1}, 10]
I need to figure out what the index of each iteration is, based on the value of the current
subdivision k, the range of the integral {a, b}, and the number of subdivisions, n. Below is the main piece of code.
tableRiemannSums[fct_, {a_, b_}, n_] := Table[{'insert index here',
leftRiemannSum[fct, {a, b}, 'insert index here'],
rightRiemannSum[fct, {a, b}, 'insert index here']},
{k, a, b - (N[b - a]/n), N[b - a]/n}]
In the above equation, the line
{k, a, b - (N[b - a]/n), N[b - a]/n}]
means the range of the table is k as k goes from 'a' to 'b - ((b - a)/n)' in steps of size '(b - a)/n'.
In each of the places where my code says 'insert index here,' I need to put the same equation. Right now, I am using 'n * k + 1' to calculate the index, which is working for positive ranges, but breaks when I have a range like {a,b} = {-1, 1}.
I think this is a fairly straightforward algebra problem, but I have been racking my brain for hours and can't find a general equation.
(I apologize if this is a duplicate question - I tried searching through the Stack overflow archives, but had a hard time summarizing my question into a few key words.)
I finally figured out how to solve this. I was over thinking the range, rather than relying on the inner functions to control it. I rewrote the function as:
tableRiemannSums[fct_, {a_, b_}, n_] := Table[{k,
leftRiemannSum[fct, {a, b}, k],
rightRiemannSum[fct, {a, b}, k]},
{k, 1, n}}]
For reference, here are the left and right sums (for anyone interested!):
leftRiemannSum[fct_, {a_, b_}, n_] :=
N[b - a]/n* Apply[Plus, Map[fct, Range[a, b - N[b - a] / n, N[b - a]/n]]]
rightRiemannSum[fct_, {a_, b_}, n_] :=
N[b - a]/n* Apply[Plus, Map[fct, Range[a + (N[b - a]/n), b, N[b - a]/n]]]
What you may want to consider is creating a function to make each line of the table. One argument to this function would be the row number.
Execute this function using MapIndexed, which will provide you a way to traverse your range as required while providing an incrementing row number.
(Create a list with the range of values, then apply your MapIndexed function to this list.)