I have some tricky question encountered on an technical exam, hope I can get an output here
In which of the following statement(s) is(are) equivalent to the condition a AND (b OR c)
a. NOT(NOT a OR NOT b) OR NOT (NOT a OR NOT c)
b. NOT(NOT a OR (NOT b AND NOT c))
c. NOT(NOT a OR NOT b) AND (NOT a OR NOT c)
d. (NOT a AND NOT b) OR NOT c
My question, if the equation is
NOT a OR NOT b
Does it give a result b or a is acceptable since it uses OR?
The answer is to write a truth table.
OK, you might be a super-logician and break these down in your head,
but the interviewer is expecting you to draw a truth table for each of the 4 parts and say which final column matches.
So your first 3 columns are a,b,c, then some intermediate columns, then a result column.
All that matters is if the result columns match.
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'm creating an encryptor/decryptor for ascii strings where I take the ascii value of a char, add 1 to it, then mod it by the highest ascii value so that I get a valid ascii char out.
The problem is the decryption.
Let's say that (a + b) % c = d
I know b, c, and d's values.
How do I get the a variables value out from that?
This is exactly the ROT1 substitution cipher. Subtract 1, and if less than lowest value (0 I assume, given how you're describing it), then add the highest value.
Using terms like "mod," while accurate, make this seem more complicated than it is. It's just addition on a ring. When you go past the last letter, you come back to the first letter and vice-versa. Once you put your head around how the math works, the equations should pop out. Basically, you just add or subtract as normal (add to encrypt, subtract to decrypt in this case), and at the end, mod "normalizes" you back onto the ring of legal values.
Use the inverse formula
a = (b - d) mod c
or in practice
a = (b - d + c) % c.
The term + c needs to be added as a safeguard because the % operator does not implement a true modulo in the negatives.
Let's assume that c is 2, d is 0 and b is 4.
Now we know that a must be 2... Or 4 actually.. or 6... Or any other even number.
You can't solve this problem, there are infinite solutions.
very new to Prolog and I'm absolutely awful at it.
I am trying to get a sum for all the marks received on assignments, and return it. The sum predicate must be of the form sumAssignments(S).
Here's the knowledge base I've made:
assignment(1, A).
assignment(2, B).
assignment(3, C).
assignment(4, D).
assignment(5, E).
...Where assignment(1, A) means that assignment 1 has a variable grade A (could be 70, could be 50, etc.).
Here is my attempt at getting the sum, just for the first two assignments for testing purposes:
sumAssignments(S) :- assignment(1, A), assignment(2, B), S=A+B.
That always returns yes. The key here is I can't use lists.
I found out that only constants belong in predicates, so instead of assignment(1, A). we would find an arbitrary number to represent 'A', in this case maybe 67 or 100: assignment(1, 67). The summation works fine after doing that.
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 have a problem to solve in either Matlab or R (preferably in R).
Imagine I have a vector A with 10 elements.
I have also a vector B with 30 elements, of which 10 have value 'x'.
Now, I want to replace all the 'x' in B by the corresponding values taken from A, in the order that is established in A. Once a value in A is taken, the next one is ready to be used when the next 'x' in B is found.
Note that the sizes of A and B are different, it's the number of 'x' cells that coincides with the size of A.
I have tried different ways to do it. Any suggestion on how to program this?
As long as the number of x entries in B matches the length of A, this will do what you want:
B[B=='x'] <- A
(It should be clear that this is the R solution.)
MATLAB Solution
In MATLAB it's quite simple, use logical indexing:
B(B == 'x') = A;