symmetric

public class symm { /* * Returns true if array A is symmetric. * Returns false otherwise. * n is the number of elements A contains. * * The running time of your algorithm is O( ). * You may add a brief explanation here if you wish. */ public static boolean symmetric( int[] A, int n ) { return symmHelper(A, n, 0); } private static boolean symmHelper(int[] A, int n, int i) { if(n==1) return true; if((n==2) && (A[i] == A[n-1-i])) return true; if((i == n-1-i) && (A[i] == A[n-1-i] )) return true; if(A[i] == A[n-1-i] && i < n/2 ) return symmHelper(A, n, i+1); return false; } } Test cases: I passed

I am learning C++ and I was wondering if I could gain some insight into the preferred way of creating binary operators that work on instances of two different types. Here is an example that I've made to illustrate my concerns: class A; class B; class A { private: int x; public: A(int x); int getX() const; int operator + (const B& b); }; class B { private: int x; public: B(int x); int getX() const; int operator + (const A& A); }; A::A(int x) : x(x) {} int A::getX() const { return x; } // Method 1 int A::operator + (const B& b) { return getX() + b.getX(); } B::B(int x) : x(x) {} int B::getX()

I have to simulate family tree in prolog. And i have problem of symetrical predicates. Facts: parent(x,y). male(x). female(y). age(x, number). Rules: blood_relation is giving me headache. this is what i have done: blood_relation(X,Y):-ancestor(X,Y). blood_relation(X,Y):-uncle(X,Y);brother(X,Y);sister(X,Y);(mother(Z,Y),sister(X,Z));(father(Z,Y),sister(X,Z));(father(Z,Y),brother(X,Z)). blood_relation(X,Y):-uncle(X,Z),blood_relation(Z,Y). and I am getting i think satisfactory results(i have double prints - can i fix this), problem is that i want that this relation be symmetrical. It is not now.

There are many cases in which JavaScript's type-coercing equality operator is not transitive. For example, see " JavaScript equality transitivity is weird ." However, are there any cases in which == isn't symmetric ? That is, where a == b is true and b == a is false ? SLaks In Javascript, == is always symmetric . The spec says : NOTE 2 The equality operators maintain the following invariants: A != B is equivalent to !(A == B) . A == B is equivalent to B == A , except in the order of evaluation of A and B . It's supposed to be symmetric. However, there is an asymmetric case in some versions of

I have a matrix in R that is supposed to be symmetric, however, due to machine precision the matrix is never symmetric (the values differ by around 10^-16). Since I know the matrix is symmetric I have been doing this so far to get around the problem: s.diag = diag(s) s[lower.tri(s,diag=T)] = 0 s = s + t(s) + diag(s.diag,S) Is there a better one line command for this? Is the workaround really necessary if the values only differ by that much? Someone pointed out that my previous answer was wrong. I like some of the other ones better, but since I can't delete this one (accepted by a user who left