% Logical operations ||, &&, and xor operator on single pairs
% of logical (boolean) values. Examples:

% demo of ||, logical OR
if 3 == 4 || 5 == 5; disp('true'); else disp('false'); end
true

% demo of &&, logical AND
if 3 == 4 && 5 == 5; disp('true'); else disp('false'); end
false

% demo xor, logical EXCLUSIVE OR
if xor(3 == 4, 5 == 5); disp('true'); else disp('false'); end
true
if xor(4 == 4, 5 == 5); disp('true'); else disp('false'); end
false

% Can also do comparisons and logical operations on vectors. 
% First set a couple vectors x and y for demos
x = [1 2 3]
x =
     1     2     3

y = [5 2 9]
y =
     5     2     9

% Vector comparisons compare all the elements of the vectors

% Which elements of x are equal to the corresponding elements of y?
x == y 
ans =
     0     1     0

% Which elements of x are not equal to the corresponding elements of y?
x ~= y
ans =
     1     0     1

% Which elements of x are greater or equal to the corresponding elements of y?
x >= y
ans =
     0     1     0

% Now do some vectorized logical operations

% What elements are true (1) in both vectors?
[0 1 0] & [1 0 1]
ans =
     0     0     0  % none of them!

% What elements are true in either vector?
[0 1 0] | [1 0 1]
ans =
     1     1     1  % of them!

% Recall our x and y vectors
x =
     1     2     3
y
y =
     5     2     9

% In which elements is x greater than y?
x > y
ans =
     0     0     0

% In which elements is x less than y?
x < y
ans =
     1     0     1

% Repeat on some random vectors
x = rand(1,6)
x =
    0.9224    0.7204    0.4806    0.7310    0.6328    0.7180
y = rand(1,6)
y =
    0.8384    0.9191    0.3769    0.9563    0.0431    0.2091

x < y
ans =
     0     1     0     1     0     0


% Set x to be a random vector with a Gaussian distribution around x=0
% (in order to get both positive and negative values). 

x = randn(1,5)
x =
   -0.7887   -2.1134    2.0821    1.8451   -0.7676

% How many positive elems are there in x?

% Well, we know how to test for positivity elementwise...
x > 0
ans =
     0     0     1     1     0

% To count the number of positive elements, just apply sum to the prev result
sum(x > 0)
ans =
     2

% Set a component of x to zero, for another deom 
x(2) = 0
x =
   -0.7887         0    2.0821    1.8451   -0.7676


% How many elems of x are not zero?
sum(x ~= 0)
ans =
     4

% Which elems of x are greater than zero
x > 0
ans =
     0     0     1     1     0

% What are the positive elements in x? Do this in pieces
x
x =
   -0.7887         0    2.0821    1.8451   -0.7676

% This will tell us which elems are positive
x > 0
ans =
     0     0     1     1     0

% Note the type of x>0: It's a vector of logical values
class(x > 0)
ans =
logical

% We can access elements of x by the vector of logical values
% First set n to be the vector of 0s and 1s marking where the 
% positive elements are
n = x > 0
n =
     0     0     1     1     0

% Now extract the positive elems of x using the logical n vector
x(n)  
ans =
    2.0821    1.8451

% Equivalently, we can do this without setting logical vector in a variable
x(x>0)
ans =
    2.0821    1.8451

% Similarly, can get the negative values of x this way...
x(x<0)
ans =
   -0.7887   -0.7676

% ...the zero values of x this way...
x(x==0)
ans =
     0
     
% ...and the nonzero values of x this way.
x(x~=0)
ans =
   -0.7887    2.0821    1.8451   -0.7676