The answer & explanation for this question is given in the attachment below.
Answer:
public static void main(String[] args) {
String ing[] = {"ten","fading","post","card","thunder","hinge","trailing","batting"};
for (String i: ing){
if (i.endsWith("ing")){
System.out.println(i);
}
}
}
Explanation:
The for-loop cycles through the entire list and the if-statement makes it so that the string is only printed if it ends with "ing"
Answer:
see my discussion as explained bellow
Explanation:
Answer:
For n = 100
For n = 200
For n = 400
For n = 800
The faster approach is A(Bx)
Explanation step by step functions:
A(Bx) is faster because requires fewer interactions to find a result: for (AB)x you have (n*n)+n interactions while for A(Bx) you have n+n, to understand why please see the step by step:
a) Function for (AB)x:
function loopcount1 = FirstAB(A,B,x)
n = size(A)(1);
AB = zeros(n,n);
ABx = zeros(n,1);
loopcount1 = 0;
for i = 1:n
for j = 1:n
AB(i,j) = A(i,:)*B(:,j);
loopcount1 += 1;
end
end
for k = 1:n
ABx(k) = AB(k,:)*x;
loopcount1 += 1;
end
end
b) Function for A(Bx):
function loopcount2 = FirstBx(A,B,x)
n = size(A)(1);
Bx = zeros(n,1);
ABx = zeros(n,1);
loopcount2 = 0;
for i = 1:n
Bx(i) = B(i,:)*x;
loopcount2 += 1;
end
for j = 1:n
ABx(j) = A(j,:)*Bx;
loopcount2 += 1;
end
end
