Answer:
Bill is building a project network that involves testing a prototype. he must design the prototype (activity 1), build the prototype (activity 2), and test the prototype (activity 3). activity 1 is the predecessor for activity 2 and activity 2 is the predecessor for activity 3. if the prototype fails testing, bill must redesign the prototype; therefore, activity 3 is a predecessor for activity 1. this is an example of
b. looping
Explanation:
- The given example is of looping because each activity is leading to another activity on the completion of some conditions.
- The answer a is not valid as it is not just an example of conditional statements rather it is a loop which will keep moving until unless reached a situation to end it.
- The option c, d an e are not right options for the given example.
Answer:
B seems like the best answer to me.
Explanation:
I believe its yes. you can just look it up
Answer:
public class MagicSquare {
public static void main(String[] args) {
int[][] square = {
{ 8, 11, 14, 1},
{13, 2, 7,12},
{ 3, 16, 9, 6},
{10, 5, 4, 15}
};
System.out.printf("The square %s a magic square. %n",
(isMagicSquare(square) ? "is" : "is not"));
}
public static boolean isMagicSquare(int[][] square) {
if(square.length != square[0].length) {
return false;
}
int sum = 0;
for(int i = 0; i < square[0].length; ++i) {
sum += square[0][i];
}
int d1 = 0, d2 = 0;
for(int i = 0; i < square.length; ++i) {
int row_sum = 0;
int col_sum = 0;
for(int j = 0; j < square[0].length; ++j) {
if(i == j) {
d1 += square[i][j];
}
if(j == square.length-i-1) {
d2 += square[i][j];
}
row_sum += square[i][j];
col_sum += square[j][i];
}
if(row_sum != sum || col_sum != sum) {
return false;
}
}
return d1 == sum && d2 == sum;
}
}
