Answer:
SyntaxError.
Explanation:
https://www.quora.com/In-Python-what-kind-of-error-is-returned-by-the-following-code-e-g-NameError-ValueError-IOError-etc-def-my_func-n1-n2-return-n1-n2-my_func-1-2-3 sorce site
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<span>an arrow ..............................</span>
Answer and Explanation:
A TM with doubly infinite tape can simulate an ordinary TM. It marks the left-hand end of the input to detect and prevent the head from moving off of that end. To simulate the doubly infinite tape TM by an ordinary TM, we show how to simulate it with a 2-tape TM, which was already shown to be equivalent in power to an ordinary
TM. The first tape of the 2-tape TM is written with the input string, and the second tape is blank. We cut the tape of the doubly infinite tape TM into two parts, at the starting cell of the input string. The portion with the input string and all the blank spaces to its right appears on the first tape of the 2-tape TM. The portion to the left of the input string appears on the second tape, in reverse order.
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;
}
}
