**8.11 (Game: nine heads and tails) Nine coins are placed in a 3-by-3 matrix with some face up and some face down. You can represent the state of the coins using a 3-by-3 matrix with values 0 (heads) and 1 (tails). Here are some examples:
0 0 0 1 0 1 1 1 0 1 0 1 1 0 0
0 1 0 0 0 1 1 0 0 1 1 0 1 1 1
0 0 0 1 0 0 0 0 1 1 0 0 1 1 0
Each state can also be represented using a binary number. For example, the preceding matrices correspond to the numbers
000010000 101001100 110100001 101110100 100111110
There are a total of 512 possibilities, so you can use decimal numbers 0, 1, 2, 3, . . . , and 511 to represent all states of the matrix. Write a program that prompts the user to enter a number between 0 and 511 and displays the corresponding matrix with the characters H and T. Here is a sample run: Enter a number between 0 and 511: 7
H H H
H H H
T T T
The user entered 7, which corresponds to 000000111. Since 0 stands for H and 1 for T, the output is correct.
0 0 0 1 0 1 1 1 0 1 0 1 1 0 0
0 1 0 0 0 1 1 0 0 1 1 0 1 1 1
0 0 0 1 0 0 0 0 1 1 0 0 1 1 0
Each state can also be represented using a binary number. For example, the preceding matrices correspond to the numbers
000010000 101001100 110100001 101110100 100111110
There are a total of 512 possibilities, so you can use decimal numbers 0, 1, 2, 3, . . . , and 511 to represent all states of the matrix. Write a program that prompts the user to enter a number between 0 and 511 and displays the corresponding matrix with the characters H and T. Here is a sample run: Enter a number between 0 and 511: 7
H H H
H H H
T T T
The user entered 7, which corresponds to 000000111. Since 0 stands for H and 1 for T, the output is correct.
import java.util.Scanner; public class ProgrammingEx8_11 { // Strategies: // Use binary number and bitwise operator to shift the bit to the right and // use masking to extract the last digit // Also see exercise 5.44 public static void main(String[] args) { Scanner input = new Scanner(System.in); System.out.print("Enter a number between 0 and 511: "); short n = input.nextShort(); int intArray[][] = new int[3][3]; short mask = 0b1; // to extract the last bit after shifting for (int i = 0; i < 9; i++) { short bit = (short) (n & mask); // extracting last bit i.e the // remainder of // division by 2 intArray[2-i / 3][2-i % 3] = bit; n = (byte) (n >> 1); // Shifting right is dividing by 2. The last // bit is the remainder of the next shift. } displayArray(intArray); } public static void displayArray(int array[][]) { for (int i = 0; i < array.length; i++) { for (int j = 0; j < array[0].length; j++) { if (array[i][j] == 1) { System.out.print("T "); } else { System.out.print("H "); } } System.out.println(""); } } }
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