Binary Nums
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ProblemAs you may know, computers work on the binary system. In this system, numbers are written using only 0s and 1s. In this question, you’ll write a program to convert binary numbers to normal (decimal) numbers. How are all numbers represented in 0s and 1s?
Here are the first few numbers written in both decimal (normal) notation and binary:
Suppose we have a binary number (a sequence of zeroes and ones) and we want to convert it to a decimal number. To do this, we multiply the digit (0 or 1) in each position of the number by an increasing power of two the further we move left. So the value of the binary number 1011001 is 1 * 2^6 + 0 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 0 * 2^1 + 1 * 2^0 = 89
Program Specifications
OutputUse System.out for output. Follow the format of the output shown above. You may notice that your program fails for large numbers. That's because of integer overflow. If you would like to get your program to work for larger binary numbers, replace int with long in your program (including using the Scanner method nextLong()). |
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SolutionDetermine how to solve the problemIf you already know how to convert from Binary Numbers to Decimal Numbers skip this section. The greatest ways to start a problem is to divide it into smaller problems and understand which parts you need to focus on. It is clear that the main problem to solve is converting binary numbers to decimal numbers. However, you will need to use the Scanner class, loops, and System.out . Let's start by trying to figure out how to convert from binary numbers to decimal numbers. Convert from Binary Numbers to Decimal NumbersSo you don't know how to convert binary numbers to decimal numbers. That's great. We can figure it out. To solve the problem, we will initially forget about writing good java code, and instead use anything simpler. We can draw diagrams, write equations, and write pseudocode. We can always write java code after we understand how to solve the problem. The first thing to do is gather information from the problem specification. In it we were told:
1 * 2^6 + 0 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 0 * 2^1 + 1 * 2^0 = 89 If we look at this carefully, we see that the right most digit of the binary number is multiplied by 2^0. The second left most digit is multiplied by 2^1. The digit to the right of that is multiplied by 2^2 and so on. Finally, all those products are summed up. However, we need to make sure we are correct. Let's check that we understand by converting some of the following numbers from binary to decimal. 0, 1, 10, 11, 101, 1000, 1101, 10101 Now, try to write some pseudocode to explain how we solved this problem. We will later shape this into Java code. 1. take the rightmost digit, if it is a 1, multiply it by 2 to the power of the current exponent and add it to the sum. 2. do the same with digits to the left of the rightmost digit. A little bit more thinking shows we need to tell what sum, and what exponent to start at. Anything multiplied by zero is zero, so we can simplify the pseudocode. Let's also add variables for clarity. exponent = 0 sum = 0 take the rightmost digit ''d'', sum = sum + d * 2^exponent do the same with digits to the left of the rightmost digit.
Convert to Java-like PseudocodeHow do I take the rightmost digit in JavaIn Java, you have some basic arithmetic operators, you can multiply (using *), divide (using /), add (+), subtract (-), and get the remainder (%). If you can define what the rightmost digit is using the above operations you can easily determine what it is. The rightmost digit of a number is the exact same thing as the remainder of a number divided by 10. For example, the remainder of 26 divided by 10 is 6. In Java, we would write 26 % 10 to do the same operation. How do I take the digits to the left of the rightmost digit in JavaSo you want to convert 321 to 32. Recall, that in Java the division of two integers ignores any fractions. So 3/2 =1, in Java code although it equals 1.5 in regular math. The way java calculates the quotient is called integer division. To chop off the rightmost digit, we simply do integer divide by 10. In Java, we would write 26 /10 and the result would be 2.
Transform the Pseudocode into Java CodeYou can continue improving the pseudocode slightly until you have replaced much of the English instructions with Java-like code. Here is an example of my result. exponent = 0 sum = 0 while (binaryNum!= 0) { rightmost = binaryNum % 10 //right most digit sum = sum + rightmost * 2^exponent binaryNum = binaryNum / 10 //do the same with the digits to the left of the rightmost digit } Now that you have some code, let's make sure that it is sound. Let's run it through the compiler. Create the Structure of the ProgramNow create a Java file. Define the class in a Java file. Name the file the same name as the class name plus the ".java" extension. Define the main method. The following code shows the result of these steps. public class BinaryNum { public static void main(String[] args) { } } Test your PseudocodeTake the code for converting numbers, and paste it into the main method. Then compile it and fix any errors. You may need to add semicolons, and declare variables. You will need to add an output statement so you can test your results. As for input, just set the binaryNum variable in the code for now (this is called hardcoded input). public class BinaryNum { public static void main(String[] args) { int binaryNum = 101; //hardcoded input int exponent = 0; int sum = 0; while (binaryNum!= 0) { rightmost = binaryNum % 10 //right most digit sum = sum + rightmost * 2^exponent; binaryNum = binaryNum / 10; //do the same with the digits to the left of the rightmost digit } System.out.println(binaryNum + "in decimal is " + sum); } } Fix your CodeFor me trying to compile my code resulted in the need to fix a few small syntactical errors. Once I ran my code, the output was strange. 0 in decimal is 4. Hmm, that cannot be right. Whenever a program does this, a computer scientist can do one of two things, only one of them is rational. He or she can get angry at the computer, and say things like -it should work. Or he or she can accept that their code is incorrect, and try to see where it is incorrect. However, it can be difficult to see where you went wrong. The same type of thinking that produced the error cannot be used to see the error. You need to think like the machine to see the error. Go through your code line by line, and determine what went wrong. You can either do this mentally, insert print statements to print out the values of the variable, or you can use a debugger. The problem with my code, was that I destroyed the value of my input. While calculating the value of the decimal number, my program eventually sets the input binaryNum to zero. The simple solution is to add another variable to store a copy of the input, and print out the original. The result is the following code public static void main(String[] args) { int binaryNum = 101; //hardcoded input int exponent = 0; int sum = 0; int rightmost; int copyOfBinNum = binaryNum; while (copyOfBinNum!= 0) { rightmost = copyOfBinNum % 10; //right most digit sum = sum + rightmost * 2^exponent; copyOfBinNum = copyOfBinNum / 10; //do the same with the digits to the left of the rightmost digit } System.out.println(binaryNum + " in decimal is " + sum); } Now test your code with different values for binaryNum. Don't forget to recompile after changing your code. Once you are satisfied, that your code correctly converts binary numbers to decimal numbers move to the next step. Implement InputThe problem specification says we need to use scanner.
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