A meeting organizer schedules a meeting by creating a meeting ____________________.

When is the base case value == anArray[mid] (where mid is the midpoint of the array) reached in a recursive binary search algorithm?

What will be an ideal response?

(The Swing part of this project is quite straightforward, but you do need to know a little about how to convert numbers from one base to another.) Write a program that converts numbers from base-10 (ordinary decimal) notation to base-2 notation. The program uses Swing to perform input and output via a window interface. The user enters a base-10 number in one text field and clicks a Convert button. The equivalent base-2 number then appears in another text field. Be sure that the two text fields are labeled. Include a Clear button that clears both text fields when clicked. Also be sure that the close-window button works correctly.

This Project has some interesting challenges even if the student knows how to convert from decimal to binary. The solution shown here is designed to work for non-negative decimal integers only and uses the successive division by 2 algorithm, which is probably the easiest to program. However, there are several details that need special attention. For example, what if the user clicks the “Convert” button before anything is entered in the decimal text field? And what if the user does not enter a valid non-negative decimal number? The solution shown here displays a message in the binary number field for the first case, but does not check the decimal field input for incorrect input (a better implementation would detect errors in data entry by catching NumberFormatException exceptions and would also print an error message if the number entered was not a non-negative integer). The algorithm used to convert decimal to binary by successive division is as follows: quotient = decimal number while(quotient is not zero) next binary digit = remainder(decimal number/2) quotient = integer part of (quotient/2) //Throw away the fractional part. Each iteration of the while loop produces one binary digit. The first iteration produces the least significant bit and each successive iteration produces the next higher significant bit, with the most significant bit produced in the last iteration. The loop ends when the quotient is zero, since all divisions after that will just add leading zeros. Note that the last digit is, necessarily, 1. The algorithm is simple to program with the modulo operator (%) and truncating integer division, but, unfortunately, the bits are obtained in the reverse order we need to print (the most significant bit must be printed first but it is the last obtained). The approach taken in the solution shown here puts the bit values in a character array, one at a time, from least significant bit to most significant bit, which makes the conversion process easy to code. Then the character array is read in reverse order, appending each character to a String so it will print in the correct order, with most significant bit on the left