Solution Manual for Invitation to Computer Science 7th Edition by Schneider Sample

Solution Manual for Invitation to Computer Science 7th Edition by Schneider Sample

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FOR MORE OF THIS COURSE AND ANY OTHER COURSES, TEST BANKS, FINAL EXAMS, AND SOLUTION MANUALS CONTACT US AT [email protected] Solutions to End-of-Chapter Exercises Chapter 1: An Introduction to Computer Science 1. There is no one correct answer. Common examples are the instructions for using a voice mail system, the instructions for opening a mail box lock, and the instructions for doing laundry. 2. A heuristic is a method for finding a reasonably close, “good enough” solution to a problem. It can be viewed as a rule-of-thumb, a method of approximation, an informal technique, or even a way to make an “educated guess.” It differs from the concept of an algorithm in that it does not guarantee to produce an optimal solution, just to make a good faith attempt to locate a reasonable one. Heuristics are often used when executing an algorithm might be too time-consuming, and we only need an approximation to the correct answer. An example of a heuristic for adding two 3-digit numbers, such as 234 + 567, might be: 1. Set the one and tens digit of both operands to 0 2. Increase the hundreds digit of the second operand by 1. These two steps result in changing the problem to the simpler one 200 + 600. 3. Add the hundreds digits, resulting in a final “answer” of 800. Now, of course, this is not the correct answer, which is 801. But the result we get may be close enough for our needs, and it is certainly a lot easier to add a single column of numbers rather than three columns of numbers. 3. One may argue that the instruction is not well-ordered, since it is unclear whether one should enter the channel first or press CHAN first. Also, it may not be effectively computable if you desire to enter a channel that is out of the DVR’s range. 4. (a) Sequential (b) Conditional (c) Sequential (d) Iterative
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5. Step 1: carry = 0, c 3 = ??, c 2 = ??, c 1 = ??, and c 0 = ?? Step 2: i = 0, all others unchanged Step 4: c 0 = 18, all others unchanged Step 5: c 0 = 8 and carry = 1, all others unchanged Step 6: i = 1, carry = 1, c 3 = ??, c 2 = ??, c 1 = ??, and c 0 = 8 Step 4: c 1 = 7, all others unchanged Step 5: carry = 0, all others unchanged Step 6: i = 2, carry = 0, c 3 = ??, c 2 = ??, c 1 = 7, and c 0 = 8 Step 4: c 1 = 1, all others unchanged Step 5: carry = 0, all others unchanged Step 6: i = 3, carry = 0, c 3 = ??, c 2 = 1, c 1 = 7, and c 0 = 8 Step 7: c 3 = 0, c 2 = 1, c 1 = 7, and c 0 = 8 Step 8: Print out 0178. 6. Replace Step 8 with the following steps: Step 8: Set the value of i to m Step 9: Repeat step 10 until either c i is not equal to 0 or i < 0 Step 10: Subtract 1 from i , moving one digit to the right Step 11: If i > 0 then print c i c i-1 . . . c 0 7 . Assume that a has n digits a n-1 , … , a 0 , and b has m digits, b m-1 , … , b 0 , with n not necessarily equal to m . Add an operation at the beginning of the algorithm that resets the two numbers to the same number of digits by adding non-significant leading zeros to the shorter one. We can then reuse the algorithm of Figure 1.2.
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