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even if lots of white space were used to separate the two booleans, Note that, last but not least, the calculation simplifies 'I ^ (l+r-l)+2'totl<r>. It is always good practice to prove properties in a goal-oriented way, working from complicated statements to simpler statements. In this case, the goal was to prove The strategy is to take the left side of this statement, since this is the more complicated side, and simplify it. In the process, we learn that the properties 0 < / and r ^ N are not relevant. Exercise 4.2. Prove

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in the same way. Prove it also using that m+2 ^m/2 <= O^ra. (That is, integer division by 2 rounds down for positive m.) D Exercise 4.3. (a) Integer division by positive number n is monotonic with respect to the atmost relation. Is it monotonic with respect to the less-than relation That is, prove or disprove that, for all integers i and j, and all positive integers n,

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(b) Is it the case that the 'if in the monotonicity of integer division by n can be replaced by an equality That is, prove or disprove that, for all integers i and j, and all positive integers n,

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(c) Compare your answers to (a) and (b) with the following properties of addition. Why is there a difference

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(i+n < j+n) = (i+n ^ j+n) = ( i ^ j ) .

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(d) Suppose n is negative. Which of the following do you expect to be true i+n ^ j+n -^ i^j, i+n ^j+n <= jXi D

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Exercise 4.4. The searching algorithm shown in Figure 4.1 is correct so long as k is assigned a value satisfying I ^ k < r. Determine which of the following assign-

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4: Implementation Issues ments meet this requirement. In the cases that the requirement is not met give an example showing how the program would not function correctly.

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Suppose that integer division is defined to round away from zero rather than towards zero. That is, suppose If this is the case, would the following assignments to k be correct k:= k:= Draw a conclusion about the safest assignment to k in the case in which it is not known whether integer division is implemented by rounding up or down. D Exercise 4.5. Rather than cutting the middle deck into two roughly equal decks, one might try to estimate where in the middle deck the value X can be found. For example, if the deck of cards contains the numbers from 0 to M in steps of roughly %, the value X might be found roughly at position Xx^. As more and more of the deck has been eliminated from the search, the array values at positions I and r-l can be used to interpolate an estimate of the position of X in the array. A suggested implementation of this idea is to use the following assignment to k: . , X-card[l] . .. k := 1+ - - x(r-l) . card[r-l] -card[l] What is wrong with this idea (Assume that real arithmetic is used to evaluate the right side and then the value is converted to an integer by rounding. The answer does not depend on how rounding is done.) D Exercise 4.6. The implementation of binary search below is taken from a textbook on developing Java software. The class Comparator, which is used in the implementation, provides a method re! ati on which compares its two arguments. Assume that array v is sorted in ascending order (possibly with duplicate entries) and c. re!ation(x,y) implements the test x < y. The notation b el : e2 denotes a conditional expression. If boolean b evaluates to true, the value of the expression is given by el, otherwise it is given by e2. (a) Suppose array v has length 2 and elements 10 and 20. Suppose also that the object o being searched for has the value 30. Trace the execution sequence of the method and identify a run-time error in the program.

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