Binary Search. Roland Backhouse February 5th, 2001

Transcription

1 1 Binary Search Roland Backhouse February 5th, 2001

2 Outline 2 An implementation in Java of the card-searching algorithm is presented. Issues concerning the correctness of the implementation are raised and resolved.

3 Implementation in Java 3 Tasks: Choose representation of each of the decks of cards. Implementation of the choice of a card in the middle deck of cards. Adding to the left and right decks. The Program.

4 Data Structures 4 The deck of cards can be represented by an array card. For simplicity, we assume that the element type is integer, and the given card X is an integer value. We let N denote the length of the array. Splitting the deck into two can be represented by returning an integer index in the range 0..N; if the returned index is l then all cards in the segment 0..l-1 should be less than X and all cards in the segment l..n-1 should be at least X. The declaration of the procedure we wish to write is thus: public int split(int[] card, int X);

5 Data Structures (Continued) 5 The left, middle and right decks can all be represented by two indices l and r. The left deck is the segment of card indexed by values in the range 0..l-1. The middle deck is the segment indexed by values in the range l..r-1. The right deck is the segment indexed by values in the range r..n-1. The indices l and r satisfy the invariant property 0 <= l <= r <= N. An empty left deck is represented by l = 0 and an empty right deck is represented by r = N.

6 Choosing a Card 6 The middle deck is empty when l = r ; choosing a card in the middle deck occurs when it is known that l < r. The choice is represented by choosing an (integer) index k in the range l..r-1. For correctness, any choice in this range will do; for efficiency, a good idea is to try to reduce the size of the middle deck by a half at each iteration. This is achieved by choosing k in the middle of the range, using the assignment: k = (l+r-1)/2 We postpone checking that this does return a value in the right range until later.

7 Adding to the Left and Right Decks 7 The final pieces of the implementation represent adding the cards in the lower deck to the left deck and adding cards in the upper deck to the right deck. These are implemented by the assignments and l = k+1 r = k respectively.

8 The Program 8 public int split(int[] card, int X); { int N = card.length; int l = 0; int r = N; /* Invariant: all cards in segment 0..l-1 are less than X * all cards in segment r..n-1 are at least X * 0 <= l <= r <= N * Bound function: r-l */ while (l < r) {int k = (l+r-1)/2; if (card[k] < X) l = k+1; else r = k; } /* All cards in segment 0..l-1 are less than X * All cards in segment l..n-1 are at least X */ return l; }

9 Verifying Correctness 9 Choosing a card One particular aspect of the Java program that might give rise to some doubt is the assignment to k: int k = (l+r-1)/2 The right side of this assignment is an integer division and not an exact division. The division operator / in Java is overloaded. The computation that is executed depends on the type of the arguments. When both its arguments are integers, the result is an integer; if both arguments are floating point values then the result is a floating point value. Can we be really sure that the assignment will set k to a value that is in the correct range; does it depend on how the exact division is rounded to an integer value? In order to be absolutely sure we need to understand properly the effect of an integer division.

10 Notation 10 It is useful to switch from the monospaced ASCII notation of Java to mathematical notation. We will use m n to denote the integer value obtained by dividing integer m by integer n. This is different from m n and m/n which both denote the real value obtained by dividing m by n. We also use := to denote the assignment operator, avoiding the confusion (and errors) that occur when the Java = symbol is mistaken for equality. In this notation the assignment is: k := (l+r 1) 2.

11 Verification Condition 11 The assignment to k is executed when 0 l < r N. The requirement is that the value assigned to k is at least l and less than r otherwise we cannot guarantee termination of the program. So what we have to verify is that l (l+r 1) 2 < r 0 l < r N. (Note: read the symbol as if.)

12 Properties of Division 12 For our purposes it is reasonable to assume the following properties of the integer division m n. 1. Dividing a multiple of n by n is exact. That is, (m n) n = m.

13 Properties of Division 13 For our purposes it is reasonable to assume the following properties of the integer division m n. 1. Dividing a multiple of n by n is exact. That is, (m n) n = m. 2. Integer division by (positive number) n is monotonic with respect to the at-most relation. That is, for integers i and j, i n j n i j.

14 Properties of Division 14 For our purposes it is reasonable to assume the following properties of the integer division m n. 1. Dividing a multiple of n by n is exact. That is, (m n) n = m. 2. Integer division by (positive number) n is monotonic with respect to the at-most relation. That is, for integers i and j, i n j n i j. 3. Integer division rounds towards 0. In particular, m n m/n 0 m 0 n. (We won t need to consider the case when either of m or n is negative.)

15 The Verification 15 Now, recall that what we have to prove is: l (l+r 1) 2 < r 0 l < r N. This is the same as showing that l (l+r 1) 2 0 l < r N and (l+r 1) 2 < r 0 l < r N.

16 The Verification (Continued) 16 l (l+r 1) 2

17 The Verification (Continued) 17 l (l+r 1) 2 = { division of multiples of 2 is exact } (2 l) 2 (l+r 1) 2

18 The Verification (Continued) 18 l (l+r 1) 2 = { division of multiples of 2 is exact } (2 l) 2 (l+r 1) 2 { division by 2 is monotonic } 2 l l+r 1 = { addition is monotonic } l r 1 = { l and r are integers. } l < r.

19 The Verification (Continued) 19 We have thus shown that l (l+r 1) 2 l < r. But So it follows that l < r 0 l < r N. l (l+r 1) 2 0 l < r N as required.

20 Exercise 1 The searching algorithm is correct so long as k is assigned a value satisfying l k < r. Determine which of the following assignments meet this requirement. In the cases that the requirement is not met give an example showing how the program would not function correctly. 20 k := l k := r k := (l+r) 2 Suppose that integer division is defined to round away from zero rather than towards zero. That is, suppose m/n m n 0 m 0 n. If this is the case, would the following assignments to k be correct? k := (l+r 1) 2 k := (l+r) 2 Draw a conclusion about the safest assignment to k in the case that it is not known whether integer division is implemented by rounding up or down.

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22 Exercises 22 Do the following exercises and hand your solutions in to your tutor. Exercises 3.1, 3.2, 3.4, 3.6.