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7.3 Divide-and-Conquer Algorithms and Recurrence Relations. If f(n) represents the number of operations required to solve the problem of size n, it follow that f satisfies the recurrence relation f(n) = af (n/b) + g(n)
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7.3 Divide-and-Conquer Algorithms and Recurrence Relations • If f(n) represents the number of operations required to solve the problem of size n, it follow that f satisfies the recurrence relation f(n) = af(n/b) + g(n) this is called a divide-and-conquer algorithms and recurrence relations • Example 1:Binary Search We introduced a binary search algorithm in section 3.1 .
The binary search • Algorithm 3: the binary search algorithm Procedure binary search (x: integer, a1, a2, …,an: increasing integers) i :=1 { i is left endpoint of search interval} j :=n { j is right endpoint of search interval} While i < j begin m := (i + j) / 2 if x > am then i := m+1 else j := m end If x = ai then location := i else location :=0 {location is the subscript of the term equal to x, or 0 if x is not found}
Divide-and-Conquer Algorithms and Recurrence Relations • Example 3: Merge Sort • Algorithm 9A Recursive Merge Sort. Procedure mergesort(L = a1, . . . , an) If n > 1 then m := n/ 2 L1 := a1, a2, . . . ,am L2 := am+1, am+2, . . ., an L := merge (mergesort(L1), mergesort(L2)) {L is now sorted into elements in nondecreasing order}
Divide-and-Conquer Algorithms and Recurrence Relations • Example 4: Fast Multiplication of Integers • There are more efficient algorithms than the conventional algorithm (described in section 3.6) for multiplying integers. • Suppose that a and b are integers with binary expansions of length 2n. • Let a = (a2n-1 a2n-2 . . . a1 a0)2 ,b = (a2n-1 a2n-2 . . . a1 a0)2 • Let a = 2nA1 + A0 , b = 2nB1 + B0 A1= (a2n-1 . . . an+1 an)2 , A0= (an-1 . . . a1 a0)2, B1 = (b2n-1 . . . bn+1 bn)2 , B0= (bn-1 . . . b1 b0)2 , • the algorithm for fast multiplication of integers is based on the fact that ab can be rewritten as • ab= (22n+2n) A1 B1 + 2n(A1 - A0) (B0 - B1 ) +(2n +1)A0B0 .
Divide-and-Conquer Algorithms and Recurrence Relations • This shows that if f(n) is the total number of bit operations needed to multiply two n-bit integers, thenf(2n) = 3f(n) + Cn • This reasoning behind this equation is as follows. • The three multiplications of n-bit integers are carried out using 3f(n)-bit operations.
Divide-and-Conquer Algorithms and Recurrence Relations • Theorem 1: let f be an increasing function that satisfies the recurrence relation f(n)= af(n/b) + c • Whenever n is divisible by b, where a≧1 ,b is an integer greater than 1, and c is a positive real number . Then f(n) is O(nlogb a) if a >1 , O(log n) if a =1 • when n=bk,where k is a positive integer, f(n) = C1nlogba +C2= C1ak+C2 where C1= f(1)+c/(a-1) and C2 = -c/(a-1).
Divide-and-Conquer Algorithms and Recurrence Relations • Example 6: Let f(n) = 5f(n/2) + 3 and f(1) =7. Find f(2k), where k is a positive integer. Also , estimate f(n) if f is an increase function. • Example 7: Estimate the number of comparisons used by a binary search. f(n) = f(n/2) + 2. • Example 8: Estimate the number of comparisons used to locate the maximum and minimum elements in a sequence using the algorithm given in example 2. f(n) = 2f(n/2) + 2.
Divide-and-Conquer Algorithms and Recurrence Relations • Theorem 2: Master Theorem Let f be an increasing function that satisfies the recurrence relation f(n)= af(n/b) + cnd Whenever n=bk, where k is a positive integer, a≧1, b is an integer greater than 1, and c and d are real numbers with c positive and d nonnegative. Then f(n) is O(nd) if a < bd, O(nd log n) if a= bd , O(n logba) if a > bd.
Divide-and-Conquer Algorithms and Recurrence Relations • Example 9: Complexity of Merge Sort In Example 3 we explained that the number of comparisons usedby the merge sort to sort a list of n elements is less than M(n), where M(n)=2 M(n/2) +n . By the Master Theorem (Theorem 2) we find that M(n) is O(n logn) , which agrees with the estimate found in section 4.4. • Example 10: Estimate the number of bit operations needed to multiply two n-bit integers using the fast multiplication algorithm described in Example 4.
Divide-and-Conquer Algorithms and Recurrence Relations • Example 11 : Estimate the number of multiplications and additions required to multiply two n x n matrices using the matrix multiplication algorithm referred to in example 5. The function for the number of multiplications and additions is f(n)=7f(n/2)+(15/4)n2.
Divide-and-Conquer Algorithms and Recurrence Relations • Example 12: The Closest-Pair Problem Consider the problem of determining the closest pair of points in a set of n points (x1,y2),. . ., (xn,yn) in the plane, where the distance between two points (xi,yi) and (xj,yj) is the usual Euclidean distance [ (xi -xj)2+ (yi - yj)2 ]0.5. • This problem arises in many applications such as determining the closest pair of airplanes in the air space at a particular altitude being managed by an air traffic controller. • How can this closest pair of points be found in an efficient way? (FIGURE 1 )
Divide-and-Conquer Algorithms and Recurrence Relations FIGURE 1 The Recursive Step of the Algorithm for Solving the Closest-Pair Problem.
Divide-and-Conquer Algorithms and Recurrence Relations FIGURE 2 Showing That There Are at Most Seven Other Points to Consider for Each Point in the Strip.
