1. 1 Efficiency of Algorithms: Logarithmic Orders Binary Search Algorithm

2. 2 Logarithmic function Definition: The logarithmic function with base b (b>0, b  1) is the following function from R + to R: log b (x) = the exponent to which b must raised to obtain x. Symbolically, log b x = y  b y = x. Property: If the base b>1, then the logarithmic function is increasing: if x 1 <x 2, then log b (x 1 ) < log b (x 2 ). Note: Logarithmic function grows very slowly, e.g., log 2 (1,024)=10, log 2 (1,048,576)=20.

3. A property of logarithmic function Proposition 1: If k is an integer and x is a real number with 2 k  x < 2 k+1, then  log 2 x  = k. Proof: 2 k  x < 2 k+1 log 2 (2 k )  log 2 (x) < log 2 (2 k+1 ) (since log 2 x increasing) k  log 2 (x) < k+1 (by definition of log 2 x)  log 2 x  = k (by definition of floor function) ■

4. 4 An application of logarithms Question: Given a positive integer n, how many binary digits are needed to represent n? Solution: Binary representation of n: 1c k-1 c k-2 …c 2 c 1 c 0 which corresponds to n = 2 k + c k-1 ∙2 k-1 + … + c 2 ∙2 2 + c 1 ∙2 + c 0 Since c i  1, n = 2 k + c k-1 ∙2 k-1 + … + c 2 ∙2 2 + c 1 ∙2 + c 0  2 k + 2 k-1 + … + 2 2 + 2 + 1 = (2 k+1 -1) / (2-1) (as a sum of geometric sequence) = 2 k+1 -1 < 2 k+1 (1) On the other hand, n = 2 k + c k-1 ∙2 k-1 + … + c 2 ∙2 2 + c 1 ∙2 + c 0 ≥ 2 k (2) Combining (1) and (2): 2 k  n < 2 k+1 (3) Based on (3) and Proposition 1, k =  log 2 n  and the number of binary digits is  log 2 n  + 1.

5. 5 Exponential and Logarithmic Orders  For all real numbers b and r with b>1 and r>0 and for all sufficiently large values of x,  log b x  x r ; ( which implies that log b x is O(x r ) )  x r  b x ( which implies that x r is O(b x ) )  For all real numbers b with b>1 and for all sufficiently large values of x, x  x log b x  x 2 (which implies that x is O(x log b x) and x log b x is O(x 2 ) )

6. Binary Search Algorithm The algorithm searches for an element x in an ascending array of elements a[1],…,a[n]. Algorithm body: index:=0, bot:=1, top:=n while (top ≥ bot and index=0) mid :=  (bot+top) / 2  if a[mid] = x then index := mid if a[mid] > x then top := mid-1 else bot := mid+1 end while Output: index (If index has the value 0 then x is not in the array; otherwise, index gives the index of the array where x is located.)

7. 7 Binary Search Algorithm: Example Suppose a[1]=Amy, a[2]=Bob, a[3]=Dave, a[4]=Erin, a[5]=Jeff, a[6]=John, a[7]=Larry, a[8]=Mary, a[9]=Mike, a[10]=Sam, a[11]=Steve, a[12]=Tom. (sorted in alphabetical order) Search for x=Erin. The table tracing the binary search algorithm: index04 bot114 top1255 mid634

8. 8 The Efficiency of the Binary Search Algorithm At each iteration, the length of the new subarray to be searched is approximately half of the previous one. If n = 2 k +m, where 0  m < 2 k, then n can be split approximately in half k times. Since 2 k  n < 2 k+1, then k =  log 2 n  (by proposition 1) Thus, the number of iterations of the while loop in a worst-case execution of the algorithm is  log 2 n  +1. The number of operations in each loop is constant (doesn’t increase with n). Thus, the binary search algorithm is O(log 2 n).