Data Structures and Algorithm Analysis - Computer Science at ...

306 Chap. 9 SearchingWe now show that this is the same as√ n∑P(need at least i probes)i=1= 1 + (1 − P 1 ) + (1 − P 1 − P 2 ) + · · · + P √ n= (P 1 + ... + P √ n) + (P 2 + ... + P √ n) +(P 3 + ... + P √ n) + · · ·= 1P 1 + 2P 2 + 3P 3 + · · · + √ nP √ nWe require at least two probes to set the bounds, so the cost is√ n∑2 + P(need at least i probes).i=3We now make take advantage of a useful fact known as Čebyšev’s Inequality.Čebyšev’s inequality states that P(need exactly i probes), or P i , isP i ≤p(1 − p)n(i − 2) 2 n ≤ 14(i − 2) 2because p(1 − p) ≤ 1/4 for any probability p. This assumes uniformly distributeddata. Thus, the expected number of probes is√ n∑ 12 +4(i − 2) 2 < 2 + 1 4i=3∞∑i=11i 2 = 2 + 1 π4 6 ≈ 2.4112Is QBS better than binary search? Theoretically yes, because O(log log n)grows slower than O(log n). However, we have a situation here which illustratesthe limits to the model of asymptotic complexity in some practical situations. Yes,c 1 log n does grow faster than c 2 log log n. In fact, it is exponentially faster! Buteven so, for practical input sizes, the absolute cost difference is fairly small. Thus,the constant factors might play a role. First we compare lg lg n to lg n.Factorn lg n lg lg n Difference16 4 2 2256 8 3 2.72 16 16 4 42 32 32 5 6.4