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

492 Chap. 15 Lower BoundsProof 1: The winner must compare against all other elements, so there must ben − 1 comparisons.✷This proof is clearly wrong, because the winner does not need to explicitly compareagainst all other elements to be recognized. For example, a standard singleeliminationplayoff sports tournament requires only n − 1 comparisons, and thewinner does not play every opponent. So let’s try again.Proof 2: Only the winner does not lose. There are n − 1 losers. A single comparisongenerates (at most) one (new) loser. Therefore, there must be n − 1 comparisons.✷This proof is sound. However, it will be useful later to abstract this by introducingthe concept of posets as we did in Section 15.2.1. We can view the maximumfindingproblem as starting with a poset where there are no known relationships, soevery member of the collection is in its own separate DAG of one element.Proof 2a: To find the largest value, we start with a poset of n DAGs each witha single element, and we must build a poset having all elements in one DAG suchthat there is one maximum value (and by implication, n − 1 losers). We wish toconnect the elements of the poset into a single DAG with the minimum number oflinks. This requires at least n − 1 links. A comparison provides at most one newlink. Thus, a minimum of n − 1 comparisons must be made.✷What is the average cost of largest? Because it always does the same numberof comparisons, clearly it must cost n − 1 comparisons. We can also considerthe number of assignments that largest must do. Function largest might doan assignment on any iteration of the for loop.Because this event does happen, or does not happen, if we are given no informationabout distribution we could guess that an assignment is made after each comparisonwith a probability of one half. But this is clearly wrong. In fact, largestdoes an assignment on the ith iteration if and only if A[i] is the biggest of the thefirst i elements. Assuming all permutations are equally likely, the probability ofthis being true is 1/i. Thus, the average number of assignments done is1 +n∑i=21i =which is the Harmonic Series H n . H n = Θ(log n). More exactly, H n is close tolog e n.How “reliable” is this average? That is, how much will a given run of theprogram deviate from the mean cost? According to Čebyšev’s Inequality, an observationwill fall within two standard deviations of the mean at least 75% of the time.For Largest, the variance isn∑i=11iH n − π26 = log e n − π26 .