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490 Chap. 15 Lower BoundsABCGDEFFigure 15.1 Illustration of using a poset to model our current knowledge of therelationships among a collection of objects. A directed acyclic graph (DAG) isused to draw the poset (assume all edges are directed downward). In this example,our knowledge is such that we don’t know how A or B relate to any of the otherobjects. However, we know that both C and G are greater than E and F. Further,we know that C is greater than D, and that E is greater than F.comparing elements in L we have at least n − m partial orders. Every such partialorder needs at least one comparison against K to make sure that K is not somewherein that partial order. Thus, any algorithm must make at least n comparisons in theworst case.✷15.2.2 Searching in Sorted ListsWe will now assume that list L is sorted. In this case, is linear search still optimal?Clearly no, but why not? Because we have additional information to work with thatwe do not have when the list is unsorted. We know that the standard binary searchalgorithm has a worst case cost of O(log n). Can we do better than this? We canprove that this is the best possible in the worst case with a proof similar to that usedto show the lower bound on sorting.Again we use the decision tree to model our algorithm. Unlike when searchingan unsorted list, comparisons between elements of L tell us nothing new about theirrelative order, so we consider only comparisons between K and an element in L. Atthe root of the decision tree, our knowledge rules out no positions in L, so all arepotential candidates. As we take branches in the decision tree based on the resultof comparing K to an element in L, we gradually rule out potential candidates.Eventually we reach a leaf node in the tree representing the single position in Lthat can contain K. There must be at least n + 1 nodes in the tree because we haven + 1 distinct positions that K can be in (any position in L, plus not in L at all).Some path in the tree must be at least log n levels deep, and the deepest node in thetree represents the worst case for that algorithm. Thus, any algorithm on a sortedarray requires at least Ω(log n) comparisons in the worst case.We can modify this proof to find the average cost lower bound. Again, wemodel algorithms using decision trees. Except now we are interested not in thedepth of the deepest node (the worst case) and therefore the tree with the leastdeepestnode. Instead, we are interested in knowing what the minimum possible is