The tournament sort is a sorting algorithm that works by...

The tournament sort is a sorting algorithm that works by building an ordered binary tree. We represent the elements to be sorted by vertices that will become the leaves. We build up the tree one level at a time as we would construct the tree representing the winners of matches in a tournament. Working left to right, we compare pairs of consecutive elements, adding a parent vertex labeled with the larger of the two elements under comparison. We make similar comparisons between labels of vertices at each level until we reach the root of the tree that is labeled with the largest element. The tree constructed by the tournament sort of 22, 8, 14, 17, 3, 9, 27, 11 is illustrated in part (a) of the figure. Once the largest element has been determined, the leaf with this label is relabeled by ??, which is defined to be less than every element. The labels of all vertices on the path from this vertex up to the root of the tree are recalculated, as shown in part (b) of the figure. This produces the second largest element. This process continues until the entire list has been sorted. How many comparisons does the tournament sort use to find the second largest, the third largest, and so on, up to the (n ? 1)st largest (or second smallest) element?

The tournament sort is a sorting algorithm that works by building an ordered binary tree. We represent the elements to be sorted by vertices that will become the leaves. We build up the tree one level at a time as we would construct the tree representing the winners of matches in a tournament. Working left to right, we compare pairs of consecutive elements, adding a parent vertex labeled with the larger of the two elements under comparison. We make similar comparisons between labels of vertices at each level until we reach the root of the tree that is labeled with the largest element. The tree constructed by the tournament sort of 22, 8, 14, 17, 3, 9, 27, 11 is illustrated in part (a) of the figure. Once the largest element has been determined, the leaf with this label is relabeled by ??, which is defined to be less than every element. The labels of all vertices on the path from this vertex up to the root of the tree are recalculated, as shown in part (b) of the figure. This produces the second largest element. This process continues until the entire list has been sorted.

How many comparisons does the tournament sort use to find the second largest, the third largest, and so on, up to the (n ? 1)st largest (or second smallest) element?

Key Concepts

Comparison Analysis

The initial tournament tree construction requires n?1 comparisons to determine the maximum element. Then, to find the second largest and subsequent elements, the algorithm performs roughly log?(n) comparisons per extraction, as only the path from the removed element's leaf to the root needs to be recomputed. Thus, after the first maximum extraction, finding each subsequent element up to the (n?1)st largest requires approximately log?(n) comparisons, leading to an overall additional cost of about (n?1)log?(n) comparisons.

Tournament Tree

A tournament tree is a complete binary tree where the leaves represent the initial elements to be sorted, and each internal node stores the result of a comparison between its two children. The winner (for example, the larger element) is propagated upward, so that the root contains the overall maximum. This structure mirrors the bracket system of a sports tournament by systematically eliminating candidates until a champion is determined.

Tournament Sort Algorithm

The tournament sort algorithm works by first building a tournament tree from the input list, using pairwise comparisons to fill the internal nodes. After the maximum element is identified at the root, it is removed (typically by replacing its leaf with a very small sentinel value) and the tree is updated along the path from that leaf to the root. This process is repeated to successively extract the second largest, third largest, and so on until the list is completely sorted.

Recomputation and Tree Updating

Once the maximum element is removed, only the branch or path of the tree affected by the removal needs to be recalculated. Since the height of the tree is proportional to log?(n) (assuming a balanced tree), each update from a removed element requires only logarithmically many comparisons. This efficient update mechanism is a critical factor in the performance of tournament sort after the first maximum is determined.

Transcript

00:01 So they want us to use the tournament sort to figure out how many comparisons we would need to find the second third all the way up to the n minus first largest element.

00:13 So we already found in the previous problem that it would take or how many it would take to get our first one.

00:23 So we don't need to really talk about that.

00:25 So let's just talk about like second third and so on.

00:27 So remember, we're assuming that n is in the form of 2 to the k for some value of k.

00:35 So let's just kind of go through some values of k and see what we get.

00:39 So let's say, well, if k is equal to 1, so it actually doesn't give us anything interesting.

00:43 So let's do 2.

00:45 So i'm just going to list 1, 2, 3 up to 4.

00:48 So we compare these elements right there for our tournament sort.

00:55 That one should be 4.

00:57 And then we compare those and that gives us 4.

00:59 So that was just finding our largest, but now let's assume that we get rid of that, replace this with negative infinity, and we want to find our second largest.

01:12 All right, well, let's see how many this would give us.

01:15 Well, we really don't need to compare three in negative infinity.

01:17 We already know that one will go up, but now we need to compare two and three.

01:22 So that would give us three up top.

01:25 So this would only really be one comparison.

01:28 Well, let's figure out what our third should be.

01:32 Well for here remember we're going to get rid of all of the threes so that's negative infinity so in this case we don't need to compare negative infinity negative infinity we just know that this blue negative infinity would get to move up you would have that then we would just compare two and negative infinity well two is going to win so again we only had one comparison there all right now what about when k is equal to three.

02:06 So maybe i should be writing that n would be equal to eight and then over here n is equal to four.

02:12 So let's just write one to eight really quickly.

02:14 So one, two, three, four, six, seven, eight.

02:19 So doing this tournament sort, so that gives us two, four, six, and eight.

02:25 Then we compare these two.

02:26 So that's four versus eight...

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