ICS 161: Design and Analysis of Algorithms
Lecture notes for January 30, 1996

Deterministic selection

Last time we saw quick select, a very practical randomized linear expected time algorithm for selection and median finding. In practice, this is all you need to use. But for theoretical purposes, it's unsatisfying to have only a randomized algorithm, and in some rare circumstances it may more important to be predictable than to be fast. So can we get a linear worst case time algorithm? We'll describe a solution invented by five people: Blum, Floyd, Pratt, Rivest, and Tarjan.

Recall that quickselect chooses a random "pivot" x, partitions the list into elements less than and greater than x, and calls itself recursively in one of the two sublists. The even quicker selection method I outlined does something similar, but chooses the pivot in a complicated way, by calling itself recursively in a random sample of the input. Our deterministic algorithm will use the same idea of choosing x by performing a recursive call.

If we could do so quickly, one good choice would simply be to let x be the median of the values. Then each recursive call would only be on a subset of half the values. But of course if we knew how to find the median, we'd be done, so finding the median is too good to hope for. Instead let's just try to get something close to the median (say within n/4 positions of it). Then each recursive call would be on a larger fraction of the input (3n/4) but this still might be good enough.

Median of medians

How can we get something close to the median, reasonably quickly? Just like the "quicker selection", Instead of finding the median of the whole set, find a median of a sample. How do we choose the sample? Medians again!

Median-of-medians algorithm:

Can we say anything about how many items are included in this last recursive call? It's easier to talk about this in terms of the elements thrown away (not included in the call).
We always throw away either L3 (the values greater than M) or L1 (the values less than M). Suppose we throw away L3.

Among the n/5 values x[i], n/10 are larger than M (since M was defined to be the median of these values).

For each i such that x[i] is larger than M, two other values in S[i] are also larger than x[i] (since x[i] is the median of S[i]).

So L3 has at least 3 elements in each of at least n/10 groups S[i], for a total of at least 3n/10 elements. By a symmetric argument, L1 has at least 3n/10 elements.

Therefore the final recursive call is on a list of at most 7n/10 elements and takes time at most T(7n/10).

This algorithm has the property we want, that each recursive call only involves a constant fraction of the input.

Deterministic selection algorithm

Before we analyze our algorithm, let's write it out more carefully in pseudocode. Also, instead of using a special algorithm to find the median in each subset S[i], let's just call the method recursively again. To prevent infinite recursion, we have to stop when L is so small that there aren't enough x[i] values to find medians of.
    if (L has 10 or fewer elements)
        sort L
        return the element in the kth position

    partition L into subsets S[i] of five elements each
        (there will be n/5 subsets total).

    for (i = 1 to n/5) do
        x[i] = select(S[i],3)

    M = select({x[i]}, n/10)

    partition L into L1<M, L2=M, L3>M
    if (k <= length(L1))
        return select(L1,k)
    else if (k > length(L1)+length(L2))
        return select(L3,k-length(L1)-length(L2))
    else return M


The pseudo-code above gives us a number of comparisons that can be found by solving the recurrence
    T(n) <= 12n/5 + T(n/5) + T(7n/10)
The 12n/5 term comes from two places: we can sort each of the sets S[i] with seven comparisons (homework 2.31), so the step in which we compute the x[i] values takes 7n/5 comparisons total. And then the step in which we partition L takes n-1 more comparisons. The other two terms come from the two recursive calls, in which we find M and then the overall return value. As we discussed earlier, the second recursive call is on a list of at most 7n/10 elements hence its T(7n/10) bound.

Actually with some more care we can do a little better: we don't really need to sort the sets S[i], just find their medians, which only requires 6n/5 comparisons. The resulting information, together with the computation of M, should already be enough to eliminate 3n/10 elements from L. So we could get a recurrence with 6n/5 in place of the 12n/5 above, and save a factor of two in the total comparisons. But since this result is mainly of theoretical interest, I've left it in the simpler and easier to understand form above.

This recurrence looks like one coming from a divide and conquer algorithm, but one which splits the problem unequal parts. But in this case the important fact is that the parts add up to less than the whole, when this happens it doesn't matter as much how equal or unequal they are.

There are two ways to analyze a problem like this. The first is the method I showed for quickselect, in which we try to form an inductive proof that something is O(n) by assuming it's cn for some specific c, expanding the right side of the recurrence, and working through the math to determine what c should be. In our case we have

    T(n) <= 12n/5 + T(n/5) + T(7n/10)

     = 12n/5 + cn/5 + 7cn/10

     = n (12/5 + 9c/10)
If this is to be at most cn, so that the induction proof goes through, we need it to be true that
    n (12/5 + 9c/10) <= cn

    12/5 + 9c/10 <= c

    12/5 <= c/10

    c <= 24
Which tells us that we can prove by induction that T(n) <= 24n (or any larger constant times n). We also need to deal with the base case but that is easy.

The second method to analyze a recurrence like this one is to draw a tree showing the sizes of the problems in each recursive call, and analyze the total size of problems on each level of the tree. The total number of comparisons can then be found by multiplying this total subproblem size by the 12/5 factor of comparisons per element in each call. The tree starts with a root problem of size n, and then each node has two subproblems, one of size 1/5 its parent, and the other of size 7/10 its parent.

        /   \
       n/5        7n/10
      /   \       /   \
    n/25 7n/50 7n/50 49n/100
    / \   / \   / \   / \
Each problem on one level is replaced by two problems on the next level down, of sizes 1/5 and 7/10 the parent. So the total size on the next level is 1/5+7/10=9/10 that of the previous level (sometimes even less when a subproblem reaches a base case and doesn't make more recursive calls).

Therefore the total number of comparisons is

    12/5 (n + 9n/10 + 81n/100 + ...)

    = 12/5 n (1 + 9/10 + (9/10)^2 + (9/10)^3 + ...)

    = 12/5 n 1/(1-(9/10))

    = 24n
As a general rule, the geometric series sum(x^i) (for i from 0 to n-1) solves to (1-x^n)/(1-x), and whenever x is less than 1 the limit of the sum as n goes to infinity becomes 1/(1-x). The sum above is just a case of this formula in which x=9/10. The same tree-expansion method then shows that, more generally, if T(n) <= cn + T(an) + T(bn), where a+b<1, the total time is c(1/(1-a-b))n.

So our deterministic selection algorithm uses at most 24n comparisons and takes O(n) time.

With a lot of work one can reduce the number of comparisons to 2.95n [see D. Dor and U. Zwick, "Selecting the Median", 6th SODA, 1995] which is a little less than twice as much as randomized selection, but much more complicated and less practical.

ICS 161 -- Dept. Information & Computer Science -- UC Irvine
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