Proof that the maximum element in a linear search is found log(n) times

Question

Consider an array $ X $ of $ m $ numbers, where each number $ X_i $ is drawn independently and uniform at random from the interval (0, 1). Let $ X_{max} = -\infty $. When we scan the array, if $ X_i > X_{max} $ then $ X_{max} \leftarrow X_i $. Assume that $ \forall i,j: X_i \neq X_j $.

I read a claim that the update step is executed in expectation at most $ log(m) $ times, or using the big-oh notation $ O(log(m)) $ times. I considered permutations of $ X $ but that seems to lead nowhere near a $ log(m) $ result. What approach should I use to validate the claim?

Is your logarithm base $2$? The number of times the value is updated is certainly random; do you mean $\log n$ times in average? — dafinguzman, Nov 09 '15 at 20:36

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

You get an update while scanning $X_i$ if and only if $X_i = \max{(X_j, j \leq i)}$.

Since your elements are taken uniformly at random, the probability of one given element of a list of size $i$ to be the maximum of this list being $\dfrac{1}{i}$, in average, you would get $H_n = 1+\dfrac{1}{2}+...+\dfrac{1}{n}$ updates for a list of size $n$.

It happens that $H_n \approx{} \ln n$, which makes the result given correct. For more info on this, see this question on Math.SE.

edited Apr 13 '17 at 12:21

Community

1

answered Nov 09 '15 at 21:02

Traklon

2,838

Isn't this result the proof that the number of updates is actually at LEAST ln $n$? – user288580 Nov 10 '15 at 21:08

Proof that the maximum element in a linear search is found log(n) times

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