Finding Min. Element in Sequence: Algo & Steps

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The discussion focuses on developing an algorithm to find the minimum element in a sequence of numbers \( A = (a_1, a_2, \dots, a_n) \) with \( n \geq 3 \) using at most \( \lceil \frac{n}{2} \rceil \) comparisons. The proposed method involves comparing the first two elements and subsequently examining either the odd or even indexed elements based on which of the first two is smaller. The participants clarify the requirement of using \( \lceil \frac{n}{2} \rceil \) comparisons, ensuring the algorithm adheres to this constraint.

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evinda
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Hello! (Smile)

We have a sequence of numbers $A=(a_1, a_2, \dots, a_n), n \geq 3$, for which, it stands the following:

$$\forall i \text{ with } 2 \leq i \leq n-1:$$

either
$$ 1. a_{i-1}<a_i \text{ and } a_i>a_{i+1}$$

or
$$2. a_{i-1}>a_i \text{ and } a_i<a_{i+1}$$

I have to write an algorithm, that finds the minimum element of $A$ with at most $\lceil \frac{n}{2} \rceil$ comparisons.

So, we have to compare the first two elements and if the first is smaller, we will just look at the elements of the odd positions, in order to find the minimum, and if the second element is smaller we will look at the elements of the even positions, right? (Thinking)

Then, how will we find the minimum of the remaining elements we are looking at? Using a while or is there an other way? (Thinking)
 
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evinda said:
I have to write an algorithm, that finds the minimum element of $A$ with at most $\lceil \frac{n}{2} \rceil$ elements.

Do you mean $\lceil \frac{n}{2} \rceil$ comparisons?
 
magneto said:
Do you mean $\lceil \frac{n}{2} \rceil$ comparisons?

Yes, I am sorry... (Blush)
 

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