MHB How many comparisons are required?

[mathmari]

Hey!

Let $S$ be a set of $n$ integers. Assume you can perform only addition of elements of $S$ and comparisons between sums. Under these conditions how many comparisons are required to find the maximum element of $S$?

I thought that we could find the maximum element as followed:

max=S(1)+S(1)
k=1
for j=2 to n
     sum=S(1)+S(j)
     if sum>max
           max=sum
           k=j
return S(k)

That means that $n-1$ comparisons are required.

Is this correct?? (Wondering)

 

[lofgran]

Hello! Your approach to finding the maximum element of $S$ using comparisons and addition is correct. However, the number of comparisons needed may not always be $n-1$. Let me explain why.

In your algorithm, you are comparing each element of $S$ to the current maximum, which is initially set to $S(1)+S(1)$. This means that for each element $S(j)$, you are performing one comparison ($S(1)+S(j)$ vs $max$). Therefore, for $n$ elements in $S$, you would need $n$ comparisons.

However, we can optimize this algorithm by using a divide and conquer approach. We can divide the set $S$ into two halves and find the maximum element in each half. Then, we can compare the maximum elements of the two halves and return the larger one as the maximum element of $S$. This approach would require only $\log_2{n}$ comparisons, which is significantly fewer than $n-1$ comparisons.

In conclusion, your approach is correct but may not always give the minimum number of comparisons needed. The optimized approach would require only $\log_2{n}$ comparisons. I hope this helps!