Difference between revisions of "Talk:TADM2E 3.11"

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(a counterexample for 3.11 b: new section)
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Why cant we just sort the input (array)? Will this not mean the smallest number between x and y is always x?
 
Why cant we just sort the input (array)? Will this not mean the smallest number between x and y is always x?
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Well, i thought the same, but we cannot. It becomes clear when you think of the interface as below:
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<math>find\_min( (x_1, x_2, .. x_n), i, j)</math>
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So, the user supplied us the sequence, and they expect us to find the minimum of a sub-sequence of the given sequence.
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--[[User:Mytreya|Mytreya]] ([[User talk:Mytreya|talk]]) 10:11, 27 September 2019 (UTC)
  
 
== a counterexample for 3.11 b ==
 
== a counterexample for 3.11 b ==

Latest revision as of 10:13, 27 September 2019

For the answer given in part 2, the space required seems to be the sum from 1 to n of n which is n(n+1)/2. This gives O(n**2) for space requirements. Doesn't it?


Why cant we just sort the input (array)? Will this not mean the smallest number between x and y is always x?

Well, i thought the same, but we cannot. It becomes clear when you think of the interface as below: $ find\_min( (x_1, x_2, .. x_n), i, j) $ So, the user supplied us the sequence, and they expect us to find the minimum of a sub-sequence of the given sequence. --Mytreya (talk) 10:11, 27 September 2019 (UTC)

a counterexample for 3.11 b

for this input = [0,1,...,98,99] and this query i = 1, j = 99.

the suggested algorithm will always span in both sides - right and left, so according to the suggested solution it will always split the task to 2. that means that if h is the height of the tree the query time is going to be O(2^h)

now h = log(n) so it's 2^log(n) = n

for O(n) cost we can simply iterate over all the values in the given range and find the minimum...