Petar Kapriš
a697f52405
To be noted: the drawing in exercise 1.14 is unfinished. I did it in a notebook, but haven't yet had the time to put it in a txt file.
10 lines
438 B
Text
10 lines
438 B
Text
Essentially, instead of us halving the time of computation,
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every time we have an even number, we instead halve the expmod call
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but then, make that call twice, in other words, instead of
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(specifically for even number):
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T(n) = O(1) + T(n/2)
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we get:
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T(n) = O(1) + 2*T(n/2), which is going to give us roughly:
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O(n)
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We get a recursion tree, growing exponentially with at every recursive call,
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of which there are O(log n), so exp(log(n)) = n.
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