Ordering Functions by Order of Growth Put the functions below in order so that each function is big-O of the next function on the list. si(n) = (1.5)" We solve this exercise by successively finding the function that grows slowest among all those left on the list. f, (n) = 8n³ +17n² +111 %3D f,(n) = (log n)° f.(n) = 2" f. (n) = log (log n) f, (n) =10000 (constant, does not increase with n) fs (n) = log(log n) (grows slowest of all the others) f (n) = (log n) (grows next slowest) f.(n) = n° (log n)' (next largest,(log n)' factor smaller than any power of n) f2 (n) = 8n² +17n² +111 (tied with the one below) f. (n) = n° +n(log n) (tied with the one above) f (n) = (1.5)" (next largest, an exponential function) f.(n) = n° (log n)’ S,(m) = 2" (n² +1) fa(n) = n² + n(log n)* 3 f.(n) = 2" (grows faster than one above since 2> 1.5) f, (n) = 2" (n² +1) (grows faster than above because of the n² +1 factor) %3D f, (n) =10000 fio (n) = n! fio (n) = n! (n! grows faster thancn for every c)
Ordering Functions by Order of Growth Put the functions below in order so that each function is big-O of the next function on the list. si(n) = (1.5)" We solve this exercise by successively finding the function that grows slowest among all those left on the list. f, (n) = 8n³ +17n² +111 %3D f,(n) = (log n)° f.(n) = 2" f. (n) = log (log n) f, (n) =10000 (constant, does not increase with n) fs (n) = log(log n) (grows slowest of all the others) f (n) = (log n) (grows next slowest) f.(n) = n° (log n)' (next largest,(log n)' factor smaller than any power of n) f2 (n) = 8n² +17n² +111 (tied with the one below) f. (n) = n° +n(log n) (tied with the one above) f (n) = (1.5)" (next largest, an exponential function) f.(n) = n° (log n)’ S,(m) = 2" (n² +1) fa(n) = n² + n(log n)* 3 f.(n) = 2" (grows faster than one above since 2> 1.5) f, (n) = 2" (n² +1) (grows faster than above because of the n² +1 factor) %3D f, (n) =10000 fio (n) = n! fio (n) = n! (n! grows faster thancn for every c)
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Proved the given ordering functions by order of growth.
why is true ?
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