Proved the given ordering functions by order of growth.why is true ? Ordering Functions by Order of GrowthPut the functions below in order so that each function is big-O ofthe next function on the list.fi(n) = (1.5)"f2 (n) = 8n +17n² +111We solve this exercise by successively finding the functionslowest among all those left on the list.thatgrowsfa(n) = (log n)²f.(n) = 2"fs(n) = log (log n)So(n) = n° (log n)'f, (n) = 2" (n² +1)fo (n) = n° + n(log n)f, (n) =10000fio (n) = 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) = 8n° +17n² +111 (tied with the one below)fa(n) = n° +n(log n) (tied with the one above)fi(n) = (1.5)" (next largest, an exponential function)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)2%3Df1o (n) = n! (n! grows faster thancn for every c)

The big O notation gives the worst case time complexity and it gives the tightest upper bound. Given…

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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