O- notation To indicate the asymptotic high end is O-notation. For agiven function g(n) and indicate by O(g(n)).O(g(n)) = {f(n): thereexist positive constants t and n0 such that 0 ? f(n) ? t *g(n) forall n ? n0 } ?-notation To indicate the asymptotic low endis ?-notation. Fora given function g(n) and indicate by ?(g(n)). ?(g(n)) ={f(n): there exist positive constants t and n0 such that 0 ? t *g(n)? f(n) for all n ? n0 } ?-notation To indicate asymptotic tight boundis ?-notation. For a given function g(n) andindicate by ?(g(n)).?(g(n))= {f(n): there exist positive constants t1,t2 and n0 such that 0 ? t1 *g(n) ? f(n) ) ? t2*g(n) for all n > n0 }The all above scenario shownin fig.

Fig. Time Complexity notations During analyzing analgorithm, mostly consider O-notation because it will give us the high end of executiontime. This is the worst case of execution time. Tocompute O-notation lower order terms need to ignore.

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Because lower order termsare relatively insignificant for large input.Let f(N) = 2 * N2 + 3 * N +5O(f(N)) = O (2 * N2 + 3 * N +5) = O(N2) Let us consider the scenario. cnt =0for edges node.edges cnt++ ifedge not in resolved dep_resolve(edge,resolved)resolved.append(node) Lets observe cnt++ execute count.When node length is 0, it will execute count 0. When node length is 1, it will execute count 1.

When node length is 2, it will execute count 2.Total number ofexecution count cnt++ is 0 + 1 + 2 +… + (N-1) = N*(N-1) / 2. So Time complexitywill be O(N2).The below table helpsto user understand the growth of several common time complexity. Thus help touser judge if the algorithm is fast enough to get an acceptance.

Length of Input (N) Worst Accepted Algorithm ?10..11?10..11 O(N!),O(N6)O(N!),O(N6) ?15..18?15..18 O(2N?N2)O(2N?N2) ?18..22?18..22 O(2N?N)O(2N?N) ?100?100 O(N4)O(N4) ?400?400 O(N3)O(N3) ?2K?2K O(N2?logN)O(N2?logN) ?10K?10K O(N2)O(N2) ?1M?1M O(N?logN)O(N?logN) ?100M?100M O(N),O(logN),O(1) Table.Several Common Time Complexity