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cs lecture nov 11

# cs lecture nov 11 - cs lecture nov 11 nov 16 practice exam...

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cs lecture nov 11 nov 16 – practice exam nov 17 – no class nov 18 – practice exam marking, no tutorial if(s==(char *)0) equal to s==0 equal to ! s. if(t[0]==’\0’) equals to !*t equals to *t==0 time complexity, big-o notation Let f(x) be a function over real numbers. O(f(x)) is a set of functions g(x)€O(f(x)) if and only if there exists M>0 and X_0 st |g(x)<=M|f(x)| for all x>x_0 ex 3x^2 €O(x^2) -> f(x) ^ g(x) proof if x>1, then |x^2 + 2| = 3X^2+2 <= 5x^2 = 5|x^2| Therefore if x_0=1 and M=5, |3x^2+2| <= M |x^2| fpr a ;; x > x_0 6sinx EO(1) proof if x>0 then |6sinx |<=6 = 6 x 1 Pick m=6, prick x_0 = 0 eg g(x) – any polynomial g(x) =a_n x^n + a_n-1 X^n-1 + … a_1 x + a_0 g(x) E O(x^n) proof if x>1 then |g(x)|<=|(a_n+…a_0)x^n| let x_0 = 1 and M = sum a_n then |g(x)| <= M |x^n| as a result if n<=m then x^n EO (x^m) (N>0 and m>0) eg x^4 EO (x^76) log_a x EO (log_b x) Proof: if x > 1 then |log_a (x)| = log_a (x) = 1/log_b (a) |log_b (x)| M=1/log_b (a) ie we can ignore the base of a logarithm we purely write the base of a logarithm in big O notation if g_0 (x) EO (f_0 (x)) and g(x) EO (f_1 (x) ) then i) g_0 (x) + g_1(x) EO (f_0(x) + f_1 (x)) ii) g_0(x) + g_1 (x) EO (f_0 (x)f_1 (x))

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cs lecture nov 11 - cs lecture nov 11 nov 16 practice exam...

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