DSAA Homework 2

Exercises

3.4

a

by the definition

3∗2n′=64∗3∗2n

2n′=2n+6

n’=n+6

c

by the definition

8n′=64∗8n

n′=64n

3.8

c

For the all values of n >
1e9ed
1, c4nlogn+c5n≤c4nlogn+c5nlogn≤(c4+c5)nlogn. Therefore, by the definition, T(n) is in O(n) for n0=1 and c=c4+c5

For all n > 1, c4nlogn+c5n≥c4n+c5n≥(c4+c5)n. Therefore, by the definition, T(n) is in Ω(n) for n0=1 and c=c4+c5.

d

For the all values of n > 29, c62n+c7n6≤(c6+c7)2n. Therefore, by the definition, T(n) is in O(n) for n0=29 and c=c4+c5.

For the all values of n > 29, c62n+c7n6≥(c6+c7)n6. Therefore, by the definition, T(n) is in Ω(n) for n0=29 and c=c4+c5.

3.11

b

limn→∞n√logn2=limn→∞n√4=∞ , so f(n) is in Ω(g(n))

e

limn→∞nlogn+nlogn=limn→∞n=∞ , so f(n) is in Ω(g(n))

h

limn→∞2n10n2=∞ , so f(n) is in Ω(g(n))

3.12

a

Θ(1)

b

Θ(n)

c

Θ(n2)

d

Θ(n2)

e

Θ(nlogn)

f

Θ(nlogn)

g

Θ(n2logn)

Projects

3.2

#include<cstdio>
#include<cstdlib>
#include<ctime>
const int maxn = 1e7 + 100;
int a[maxn];
int random(int n){
return rand() % n;
}
int SeSearch(int val, int n){
for(int i = 0; i < n; i++)
if(a[i] == val) return i;
return -1;
}
int BiSearch(int val, int n){
int l = 0, r = n;
while(r - l > 1){
int m = (l + r) / 2;
if(a[m] > val) r = m;
else l = m;
}
return a[l] == val ? l : -1;
}
int main(){
for(int i = 0; i < maxn; i++) a[i] = i;
const int T = 100000;
int tmp;
clock_t begin, end;
for(int n = 10, k = 1; k <= 7; n *= 10, k++){
printf("i:%d ", k);
begin = clock();
for(int i = 0; i < T; i++)
tmp = SeSearch(random(n), n);
end = clock();
printf("%.2lf ", (double)(end - begin) / CLOCKS_PER_SEC);

begin = clock();
for(int i = 0; i < T; i++)
tmp = BiSearch(random(n), n);
end = clock();
printf("%.2lf\n", (double)(end - begin) / CLOCKS_PER_SEC);
}
return 0;
}

i	Sequential	Binary
1	0.02	0.01
2	0.02	0.02
3	0.17	0.02
4	1.57	0.03
5	5.50	0.03
6	5.30	0.03
7	5.44	0.03

So binary search always faster than sequential search.