asked on

analysis of algoritm

P1. Write down a computer program to compute the values of the following group of functions:

a) f(n)=lg100 (n) b) f(n)=20*n3 c)f(n)=lglglg(n)=lg(3) (n)
g(n)=n0.05 g(n)=(1.01) n g(n)=lg* (n))

Note that, the domain of f(n) and g(n) are positive natural numbers; n=1,2,3,… . You will compute and display the values f(n) and g(n); and your program will terminate when one of the functions becomes greater than the other for five consecutive n values.

i dont understand what he means at the last sentence. " when one of the functions becomes greater than the other for five consecutive n values."

can anybody help me to understand the problem ?

mosphat

I believe it means this:
one of those functions returns a value that is greater than the other five for a certain n. When that same function returns a value that is greater than the other five for n+1, n+2, n+3 and n+4, the program should terminate.

HTH,
Ruud

xinex

ASKER

can you explain more,
if its possible can you write the if statment for this condition

mosphat

Made a typo in my first post: 'other five' should be just 'other'.

You start at n = 1, Then you go n = n + 1. Sooner or later you'll have 5 n's in a row (e.g. 6,7,8,9,10 or 40,41,42,43,44) where f(n) is greater than g(n). Or 5 n's where g(n) is greater than f(n).

Good luck writing that program. Is it homework?

xinex

ASKER

yes lab. homework

i write small program,

      int main( int argc, char *argv[]){

            int i = 1;

            /* Preliminary 1 - a */
            printf("Preliminary 1\n");

            while( i <= 20 ){
                  printf("%d\n",i);
                        printf("a-) %.2f %.2f\n",pow(log10(i),100),pow(i,0.05));
                        printf("b-) %.2f %.2f\n",20*pow(i,3),pow(1.01,i));
                        printf("c-) %.2f %.2f\n",pow(log10(i),3),log(i));

                  printf("\n");
                  i+=1;
            }

            return 0;
      }

from the output can you explain which value is satisfy

Preliminary 1
a-) 0.00 1.00
b-) 20.00 1.01
c-) 0.00 0.00

a-) 0.00 1.04
b-) 160.00 1.02
c-) 0.03 0.69

a-) 0.00 1.06
b-) 540.00 1.03
c-) 0.11 1.10

a-) 0.00 1.07
b-) 1280.00 1.04
c-) 0.22 1.39

a-) 0.00 1.08
b-) 2500.00 1.05
c-) 0.34 1.61

a-) 0.00 1.09
b-) 4320.00 1.06
c-) 0.47 1.79

a-) 0.00 1.10
b-) 6860.00 1.07
c-) 0.60 1.95

a-) 0.00 1.11
b-) 10240.00 1.08
c-) 0.74 2.08

a-) 0.01 1.12
b-) 14580.00 1.09
c-) 0.87 2.20

a-) 1.00 1.12
b-) 20000.00 1.10
c-) 1.00 2.30

a-) 57.74 1.13
b-) 26620.00 1.12
c-) 1.13 2.40

a-) 2039.10 1.13
b-) 34560.00 1.13
c-) 1.26 2.48

a-) 48563.57 1.14
b-) 43940.00 1.14
c-) 1.38 2.56

a-) 838134.28 1.14
b-) 54880.00 1.15
c-) 1.51 2.64

a-) 11068842.87 1.14
b-) 67500.00 1.16
c-) 1.63 2.71

a-) 116674612.55 1.15
b-) 81920.00 1.17
c-) 1.75 2.77

a-) 1014749836.62 1.15
b-) 98260.00 1.18
c-) 1.86 2.83

a-) 7478278499.90 1.16
b-) 116640.00 1.20
c-) 1.98 2.89

a-) 47718520380.19 1.16
b-) 137180.00 1.21
c-) 2.09 2.94

a-) 268368134667.23 1.16
b-) 160000.00 1.22
c-) 2.20 3.00

(20 iteration)

ASKER CERTIFIED SOLUTION

mosphat

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