Answer:
see explaination
Explanation:
MATLAB script:
% MATLAB script that calculates reciprocal Fibonacci constant Ψ
% Initial Values
a0 = 1;
a1 = 1;
% Looping variable
i = 2;
% Reading n value from user
n = input(' Enter n value: ');
% Initializing sum
sum = (1/a0) + (1/a1);
% Loop till i reaches n value
while i <= n
% Finding term in Fibnocii series
a2 = a0 + a1;
% Accumulating Sum
sum = sum + (1/a2);
% Updating previous terms
a0 = a1;
a1 = a2;
% Incrementing loop variable
i = i + 1;
endwhile
% Printing result
printf("\n Reciprocol Fibnocii Constant: %f \n", sum);
See attachment for sample output
Answer:
See explaination for the program code
Explanation:
The code below
Pseudo-code:
//each item ai is used at most once
isSubsetSum(A[],n,t)//takes array of items of size n, and sum t
{
boolean subset[n+1][t+1];//creating a boolean mtraix
for i=1 to n+1
subset[i][1] = true; //initially setting all first column values as true
for i = 2 to t+1
subset[1][i] = false; //initialy setting all first row values as false
for i=2 to n
{
for j=2 to t
{
if(j<A[i-1])
subset[i][j] = subset[i-1][j];
if (j >= A[i-1])
subset[i][j] = subset[i-1][j] ||
subset[i - 1][j-set[i-1]];
}
}
//returns true if there is a subset with given sum t
//other wise returns false
return subset[n][t];
}
Recurrence relation:
T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times
T(1)=1
solving recurrence:
T(n)=T(n-1)+t
T(n)=T(n-2)+t+t
T(n)=T(n-2)+2t
T(n)=T(n-3)+3t
,,
,
T(n)=T(n-n-1)+(n-1)t
T(n)=T(1)+(n-1)t
T(n)=1+(n-1)t = O(nt)
//so complexity is :O(nt)//where n is number of element, t is given sum
Answer:
ok I know it's not for me but for who?
Explanation:
Just asking. And yes everyone deserves a second chance. Even liars?
please don't take my question wrongly have an above average day!
