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natima [27]
3 years ago
12

Write a function isPrime of type int -> bool that returns true if and only if its integer parameter is a prime number. Your f

unction need not behave well if the parameter is negative.

Computers and Technology
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

import math

def isPrime(num):

 

   if num % 2 == 0 and num > 2:

       return False

   for i in range(3, int(math.sqrt(num)) + 1, 2):

       if num % i == 0:

           return False

   return True

Explanation:

The solution is provided in the python programming language, firstly the math class is imported so we can use the square root method. The first if statement checks if the number is even and greater than 2 and returns False since all even numbers except two are not prime numbers.

Then using a for loop on a range (3, int(math.sqrt(num)) + 1, 2), the checks if the number evenly divides through i and returns False otherwise it returns True

see code and output attached

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You are given a list of n positive integers a1, a2, . . . an and a positive integer t. Use dynamic programming to design an algo
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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

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