2022-04-25
Each composite number (n) can be considered as the product of two numbers. If we consider every pair of natural numbers (x,y) where 1 < x <= squareroot(n) and 1 < y <=n.
Checking if the product of the pair (x*y) == n is a problem that takes O(|x|+|y|)^k time (polynomial time)
This makes the given problem a composite NP problem
(Edited: 2022-04-25) Each composite number (n) can be considered as the product of two numbers. If we consider every pair of natural numbers (x,y) where 1 < x <= squareroot(n) and 1 < y <=n.
Checking if the product of the pair (x*y) == n is a problem that takes O(|x|+|y|)^k time (polynomial time)
This makes the given problem a composite NP problem
The language L= {<n, x> | n% x ==0} for composite numbers belong to P since division is performed in polynomial time. The algorithm takes n and x values as inputs and checks for divisibility condition n%x==0. Since the algorithm takes polynomial time (where n is the length of number in binary), it is said to be in P and hence is in NP.
The language L= {<n, x> | n% x ==0} for composite numbers belong to P since division is performed in polynomial time. The algorithm takes n and x values as inputs and checks for divisibility condition n%x==0. Since the algorithm takes polynomial time (where n is the length of number in binary), it is said to be in P and hence is in NP.
We define the language L as {<n, x> | n% x ==0} for composite numbers that belong in P. There is an algorithm that takes polynomial time where there exists n and x such that n%x==0. Since division is in polynomial time, we define L to be in P. Therefore, COMPOSITE is in NP.
We define the language L as {<n, x> | n% x ==0} for composite numbers that belong in P. There is an algorithm that takes polynomial time where there exists n and x such that n%x==0. Since division is in polynomial time, we define L to be in P. Therefore, COMPOSITE is in NP.
The product of two other numbers is a composite number (n)
Let's consider 1<x<=n and 1<y<=root(n), if x, y are a pair of natural numbers.
If the product of each pair (x*y) == n is a problem that takes O(|x|+|y|)^k time (polynomial time) which makes the given problem a composite NP problem.
The product of two other numbers is a composite number (n)
Let's consider 1<x<=n and 1<y<=root(n), if x, y are a pair of natural numbers.
If the product of each pair (x*y) == n is a problem that takes O(|x|+|y|)^k time (polynomial time) which makes the given problem a composite NP problem.
It can be proved to be in NP if the verification could be solved in polynomial time. We can check if n is composite by passing n and x as inputs to an algorithm A which checks if n%x==0 and division has a time complexity of n(length of binary n) which is polynomial making to be in p, resulting in COMPOSITE belonging to NP.
It can be proved to be in NP if the verification could be solved in polynomial time. We can check if n is composite by passing n and x as inputs to an algorithm A which checks if n%x==0 and division has a time complexity of n(length of binary n) which is polynomial making to be in p, resulting in COMPOSITE belonging to NP.
L= {<n, x> | n% x ==0} for composite numbers to belong in P. The values n and x can be passed to the algorithm to check if n%x==0. If the algorithm takes polynomial time to do the division(where n is the length of number in binary), then it is said to be in P and hence COMPOSITE is in NP.
L= {<n, x> | n% x ==0} for composite numbers to belong in P. The values n and x can be passed to the algorithm to check if n%x==0. If the algorithm takes polynomial time to do the division(where n is the length of number in binary), then it is said to be in P and hence COMPOSITE is in NP.
Using the division algorithm, we can divide a
composite number represented by a binary string with another number within polynomial time.
L = {<n, x> | n % x == 0} for composite numbers to be in P
As division can be done in polynomial time, L belongs to P and NP.
Using the division algorithm, we can divide a
composite number represented by a binary string with another number within polynomial time.
L = {<n, x> | n % x == 0} for composite numbers to be in P
As division can be done in polynomial time, L belongs to P and NP.
2022-04-26
Each composite number (n) is the product of two other numbers => consider every pair of natural numbers (x,y) where 1<x<=root(n) and 1<y<=n Checking if the product of each pair (x*y) == n is a problem that takes O(|x|+|y|)^k time (polynomial time) This makes the given problem a composite NP problem
Each composite number (n) is the product of two other numbers => consider every pair of natural numbers (x,y) where 1<x<=root(n) and 1<y<=n Checking if the product of each pair (x*y) == n is a problem that takes O(|x|+|y|)^k time (polynomial time) This makes the given problem a composite NP problem
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