2022-04-21
Using the division algorithm, we can divided a
composite number represented as a binary string by
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.
(Edited: 2022-04-21) <pre>Using the division algorithm, we can divided a
composite number represented as a binary string by
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.</pre>
((resource:WhatsApp Image 2022-04-21 at 9.54.40 PM.jpeg|Resource Description for WhatsApp Image 2022-04-21 at 9.54.40 PM.jpeg))
If the verification is solved in polynomial time, it can be proved that COMPOSITE is in NP.
We rewrite COMPOSITE = {<n> | <n, x> ∈ L}, L = {<n, x> | n % x == 0}.
The algorithm used in L takes O(n) time and thus proves that L is in P and COMPOSITE in NP
If the verification is solved in polynomial time, it can be proved that COMPOSITE is in NP.
We rewrite COMPOSITE = {<n> | <n, x> ∈ L}, L = {<n, x> | n % x == 0}.
The algorithm used in L takes O(n) time and thus proves that L is in P and COMPOSITE in NP
2022-04-22
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.
COMPOSITE = {<n> | <n,x> ∈ L}
L = {<n, x> | n % x == 0}
The COMPOSITE is in NP is the verification is solved in polynomial time.
L runs in order of O(n) so L is in P.
Thus, COMPOSITE is in NP.
(Edited: 2022-04-22) COMPOSITE = {<n> | <n,x> ∈ L}
L = {<n, x> | n % x == 0}
The COMPOSITE is in NP is the verification is solved in polynomial time.
L runs in order of O(n) so L is in P.
Thus, COMPOSITE is in NP.
To show COMPOSITE is in NP, the algorithm must run in
polynomial time.
COMPOSITE = {<n> | <n,x> ∈ L}
L = {<n, x> | n% x ==0}
L runs in O(n) time, thereby proving that L is in P and
COMPOSITE in NP.
To show COMPOSITE is in NP, the algorithm must run in
polynomial time.
COMPOSITE = {<n> | <n,x> ∈ L}
L = {<n, x> | n% x ==0}
L runs in O(n) time, thereby proving that L is in P and
COMPOSITE in NP.
2022-04-24
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.
L= {<n, x> | n% x ==0} for composite numbers to belong in P. An algorithm can check if n%x==0 for n and x passed as inputs. If the algorithm takes polynomial time to check, it is said to be in P. Therefore, COMPOSITE is in NP.
L= {<n, x> | n% x ==0} for composite numbers to belong in P. An algorithm can check if n%x==0 for n and x passed as inputs. If the algorithm takes polynomial time to check, it is said to be in P. Therefore, COMPOSITE is in NP.
As shown in the slides, to prove that COMPOSITE is in NP, we need to show that COMPOSITE has a verification algorithm that runs in polynomial time. The language can be written as follows: L = {<n, x> | n % x == 0}. Since we can do division in polynomial time, we can determine L to be in P, which also means COMPOSITE belongs to NP.
As shown in the slides, to prove that COMPOSITE is in NP, we need to show that COMPOSITE has a verification algorithm that runs in polynomial time. The language can be written as follows: L = {<n, x> | n % x == 0}. Since we can do division in polynomial time, we can determine L to be in P, which also means COMPOSITE belongs to NP.
Let COMPOSITE be the set of binary strings x which when viewed as a natural number are evenly divisible by some other natural number >1
We say a language L belongs to NP if there exists a two input polynomial-time algorithm A and a constant c such that
L={x∈{0,1}⋆:∃y,∣∣y∣∣=O(∣∣x∣∣c) and A(x,y)=1}
A (n,x) can be given as
L = { <n,x> | n%x ==0}
So for composite we get that n is a composite number and A( n,x) = 1 ,
we can also argue that using a the grade school division algorithm this can be obtained within polynomial time. So Composite is in NP.
Let COMPOSITE be the set of binary strings x which when viewed as a natural number are evenly divisible by some other natural number >1
We say a language L belongs to NP if there exists a two input polynomial-time algorithm A and a constant c such that
L={x∈{0,1}⋆:∃y,∣∣y∣∣=O(∣∣x∣∣c) and A(x,y)=1}
A (n,x) can be given as
L = { <n,x> | n%x ==0}
So for composite we get that n is a composite number and A( n,x) = 1 ,
we can also argue that using a the grade school division algorithm this can be obtained within polynomial time. So Composite is in NP.
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