MHB Does Algorithm Terminate when Input is not in the Finite Set?

[agapito]

A theorem states: Any finite set of natural numbers is effectively decidable. I understand the construction of the (brute force) algorithm to check every input for inclusion in the set. However, if the input is not included in the set, the algorithm would not terminate, and we require algorithms to terminate. What am I missing here?

 

[I like Serena]

A theorem states: Any finite set of natural numbers is effectively decidable. I understand the construction of the (brute force) algorithm to check every input for inclusion in the set. However, if the input is not included in the set, the algorithm would not terminate, and we require algorithms to terminate. What am I missing here?

Hi agapito, welcome to MHB, (Wave)

When you say 'a theorem states', can you please clarify which theorem you're referring to?
And when you say 'Any finite set of natural numbers is effectively decidable', can you clarify what you mean by natural numbers being decidable? As I see it, a number does not decide anything.

 

[agapito]

Hi agapito, welcome to MHB, (Wave)

When you say 'a theorem states', can you please clarify which theorem you're referring to?
And when you say 'Any finite set of natural numbers is effectively decidable', can you clarify what you mean by natural numbers being decidable? As I see it, a number does not decide anything.

1.- Theorem is from my textbook in section named "Effectively decidable properties and sets". I have quoted it in its entirety
2.- Decidability means that an algorithm can be constructed to check whether a given number belongs to the set.

I am unclear in the case that a number is input to the algorithm that does not belong to the set, it seems to me like the algorithm would not terminate as it's supposed to.

I certainly appreciate your help with this.

 

[HOI]

What ever number is given, you only need to check it against the "n" numbers in the finite set. If the number is in the set that might be determined in less than n comparisons. If the number is not in the set that will be determined in n comparisons.

 

[agapito]

What ever number is given, you only need to check it against the "n" numbers in the finite set. If the number is in the set that might be determined in less than n comparisons. If the number is not in the set that will be determined in n comparisons.

Many thanks for the clarification, very helpful to me. am