They're not different sizes, they are simply made up of different logical constructs.
Given any finite N-bit string, a construct C can be defined such that 1 <= |{C}| <= N!
Therefore given an initial string of size N and a second (compressed) string of size M, if M! = N then a bijection is possible
They're not different sizes, they are simply made up of different logical constructs.
Given any finite N-bit string, a construct C can be defined such that 1 = |{C}| = N!
Therefore given an initial string of size N and a second (compressed) string of size M, if M! = N then a bijection is possible
They're not different sizes, they are simply made up of different logical constructs. Given any finite N-bit string, a construct C can be defined such that 1 <= |{C}| <= N! Therefore given an initial string of size N and a second (compressed) string of size M, if M! = N then a bijection is possible
They're not different sizes, they are simply made up of different logical constructs.
Given any finite N-bit string, a construct C can be defined such that 1 |{C}| N!
Therefore given an initial string of size N and a second (compressed) string of size M, if M! = N then a bijection is possible
They're not different sizes, the are simply made up of different logical constructs.
Given any finite N-bit string, a construct C can be defined such that 1
Therefore given an initial string of size N and a second (compressed) string of size M, if M! = N then a bijection is possible.