Just the expression a^i b^j c^k would be a regular expression, however the requirement
(i=1)->(j=k) generates a subset of the language denoted as [a b^j c^j ]. Here if a=x, b^1=y, c^1=z
then the string xyz is accepted by the language. Can y be pumped?
Data Mining Hw 6
Datamining Grocery Store Purchases:
This kind of datamining problem is probably the most cited out of the entire discipline of the field. The
procedure described was intuitive and elegant. The author simply chose to
Rev( , ).
Rev([H|T], Res):Rev(T, TREs),
Append(TRes, [H], Res).
TailRecursion has answer at boundary call and beams the result back up the levels of recursion.
Answer is built in a parameter.
M is N -1.
If A1 > B1 -> A > B
If A1 < B1 -> B > A
If A1 = B1 -> same steps with next digit (A=B if this is last digit)
A1 ^ B1 = A1=B1 -> next digit
If(A1 v B1)=true ^ (A1 v `B1)=false then B > A
If(A1 v B1)=true ^ (`A1 v B1)=false then B < A
If (A1 v B1) = fals
Jozef Pliszka 810 261 632
Intro to Theory of Computing HW 1
a. Let the set S= cfw_1,2,3 and the relation R contain cfw_(1,1) (1,2) (2,2) (2,3) (3,3) (3,2)
Then the relationship is reflexive because every element is paired with itself. Symmetric