for odd palindromes we are allowed to choose the symbol in the very center of the string.
Palindrome contraints :
even : n/2
odd : (n-1)/2
the total number of binary even-length palindromes grows exponentially in the length n
counting argument
when disputes can be settled by counting the number of possibilities
DFA's can't identify palindromes --> the palindrome structure only manifests itself after we have processed half of the string (only if it has at least 2^(n/2) distinct internal states --> exponential growth of states)
DFA struggle to identify palindromes :
they must process input symbols sequentially from left to right in one go
they can only read input symbols (no memory)
square symbol
empty paper
Identify a palindrome by an iterative checking procedure. Will be n/2 iterative application of this one-sided tests.
N(k) = 2k+3 for k in {0,2,4,6, .. n } . At most N = 0.5n^2 + 5/2 n
worst case number of steps (palindrome) scales quadratically
Turing machine model components :
the finite state control (each instant == 1 state)
the tape (consists of tape squares)
the tape head (current position)
In which optic are DFA similar to a turing machine
finite resources are enough to specify
can perform computations on a input strings with varying length
difference :
They can loop back and forward -> longer run times
halting states
accept, reject states, if the machine enters either of these it stops immediately
Turing machine (TM)
A) finite
B) internal states
C) input alphabet
D) tape alphabet
E) transition
F) start
G) accept
H) reject
Turing machine
A) accepts or rejects
B) read/write
C) internal states
D) input alphabet
E) tape alphaebt
F) transition
G) start
H) accept
I) reject
Transition function TM δ(p,a) = (q,b,D). WHere p,q are states and a,b are tape alphabet symbols
δ(p,a)=(q,b,D)
p,q are states
a,b are tape alphabet
D the direction of the tape
A) a,b,D
a computing step is determined by the content of the tape and the transition function