We want to show that this automaton accepts exactly those strings with an odd nu
ID: 3784178 • Letter: W
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We want to show that this automaton accepts exactly those strings with an odd number of 1's, or more formally: (A,w) = B if and only if w has an odd number of 1's.
5. Here is the transition function of a simple, deterministic automaton with start state A and accepting state B 0 1 AAB We want to show that this automaton accepts exactly those strings with an odd number of l's, or more formally: 6(A, w) B if d only if w has an odd number of l's w Here, S is the extended transition function of the automato that is, o(A, is the state that the automaton is in after processing input s w. The proof of the stalemenu above is an induction on the length of w, Below, we give the proof with reasons missing. You must give a reason for each step, and then demonstrate your understanding of the proof by classifying your reasons into the following three categories: A) Use of the inductive hypothesis. Reasoning about properties of deterministic finite automata, e.g., that if string Reasoning about proporties of binary strings (strings of 0's and l's e.g., that every string is longer than any of its proper substrings. Basis w 0) w because (ASE) A because E has an even number of 1's because duction (w n 0) There are two cases (a) when w X1 and (b) when w x00 because Case (a In case (a), w has an odd number of l's if and only ifx has an even number of l's because In case (a), OCA,x) A if and only if w has an odd number of 1's because In case (a), 0(A,w) Bif and only if w has an odd number of l's because Case (b In case (b), w has an odd number of 1's if and only if x as an odd number of l's because In case (b)T 50A,x) Bif and only if w has an odd number of 1's because (10) In case (b), DIA, Bif and only if w has an odd number of l's because a) (7) for reason C b) (1) for reason A c) (8) for reason A d) (8) for reason CExplanation / Answer
BASIS: |w| = 0
(1)
w = because we have assumed the length of the string in the basis as 0 i.e. |w| = 0
(2)
(A,) = A because ->
for the same reason explained above. |w| = 0. Any transaction on state A with empty string will result in
state A itself.
(3)
has an even number of 0's because
This is again due to |w| = 0.
it has even number of 0's, namely zero 0's.
Induction (|w| = n > 0)
(4)
There are two cases: (a) when w = x1 and (b) when w = x0 because ->
The given automata ends on two conditions and the FINAL state is B.
2 conditions are -> State A with 1 as input will reach B and State B with 0 as input will stay at B.
Here w = x1 means -> x can be anything followed by 1 which is the last element.
Similarly, w = x0 -> x can be anything followed by 0 which is the last element.
Since the automata only accepts odd number of 1's, we have 2 cases based on w=x1 and w=x0.
(5)
In case (a), w has an odd number of 1's if and only if x has an even number of 1's because
case (a) is w = x1.
which means w is any string with 1 as the last element.
Since we already have 1 in the w, x has to contain even number of 1's.
(6)
In case (a), (A,x) = A if and only if w has an odd number of 1's because ________
Since w = x1 and the end state is B, final tranition should end on B.
If w has odd number of 1's, it means x has even number of 1's because w = x1.
Since x has even number of 1's, (A,x) should end with the same state it started and hence its A.
(7)
In case (a), (A,w) = B if and only if w has an odd number of 1's because ________
Since w = x1 and the end state is B, final tranition should end on B and it is as per the given automata and acceptance criteria.
Similarly you can answer for the rest of questions.
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