Prove L = { 0m1n2n | m,n are natural numbers } is not regular
Answer:
If the language is infinite it may or may not be regular (means not regular). If the infinite language has the Finite Autometa then it is regular and that finite autometa should have loop (repeated pattern) to accept the infinite language.
The given language is L={0m1n2n | m,n are natural numbers}
This is infinite language.
The language accepts the strings like { 001122, 00111222, 0011222,............}
These strings have no repeated pattern means 01212,01012........
so,the given language has no loops so we can say that it is not regular
Prove that the language L = {0^n1^m0^n | m, n greaterthanorequalto 0} is not regular.
Prove that each of the following languages is not regular A) L= {a^n b^m c^k : k = 2n + 3m and n, m, k ≥ 0} B) L = {a^n : n is a power of 5}
(a) Prove that, for all natural numbers n, 2 + 2 · 2 2 + 3 · 2 3 + ... + n · 2 n = (n − 1)2n+1 + 2. (b) Prove that, for all natural numbers n, 3 + 2 · 3 2 + 3 · 3 3 + ... + n · 3 n = (2n − 1)3n+1 + 3 4 . (c) Prove that, for all natural numbers n, 1 2 + 42 + 72...
Prove by Induction
24.) Prove that for all natural numbers n 2 5, (n+1)! 2n+3 b.) Prove that for all integers n (Hint: First prove the following lemma: If n E Z, n2 6 then then proceed with your proof.
Prove that the language is regular or not. {a^nb^m | n >= m and m <= 481}
Please use induction to prove the following question for all
natural numbers n.
(d) Prove that vns įt<2vn.
Prove that the following language is regular or explain why it is nonregular: L = {amb" (m is odd and n is even) OR (m is even and n is odd))
5. Prove that the following languages are not regular: (a) L = {a"bak-k < n+1). (b) L-(angla": kメn + 1). (c) L = {anglak : n = l or l k} . (d) L = {anb : n2 1} L = {w : na (w)关nb (w)). "(f) L = {ww : w E {a, b)'). (g) L = {w"www" : w E {a,b}*}
Prove that is L is regular and h is a homomorphism, then h(L) is regular.
1. (Induction.) Consider the following program, called Ackbar(m,n). It takes in as input any two natural numbers m, n, and does the following: (i) If m-0, Ackbar(0, n) = n + 1. (ii) If n-0, Ackbar(rn,0) is equal to Ackbar(m-1, 1). iii) Otherwise, if n, m > 0, then Ackbar(m, n) can be found by calculating Ackbar(m - 1, Ackbar(m,n 1)) Here's a handful of calculations to illustrate this definition: Ackbar(1,0)-Ackbar(0,1) = 1 + 1-2 Ackbar (1, 1) Ackbar (0,...