Structural Transformation of Transition Economies Post USSR Dissolution
This paper delves into the growth and structural changes in former Soviet Union countries post the dissolution of the USSR in 1991. It introduces the Economic Transformation Database of Transition Economies, compares structural changes in FSU countries with others, and highlights challenges faced by independent FSU nations. The Economic Transformation Database tracks employment and value-added output in these countries, providing valuable insights.
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Regular Languages Closure Properties, Take 2 CS 154, Omer Reingold
Reverse Theorem for Regular Languages Theorem: The reverse of a regular language is also a regular language If a language can be recognized by a DFA that reads strings from right to left, thenthere is an normal DFA that accepts the same language Proof? Given a DFA for a language L, reverse its arrows and flip its start and accept states, getting an NFA. Convert that NFA back to a DFA!
Using NFAs in place of DFAs can make proofs about regular languages much easier! Remember this on homework/exams!
Regular Languages are closed under concatenation Concatenation: A B = { vw | v A and w B } Given DFAs M1 and M2, connect the accept states of M1 to the start states of M2 1 1 1 0 0 0 0,1 0,1 0,1 1 1 1 0 0 0 0 0 0 1 1 1 L(N) = L(M1) L(M2)
Regular Languages are closed under star A* = { s1 sk| k 0 and each si A } Let M be a DFA, and let L = L(M) We can construct an NFA N that recognizes L* 1 0 0,1 1 0 0 1
Formally, the construction is: Input: DFA M = (Q, , , q1, F) Output: NFA N = (Q , , , q0, F ) Q = Q {q0} F = F {q0} { (q,a)} {q1} {q1} if q Q and a if q F and a = if q = q0 and a = if q = q0and a else (q,a) =
Regular Languages are Closed Under Star How would we prove that this NFA construction works? Want to show: L(N) = L* 1. L(N) L* 2. L(N) L*
1. L(N) L* Assume w = w1 wk is in L* where w1, ,wk L We show N accepts w by induction on k Base Cases: k = 0 k = 1 Inductive Step: Assume N accepts all strings v = v1 vk L*, vi L Let u = u1 ukuk+1 L* , uj L (w = ) (w L) Since N accepts u1 uk (by induction) and M accepts uk+1, N also accepts u (by construction)
2. L(N) L* Assume w is accepted by N; we want to show w L* If w = , then w L* I.H. N accepts u and takes at most k -transitions u L* u L(N), so By I.H. u L* u Let w be accepted by N with k+1 -transitions. Write w as w=uv, where v is the substring read after the last - transition v v L w = uv L* accept
Closure Properties for Regular Languages Union: A B = { w | w A or w B } Intersection: A B = { w | w A and w B } Complement: A = { w * | w A } Reverse: AR = { w1 wk | wk w1 A, wi } Concatenation: A B = { vw | v A and w B } Star: A* = { s1 sk | k 0 and each si A } Theorem: if A and B are regular then so are: A B, A B, A, AR , A B, and A*
