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Let B be the language of properly nested parentheses and brackets. For example,is in B but []is not. Show that B is in L.

Short Answer

Expert verified

Every pair of matching parentheses or the brackets must be of the same kind.

Step by step solution

01

Introduce nested parenthesis and L languages

"Nested" parentheses are parentheses that are contained within another parenthesis. The simplification procedure is the same as it was in the previous page simpler examples, but we must be a little more cautious as we work our way through the grouping symbols.

02

B is in L. 

First, repeat the preceding Exercise technique, replacing [and] with (and). After that, locate matching pairs by repeating the technique for each left parenthesis or brackets in input (by starting the count at that point).

Reject those that are of a different sort and accept the rest.

03

Final answer

Therefore, when completing the previously mentioned technique on each leftparenthesis/bracket, we employ an extra counter to keep track of our progress by counting how far we moved."

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Most popular questions from this chapter

An undirected graph is bipartite if its nodes may be divided into two sets so that all edges go from a node in one set to a node in the other set. Show that a graph is bipartite if and only if it doesnโ€™t contain a cycle that has an odd number of nodes. Let BIPARTITE={(G)|Gisabipartitegraph}. Show that BIPARTITEโˆˆ NL.

For any positive integer x, let xR be the integer whose binary representation isthe reverse of the binary representation of x. (Assume no leading in the binary representation of x.) Define the function R+:Nโ†’NwhereR+(x)=x+xR.
a. Let A2=โŸจx,yโŸฉ|R+(x)=y.

ShowA2โˆˆL.
b. Let A2=โŸจx,yโŸฉ|R+(R+(x))=y. ShowA3โˆˆL.

Show that ANFA is NL-complete.

The cat-and-mouse game is played by two players, "Cat" and "Mouse," on an arbitrary undirected graph. At a given point, each player occupies a node of the graph. The players take turns moving to a node adjacent to the one that they currently occupy. A special node of the graph is called "Hole." Cat wins if the two players ever occupy the same node. Mouse wins if it reaches the Hole before the preceding happens. The game is a draw if a situation repeats (i.e., the two players simultaneously occupy positions that they simultaneously occupied previously, and it is the same player's turn to move).
HAPPY-CAT={<G,c,m,h>G,c,m,hAre respectively a graph and positions of the Cat, Mouse, and Hole, such that Cat has a winning strategy if Cat moves first}.
Show thatHAPPY-CAT is in P. (Hint: The solution is not complicated and doesn't depend on subtle details in the way the game is defined. Consider the entire game tree. It is exponentially big, but you can search it in polynomial time.)

For each n, exhibit two regular expressions,Rโ€‰โ€‰andโ€‰โ€‰S , of length poly(n), whereL(R)โ‰ L(S), but where the first string on which they differ is exponentially long. In other words,L(R)โ€‰andโ€‰โ€‰L(S) must be different yet agree on all strings of length up to2nd for some constant ฮต>0.

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