Extend and implement the Dynamic Programming Algorithm in the Tast example of the relevant slides so that it prints a parenthesized expression yielding the goal "No parenthesization possible." if no such parenthesization exists. Input format (no need to check its validity): The first line contains a positive integer (say n). The symbols: The second line contains at least n characters (if it contains more, only the first n are relevant). These characters are the symbols of the alphabet. The last symbol is the "goal". The operation: The following n lines contain at least n characters each (if a line contains more characters, only the first n are relevant). This is an nxn matrix containing the results of the operations. Expression: The last line of the input contains a sequence of characters, consisting of the valid symbols for which you have to seek for a parenthesization. A sample input: 3 abc ас baa caa caaccb and corresponding possible outputs (not exhaustive): (c((aa)((cc)b)) (c(a(a((cc)b)))) (((ca)a) ((cc)b)) ((ca)a) ((cc)b))

Extend and implement the Dynamic Programming Algorithm in the Tast example of the relevant slides so that it prints a parenthesized expression yielding the goal "No parenthesization possible." if no such parenthesization exists. Input format (no need to check its validity): The first line contains a positive integer (say n). The symbols: The second line contains at least n characters (if it contains more, only the first n are relevant). These characters are the symbols of the alphabet. The last symbol is the "goal". The operation: The following n lines contain at least n characters each (if a line contains more characters, only the first n are relevant). This is an nxn matrix containing the results of the operations. Expression: The last line of the input contains a sequence of characters, consisting of the valid symbols for which you have to seek for a parenthesization. A sample input: 3 abc ас baa caa caaccb and corresponding possible outputs (not exhaustive): (c((aa)((cc)b)) (c(a(a((cc)b)))) (((ca)a) ((cc)b)) ((ca)a) ((cc)b))

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Question

Relevant slide part is added too. Please use c++ as the language. Thanks

Extend and implement the Dynamic Programming Algorithm in the last example of the relevant slides so that it prints a parenthesized expression yielding the goal or
"No parenthesization possible." if no such parenthesization exists.
Input format (no need to check its validity):
The first line contains a positive integer (say n).
The symbols: The second line contains at leastn characters (if it contains more, only the first n are relevant). These characters are the symbols of the alphabet.
The last symbol is the "goal".
The operation: The following n lines contain at least n characters each (if a line contains more characters, only the first n are relevant). This is an nxn matrix containing the
results of the operations.
Expression: The last line of the input contains a sequence of characters, consisting of the valid symbols for which you have to seek for a parenthesization.
A sample input:
3
abc
aac
baa
саа
caaccb
and corresponding possible outputs (not exhaustive):
(C((aa)((cc)b))
(c(a(a((cc)b))))
(((ca)a)((cc)b))
(((ca)a)((cc)b))

• Let us define a binary operation ® on three symbols a, b, c according
to the following table; thus a b = b , b® a = c , and so on. Notice
that the operation defined by the table is neither associative nor
commutative.
a
b
C
a
b
a
a
Describe an efficient algorithm that examines a string of these
symbols, say bbbbac , and decides whether or not it is possible to
parenthesize the string in such a way that the value of the resulting
expression is p = a. For example, on input bbbbac your algorithm
should return yes because ((b® (b® b)) ® (b ® a)) ® c = a.

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