Please use C++ 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 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 aac 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))
Please use C++ 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 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 aac 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)
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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
Please use C++
Extend and implement the Dynamic Programming
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
aac
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))
Transcribed Image Text:• 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.
![Solution
• For 1 <isj< n, we let T[i,j] C {a, b, c} denote the set of symbols e for which
there is a parenthesization of x;..X; yielding e. We let e ® b denote the product
of e and b using the table. p e {a, b, c} is the symbol we are considering.
• Pseudocode
for i +1 to n do
T[i, i] + X;
for s +1 to n-1 do
for i +1 to n-s do
T[i,i + s] + Ø
for k +i to i+s-1 do
for each e e T[i,k] do
for each b e T[k + 1,i + s] do
T[i,i+s] + T[i,i+s] Ue ® b
if p e T[1,n] then
return yes
else
return no](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1970764-eba4-433f-a458-68ccd941ae3c%2F03eb603d-c8a7-4f51-ad14-724e4fbcbe15%2F7wftmg_processed.png&w=3840&q=75)
Transcribed Image Text:Solution
• For 1 <isj< n, we let T[i,j] C {a, b, c} denote the set of symbols e for which
there is a parenthesization of x;..X; yielding e. We let e ® b denote the product
of e and b using the table. p e {a, b, c} is the symbol we are considering.
• Pseudocode
for i +1 to n do
T[i, i] + X;
for s +1 to n-1 do
for i +1 to n-s do
T[i,i + s] + Ø
for k +i to i+s-1 do
for each e e T[i,k] do
for each b e T[k + 1,i + s] do
T[i,i+s] + T[i,i+s] Ue ® b
if p e T[1,n] then
return yes
else
return no
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