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erma4kov [3.2K]
3 years ago
5

C++ a. Write a program that uses the function isPalindrome given in example 6-6 (Palindrome). Test your program on the followinn

g strings: "madam", "abba", "22", "67876", "444244", and "trymeuemyrt" b. Modify the function isPalindrome of example 6-6 so that when determining weather a string is a palindrome, cases are ignored, that is, uppercase and lowercase letters are considered the same. Example 6-6: bool isPalindrome (string str) { int length = str.Length(); for (int i = 0; i < length / 2; i++) if (str[i] != str[length - 1 - i]; return false; return true; }
Computers and Technology
1 answer:
Whitepunk [10]3 years ago
5 0

Answer:

#include <iostream>

#include <array>

using namespace std;

bool isPalindrome(string str)  

{  

int length = str.length();  

for (int i = 0; i < length / 2; i++)  

 if (toupper(str[i]) != toupper(str[length - 1 - i]))

  return false;  

return true;  

}

int main()

{

array<string, 6> tests = { "madam", "abba", "22", "67876", "444244", "trymEuemYRT" };

for (auto test : tests) {

 cout << test << " is " << (isPalindrome(test) ? "" : "NOT ") << "a palindrome.\n";

}

}

Explanation:

The toupper() addition forces characters to uppercase, thereby making the comparison case insensitive.

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Determine whether or not the following pairs of predicates are unifiable. If they are, give the most-general unifier and show th
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Answer:

a) P(B,A,B), P(x,y,z)

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }.

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q.

c. Older(Father(y),y), Older(Father(x),John)

Thus , mgu ={ y/x , x/John }.

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B.

e) P(f(x), x, g(x)), P(f(y), A, z)    

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

Explanation:  

Unification: Any substitution that makes two expressions equal is called a unifier.  

a) P(B,A,B), P(x,y,z)  

Use { x/B}  

=> P(B,A,B) , P(B,y,z)  

Now use {y/A}  

=> P(B,A,B) , P(B,A,z)  

Now, use {z/B}  

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }  

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q  

c. Older(Father(y),y), Older(Father(x),John)  

Use {y/x}  

=> Older(Father(x),x), Older(Father(x),John)  

Now use { x/John }  

=> Older(Father(John), John), Older(Father(John), John)  

Thus , mgu ={ y/x , x/John }  

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))  

Use { y/x }  

=> Q(G(x,z),G(z,x)), Q(G(x,x),G(A,B))

Use {z/x}  

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B  

e) P(f(x), x, g(x)), P(f(y), A, z)  

Use {x/y}  

=> P(f(y), y, g(y)), P(f(y), A, z)  

Now use {z/g(y)}  

P(f(y), y, g(y)), P(f(y), A, g(y))  

Now use {y/A}  

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

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Answer:

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Explanation:

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All mentioned IDEs such as Eclipse, Visual Studio, and Xcode support multiple high-level programming languages except Linux.

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While Linux is not an IDEs, it is an open-source operating system based on Linux kernel.

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