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Alex_Xolod [135]
3 years ago
12

You are given a string of n characters s[1 : : : n], which you believe to be a corrupted text document in which all punctuation

has vanished (so that it looks something like itwasthebestoftimes...). You wish to reconstruct the document using a dictionary, which is available in the form of a Boolean function dict(): for any string w, dict(w) = true if w is a valid word false otherwise . (a) Give a dynamic programming algorithm that determines whether the string s[] can be reconstituted as a sequence of valid words. The running time should be at most O(n2), assuming calls to dict take unit time. (b) In the event that the string is valid, make your algorithm output the corresponding sequence of words.
Computers and Technology
1 answer:
alexdok [17]3 years ago
8 0

Answer: provided in the explanation section

Explanation:

Given that:

Assume D(k) =║ true it is [1 : : : k] is valid sequence  words or false otherwise

  • To determine D(n)

now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.

So, from here we have that D(k) is given by the following recorance relation:

D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise

Algorithm:

Valid sentence (s,k)

D [1 : : : k]             ∦ array of boolean variable.

for a ← 1 to m

do ;

d(0) ← false

for b ← 0 to a - j

for b ← 0 to a - j

do;

if D[b] ∧ DICT s([b + 1 : : : a])

d (a) ← True

(b). Algorithm Output

      if D[k] = = True

stack = temp stack           ∦stack is used to print the strings in order

c = k

while C > 0

stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c

P = W (p) - 1

output stack

= 0 =

cheers i hope this helps !!!

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If the object instance is created in a user program, then the object instance can access both the public and private members of
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Answer:

False

Explanation:

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<u>Following are the example is given below in C++ Language </u>

#include<iostream>   // header file

using namespace std;  

class Rectangle

{    

   private:  

       double r; // private member  

   public:      

       double  area()  

       {     return 3.14*r*r;  

       }        

};  

int main()  

{    

  Rectangle r1;// creating the object  

   r1.r = 3.5;  

 double t= r1.area(); // calling

cout<<" Area is:"<<t;  

   return 0;  

}  

Output:

compile time error is generated

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#include<iostream>   // header file

using namespace std;  

class Rectangle

{    

   private:  

       double r; // private member  

   public:      

       double  area()  

       {    

r1=r;

double t2=3.14*r2*r2;

return(t2); // return the value  

       }        

};  

int main()  

{    

  Rectangle r1;// creating the object  

   r1.r = 1.5;  

 double t= r1.area(); // calling

cout<<" Area is:"<<t;  

   return 0;  

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Therefore the given statement is False

5 0
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Answer:

<em>The programming language is not stated;</em>

<em>I'll answer using C++</em>

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cin>>side1>>side2>>side3;

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  cout<<"Right Angled";

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return 0;

}

Explanation:

The following line declares the three variables

int side1, side2, side3;

The next line prompts user for input of the three sides

cout<<"Enter the three sides of the triangle: "<<endl;

The next line gets user input

cin>>side1>>side2>>side3;

The following if condition checks if any of user input is negative or 0

<em> if(side1<=0 || side2 <= 0 || side3 <= 0) { </em>

<em>  cout<<"Invalid Inputs"; </em>

<em> } </em>

If otherwise

else {

The following if condition assumes that side1 is the largest and test using Pythagoras Theorem

<em>if(abs(pow(side1,2) - (pow(side2,2) + pow(side3, 2)))<0.001) { </em>

<em>   cout<<"Right Angled"; </em>

<em>  } </em>

The following if condition assumes that side2 is the largest and test using Pythagoras Theorem

<em>  else if(abs(pow(side2,2) - (pow(side1,2) + pow(side3, 2)))<0.001) { </em>

<em>   cout<<"Right Angled"; </em>

<em>  } </em>

The following if condition assumes that side3 is the largest and test using Pythagoras Theorem

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<em>   cout<<"Right Angled"; </em>

<em>  } </em>

If none of the above conditions is true, then the triangle is not a right angles triangle

<em>  else { </em>

<em>   cout<<"Not Right Angled"; </em>

<em>  } </em>

}

return 0;

Download cpp
4 0
3 years ago
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