Apple's redesigned OS X reportedly debuting next month

For the past few years, Apple's desktop software has played second fiddle to iOS at its annual developer conference WWDC. According to 9to5Mac's Mark Gurman, who is seen as a reputable source for Apple news, that won't be the case next month.

It's long been speculated that OS X 10.10 will see a big refresh this year. Gurman states that this "end-to-end redesign" will be Apple's most comprehensive since launching OS X over a decade ago. Although the new look will reportedly take design cues from last year's radical rethink of iOS, these aesthetic changes will not change the fundamentals of OS X.

OS X WILL LOOK MORE LIKE IOS, BUT REMAIN DISTINCT

Codenamed "Syrah," OS X 10.10 will reportedly have sharper corners, more white space, and iOS 7-esque toggles. Apple will also apparently continue its trend of bringing certain features from iOS into its desktop OS. Apple has repeatedly reassured its users that its two operating systems will remain distinct, despite the past few years seeing the addition of an iOS-like Launchpad, Mac App Store, and applications that began on mobile like Maps.

As for what's in store for iOS 8, Gurman builds on his previous rumors, saying that new apps like Healthbook, TextEdit, and Preview are on the way, along with changes to Siri and iCloud. He notes that "some of the notable features in the works" could potentially be pushed back to a later release, as a number of employees that usually work on iOS have been moved over to work on getting OS X ready for its WWDC unveil. The rumor is similar to those spread last year that Apple's OS X team was being tapped to work on iOS 7.

A NEW APPLE TV IS LESS LIKELY

Despite earlier earmarking a new Apple TV for the first half of this year, 9to5Mac is now stating that "the product's debut seems to have been pushed back." According to the site's sources, both a new set-top box and a "thorough revamp" of the Apple TV software are in the works, but the two projects are not explicitly tied together. That means we could see new software coming to older Apple TVs. Gurman says the new set-top box, which has been the source of major speculation this year, is a major priority for Apple. WWDC officially kicks off a month from now on June 2nd, and we'll be covering all the news from its opening keynote live.

Write a C++ program to find the optimal binary search tree for the given set

Optima binary search tree:-

We can construct [1/(n+1)](2n_cn) binary search trees for a given set which will be a time consuming process.in order to reduce that we will solve this problem by using dynamic programming.

Formulas for obst

    c[i][i]=0.0;

    r[i][i]=0;

                wt[i][i]=q[i];

           

                wt[i][j]=p[j]+q[j]+wt[i][j-1];

                c[i][j]=c[i][k-1]+c[k][j]+wt[i][j];

                r[i][j]=k;

            To construct a tree  consider the last row first column of  a tree

            T[i][j]=t[i][k-1],t[k][j];

Algorithm

void con_obst(void)

{

      int i,j,k,min;

      for(i=0 to n)

      { //Initialisation

                c[i][i]=0.0;

                r[i][i]=0;

                w[i][i]=q[i];

                j=i+1;

                k=j;

                w[i][j]=p[j]+q[j]+w[i][j-1];

                c[i][j]=c[i][k-1]+c[k][j]+w[i][j];

                r[i][j]=k;

      }

      //for j-i=2,3,4….,n

      for(j=2 to n)

      {

                for(i=0;i<=n-j;i++)

                {

                          j=j+i;

                          w[i][j]=p[j]+q[j]+w[i][j-1];

                          c[i][j]=9999;

                          for(k=i+1 to j)

                          {

                                    if(c[i][j]>(c[i][k-1]+c[k][j]))

                                    {

                                          c[i][j]=c[i][k-1]+c[k][j];

                                          r[i][j]=k;

                                    }

                          }

                          c[i][j]+=w[i][j];

                          j=j-i;

                }

                cout<<endl;

      }

Analysis:- The time complexity is O(n³)

Program

#include<iostream>

using namespace std;

//#include<conio.h>

//#include<stdio.h>

class obst

{

public:

void obstm(int);

void print(int,int);

public:

float p[20],q[20],wt[20][20],c[20][20];

int r[20][20],i,j,t[20][20],k;

};

void obst::obstm(int n)

{

//      int i;

      cout<<"\n****** PROGRAM FOR OBST ******\n";

//      cout<<"\nEnter the no. of nodes : ";

  //    cin>>n;

      cout<<"\nEnter the p values ";

      for(i=1;i<=n;i++)

      {

                cout<<"\np["<<i<<"]  :";

                cin>>p[i];

      }

      cout<<"\nEnter the q values ";

      for(i=0;i<=n;i++)

      {

                cout<<"\nq["<<i<<"]  :";

                cin>>q[i];

      }

//      int i,j;

      for(i=0;i<=n;i++)

      { //Initialisation

                c[i][i]=0.0;

                r[i][i]=0;

                wt[i][i]=q[i];

                j=i+1;

                k=j;             

                wt[i][j]=p[j]+q[j]+wt[i][j-1];

                c[i][j]=c[i][k-1]+c[k][j]+wt[i][j];

                r[i][j]=k;

      }

      //for j-i=2,3,4….,n

      for(j=2;j<=n;j++)

      {

                for(i=0;i<=n-j;i++)

                {

                          j=j+i;

                          wt[i][j]=p[j]+q[j]+wt[i][j-1];

                          c[i][j]=9999;

                          for(k=i+1;k<=j;k++)

                          {

                                    if(c[i][j]>(c[i][k-1]+c[k][j]))

                                    {

                                          c[i][j]=c[i][k-1]+c[k][j];

                                          r[i][j]=k;

                                    }

                          }

                          c[i][j]+=wt[i][j];

                          j=j-i;

                }

                cout<<endl;

      }

      cout<<"\n\nOptimal BST is :: ";

      cout<<"w values are\n";

      for(i=0;i<=n;i++)

      {

      k=0;

      for(j=i;j<=n;j++,k++)

      cout<<wt[k][j]<<"\t";

      cout<<"\n";

      }

      cout<<"c values are \n";

      for(i=0;i<=n;i++)

      {

      k=0;

      for(j=i;j<=n;j++,k++)

      cout<<c[k][j]<<"\t";

      cout<<"\n";

      }

      cout<<"r values are \n";

      for(i=0;i<=n;i++)

      {

      k=0;

      for(j=i;j<=n;j++,k++)

            {

            t[k][j]=r[k][j];

      cout<<r[k][j]<<"\t";

}

      cout<<"\n";

      }

}

void obst::print(int i,int j)

{

            int k;

      if(i>=0&&i!=j)

      {

            k=t[i][j];

      cout<<"\n Left child of   "<<k<<" is   "<<t[i][k-1];

      cout<<"\n Right child of  "<<k<<" is   "<<t[k][j];

      print(i,k-1);

      print(k,j);

}

      return;

}

int main()

{

int n;

cout<<"enter n";

cin>>n;

obst o;

o.obstm(n);

o.print(0,n);

return 0;

}

Enter the no. of nodes : 4

Enter the p values

p[1]  :3

p[2]  :3

p[3]  :1

p[4]  :1

Enter the q values

q[0]  :2

q[1]  :3

q[2]  :1

q[3]  :1

q[4]  :1

Output

Optimal BST is :: w values are

2       3       1       1       1

8       7       3       3

12      9       5

14      11

16

c values are

0       0       0       0       0

8       7       3       3

19      12      8

25      19

32

r values are

0       0       0       0       0

1       2       3       4

1       2       3

2       2

2

 Left child of   2 is  :: 1

 Right child of  2 is  :: 3

 Right child of  3 is  :: 4

Output-2

p[1]  :1

p[2]  :4

p[3]  :2

p[4]  :1

Enter the q values

q[0]  :4

q[1]  :2

q[2]  :4

q[3]  :1

q[4]  :1

Optimal BST is :: w values are

4       2       4       1       1

7       10      7       3

15      13      9

18      15

20

c values are

0          0          0          0          0

7          10        7          3

22        20        12

32        27

39

r values are

0       0       0       0       0

1       2       3       4

2       2       4

1       2

1

 Left child of   1 is   0

 Right child of  1 is   2

 Left child of   2 is   0

 Right child of  2 is   4

 Left child of   4 is   3

 Right child of  4 is   0

 Left child of   3 is   0

 Right child of  3 is   0