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Resources for Code Optimization

Agner Optimizing Assembly Code

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Sum Of Large Numbers In C

April 20, 2016 by ibia Leave a Comment

Write a program which can do sum of large numbers having up to 1000 digits

Solution by Arjun Suresh

#include <stdio.h>
#include <string.h>
#include <ctype.h>
 
void print(char *a, int l);
int read(char *a);
int sum(char *a, char* b, char * c, int al, int bl); 
int main(void) {
	char a[1000], b[1000], c[1000]; //a and b hold the input numbers and c hold the output number. Each array entry is a digit 
	int al, bl; //for storing the number of digits of the two input numbers
	printf("Enter the first number ");
	al = read(a);
	printf("Enter the second number ");
	bl = read(b);
	int l = sum(a, b, c, al, bl);
	printf("sum = ");
	print(c, l);
	return 0;
}
 
int sum(char *a, char* b, char * c, int al, int bl)
{
	int l = al > bl? al:bl;//l stores the no.of digit of output which may become l+1 due to carry from MSB
        int i = l;
        char carry = 0;
        while(al > 0 && bl > 0)
        {
                char val = a[--al]  + b[--bl] + carry;
                c[i--] = val % 10;
                carry =   val / 10;
        }
        while(al > 0)//If a has more digits than b
        {
                char val = a[--al]  + carry;
                c[i--] = val % 10;
                carry =   val / 10;
        }
        while(bl > 0)//If b has more digits than a
        {
                char val =  b[--bl] + carry;
                c[i--] = val % 10;
                carry =   val / 10;
        }
        c[0] = carry;//Assigning the final carry
	return l+1;//Return the no. of digits of output
} 
 
int read(char *a)
{
	char c;
	int i = 0;
	do
	{
		c = getchar();
	}
	while(isspace(c)); //Reading and discarding any whitespace char typed
	while(isdigit(c))
	{
		a[i++] = c - 48; //getchar returns the ASCII. So, for 1 it returns 49. Subtracting 48, we get the actual int value entered via keyboard 
		c = getchar();
	}
	return i;//Return the no. of digits read
}
 
void print(char *a, int l)
{
	int i;
	if(a[0] != 0)
		printf("%d", a[0]);//If final carry is 0, ignore it
	for(i = 1; i < l; i++)
	{
		printf("%d", a[i]);
	}
	printf("\n");
}

#include <stdio.h>

#include <string.h>

#include <ctype.h>

void print(char *a, int l);

int read(char *a);

int sum(char *a, char* b, char * c, int al, int bl);

int main(void) {

char a[1000], b[1000], c[1000]; //a and b hold the input numbers and c hold the output number. Each array entry is a digit

int al, bl; //for storing the number of digits of the two input numbers

printf("Enter the first number ");

al = read(a);

printf("Enter the second number ");

bl = read(b);

int l = sum(a, b, c, al, bl);

printf("sum = ");

print(c, l);

return 0;

}

int sum(char *a, char* b, char * c, int al, int bl)

{

int l = al > bl? al:bl;//l stores the no.of digit of output which may become l+1 due to carry from MSB

int i = l;

char carry = 0;

while(al > 0 && bl > 0)

{

char val = a[--al] + b[--bl] + carry;

c[i--] = val % 10;

carry = val / 10;

}

while(al > 0)//If a has more digits than b

{

char val = a[--al] + carry;

c[i--] = val % 10;

carry = val / 10;

}

while(bl > 0)//If b has more digits than a

{

char val = b[--bl] + carry;

c[i--] = val % 10;

carry = val / 10;

}

c[0] = carry;//Assigning the final carry

return l+1;//Return the no. of digits of output

}

int read(char *a)

{

char c;

int i = 0;

{

c = getchar();

}

while(isspace(c)); //Reading and discarding any whitespace char typed

while(isdigit(c))

{

a[i++] = c - 48; //getchar returns the ASCII. So, for 1 it returns 49. Subtracting 48, we get the actual int value entered via keyboard

c = getchar();

}

return i;//Return the no. of digits read

}

void print(char *a, int l)

{

int i;

if(a[0] != 0)

printf("%d", a[0]);//If final carry is 0, ignore it

for(i = 1; i < l; i++)

{

printf("%d", a[i]);

}

printf("\n");

}

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The basis of this is that your ID must be in the list of authorized hosts in the server. So, for password-less login to work, generate your public key, and append it to the authorized_hosts file in the server.

To create a public key, from terminal do

ssh-keygen -t rsa

1	ssh-keygen -t rsa

To copy to authorized_hosts file do

cat ~/.ssh/id_rsa.pub | ssh serveruser@server &#039;cat &amp;gt;&amp;gt; ~/.ssh/authorized_keys&#039;

1	cat ~/.ssh/id_rsa.pub \| ssh serveruser@server 'cat &gt;&gt; ~/.ssh/authorized_keys'

Test by

ssh serveruser@server

1	ssh serveruser@server

If no password is asked, everything worked fine.
You can also do the following command to copy the key

ssh-copy-id serveruser@server

1	ssh-copy-id serveruser@server

4,368 total views, no views today

Useful Links for Programming Interviews

April 19, 2016 by ibia Leave a Comment

6,354 total views, no views today

Some Reduction Inferences

April 19, 2016 by ibia Leave a Comment

P, NP, NPC, NPH

$P$ is a subset of $NP$, but whether $P = NP$ is still not known
$NPC$ is a subset of $NP$ (only if $P \neq NP$, $NPC$ is a proper subset of $NP$)
$NPC$ is a proper subset of $NPH$

Reduction

Reducing a problem $A$ to problem $B$ means converting an instance of problem $A$ to an instance of problem $B$. Then, if we know the solution of problem $B$, we can solve problem $A$ by using it.

Consider the following example:

You want to go from Bangalore to Delhi and you can get a ticket from Mumbai to Delhi. So, you ask your friend for a lift from Bangalore to Mumbai. So, the problem of going from Bangalore to Delhi got reduced to a problem of going from Mumbai to Delhi.

Some Inferences:

Consider problems $A$, $B$ and $C$.

Assume all reductions are done in polynomial time.

Two Important Points:

If $A$ is reduced to $B$ and $B$ $\in$ class $X$, then $A$ cannot be harder than $X$, because we can always do a reduction from $A$ to $B$ and solve $B$ instead of directly solving $A$. This reduction thus gives an upper bound for the complexity of $A$.
If $A$ is reduced to $B$ and $A$ $\in$ class $X$, then $B$ cannot be easier than $X$. The argument is exactly same as for the one above. This reduction is used to show if a problem belongs to $NPH$ – just reduce some known $NPH$ problem to the given problem. This reduction thus gives a lower bound for the complexity of $B$.

Somethings to take care of:

If $A$ is reduced to $B$ and $B$ $\in$ $NPC$, then $A$ $\in$ $NP$
- Here, we cannot say $A$ is $NPC$. All we can say is $A$ cannot be harder than $NPC$ and hence $NP$ (all $NPC$ problems are in $NP$). To belong to $NPC$, all $NP$ problems must be reducible to $A$, which we cannot guarantee from the given statement.
If $A$ is reduced to $B$, $C$ is reduced to $A$ , $B \in NP $ and $C \in$ $NPC$, then $A$ $\in$ $NPC$
- Here, the first reduction says that $A$ is in $NP$. The second reduction says that all $NP$ problems can be reduced to $A$ (because $C$ is in $NPC$, by definition of $NPC$, all problems in $NP$ are reducible to $C$ and now $C$ is reduced to $A$). Hence, the two necessary and sufficient conditions for $NPC$ are satisfied and $A$ is in $NPC$
If $A$ is reduced to $B$ and $B$ $\in$ $NPH$, then $A$ $\in$ $?$
- Here we can’t say anything about $A$. It can be as hard as $NPH$, or as simple as $P$.

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Exams I Can Attempt Along with GATE

April 18, 2016 by ibia Leave a Comment

UGC NET: Happens in June and December but restricted to Master students (including MCA). Syllabus is bit more than GATE but difficulty level is slightly lower
TIFR: For those interested in Theoretical Computer Science. Happens in December.
PGEEE: IIITH is comparable to old IITs. Even if you are not financially sound you should consider taking loan and joining here as you can easily pay back loan after studying here.
BITS HD
BARC
ISRO:
Other PSU exams here

17,726 total views, no views today

C Coding Questions

April 15, 2016 by ibia Leave a Comment

Consider the following program:

#include <stdio.h> int main() { int a[10][20][30]={0}; printf("%ld",&a+1 - &a); return 0; }

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#include <stdio.h>
int main()
{
    int a[10][20][30]={0};
    printf("%ld",&a+1 - &a);

    return 0;
}

What is the output of this program?

Consider the following program:

#include <stdio.h> 
int main()
{
    int a[10][20][30] = {0};
    int *b = a;
    int *c = a+1;
    printf("%ld", c-b);
 
    return 0;
}
What is the output of this program? (You may ignore compiler warnings)

#include <stdio.h>

int main()

{

int a[10][20][30] = {0};

int *b = a;

int *c = a+1;

printf("%ld", c-b);

return 0;

}

What is the output of this program? (You may ignore compiler warnings)

What is the output of the following program?

#include <stdio.h>
#include <stdlib.h>
int* fun();
 
int main()
{
    int *a = fun();
    printf("%d",*a);
 
    return 0;
}
int* fun()
{
    int *a =(int*) malloc(sizeof(int));
    *a = 10;
    return a;
}

#include <stdio.h>

#include <stdlib.h>

int* fun();

int main()

{

int *a = fun();

printf("%d",*a);

return 0;

}

int* fun()

{

int *a =(int*) malloc(sizeof(int));

*a = 10;

return a;

}

What is the output of the following program?

#include <stdio.h> int main() { int *a = fun(); printf("%d",a); return 0; } int fun() { int a = 10; return a; }

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#include <stdio.h>
int main()
{
    int *a = fun();
    printf("%d",a);

    return 0;
}
int fun()
{
    int a = 10;
    return a;
}
What is the output of the following program?

#include <stdio.h> #include <string.h> int main() { char string[] = "Hello"; printf("%zu %zu",sizeof(string),strlen(string)); return 0; }

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#include <stdio.h>
#include <string.h>

int main()
{
    char string[] = "Hello";
    printf("%zu %zu",sizeof(string),strlen(string));

    return 0;
}
What is the output of the following program?

#include <stdio.h> int main() { float a = 0.5; if(a == 0.5) printf("Yes"); else printf("No"); return 0; }

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#include <stdio.h>
int main()
{
    float a = 0.5;
    if(a == 0.5)
        printf("Yes");
    else
        printf("No");

    return 0;
}
What is the output of the following program?

#include <stdio.h> #include <string.h> void foo(char *); int main() { char a[100] = {0}; printf("%zu %zu",sizeof(a),strlen(a)); return 0; }

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#include <stdio.h>
#include <string.h>
void foo(char *);

int main()
{
    char a[100] = {0};
    printf("%zu %zu",sizeof(a),strlen(a));

    return 0;
}
What is the output of the following program? (Assume input is 10)

#include <stdio.h> int main() { int a; printf("%d",scanf("%d",&a)); return 0; }

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#include <stdio.h>
int main()
{
    int a;
    printf("%d",scanf("%d",&a));

    return 0;
}

If the binary equivalent of 5.375 in normalised form is 01000000 10101100 00000000 00000000, what will be the output of the following program?

#include <stdio.h>
 
int main()
{
    float a=5.375;
    char *p;
    int i;
    p = (char*)&a;
    for(i=0; i<2; i++)
        printf("%x", (unsigned char)(p[i]^p[3-i]));
 
    return 0;
}

#include <stdio.h>

int main()

{

float a=5.375;

char *p;

int i;

p = (char*)&a;

for(i=0; i<2; i++)

printf("%x", (unsigned char)(p[i]^p[3-i]));

return 0;

}

What is the output of the following program?

#include <stdio.h> int main() { char str[] = {'a','b','c','\0'}; str[0] -= 32; printf("%s",str); return 0; }

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#include <stdio.h>
int main()
{
    char str[] = {'a','b','c','\0'};
    str[0] -= 32;
    printf("%s",str);

    return 0;
}

Questions from Datatypes

Consider an implementation where int is 4 bytes and long int is 8 bytes. Which of the following initializations are correct?

#include <stdio.h>
int main()
{
   long int a = 0x7fffffff * 0x7ffffff;
   long int b = 0x7ffffffff * 0x7ffffff;
   long int c = 0x7fffffff * 0x7fffffff;
   long int d = 0x7fffffff * 0x7fffffffl;
   printf("a = %ld, b = %ld, c = %ld, d = %ld\n", a, b, c, d);
   return 0;
}

#include <stdio.h>

int main()

{

long int a = 0x7fffffff * 0x7ffffff;

long int b = 0x7ffffffff * 0x7ffffff;

long int c = 0x7fffffff * 0x7fffffff;

long int d = 0x7fffffff * 0x7fffffffl;

printf("a = %ld, b = %ld, c = %ld, d = %ld\n", a, b, c, d);

return 0;

}

Consider an implementation where int is 4 bytes and long int is 8 bytes. What will be the output of the following code?

#include <stdio.h>
int main()
{
   int i = 0;
   size_t a = sizeof i, b = sizeof (long);
   printf("a = %zd, b = %zd\n", a, b); //If %zd is given the compiler will automatically give it the correct type whether short, long or normal. This is useful for special data types like size_t whose size is implementation specific
   return 0;
}

#include <stdio.h>

int main()

{

int i = 0;

size_t a = sizeof i, b = sizeof (long);

printf("a = %zd, b = %zd\n", a, b); //If %zd is given the compiler will automatically give it the correct type whether short, long or normal. This is useful for special data types like size_t whose size is implementation specific

return 0;

}

What will be the output of the following code?

#include <stdio.h> int main() { unsigned int a = 5; if(a > -1) printf("5 is > -1\n"); return 0; }

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#include <stdio.h>
int main()
{
   unsigned int a = 5;
   if(a > -1)
      printf("5 is > -1\n");
   return 0;
}
What will be printed by the following code?

#include <stdio.h> int main() { char buff[255] = "abc\ pee"; printf("%s", buff); return 0; }

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#include <stdio.h>

int main()
{
   char buff[255] = "abc\
pee";
   printf("%s", buff);
   return 0;
}
What will be printed by the following code?

#include <stdio.h> int main() { char buff[255] = "\0"; printf("%s", buff); return 0; }

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#include <stdio.h>

int main()
{
    char buff[255] = "\0";
    printf("%s", buff);
    return 0;
}

What will be the output of the following code? (Assume a 64 bit machine/compiler)

#include <stdio.h>
int main()
{
    char *p = "Hello World";
    char q[] = "Hello World";
    printf("%zd %zd", sizeof p, sizeof *p);
    printf("\n");
    printf("%zd %zd", sizeof q, sizeof *q);
    return 0;
}

#include <stdio.h>

int main()

{

char *p = "Hello World";

char q[] = "Hello World";

printf("%zd %zd", sizeof p, sizeof *p);

printf("\n");

printf("%zd %zd", sizeof q, sizeof *q);

return 0;

}

What will be the output of the following code?

#include <stdio.h>
int main()
{
   {
      char a = 5;
      int b = 5;
      if(a == b)          
         printf("char and int compared equal\n");    }
   {
      int a = 5;
      long int b = 5;
if(a == b)
         printf("int and long compared equal\n");
   }
   {
      float a = 5.0;
      double b = 5.0;
      if(a == b)
         printf("float and double compared equal\n");
   }
   {
      float a = 5.2;
      double b = 5.2;
      if(a == b)
         printf("float and double again compared equal\n");
   }
   {
      float a = 5.2;
      if(a == 5.2)
         printf("float compared equal with constant\n");
   }
   {
      double a = 5.2;
      if(a == 5.2)
         printf("double compared equal with constant\n");
   }
   return 0;
}

#include <stdio.h>

int main()

{

char a = 5;

int b = 5;

if(a == b)

printf("char and int compared equal\n"); }

{

int a = 5;

long int b = 5;

if(a == b)

printf("int and long compared equal\n");

}

{

float a = 5.0;

double b = 5.0;

if(a == b)

printf("float and double compared equal\n");

}

{

float a = 5.2;

double b = 5.2;

if(a == b)

printf("float and double again compared equal\n");

}

{

float a = 5.2;

if(a == 5.2)

printf("float compared equal with constant\n");

}

{

double a = 5.2;

if(a == 5.2)

printf("double compared equal with constant\n");

}

return 0;

}

Questions from Pointers

What will be the output of the following code?

#include <stdio.h> int main() { int a = 5; int* p = &a; printf("%d", ++*p); }

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#include <stdio.h>

int main()
{
  int a = 5;
  int* p = &a;
  printf("%d", ++*p);
}
What will be the output of the following code?

#include <stdio.h> int main() { char a[] = "Hello World"; char* p = &a; printf("%s", p+2 ); }

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#include <stdio.h>

int main()
{
  char a[] = "Hello World";
  char* p = &a;
  printf("%s", p+2 );
}
What will be the output of the following code?

#include <stdio.h> int main() { int a; int* p = &a; printf("%zu", sizeof( *(char*)p )); }

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#include <stdio.h>

int main()
{
  int a;
  int* p = &a;
  printf("%zu", sizeof( *(char*)p  ));
}

Is the following code legal?

#include <stdio.h>
 
int main()
{
  int a = 1;
  if((char*)  &a)
  {
    printf("My machine is little endian");
  }
  else
  {
     printf("My machine is big endian\n");
  }
}

#include <stdio.h>

int main()

{

int a = 1;

if((char*) &a)

{

printf("My machine is little endian");

}

else

{

printf("My machine is big endian\n");

}

What will be the output of the following code?

#include <stdio.h> int main() { int *a = (int*) 1; printf("%d", a); }

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#include <stdio.h>

int main()
{
int *a = (int*) 1;
printf("%d", a);
}
What will be the output of the following code?

#include <stdio.h> int main() { int a = 1, *p, **pp; p = &a; pp = p; printf("%d", **pp); }

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#include <stdio.h>

int main()
{
  int a = 1, *p, **pp;
  p = &a;
  pp = p;
  printf("%d", **pp);
}
What will be the output of the following code?

#include <stdio.h> int main() { int a = 1, *p, **pp; p = &a; pp = p; printf("%d", *pp); }

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#include <stdio.h>

int main()
{
  int a = 1, *p, **pp;
  p = &a;
  pp = p;
  printf("%d", *pp);
}

Assuming a little endian machine, what will be the output of the following program?

#include <stdio.h>
 
fun(int a)
{
  char *arr[] = {"0000", "0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
  unsigned char* p = (unsigned char*) &a ;
  p+=3;
  int i;
  for(i = 0; i < sizeof a; i++) { int d = (*p)>>4; 
    printf("%s", arr[d]);
    d = (*p) & 0xf;
    printf("%s ", arr[d]);
    p--;
  }
}
 
int main()
{
 int a;
 scanf("%d", &a);
 fun(a);
}

#include <stdio.h>

fun(int a)

{

char *arr[] = {"0000", "0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};

unsigned char* p = (unsigned char*) &a ;

p+=3;

int i;

for(i = 0; i < sizeof a; i++) { int d = (*p)>>4;

printf("%s", arr[d]);

d = (*p) & 0xf;

printf("%s ", arr[d]);

p--;

}

int main()

{

int a;

scanf("%d", &a);

fun(a);

}

Some codes

What is the following function doing?

int foo(int n) { int sum = 0; while(n > 0) { n = n & n-1; sum++; } return sum; }

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int foo(int n)
{
    int sum = 0;
    while(n > 0)
    {
        n = n & n-1;
        sum++;
    }
    return sum;
}

What is the following function doing?

int foo(int a, int b)
{
    int c = a, d = b;
    while(a != b)
    {
        if(a &lt; b)
            a = a+c;
        else
            b = b+d;
    }
    return a;
}

int foo(int a, int b)

{

int c = a, d = b;

while(a != b)

{

if(a < b)

a = a+c;

else

b = b+d;

}

return a;

}

What is the following function doing?

int foo( int a, int b) { int c = a-b; c = c&(0x80000000); return (!c)*a +(!!c)*b; }

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int foo( int a, int b)
{
    int c = a-b;
    c = c&(0x80000000);
    return (!c)*a +(!!c)*b;
}

What is the following function doing?

unsigned fun(unsigned a, unsigned b)
{
    int i;
    unsigned  j = 0;
    for(i = 0; i &lt; 32; i++)
    {
        j &lt;&lt;= 1;
        j += !!(a &amp; 0x80000000);
        a &lt;&lt;= 1; if(j &gt;=b)
        {
            j -= b;
            a++;
        }
    }
    return a;
}

unsigned fun(unsigned a, unsigned b)

{

int i;

unsigned j = 0;

for(i = 0; i < 32; i++)

{

j <<= 1;

j += !!(a & 0x80000000);

a <<= 1; if(j >=b)

{

j -= b;

a++;

}

return a;

}

What is the following function doing?

unsigned fun(unsigned int a)
{
    unsigned int i, x = 0, y = 0, z = 0;
    for(i = 0; i &lt; 16; i++)
    {
       	y &lt;&lt;= 2;
       	y += !!(a &amp; 0x80000000) &lt;&lt; 1;
       	y += !!(a &amp; 0x40000000);
       	a &lt;&lt;= 2;
       	x = x + (x&amp;1);
       	x &lt;&lt;= 1;
       	z &lt;&lt;= 1;
       	if(x + 1 &lt;= y)
       	{
            x++;
            z++;
	    y-=x;
        }
    }
    return z;
}

unsigned fun(unsigned int a)

{

unsigned int i, x = 0, y = 0, z = 0;

for(i = 0; i < 16; i++)

{

y <<= 2;

y += !!(a & 0x80000000) << 1;

y += !!(a & 0x40000000);

a <<= 2;

x = x + (x&1);

x <<= 1;

z <<= 1;

if(x + 1 <= y)

{

x++;

z++;

y-=x;

}

return z;

}

Write the code to dynamically allocate a 2-D array of size m x n.
Declare a pointer to a function accepting an integer and returning void.
Write the condition so that the below code outputs Hello World.

#include <stdio.h> int main() { if() { printf("Hello "); } else { printf("World\n"); } return 0; }

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#include <stdio.h>
int main()
{
    if()
    {
printf("Hello ");
    }
    else
    {
printf("World\n");
    }

    return 0;
}
Write a one line code to check if a number is a power of 2.
Write a one line code to invert the last four bits of an integer.

11,812 total views, 1 views today

CCMT GATECSE Cutoff

April 14, 2016 by ibia 1 Comment

The last rank scores shown gives an idea of the preference last year students gave. Most of this will be based just on placement preference. The following sheets only show the scores for CS/IT departments and for other branches please see the full PDF of cutoffs given at the bottom of the sheet. This sheet shows the cutoff for General, OBC, SC, ST and PWD in each of those as well.

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New IITs GATECSE Cutoff

April 14, 2016 by ibia Leave a Comment

13,449 total views, 3 views today

CMI/ISI Admissions

April 14, 2016 by ibia Leave a Comment

5,917 total views, no views today

ibia

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Reduction

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C Coding Questions

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Solution by Arjun Suresh

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Reduction

Consider the following example:

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Two Important Points:

Somethings to take care of:

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