You are asked to give an implementation of the ADT set using an array of bits and having the following functions:
- void toempty(SET *S)
- int empty(SET S)
- int find(int x, SET S)
- void add(int x, SET *S)
- void removeS(int x, SET *S)
- void intersect(SET S1, SET S2, SET *R)
- void unionS(SET S1, SET S2, SET *R)
- void complement(SET S, SET *R)
- void display(SET S)
Here, we assume that N ≤ 32. A set is represented by a long integer (without sign), with the convention $$i \in e \rightarrow$$ the $$i$$th bit of e is 1.
This implementation of sets is a classic of C courses, since it represents one of the rare "admirable" uses of operators on bits:
- a & b: calculation of a number whose ith bit is the conjunction of the ith bit of a with the ith bit of b.
- a | b: calculation of a number whose ith bit is the disjunction of the ith bit of a with the ith bit of b.
- ~ a: calculation of a number whose ith bit is the opposite of the ith bit of a.
- a << n: calculation of a number whose bits are those of a, shifted by n positions to "the left" (the side of the "most significant"). The n bits on the left are lost. ( i.e. on the right enter n bits to 0).
- a >> n: calculation of a number whose bits are those of a, shifted by n positions to "the right" (the side of the units). The n right bits are lost. On the left enter:
- if a is of an unsigned type, enter n bits 0,
- if a is of a signed type, n copies of the leftmost bit of a.
The basic idea is simple: store the binary information "i belongs to e" in the bits of a long integer rather than in the bytes of an array of bytes.
Difficulty level
This exercise is mostly suitable for students
#include <stdio.h>
#include <stdlib.h>
#define N 32
typedef unsigned long SET;
void toempty(SET *e)
{
*e = 0;
}
int empty(SET e)
{
return e == 0;
}
int find(int x, SET e)
{
return (e & (1L << x)) != 0;
}
void add(int x, SET *e)
{
*e |= (1L << x);
}
void removeS(int x, SET *e)
{
*e &= ~(1L << x);
}
void intersect(SET e1, SET e2, SET *r)
{
*r = e1 & e2;
}
void unionS(SET e1, SET e2, SET *r)
{
*r = e1 | e2;
}
void complement(SET e, SET *r)
{
*r = ~e;
}
void display(SET e)
{
int i;
printf("[ ");
for (i = 0; i < N; i++)
if (find(i, e))
printf("%d ", i);
printf("]");
}
int main()
{
SET a, b, c;
int i;
toempty(&a);
for (i = 0; i < 32; i += 2)
add(i, &a);
for (i = 30; i >= 0; i -= 3)
add(i, &a);
printf(" a : ");
display(a);
printf("\n");
toempty(&b);
printf("a empty? %s\n", empty(a) ? "yes" : "no");
printf("b empty? %s\n", empty(b) ? "yes" : "no");
for (i = 0; i < 32; i += 5)
add(i, &b);
printf(" b : ");
display(b);
printf("\n");
complement(b, &c);
printf(" ~b : ");
display(c);
printf("\n");
intersect(a, b, &c);
printf("a.b : ");
display(c);
printf("\n");
unionS(a, b, &c);
printf("a+b : ");
display(c);
printf("\n");
for (i = 0; i < 32; i += 3)
removeS(i, &c);
printf("c : ");
display(c);
printf("\n");
return 0;
}
Back to the list of exercises
Looking for a more challenging exercise, try this one !!
Sorting using the counting sort algorithm