Every subset of a set {0, 1, 2, . . . , n − 1} can be represented as an n bit integer whose one bits indicate which elements belong to the subset. This is an efficient way to represent sets, because every element requires only one bit of memory, and set operations can be implemented as bit operations.

For example, since int is a 32-bit type, an int number can represent any subset of the set {0, 1, 2, . . . , 31}. The bit representation of the set {1, 3, 4, 8} is 00000000000000000000000100011010, which corresponds to the number 28 + 24 + 23 + 21 = 282.

The following code declares an int variable x that can contain a subset of {0, 1, 2, . . . , 31}. After this, the code adds the elements 1, 3, 4, and 8 to the set and prints the size of the set.

int x = 0;

x |= (1<<1);

x |= (1<<3);

x |= (1<<4);

x |= (1<<8);cout << __builtin_popcount(x) << “\n”; // 4

Then, the following code prints all elements that belong to the set:

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

if (x&(1<<i)) cout << i << ” “;

}

// output: 1 3 4 8

Set Operations Table shows how set operations can be implemented as bit operations. For example, the following code first constructs the sets x = {1, 3, 4, 8} and y = {3, 6, 8, 9} and then constructs the set z = x ∪ y = {1, 3, 4, 6, 8, 9}:

int x = (1<<1)|(1<<3)|(1<<4)|(1<<8);

int y = (1<<3)|(1<<6)|(1<<8)|(1<<9);

int z = x|y;

cout << __builtin_popcount(z) << “\n”; // 6

The following code goes through the subsets of {0, 1, . . . , n − 1}:

for (int b = 0; b < (1<<n); b++) {

// process subset b

}

Then, the following code goes through the subsets with exactly k elements:

for (int b = 0; b < (1<<n); b++) {

if (__builtin_popcount(b) == k) {

// process subset b

}

}

Finally, the following code goes through the subsets of a set x:

int b = 0;

do {

// process subset b

} while (b=(b-x)&x);

C++ Bitsets The C++ standard library also provides the bitset structure, which corresponds to an array whose each value is either 0 or 1. For example, the following code creates a bitset of 10 elements:

bitset<10> s;

s[1] = 1;

s[3] = 1;

s[4] = 1;

s[7] = 1;

cout << s[4] << “\n”; // 1

cout << s[5] << “\n”; // 0

The function count returns the number of one bits in the bitset:

cout << s.count() << “\n”; // 4

Also bit operations can be directly used to manipulate bitsets:

bitset<10> a, b;

// …

bitset<10> c = a&b;

bitset<10> d = a|b;

bitset<10> e = a^b;