A set which containing $and pi are the end blocks are the finite and without these are infinite
Enumerating subsets requires that each element in the set be mapped to a bit value. Combining these bit values (with logical OR) gives you the subset. Each subset will then have a value in the range 0 to 7, giving 8 subsets in all: {}, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}. However, you cannot simply count from 0 to 7 to enumerate these subsets because this is highly inefficient. In a set of 3 it's not a significant loss, but in a larger set the inefficiency is substantial. For instance, in a set of just 64 elements, you must count from 0 to 9,223,372,036,854,775,809 in order to enumerate the 64 subsets that have just one element. If that doesn't drive the point home, then nothing will.In order to correctly enumerate subsets you must use the Banker's sequence. Each sequence varies depending upon how many elements are in the set, however the sequence allows you to quickly enumerate one set after another in a logical sequence. It also allows you to start from anywhere in the sequence and continue from that point on, thus making it possible to quickly enumerate all sets with a specific number of elements.The following code is quite involved but includes several auxiliary functions that help speed up the main code at the bottom. Some examples are included to demonstrate different usages, including the enumeration of subsets of {x, y, z}, subsets of a set of 5 elements and subsets of a subset of a 9 element set. Note that the final example (all subsets of a 32 element set) is commented out because it will take a seriously long time to enumerate all the possible subsets. However, the code is flexible enough to allow you to narrow the search down to just those subsets with a specific number of elements.Note that the two main functions to focus upon are the find_first_subset() and find_next_subset() functions. These two functions are the driving force behind the Banker's sequence.#include // Returns the count of set bits in the given subset.// E.g., 01010010 returns 3.unsigned int count_set_bits( register unsigned int subset ){// An alternative method is to logical AND the bitset with 0x1// then right shift the bitset by 1 bit. That requires an average// of 32 separate operations (worst case is 64). This method always// uses 17, thus always returns in constant time. For unsigned// short, eliminate the last shift. For char, eliminate the last// two shifts.subset -= (( subset >> 1 ) & 0x55555555 );subset = ((( subset >> 2 ) & 0x33333333 ) + ( subset & 0x33333333 ));subset = ((( subset >> 4 ) + subset ) & 0x0f0f0f0f );subset += ( subset >> 8 );subset += ( subset >> 16 );return( subset & 0x0000003f );}// Returns all set bits from the given subset's MSB down.// E.g., 00100010 returns 00111111.// The inversion of the return value can be used to mask// bits that are greater than the MSB in another subset.unsigned int fill_down( register unsigned int subset ){subset |= ( subset >> 1 );subset |= ( subset >> 2 );subset |= ( subset >> 4 );subset |= ( subset >> 8 );subset |= ( subset >> 16 );return( subset );}// Returns the least-significant set bit in the given subset.// E.g., 00100110 returns 00000010.unsigned int get_LSB( register unsigned int subset ){return( subset ^ ( subset & ( subset - 1 )));}// Returns the most-significant set bit in the given subset.// E.g., 00100110 returns 00100000.unsigned int get_MSB( register unsigned int subset ){subset = fill_down( subset );return( subset & ~( subset >> 1 ));}// Returns the first subset of subset_count within the given set.// The return value can be passed to get_next_subset().unsigned int get_first_subset( const unsigned set, unsigned int subset_count ){// Initialise the subset.unsigned int subset = 0;// Check validity of parameters.if( subset_count && subset_count
A computer program.
The defining characteristic of FA is that they have only a finite number of states. Hence, a finite automata can only "count" (that is, maintain a counter, where different states correspond to different values of the counter) a finite number of input scenarios.There is no finite automaton that recognizes these strings:The set of binary strings consisting of an equal number of 1's and 0'sThe set of strings over '(' and ')' that have "balanced" parenthesesThe 'pumping lemma' can be used to prove that no such FA exists for these examples.
Each element of C can either be in a particular subset or not in it - two options for each element of C. So if there are n elements in C, then the number of subsets of C is 2n. These include the null set and C itself.
A subset of some set X is, by definition, any set whose elements are entirely contained in X. So the answer is yes. As an example, take your infinite set, and select 3 or 10 or any finite number of your favorite elements in this set. The set of your chosen elements is a finite subset of the infinite set.
Yes,an empty set is the subset of every set. The subset of an empty set is only an empty set itself.
Yes the null set is a subset of every set.
No. The empty is the a subset of every set and every set is a subset of itself.
The null set. It is a subset of every set.
A finite set, consisting of N elements, will have 2N subsets.
It isn't. The empty set is a subset - but not a proper subset - of the empty set.
The null set. Every set is a subset of itself and so the null set is a subset of the null set.
Every set contains the empty set. Every set is a subset of itself.
The empty set!
The empty set is a subset of all sets. No other sets have this property.
A is a subset of a set B if every element of A is also an element of B.