This is an example of the permutations of 3 string items (apple, orange, cherry):
// Create the initial vector of 3 elements (apple, orange, cherry) ICombinatoricsVector<String> originalVector = Factory.createVector(new String[] { "apple", "orange", "cherry" }); // Create the permutation generator by calling the appropriate method of the Factory class Generator<String> gen = Factory.createPermutationGenerator(originalVector); // Print the result for (ICombinatoricsVector<String> perm : gen) System.out.println(perm);
And the result
CombinatoricsVector=([apple, orange, cherry], size=3) CombinatoricsVector=([apple, cherry, orange], size=3) CombinatoricsVector=([cherry, apple, orange], size=3) CombinatoricsVector=([cherry, orange, apple], size=3) CombinatoricsVector=([orange, cherry, apple], size=3) CombinatoricsVector=([orange, apple, cherry], size=3)
The permutation may have more elements than slots. For example, the three possible permutation of 12 in three slots are: 111, 211, 121, 221, 112, 212, 122, and 222.
Let's generate all possible permutations with repetitions of 3 elements from the set of apple and orange:
// Create the initial vector of 2 elements (apple, orange) ICombinatoricsVector<String> originalVector = Factory.createVector(new String[] { "apple", "orange" }); // Create the generator by calling the appropriate method in the Factory class. // Set the second parameter as 3, since we will generate 3-elemets permutations Generator<String> gen = Factory.createPermutationWithRepetitionGenerator(originalVector, 3); // Print the result for (ICombinatoricsVector<String> perm : gen) System.out.println( perm );
And the result
CombinatoricsVector=([apple, apple, apple], size=3) CombinatoricsVector=([orange, apple, apple], size=3) CombinatoricsVector=([apple, orange, apple], size=3) CombinatoricsVector=([orange, orange, apple], size=3) CombinatoricsVector=([apple, apple, orange], size=3) CombinatoricsVector=([orange, apple, orange], size=3) CombinatoricsVector=([apple, orange, orange], size=3) CombinatoricsVector=([orange, orange, orange], size=3)
A simple k-combination of a finite set S is a subset of k distinct elements of S. Specifying a subset does not arrange them in a particular order. As an example, a poker hand can be described as a 5-combination of cards from a 52-card deck: the 5 cards of the hand are all distinct, and the order of the cards in the hand does not matter.
For example, let's generate 3-combination of the set (red, black, white, green, blue).
// Create the initial vector ICombinatoricsVector<String> initialVector = Factory.createVector(new String[] { "red", "black", "white", "green", "blue" }); // Create a simple combination generator to generate 3-combinations of the // initial vector Generator<String> gen = Factory.createSimpleCombinationGenerator(initialVector, 3); // Print all possible combinations for (ICombinatoricsVector<String> combination : gen) { System.out.println(combination); }
And the result of 10 combinations
CombinatoricsVector=([red, black, white], size=3) CombinatoricsVector=([red, black, green], size=3) CombinatoricsVector=([red, black, blue], size=3) CombinatoricsVector=([red, white, green], size=3) CombinatoricsVector=([red, white, blue], size=3) CombinatoricsVector=([red, green, blue], size=3) CombinatoricsVector=([black, white, green], size=3) CombinatoricsVector=([black, white, blue], size=3) CombinatoricsVector=([black, green, blue], size=3) CombinatoricsVector=([white, green, blue], size=3)
As an example. Suppose there are 2 types of fruits (apple and orange) at a grocery store, and you want to buy 3 pieces of fruit. You could select
Generate 3-combinations with repetitions of the set (apple, orange).
// Create the initial vector of (apple, orange) ICombinatoricsVector<String> initialVector = Factory.createVector(new String[] { "apple", "orange" }); // Create a multi-combination generator to generate 3-combinations of // the initial vector Generator<String> gen = Factory.createMultiCombinationGenerator(initialVector,3); // Print all possible combinations for (ICombinatoricsVector<String> combination : gen) { System.out.println(combination); }
And the result of 4 multi-combinations
CombinatoricsVector=([apple, apple, apple], size=3) CombinatoricsVector=([apple, apple, orange], size=3) CombinatoricsVector=([apple, orange, orange], size=3) CombinatoricsVector=([orange, orange, orange], size=3)
A set A is a subset of a set B if A is "contained" inside B. A and B may coincide. The relationship of one set being a subset of another is called inclusion or sometimes containment.
Examples:
All subsets of (1, 2, 3) are:
And code which generates all subsets of (one, two, three)
// Create an initial vector/set ICombinatoricsVector<String> initialSet = Factory.createVector(new String[] { "one", "two", "three" }); // Create an instance of the subset generator Generator<String> gen = Factory.createSubSetGenerator(initialSet); // Print the subsets for (ICombinatoricsVector<String> subSet : gen) { System.out.println(subSet); }
And the result of all 8 possible subsets
CombinatoricsVector=([], size=0) CombinatoricsVector=([one], size=1) CombinatoricsVector=([two], size=1) CombinatoricsVector=([one, two], size=2) CombinatoricsVector=([three], size=1) CombinatoricsVector=([one, three], size=2) CombinatoricsVector=([two, three], size=2) CombinatoricsVector=([one, two, three], size=3)
Version 2.0 of the combinatoricslib supports sets with duplicates. For example, if the original vector contains duplicates like (a, b, a, c), then the result will contain 14 subsets (instead of 16):
() (a) (b) (a, b) (a, a) (b, a) (a, b, a) (c) (a, c) (b, c) (a, b, c) (a, a, c) (b, a, c) (a, b, a, c)
If you still would like to treat the set with duplicates as not identical,
you should create a generator and set the second parameter of the method
Factory.createSubSetGenerator()
as false
. In this
case all 16 subsets will be generated.
Note. If the initial vector contains duplicates then the method
getNumberOfGeneratedObjects
won't be able to return the number
of the sub sets/lists. It will throw a runtime exception
In number theory, a partition of a positive integer n is a way of writing n as a sum of positive integers. Two sums that differ only in the order of their summands are considered to be the same partition; if order matters then the sum becomes a composition. A summand in a partition is also called a part.
WARNING! Be careful because number of all partitions can be very high even for not great given N.
The partitions of 5 are listed below:
The number of partitions of n is given by the partition function p(n). In number theory, the partition function p(n) represents the number of possible partitions of a natural number n, which is to say the number of distinct (and order independent) ways of representing n as a sum of natural numbers.
Let's generate all possible partitions of 5:
// Create an instance of the partition generator to generate all // possible partitions of 5 Generator<Integer> gen = Factory.createPartitionGenerator(5); // Print the partitions for (ICombinatoricsVector<Integer> p : gen) { System.out.println(p); }
And the result of all 7 integer possible partitions
CombinatoricsVector=([1, 1, 1, 1, 1], size=5) CombinatoricsVector=([2, 1, 1, 1], size=4) CombinatoricsVector=([2, 2, 1], size=3) CombinatoricsVector=([3, 1, 1], size=3) CombinatoricsVector=([3, 2], size=2) CombinatoricsVector=([4, 1], size=2) CombinatoricsVector=([5], size=1)
A composition of an integer n is a way of writing n as the sum of a sequence of (strictly) positive integers. Two sequences that differ in the order of their terms define different compositions of their sum, while they are considered to define the same partition of that number.
The sixteen compositions of 5 are:
Compare this with the seven partitions of 5:
Example. Generate compositions all possible integer compositions of 5.
// Create an instance of the integer composition generator to generate all possible compositions of 5 Generator<Integer> gen = Factory.createCompositionGenerator(5); // Print the compositions for (ICombinatoricsVector<Integer> p : gen) { System.out.println(p); }
And the result
CombinatoricsVector=([5], size=1) CombinatoricsVector=([1, 4], size=2) CombinatoricsVector=([2, 3], size=2) CombinatoricsVector=([1, 1, 3], size=3) CombinatoricsVector=([3, 2], size=2) CombinatoricsVector=([1, 2, 2], size=3) CombinatoricsVector=([2, 1, 2], size=3) CombinatoricsVector=([1, 1, 1, 2], size=4) CombinatoricsVector=([4, 1], size=2) CombinatoricsVector=([1, 3, 1], size=3) CombinatoricsVector=([2, 2, 1], size=3) CombinatoricsVector=([1, 1, 2, 1], size=4) CombinatoricsVector=([3, 1, 1], size=3) CombinatoricsVector=([1, 2, 1, 1], size=4) CombinatoricsVector=([2, 1, 1, 1], size=4) CombinatoricsVector=([1, 1, 1, 1, 1], size=5)
It is possible to generate non-overlapping sublists of length n of a given list
For example, if a list is (A, B, B, C), then the non-overlapping sublists of length 2 will be:
To do that you should use an instance of the complex combination generator
// create a vector (A, B, B, C) ICombinatoricsVector<String> vector = Factory.createVector(new String[] { "A", "B", "B", "C" }); // Create a complex-combination generator Generator<ICombinatoricsVector<String>> gen = new ComplexCombinationGenerator<String>(vector, 2); // Iterate the combinations for (ICombinatoricsVector<ICombinatoricsVector<String>> comb : gen) { System.out.println(ComplexCombinationGenerator.convert2String(comb) + " - " + comb); }
And the result
([A],[B, B, C]) - CombinatoricsVector=([CombinatoricsVector=([A], size=1), CombinatoricsVector=([B, B, C], size=3)], size=2) ([B, B, C],[A]) - CombinatoricsVector=([CombinatoricsVector=([B, B, C], size=3), CombinatoricsVector=([A], size=1)], size=2) ([B],[A, B, C]) - CombinatoricsVector=([CombinatoricsVector=([B], size=1), CombinatoricsVector=([A, B, C], size=3)], size=2) ([A, B, C],[B]) - CombinatoricsVector=([CombinatoricsVector=([A, B, C], size=3), CombinatoricsVector=([B], size=1)], size=2) ([A, B],[B, C]) - CombinatoricsVector=([CombinatoricsVector=([A, B], size=2), CombinatoricsVector=([B, C], size=2)], size=2) ([B, C],[A, B]) - CombinatoricsVector=([CombinatoricsVector=([B, C], size=2), CombinatoricsVector=([A, B], size=2)], size=2) ([B, B],[A, C]) - CombinatoricsVector=([CombinatoricsVector=([B, B], size=2), CombinatoricsVector=([A, C], size=2)], size=2) ([A, C],[B, B]) - CombinatoricsVector=([CombinatoricsVector=([A, C], size=2), CombinatoricsVector=([B, B], size=2)], size=2) ([A, B, B],[C]) - CombinatoricsVector=([CombinatoricsVector=([A, B, B], size=3), CombinatoricsVector=([C], size=1)], size=2) ([C],[A, B, B]) - CombinatoricsVector=([CombinatoricsVector=([C], size=1), CombinatoricsVector=([A, B, B], size=3)], size=2)