Combinatorics

University of Akron Programming Team

9/23/2011

PermutationsWays of ordering a set of items.

- OR -

Counting PermutationsDepends on the size of the set S

of items.◦|S| = 1 1 Permutation◦|S| = 2 2 Permutations◦|S| = 3 ?

Counting Permutations

|S| = 3 6 Permutations

Counting Permutations|S| = 3

◦Any of the three items can go in the 1st spot. Any of the remaining two items can go in

the 2nd spot Any of the remaining one items can go into the 3rd

spot.

◦3 options * 2 options * 1 option 6 total options

Counting PermutationsIn general, there are |S|! (factorial)

permutations.

Knowing how quickly factorials grow lets us know whether enumerating all the permutations of a set is reasonable within the confines of a programming competition.

Count ms / perm in 15s

6! = 720 20.83

7! = 5,040 2.976

8! = 40,320 0.372

9! = 362,880 0.0413

10! = 3,628,800

0.00413

11! = 39,916,800

0.0003757

Generating PermutationsBottom Up

If you knew all the permutations of the set {A, B, C}, could you utilize that to quickly generate the permutations of {A, B, C, D}?◦Permutations of {A, B, C}

(A, B, C) (A, C, B) (B, A, C) (B, C, A) (C, A, B) (C, B, A)

Generating PermutationsBottom Up

Easier example: If we know all the permutations of {A}, can we generate all the permutations of {A, B}?◦Permutations of {A}

(A)

◦Let’s add B to the existing permutation. Two options: Add B to the right of A

(A, B) Add B to the left of A

(B, A)

Generating PermutationsBottom Up

Permutations of {A, B}◦ (A, B)◦ (B, A)

Let’s generate the permutations of {A, B, C}◦Using (A, B) as a starting point

Add C to the right: (A, B, C) Add C in the middle: (A, C, B) Add C on the left: (C, A, B)

◦Using (B, A) as a starting point Add C to the right: (B, A, C) Add C to the middle: (B, C, A) Add C to the left: (C, B, A)

Generating PermutationsBottom Up

BottomUpPermutations(List list)◦List<List> results◦Add 1st element of list (as a new list) to results◦for(i = 2 to |list|)

{ List<List> nextLengthResults for(List permutation in results)

{ Add the ith elemnt of list to each position in permutation}

results = nextLengthResults }

◦return results

Generating PermutationsBottom Up – CODE!

public static <T> List<List<T>> BottomUp(List<T> items){

List<List<T>> results = new ArrayList<List<T>>();

List<T> initial = new ArrayList<T>();initial.add(items.get(0));results.add(initial);

for(int i = 1; i < items.size(); i++){List<List<T>> nextLengthResults = new ArrayList<List<T>>();for(List<T> permutation: results){for(int j = 0; j <= permutation.size(); j++){// Add the ith item to the jth position & add that to the nextLengthResultsArrayList<T> tempPerm = new ArrayList<T>(permutation);tempPerm.add(j, items.get(i));nextLengthResults.add(tempPerm);}}

results = nextLengthResults;}

return results;}

Generating PermutationsSpecial Orderings

Minimum Change◦ Each consecutive permutation differs by only one

swap of two items. (1 2 3) (1 3 2) (3 1 2) (3 2 1) (2 3 1) (2 1 3)

Lexicographic order◦ Consider the input list to be in “alphabetic order.”

Then the lexicographic order gives all permutations in combined alphabetic order

◦ Input list: (A B C) (A B C) (A C B) (B A C) (B C A) (C A B) (C B A)

SubsetsPick as many or few items from

this set as you’d like:

Subsets

Counting SubsetsDepends on the size of the set S

of items.◦|S| = 0 1 Subset◦|S| = 1 2 Subsets◦|S| = 2 4 Permutations◦|S| = 3 ?

Counting Subsets

|S| = 3 8 Subsets

Counting Subsets|S| = 3

◦Item 1 can either be part of the subset or not. 2 options

◦Item 2 can either be part of the subset or not. 2 options * 2 options = 4 options

◦Item 3 can either be part of the subset or not. 4 options * 2 options = 8 options

Counting SubsetsIn general, there are 2|S| subsets

(exponential).

Knowing how quickly exponentials grow lets us know whether enumerating all the subsets of a set is reasonable within the confines of a programming competition.

Count ms / perm in 15 s

26 = 64 234.375

210 = 1,024 14.648

214 = 16384 0.9155

218 = 262144 0.05722

222 = 4194304 0.003576

226 = 67108864 0.0002235

Generating SubsetsBottom Up

BottomUpSubsets(List list)◦ If list has 0 elements

return {Ø}

◦ results := new List<List>◦head := first element of the list◦headlessList := list with head removed◦ for(List subset in BottomUpSubsets(headlessList))

{ Add subset to results Add subset + head to results }

◦ return results

Generating SubsetsBottom Up – CODE!

public static <T> List<Set<T>> bottomUp(Set<T> originalSet){

List<Set<T>> result = new ArrayList<Set<T>>();if(originalSet.size() == 0){

result.add(new HashSet<T>());return result;

}

List<T> list = new ArrayList<T>(originalSet);T first = list.get(0`);Set<T> remainder = new HashSet<T>(list.subList(1, list.size()));

for (Set<T> without : bottomUp(remainder)){

Set<T> with = new HashSet<T>(without);with.add(first);result.add(without);result.add(with);

}

return result;}

Generating SubsetsAnother Approach

Can take advantage of bit representations of integers.

Consider S = (A, B, C), using ints in [0, 2|

S|-1]◦0 000 _ _ _◦1 001 _ _ C◦2 010 _ B _◦3 011 _ B C◦4 100 A _ _◦5 101 A _ C◦6 110 A B _◦7 111 A B C

Counting TopicsBinomial coefficients

◦n choose k “k member committee from n people” Alternate notation nCk There are n! / (n-k)!k!) ways. nCk = (n-1)C(k-1) + (n-1)C(k) Ex: “Num paths from (0, 0) to (10, 10) in

plane only making steps in the positive directions.”

◦Pascals Triangle relationship◦Coefficients on (a+b)n

Counting TopicsStirling numbers

◦ First kind – permutations on n with exactly k cycles.

◦ Second kind – ways to partition a set of n objects into k groups.

Catalan numbers◦ Number of ways to balance n sets of parentheses

Cn = 1/(n+1) * (2nCn)

Eulerian Numbers◦ Number of permutations of length n with k

ascending sequences.Solving recurrence relations for closed form

solutions.

Other Combinatorics ProblemsPermutations with duplicate

elements◦(A, A, B)

(A, A, B) (A, B, A) (B, A, A)

“Strings” of length n on string s◦Length 2 over “ab”

“aa”, “ab”, “ba”, “bb”

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