1. Combinatorics

2. Combinatorics is a branch of mathematics concerning the study of finite or countable discrete structures
In the past, ad-hoc solutions were available.
Now, ....

3. Counting Techniques product rule sum rule Inclusion-Exclusion Formula
5 shirts * 4 pants = total 20 combinations
sum rule
if any one is missing from the closet, it is one of 9 clothing pieces
Inclusion-Exclusion Formula
|A  B| = |A| + |B| - |A  B|
|A  B  C | = |A| + |B| + |C| - |A  B| - |B  C| - |B  C| + |A  B  B |
eliminates double counting
Bijection
map a member of a set to a member of target set that is to be counted

4. More counting techniques
Permutation
an arrangement of n items where every item appears exactly once - n!=∏ni=1 i
arrange three letters word using a, b, c
abc, acb, bac, bca, cab, cba, - 6 words (3!)
what if letters are reused? (strings)
aaa, abb, n∏r= nr
Subsets
there are 2n subsets of n items
a, b, c, ab, bc, ac, abc, and an empty set – 8 subsets

5. Recurrence Relations Recursively defined structures
tree, list, WFF, ...
divide conquer algorithms work on them
recurrence relation
an = an  an = n
an = 2an  an = 2n
an = nan  an = n!

6. Binomial Coefficient nCk - choose k things out of n easy one
how many ways to form a k-member committee from n people
Paths across a grid of n X m

7. Binomial Coefficient (2)
coefficient of (a+b)n
(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3
(a + b)3 = aaa + 3aab + 3abb + bbb
think the other way
what is the coeff of akbn-k ?
how many ways to choose k a-terms out of n

8. Pascal’s Triangle 4C3 = 3C2 + 3C3 4C0 4C1 4C2 4C3 4C4
Sum=2n 1 2 4 8 16 32
n=0
n=1
n=2
n=3
n=4
n=5
4C3 = 3C2 + 3C3
4C0
4C1
4C2
4C3
4C4
nCk = (n-1)C(k-1) + (n-1)Ck

9. Binomial Coefficient(3) k
n-1 n … …
case1: this is included
same as choosing k-1 from n-1
case2: not
must choose k from n-1
nCk = (n-1)C(k-1) + (n-1)Ck
(n-k)!k! can cause overflow n k n!
(n-k)!k! =

10. Stop the recurrence good ending points nCk = (n-1)C(k-1) + (n-1)Ck
nC1 = n
nCn = 1

11. safeness time complexity what about 3C0, 2C5, (-3)C5
int bc(int n, int k) {
if (k == 1) return(n);
return bc(n-1, k-1) + bc(n-1, k); }
nCk = (n-1)C(k-1) + (n-1)Ck
safeness
what about 3C0, 2C5, (-3)C5
don’t we need else part?
time complexity
one comparison and one addition (let it be C1)
let recursion overhead be C2
T(n, k) = T(n-1, k-1) + T(n-1, k) + C1 + C2

12. Fibonacci Numbers F0 = 0 F1 = 1 Fn = Fn-1 + Fn-2 for n  2
int fb(int n) {
if (n < 0) ????
if (n < 2)) return(n); // f(0) =0, f(1) = 1
else return fb(n-1) + fb(n-2); }

13. Eulerian Numbers
the number of permutations of the numbers 1 to n in which exactly k elements are greater than the previous element
review Stirling Numbers, Partitions
0..9
<0..8> 에 9를 넣을 수 있는 위치는 마지막 위치만 가능
> k=3
9를 추가해서 k=4로 만들 수 있는 경우의 수는?

14. Induction prove/disprove that Tn=2n-1 n … Tn Tn = 2Tn-1 + 1, T0=0 1 21 2 3 4 5 6 7 … Tn 15 31 63
127
prove/disprove that Tn=2n-1

15. How Many Pieces of Land
Let P(n) be the number of lands when n points are drawn on circle  Now assume knowing P(n-1), want to know P(n) by adding the n-th point  when the n-th point draw a line to the k-th point (1 <= k <= n-1)  on the LHS on k-th point, there is k-1 points between k-th and n-th point  on the RHS on k-th point, there is (n-1)-k+1 points between k-th and n-th point  the points in these 2 sets(LHS, RHS) are completely connected,  so there are (k-1)*(n-k) lines crossed by the newly drawn line,  creating (k-1)*(n-k)+1 new lands......  So, drawing lines from newly added n-th point to all n-1 old points,  will create  summation[(k-1)(j-k)+1] {k = 1..j (j = n-1)} new lands  let says the summation be F(j)  P(n)= P(n-1) + F(n-1)  = P(n-2) + F(n-2) + F(n-1)  .....  = P(0) + F(1) + F(2) F(n-1)  = 1 + summation(F(i)) {i = 1..n-1}  by using formula of summation to simplify the equatoin(what a difficult job.....)  I get  answer = (i - 1) * i * (i * i - 5 * i + 18) / ; 

16. Tower of Hanoi