COMP 2711H: Lecture 9

Date: 2024-09-23 09:11:39

Reviewed: 2024-10-07 03:13:46

Topic / Chapter: Principle of Inclusion & Exclusion

summary

❓Questions

Notes

Functions

Derangements
- for $[n] = {1, 2, 3, \dots, n}$ , how many $π$ s.t. $\forall i \in [n], π [i] \neq = i$
- $d_{1} = 0; d_{2} = 1;$
- $d_{n} = (n - 1) \cdot [d_{n - 1} + d_{n - 2}]$
  - choose $i$ first, then do casework
- what if: exactly $k$ people, among $n$ , must be in position?
  - how many such $π$ exist?
  - or: exactly $k$ position $i_{1}, \dots, i_{k}$ s.t. $π [i_{j}] = i_{j}$
- $d [n, k] = (k n) d_{n - k}$
  - i.e. deranging $n - k$ people
    - and choosing $k$ people to remain seated (in their own)
  - and for $k$ remaining people: only one way
No. of functions
- how many functions from $[n]$ to $[m]$ are onto?
  - or: surjective (French)
  - i.e. all elements of $[m]$ has mapping?
- first, how many functions w/ $m$ possible input and $n$ possible output exist?
  - if $m = n = 1$ , there is only 1
  - if $m = 3; n = 2$ , there are $n^{m} = 2^{3} = 8$
- and subtract no. of non-onto functions
  - i.e. all mapped to $1$ , or $2$ only
  - = 2 cases
  - $2^{3} - 2 = 6$ onto functions
- how about a recursive formula?
  - first: $m$ choices for input $n$ , let's say $i$ was chosen
    - then, later ones: can either include $i$ or not
  - $O (n, m) = m \cdot (O (n - 1, m) + O (n - 1, m - 1))$
  - or: consider the reverse mapping. What's mapped to $m$ ?
    - i.e. subset of $[n]$ mapped to $m$
    - assume: $k$ inputs are mapped to $m$
    - i.e. $(k n)$ possibilities
  - $O (n, m) = \sum_{k = 1}^{n} (k n) O (n - k, m - 1)$
- also: it's the same as dividing $[n]$ into $m$ non-empty sets
  - based of the output they are mapped to
- let $s (n, m)$ : no. of ways of dividing $[n]$ into $m$ non-empty sets
  - without order
  - 👨‍🏫 same as the ball-and-box problem!
  - if the ball is in its own box:
    - $s (n - 1, m - 1)$
  - if the ball is sharing its bow with others
    - $s (n - 1, m)$
  - $s (n, m) = s (n - 1, m - 1) + s (n - 1, m)$
Problem 1: how many paths do not go through marked segments?
- i.e. segment $A B$ , $C D$ , $EF$ , $H G$
- let $α [i, j]$ : no. of paths to point $(i, j)$
- $α [0, 0] = 1$
- $α [i, j] = path not forbidden \cdot α [i - 1, j] + path not forbidden \cdot α [i, j - 1]$
- but, how many (invalid) paths go through at most exactly $2$ segments?
- simply: count each cases manually
  - going through $A B, C D$ and $C D, EF$
- 👨‍🏫 is there a more elegant formula?
  - $α [i, j, k]$ : no. of paths to point $(i, j)$ , including exact $k$ paths
  - $α [i, j, k] = \neg forbidden \cdot α [i - 1, j, k] + \neg forbidden \cdot α [i, j - 1, k] + forbidden \cdot α [i - 1, j, k - 1] + forbidden \cdot α [i, j - 1, k - 1]$

Principle of Inclusion & Exclusion (PIE)

Principles of inclusion & exclusion
- how many no. $[1..1000]$ are divisible by $5$ ?
  - $1000/5 = 200$
- how many no. $[1..1000]$ are divisible by $2$ ?
  - $1000/2 = 500$
- how many no. $[1..1000]$ are divisible by both $2, 5$ ?
  - i.e. multiple of $10$
  - $1000/10 = 100$
- how many no. $[1..1000]$ are divisible by $2∣∣5$ ?
  - no double counting!
  - $200 + 500 - 100 = 600$
- same as: $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣ - ∣ A B ∣$
- how many no. $[1..1000]$ are divisible by $2∣∣5∣∣7$ ?
  - same as: $∣ A \cup B \cup C ∣ = ∣ A ∣ + ∣ B ∣ + ∣ C ∣ - ∣ A B ∣ - ∣ BC ∣ - ∣ A C ∣ + ∣ A BC ∣$
    - i.e. eliminate double counting within double-counted elements!
    - ( $A B, BC, A C$ )
  - $200 + 500 + 142 - 100 - 28 - 71 + 14 - = 600$
- how many no. $[1..7000]$ are divisible by $2∣∣5∣∣7$ ?
  - same as: $∣ A \cup B \cup C ∣ = ∣ A ∣ + ∣ B ∣ + ∣ C ∣ - ∣ A B ∣ - ∣ BC ∣ - ∣ A C ∣ + ∣ A BC ∣$
    - i.e. eliminate double counting within double-counted elements!
    - ( $A B, BC, A C$ )
  - $2500 + 1400 + 1000 - 700 - 200 - 500 + 100 = 3600$
- how about $k$ sets?
  - finding the pattern $∣ i = 1 ⋃ k A_{i} ∣ = \sum ∣ A_{i} ∣ - i < j \sum ∣ A_{i} A_{j} ∣ + i < j < j_{2} \sum ∣ A_{i} A_{j} A_{j_{2}} ∣ + \dots + + (- 1)^{k + 1} \sum ∣ A_{i} A_{j} \dots A_{k} ∣$
- let $∣ ⋃_{j = 1}^{t} A_{i_{j}} ∣ := ω (i_{1}, i_{2}, \dots, i_{t})$
- $ω (t) := \sum_{i_{1}, \dots i_{t}} ∣ A_{i_{1}} A_{i_{2}} \dots A_{k} ∣$
- then $∣ ⋃_{i = 1}^{k} A_{i} ∣ = \sum_{t = 1}^{k} (- 1)^{t + 1} ω (t)$
- 👨‍🏫 how can we prove it without induction?
  - induction: painful...
- proof: let $x \in ⋃_{i = 1}^{k} A_{i}$ , suppose $x$ is in exactly $r$ sets
  - 👨‍🏫: ensure you count $x$ only once for both sides!
  - if $r = 1$ : it only gets added once in $ω (1)$
  - if $r = 2$ : it only gets counted once in $ω (1) - ω (2)$
  - for general $r$ : gets added
    - no. of being added: $(1 r)$ in $ω (1)$ , $(2 r)$ in $ω (2)$
    - $(1 r) - (2 r) + (3 r) + \dots + (- 1)^{r + 1} (r r)$
  - 👨‍🏫 as proven, there are equal no. of even subsets & odd subsets (including $\emptyset$ )
    - but we don't have $\emptyset$ in out equation (which could have been subtracted)
    - thus, $(1 r) - (2 r) + (3 r) + \dots + (- 1)^{r + 1} (r r) = 1$
      - from $- 1 + (1 r) - (2 r) + (3 r) + \dots + (- 1)^{r + 1} (r r) = 0$
Problem 2: derangements
- how many permutations $π$ of $1.. n$ exist s.t. $\forall i, π [i] \neq = i$ ?
- $A_{i}$ : set of perms $π$ s.t. $π [i] = i$
- solution: $∣ A_{1}^{c} \cap A_{2}^{c} \cap \dots A_{n}^{c} ∣$
  - not in position for all $i \in [1, n]$
- and, as we have total $n!$ perms
  - $∣ A_{1}^{c} \cap A_{2}^{c} \cap \dots \cap A_{n}^{c} ∣ = n! - ∣ A_{1} \cup A_{2} \cup \dots \cup A_{n} ∣$
  - $∣ ⋃_{j = 1}^{t} A_{i_{j}} ∣ = ?$
    - $= n! + \dots$ as there are $n \times (n - 1)!$ permutations
    - $= n! - (2 n) (n - 2)! + \dots$
    - $= n! - (2 n) (n - 2)! + (3 n) (n - 3)! + \dots$
  - $∣ A_{i} ∣ = (n - 1)!$ (as there are $n$ )
  - $∣ A_{i} A_{j} ∣ = (n - 2)!$
    - to be subtracted
  - $∣ A_{i} A_{j} A_{k} ∣ = (n - 3)!$
  - $∣ ⋃_{j = 1}^{t} A_{i_{j}} ∣ = \sum_{i = 1}^{n} (- 1)^{i + 1} (i n) (n - i)!$ $d_{n} n \to \infty lim \frac{( - 1 ) ^{i}}{i !} n \to \infty lim \frac{d _{n}}{n !} = n! - i = 1 \sum n (- 1)^{i + 1} (i n) (n - i)! = n! - i = 1 \sum n (- 1)^{i + 1} \frac{n !}{( n - i )! i !} (n - i)! = n! - i = 1 \sum n (- 1)^{i + 1} \frac{n !}{i !} = n! - \frac{n !}{1 !} + \frac{n !}{2 !} + \dots = i = 0 \sum n (- 1)^{i} \frac{n !}{i !} = n! i = 0 \sum n (- 1)^{i} \frac{1}{i !} = 1/ e = 1/ e or: e^{- 1}$
- as $n$ converges to infinity, probability of picking a derangement: $1/ e$
- $⋃ A_{i}^{c} = ∣ U ∣ + \sum_{t = 1}^{k} (- 1)^{t} w (t)$
  - $-$ : used because exponent of $- 1$ inside sum changed
  - according to PIE
Problem 3: how many functions from $[n]$ to $[m]$ are onto?
- i.e. all element w/ at least 1 mapping?
- let $A_{1}$ : set of functions not covering $1$
- let $A_{i}$ : set of functions not covering $i$
- $∣ A_{1}^{c} \cap A_{2}^{c} \cap \dots \cap A_{m}^{c} ∣ = m^{n} - ∣ ⋃_{i = 1}^{m} A_{i} ∣$
  - $∣ A_{i} ∣ = (m - 1)^{n}$
  - $∣ A_{i} A_{j} ∣ = (m - 2)^{n}$
  - 👨‍🏫 we can do so because we know that size of $A_{i}$ are all same
  - ... and so on; thus: $∣ i = 1 ⋃ m A_{i} ∣ = m \cdot (m - 1)^{n} - (2 m) \cdot (m - 2)^{n} + (3 m) \cdot (m - 3)^{n} = i = 1 \sum m (- 1)^{i + 1} (i m) (m - i)^{n}$
- and, as above is the set of non-onto function $O (n, m) = m^{n} + i = 1 \sum m (- 1)^{i} (i m) (m - i)^{n} = i = 0 \sum m (- 1)^{i} (i m) (m - i)^{n}$

COMP 2711H: Honors Discrete Mathematical Tools for Computer Science