COMP9020 – Foundations of Computer Science

FINAL Exam – Term 3 2022

From Question 5 to Question 10

**With my own solution (might be incorrect)**



# Question 5

![image.png](/static/img/8e2ec7362acb9d23df7bdf3f1f162615.image.webp)

## 5.1

:::info{title="Question"}
Prove or disprove:

G has an Eulerian path
:::

Here is the degree list for all of the vertices:

- vertex 0: 3
- vertex 1: 2
- vertex 2: 2
- vertex 3: 4
- vertex 4: 4
- vertex 5: 2
- vertex 6: 2
- vertex 7: 3

We could find that there are 2 vertices (0 and 7) whose degree is odd. This could prove that there is an Eulerian path in G.

For example:

0 → 2 → 3 → 1 → 0 → 4 → 5 → 7 → 6 → 4 → 3 → 7

## 5.2

:::info{title="Question"}
Prove or disprove:

G has a Hamiltonian cycle
:::

If it has a Hamiltonian cycle, it means that no matter which vertex I start, I could find a cycle. Because this cycle should get every vertex.

Divide these vertices into 2 subsets: {0, 1, 2, 3} and {4, 5, 6, 7}. Edges (0, 4), (3, 4) and (3, 7) are linking these 2 subsets.

Set start point at vertex 1, and we could go to vertex 0 or vertex 3.

If we go to vertex 0, we could go to vertex 2 or vertex 4 for next step.

If we go to vertex 2, we must go to vertex 3 for next step, but it means that we could not come back from {4, 5, 6, 7} because we should use vertex 0 or vertex 3 to go back and get vertex 1 at the last step.

So we should return, and go to vertex 4 when we are at vertex 0. However, after a trip in {4, 5, 6, 7}, we can only go back by vertex 3, and then we must choose to go to vertex 2 and get vertex 0. This is also a bad path.

So we return again, and go to vertex 3 when we are at vertex 1. However, it is same when we come back from {4, 5, 6, 7}. we could only choose vertex 0 to come back, but we have to go to vertex 2 and reach the vertex 3 in the end.

In conclusion, after the path finding, we could prove that this graph does not have a Hamiltonian cycle.

## 5.3

:::info{title="Question"}
Prove or disprove:

The chromatic number of G is equal to the clique number of G.  That is, χ(G) = κ(G).
:::

The clique number should be the vertex number of the smallest complete sub-graph in G. 

In graph G, we could only find 2-clique (an edge and 2 vertices at 2 sides). There is not any tri-angle (3-complete graph) in G. Therefore, $κ(G) = 2$.

Set vertex 0 in BLACK. And we could find that:

- vertex 0: BLACK
- vertex 1: WHITE
- vertex 2: WHITE
- vertex 3: BLACK
- vertex 4: WHITE
- vertex 5: BLACK
- vertex 6: BLACK
- vertex 7: WHITE

2 colors(black & white) could color graph G. Therefore, $χ(G) = 2$.

We could prove that:

$$
χ(G) = κ(G)
$$


# Question 6

Let Σ = {a,b,c}

Consider the function del:Σ* → Σ* which works as follows:

For w∈Σ*, del(w) is the word that results from removing all c's from w.

For example:

- del(aabbcc) = aabb
- del(ccbbaacc) = bbaa
- del(bacabc) = baba
- del(cccc) = λ
- del(bbbb) = bbbb

## 6.1

:::info{title="Question"}
Give a recursive definition of the function del:Σ* → Σ*
:::

- $del(\lambda) = \lambda$
- $del(cx) = del(x)$
- $del(mx) = m + del(x)$ if $m \in {a,b}$

``` Python
def del(w):
    if not w:
        return w
    if w[0] == 'c':
        return del(w[1:])
    else:
        return w[0] + del(w[1:])
```

## 6.2


:::info{title="Question"}
Let Σ={a,b,c}.

For w∈Σ*, let P(w) be the proposition that:

For all $v \in Σ^*$: $del(wv) = del(w)del(v)$

Prove, using your definition of del from part (a), that P(w) holds for all w∈Σ*.
:::


When $w = \lambda$, P(w) could be:

For all $v \in Σ^*$: $del(v) = del(v)$

It is true.


When $w = w_1w_2...w_k$,

Set: $Σ' = {a,b}$, and single element $w_i$ could be: $Σ'^*c^*Σ'^*$.

$del(wv) = del((Σ'^*c^*Σ'^*)^*v) = (Σ'^*Σ'^*)^*del(v)$

In addition, $del(w) = (Σ'^*Σ'^*)^*$

Therefore, we could prove that:

P(w) holds for all w∈Σ*.


# Question 7

Consider the following two pieces of code.

The first moves the minimum element of a list to its head

```
extractMin(L):	 
    if length(L) <= 1: return L
 	 
    x = L.head	 
    L2 = extractMin(L.next)	 
    if x < L2.head:	 
        return L	 
    else:	 
        L3 = (x, L2.next)	 
        return (L2.head, L3)
```

The second sorts a list by repeatedly extracting the minimum element using `extractMin`

```
mySort(L):	 
    i = 0	 
    L2 = empty
    L3 = L	 
 	 
    while i < length(L):	 
        L3 = extractMin(L3) 
        L2 = (L3.head, L2)	 
        L3 = L3.next	 
        i = i+1	 
 	 
    return L2
```

## 7.1


:::info{title="Question"}
Find an asymptotic upper bound for T(n): the running time of the worst-case of `extractMin(L)` where L is a list with n elements.

You may assume that creating a list is an elementary operation, as is checking the length of L.
:::

The worst case is that:

for every element in position `i`, it is the least element compared with all elements before. function should move `n-i` times to make it to the top of this list.

$$
T(n) = T(n-1) + O(n)
$$

And the sum is:

$$
T(n) = O(n) + O(n-1) + ... + O(1) = O(\frac{n(n+1)}{2}) \\
T(n) = O(n^2)
$$


## 7.2

:::info{title="Question"}
Find an asymptotic upper bound for R(n): the running time of the worst-case of `mySort(L)` where L is a list of n elements.

You may assume that creating a list is an elementary operation, as is checking the length of L.  You may assume the running time of `extractMin` on a list of length n is T(n) [if not previously calculated].
:::

We have proved that the worst-case of `extractMin(L)` is: $O(n^2)$.

And for `mySort(L)`, it use $O(n)$ time to execute `while` command, and every loop it executes `extractMin(L)` to calculate the least element.

Therefore, the worst upper bould for $R(n) = O(n^2 \times n) = O(n^3)$.


# Question 8

A spaceship is full of crewmates and imposters.  Crewmates always tell the truth, and imposters always lie.  Everyone is either a crewmate or an imposter.  Three members of the spaceship, Red, Green and Blue, make the following statements:

- Red: Blue is not an imposter
- Green: At least one of Blue or Green is a crewmate
- Blue: Green is an imposter

You would like to know if at least one of Red or Blue is an imposter.

## 8.1

:::info{title="Question"}

Explain how this problem can be modelled as a problem in Propositional Logic.  In particular, clearly define:

- The propositional variables you are using and what propositions they represent
- Any formulas that are appropriate, and what propositions they represent
- The Propositional Logic problem to be solved
:::

We could define that:

- $R$：Red is crewmate
- $G$: Green is crewmate
- $B$: Blue is crewmate

According to the statement of Red, if Red is crewmate, we could find that Blue is crewmate. If Red is not crewmate, Blue is not crewmate. That means:

$$
R \rightarrow B\\
\neg R \rightarrow \neg B \\
R \leftrightarrow B
$$

According to the statement of Green, there is at least one crewmate between Green and Blue.

$$
\neg B \vee \neg G

$$

According to the statement of Blue, if Green is a crewmate, Blue is an imposter; if Green is an imposter, we could find that this statement is true, and Blue is crewmate. That means:

$$
\neg G \rightarrow B \\
G \rightarrow \neg B \\
\neg G \leftrightarrow B
$$


## 8.2

:::info{title="Question"}
Using the Propositional Logic model of part (a), solve the given problem.
:::

We have that:

- $R \leftrightarrow B$
- $\neg G \leftrightarrow B$
- $\neg B \vee \neg G$

According to first two statement, we could find that:

- $R \leftrightarrow \neg G$

Assum: R is true. 

- According to $R \leftrightarrow B$, B is true.
- According to $R \leftrightarrow \neg G$, G is false.
- $\neg B \vee \neg G$ is true. However, it should be false because Green is imposter.

Assum: R is false.

- According to $R \leftrightarrow B$, B is false.
- According to $R \leftrightarrow \neg G$, G is true.
- $\neg B \vee \neg G$ is true.

Therefore, there is a solution for this spaceship:

- Red is imposter
- Green is crewmate
- Blue is imposter


# Question 9

## 9.1


:::info{title="Question"}

Suppose you have:

- x indistinguishable red balls
- y indistinguishable blue balls
- z indistinguishable green balls

Show that there are:

$$
\binom{x+y+z}{z}\binom{x+y}{y}
$$

ways of placing the balls into x+y+z distinguishable boxes with exactly one ball per box.
:::

We could use 3 steps to place all the balls:

1. select z empty boxes and place all of the green balls.
2. select y empty boxes and place all of the blue balls.
3. place all of the red balls into the rest boxes.

Because every ball is indistinguishable, we do not need to consider the order of placing between same color balls.

For the first step, we have $T_1$ ways:

$$
T_1 = C_{x+y+z}^{z} = \binom{x+y+z}{z}
$$

For the second step, we have $T_2$ ways:

$$
T_2 = C_{x+y}^{y} = \binom{x+y}{y}
$$

And the last step do not need any extra calcuation. Because after step 1 and step 2, there is just one option for step 3 to put balls.

In addition, the selection in step 1 could effect selection in step 2. Because every box is distinguishable. Therefore, every selection is unique.

So we could prove that the way of placing balls are:

$$
T_1 \times T_2 = \binom{x+y+z}{z}\binom{x+y}{y}
$$


## 9.2

:::info{title="Question"}
Show that for all x,y,z ∈ ℕ:

$$
\binom{x+y+z}{z}\binom{x+y}{y} = \binom{x+y+z}{x}\binom{y+z}{y}
$$
:::

If we have the same question in 9.1, we could change the steps in another way:

1. select x empty boxes and place all of the red balls.
2. select y empty boxes and place all of the blue balls.
3. place all of the green balls into the rest boxes.

Because every ball is indistinguishable, we do not need to consider the order of placing between same color balls.

For the first step, we have $T_3$ ways:

$$
T_3 = C_{x+y+z}^{x} = \binom{x+y+z}{x}
$$

For the second step, we have $T_4$ ways:

$$
T_4 = C_{y+z}^{y} = \binom{y+z}{y}
$$

And the last step do not need any extra calcuation. Because after step 1 and step 2, there is just one option for step 3 to put balls.

In addition, the selection in step 1 could effect selection in step 2. Because every box is distinguishable. Therefore, every selection is unique.

So we could prove that the way of placing balls are:

$$
T_3 \times T_4 = \binom{x+y+z}{x}\binom{y+z}{y}
$$

We use different ways to solve same question, and the result should be equal.

Therefore, it should be:

$$
T = T_1 \times T_2 = T_3 \times T_4
$$

It means that:

$$
\binom{x+y+z}{z}\binom{x+y}{y} = \binom{x+y+z}{x}\binom{y+z}{y}
$$


# Question 10


Let $Σ = \{a,b,c\}$

Let $p(k, n)$ be the probability that a word (over Σ) of length n, drawn uniformly at random from the set of all words of length n, contains exactly k c's.


## 10.1

:::info{title="Question"}
What is p(0,n)? 

Justify your answer
:::

For every character in a n-word, the possibility that it is not 'c' is $\frac{2}{3}$.

Therefore, $p(0,n) = (\frac{2}{3})^n$.

## 10.2

:::info{title="Question"}
What is p(n,n)? 

Justify your answer
:::

For every character in a n-word, the possibility that it is 'c' is $\frac{1}{3}$.

Therefore, $p(n,n) = (\frac{1}{3})^n$.

## 10.3

:::info{title="Question"}
Give, with justification, an expression for p(k,n), either:

- in terms of combinatorial functions, or
- recursively, in terms of p(k', n') where k'≤k and n'≤n
:::

We use combinatorial functions to solve this question.

For a n-word, we should select k characters to set 'c', and other characters are not 'c'.

Therefore, we could find that:

$$
p(k,n) = \binom{n}{k} \times (\frac{1}{3})^k \times 1 \times (\frac{2}{3})^{n-k}
$$

It is:

$$
p(k,n) = \binom{n}{k} \times \frac{2^{n-k}}{3^k}
$$


By the way, the recursive way is:

$$
p(k,n) = \frac{1}{3}p(k-1,n-1) + \frac{2}{3}p(k, n-1)
$$

## 10.4


:::info{title="Question"}
What is the expected number of c's in a word of length n, drawn uniformly at random from the set of all words (over Σ) of length n?
:::

For every character, $c_i$, in n-word, the possibility of 'c', $p_i$, is $\frac{1}{3}$.

The expected number of c is:

$$
\sum_{i=1}^n p_i \times length(w_i) = \frac{1}{3} \times 1 + \frac{1}{3} \times 1 ... + \frac{1}{3} \times 1 = \frac{n}{3}
$$

COMP9020 Sample Exam 2022 T3 Part 2

COMP9020 – Foundations of Computer Science

FINAL Exam – Term 3 2022

From Question 1 to Question 4

**With my own solution (might be incorrect)**



# Part 1

It seems that I can't prove this question perfectly.

It is confusing.

## 1.1

:::info{title="Question"}
Prove that for all $m,n,p ∈ ℕ$ :

If $lcm(m,p) = lcm(n,p)$, then $lcm(m+n,p) \le lcm(n,p)$.
:::

I can't prove it because I think it is incorrect. $lcm(m+n,p)$ should greater or equal $lcm(n,p)$.

The counter example is in Part 1.2



## 1.2

:::info{title="Question"}

Find a counter example to show that it is not the case that for all $m,n,p ∈ ℕ$ :

If $lcm(m,p) = lcm(n,p)$, then $lcm(m+n,p) = lcm(n,p)$.
:::

Assum:
- $m = 1$
- $n = 4$
- $p = 8$

Then:
- $lcm(m,p) = 8$
- $lcm(n,p) = 8$
- $lcm(m+n,p) = 40$

Therefore:

$lcm(m+n, p) \ne lcm(n,p)$


## 1.3

:::info{title="Question"}

Prove that for all $m,n,p ∈ ℕ$ :

If $lcm(m,p) = lcm(n,p)$, then $lcm(m+n,p) | lcm(n,p)$.
:::

# Part 2

## 2.1

:::info{title="Question"}

Prove, or find a counter example to disprove, for all sets A, B, C:

$A \cap (B \backslash C)^C=(A \cap C) \cap (A \cap B^C)$
:::

Definition of \\: $A\backslash B = A∩Bᶜ$:

$$
A \cap (B \backslash C)^C = A \cap (B \cap C^C)^C
$$

De Morgan's, C over ∩: $(A∩B)^C = A^C \cap B^C$:

$$
A \cap (B \cap C^C)^C = A \cap (B^C \cup C)
$$

Distributivity of ∩ over ∪: $A∩(B∪C) = (A∩B)∪(A∩C)$:

$$
A \cap (B^C \cup C) = (A \cap B^C) \cup (A \cap C)
$$

Therefore, it is incorrect.


## 2.2

:::info{title="Question"}
Let Σ = {0,1}.

Prove, or find a counter example to disprove that for all X,Y ⊆ Σ*:

$(X^*Y)^* \subseteq (X \cup Y)^*$
:::

Assum:

$(X^*Y)^* = w_1w_2...w_i$

Every element $w_i = x_{i1}x_{i2}...x_{ik}y_i$

and it could also be that:

$x_{i1}x_{i2}...x_{ik}y_i = (x_i \cup \empty)^* \cup (\empty \cup y_i) \subseteq (X \cup Y)^*$

Therefore, we could find that every element $w_i$ in string $(X^*Y)^*$ could be restructed by $(X \cup Y)^*$. 

Proved:

$$
(X^*Y)^* \subseteq (X \cup Y)^*
$$

# Part 3

Let F denote the set of all well-formed formulas.  Define the relation $R ⊆ F×F$ as follows:

$$
(\varphi, \psi) \in \mathbb{R} \text{ if and only if } (\varphi \rightarrow \psi) \text{ is satisfiable}
$$

## 3.1

:::info{title="Question"}
Prove that for all $\varphi, \psi \in F$,

If $\varphi \models \psi$ then $(\varphi, \psi) \in R$.
:::

Because $\varphi \models \psi$, when $\varphi$ is true, the $\psi$ is true. 

$$
\varphi \rightarrow \psi = \neg \varphi \lor \psi
$$

When $\varphi = true$, $\neg \varphi \lor \psi = true$.

When $\varphi = false$, $\neg \varphi \lor \psi = true$.

Because $\varphi \rightarrow \psi$ could be $true$, it means that $\varphi \rightarrow \psi$ is satisfiable.

Therefore, $(\varphi, \psi) \in R$.


## 3.2
:::info{title="Question"}
Prove or disprove:

R is reflexive

You may use the result of part (a) [without proof] if it helps
:::

If we want to prove that R is reflexive, we should prove that for every element $\varphi \in F$, $(\varphi, \varphi) \in R$.

It is easy to find that:

When $\varphi$ is $true$, $\varphi$ is $true$.

It means that:

$$
\varphi \models \varphi
$$

According to the result of part (a), if $\varphi \models \varphi$ then $(\varphi, \varphi) \in R$.

Therefore, we could prove that R is reflexive.

## 3.3

:::info{title="Question"}
Prove or disprove:

R is transitive

You may use the result of part (a) [without proof] if it helps
:::

Assum:

- $X \models Y$
- $Y \models Z$

According to the result of part (a), we could find that:

$$
(X,Y) \in R \\
(Y,Z) \in R
$$

Because $X \models Y$, $Y models Z$, when $X = true$, we could find that $Y = true$ and $Z = true$.

Therefore, we could find that $X \models Z$.

According to the result of part (a), we could find that:

$$
(X,Z) \in R
$$

Because:

$$
(X,Y) \in R \\
(Y,Z) \in R \\
(X,Z) \in R
$$

We could prove that R is transitive.


# Part 4

## 4.1

:::info{title="Question"}
Arrange the following functions in increasing order of asymptotic complexity (i.e. if f ∈ O(g) then f appears before g in the list

(For recursively defined functions, assume T(0) = T(1) = 1).

- $T(n) = T(n/2) + log(n)$
- $T(n) = 4T(n/4) + 4n$
- $T(n) = T(n - 2) + 2n$
- $T(n) = T(n/3) + \sqrt{3n}$
- $\sqrt[3]{n^2+2n-1}$
- $\frac{100}{n}+\frac{n}{100}$
:::

- $T(n) = T(n/2) + log(n)$

    Use Main Theorem, we could find that:
    $$
    a = 1 \\
    b = 2 \\
    f(n) = log(n) \\
    n^{log_ba} = n^{log_21} = n^0 = 1
    $$
    
    Because $f(n) = O(log(n)) = O(n^{log_21}log^1(n))$, we could prove that:
    
    $T(n) = O(n^{log_21}log^{1+1}(n)) = O(log^2(n))$.
    
- $T(n) = 4T(n/4) + 4n$

    Use Main Theorem, we could find that:
    $$
    a = 4 \\
    b = 4 \\
    f(n) = 4n \\
    n^{log_ba} = n^{log_44} = n^1 = n
    $$

    Because $f(n) = 4n = O(n) = O(n^{log_ba}log^0(n))$, we could prove that:
    
    $T(n) = O(n^{log_ba}log^{0+1}(n)) = O(nlog(n))$.
    
- $T(n) = T(n - 2) + 2n$

    Every time the $n$ decrease by 2. Therefore, the recursion time is $n/2$.
    
    Every recursion we have to execute $2n$ times.
    
    Therefore, $T(n) = \frac{n}{2} \times 2n = O(n^2)$.

- $T(n) = T(n/3) + \sqrt{3n}$

    Use Main Theorem, we could find that:
    
    $$
    a = 1 \\
    b = 3 \\
    f(n) = \sqrt{3n} \\
    n^{log_ba} = n^{log_31} = n^0 = 1
    $$
    
    Use Main Theorem Case 3, we could prove that:
    
    $T(n) = O(n^{0.5})$.

- $\sqrt[3]{n^2+2n-1}$

    We could find that:$\sqrt[3]{n^2+2n-1} = O(n^{2/3})$

- $\frac{100}{n}+\frac{n}{100}$

    We could find that: $\frac{100}{n}+\frac{n}{100}= O(n + 1/n) = O(n)$ 


This is the list of functions:

1. $O(log^2(n))$: $T(n) = T(n/2) + log(n)$
2. $O(n^{1/2})$: $T(n) = T(n/3) + \sqrt{3n}$
3. $O(n^{2/3})$: $\sqrt[3]{n^2+2n-1}$
4. $O(n)$: $\frac{100}{n}+\frac{n}{100}$
5. $O(nlogn)$: $T(n) = 4T(n/4) + 4n$
6. $O(n^2)$: $T(n) = T(n - 2) + 2n$

## 4.2


:::info{title="Question"}

Consider the function f:ℕ→ℕ defined as:

$$
f(n) = 3n + 2 \text{   (if n is even)} \\
f(n) = n^2 \text{   (if n is odd)}
$$

Which of the following properties does f satisfy?

Select all that apply:

- f∘f (n) = (f(n))² for all n
- f(n) ∈ O(n²)
- f is injective
- f(n) ≤ f(n+2) for all n
- f is surjective
- f(n) ≤ f(n+1) for all n
- f(n) ∈ O(n)
:::

Correct:

- f(n) ∈ O(n²)
- f is injective
- f(n) ≤ f(n+2) for all n

COMP9020 Sample Exam 2022 T3 Part 1

Test started Apr 17, 2024 6:00 PMtoApr 24, 2024 6:00 PM




## 1.1

### Question


A quiz consists of 10 multiple choice questions, each question has 4 options.  You score +1 for each correct answer, and -0.5 for each incorrect answer.

Suppose you choose an answer uniformly at random for each question.  What is your expected score?

Enter your answer to 2 d.p. (answers within ±0.01 will be marked correct)

### Answer

对于每一道题，有0.25的概率答对，0.75的概率答错，因此每一道题的期望得分是：

$$
E(X) = 0.25 \times 1 + 0.75 \times (-0.5) = -0.125
$$

对于每一道题，它的期望都是一致的。因此，十道题的得分期望是：

$$
E(X_1+...+X_10) = E(10 \times X_1) = 10\times E(X_1) = -1.25
$$

## 1.2

### Question

Suppose A, B, C, D are events with A⊥B, B⊥C and C⊥D.

True or false: It must be the case that D⊥A.

### Answer

当事件 A, B, C, D 之间存在相互独立的关系时，如 A⊥B, B⊥C, C⊥D，这只意味着某些特定事件之间独立。独立性不具有传递性，即如果事件 A 独立于事件 B，并且事件 B 独立于事件 C，这并不意味着事件 A 必然独立于事件 C。

同理，尽管 B 独立于 C，且 C 独立于 D，我们不能推断出 A 必然独立于 D，除非有额外的信息表明这种独立性成立。这种独立关系不是通过中介事件自动传递的。因此，没有具体证据表明 A 和 D 独立，我们不能假设它们独立。

FALSE

## 1.3

### Question

Suppose we have three urns with black and white balls distributed as follows:

- Urn A has 10 black balls and 20 white balls
- Urn B has 10 black balls and 10 white balls
- Urn C has 10 black balls and 1 white ball.

Suppose we choose an urn (uniformly at random) and draw a ball (uniformly at random) from that urn. What is the probability that the ball is white?

### Answer

分别计算。

1. 抽中A中的白球：

$$
\frac{1}{3} \times \frac{20}{30} = \frac{2}{9}
$$

2. 抽中B中的白球：

$$
\frac{1}{3} \times \frac{10}{20} = \frac{1}{6}
$$

3. 抽中C中的白球：

$$
\frac{1}{3} \times \frac{1}{11} = \frac{1}{33}
$$

总概率即：

$$
\frac{2}{9} + \frac{1}{6} + \frac{1}{33} = \frac{83}{198}
$$

## 1.4

### Question

Let P be a probability distribution, and A and B events such that P(A) ,P(B), P(A|B) and P(B|A) are all non-zero.

True or false: $P(A) = P(B)$ if and only if $P(A|B) = P(B|A)$

### Answer

我们可以来进行一下推理。


$$
P(A|B) = \frac{P(A \cap B)}{P(B)}
$$

$$
P(B|A) = \frac{P(B \cap A)}{P(A)}
$$

由于 所有变量均不为0，因此充分性和必要性都得以证明。

TRUE

## 2.1

### Question

True or false:

For any random variable X, the probability that X>E(X) is 1/2

### Answer

X可以是非对称分布。

FALSE

## 2.2

### Question

One way to simulate a die roll with a fair coin is as follows. Toss the coin three times:

- If the coin shows HHH: Output 1
- If the coin shows HHT: Output 2
- If the coin shows HTH: Output 3
- If the coin shows HTT: Output 4
- If the coin shows THH: Output 5
- If the coin shows THT: Output 6
- If the coin shows TTH or TTT: Repeat the process.

It was shown in lectures that each of the six outputs is equally likely, with probability 1/6.

A trial consists of three coin tosses. Note that if the coin shows TTH or TTT there will be more than one trial.

What is the expected number of trials before an output is produced?
Enter your answer to 3 d.p.s. An answer within ±0.001 will be correct.

### Answer

我们先对这个期望进行计算。

$$
E(X_n) = \sum_{i=1}^{∞} i \times \frac{6}{8} \times (\frac{2}{8})^{i-1} \\
=\sum_{i=1}^{∞} i \times \frac{3}{4} \times (\frac{1}{4})^{i-1} \\
=\sum_{i=1}^{∞} \frac{3i}{4^i} \\
=\sum_{i=1}^{∞} \frac{3(i-1) + 3}{4^{i-1} \times 4} \\
=\sum_{i=1}^{∞} \frac{1}{4} \times [\frac{3(i-1)}{4^{i-1}}+\frac{3}{4^{i-1}}] \\
=\frac{1}{4}\sum_{i=1}^{∞}\frac{3(i-1)}{4^{i-1}} + \frac{1}{4}\sum_{i=1}^{∞} \frac{3}{4^{i-1}} \\
=\frac{1}{4}E(X_n)+3 \times \sum_{i=1}^{∞}\frac{1}{4^i}
$$

这个几何级数 $\sum_{i=1}^{∞}\frac{1}{4^i}$ 收敛于 $\frac{1}{3}$。因此，原式可表达为：

$$
E(X_n) = \frac{1}{4}E(X_n) + 3 \times \frac{1}{3}
$$

计算得到：

$$
E(X_n) = \frac{4}{3}
$$

另外，GPT给出了一种更简单的证明思路：

![image.png](/static/img/c902fc38f4b6f4763c7282bdc645f683.image.webp)

因此，答案为：

1.333

## 2.3

### Question

Suppose two normal, six-sided dice are rolled, a red die and a black die.

Let:

- A be the event that both dice show 4 or higher;
- B be the event that the red die shows 4 or lower; and
- C be the event that the black die shows 4.

True or false: A and (B∩C) are independent events.

### Answer

判断是否为独立事件，最简单的方法就是判断它们同时发生的概率是否等于它们各自发生的概率的乘积，即：

$$
P(A \cap (B \cap C)) = P(A) \times P(B \cap C)
$$

首先我们可以意识到，B和C确实是独立事件，它们是不同的骰子，因此互不影响。

$$
P(A) = \frac{3}{6} \times \frac{3}{6}= \frac{1}{4}
$$

$$
P(B) = \frac{4}{6} = \frac{2}{3}
$$

$$
P(C) = \frac{1}{6}
$$

因此，我们可以得到：

$$
P(B \cap C) = \frac{2}{3} \times \frac{1}{6} = \frac{1}{9}
$$

同时，我们可以注意到，A、B、C同时发生时，需要两个骰子都显示4。

则：

$$
P(A \cap (B \cap C)) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} = P(A) \times P(B \cap C)
$$

因此，答案为：

TRUE

## 2.4

### Question


You are in a maze with 4 rooms. To navigate the maze you decide to wander randomly with the following strategy:

- In room A you will exit with probability 1/3, move to room B with probability 1/3, and move to room C with probability 1/3.
- In room B you will move to room D with probability 1/3, and room A with probability 2/3.
- In room C you will move to room D with probability 2/3, and room A with probability 1/3.
- In room D you will exit with probability 1/3, move to room B with probability 1/3, and move to room C with probability 1/3.

Starting from Room A, what is the probability that you will exit the maze from Room D?

Enter your answer to 3 d.p.s. An answer within ±0.001 will be correct.

### Answer

我们设置：$P_X$ 为从房间 $X$ 开始逃脱的概率。

我们直接将从A离开的可能性置为0（因为它没有去任何其他的房间），从D离开的可能性置为1（我们要找的）。

那么：

$$
P_A = \frac{1}{3} \times 0 + \frac{1}{3} P_B + \frac{1}{3} P_C
$$

$$
P_B = \frac{2}{3} P_A + \frac{1}{3} P_D
$$

$$
P_C = \frac{1}{3} P_A + \frac{2}{3} P_D
$$

$$
P_D = \frac{1}{3} \times 1 + \frac{1}{3} \times P_B + \frac{1}{3} \times P_C
$$

联立方程代入解得 $P_A = \frac{1}{3} = 0.333$

COMP9020 Quiz 9

- Random Variables and Expectations
- Linearity of Expectation
- Expected Time to Success
-  Standard Deviation and Variance



# 随机变量与期望 Random Variables and Expectatiions

## 随机变量

整数随机变量X是一个从 $ Ω $ 到 $ ℤ $ 的函数。换句话说，它将每一个结果都与一个数值关联起来。

- 随机变量通常用 X, Y, Z 等字母表示。
- 我们以自然的方式将算术运算扩展到随机变量上。

给定随机变量 $X: Ω → ℤ$ 和随机变量 $Y: Ω → ℤ$ 以及整数 $k$，我们可以将 X、Y 和 k 组合起来，得到以下在所有 $ω ∈ Ω$ 上的函数：

- 变量加法：$X + Y ：ω ↦ X(ω) + Y(ω)$
- 变量乘法：$X · Y ：ω ↦ X(ω) · Y(ω)$
- 标量加法：$X - k ：ω ↦ X(ω) - k$
- 标量乘法：$kX ：ω ↦ k · X(ω)$

![image.png](/static/img/f57e86fa1a3c2b75211e040e50f29945.image.webp)

## 期望

随机变量 X 的期望值（通常称为“期望”或“平均值”）是通过下面的公式计算的：

$$
E(X) = \sum_{k \in Z}P(X=k) \times k
$$

期望是一个真正普遍的概念；它是所有决策、估计盈亏、在风险下行动的基础。在概率论概念出现之前，期望值这一基本概念就已经存在了。


# 期望值的线性 Linearity of expectation

对于任意随机变量 X 和 Y，以及整数 k，有：

- $E(X+Y) = E(X) + E(Y)$
- $E(k \times X) = k \times E(X)$

![image.png](/static/img/3261f8215dfd47b23f4531ecdd927941.image.webp)

## 观察可得 Observations

如果 $X_1$，$X_2$，...，$X_n$ 是独立同分布的随机变量，那么：

$$
E(X_1+X_2+...+X_n) = E(nX_1) = nE(X_1)
$$

当随机变量独立同分布时，它们和的期望值等于任意一个随机变量期望值的n倍。

尽管 $X_1+X_2+...+X_n$ 与 $nX_1$ 期望值相同，但这两个表达式代表的随机变量本身是非常不同的。

![image.png](/static/img/576a4c257fe0aa75721be06467249c64.image.webp)


# 成功的期望时间 Expected Time to Success

几何级数 & 递归

![image.png](/static/img/d67fa39d432e275dafd6cc8e0e62dbf7.image.webp)

一个更简单但同样有效的论证是，你预期在一次抛掷中得到“一半”的正面，所以你应该期望在两次抛掷中得到一个“完整”的正面。

## 成功与期望值

问题：高成功概率是否导致高期望值？

一般来说，不是这样。例如：

- 购买更多的彩票可以增加中奖的机会，但是中奖的期望却降低了。

## 赌徒陷阱 Gambler's ruin

很多所谓的“获胜系统”声称提供一个获胜策略，实际上它们提供了一个频繁获得适度胜利的方案，但代价是偶尔会有非常大的损失。

事实上，有一个正式的定理表明，没有任何系统可以将一个“不公平”的游戏转变为一个“公平”的游戏。在决策理论中，“不公平”指的是那些每个个别投注都有负期望值的游戏。

很容易验证，无论是在轮盘赌上的个别投注，购买彩票，或者任何商业提供的游戏上的投注，都具有负的期望值。


# 标准差与方差

对于随机变量X，其期望值为 $μ = E(X)$，标准差为：

$$
\sigma = \sqrt{E((X-μ)^2)}
$$

方差为：

$$
\sigma^2
$$

标准差和方差度量了随机变量的值分散程度。方差越小，我们越能确信对于随机选择的 $ω$，有：

$$
X(ω) = E(X)
$$

并且，方差可以计算为：

$$
E((X-μ)^2) = E(X^2) - μ^2
$$

![image.png](/static/img/877297034b87be3d2de1850da18bef9a.image.webp)

COMP9020 10.1 Statistics

quiz8



## 1.1

### Question

In a class of 200:

- 80 are funny
- 130 are clever
- 42 are neither funny nor clever
How many are both funny and clever?

### Answer


$$
|F \cup C| = |F| + |C| - |F \cap C|
$$

因此，答案为：52

## 1.2

### Question

How many integers are there between 1 and 1000 which are divisible by 6 or 15 but not both?

### Answer
$$
|A \cup B| = |A| + |B| - |A \cap B|
$$

- 集合A的大小，即1到1000之间可以被6整除的数的数量。（199）
- 集合B的大小，即1到1000之间可以被15整除的数的数量。（66）
- 集合A和B的交集的大小，即1到1000之间可以被30整除的数的数量。（33）


因此，答案为：166（因为33被计算了两次，所以199+66-33*2）

## 1.3

### Question

Let S be a set of size n, and fix a constant k>1 (i.e k does not depend on n). Which of the following gives the best asymptotic upper bound for the number of subsets of S of size k?

### Answer

集合不考虑顺序，因此只需要在n里选出k个元素，即 $(^n_k)$。

$$
(^n_k)=\frac{(n-k)!}{k!}
$$

同时，由于k是常数，所以直接忽略。且：$(n-k)! \rightarrow O(n^k)$

因此，答案为：$O(n^k)$。

## 1.4

### Question

Suppose I place 2 red balls, 3 green balls, and 4 blue balls into a hat, and draw two balls without replacement.

What is the probability they are both blue?

### Answer

第一个球是蓝球的概率是4/9，满足第一次的情况下出现第二个蓝球的概率是4-1/9-1=3/8

因此，答案为：1/6

## 2.1

### Question

How many ways are there of choosing an ordered sequence of three letters (with no replacements) from 'SYDNEY'?

### Answer

计算三个字母的有序序列的总数，我们需要分四种情况来考虑：

- 序列中没有"Y"。
- 序列中有一个"Y"。
- 序列中有两个"Y"。

第一种情况，我们要从"SDNE"中选择三个，并组成有序序列，共有：$C_4^3 \times A^3_3 = 4 \times 6 = 24$。

第二种情况，我们要从"SDNE"中选择两个，并和另一个Y组成有序序列，共有：$C_4^2 \times A_3^3 = 6 \times 6 = 36$。

第三种情况，我们要从"SDNE"中选择一个，和另外两个"Y"组成有序序列。两个Y构成插槽，因此实际上该字母能选择位置只有3个，共有：$C_4^1 \times 3 = 12$。


因此，答案为：72


## 2.2

### Question

How many sequences of 2n coin flips have exactly n heads and n tails?

### Answer

我们只需要考虑，在这2n个硬币中，选择n个正面朝上即可，我们不需要考虑它的顺序或者别的东西。

因此，答案为：$(^{2n}_n)$

## 2.3

### Question

How many ways are there of placing 6 indistinguishable balls into 3 distinguishable boxes so that each box contains at least 1 ball?

### Answer

六个球是等价的，三个盒子不等价。我们可以先在每个盒子中都放一个球，把问题转化为：

现在有3个球，放入三个不同的盒子中，有几种放法？

有三种情况：

1. 三个球在一个盒子里（3种）
2. 有一个盒子是空的，另外两个盒子2&1分布（3*2）
3. 三个球在三个盒子里（1种）

因此，答案为：10

## 2.4

### Question

Suppose I place 2 red balls, 3 green balls, and 4 blue balls into a hat, and draw two balls without replacement.

What is the probability that the second ball is blue, given that the first ball is blue?

### Answer

感觉跟前面那题没区别，答案为3/8