hw08-sol

CS 70 Discrete Mathematics and Probability Theory Fall 2023 Tal, Rao HW 08 1 Countability Basics Note 11 (a) Is f : N → N , defined by f ( n ) = n 2 , an injection (one-to-one)? Briefly justify. (b) Is f : R → R , defined by f ( x ) = x 3 + 1, a surjection (onto)? Briefly justify. (c) The Bernstein-Schroder theorem states that, if there exist injective functions f : A → B and g : B → A between the sets A and B , then a bijection exists between A and B . Use this to demonstrate that ( 0 , 1 ) and R + = ( 0 , ∞ ) have the same cardinality by defining appropriate injections. Solution: (a) Yes. One way to illustrate is by drawing the one-to-one mapping from n to n 2 . More formally we can show that the preimage is unique by showing that m ̸ = n = ⇒ f ( m ) ̸ = f ( n ) . Alternative 1: Suppose m ̸ = n . Then without loss of generality let m > n . Thus implies that f ( m ) = m 2 > n 2 = f ( n ) and hence f ( m ) ̸ = f ( n ) . Alternative 2: We’ll do a proof by contraposition. f ( m ) = f ( n ) = ⇒ m = n . f ( m ) = f ( n ) = ⇒ m 2 = n 2 = ⇒ m 2 − n 2 = 0 = ⇒ ( m − n )( m + n ) = 0 = ⇒ m = ± n Since n can’t be negative, we have an injection. (b) Yes. For any y ∈ R , there exists a corresponding value x ∈ R such that y = f ( x ) . To see this take x = 3 √ y − 1. (c) The injection from ( 0 , 1 ) to ( 0 , ∞ ) is trivial; consider the function f : ( 0 , 1 ) → ( 0 , ∞ ) given by f ( x ) = x . It is easy to see that f is injective. For the other way, consider the function g : ( 0 , ∞ ) → ( 0 , 1 ) given by g ( x ) = 1 x + 1 . To see that g is injective, suppose g ( x ) = g ( y ) . Then 1 x + 1 = 1 y + 1 = ⇒ x = y . Hence g is injective. Thus we have an injective function from ( 0 , 1 ) to ( 0 , ∞ ) and an injective function from ( 0 , ∞ ) to ( 0 , 1 ) . By the Bernstein-Schroder theorem there exists a bijection from ( 0 , 1 ) to ( 0 , ∞ ) and hence they have the same cardinality. CS 70, Fall 2023, HW 08 1

2 Unprogrammable Programs Note 12 Prove whether the programs described below can exist or not. (a) A program P ( F , x , y ) that returns true if the program F outputs y when given x as input (i.e. F ( x ) = y ) and false otherwise. (b) A program P that takes two programs F and G as arguments, and returns true if F and G halt on the same set of inputs (or false otherwise). Solution: (a) P cannot exist, for otherwise we could solve the halting problem: def halt(F, x): def Q(x): F(x) return 0 return P(Q, x, 0) Halt defines a subroutine Q that first simulates F and then returns 0, that is Q ( x ) returns 0 if F ( x ) halts, and nothing otherwise. Knowing the output of P ( F , x , 0 ) thus tells us whether F ( x ) halts or not. (b) We solve the halting problem once more: def Halt(F, x): def Q(y): loop def R(y): if y == x: F(x) else: loop return not P(Q, R) Q is a subroutine that loops forever on all inputs. R is a subroutine that loops forever on every input except x , and runs F ( x ) on input x when handed x as an argument. Knowing if Q and R halt on the same inputs boils down to knowing whether F halts on x (since that is the only case in which they could possibly differ). Thus, if P ( Q , R ) returns “True”, then we know they behave the same on all inputs and F must not halt on x , so we return not P(Q,R) . CS 70, Fall 2023, HW 08 2

if n not in N : error x = g(n) if F(x) = x: return n else: return n+1 return g(FindFixedPointInN(F’)) Since g is a bijection, g − 1 exists, and if g ( n ) = x , then n = g − 1 ( x ) . If there is some x such that F ( x ) = x , then n = g − 1 ( x ) will be a fixed point for F ′ . Thus, if F has a fixed point, FindFixedPointInN ( F ′ ) will return some n where g ( n ) is a fixed point for F . On the other hand, if F ( x ) ̸ = x for every input x , then F ′ ( n ) = n + 1 ̸ = n for every n , which means F ′ will also not have a fixed point. Thus, our program solves the problem in part (b), which is already shown to be uncomputable. This means that this problem is also uncom- putable. 4 Probability Warm-Up Note 13 (a) Suppose that we have a bucket of 30 green balls and 70 orange balls. If we pick 15 balls uniformly out of the bucket, what is the probability of getting exactly k green balls (assuming 0 ≤ k ≤ 15) if the sampling is done with replacement, i.e. after we take a ball out the bucket we return the ball back to the bucket for the next round? (b) Same as part (a), but the sampling is without replacement, i.e. after we take a ball out the bucket we do not return the ball back to the bucket. (c) If we roll a regular, 6-sided die 5 times. What is the probability that at least one value is observed more than once? Solution: (a) Let A be the event of getting exactly k green balls. Then treating all balls as distinguishable, we have a total of 100 15 possibilities to draw a sequence of 15 balls. In order for this sequence to have exactly k green balls, we need to first assign them one of ( 15 k ) possible locations within the sequence. Once done so, we have 30 k ways of actually choosing the green balls, and 70 15 − k possibilities for choosing the orange balls. Thus in total we arrive at P [ A ] = ( 15 k ) · 30 k · 70 15 − k 100 15 = 15 k 3 10 k 7 10 15 − k . (b) Using a similar approach, there are a total of 100 · 99 ··· 86 = 100! ( 100 − 15 ) ! = 100! 85! ways to grab 15 balls. CS 70, Fall 2023, HW 08 4

We still want k green balls and 15 − k orange balls, which we can select in 15 k · ( 30 · 29 ··· ( 30 − ( k − 1 ))) · ( 70 · 69 ··· ( 70 − ( 15 − k − 1 ))) = 15 k 30! ( 30 − k ) ! · 70! ( 70 − ( 15 − k )) ! ways. Here, we have ( 15 k ) possible locations for the k green balls, and we use permutations rather than combinations to account for the fact that we are picking balls without replacement. Since our sample space is uniform, the probability is thus P [ A ] = ( 15 k ) 30! ( 30 − k ) ! · 70! ( 70 − ( 15 − k )) ! 100! ( 100 − 15 ) ! = ( 15 k ) 30! ( 30 − k ) ! · 70! ( 55 + k ) ! 100! 85! . Alternative Solution: Instead of considering a probability space where each ball is distinct, we can also consider a probability space where each ball is indistinguishable, and order does not matter. Intuitively, this is equivalent to the previous solution, since the numerator and denominator are accounting for order in the exact same way; considering the same situation without order and with indistinguishable balls is equivalent to dividing both the numerator and denominator by 15! for the number of arrangements of the 15 balls we select. With this in mind, we note that the size of the sample space is now ( 100 15 ) , since we are choosing 15 balls out of a total of 100. To find | A | , we need to be able to find out how many ways we can choose k green balls and 15 − k orange balls. This means that we have | A | = ( 30 k )( 70 15 − k ) , since we must select k green balls out of 30 total, and 15 − k orange balls out of 70 total. Putting this together, we have P [ A ] = ( 30 k )( 70 15 − k ) ( 100 15 ) . (c) Let B be the event that at least one value is observed more than once. We see that P [ B ] = 1 − P [ B ] . So we need to find out the probability that the values of the 5 rolls are distinct. We see that P [ B ] simply the number of ways to choose 5 numbers (order matters) divided by the sample space (which is 6 5 ). So P [ B ] = 6! 6 5 = 5! 6 4 . And P [ B ] = 1 − 5! 6 4 . 5 Five Up Note 13 Say you toss a coin five times, and record the outcomes. For the three questions below, you can assume that order matters in the outcome, and that the probability of heads is some p in 0 < p < 1, but not that the coin is fair ( p = 0 . 5). CS 70, Fall 2023, HW 08 5

(e) We will use the complementary event, which is the event of getting no heads. The probability of getting no heads is the probability of getting all tails. This event has a probability of 1 2 5 by a similar argument to the previous part. Since we are asking for the probability of getting at least one heads, our final result is: 1 − 1 2 5 . (f) To have more heads than tails is to claim that we flip at least 3 heads. Since each outcome in this probability space is equally likely, we can divide the number of outcomes where there are 3 or more heads by the total number of outcomes to give us: ( 5 3 ) + ( 5 4 ) + ( 5 5 ) 2 5 = 1 2 Alternatively, we see that for every sequence with more heads than tails we can create a corre- sponding sequence with more tails than heads by “flipping” the bits. For example, a sequence HTHHT which has more heads than tails corresponds to a flipped sequence THTTH which has more tails than heads. As a result, for every sequence with more heads there’s a sequence with more tails. Thus, the probability of having a sequence with more heads is 1 / 2. (g) By using the same idea of independence we get for HHHTT: 2 3 × 2 3 × 2 3 × 1 3 × 1 3 = 2 3 3 5 For HHHHT, we get: 2 3 × 2 3 × 2 3 × 2 3 × 1 3 = 2 4 3 5 (h) Similar to the unbiased case, we will first find the probability of the complement event, which is having no heads. The probability of this is 1 3 5 , which makes our final result 1 − 1 3 5 . (i) In this case, since we are working in a nonuniform probability space (getting 4 heads and 3 heads don’t have the same probability), we need to separately consider the events with different numbers of heads to find our result. Thus for each 3 ≤ i ≤ 5, we need to count the total number of outcomes with exactly i heads and multiply it by the probability of achieving any of those outcomes. This yields: P [ ≥ 3 Heads ] = 5 3 2 3 3 1 3 2 + 5 4 2 3 4 1 3 1 + 5 5 2 3 5 1 3 0 = 5 3 2 3 3 1 3 2 + 5 4 2 3 4 1 3 + 5 5 2 3 5 . CS 70, Fall 2023, HW 08 7