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Question Description

### Problem One. Walks in Graphs of k-kool strings.

A 3-bit string is a string made up of 3 characters, each a 0 or a 1. Suppose you’d like to write out, in one string, all eight of the 3-bit strings in any convenient order.

For example, if you wrote out the 3-bit strings in the usual order starting with 000, 001, 010, . . . , you could concatenate them together to get a length 3 times 8 = 24 24-string that started 000001010. . . .

But you can get a shorter string containing all eight 3-bit strings by starting with 00010. . . . In which Now 000 is present as bits 1 through 3, and 001 is present as bits 2 through 4, and 010 is present as bits 3 through 5, . . . .

- Say a string is 3-kool if it contains every 3-bit string as 3 consecutive bits somewhere in it. Find a 3-kool string of length 10, and explain why this is the minimum length for any string that is 3-kool.
- Explain how any walk that includes every edge in the graph shown in Figure determines a string that is 3-kool. Find the walk in this graph that determines your 3-kool string from part 1.
- Explain why a walk in the graph of Figure above that includes every every edge
provides a minimum length 3-kool string.*exactly once* - The situation above generalizes to k >=2. Namely, for each k there is a digraph,

$B$k , such that the nodes are V(k )=0 , 1 } k , and any walk throughk that contains every edge exactly once determines a minimum length , (k+1)-kool bit-string. What is this minimum length? - Define the transitions of Bk. Verify that the in-degree and out-degree of every vertex is even, and that there is a positive path from any vertex to any other vertex (including itself) of length at most k.

### Problem Two.

Let x =21212121; y = 12121212: Use the Euclidean algorithm to find the GCD of x and y. Show all steps.

### Problem Three. The Bug on a Rug with Shahrzaad.

There is a bug on the edge of a 1-meter rug. The bug wants to cross to the other side of the rug. It crawls at 1 cm per second. However, at the end of each second, a malicious first-grader named Shahrzaad stretches the rug by 1 meter. Assume that her action is instantaneous and the rug stretches uniformly. Thus, here’s

what happens in the first few seconds:- The bug walks 1 cm in the first second, so 99 cm remain ahead.
- Shahrzaad stretches the rug by 1 meter, which doubles its length. So now there are 2 cm behind the bug and 198 cm ahead.
- The bug walks another 1 cm in the next second, leaving 3 cm behind and 197 cm ahead.
- Then Shahrzaad strikes, stretching the rug from 2 meters to 3 meters. So there

are now 3 times 3/2 = 4.5 cm behind the bug and 197 *3/2= 295.5 cm ahead. - The bug walks another 1 cm in the third second, and so on.

**Your job is to determine this poor bug’s fate.**(a) During second i , what fraction of the rug does the bug cross?

$H$

(b) Over the first n seconds, whatof the rug does the bug cross altogether?**fraction**

Express your answer in terms of the Harmonic numbern = 1 1 + 1 2 + 1 3 + . . . + 1 n

(c) Our currently known universe is thought to be about⋅ light years in diameter.10 10

How many universe diameters must the bug travel to get to the end of the rug?

(This distance is NOT the inflated distance caused by the stretching but only the actual walking done by the bug).### Problem Four. HT I wins, TH II wins, if not repeat.

To determine which of two people gets a prize, a coin is flipped twice. If the flips are a Head and then a Tail, the first player wins. If the flips are a Tail and then a Head, the second player wins. However, if both coins land the same way, the flips don’t count and whole the process starts over.Assume that on each flip, a Head comes up with probability p, regardless of what happened on other flips.

- Use the four step method to find a simple formula for the probability that the first player wins.
- What is the probability that neither player wins?

Suggestions: The tree diagram and sample space are infinite, so you’re not going to finish drawing the tree. Try drawing only enough to see a pattern. Summing all the winning outcome probabilities directly is difficult. However, a neat trick solves this problem and many others.

*Hint: Let s be the sum of all winning outcome probabilities in the whole tree. Notice that you can write the sum of all the winning probabilities in certain subtrees as a function of s. Use this observation to write an equation in s and then solve.*### Problem 5.

If we raise an irrational number to an irrational power, can the result be rational?

$2$

Show that it can by considering2 and arguing by cases.(Note that2 is not rational, but …)