$$ 1 \leftarrow \begin{array}{l} ext { Suppose you want to find the solution to a traveling salesman } \ ext { problem with } 19 ext { cities. } \ ext { a. How many Hamiltonian circuits will need to be tested? } \ ext { b. With a computer that can test } 1 ext { million circuits per second, } \ ext { how long will the job take? } \ ext { c. Put the answer to part b in perspective by comparing it to } \ ext { some well-known length of time. } \ ext { a. About } \square ext { circuits will need to be tested. } \ ext { (Use scientific notation. Round to the nearest tenth as needed. } \ ext { Use the multiplication symbol in the math palette as needed.) }\end{array} $$

Question

$$ 1 \leftarrow \begin{array}{l}	ext { Suppose you want to find the solution to a traveling salesman } \ 	ext { problem with } 19 	ext { cities. } \ 	ext { a. How many Hamiltonian circuits will need to be tested? } \ 	ext { b. With a computer that can test } 1 	ext { million circuits per second, } \ 	ext { how long will the job take? } \ 	ext { c. Put the answer to part b in perspective by comparing it to } \ 	ext { some well-known length of time. } \ 	ext { a. About } \square 	ext { circuits will need to be tested. } \ 	ext { (Use scientific notation. Round to the nearest tenth as needed. } \ 	ext { Use the multiplication symbol in the math palette as needed.) }\end{array} $$

UpStudy AI · Accepted Answer

a. We can model the traveling salesman problem with $$ n $$ cities by noting that a Hamiltonian circuit is a route that starts at one city, visits all others exactly once, and returns to the starting city. Since rotations and reversals of the circuit represent the same circuit, the total number of distinct Hamiltonian circuits is given by

$$
\frac{(n-1)!}{2}.
$$

For $$ n = 19 $$ cities, the number of circuits is

$$
\frac{18!}{2}.
$$

We know that

$$
18! \approx 6.402 \times 10^{15},
$$

so dividing by 2 gives

$$
\frac{18!}{2} \approx 3.201 \times 10^{15}.
$$

Rounded to the nearest tenth in scientific notation, this is

$$
3.2 \times 10^{15}.
$$

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b. If a computer can test $$ 1 \times 10^6 $$ circuits per second, the time $$ T $$ in seconds needed to test $$ 3.2 \times 10^{15} $$ circuits is

$$
T = \frac{3.2 \times 10^{15}}{1 \times 10^6} = 3.2 \times 10^9 \text{ seconds}.
$$

To convert seconds to years, we use the conversion factor of approximately $$ 3.16 \times 10^7 $$ seconds per year:

$$
\text{Time in years} = \frac{3.2 \times 10^9 \text{ seconds}}{3.16 \times 10^7 \text{ seconds/year}} \approx 101.3 \text{ years}.
$$

Rounded, the computation would require about 101 years.

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c. To put the answer in perspective, 101 years is slightly more than a century. This means that even a modern computer testing one million circuits every second would take over 100 years to exhaustively search all Hamiltonian circuits for 19 cities—a time span comparable to the entire modern era of many nations and far beyond a human lifetime.
a. Approximately $$ 3.2 \times 10^{15} $$ circuits need to be tested. b. Testing at 1 million circuits per second would take about 101 years. c. 101 years is roughly the same as a century, which is longer than most human lifetimes.

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