## Exponential Complexity Homework Question

Im currently learning about algorithm analysis in my class and I came across this homework question. I really don’t know where to start or what the question is asking for. Can someone guide me in the right direction? Thanks.

Suppose the number of operations required by a particular algorithm is exactly T(n) = 2^​n​ and our 1.6 Ghz computer performs exactly 1.6 billion operations per second. What is the largest problem, in terms of n, that can be solved in under a second? In under a day?

What I know so far:

The computer can perform 1,600,000,000 operations per second, so 13,824,000,000 operations in 1 day.

## Calculus 1 Courses’s Optimization Homework Problem, Produces Strangely Complex Value.

I am a Calculus 1 student and I have an optimization word-problem that is giving me a lot of trouble.

It has two variables. I have found the value for $$y$$, but when I plugged it into the equation and tried to solve for the $$x$$ I couldn’t find it’s value. I used Symbolab to solve it, but it came up with a decimal number that’s extremely complicated when written as a fraction. My professor has given us very complicated problems before, but the complexity of this number is such that I feel like it’s very likely I did something wrong.

I have checked other parts of my work with Symbolab and I am still not sure where I went wrong, but I would really appreciate it if you would take a look and determine if there are any parts that don’t look right to you.

An oil refinery is located on the north bank of a straight river which is $$2$$km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river $$6$$km east of the refinery. The cost of laying pipe is $$400,000$$ per km over land to a point $$P$$ on the north bank and $$800,000$$ per km under the river to the tanks. To minimize the cost of the pipeline, where should $$P$$ be located?

$$P=$$ The area where the pipeline enters the river.

$$x=$$ The horizontal distance between the oil refinery and the storage tanks.

$$y=$$ The euclidean distance between $$P$$ and the storage tanks.

The Pythagorean theorem states $$2^2+(6-x)^2=y^2$$.

The cost of the pipeline is $$C = 400,000x+800,000y$$.

finding y:

$$4+(6-x)^2 = y^2 \to y= \pm \sqrt{4+(6-x)^2}$$

Differentiating $$C = 400,000x+800,000y$$:

$$\frac{d}{dx}\sqrt{4+(6-x)^2}=\frac{1}{2\sqrt{4+(6-x)^2}}\cdot-2(6-x)=\frac{-(6-x)}{\sqrt{4+(6-x)^2}}$$

$$\frac{d}{dx}800,000\sqrt{4+(6-x)^2}=[\frac{-(6-x)}{\sqrt{4+(6-x)^2}}\cdot800,000] = \frac{-800,000(6-x)}{\sqrt{4+(6-x)^2}}$$

$$\frac{d}{dx}400,000x+800,000\sqrt{4+(6-x)^2}=400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}}$$

Setting $$400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$$ and solving for $$x$$.

$$x=4.84530..$$

I’m not completely sure how to write the fraction out with math notation here because a single square root seems to cover part of the numerator and all of the denominator, but you can see it if you plug $$400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$$ into Symbolab’s “solve for” calculator.

## Urgent!!! Homework for C++ class. my teacher won’t let us use strings or other following commands. Thanks

For this assignment, you are to write a C++ program (that compiles and runs) that will ask a person to type in their first name, read their name, and then display “Hello ‘first name’, how are you today?” where ‘first name’ is the first name of the person that was typed or input. For example, your output should look something like the following: Hello Fred, how are you today? For this assignment, you are restricted to commands that we have covered in class (i.e., you cannot use arrays, strings, or anything else that we have not covered). As a hint, you may want to use a special keyboard character to designate the end of the name.

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## Homework Help: How to prove that a language is not regular

Let $$L_{\pi}$$ be the language consisting of prefixes of the decimal expansion of $$\pi$$: $$L_\pi = \{3, 31, 314, 3141, 31415, 314159, \ldots\}.$$ Prove that Lπ is not DFA-recognizable. You may use the fact that $$\pi$$ cannot be written as a repeating decimal, that is, there are no sequences of digits $$d_1, d_2, \ldots,d_m$$ and $$e_1, e_2, \ldots,e_n$$ such that

$$\pi = d_1.d_2\cdots d_ne_1e_2\cdots e_n=d_1.d_2 \cdots d_m(e_1\cdots e_n)(e_1\cdots e_n)(e_1\cdots e_n)\cdots.$$

This is how I started my proof:

Assume $$L_\pi = L(M)$$ for some DFA $$M$$. We’ll construct a diabolical string $$x$$ such that $$x\notin L_\pi$$, but $$M$$ accepts $$x$$.

Let $$x = d_1.d_2\cdots d_ne_1e_2\cdots e_n=d_1.d_2 \cdots d_m(e_1\cdots e_n)(e_1\cdots e_n)(e_1\cdots e_n)\cdots$$; clearly $$x\in L$$.

Go through $$n + 1$$ states, only $$n$$ states in $$M_1$$ so $$q_i = q_j$$ for some $$i, j$$.

I based my proof off of this: https://courses.cs.cornell.edu/cs2800/wiki/images/2/29/Sp19-lec24-pumping-board.pdf

I’m stuck at the part where I’m trying to make a contradiction.