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**fusilli_jerry89**- Replies: 2

Suppose we are given a list of n people who want to fly to the moon (and return). Person i

has priority pi and they weigh wi pounds. You may assume all of the priorities are different.

Our spaceship can carry at most k pounds which is, unfortunately, smaller than the total

weight of all the people. If priority pi > pj then if we take person j we must take person i.

Describe an algorithm that runs in O(n) (linear) time that finds the largest set of people we can fly to

the moon without exceeding the weight limit.

We fly everyone with priority greater than p(m).

Hint: Use recursion. How can you use the linear time k-select algorithm to insure that the

subproblems are small?

k-select returns the kth smallest element in linear time. (so if you have 1,2,3,4,5,6,7,8,9 kselect(3) will return 3)

**fusilli_jerry89**- Replies: 1

I am having a little trouble understanding this question. Maybe i'm too tired of studying.. Thanks ahead of time!

For each of the linear maps R^2 → R^2, find a basis of R2 consisting of eigenvectors

of the linear map, or explain why this is not possible.

(a) The dilation D : R^2 → R^2 given by D(u) = −3u for all u ∈ R^2

**fusilli_jerry89**- Replies: 1

An n by n maze is randomly generated with numbers from 1 to n. Fo example: a 5 by 5 maze might look like the following:

23125

55132

54214

13542

14341

In order to solve the maze, you have to start from one of the top numbers, and get to one of the bottom numbers in n-1 steps (4 in this case). (You can only move down, and diagonal downwards). In addition, you must visit each number from 1 to n once and only once. (You must visit 1,2,3,4,5 in the example above). For example, a solution in the example above might be 2,5,4,3,1 or 5,2,1,4,3.

What is the possibility of a maze with no solutions?

I know that in order to get the answer to this we have to take (1/n)^n² (probability for each maze) and multiply it by the total of number of mazes without solutions. I can see that we can add up all the mazes without a certain number (i.e. a maze with no 2's will yield no solutions), but how do I find out how many mazes there are with no solutions otherwise?

Thanks!

Think I found a solution. Please let me know if it's valid.

First I concluded that the average number would be 13. Hence, the average group of 3 would be 13 times 3 = 39. In order to have an average of 39, either all numbers have to be 13, or at least one group has to add up to over 39.. thus at least 39.

**fusilli_jerry89**- Replies: 2

A wheel of fortune has the integers from 1 to 25 placed on it in a random manner.

Show that regardless of how the numbers are positioned on the wheel, there are three

adjacent numbers whose sum is at least 39.

**fusilli_jerry89**- Replies: 1

∫∫ sin(pi(x²))/x dxdy

where inner integral is from √(y) to 1 and outer integral is from -1 to 1

wow... I just got it lol. Sorry for wasting space.

I realize now that the prof chose only 3/4 possible A's and so since all A's have to be used up to be a function, and (T,F) is not used in his example, then it is not a function. Is this reasoning correct? Thx (P.S. I think a discontinuous function is not a function in logic)

**fusilli_jerry89**- Replies: 2

heya, I was just wondering if someone could explain this to me because I don't get it.

My logic prof said that there is a function f : {T,F} x {T,F} -> {T,F} (where A -> B means A is independent and B is dependent).

He says that f = {((F,F),F),((F,T),T),((T,T),T)} and that this is not a function because there is no pair whos 1st element is (T,F). Did my prof just chose 1 example out of many possible ones from A to B ? And if he did, does this mean other examples you can pull out would be functions? Also, why is it not a function if there is no pair whos 1st element is (T,F) ?

Thx if you understand lol

ooooooooooooh I get it now. Thanks!

But as another example, say we are given an equation for an arbitrary element in a set X say x = 5a + 3 (where a is an int) and a set Y with an arbitrary element y = 5b - 2. So I have to prove that the element x is in the set Y and the element y is in the set X right?

**fusilli_jerry89**- Replies: 5

Prove or disprove each of the following statements for all subsets A and B of a universal set

U :

a. P (A ∩ B) = P (A) ∩ P (B).

Yeah I was just wondering where to start with this problem..

10 mins till exam..

k but if we assume 2^t divides n and try to prove that 2^t divides x + y10^t , can't we just say that since n = x + y10^t and 2&t divides n than it must divide x + y10^t because that's what n is?

Or must I do more than this? lol

ok so I said that n=the number, and the number t is the amount of numbers at the end of n. I call the numbers in t x. so n = x + (n - x) (where x = last digits). So now I must show that 2^t | n implies 2^t | the last digits and 2^t | the last digits implies 2^2 | n.

So n = x + y10^t (where y is the first digits and x is the last digits)

So now I think I must show that 2^t divides n and 2^t divides (x + y10^t).

Now I am stuck..

if u take half of an integer and keep adding half of that and half of that to infinite you will get the integer.

For example: 64 can be written 32+16+8+4+2+1+.5+.25+.125+....................etc.

This will equal 64 since the limit is 64.

So how am i supposed to find how many powers of 2 are in 10^t ?

so im a bit rusty:

yer asking (10^t)/2^x ; what is x right? Ok well I forgot how to divide powers with different bases lol.

**fusilli_jerry89**- Replies: 9

Consider the following theorem:

A positive integer n is divisible by 2^t

if and only if the number consisting of the

last t digits of n is divisible by 2^t

For instance 68 is divisible by 2^2 = 4, and hence 38168 is divisible by 4. On the other hand,

0260 is not divisible by 2^4 = 16, and so 10440260 is not divisible by 16.

Prove the theorem using a direct proof. Hint: to prove a biconditional, you need to

prove two implications.

Yeah Im confused..

wat if the velocity is constantly changing tho? Acceleration?

**fusilli_jerry89**- Replies: 0

dV/dt = -a*sqrt(2gh)

Suppose a tank is cylindrical with height 6 ft and radius 2 ft and a hole at the bottom of the tank is circular with radius 1 inch. If we take g = 32 ft/s^2, show that h satisfies the differential equation dh/dt = -sqrt(h)/72.

This first part is easy so I don't need help on that.

The next question is: solve this equation to find the height of water at time t.

Integrating we get: 2sqrt(h) = -t/72 + C

and eventually: h(t) = t²/144² - Ct/72 + C²

C² must equal 6 since the height is 6 at t = 0 so we get:

h(t) = T²/20736 - sqrt(6)t/72 + 6

Is this correct? Thx.

If this is correct, could we also go like this when integrating?

∫h-½ dh = -1/72 ∫dt

(from 6 to h) (0 to t)

I end up getting the exact same thing.

**fusilli_jerry89**- Replies: 0

Im trying to solve a physics problem and I am doing a dipole problem. The solution says that the key to the problem is that 1/[x² + (z+(d/2)²]^(3/2) is approx. equal to [1/(z²+x²)^(3/2)][1-(3zd)/(2(x²+z²))]. How is this?

**fusilli_jerry89**- Replies: 2

Consider a water tank shaped like an inverted right circular cone of heaight 1m and top radius 1m. Let h(t) denote the height of the water level of the tank in metres with time in days. Water flow into the tank at a constant rate of 0.01 m^3 per day. Water evaporates from the tank at a rate of 0.01 A in m^3 per day where A is the area of the water surface. When h = 0.2, how fast is the water level rising?

The book says that we say the Volume = 0.01t - 0.01(pi)h²*t

I get why 0.01 times the time will give us the added volume because it is at a constant rate. But how can the subtracting volume be equal to 0.01 times pi times h² times time? Becuz the subtracting part is not constant, so how can multiplying it by a constant time give us the correct answer? Unless the book is wrong.

**fusilli_jerry89**- Replies: 0

Quick question. On my notes the teacher says a=g - (b/m)v goes to v = v(T)[1-e^(-bt/m)]. I see that it must be integration, but how do you do it? thx