I am not good at probability and I know it's not a coding problem directly. But I wish you would help me with this. While I was solving a computation problem I found this difficulty:
Problem definition:
The Little Elephant from the Zoo of Lviv is going to the Birthday
Party of the Big Hippo tomorrow. Now he wants to prepare a gift for
the Big Hippo. He has N balloons, numbered from 1 to N. The i-th
balloon has the color Ci and it costs Pi dollars. The gift for the Big
Hippo will be any subset (chosen randomly, possibly empty) of the
balloons such that the number of different colors in that subset is at
least M. Help Little Elephant to find the expected cost of the gift.
Input
The first line of the input contains a single integer T - the number
of test cases. T test cases follow. The first line of each test case
contains a pair of integers N and M. The next N lines contain N pairs
of integers Ci and Pi, one pair per line.
Output
In T lines print T real numbers - the answers for the corresponding test cases. Your answer will considered correct if it has at most 10^-6 absolute or relative error.
Example
Input:
2
2 2
1 4
2 7
2 1
1 4
2 7
Output:
11.000000000
7.333333333
So, Here I don't understand why the expected cost of the gift for the second case is 7.333333333, because the expected cost equals Summation[xP(x)] and according to this formula it should be 33/2?
Yes, it is a codechef question. But, I am not asking for the solution or the algorithm( because if I take the algo from other than it would not increase my coding potentiality). I just don't understand their example. And hence, I am not being able to start thinking about the algo.
Please help. Thanks in advance!
There are three possible choices, 1, 2, 1+2, with costs 4, 7 and 11. Each is equally likely, so the expected cost is (4 + 7 + 11) / 3 = 22 / 3 = 7.33333.
Related
Hello Stack Overflow Community,
I'm attempting to solve this problem:
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1040
The problem is to find the best path based on capacity between edges. I get that this can be solved using Dynamic Programming, I'm confused by the example they provide:
According to the problem description, if someone is trying to get 99 people from city 1 to 7, the route should be 1-2-4-7 which I get since the weight of each edge represents the maximum amount of passengers that can go at once. What I don't get is that the description says that it takes at least 5 trips. Where does the 5 come from? 1-2-4-7 is 3 hops, If I take this trip I calculate 4 trips, since 25 is the most limited hop in the route, I would say you need 99/25 or at least 4 trips. Is this a typo, or am I missing something?
Given the first line of the problem statement:
Mr. G. works as a tourist guide.
It is likely that Mr. G must always be present on the bus, thus the equation for the number of trips is:
x = (ceil(x) + number_of_passengers) / best_route
rather than simply:
x = number_of_passengers / best_route
or, for your numbers:
x = (ceil(x) + 99) / 25
Which can be solved with:
x == 4.16 (trips)
Given N integers in the form of Ai where 1≤i≤N, the goal is to find the M that minimizes the sum of |M-Ai| and then report that sum.
For example,
Sample Input: 1 2 4 5
Sample Output: 6
Explanation: One of the best M′s you could choose in this case is 3.
So the answer = |1−3|+|2−3|+|4−3|+|5−3| = 6.
The approach I used is sort the given input and take the middle number as M.
But I was not able to solve all the test cases. I am unable to find any other approach for this question. Where did I go wrong?(Please help me this question has been bugging me from the past 2 days.Thanks)
Can M be any real number or must it be an integer?
If there are no constraints on M your algorithm must work fine.
If M must be an integer then you have to choose M among floor(The Middle Number) and ceiling(The Middle Number).
In which language did you code up the algorithm?
I need to do this assignment. I just don't know how it works. The question is.
Modify the function roll() from the lecture in a way that it simulates a single n-sided die where the side with the highest number shows up twice as often as all other sides. Functions you may find useful are ?, c(), min(), max(), length(), sort() and rep().
And the function goes.
roll <- function( num = 1:6, rolls = 1) {
dice <- sample(num, size = rolls, replace = TRUE)
return(dice)
}
I'm pretty sure that i have to use the 'prob'-parameters in the sample-Function but i don't know how.
You can do it without the prob argument by thinking about what kind of fairly-weighted (all faces equally probable) die would give the results you want.
sample(1:6, 1) gives you a single sample from an unbiased six-sided die. What you seem to want in this instance is equivalent to a seven-sided die with two sixes. Which would be...
sample(c(1:6,6),1)
That's an equal change of 1 to 5, and double the chance of a 6.
> table(sample(c(1:6,6),7000,replace=TRUE))
1 2 3 4 5 6
972 1018 1016 980 1018 1996
Its not clear to me whether "the highest number shows up twice as often as all other sides" means "all the other sides put together". In which case you want to sample from a 10-sided die with 1 to 5 plus 5 sixes:
sample(c(1:5, rep(6,5)),1)
That's an equal chance of either getting 1 to 5 OR 6.
> table(sample(c(1:5, rep(6,5)),10000,replace=TRUE))
1 2 3 4 5 6
1012 961 943 1018 1026 5040
Generalise to N and write your function.
You are right, the prob-Parameter is useful here (eventhough you could do without).
Here are the steps you have to complete:
Find out which of the entries in num is largest (dont assume that it is the last)
You need the index (="position") of that entry.
Calculate which probability each entry except the largest one would have. Example: If n=6 then each prob is 1/7 with the exception of the last which has 2/7.
Make a vector containing these probabilities in the right positions. You already know the position of the largest, so you would put the doubled prob in that position.
Give the prob to sample().
Test! Run it many times to find out if the largest is really approx. double as often.
I can have any number row which consists from 2 to 10 numbers. And from this row, I have to get geometrical progression.
For example:
Given number row: 125 5 625 I have to get answer 5. Row: 128 8 512 I have to get answer 4.
Can you give me a hand? I don't ask for a program, just a hint, I want to understand it by myself and write a code by myself, but damn, I have been thinking the whole day and couldn't figure this out.
Thank you.
DON'T WRITE THE WHOLE PROGRAM!
Guys, you don't get it, I can't just simple make a division. I actually have to get geometrical progression + show all numbers. In 128 8 512 row all numbers would be: 8 32 128 512
Seth's answer is the right one. I'm leaving this answer here to help elaborate on why the answer to 128 8 512 is 4 because people seem to be having trouble with that.
A geometric progression's elements can be written in the form c*b^n where b is the number you're looking for (b is also necessarily greater than 1), c is a constant and n is some arbritrary number.
So the best bet is to start with the smallest number, factorize it and look at all possible solutions to writing it in the c*b^n form, then using that b on the remaining numbers. Return the largest result that works.
So for your examples:
125 5 625
Start with 5. 5 is prime, so it can be written in only one way: 5 = 1*5^1. So your b is 5. You can stop now, assuming you know the row is in fact geometric. If you need to determine whether it's geometric then test that b on the remaining numbers.
128 8 512
8 can be written in more than one way: 8 = 1*8^1, 8 = 2*2^2, 8 = 2*4^1, 8 = 4*2^1. So you have three possible values for b, with a few different options for c. Try the biggest first. 8 doesn't work. Try 4. It works! 128 = 2*4^3 and 512 = 2*4^4. So b is 4 and c is 2.
3 15 375
This one is a bit mean because the first number is prime but isn't b, it's c. So you'll need to make sure that if your first b-candidate doesn't work on the remaining numbers you have to look at the next smallest number and decompose it. So here you'd decompose 15: 15 = 15*?^0 (degenerate case), 15 = 3*5^1, 15 = 5*3^1, 15 = 1*15^1. The answer is 5, and 3 = 3*5^0, so it works out.
Edit: I think this should be correct now.
This algorithm does not rely on factoring, only on the Euclidean Algorithm, and a close variant thereof. This makes it slightly more mathematically sophisticated then a solution that uses factoring, but it will be MUCH faster. If you understand the Euclidean Algorithm and logarithms, the math should not be a problem.
(1) Sort the set of numbers. You have numbers of the form ab^{n1} < .. < ab^{nk}.
Example: (3 * 2, 3*2^5, 3*2^7, 3*2^13)
(2) Form a new list whose nth element of the (n+1)st element of the sorted list divided by the (n)th. You now have b^{n2 - n1}, b^{n3 - n2}, ..., b^{nk - n(k-1)}.
(Continued) Example: (2^4, 2^2, 2^6)
Define d_i = n_(i+1) - n_i (do not program this -- you couldn't even if you wanted to, since the n_i are unknown -- this is just to explain how the program works).
(Continued) Example: d_1 = 4, d_2 = 2, d_3 = 6
Note that in our example problem, we're free to take either (a = 3, b = 2) or (a = 3/2, b = 4). The bottom line is any power of the "real" b that divides all entries in the list from step (2) is a correct answer. It follows that we can raise b to any power that divides all the d_i (in this case any power that divides 4, 2, and 6). The problem is we know neither b nor the d_i. But if we let m = gcd(d_1, ... d_(k-1)), then we CAN find b^m, which is sufficient.
NOTE: Given b^i and b^j, we can find b^gcd(i, j) using:
log(b^i) / log(b^j) = (i log b) / (j log b) = i/j
This permits us to use a modified version of the Euclidean Algorithm to find b^gcd(i, j). The "action" is all in the exponents: addition has been replaced by multiplication, multiplication with exponentiation, and (consequently) quotients with logarithms:
import math
def power_remainder(a, b):
q = int(math.log(a) / math.log(b))
return a / (b ** q)
def power_gcd(a, b):
while b != 1:
a, b = b, power_remainder(a, b)
return a
(3) Since all the elements of the original set differ by powers of r = b^gcd(d_1, ..., d_(k-1)), they are all of the form cr^n, as desired. However, c may not be an integer. Let me know if this is a problem.
The simplest approach would be to factorize the numbers and find the greatest number they have in common. But be careful, factorization has an exponential complexity so it might stop working if you get big numbers in the row.
What you want is to know the Greatest Common Divisor of all numbers in a row.
One method is to check if they all can be divided by the smaller number in the row.
If not, try half the smaller number in the row.
Then keep going down until you find a number that divides them all or your divisor equals 1.
Seth Answer is not correct, applyin that solution does not solves 128 8 2048 row for example (2*4^x), you get:
8 128 2048 =>
16 16 =>
GCD = 16
It is true that the solution is a factor of this result but you will need to factor it and check one by one what is the correct answer, in this case you will need to check the solutions factors in reverse order 16, 8, 4, 2 until you see 4 matches all the conditions.
This is more of a maths question than programming but I figure a lot of people here are pretty good at maths! :)
My question is: Given a 9 x 9 grid (81 cells) that must contain the numbers 1 to 9 each exactly 9 times, how many different grids can be produced. The order of the numbers doesn't matter, for example the first row could contain nine 1's etc. This is related to Sudoku and we know the number of valid Sudoku grids is 6.67×10^21, so since my problem isn't constrained like Sudoku by having to have each of the 9 numbers in each row, column and box then the answer should be greater than 6.67×10^21.
My first thought was that the answer is 81! however on further reflection this assumes that the 81 numbers possible for each cell are different, distinct number. They are not, there are 81 possible numbers for each cell but only 9 possible different numbers.
My next thought was then that each of the cells in the first row can be any number between 1 and 9. If by chance the first row happened to be all the same number, say all 1s, then each cell in the second row could only have 8 possibilites, 2-9. If this continued down until the last row then number of different permutations could be calculated by 9^2 * 8^2 * 7^2 ..... * 1^2. However this doesn't work if each row doesn't contain 9 of the same number.
It's been quite a while since I studied this stuff and I can't think of a way to work it out, I'd appreciate any help anyone can offer.
Imagine taking 81 blank slips of paper and writing a number from 1 to 9 on each slip (nine of each number). Shuffle the deck, and start placing the slips on the 9x9 grid.
You'd be able to create 81! different patterns if you considered each slip to be unique.
But instead you want to consider all the 1's to be equivalent.
For any particular configuration, how many times will that configuration be repeated
due to the 1's all being equivalent? The answer is 9!, the number of ways you can permute the nine slips with 1 written on them.
So that cuts the total number of permutations down to 81!/9!. (You divide by the number of indistinguishable permutations. Instead of 9! indistinguishable permutations, imagine there were just 2 indistinguishable permutations. You would divide the count by 2, right? So the rule is, you divide by the number of indistinguishable permutations.)
Ah, but you also want the 2's to be equivalent, and the 3's, and so forth.
By the same reasoning, that cuts down the number of permutations to
81!/(9!)^9
By Stirling's approximation, that is roughly 5.8 * 10^70.
First, let's start with 81 numbers, 1 through 81. The number of permutations for that is 81P81, or 81!. Simple enough.
However, we have nine 1s, which can be arranged in 9! indistinguishable permutations. Same with 2, 3, etc.
So what we have is the total number of board permutations divided by all the indistinguishable permutations of all numbers, or 81! / (9! ** 9).
>>> reduce(operator.mul, range(1,82))/(reduce(operator.mul, range(1, 10))**9)
53130688706387569792052442448845648519471103327391407016237760000000000L