Closed. This question is off-topic. It is not currently accepting answers.
Want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 10 years ago.
Improve this question
Im truggling to find a closed form for the number of k-permutations of a set S of cardinality n.
The combinations should regard ordering, but no repitions.
Example:
|S| = n = 3
S = {a,b,c}
k = 2
{a,b}
{b,a}
{b,c}
{c,b}
{a,c}
{c,a}
Anyone could help me out how to compute the number of viable permutations (and not the permutations itself)?
What I've tried:
I've read through different material and found out, that including repitititions it is
O(n) = n^k
My initial though was, that I need to eliminiate the permutations like
{a,a}
{b,b}
{c,c}
But I struggle finding a closed form for the number of perceivable repititions.
You are looking for the number of k-permutations of a set S of cardinality n.
The formulae is well known : n!/(n-k)!
Pseudo-proof :
for the 1st element, you are able to choose among the n elements of S ;
for the 2nd, only among : n-1, because you don't want doublons ;
...
for the ith, only among : n-(i-1) ;
...
for the kth, only among : n-(k-1) ;
So, finally :
n * (n-1) * ... * (n-i) * ... * (n-k+1)
= n! / (n-k)!
Related
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 2 years ago.
Improve this question
Language of odd length and cannot contain length multiple of 3 over
šŗ={š,š}
OK to have an odd length that's not a multiple of 3, we can have a "head" part that generates strings of length 6n, and then a "tail" part that generates strings of length 1 and 5 (but not 7). Any odd number that's not a multiple of 3 can be written as 6n + 1 or 6n + 5 for n >= 0. So, let's get crackin.
r = ((a + b)^6)*((a + b) + (a + b)^5)
I have used a shorthand notation here which you can omit; basically, s ^ n stands for the regular expression s repeated n times. As long as n is a definite number this can be written out, just makes the regular expression a little longer.
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 3 years ago.
Improve this question
What does it mean in MATLAB when I do:
array = array (1:number)
and what does it mean:
array = array(indexes_array)
and finally, what does it mean:
array = array(indexes_array,:)
Answers, according to MATLAB docs:
1)
array = array (1:number)
The colon notation in MATLAB provides an easy way to extract a range of elements from v:
v(3:7) % Extract the third through the seventh elements
ans =
9 4 2 11 7
2)
array = array(indexes_array)
array is reorganized according to indexes_array order, assuming that indexes_array is composed of indexes.
3)
array = array(indexes_array,:)
just the same as number 1), array is reorganized according to all the rows of indexes_array, ignoring the columns.
References:
https://www.mathworks.com/help/matlab/math/array-indexing.html
https://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 4 years ago.
Improve this question
i have just wasted my 2 hours only for solving this programming question.if any one knows the trick of doing then please share it.question is given below.
You have given an array A having N integers. Let say G is the product of all elements of A.You have to find the number of distinct prime divisors of G.
Input Format
The first argument given is an Array A, having N integers.
Output Format
Return an Integer, i.e number of distinct prime divisors of
G.
Constraints
1 <= N <= 1e5
1 <= A[i] <= 1e5
For Example
Input:
A = [1, 2, 3, 4]
Output:
2
Explanation:
here G = 1 * 2 * 3 * 4 = 24
and distinct prime divisors of G are [2, 3]
Since this seems to be a homework question, I'll give some pushes in the right direction.
Theoretically, this is a very simple thing to do. Anyone can write code that loops through an array and multiplies its elements. It is also very easy to find pseudocode (or even real code) for factorizing a number into its prime factors.
However, this approach will not work here, since we will be dealing with HUGE numbers. The maximum value of G, given your constraints, is (10āµ)^(10āµ) = 10āµā°ā°ā°ā°ā°. This by far exceeds the number of electrons in the observable universe. We cannot factorize such huge numbers.
But luckily we don't need to know the value of G. We are only required to calculate it's prime factors, but we don't need to know the value of G to do so. So instead you will have to factorize the individual numbers in the array. I would recommend something like this code:
factors = set()
for num in A:
f = factorize(num) # Function that returns the set of prime factors in num
factors |= f # Add all elements in f to factors
You were only interested in the distinct prime factors, so using a set will take care of that. Just add everything, and it will automatically throw away any duplicates. factorize(x) is a function you will need to write that takes a number as argument and return the set of prime factors.
Good luck!
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 5 years ago.
Improve this question
i want to design a DFA of an alphabet {x,y,z} which accepts words with a number of 'z' multiples of 3 (eg "xzyyxzzyy")
Does anybody knows how? or which is the language that accepts it?
You will need three states to keep track of the number of z's seen, modulo three; the states will cycle through each other on input z, and the one for #z(w) = 0 (mod 3) will be the only accepting state.
To allow any arbitrary x's and y's, each state can loop to itself on these inputs.
You might use q0, q1 and q2 for the states, making q0 the initial state and only accepting state. Then, you have three transitions f(qi, z) = wh where j = i + 1 (mod 3), three transitions f(q, x) = q and three transitions f(q, y) = q for a total of nine transitions.
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 9 years ago.
Improve this question
Interesting thought question for you guys. Given an array of length n, if I were to pick two random indexes in this array, a and b on average how far apart would they be? As in how many steps would I have to take to walk from a to b. There are no restrictions so there's a chance I pick the same index for both, and there's a chance a and b are at opposite ends of the array.
I've thought about this for a while, my initial idea being they're on average n/2(ish) apart, but I think this hunch is incorrect. An index chosen in the center of the array at most would have to walk n/2 places to find its corresponding second choice, whereas only at the ends of the array would the second choice ever be around n distance away.
Thanks!
After scribbling some grids of possible distances for the first few values of n, I think the exact result is in fact given by:
f(n) = (nĀ² - 1) / 3n
Choosing two places in an array is equivalent to splitting the array up into 3 sections. The average size of each of those sections will be n/3 so the average distance between the two points is also n/3.
Using a monte carlo method in python:
from collections import defaultdict
import random
sample = [abs(random.choice(range(0,10)) - random.choice(range(0,10))) for i in range(0,10000)]
avg = float(sum(sample) / len(sample))
print ("Average: %f" % avg)
freq = defaultdict(int)
for s in sample:
freq[s] += 1
scale = 40.0 / max(freq.values())
for i in range(0,10):
print ("%d : %s" % (i, "#" * int(freq[i] * scale)))
Output:
Average: 3.293700
0 : ######################
1 : ########################################
2 : ####################################
3 : ###############################
4 : ##########################
5 : ######################
6 : #################
7 : #############
8 : #########
9 : ####
So, looks like it's n/3 - but it's not evenly distributed.
There is an easy way to know: for all the couples (a, b), computer their distance. Knowing that all the couples (a, b) have the same probability of appearance, you will just need to do the average of those distances in order to answer your question.