binomial-coefficients

问题 I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it. Thanks of @LutzL's suggestion. I thought of stirling's approximation before. 回答1: The O -part should be easy. Choosing exactly n /2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not. The Ω -part is harder. In fact, plotting 4n / binomial(2 n

问题 I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it. Thanks of @LutzL's suggestion. I thought of stirling's approximation before. 回答1: The O -part should be easy. Choosing exactly n /2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not. The Ω -part is harder. In fact, plotting 4n / binomial(2 n

问题 I'm currently computing the binomial coefficient of two natural numbers by write a tail recursion in Scala. But my code has something wrong with the dividing numbers, integer division by k like I did as that will give you a non-zero remainder and hence introduce rounding errors. So could anyone help me figure it out, how to fix it ? def binom(n: Int, k: Int): Int = { require(0 <= k && k <= n) def binomtail(n: Int, k: Int, ac: Int): Int = { if (n == k || k == 0) ac else binomtail(n - 1, k - 1,

问题 I have a problem with calculating OR confidence intervals from a glm in the latest version of R, but I have not had this issue before. With any glm where family="binomial" , no matter how simple the model is, it will easily allow me to extract the summary and exp(coef(model)), however when I try to extract the confint() or exp(confint(model)) , the "Waiting for profiling to be done..." message is displayed and nothing happens (I've waited up to 10 mins then cancelled the procedure, this

问题 I have a problem with calculating OR confidence intervals from a glm in the latest version of R, but I have not had this issue before. With any glm where family="binomial" , no matter how simple the model is, it will easily allow me to extract the summary and exp(coef(model)), however when I try to extract the confint() or exp(confint(model)) , the "Waiting for profiling to be done..." message is displayed and nothing happens (I've waited up to 10 mins then cancelled the procedure, this

问题 Can this be done without a loop? import numpy as np n = 10 x = np.random.random(n+1) a, b = 0.45, 0.55 for i in range(n): x = a*x[:-1] + b*x[1:] I came across this setup in another question. There it was a covered by a little obscure nomenclature. I guess it is related to Binomial options pricing model but don't quite understand the topic to be honest. I just was intrigued by the formula and this iterative update / shrinking of x and wondered if it can be done without a loop. But I can not

问题 Can this be done without a loop? import numpy as np n = 10 x = np.random.random(n+1) a, b = 0.45, 0.55 for i in range(n): x = a*x[:-1] + b*x[1:] I came across this setup in another question. There it was a covered by a little obscure nomenclature. I guess it is related to Binomial options pricing model but don't quite understand the topic to be honest. I just was intrigued by the formula and this iterative update / shrinking of x and wondered if it can be done without a loop. But I can not

问题 I have a counting algorithm for which I am trying to get a general big-o description. It is horribly nested and horribly exponential. Here it is: 1. For each T_i in T 2. For k = 1 to max_k 3. For each of 2^k*(n choose k) items 4. For each t in T_i 5. check if the item is in t...etc. Here is a line-by-line idea of each running time This is a simple partitioning and I'm going to just give it a constant c1. max_k is a small number, always less than n, perhaps around 4 or 5. I will use k below.

问题 I am trying to find the sum of the first r binomial coefficients for a fixed n. (nC1 + nC2 + nC3 + ... + nCr) % M where r < = n. Is there an efficient algorithm to solve this problem ? 回答1: Note that the "first" binomial coefficient for fixed n is nC0 . Let f(n) = nC0 + nC1 + ... + nC(r-1) . Using the "Pascal's triangle" identity, nCk = (n-1)C(k-1) + (n-1)Ck we have nC0 + nC1 + nC2 + ... + nC(r-1) = (n-1)C(-1) + (n-1)C0 + (n-1)C0 + (n-1)C1 + (n-1)C1 + (n-1)C2 + ... + (n-1)C(r-2) + (n-1)C(r-1)

问题 ... preferably in Java. Here is what I have: //x choose y public static double choose(int x, int y) { if (y < 0 || y > x) return 0; if (y == 0 || y == x) return 1; double answer = 1; for (int i = x-y+1; i <= x; i++) { answer = answer * i; } for (int j = y; j > 1; j--) { answer = answer / j; } return answer; } I'm wondering if there's a better way of doing this? 回答1: choose(n,k) = n! / (n-k)! k! You could do something like this in O(k): public static double choose(int x, int y) { if (y < 0 ||