![]() This way the method does slightly less work and can work with more numbers without overflowing, while still being as concise as before. That said you can optimize this version a bit by using the fact that x! / y! is the product of the numbers from k+1 to n (using my suggested Product method): return Product(k+1, n) * Factorial(n-k) This seems to be a less optimized reimplementation of Permutation.Choose, I don't see why you'd need this method twice. Or you could define it using LINQ's Aggregate method (you did tag this "functional programming" after all).Īlso I'd put the main logic into a method int Product(int from, int to) which multiplies the numbers from from to to and then just call that as Product(1, n) to define Factorial. The backwards-counting while-loop seems harder to understand than a plain forward-counting for-loop to me. Though of course that depends on how it is used. It might be nicer to have a static method which returns an enumerable of all the permutations instead of having to call the Next method all the time. I don't see why you use GetValue instead of here. You should make ApplyTo generic, so you don't need to cast the returned array back to the original type. I wasn't strong enough to read through second class, so probably some my points will be applicable there also.Ī couple of things that I've noticed (though I haven't worked through all of the code in detail): public Object ApplyTo(Object arr)Īgain I don't know how this is going to be used, but this looks suspiciously as if you were doing something like Foo myFoos = (Foo) myPermutation.ApplyTo(oldFoos), which is bad. Why have you commented out Linq namespace? Use it at least here: this.data = new int When n r this reduces to n, a simple factorial of n. Return Factorial(n) / (Factorial(k) * Factorial(n - k)) Permutation The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed. Public static int Combination(int n, int k) IEnumerator enumerator = lists.GetEnumerator() Įnumerator = lists.GetEnumerator() Įnumerator = enumStack.Pop() as IEnumerator Public static ArrayList Generate(Object lists)ĪrrayList accum = new ArrayList(seqCount) public static IEnumerable> CartesianProduct(this IEnumerable> sequences) Int permNum = new Random().Next(Util.Combination(list.Length, choose)) Public static ArrayList Generate(Object list, int choose, bool random)Ĭombination c = new Combination(list.Length, choose) Return false // duplicate or not lexicographicĬombination ans = new Combination(this.n, this.k) įor (i = 0 i 0 & ans.data = this.n - this.k + i -i) Int i = left + 1 // Step #4 - order the tail Int temp = result.data // Step #3 - swap and Right = result.order - 1 // Step #2 - find right first value > left If ((left = 0) & (this.data > this.data)) Copyright © 2013, Greg Baker.How can this code be improved? using System įor (int i = 0 i result.data) & (left >= 1)) Row \(n\) is the coefficients of the expansion of \((x+y)^n\).The pattern holds for any \(n\) and \(r\): there are \(n\) ways to choose the first item, \(n-1\) for the second, and so on.We found the number of 5-permutations of the 52 cards earlier: \(52\cdot 51\cdot 50\cdot 49\cdot 48\).Obviously these are integers with \(0\le r \le n\).We will write \(P(n,r)\) for the number of \(r\)-permutations of \(n\) elements.(“How many permutations are there of these 6 things?” is asking about 6-permutations.) If we don't specify an \(r\), then we mean all of the elements.“too” is not a permutation of those values, since one element is included twice.Normally it is done without replacement, to form the. ![]() these are different 3-permutations of the 26 lowercase letters: “ate”, “fog”, “ear”, “wqx”. Permutations and combinations are the various ways in which objects from a given set may be selected. An \(r\)-permutation is a selection of \(r\) objects.A permutation is a selection of objects in a particular order.For the second question, A wins, then B, then C is a different outcome than B then C then A.… how many ways are there for the gold, silver, and bronze medals to be given out?. ![]() ![]() … how many ways are there to choose the ones that place in the top three?.These questions should get different answers: Eight people are running a race…. ![]()
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