![]() ![]() 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) Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. This seems to be a less optimized reimplementation of Permutation.Choose, I don't see why you'd need this method twice. Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. ![]() 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. Solution: The solution to this problem involves counting the number of permutations of 7 distinct objects, taken 3 at a 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 Return Factorial(n) / (Factorial(k) * Factorial(n - k)) 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)) No Repetition: for example the first three people in a running race. P osition' Permutations There are basically two types of permutation: Repetition is Allowed: such as the lock above. With $3$ people and $3$ awards, there would be $3^3=27$ ways to give these awards to the people.How can this code be improved? using System įor (int i = 0 i result.data) & (left >= 1)) To help you to remember, think ' P ermutation. If we can give multiple awards to each person, we have replacement. Applied to your problem, imagine a variation in which instead of ranks we were giving $A,B,$ and $C$ awards. But since we put it back, on the next draw we also have $3$ possibilities! Thus, if we draw $k$ times, we have $3^k$ ways to pull the names. For the first name we pull, we have $3$ possible names to pull. How many ways are there to pull the names? Say we have $3$ names. This occurs when you allow replacement of the objects! Imagine you're pulling names from a hat, but every time you pull a name and read it, you put it back. The $3^3=27$ answer comes in a slightly different variation which involves neither permutations or combinations. To arrange $n$ items in a row (which can be accomplished in $n!$ ways) is equivalent to picking $k$ of $n$ items to arrange in a row (which can be accomplished in $\frac=3!$ - exactly the result we were looking for!
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