用sencha touch 2做一个不错的路由器(Make a nice router with sencha touch 2)

 通常， 折叠是在称为monoid [wiki]的代数结构上完成的。 
 
 幺半群是代数结构。 它在具有关联函数f的集合S上定义（模数不是关联函数）。 此外，应该有一个元素e∈S ，称为“* identity”元素。 f（e，x）= x和f（x，e）= x的元素。 
 
 人们可以证明，对于一个幺半群总是有一个单位元素（所以不可能构造一个幺半群，其中两个元素是单位元素。证明如下：说有两个元素a和b是中性元素，有一个≠b ，然后f（a，b）= a （上面的定义），但f（a，b）= b也成立，因为函数只能返回一个元素，而a≠b ，我们已经达成了矛盾。因此可以得出结论： a = b 。 
 
 我们不能为模运算（以及自然，整数，...）数定义一个monoid，因为运算符应该是关联的 ，对于这些集，它不能保持： （a％b）％c = a％ （b％c） 。 
 
 回答你的问题（一般来说）： 
 
 
 
 模数运算符的标识元素是什么？ ％（身份，a）= a 
 
 
 如果我们采用自然的，积分的，理性的，真实的等数字，那么模数就没有一个恒等元素。 
 
 
 
 是否有任何特定属性，运营商必须遵循这些属性才能拥有身份？ 
 
 
 简单地说，存在元素e∈S ，使得f（x，e）= x 。 我们可以从每个函数“构造”这样的函数，只需选择一个元素e∈S ，然后将其定义为： f'（x，y）= x如果y = e且f'（x，y）= f（ x，y）否则。 
 
 
 
 如果运算符具有标识，是否有通用的方法来查找运算符的标识。 
 
 
 由于上面的内容表明我们可以为每个函数构造这样的函数，因此据我所知，没有一般方法可以为运算符的标识元素提供资金。 

Typically a fold is done over an algebraic structure called a monoid [wiki].
 
A monoid is an algebraic structure. It is defined over a set S with an associative function f (modulo is not an associative function). Furthermore there should be an element e ∈ S that is called the "*identity" element. An element such that f(e,x) = x and f(x,e) = x.
 
One can proof that for a monoid there is always one identity element (so it is impossible to construct a monoid where two elements are identity elements. The proof is a follows: say there are two elements a and b that are neutral elements, with a ≠ b, then f(a,b) = a (definition above), but f(a,b)=b also holds, since a function can only return one element, and a≠ b, we have reached a contradiction. We can furthermore thus conclude that a = b.
 
We can not define a monoid for the modulo operation (and natural, integral, ...) numbers, since the operator should be associative, and for these sets, it does not hold that: (a % b) % c = a % (b % c).
 
To answer your questions (in general):
 
 
 
What is identity element for modulus operator if it exists? %(identity, a) = a
 
 
If we take natural, integral, rational, real, etc. numbers, then modulo does not have an identity element.
 
 
 
Are there any specific properties that the operator must follow to have an identity?
 
 
Simply that there is an element e ∈ S such that f(x, e) = x. We can "construct" such function from every function, by simply picking an element e ∈ S, and then define it as: f'(x,y) = x if y = e and f'(x,y) = f(x,y) otherwise.
 
 
 
Is there a general way to find the identity of an operator if it has an identity.
 
 
Since the above shows that we can construct such function for every function, there are no general methods as far as I know to fund the identity element of an operator.