588.txt

This complete matlab for neural network

发信人: GzLi (笑梨), 信区: DataMining
标  题: [合集]对偶性问题
发信站: 南京大学小百合站 (Tue Oct 29 18:53:02 2002), 站内信件

helloboy (hello) 于Mon Oct 28 10:34:04 2002提到：

如果定义一套转换规则，将线性代数的定理进行改写。
* -> +
+ -> min
1->0
0->max(最大的正数)

A+B -> min( A, B)

A*B -> AoB= min(A(i,k)+B(k,j))
             k

0矩阵-> [max max.... max]
        [max   .........]
        [....           ]
        [............max]

 E（单位）矩阵 -> [0 max...  . max]
                  [max 0 .........]
                  [....           ]
                  [...........  .0]

那么请问

-(减，求负) ->  ?

A-B ->  A?B 


fpzh (fpzh) 于Mon Oct 28 11:05:39 2002提到：

是要用+,min,max,0来描述“-(减，求负) ->  ? ”和“A-B ->  A?B”吗。就是只能使
用那4种运算吗

“-(减，求负) ->  ? ”是什么意思

这种东西有什么用吗



helloboy (hello) 于Mon Oct 28 12:08:24 2002提到：

我在我的论文推导要用到.
不是,我是说如果
*->+
+->min
1->0
0->max
线形代数很多定理都可以转换过来.
例如下面讲的0矩阵和单位矩阵等等。
就是在 求负运算 - 这里转不不过来
- -> ?

利用普通代数里 a-0= a
对应   a?max =a
 好象? 可以取为min操作.
但是-是非对称的. a-0 <> 0-a
而min是对称的.
min(a,b)=min(b,a)
所以min不对.那应该是什么呢?
                                     



fpzh (fpzh) 于Mon Oct 28 14:13:48 2002提到：

我还没有仔细考虑，但我觉得靠+，min,0，max是做不出'-'的，你是想再换一种或加一
种运算吗



helloboy (hello) 于Mon Oct 28 14:59:19 2002提到：

是啊，没有限制哪种运算，只要能给置换掉原来的-,
线性代数的定理在置换后还可以成立


fpzh (fpzh) 于Mon Oct 28 20:25:12 2002提到：

能不能引入-1呢



helloboy (hello) 于Mon Oct 28 22:15:48 2002提到：

可以啊。你说是什么运算啊？


fervvac (高远) 于Tue Oct 29 04:55:09 2002提到：

1. It is not dual, it is homo??? (TONG GOU) in algebra

2. You can refer to (advanced) algebra for a more theoretic desc treatment of 
the topic.  You will see everything is  integrated ni ely
in that theory!

3. I am not sure what is your 0, but for any b != 0, your a+b <> 0.
That is, there is no reverse element for your +. Thats not conforming to
the requirement of a field, where each non-zero element has a reverse 
element such that a + (-a) = 0



helloboy (hello) 于Tue Oct 29 08:39:12 2002提到：

thx.
0 -> MAX(最大的正数）
+ -> min

So
a+b -> min (a,b)
a+0 -> min( a, MAX) =a

这些在影射后都是对的。
So I want to ask
a - b -> ?(a,b)

同样
a - 0 =a -> ?(a,MAX)=a

我就是说?其实可以取min运算。
但是原来代数里面的-号是非对称的。a-b <> b-a
但 - ->min的话，
a - b -> min(a, b)=min(b,a)



fervvac (高远) 于Tue Oct 29 15:32:06 2002提到：

In the highest level, you want to find a "mapping" between your algebra syste
m and the existing algebra system on matrix. 

Basically your system should be homo*** with the matrix one, so that you can use
existing algorithms / operation / theorems for the matrix system. To that end,
you have to define what is the reverse element for any element in your system.
In fact, - is not a valid operation. a - b should be a + (b -1)), where b-1 is
the reverse element w.r.t. + (sorry, should be b^-1).

If you define + to be min, there is no reverse element for any non-zero 
element. That means your system is totally different from the martrix one
due to your definition of + operator. That beats the purpose of your approach.



fpzh (fpzh) 于Tue Oct 29 17:50:37 2002提到：

fervvac兄的理论水平很高啊，pf pf

要想农这动西，还要好好看看袋鼠系统啊