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(w * w^R)^R = w*w^R 的证明

[英]Proof of (w * w^R)^R = w*w^R

Question: if w exist in T*, prove that (w * w^R)^R = w * w^R问题:如果w存在于T*中,证明(w * w^R)^R = w * w^R

Hi, I am new to theories of algorithms, and I am having trouble understanding how to prove this, if someone can hint me towards the right direction that would be much appreciated.嗨,我是算法理论的新手,我无法理解如何证明这一点,如果有人能向我暗示正确的方向,我将不胜感激。

Note: R means the string is reversed, ex: (abc)^R = cba注意: R 表示字符串反转,例如:(abc)^R = cba

Also Note: * means concatenation so (abc * def) = abcdef另请注意: * 表示连接所以 (abc * def) = abcdef

In general case , try using group theory (you even have a hint : "Also Note: * means concatenation so (abc * def) = abcdef "):一般情况下,尝试使用群论(您甚至有一个提示:“另请注意:* 表示连接所以(abc * def) = abcdef ”):

  a, b, c, ...          - (characters)    - group's elements (generators in fact)
  a * b       == ab     - (concatenation) - group's operation
  ε                     - (empty string)  - group's 1 
  a..z**-1    == z..a   - rule for the item reversing; a**-1 == a

So far, so good, string reversing is **-1 operation, for any ab...yz item we have:到目前为止,很好,字符串反转是**-1操作,对于我们有的任何ab...yz项:

  (ab...yz)((ab..yz)^R) == ab..yz * zy ..ba == ab..yzzy..ba

    since zz == z * z == ε (z == z**-1) 

  we have

    ab..yzzy..ba == 
    ab..yy..ba == 
    ab..ba ==
    ε

Your theorem is quite easy then: change string reversing to **-1 and have那么你的定理很简单:将字符串反转更改为**-1并有

  (w * w**-1)**-1=(w**-1)**-1 * w**-1 == w * w**-1 

For this particular case the group we've built may be an overkill, however, it can be very useful when'll be solving the similar problems.对于这种特殊情况,我们建立的组可能有点矫枉过正,但是,它在解决类似问题时非常有用。

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