雙語暢銷書《艾倫圖靈傳》第5章:解謎接力賽(40)

Nevertheless, something could still be done.

但是這並不是徒勞，

Suppose that at some point in the process of running through all the rotor positions, the set-up happens to be:

假設在運行過程中的某一點，機器的狀態變成這樣：

Then it may be asked whether the letters GFGCORL could, or could not, be obtained from GENERAL by the effect of a plugboard swapping.

然後我們可以考察，GNGCORL是否有可能通過配線板而編碼成GENERAL。

In this example, the answer is 'no', since no swapping could leave the first G unchanged, but swap the second G into an N;

在這個例子中，顯然是不行的，因爲沒有一種對換，可以使第一個G保持不變，而第二個G變成N，

no swapping could turn the first E of GENERAL into an F and the second into a C.

也沒有一種對換，可以把第一個E變爲F，卻把第二個E變成C。

Furthermore, no swapping could change the R of GENERAL into an O, but then change the A into an R.

當然，也沒有一種對換，可以把R變成O，但卻把A變成R。

Any one of these observations suffices to rule out that particular rotor position.

總之，以上任何一項觀察，都可以排除掉這種盤位。

One way of thinking of this question is in terms of consistency.

這個問題的一個思路，就是相容性。

Having fed the cipher-text into the Enigmas, is the output consistent with the known plain-text, in that it differs only by virtue of swapping?

把密文輸入謎機，輸出是否與明文只差一組對換？

From this point of view, the correspondences (OR) and (RA), or (EF) and (EC), are contradictions.

從這一點看，上例中的（OR）和（RA），或（EF）和（EC），就會引起矛盾。

A single contradiction is enough to eliminate all the billions of possible plugboards, on this hypothetical machine.

只要有一個矛盾，就可以排除幾十億種配線關係。