Re: double encyphering with an enigma

From: Geoff Sullivan (qkcjp2_at_hotmail.com)
Date: 02/27/04 

Date: 27 Feb 2004 12:24:49 -0800

David Hamer <dhamer@bletchleypark.org.uk> wrote in message news:<403E3126.101D6B1@bletchleypark.org.uk>...

> not 17,756 [26^3] but 16,900 [26*25*26]. The general formula
> for wheels with multiple turnover notches [e.g. navy wheels
> VI-VIII] is 26.(26/m)-1.26/f where m and f are the number of
> turnover notches on the middle and 'fast' [right hand] wheels
> respectively. If 2-notch wheels are used the period can be as
> short as 4056.

In general, if m and f are factors of 26.
My M4 Enigma has 4 notches on the middle wheel and two notches on
the fast wheel and it produces a sequence of 3718 alphabets.
(26x11x13)

Geoff

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