Ditto here. I’ve read all of Holmes several times, except for VoF, which I started once and never finished, and have never looked at since.
That assumes one key per message. What if they’ve agreed that the key will initially be 100, but after ever fifth word, they add five to the key?
Yes. Or they could have a page of numbers and add a new number to each 5 digit group. And pages of numbers were available. But remember: they are doing this by hand, and it’s being done by office workers, not specialists whose friends may be killed if they get it wrong.
Even now, most office workers have trouble doing accurate addition without a calculator.
Before the invention of calculators, most people were a LOT better at mental arithmetic, and arithmetic with a pencil and paper.
I’m going to have to be nitpicky here and note that while this is true for modern-day “International Morse”, a 19[sup]th[/sup]-century U.S. telegram would have been sent using the much different American Morse, which used significantly different dot-and-dash sequences for the numerals than that used by International Morse.
That’s just a crude way of making the key longer. Cryptanalysis usually assumes Kerckhoff’s Principle – the enemy knows your algorithm but doesn’t know your key. So whether he has to guess one long number or two short ones is irrelevant. I estimated 1-2 days to exhaustively search for a 5 digit key. The numbers you’re currently using are shorter than the 5-digit key you mentioned earlier. A 3 digit primary key and one digit secondary key is actually 10 times easier to crack than a single 5 digit key, and could be done in a couple of hours with a team of 100. You’d really want to use a 5 digit primary key and a long, maybe 5 digit, secondary key, adding them mod 100,000. But of course, this is all based on modern understanding of ciphers, which didn’t exist in the 19th century.