I was time traveling with Fred last week - you know Fred, he's a nice guy - when we got separated. His hand-held automatic chronological portal manipulation and guidance unit (HHACPMGU) (we're working on a better name) malfunctioned! We were eating ice cream with Richard Feynman and talking about our latest visit to Sumer when Fred's pocket vibrated, the wormhole opened, and he was gone!

I raced home - time, not space - because our contingency plan was always to leave a message is a particular secure location so that the other could retrieve it in our home time. Fred wasn't waiting for me so I knew his remote couldn't get him home and he needed me to come save him. I dug in the exact coordinates that we'd agreed upon and found a small metal box. It contained a sheet of metal with an encoded message engraved on its surface. I had tried to tell Fred that there was no way that anybody would ever find the message should we ever need to send one but he was always more paranoid that someone would discover our time traveling secrets.

Now, because Mr Paranoid went and got himself stuck in the past, I have to try to decode this thing so I can go rescue him. Knowing him, it isn't even a direct message. It's probably just something related to where / when he is. It doesn't matter how long it takes to figure out - time traveler, remember? - but I have to get it done or Fred is stuck!

When and where should I meet Fred?




Addition 1:

I have a busy weekend planned so I wanted to get this information out there before it started. You'd think I could free up some time with my HHACPMGU but you have to be really careful about not running into yourself in the past. Fred and I used to travel with Kevin until The Incident. That took forever to clean up. Poor Sally, too, left with all those kids... Anyway, that's why I don't use the HHACPMGU for such trivial things so I don't have as much time to try decryption this weekend. I hope you can still help!

I found another box under the first one with a plaintext message! I guess Fred knew I would never solve this without at least knowing where to start. The message told me that the decrypted text would only include uppercase alphanumeric characters. I probably would have guessed that by myself so it's a good thing that the message also said that the encrypted version is longer than the decrypted version. It didn't say how much longer, though, because I guess Fred is more of a paranoid jerk than I thought.

Addition 2:

I manged to find some time this weekend - despite all that's happening - so I could check in on the progress. A comment from Ian MacDonald got my attention. I did a quick analysis of character frequency and I got some weird numbers. There is something definitely going on here...

A 17%, B 22%, C 4%, D 4%, E 4%, F 3%, G 4%, H 3%, I 3%, J 3%, K 2%, L 2%, M 3%, N 3%, O 2%, P 2%, Q 2%, R 3%, S 2%, T 2%, U 2%, V 2%, W 2%, X 1%, Y 2%, Z 1% Not only do these not line up at all with a typical character distribution, but there are way too many As and Bs

Addition 3:

An amazing thing happened last night. A Western Union delivery man shows up on my doorstep with an old, weathered envelope. He says that they've been holding on to this letter for a long time with specific intructions on when and where to deliver it. In fact, the office even had a bet going on whether or not I'd even be here. The delivery guy had lost the bet, it seems. I figure this must be a letter from Fred so I rip open the package and pull out a single sheet of paper with the following message:

I am a programmer. I use zero-based numbering. Z = 25

Well, I'm not a programmer, Fred! Now I have to go lookup what you're talking about. I was going to ask the Western Union delivery guy exactly how long they've had this letter as that would answer half my question but he was already gone. I suppose he was upset about losing the bet.

Ah, well. Back to work, I suppose. How does this work with the "every third letter" pattern that Tryth found?

  • 8
    $\begingroup$ I will only help if you give me your time travel secrets $\endgroup$
    – kaine
    Commented Apr 3, 2015 at 15:14
  • 10
    $\begingroup$ @kaine Not again, I won't! You tricked me last time and then you tried to conquer the world. I suppose you don't remember that because I had to go back and change it but I certainly remember and I've learned my lesson. $\endgroup$ Commented Apr 3, 2015 at 15:30
  • 3
    $\begingroup$ I suppose that HHACPMGU is somehow a key $\endgroup$
    – leoll2
    Commented Apr 3, 2015 at 15:32
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    $\begingroup$ @leoll2 The only thing I could get Fred to agree to about his insistence on encryption is that it must be entirely self-contained and require no other knowledge. We weren't sure when we would recover the message and didn't want to use some knowledge that we hadn't yet acquired. No, I'm sure that there is no key. The only information you need to solve it is in the block quote text. $\endgroup$ Commented Apr 3, 2015 at 16:00
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    $\begingroup$ Every third letter of the ciphertext is an A or a B. I find that hard to attribute to coincidence... $\endgroup$
    – Tryth
    Commented Apr 5, 2015 at 0:11

1 Answer 1


The way to decode the cipher is:

Each 3-letter sequence represents a single uppercase letter and can be decoded as such:

If third letter is A, add the 0-based values of the first two letters (modulo 26), convert back to a letter.
Else (third letter is B), subtract the 0-based value of the second letter from the 0-based value of the first (modulo 26), convert back to a letter.

(0-based values = 0=A,...,25=Z)

The raw plaintext is:


The solution is, presumably:

Place: Gettysburg, Pennsylvania
Time: November 19, 1863

Bonus - how I found the solution:

All three paragraphs have length divisible by 3, suggesting that each 3-letter sequence represents a single unit.

Doing frequency analysis for all 3-letter sequences showed that the two most frequent sequences were UQB (20,16,1) and PLB (15,11,1) - both ended with B(1) and had a difference of 4 (E) between the first two letters, suggesting I was on to something.

The rest was just plugging the values generated by the above algorithm into the 3-letter sequences ending with B, getting much of the letters and easily figuring out what the rule for sequencing ending with A is.

  • 2
    $\begingroup$ Fred will be saved! Thank you so much for not leaving him stranded! I'm going to fire up the HHACPMGU tonight and go get him. We have a very important mission after I pick him up. You'll know we succeeded if Abraham Lincoln becomes the first president to serve three terms and leads America into a new age of enlightenment and peace. $\endgroup$ Commented Apr 7, 2015 at 15:11
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    $\begingroup$ @EngineerToast Dang it.... you failed us. $\endgroup$
    – kaine
    Commented Apr 22, 2015 at 14:06
  • $\begingroup$ @EngineerToast Don't. Play. With. PAST!!! $\endgroup$
    – EKons
    Commented Aug 30, 2016 at 5:51

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