Challenge Six - second hint

How are you getting on? You have the vowels from our last hint. TO get some fo the other heiroglyphs take a look at the menu bar. The heorglyphs spell out the following words:

HOME JOURNAL NOTEBOOK ENTRY FORM FAME LINKS INFO

so you should be able to get quite a few more letters. Note that, for example, we have used two different glyphs for N in the menu bar, but Aunt Aggie only used one of them for N in the Journal.

Once you have deciphered a reasonable number of the glyphs you might want to work out which transposition cipher she used. You can at least work out how many columns you should use when writing out the text in English. A bit of experiment with the first few rows will then get you a long way.

Good luck! We still don't have a hundred correct solutions this week so you can really improve your position on the leader board with a solution.

Challenge Five Solution Revisited

Well done to everyone who tried Challenge Five. Our German friends were devious in their choice of cipher, but not devious enough to foil Aunt Aggie, nor to fool many of you.

It was a transposition cipher, based on the keyword STONE. So the text was arranged in five columns headed STONE and the columns rearranged to give ENOST. To decipher it you needed to rearrange them back again.

The first clue was that the number of characters was 695=5x139 (not 645 characters as we previously asserted which would have led to a more complicated solution - see the original solution below). Hence the keyword must have length 5 or 139. While a 139 character keyphrase is possible it is not likely, and in this case a five letter keyword has been used. Fortunately the text was already grouped in 5 letter blocks making it easy to experiment with the rearrangements.

There are precisely 5!=5x4x3x2x1 possible rearrangements of the columns, so if you know German a brute force attack would work, trying each of the 120 possibilities in turn.

A better approach is to look for a crib, a word or part of a word you expect to see. Aunt Aggie was looking for SUEVIC, and it wasn't too hard to see the block CUVIE suggesting we should rearrange it as UEVIC. This cracks the code.

Challenge Six - first hint

Many (not all) of the heiroglyphs can be found on the website. You might be able to work out what they stand for from that, though transliterating heiroglyphs is not an exact science and in particular vowels can be muddled. You could also try frequency analysis on the heiroglyphs to sort out which ones stand for E and T. To get you started here is a table of the vowels as they are used in the Journal:

More later,

Challenge Six

Looks like Aunt Aggie learnt a thing or two from her German buddies. This one had me stumped for a while. I figure I have a pretty good idea what the heiroglyphs stand for but even then it still looks like garbage - I reckon this one is double encrypted. You'll need to translate the heiroglyphs into English then figure out what sort of cipher the old bird has used. At least the first step is pretty straightforward, but I'm pretty sure this one will slow down even the fastest cryptographers out there!

Challenge Five Winners

Congratulations to:

Timothy Gale of The Leventhorpe School,
Paul Jefferys of Berkhamsted Collegiate School,
Andrew Binnie (Got Root) of Linlithgow Academy,
Lauren Willlcox (The Nosy Clan) of Sutton High School,
John Richardson of King's School.

Each of these teams wins a fifty pound cheque.

Challenge Five Solution

Well done to everyone who tried Challenge Five. Our German friends were devious in their choice of cipher, but not devious enough to foil Aunt Aggie, nor to fool many of you.

It was a transposition cipher, based on the keyword STONE. So the text was arranged in five columns headed STONE and the columns rearranged to give ENOST. To decipher it you needed to rearrange them back again.

The first clue was that the number of characters was 645=5x3x43. Hence the keyword must have length 3, 5, 15, 43, 129 or 215. While a 43, 129 or 215 character keyphrase is possible it is not likely, and 3 characters is not enough. In this case a five letter keyword has been used. Fortunately in this case the text was already grouped in 5 letter blocks making it easy to experiment with the rearrangements.

There are precisely 5!=5x4x3x2x1 possible rearrangements of the columns, so if you know German a brute force attack would work, trying each of the 120 possibilities in turn.

A better approach is to look for a crib, a word or part of a word you expect to see. Aunt Aggie was looking for SUEVIC, and it wasn't too hard to see the block CUVIE suggesting we should rearrange it as UEVIC. This cracks the code.

Challenge Five

How are you getting on. The key to this one is to find the crib and to work out the length of the anagram rule. Like it says in the notes, finding the length shouldn't be too hard, since it must be a divisor of the number of letters in the message, which in this case is 139*5. Since a key length of 139 is pretty unlikely a good guess would be 5, so the text is blocked into its anagram groups. Next you should look for a crib. Aunt Aggie seems to think that the word SUEVIC is going to appear somewhere, so you should look for anagrams of parts of the word. I'll say about more about this later. Keep working on it.

By the way, your task is to decrypt the message not to translate it! Your solution should be letter for letter as the original message, with one small proviso. Don't use accents and do use a single s in place of the German double s character, just as we did in the text version of the Challenge.

Challenge Four Winners

Congratulations to our cash prize winners for Challenge 4. They are

Thomas FitzMaurice, Alex Collins, Annabelle Hughes, Alice Slight (the Comberton Crackers) of Comberton Village College
Sarah Wynne of Guildford High School
Saajan Singh Chana of St Albans School
Harry Thimont, Robbie Parks, Eric Young of The London Oratory School
Henry Carruthers, Carl Sitch, Chloe Perfitt, Hayley Flecher (9_1 AT SANCROFT ) of Archbishop Sancroft High School

Each team will receive a fifty pound cheque.

Challenge Five

Is this the key to the whole mystery? Looks like Aunt Aggie had spirit. Looks like she isn't going to let the Babylon Stone go so easy. Figure she plans on stealing it back, and the first step is to crack the code protecting the Germans' shipping plan.

She already told us a lot about this message in her last diary entry so she must have a pretty good idea how to crack it. Frequency analysis won't help. Its a transposition cipher, or anagram cipher as she seems to call it. In any case I wrote a few notes for you on these things in case it helps. You can download them from here. Good luck. The deadline is 12.01 am on Thursday November 20th, maybe I'll give you some more hints later.

Challenge Four Solution

Well, how did you get on? It seems a long time since we last spoke, and I guess you guys have had a vacation, but frim the looks of things you were pretty busy. You can view the solution to Challenge 4 in Aggie's Journal.

Aggie hit on the idea of a keyword cipher this time. using the keyphrase BABYLONSTONE. Strking out duplicate letters this becomes BAYLONSTE and continuing the alphabet from F gives an encryption scheme of BAYLONSTEFGHIJKMPQRUVWXZCD, i.e., a is encrypted as B, b as A, c as Y and so on. As usual frequency analysis gives the game away. E only appeared 7 percent of the time so we can assume some kind of subsitution is taking place, and with O appearing 11 percent f the time its a fair guess that it substitutes for e.

Several of you were tripped up or confused by the phrase "m will" in the text. Some of you even came up with fancy theories about it like, "Since m should be encrypted by I perhaps it shold have read "i will", but the truth is much simpler. M is an initial. Remeber this is a journal and people often use abbreviations when writing down their secret thoughts.

Winners and the Hall of fame to be announced later today, and the next challenge goes up at 4pm. As usual the site will be down for maintenance for a little while beforehand.