What is the answer to this Microsoft PDC challenge?

Question

On today video9.msdn.com , the PDC guys set out to decrypt this code:

2973853263233233753482843823642933243283 6434928432937228939232737732732535234532 9335283373377282333349287338349365335325 3283443783243263673762933373883363333472 8936639338428833535236433333237634438833 3275387394324354374325383293375366284282 3323383643473233852922933873933663333833 9228632439434936334633337636632933333428 9285333384346333346365364364365365336367 2873353883543533683523253893663653393433 8837733538538437838338536338232536832634 8284348375376338372376377364368392352393 3883393733943693253343433882852753933822 7533337432433532332332328232332332932432 3323323323323336323333323323323327323324 2873323253233233233892792792792792792792 7934232332332332332332332733432333832336 9344372376326339329376282344 

Decrypt it and win a T-shirt. (Hmm, I know, hoped for a free trip to PDC.)

I notice some interesting patterns in this code, for example, pattern 332 towards the end, but I don't understand where to go from here. They said the answer is a text question.

Any ideas on decrypting this code?

Answer 1

I am still doing this - there is no answer or even a clear indication yet, but some of these random variations of facts may be useful to someone.

Meta: Is there a way to mark "read more" in the answer? Sorry in advance for scrolling this answer.

The code is 708 characters long. Initial factorization: 2 2 3 59. If they are not complicated, filling the ends, the size of the piece should be 1, 2, 4, 6 or 12; higher factors are stupid. This assumes, of course, that the code is based on concatenated pieces, which may not be.

Mike Stone suggested piece size 3. Here's the distribution for this:

  Number of distinct chunks: 64
     Number of chunks: 236 (length of message)

       275: ###
       279: #######
       282: ####
       283: #
       284: ####
       285: ##
       286: #
       287: ###
       288: #
       289: ###
       292: #
       293: ####
       297: #
       323: ################################
       324: #######
       325: #######
       326: ####
       327: ####
       328: ##
       329: #####
       332: ###
       333: ############
       334: ###
       335: ######
       336: ###
       337: #
       338: ####
       339: ###
       342: #
       343: ##
       344: ###
       345: #
       346: ###
       347: ##
       348: ###
       349: ###
       352: ####
       353: #
       354: ##
       363: ##
       364: #######
       365: #####
       366: #####
       367: ##
       368: ###
       369: ##
       372: ###
       373: ##
       374: ##
       375: ###
       376: #######
       377: ####
       378: ##
       382: ###
       383: ###
       384: ###
       385: ####
       387: ##
       388: ######
       389: ##
       392: ###
       393: ####
       394: ###
       449: #

If it is base64 encoded, we might have something;), but my gut tells me that there are too many clear fragments of length 3 for plain text in English. Indeed, this is an odd image for the character "323".

A little more interesting is the size of piece 2:

  Number of distinct chunks: 49
     Number of chunks: 354 (length of message)

       22: ##
       23: ###########################
       24: #####
       25: ######
       26: #
       27: ######
       28: ##########
       29: ####
       32: ######################################
       33: #####################################################
       34: ############
       35: #########
       36: ###############
       37: #############
       38: ####################
       39: ####
       42: ##
       43: ############
       44: ###
       45: #
       46: #
       47: #
       49: ##
       52: #
       53: ##########
       54: ##
       62: #
       63: ##############
       64: ####
       65: ###
       66: ##
       67: ##
       68: #
       72: ###
       73: #############
       74: #
       75: ####
       76: #####
       77: #
       79: ####
       82: ######
       83: ############
       84: #####
       85: ####
       88: ####
       89: #
       92: ##########
       93: #################
       94: ##

Regarding the frequency of letters, this is a good strategy, but remember that the text is likely to contain spaces and punctuation marks. Space may be the most common character to date!

Meta: This question repeatedly asks a question found elsewhere. Is this considered homework? :)

Answer 2

Well, based on the 332 pattern that you specified, and the fact that the number of numbers is divisible by 3 and that some of the first three-digit groups have matches ... maybe every 3 digits are a character, Get a distribution of coincidence of numbers for all three-digit groups, and then see if this distribution looks like a distribution of common letters.

If so, each three-digit code can be matched with a character, and you can get many characters filled in for you this way, and then just see if you can fill in the blanks of less common letters, which might not match the distribution.

A quick Google search showed this source for frequency distribution in English.

This, of course, may not be fruitful, but it is a good first attempt.

Answer 3

I wrote C # code to scan the encryption and returned me some statistics. Here are some interesting results:

With block size 3,

• There are 236 pieces.

• There are 172 duplicates.

• Code 323 shows a whopping 29 times!

• Code 333 is displayed 11 times.

• All other codes are displayed 7 times or less.

• 35 pieces start at 2.

• 200 pieces start at 3. (Interesting!)

• 1 fragment begins with 4.

• Despite the cipher containing 2s, 3s, 4s, 5s, 6s, 7s, 8s, and 9s, chunks start with only 2 and 3, with the exception of 1 block, which starts with 4.

• No 0.

• No 1s.

• There are 115 2s.

• There are 293 3s.

• There are 56 4s.

• There are 38 5s.

• There are 49 6s.

• There are 52 7s.

• There are 63 8s.

• There are 42 9s.

I would describe 323 views as very irregular. I also assume that the fact that all pieces start with 3 or 2 (with the exception of 1 kind of 4 pieces) is also very irregular.

I did the same analysis using chunks 2, 4, and 8, and the results look more or less random. At the moment, I am inclined to 3 pieces.

Answer 4

I would say that anyone who finds an answer should keep it to himself, and instead of posting, he should simply add a note that you can read by reading a specific URL to find it or send someone an email or something even if they want to know the answer to it. At a time when Channel9 says it’s broken or sends a response on its own, post it here, but until then just discuss and think. Much better for the brain.