Two Timer

David Wilson


The entire trick to the puzzle rests in understanding the relationship between the two files, which is hinted through the flavortext. The two files are encrypted using a one-time pad--that is, they have each been (bitwise) XORed with a random string. Normally, breaking the encryption of a one-time pad is absolutely impossible (see the man's last line). However, in this case, both files have been encrypted using the same one-time pad (hence the constant references to two-timing).

Since each encryption is simply the XOR of the message and the pad, the XOR of the two files should give (m1 XOR pad) XOR (m2 XOR pad) = (m1 XOR m2). Once this value is obtained, one can decrypt the messages by guessing individual parts. For example, if one assumes the messages are in ASCII, spaces are noticeable. Assuming ASCII English, and may then try XORing " the " at various points in the string to see at which points a reasonable value is obtained for the other message. The other message text can then be used to guess additional words. A decryption might proceed thusly:

guess " the ":

??is is the ???????????????????????????????????
?? the avera???????????????????????????????????

guess "avera"⇒"average ":

??is is the tot????????????????????????????????
?? the average ????????????????????????????????

guess "tot"⇒"total ":

??is is the total ?????????????????????????????
?? the average of ?????????????????????????????

At this point one may guess "their", particularly if one did not surround "the" with spaces in the first test. Assuming one does not, at this point one might try other common words, such as " of ":

??is is the total ?????? of ???????????????????
?? the average of ??????weig???????????????????

guess "weig"⇒" weight " or " weights " (the latter works better):

??is is the total ?????t of five???????????????
?? the average of ????? weights ???????????????

A space after "five" gives an "i", which combined with the positioning of spaces might lead one to guess "is " or "if ":

This is the total ?????t of five Br????????????
if the average of ????? weights is ????????????
This is the total ?????t of five Gr????????????
if the average of ????? weights if ????????????

A few more guesses ("total weight," perhaps, since "average of ...weights" is mentioned. After that, one might be tempted to try units of weight, upon which time one would find that "pounds" works...):

This is the total weight of five Br???sh peop??
if the average of their weights is ??? pounds??

...which quickly yields:

This is the total weight of five British people
if the average of their weights is 168 pounds. 

The total weight would be 840 pounds. However, note the significance of the fact that British people are being weighed; in the UK, personal weight is typically measured in stone ("stone" being singular or plural), where one stone is exactly fourteen pounds. Thus, the final answer is SIXTY STONE.