You always have to look for an easy way out when attempting to complete anything. Even though accomplishments are treasured more when achieving them was difficult, it is sometimes better to try to take it easy. In the case of breaking these codes, achieving the information by alternative means is much easier than spending days trying to decode it. So, with "The Crypts", I executed much espionage. For the most part, I focused on Samir, since he was the most easily accessible. I have been in his room many times, and when he steps out, I get to work. I looked through every paper that he has in his room, yet I couldn't find anything. I booted up his computer, and checked in as many files as I had time to do. This yielded the same negative results. My conclusion was that someone else in the group held the information that I needed. I moved on. The only other technique that I used was to check in Nicole and Jessies' notebook for the class one day at lunch. This also provided no new information, and I kind of felt bad, yet I was soon over it. In all of the time that I invested in attempting to gain vital secrets, I failed, and wasted my time.
Now the only thing left was to straight up attempt to solve it using cryptanalysis techniques. From the format of the enciphered text, It resembles normal paragraphs, in which a simple substitution was made. This is ridiculously simple, and I knew that I was more complex than that, but It was a start. I assumed that the letters by themselves were either "a" or "I". After assuming this, I plugged these letter back into other words. I discovered some words, however, when using the letters discovered in the words I found in other groups of letters yielded only negative results. At least I could disregard the idea that a simple substitution was performed.
In my next evaluation, I put all of the letters together, not worrying about the spacings, and checked the letter frequency of them. I found there to be no striking significance worth worrying about in the frequencies of the letters. Next I looked for the least frequent letters, however this also proved to be fruitless for me. I only had time for a couple of other tests. I reversed the order of all of the letters and tried my substitution method again, however I felt like it was a waste of my time, and it really was. I performed a couple of trivial transposition techniques involving looking at it in columns reading up to down, and accomplished nothing by doing so. In all of my attempts at breaking THE CRYPT's code, I never could crack the hidden meaning.
Due to the short span in time allotted for us to break the codes, and with all of the other papers and midterms occupying most of our time, I don't believe that it is possible for anyone to have broken any of the codes. This doesn't mean that they can't be broken, since I believe that any man made code is breakable, you just need enough time.
In an attempt to decipher the cryptogram of "The Null Set", I resorted only to the techniques of cryptanalysis, foregoing the espionage part, which may have proven useful otherwise. I made an assumption that the cipher was a simple substitution, where each pair of symbols stood for one letter, based on the fact that the cryptogram employed only 6 symbols, and the most rational way to represent 26 letters with that amount of symbols was to pair them up. The pairing up yielded 36 different combinations, all of which, unfortunately, were present in the cryptogram. To me that meant that either there were a lot of errors made, when the members of "The Null Set" were typing in the symbols or that some of the combinations served as punctuation signs or signaling signs or dummy variables. I counted the "letters" on one page and made a distribution chart. From that I found out that the most frequent combination was 3/4 (101), which I then presumed to be the letter "e". I then counted all the letters that appeared after e, knowing that the most frequent digraph with e is "er". That way I discovered that the combination ߣ might stand for the letter "r", and the overall frequency of this combination (35, 6th most frequent) correlated rather closely with the frequency of "r" in English (7th most frequent). I made several other assumptions based on similar tactics, namely--ߏ or (c) became "s" or "t", (c)-"u" or "y", and 3/4--"d". However, when I tried to make substitutions with those letters, they did not lead to any conclusive results, simply because these substitutions were not enough to make guesses about the words that contained them.
What complicated the decipherment more was that the spaces of the plaintext were omitted in the cryptogram, and I could not try to reconstruct the key based on the one-letter or two-letter words. However, when the method with letter frequencies had failed, I took another approach. I had noticed repeated combinations of letters throughout the text-apparently, words. One of the most frequent longer words (6 letters) was the string of symbols "(c)(c) (c)3/4 (c) (c) (c) 3/4" (if my finding of the letter "e" was correct, the word ended with an "e")-it appeared in the cryptogram 31 times. The fact, that almost all letters in this word started with the symbol (c), alarmed me; maybe, the string served as a signal of some sort, but a signal would not normally be this long. Also, on the first page, the seven-letter sequence " ߏ (c) (c) 3/4 (c) " appeared 8 times (and once in the rest of the text), most of the time, preceded by the letter with the symbols , which, however, did not appear to be part of the word.
A lot of times an "e" was added on to the end of this word, sometimes, at the end of the line, which made me guess that the word may have been a noun taken from Latin, as those often form plurals by an addition of "e", but I could not think of a suitable word ("formula" did not fit, because the 3rd and 6th letters in the word are the same). I highlighted a lot of similarly repeating patterns of letters-some of them very long (14 sets of symbols), which probably meant that those were repeating phrases. However, since I did not even know the subject of the plaintext, I could not make any guesses on the words that I've selected.
It was very frustrating to look at the text and see words emerging and not being able to get the slightest clue on their meaning. I felt that if I was given a single word in the cryptogram, I would be able to decipher the whole text. Nevertheless, I have failed to do that, perhaps due to the time-constriction, or due to the minimal amount of experience or due to the lack of sufficient patience, required for this type of task. "The Null Set" seems a rather easy cryptogram to decipher if given any clues (which I could possibly have obtained through espionage) or if done by a better cryptanalyst than me. Even though I did not succeed, I support Poe's claim that a human would be able to break this "code".
The same phenomenon that had the power to conquer Superman had a similar effect on me. My attempt to break the "Kryptonite" code yielded very interesting results, but sadly none of these led to the actual decipherment of the code. However, given more time and knowledge in cryptanalysis, it is feasible that someone can break this code.
The initial observations I made about the code dealt with its characters, their magnitude and frequencies, some specific combinations, and patterns throughout the encoded text. For instance, the first thing I did was to identify the different symbols used in the code. Each letter in the English alphabet was utilized (26 in all), including the symbols !, ?, *, @, and **. There are 84 characters per row on average and 111 rows, meaning that there are approximately 9324 characters in the entire ciphered text sample. To gain some quantitative data, I examined a sample composed of 10 lines of the enciphered text amounting to 844 characters. By counting the frequencies of each individual symbol within this sample size, approximations of these frequencies throughout the entire enciphered text can be made. The frequencies of each character based on a sample size of 844 characters are approximately as follows:
A: 1.3% B: .59% C: .71% D: 1.78% E: 1.3% F: .83%
G: 1.54% H: 1.54% I: .83% J: .83% K: 1.18% L: 1.78%
M: .95% N: .71% O: 1.54% P: 1.78% Q: .59% R: 1.07%
S: 1.18% T: .71% U: 1.30% V: .47% W: .95% X: .47%
Y: 1.07% Z: .36% @: 16.82% * : 15.76% ** : 6.28% ! : 13.86%
+ : 2.01% ? : 17.89%
With over 9,000 characters encoding plain text of approximately 500 words, it is likely that 3 characters encode one letter in the English alphabet. Therefore, the frequencies of each character by itself are not very significant.
Therefore, combinations of different characters were taken into account. Several observations can be made about the code regarding these combinations. For instance, the sequence " +?** " occurs approximately 112 times throughout the entire coded text. Furthermore, every occurrence of this sequence is preceded by a vowel. Another such character combination involves the @ character, which always follows a vowel and is never found by itself. Observations about the vowels themselves can be made also. For instance, vowels are always separated from each other; never can a combination of two vowels in a row be found.
It seemed logical to link these observations together in some sort of a pattern, and maybe form hypotheses about the rules of this code. For instance, when considering the character frequencies in the encoded text, it may be hypothesized that the non-letter characters (!, ?, *, @, and **) encode placeholders or punctuation marks since they occur in such high frequencies compared to the frequencies of capital letters. Also, it may be assumed from the way this code is presented that the characters were strung in a continuous sequence without adhering to some sort of structure. Since each row contains anywhere from 82 to 86 characters that are not separated by spaces, it seems logical to assume that they are merely strung together continuously. This makes it quite confusing at first glance and also more difficult to decode.
The attempts to decipher the "Kryptonite" code yielded interesting results and observations, but nothing substantial enough to actually succeed in discovering the plain text it encoded.
On one hand, it seems more difficult to decipher than our own code, "Enigma," since it is merely a consecutive string of characters. However, Enigma only utilizes 10 different characters compared to Kryptonite's 31.
The text that we enciphered for our Enigma code is an excerpt from The Book of Table Talk by Charles Knight. The technique we used to decode this message is a simple substitution of letters for numbers of 2 digits. The key is as follows:
A - 22 B - 39 C - 51 D - 74 E - 19 F - 68 G - 24 H - 91 I - 46 J - 17 K - 71 L - 36 M - 27 N - 89 O - 54 P - 43 Q - 12 R - 08 S - 70 T - 03 U - 99 V - 44 W - 67 X - 11 Y - 28 Z - 13
To make the code more confusing, however, we chose to present it in a misleading structure. The first letter of the plain text is encoded by 03, starting at the upper left-hand corner of the encoded text. From that point proceeding down the twenty rows, every two digits code for the subsequent letter of the plaintext. Therefore, the beginning of the coded text reads 03, 91, 19 down the first three rows and begins the message with the word "The." The two-digit numbers for the first twenty plaintext letters were recorded in sequence down the 20 rows. Then the code proceeds in an upward direction for 20 rows and winds back in a downward direction again. This winding method of recording the characters of our code was meant to confuse the person trying to decipher it. Once all these numbers were recorded, they were divided so that each page contained 1200 individual numbers. These numbers were split into 12 columns (each column containing five-digit groups) and 20 rows on each page. In order to adhere to this structure of 12 columns and 20 rows of 5-digit groups, several placeholder symbols were needed to fill the spaces remaining on the last page once the plaintext was completely deciphered. Since we encoded three pages of the book(197, 198, 199), and each page was encoded separately, the dummy variables appear three times in our code. To someone who knows the key, this signals, that the remainder of the information starts on the next page or, if it's the last page of the cipher, that the plaintext ends. These "dummy" numbers were merely any two-digit numbers that weren't used to encode a specific letter.
Overall, the science of enciphering and deciphering codes can range from very simple to very complex. For instance, the basic rules and structure of a code may seem simple to the person who creates it. Those who attempt to break this code, however, may find it tremendously difficult. Such is the case with our attempts to break the Kryptonite, Null Set, and Crypt codes. Though they may be fundamentally simple codes, we found their decipherment very difficult. On the other hand, our own code, Enigma, seems easy to us because we are its creators and thus know the key. Therefore, cryptography is definitely a science that exercises intuition, patience, and analyzing ability. With these three attributes, enough time, and perhaps a hint or two, any code written by man can presumably be cracked.
Back to the group projects page.