Encryption 2

Encryption 2

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//////////////////////////////////////////// Encryption 2 Article \\\\\\\\\\\ ’ If spoilers exist, I’d ask ’ that the admins amend this article to ’ their approval rather than remove it. ’ ’ I only leave spoilers, because, it only ’ hurts the person who just uses it just to ’ complete the challenge without getting a ’ grasp on the material. Otherwise, spoilers ’ can help just as much as hints. ’ ’ Read, Learn, Apply ’ -Sent ////////////////////////////////////////////

Overview: This is an excellent challenge for several reasons…

1. It provides a great example of a very basic cypher.
2. It gives those of you, whom are new to cryptology a foundation to build on.

I’ll elaborate…

Tools: PolyCrypt: http://tws.serveftp.org/hbh/tools/PolyCrypt.zip

Let’s Begin:

Our solution in cyphered form which we are given:

05171606.05161220.16’12.1810161118.0510.132005.1605.241313.22101220.100405;.05171606.05161220. 16’12.1810161118.0510.0605241121.0409.241121.0617100405;.16.22072422142021.05171606!.

To an untrained eye this looks like scrambled numbers/punctuation. So to begin, we start a quick lesson on basic cryptography. ->> The best cryptographers not only need to possess skills in mathematics, but language as well. <<- This includes learning the most common: vowel, consonants, two-letter, three-letter words, articles. Also, how sentences are constructed. Basically study the English language statistically speaking.

In this article we will learn fundamental cryptography in Substitution Encryption. You will also begin to notice many commonalities within our language.

To start, examine the cyphered string. Notice how between each “.” is an equal amount of numbers. Now we’re not sure what the solution really contains: numbers? letters? both? symbols? punctuation? … but we’ll guess that the numbers mean something.

To figure that out, we need to somehow evaluate each letter in the alphabet to a number. EASY, since there’s only 26.

Assigned Point Value System: Screenshot: http://tws.serveftp.org/hbh/articles/encryption/pvs.JPG

No take a look back at the cyphered text. It begins with [05171606]. Well according to our point value system, there is no “zeroth” position, so let’s

try splitting it up in two’s.

We get [05 17 16 06]. This looks logical. So take each pair and write in the letter which corresponds with the number on our point system.

->> If you did it right, you’d get [EQPF]

Now apply this to the entire cypher:

->> If done correctly, you’d get: (POSSIBLE SPOILER! ADMINS PLEASE EDIT IF NOT SATISFIED)

(CAPITALIZED) EQPF.EPLT.P’L.RJPKR.EJ.MTE.PE.XMM.VJLT.JDE;.EQPF.EPLT.P’L.RJPKR.EJ.FEXKU.DI.XKU.FQJDE;.P.VGXVNTU.EQPF!.

(lowercase) eqpf.eplt.p’l.rjpkr.ej.mte.pe.xmm.vjlt.jde;.eqpf.eplt.p’l.rjpkr.ej.fexku.di.xku.fqjde;.p.vgxvntu.eqpf!.

Use either one, I prefer looking at it capitalized, but at the end we’ll have to change that.

OK! Halfway done! Now here’s where a little bit of knowledge about language statistics comes in hand.

A substitution cypher is quite simply: A system in which taking an existing alphabet and assigning new letters to existing ones.

Lets now open PolyCrypt and select the “Substitution” tab. Paste our new cypher in the top box.

Our goal is to figure out which letters correspond with the right letter to make this cypher readable. I’ll start you off.

The solution couldn’t be a complete sentence with all these periods in it; It would just be fragments. So let’s assign “.” to a new letter. In this case, lets assign it to “ “; a space. I definitely looks like it would apply.

In PolyCrypt, for each char you put into the Original alphabet text, you assign it a new char in the New alphabet text.

So put: (without quotes) Orig alphabet: “.” New alphabet: “ “

Click “Decrypt” and watch the new text appear with all “.” changed to “ “.

Now… The most common vowel used in the English language is “E”. The most common consonant is “T”. The most word is “THE”. There are so many statistics that can be recognized, and we never even realize it when speaking. It really is quite fascinating if do a little

research on it.

Anyway… In the entire string, we see this [EQPF.EPLT] repeatedly. So lets take an educated guess and say that:

E = T & Q = H (Most common letter found following a “t”)

Fill in the new variables, and decrypt!

Screenshot: http://tws.serveftp.org/hbh/articles/encryption/enc2_2.JPG

Now try to finish the rest of this challenge using logic by examining the cryptic words. To solve the challenge, capitalization and punctuation must be correct.

Good Luck. Hope this helped. Comment if you’d like; I’d like to write more for HBH.

-Sent