Good evening, members of EC! I would like to present you a little challenge, a game if you wish, in which you, the random member who is reading this has to crack my uncrackable encryption algorithm I have created in about 30 minutes. It uses two very simple formulas, formulas that you will (partially) find down below. How to crack the uncrackable? Well, it's quite simple: Complete each formula Take each value individually and apply one of the formulas Take the result and apply the other formula The result is the ASCII value of a certain letter, convert it Concatenate the letters into a coherent message Post the message bellow! The two formulas are incomplete, that means there is one unknown factor in each one. That factor was designated the letter n and it is your job to figure out what number goes there. Formula 1: Formula 2: NOTE: n from the first formula is different to the n in the second formula. And, in order to make your life easier, I have attached the ASCII table at the end of this post so that you don't have to go searching for it on the internets, although it is pretty easy to find. The encrypted message is: Spoiler 2618.583 7187.25 7058.333 6188.583 7581 5488.583 7849.333 7985.25 6804 5488.583 7849.333 6431.25 7187.25 7058.333 7714.583 1129.333 597.3333 7581 5488.583 7058.333 5833.333 7187.25 6930.583 597.3333 2777.25 6930.583 7317.333 7849.333 8540.583 597.3333 2618.583 6804 7187.25 7714.583 5950.583 7849.333 7714.583 597.3333 6930.583 5950.583 6930.583 5602.333 5950.583 7581 635.25 98.58334 58.33333 98.58334 58.33333 4620.583 7187.25 7985.25 597.3333 6309.333 5488.583 8122.333 5950.583 597.3333 7714.583 7985.25 5717.25 5717.25 5950.583 5950.583 5833.333 5950.583 5833.333 597.3333 6431.25 7058.333 597.3333 5717.25 7581 5488.583 5717.25 6678.583 6431.25 7058.333 6188.583 597.3333 7849.333 6309.333 5950.583 597.3333 5717.25 7187.25 6930.583 7317.333 6804 5950.583 7849.333 5950.583 6804 8540.583 597.3333 5717.25 7581 5488.583 5717.25 6678.583 5488.583 5602.333 6804 5950.583 597.3333 5950.583 7058.333 5717.25 7581 8540.583 7317.333 7849.333 6431.25 7187.25 7058.333 597.3333 5488.583 6804 6188.583 7187.25 7581 6431.25 7849.333 6309.333 6930.583 597.3333 7849.333 6309.333 5488.583 7849.333 597.3333 6431.25 7714.583 597.3333 7714.583 7187.25 597.3333 6804 5488.583 6930.583 5950.583 597.3333 5488.583 7058.333 5833.333 597.3333 7985.25 7714.583 5950.583 6804 5950.583 7714.583 7714.583 597.3333 6431.25 7849.333 597.3333 6309.333 5488.583 7714.583 597.3333 7058.333 7187.25 597.3333 7058.333 5488.583 6930.583 5950.583 635.25 597.3333 3024 5950.583 7581 5950.583 1129.333 597.3333 6309.333 5488.583 8122.333 5950.583 597.3333 5488.583 597.3333 5717.25 7187.25 7187.25 6678.583 6431.25 5950.583 635.25 98.58334 58.33333 98.58334 58.33333 3733.333 7187.25 7714.583 7849.333 597.3333 8540.583 7187.25 7985.25 7581 597.3333 5488.583 7058.333 7714.583 8260.583 5950.583 7581 597.3333 6431.25 7058.333 597.3333 5488.583 597.3333 7581 5950.583 7317.333 6804 8540.583 597.3333 7849.333 7187.25 597.3333 7849.333 6309.333 5950.583 597.3333 7849.333 6309.333 7581 5950.583 5488.583 5833.333 597.3333 7714.583 7187.25 597.3333 5950.583 8122.333 5950.583 7581 8540.583 7187.25 7058.333 5950.583 597.3333 6678.583 7058.333 7187.25 8260.583 7714.583 597.3333 6309.333 7187.25 8260.583 597.3333 5488.583 8260.583 5950.583 7714.583 7187.25 6930.583 5950.583 597.3333 8540.583 7187.25 7985.25 597.3333 5488.583 7581 5950.583 1234.333 98.58334 58.33333 98.58334 58.33333 2464.583 7058.333 5833.333 7581 5950.583 8260.583 1129.333 597.3333 7187.25 7985.25 7849.333 1234.333 NOTE: The spaces between each number separates individual characters. Here is an example: 2940.583 7187.25 7187.25 5833.333 597.3333 6804 7985.25 5717.25 6678.583 635.25 = Good luck! I am curious to see who will manage to crack the algorithm and take over the world. Let the games begin.
I'm attempting it, but I'm having trouble understanding the steps. Do I solve for n and get a single value? Or do I plug a number in and get different values?
You have to figure out the two n's from the example I gave you. The order in which you apply those two formulas is important.
Is there any kind of rounding going on or if I pass 2940.583 through both functions should I get 71 exactly?
Spoiler Congratulations, random Empty Closets member! You have succeeded in cracking the completely crackable encryption algorithm that is so lame and useless it has no name! Here, have a cookie! Post your answer in a reply to the thread so everyone knows how awesome you are. Andrew, out.
I am definitely not an accomplished cryptanalyst, but at first blush, your encryption scheme looks trivial to break. Still, I love people thinking about cryptography, and I love a recreational math challenge. Can you please describe the algorithm more carefully? Or maybe show the steps for encrypting a given plaintext? It seems like your encryption scheme is a stream cipher (over ASCII, so block size is one byte). The operations are E_1(x, n) = x^(6/n) and E_2(x, n) = xn/12. I imagine that x is a byte of plaintext, and n is a byte of the key. What is the key length?
I assume Night Rain has the correct answer. I'm disappointed that I wasn't able to figure it out but am curious to know how to solve this.
I'm hoping that AndreiH will answer. I'm not sure Night Rain's answer is correct, but I don't know how to check it. One important thing to note in this encryption scheme is that it's very weak because none of the operations are based on mathematical problems that are hard to reverse. E_1 is just an exponentiation reversed by a root taking, and E_2 is a multiplication reversed by a division.
For some reason, the images aren't loading... Randomly trying to not make this post completely useless... *hits you back with an encryption* My thoughts on this thread - e61bae0bc7a8810dd181332ce627f091 SHA-1 if you feel like it - a275da9fb9a276bc2fb914ba3ab0d65c28648ee3 Have fun.
Figured it out. One of my n values isn't an integer, but it works! C-o-n-g blah blah. it's pretty much what Night Rain wrote. (too lazy to type it all out) Although one my my n values was not an integer, I was able to store it as a letter in my graphing calculator and use it to find the answers. This was very hard, but I'm relieved I did it. Congrats to Night Rain for getting it first.
Both values are integers. Solution: Spoiler Formula 1: Formula 2: The order in which the encryption occurred is f1 then f2, so from that relation we have a decryption formula: Take the first number in the encrypted message and decrypt it: Repeat the process for all of the characters. Congratulations to Eponine and Night Rain. They both get a cookie.
After seeing how easily Night Rain figured out the answer, I have decided to create a better algorithm (a real one this time) and sure enough, I'm working on it. I'm curious to see who will crack this one.
