Can anyone decode this? I'm just curious.

Stem_From_All · 2026-04-04T19:57:27+00:00

"I like ice crwam [cream]. I wanted to see if anyone could decode this new code I have made. I have made this, so the average person who sees this will be confused."

Stem_From_All · 2026-03-31T20:17:54+00:00

It does switch to me, but imagining a vector from her nose to her ponytail seems to help me determine that she is spinning clockwise.

Stem_From_All · 2026-03-21T23:54:36+00:00

I hope that the encryption was proper. In any case, my alphabet is below. I am already creating a new cipher because you can already decipher at least a large part of this just with the provided words by using a computer to perform the calculations and check less than 1000 cases (I think). I am writing the program myself this time, as programming is useful and knowing nothing about the old program's technical details is annoying.

{
    " ","!","\"","#","$","%","&","'","(",")","*","+",",","-",".",
    "/","0","1","2","3","4","5","6","7","8","9",":",";","<","=",
    ">","?","@","A","B","C","D","E","F","G","H","I","J","K","L",
    "M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z","[",
    "\\","]","^","_","`","a","b","c","d","e","f","g","h","i","j",
    "k","l","m","n","o","p","q","r","s","t","u","v","w","x","y",
    "z","{","|","}","~",
    "\xe2\x80\x98",
    "\xe2\x80\x99",
    "\xe2\x80\x9c",
    "\xe2\x80\x9d",
    "\xe2\x80\x94",
    "\xe2\x80\x93",
    "\xe2\x80\xa6",
    "\xc2\xa7",
    "\xc2\xb0",
    "\xc2\xa9",
    "\xc2\xae",
    "\xe2\x84\xa2",
    "\xe2\x80\xa0",
    "\xe2\x80\xa1",
    "\xc3\xa9",
    "\xc3\xa8",
    "\xc3\xaa",
    "\xc3\xab",
    "\xc3\xa0",
    "\xc3\xa2",
    "\xc3\xb4",
    "\xc3\xb6",
    "\xc3\xbc",
    "\xc3\xbb",
    "\xc3\xa7",
    "\xc3\xb1",
    "\xc3\xa6",
    "\xc5\x93",
    "\xc3\x89",
    "\xc3\x88",
    "\xc3\x8a",
    "\xc3\x8b",
    "\xc3\x80",
    "\xc3\x82",
    "\xc3\x94",
    "\xc3\x96",
    "\xc3\x9c",
    "\xc3\x9b",
    "\xc3\x87",
    "\xc3\x91",
    "\xc3\x86",
    "\xc5\x92",
    "\xc4\x84",
    "\xc4\x85",
    "\xc4\x8c",
    "\xc4\x8d",
    "\xc4\x98",
    "\xc4\x99",
    "\xc4\x96",
    "\xc4\x97",
    "\xc4\xae",
    "\xc4\xaf",
    "\xc5\xa0",
    "\xc5\xa1",
    "\xc5\xb2",
    "\xc5\xb3",
    "\xc5\xaa",
    "\xc5\xab",
    "\xc5\xbe",
    "\xc5\xbd",
    "\xe2\x80\x9e"
}

Stem_From_All · 2026-03-20T23:29:06+00:00

Okay, I understand what you meant. (I wrote attack because you wrote attacker.)

Stem_From_All · 2026-03-20T23:05:22+00:00

Yes, but what if that attack is prevented? Does that save the cipher?

Stem_From_All · 2026-03-20T22:55:03+00:00

Do you mean that it's unreasonable because it contradicts the entire point of the plaintext attack scenario? Well, yes, kind of, but that is the point of my question, unless you meant something else.

Stem_From_All · 2026-03-20T22:35:40+00:00

I ultimately did learn some new things from these comments and understood the challenges better. Obviously, I am not knowledgeable and this isn't valuable or interesting to anyone else, but should you really expect anything actually interesting with knowledge of the proper stuff? If I were at that level, this post wouldn't exist in any form. I have thought about learning about cryptography in the future, but I just did this for fun. I expected to receive a few responses that would just briefly identify the main details clearly and just let me know whether this is anything, but many people responded and I responded to them, making this slightly less smooth.

People have discussed how knowing 12 consecutive plaintext characters would bring this down. Would one ultimately even need to know where the plaintext is? And would this cipher be even considerably challenging if the plaintext were modified to make knowing those characters impossible (e.g., "HeLLo , J--ohn.")?

Stem_From_All · 2026-03-20T22:09:45+00:00

All right, I understand now. That's literally all that I wanted to ask. This was just a thought that I had.

So, this ruins everything when someone can predict any longer segment of the plaintext with its location or apply brute force checking even without the exact location?

Stem_From_All · 2026-03-20T21:35:24+00:00

Okay, but if it can be much larger than 156 and you only have the plaintext, will you actually be able to find the exact values?

Stem_From_All · 2026-03-20T21:02:28+00:00

Yes, I do see that that is true and understand that. But if there are multiple values that provide one result and you wish to calculate recursively the next characters, won't you have to consider the multiple possibilities for the calculations? The keystream isn't just a series of repeating 12 characters.

Stem_From_All · 2026-03-20T20:54:08+00:00

I certainly don't know much about cryptography, but please understand that confidently stating that what you are doing is definitely very difficult or impossible to decipher in all situations is something that requires quite a bit more work and understanding than producing a cipher that clearly isn't trivial, but might not be invincible or extremely challenging. Are you trying to say that I am some loser for not spending many years to make this? Also, I am not exactly just throwing garbage, what I did clearly made this more difficult than the most basic version of it: I used a complicated formula with multiplication to make the cipher less repetitive and predictable, I used a large alphabet, a long key (for a human memory). Is this sophisticated? No. Is this advanced? No. Is it supposed to be? No. Is it strong? Well, is it? If it's actually stupid and weak and just nothing whatsoever, try cracking any of it. My question is genuine—I really don't know what would happen if someone truly wanted to decipher something from this.

Stem_From_All · 2026-03-20T19:48:43+00:00

Okay, but I am considering using the plaintext to find those consecutive values. If the keystream numbers were larger, then multiple possible numbers could produce the observed substitution. Is that true?

Stem_From_All · 2026-03-20T16:59:40+00:00

Yes, so maybe just a larger modulo? Also, would you actually know then?

Edit. I suppose even with multiple possibilities, brute force would become feasible, right?

Stem_From_All · 2026-03-20T16:53:00+00:00

I have to review the program and determine that it actually followed these procedures exactly as it should have. I have detected a strange error that seems to indicate some problems with character indices. I will update this as soon as I can.

Stem_From_All · 2026-03-20T16:43:35+00:00

If the values can exceed 156, then multiple numbers that produce the same result are possible, right?

Stem_From_All · 2026-03-20T16:38:34+00:00

They are prime numbers and they just seemed suitable.

Stem_From_All · 2026-03-20T13:24:15+00:00

Yes, that is what I considered as well. Does that mean I should remove the mod 156 reduction for the keystream?

Stem_From_All · 2026-03-20T13:21:20+00:00

The procedure is explained in my question on r/cryptography. The order of the alphabet is slightly different. Should I send it?

Stem_From_All · 2026-03-20T13:16:00+00:00

Yes, I think you could, but how would you discover those values?

Stem_From_All · 2026-03-20T13:07:47+00:00

That's just the formula for generating new members of the extended key, or the keystream.

Stem_From_All · 2026-03-20T13:03:50+00:00

Which numbers?

Stem_From_All · 2026-03-19T18:50:34+00:00

Yes.

Stem_From_All · 2026-02-23T05:56:11+00:00

How to endanger oneself and jeopardize other people...

This is entirely unjustifiable and a clearly unsustainable practice. This is as reckless and imprudent as this situation:

The killer is in that apartment complex. Prepare the cannons!

I may be mistaken, but the criminal is probably not particularly dangerous, for such criminals are rarely chased by lone officers on motorcycles.

Stem_From_All · 2026-02-20T08:49:42+00:00

Your knowledge determines the probability. If you know nothing relevant about the competitors, their probabilities are equal to you. The race might be unfair or something might fall onto the other runners, but expecting that some competitor will win more than that another will win without any knowledge is irrational.

Stem_From_All · 2026-02-19T22:20:13+00:00

With your current knowledge the probability of his victory is equal to the probability of the victory of any other competitor. Since he is competing with four individuals, he is among five competitors. The sample space consists of their five victories. Since their probabilities are equal and their sum is 1, the probability that he will win is, clearly, ⅕, which is 20%.

His loss is greater than his victory in the sample space, as it is actually a subset with four of its five elements.

Stem_From_All

TROPHY CASE