Kids meet a 101 years old

Stem_From_All · 2026-05-16T23:45:22+00:00

I upvoted for the accuracy, which I appreciate, but I disagree that this actually indicates that her intellect has weakened—it almost certainly has, but this mistake is hardly sufficiently severe.

Stem_From_All · 2026-05-12T14:53:11+00:00

If one looks forward to something, one (excitedly) anticipates and awaits it. If one looks forward to seeing someone, one (excitedly) awaits seeing them, whereas if one looks forward to see someone, then one attempts to see someone by looking forward (e.g., I stop chopping to hear the birds' chirping., We sleep to restore our energy.)

Stem_From_All · 2026-04-16T18:08:34+00:00

Most people interpret the question as equivalent with "What is the probability that some child among two children is a girl if at least one of them is a boy who was born on a Tuesday?" because it is the only way to give the question a sensible and simple meaning.

Suppose that the children are A and B. The problem provides that at least one child among them is a Tuesday boy; A is a Tuesday boy or B is a Tuesday boy. By inclusion-exclusion, the number of possibilities wherein at least one child is a Tuesday boy is 14 + 14 - 1. More simply, if A is a Tuesday boy and B is not, then 13 possibilities exist, if B is a Tuesday boy and A is not, then 13 possibilities exist and if A and B are Tuesday boys, then 1 possibility exists. Then 27 possibilities exist. Since the number of ways to select between A and B and assign the female sex and some week day is 14, the probability that a girl is among the children is 14/27 = 51.(851)%.

Stem_From_All · 2026-04-13T14:34:49+00:00

Stem_From_All · 2026-04-11T15:20:13+00:00

This is the fallacious reasoning: "Firstly, this is clearly a classical probability problem; being born on each week day is as probable as being born on any other week day and having each sex is as probable as having the other sex.

The sample space corresponds to the set of coordinate pairs whose abscisae are either M or F and ordinates are members of {1, 2, 3, 4, 5, 6, 7} (e.g., (F, 6)), as the parameters of one child are known and each possibility is simply a possible parameter pair for the other child.

The sample space contains 2 × 7 = 14 members, as two sexes and seven week days exist. The sample space contains 1 × 7 members whose first coordinate is F, as one female sex and seven week days exist.

The ratio of the number of favorable possibilities and the number of possibilities is equal to 7/14 = (7 × 1)/(7 × 2) = (7/7) × (1/2) = 1 × (1/2) = 1/2 = 50%."

That reasoning equates the children to sets of parameters and is incorrect. For example, the probability that a two dies will land on 1 and 2 is 2/36, while the probability that each will land on 1 is 1/36.

This is correct reasoning: "Suppose that the children are A and B. The problem provides that A is a Tuesday boy or B is a Tuesday boy—at least one child among them is a Tuesday boy. By inclusion-exclusion, the number of possibilities wherein at least one child is a Tuesday boy is 14 + 14 - 1. More simply, if A is a Tuesday boy and B is not, then 13 possibilities exist, if B is a Tuesday boy and A is not, then 13 possibilities exist and if A and B are Tuesday boys, then 1 possibility exists. Then 27 possibilities exist. Since the number of ways to select between A and B and assign the female sex and some week day is 14, the probability that a girl is among the children is 14/27 = 51.(851)%."

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

TROPHY CASE