Yes! Thank you very much! It is equivalent to shifting the green points in the x-direction to x=1. My mistake. The solution to the puzzle at 1:09 is indeed not unique. Now the video itself becomes a meme that self-referential puzzles are hard (even to make).
yeah, I got those two solutions: A, C, B; D, C, B; my process for solving was just to write out the situation in formal logic. here's what I got (with the restriction that a_n, b_n, c_n, and d_n are mutually exclusive): a₁⇒¬a₂∧¬a₃ b₁⇒¬a₂∧¬a₃ c₁⇒¬a₁∧a₂∧a₃ d₁⇒¬a₁∧¬a₂∧¬a₃ a₂⇒d₃ b₂⇒c₃ c₂⇒b₃ d₂⇒a₃ a₃⇒¬b₁∧¬b₂∧¬b₃ b₃⇒¬b₁∧¬b₂ c₃⇒b₁∧b₂ d₃⇒b₁∧b₂∧b₃ if you work it out, the first thing's you'll find are that a₁ and b₁ are impossible, and from that a₂, d₃, a₃, b₂, c₃ are impossible. That leaves only the two aforementioned solutions left.
Q1 = A, Q2 = C, Q3 = B is also consistent no?
Yes! Thank you very much! It is equivalent to shifting the green points in the x-direction to x=1. My mistake.
The solution to the puzzle at 1:09 is indeed not unique.
Now the video itself becomes a meme that self-referential puzzles are hard (even to make).
yeah, I got those two solutions: A, C, B; D, C, B;
my process for solving was just to write out the situation in formal logic. here's what I got (with the restriction that a_n, b_n, c_n, and d_n are mutually exclusive):
a₁⇒¬a₂∧¬a₃
b₁⇒¬a₂∧¬a₃
c₁⇒¬a₁∧a₂∧a₃
d₁⇒¬a₁∧¬a₂∧¬a₃
a₂⇒d₃
b₂⇒c₃
c₂⇒b₃
d₂⇒a₃
a₃⇒¬b₁∧¬b₂∧¬b₃
b₃⇒¬b₁∧¬b₂
c₃⇒b₁∧b₂
d₃⇒b₁∧b₂∧b₃
if you work it out, the first thing's you'll find are that a₁ and b₁ are impossible, and from that a₂, d₃, a₃, b₂, c₃ are impossible. That leaves only the two aforementioned solutions left.
@@thezipcreator D, D, A doesn't work.
@@elorating edited my comment and fixed it before you said that lol (I assume you probably saw the old version in the notification)
It doesn't work
Teacher: the test isn’t that confusing
The test: 1:15
🤣
There are fourteen minus three e in this sentence.
The meme has five e in it.
@@ron-math hahahahaha.
Interesting :)
Please read aloud the questions so I can just listen.
Cool