If you're given just a⊕b and b⊕c, then you can calculate (a⊕b) ⊕ (b⊕c) = a ⊕ (b ⊕ b) ⊕ c (since ⊕ is associative) = a ⊕ 0 ⊕ c (since X⊕X=0 for any X) = a ⊕ c (since X⊕0=X for any X) so in effect when you're given a⊕b, b⊕c, a⊕c you've only been given two numbers (because the last one is redundant). So (assuming a,b,c are 8 bits as ...


This is not possible. Consider the two cases where a, b and c are all true or all false. Now in both cases we have a⊕b = b⊕c = a⊕c = false And more generally, $(¬a)⊕(¬b)=a⊕b$, so if $a,b,c$ is a solution, then so is $¬a,¬b,¬c$. (from Klaus Draeger)


Note that once an alien is in the spot matching its number, it will never swap with another alien again. Every step, one alien moves into its spot, so the number of aliens in the right spot increases by 1 every step. This number can't exceed the number of aliens, which is finite, so there must be a finite number of steps.


Why, her name was Explanation: First, what everyone has deduced so far: Then, How do you like them apples?


Your question with the XOR operator mathematically reduces to: In the $\mathbb{F}_{2^m}$ field (field of characteristic 2 whose elements can be represented as sequences of $m$ bits, and where addition is bitwise XOR), is the following matrix: $$ M = \left( \array{1\ 1\ 0\\1\ 0\ 1\\0\ 1\ 1}\right) $$ invertible ? Indeed, for a dimension-3 vector $V = (x, y, ...


I'm pretty sure WeShall has it, but I'd like to point out that it might be as high as if we assume that




1.Split the input into blocks of size 3. Make the following encryption : AAA : BBBBB AAB : BBBAB ABA : BBABB ABB : BABBB BAA : ABBBB BAB : BABAB BBA : ABBAB BBB : ABABB This leaves us with what to do with the last 1 or two possible characters. This is easily accomplished with: A : A B : B AA : ABA AB : AB BA : BA BB : BB It is easy to see that this ...


I believe the answer is Each of the four lines relates to


No, since, A ⊕ B = B ⊕ C = A ⊕ C = 0 can be either A = B = C = 0 or A = B = C = 1


This might not qualify as a blue-eyes puzzle because it does not use common knowledge, but it involves chains of deductions based on nothing happening for a particular amount of time: $n$ villagers wear either black or white hats. They sit in a line, so that each villager can see all the hats in front of them, but not the hats behind them. If a villager ...




First, let's change our interpretation of the rational numbers as As you may see, if we start with $2^a3^b$ Next What's so special about it? Here, we will get The next part is actually Leaving the "final" number Hence


The angels should play the game, as long as the formula for the computable function must be fixed in advance and finite. Proof: let F be the shortest encoding of the formula (length l(F)) using a set of symbols S of size n(S). Then the guessing angel can simply count up from 0 in base n(S), and some time before n(S)l(F) they will encounter the formula. (...


The question was: Justification: Then the four answers were:


You are I am, in a way, half of eight but not four - not that straight. I am, in a way, half of infinite my location is always definite. I can mean not only not positive for a logician but also almost or similar for a mathematician My origin is of Spain or Portuguese its so easy, so please figure me out of these. Just noticed the title "You are almost ...


You are using an unusual real-world device called Calculations: And for the question:


(as an extension of GOTO 0's answer) This is also not possible if you consider these as binary numbers, you cannot determine the values of a, b, or c. For instance: a = 0b11001111010 b = 0b11001111101 c = 0b11001111001 All three of the xors for these binary representations compress the results down to at most three bits (the right-most three). Anywhere ...


I guess that the answer you are looking for is ...


I'm not convinced this is the final solution (since it does not seem to use all the poster information):


The answer is: Why? Of course, we can cross-validate:


Converting these UTF-8 characters to binary gives the below 7x8 array of bits. Since the bottom is all ones, and you said start at the bottom, the intention must be that you can walk along ones while zeroes are walls. The solution is to stay in the third column from the left and head up, except to jog around a zero in the second row from the top. 01110111 ...


Here is the answer: This was pretty easy though.


Using Jonathan Allan's results, I suspect the final answer to be How I reached this:


There are only a few cases, let's perform a quick inspection: Let $$a⊕b=k_1$$ $$b⊕c=k_2$$ $$c⊕a=k_3$$ , we have \begin{matrix} a & b & c & &k_1 & k_2 & k_3\\ -&-&-&-&-&-&-&\\ 0 & 0 & 0 & \rightarrow &0 & 0 & 0\\ 0 & 0 & 1 & \rightarrow &0 & 1 & 1\\ 0 &...


What you seem to be asking (assuming the squares are completely filled), is to find an Hamiltonian path on a connected subset of the square grid graph. Defintions (lifted from Wikipedia) A Hamiltonian path (or traceable path) is a path in an undirected or directed graph that visits each vertex exactly once. A square grid graph is the graph whose ...


A Limited Word™ is: For example: In detail:


I've given up and tried to bruteforce it - with little regards to what could happen to my machine. I'm importing every builtin module and evaluating the input on every function they've got. As I don't know a list of builtins, I'm using the help("modules") one, the pkgutil.iter_modules() one, and the sys.builtin_module_names one. import pkgutil ...


The guessing angel can win in finite time with 0 integer queries. The computable functions are countable. Enumerate them and guess them in order.


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