XOR
I have a sort of weird question. Say I have three numbers, A, B, and C. All three numbers are between 0 and 255. Say I have:
A xor B=205
A xor C=82
B xor C=159
Can you find A, B, and C? If so, please tell me how.
XOR
Forum
» General Discussion / Open Topic
» XOR
Summary:
weird question
I'm not quite sure what you're asking. Are you talking about bitwise xor, so (A xor B)=205=%10101011 and (A xor C)=82=%01010010 and (B xor C)=159=%10011111?
It's generally possible to determine A, B, and C (not necessarily unique), but those conditions you gave aren't possible.
Let $A = A_1A_2A_3A_4A_5A_6A_7A_8, B = B_1...B_8, C = C_1...C_8$. To find A, B, and C you have to find them one binary digit at a time. Starting with the left-most bit, we have that $A_1 xor B_1 = 1, A_1 xor C_1 = 0, B_1 xor C_1 = 1$. Now since A1 xor C1 = 0, that means A1=C1. Since A1 xor B1 = 1 then A1 != B1, and since B1 xor C1 = 1 then B1 ≠ C1. This is satisfied by either (A1, B1, C1) = (1, 0, 1) or (0, 1, 0).
Moving on to bit two, we have $A_2 xor B_2 = 0, A_2 xor C_2 = 1, B_2 xor C_2 = 0$. These yield that A2=B2, B2=C2 so that means A2=C2, but since A2 xor C2 = 1, A2≠C2 which leads to a contradiction so the particular configuration you gave is impossible.
right, I know it will give two results. I will have a way to know which is the right one. I just need to know how to find the two solutions.
EDIT: Also recall that A is the same in A xor B, and A xor C. The same goes for B and C.
The first bit gives two results, and they don't affect the other bits. The problem is the second set of bits. There don't exist any combinations of A2, B2, C2 such that (A2 xor B2 = 0) and (A2 xor C2 = 1) and (B2 xor C2 = 0).
wait a second, this is the correct binary for 205:
205=%11001101
also, click options on my post, and then edit. In hidden text, are the three solutions. So I know that they are accurate when I used xor on them.
wait a second, this is the correct binary for 205:
205=%11001101
Whoops, that could be why I thought it was impossible :)
Do you want me to list all 256 possibilities?
What do you mean? For every binary digit there are exactly 2 combinations of A/B/C bits, and since each bit is independent from the others there are 2^8=256 distinct possible sets of A/B/C satisfying those conditions you gave.
If you're asking about my methods, I just repeated the process in my second post 8 times.
random: For some reason the auto-spell-checker thing doesn't mark "kinda" or "wanna" as incorrect, but it does say "combinations" is spelled wrong.
Removed due to making the forum too large.
Two questions:
1) What would the checksum be checking for?
2) What's this being used for?
Well, I am compressing every three bytes into two:
A XOR B=D
A XOR C=E
D XOR E=(A XOR B) XOR (A XOR B$)
The checksum would be to make sure the decompressed file is the right one.
So basically you are going to compress three bytes into two, then add a third byte as a check sum, for a total compression of… zero bytes?
Okay, I have a new problem/challenge. Say I have X number of bytes (numbers between 0 and 255). Note that X is an even number. For simplicity, I will use 4 for X. Now then, these example numbers:
| Number | A-S | Output |
|---|---|---|
| 37 | 0 | 241 |
| 241 | 241 | |
| 232 | 195 | 64 |
| 131 | 131 |
Let me explain. Every odd number, eg. 1 and 3, is labeled A. Every even number, eg. 2 and 4, is labeled B. The very first byte is labeled S. In the second column, You see that I took whatever A was, and subtracted S. In the third column, I took the A and B pairs, and did XOR logic on them, resulting in the output.
I now need to get the input. You will be able to use S, and any of the output numbers to find the input numbers. Please tell me how you got the solutions. This needs to work for all of the numbers, not just the first two. I know that this is possible.
Egh, your table was a mess to sort through. For a while, I was looking at the words "odd" and "even" as though they were describing the numbers in the cells, rather than the row numbers. To me, something like this would make more sense…
|
|
…and then the question is, "How do we get B, C, and D if A, E, and F are all that we are given?"
For starters, we know that B must be E. But, in order to find C in your example, we'll need to find a combination of X and Y such that (X-37) xor Y is 64. The possibilities, which are infinite, are such that the difference between X and Y is 101. Good luck getting the program to figure out which one is right; this can be compared to taking the number 63 and then asking what the addends were. Sure, we might have plugged in 18 + 45 at some point in time, but there's no way to restore our initial inputs exactly, unless they were written down on the side. I leave it up to you what this entails.
Well, on my example, The S would have been recorded.
S… A… We still know it to be 37, and it is still not enough to make headway towards finding C or D.
If we act upon a number of objects with a fixed set of properties, then we are either going to A) lose track of those properties, B) swap them, which automatically creates ancestral ambiguity (unless a new fact-checking variable is added), or possibly C) duplicate them between elements, but then that would give us something else to memorize.
But, if we were to displace an item from the set, and it retained a unique quality that wasn't preserved by a different resource, then all hope of bringing everything back to the way it started is lost. There's no approach that will settle this in favor of compression, because if there were, we could do it again on the set, ad infinitum. That's all there is to it.
/forum/t-122587/xor#post-