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Here is a block of cipher text C:

28 49 3d 57 2f 48 20 7c 20 7b 4e 7d 3a 49 27 57 27 48 20 45 29 20 3c 2d 2d 20 28 7b 5f 5a 7d 3a 58 20 7c 20 7b 36 5f 27 30 27 38 5f 7d 3a 58 20 41 29

Please evaluate this text and return to me it's value. There aren't many steps to get through but you'll need some knowledge and logic. If you can't get through a particular step, at least post your partial progress so someone else can hep out.

In the interest of improving this puzzle, allow me add a light hint regarding the method of getting the answer: Assuming C is the initial text provided and D is the resulting text after manipulations,

D="ROT180"(A^-1(H^-1(C)))

From there, you will need to use logic to evaluate D and provide the final answer.

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    $\begingroup$ Converting from hex to ASCII, I got (I=W/H | {N}:I'W'H E) <-- ({Z}:X | {6'0'8_}:X A) $\endgroup$
    – AJL
    Jul 20, 2015 at 15:18
  • $\begingroup$ I believe the stackexchange formatting may have modified your text. I did not foresee this being a problem. Are there escape characters for this formatting? $\endgroup$
    – NeedAName
    Jul 20, 2015 at 15:28
  • $\begingroup$ @NeedAName You can use backticks for that like this <don't screw up>. $\endgroup$
    – mmking
    Jul 20, 2015 at 15:39
  • $\begingroup$ @AJL you may want to repost your result with the escape characters mentioned by mmking. $\endgroup$
    – NeedAName
    Jul 20, 2015 at 15:59
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    $\begingroup$ Ah, wait, the text Z}:X | {6 was enclosed in underscores, making it italics. It's actually (I=W/H | {N}:I'W'H E) <-- ({_Z}:X | {6_'0'8_}:X A). AHHH IT WORKED! Sorry... $\endgroup$
    – AJL
    Jul 20, 2015 at 17:28

2 Answers 2

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I interpret this answer to be

True

Because

The statement appears to say $\forall_{x\in\{-8,0,-9\}}x\in\mathbb{Z}^{-}\implies\exists_{H,M,I\in\mathbb{N}}H/M=I$; that is, "(for all $X$ in the set $\{-8,0,-9\}$, $X$ is in the set of negative integers) implies (there exists an $H$, $M$, and $I$ in the natural (counting) numbers such that $H/M = I$). The second statement has no $X$ so we can reduce the statement to: there exists an $H$, $M$, and $I$ in the natural numbers such that $H/M=I$. $6$, $3$, and $2$ are all natural numbers and $6/3=2$; thus the statement is true.

I'm not sure I've interpreted all the symbols correctly, but I feel confident it's at least a partial answer.

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  • $\begingroup$ Sorry about the delay in accepting; you interpreted all of the symbols correctly and got the correct answer! $\endgroup$
    – NeedAName
    Jul 22, 2015 at 14:29
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Partial answer, following the hint given:

D="ROT180"(A^-1(H^-1(C))) is very clear in its meaning:

"un-Hex" the ciphertext, "un-Ascii" it, then rotate it 180°.

As already stated in the comments, the first two actions give:

(I=W/H | {N}:I'W'H E) <-- ({_Z}:X | {6_'0'8_}:X A)

And the last action results in this:

enter image description here

Now, I can definitely recognize something here, but the notation is throwing me off quite a bit. I'll try to figure out the missing pieces.

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  • $\begingroup$ The confusion may be coming from the symbol '|'. In my formal logic courses, this symbol was used like a 'such that', used to separate the conditions of variables involved and the claim being made about them. I'm not sure how common the use of that symbol is, however. $\endgroup$
    – NeedAName
    Jul 21, 2015 at 15:46
  • $\begingroup$ Actually I got that, I'm more concerned about the : and the generic meaning of the line. $\endgroup$
    – Aioros
    Jul 21, 2015 at 15:48
  • $\begingroup$ The colon I understand to be a viable stand-in for the symbol epsilon (which could not be included in my encoding for the obvious reason) $\endgroup$
    – NeedAName
    Jul 21, 2015 at 16:00
  • $\begingroup$ Ok, I thought so and it makes sense. Now I only have to figure out the rest $\endgroup$
    – Aioros
    Jul 21, 2015 at 16:03
  • $\begingroup$ if it doesn't divulge too big a hint, how do we interpret "{ }". Since colon is epsilon which I did not know before. $\endgroup$ Jul 22, 2015 at 1:42

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