6
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I have two operators! They are very close, brothers, actually! Though they are like each other's negative.They are named $\mathfrak{C}^+$ and $\mathfrak{C}^-$.

No one knows what's going on in their heads. It is now crucial! These two operators have concocted a plan to overthrow the common core! The common core government, however, have kidnapped $\mathfrak{C}^+$ and $\mathfrak{C}^-$!

This is a problem. Their mother, $\mathfrak{C}^\pm$, is worried to death. Their father, $\Gamma$, first needs to overthrow the tyrannous common core, but, to do so, he needs his children's plans! If only... if only we could find out what is going on in their heads!!

But news! Their father received a ransom from the common core; in exchange for their children, $\mathfrak{C}^\pm$ and $\Gamma$ must pay \$$g_{64}$! The ransom was penned in the two children's writing, and hurriedly scribbled in the margins is some I/Os. This is the key! Since $\mathfrak{C}^\pm$ can tell what her children are talking about if they tell her, she can figure out what their plan is!

So everything seems all figured out. Except for one thing. No one can find what functions $\mathfrak{C}^+$ and $\mathfrak{C}^-$ are. This is where you come in.

Given the I/O sheet, find out $\mathfrak{C}^+$'s function rule is! From there, the mother will decipher the hidden clues in the message. (The other child did not pen the letter)


Objective: Find the function rule for $\mathfrak{C}^+$. That is all.

Here is the sheet: $$X\subseteq Y \implies \mathfrak{C}^+(X,Y)=Y$$

$$\mathfrak{C}^+(1,2) = \sim6$$ $$\mathfrak{C}^+(1,5) = 9$$ $$\mathfrak{C}^+(1,8) = 8$$ $$\mathfrak{C}^+(2,3) = \sim6$$ $$\mathfrak{C}^+(2,7) = \sim6$$ $$\forall x\in\{0,4,5,6,8,9\}[\mathfrak{C}^+(2,x)=8]$$ $$\mathfrak{C}^+(4,7) = 9 \setminus \{\bot,\bot,\bot,\bot,\top,\bot,\top\}$$


Hints

Ignore the story. It's just for colour.

z1110111
o0010010
t1011101
t1011011
f0111010
f1101011
s1101111
s1010010
e1111111
n1111011

The numbers and letters are sets, which contain either $\top$s or $\bot$s; $\{\bot\}\subseteq\{\top\}$; think about operations of these symbols corresponding to numbers.
time is like a drug (LSD), except without the ex's and PT's (switch 'em)


I believe I have put up sufficient information. Please inform me if this belief is incorrect.

Good luck!!!

Edit: An apology

A key part of the riddle failed to display in mathjax; I have denoted some numbers with a tilde, but did not display. Sorry!

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  • $\begingroup$ What does $\{\bot,\bot,\bot,\bot,\bot,\bot,\top\}$ mean? How can a set have the same element 6 times? $\endgroup$ – Rand al'Thor Jun 7 '15 at 22:38
  • $\begingroup$ @randal'thor I use a loosely-defined set. It's more of an array than anything else. $\endgroup$ – Conor O'Brien Jun 7 '15 at 22:46
  • $\begingroup$ Also, you say $\{\bot\}\subseteq\{\top\}$; how can this be true if the sets contain only a single element each and the elements are different? Is $\subseteq$ used in a non-standard sense as well, i.e. not with the usual 'subset' definition but as an abstract operator between sets/arrays? $\endgroup$ – Rand al'Thor Jun 7 '15 at 23:01
  • $\begingroup$ @randal'thor you got it ;). It is related to $\subseteq$ in the context of sets, but its not the same. $\endgroup$ – Conor O'Brien Jun 7 '15 at 23:02
  • $\begingroup$ Also, should "lcd" be "lsd"? I've heard of LSD drugs but never LCD ones... $\endgroup$ – Rand al'Thor Jun 8 '15 at 22:57
5
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This is about

seven-segment numerical displays. $\mathfrak{C}^+$ returns the union of the segments that are active in its inputs. The numbers in the I/Os are simply shorthand for the set of active segments used to form those digits. In the case of outputs that don't quite match any digits, approximate matches are used. $\sim6$ represents a left-to-right-mirrored 6, and the 9 in the final equation has the bottom segment missing (but is still recognizably a 9).

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  • $\begingroup$ Please see my revision. You're very close! (actually, you're pretty much right--just revise to remove confusion ;)) $\endgroup$ – Conor O'Brien Jun 10 '15 at 20:55
  • $\begingroup$ Is the last equation also wrong? It looks like (if I'm right about what's going on) it should be $$\mathfrak{C}^+(4,7) = 9 \setminus \{\bot,\bot,\bot,\bot,\top,\bot,\top\}$$. $\endgroup$ – user1618143 Jun 11 '15 at 20:00
  • $\begingroup$ Haha, yes, that seems to be the case. Another mistake on my part; terribly sorry. $\endgroup$ – Conor O'Brien Jun 11 '15 at 20:40
  • $\begingroup$ Very good work! Kudos! $\endgroup$ – Conor O'Brien Jun 11 '15 at 23:29
2
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Partial answer

For ease of notation, let's use $0$ to represent $\bot$ and $1$ to represent $\top$. (This is fairly standard in logic theory anyway.) Then $\{\bot\}\subseteq\{\top\}$ makes sense, interpreting $\subseteq$ as simply $\leq$.

The 'numbers' being operated on by $\mathfrak{C}^+$, which we know are actually 'sets' or rather arrays, are now given by part of the spoilertagged hint: we interpret

z1110111
o0010010
t1011101
t1011011
f0111010
f1101011
s1101111
s1010010
e1111111
n1111011

as

0 = 1110111 (or $\{\top,\top,\top,\bot,\top,\top,\top\}$)
1 = 0010010 (or $\{\bot,\bot,\top,\bot,\bot,\top,\bot\}$)
2 = 1011101
3 = 1011011
4 = 0111010
5 = 1101011
6 = 1101111
7 = 1010010
8 = 1111111
9 = 1111011

We interpret $\subseteq$ to mean a componentwise $\leq$, i.e. an array of seven digits is $\subseteq$ another iff each digit in the first is $\leq$ the corresponding digit in the second. So for instance $5\subseteq6$, $3\subseteq9$, and everything is $\subseteq8$. This means $\mathfrak{C}^+(1,8)=8$ is already contained in the rule $X\subseteq Y \implies \mathfrak{C}^+(X,Y)=Y$.

Everything down to here has been confirmed by the OP.

All the examples we've been given of $\mathfrak{C}^+(X,Y)$ satisfy the following rule:

write X and Y as a string of digits ($0$s and $1$s), and sum the digits in each column where possible (so that $(0,1)\rightarrow1$, $(1,0)\rightarrow1$, $(0,0)\rightarrow0$). When the column contains two $1$s, the resulting digit in the new number is also $1$ unless the preceding column contains two $0$s, in which case we have $(01,01)\rightarrow10$ ... except in $\mathfrak{C}^+(1,5) = 9$ and $\mathfrak{C}^+(1,8) = 8$ and $\mathfrak{C}^+(4,7)=9\setminus\{\bot,\bot,\bot,\bot,\bot,\bot,\top\}=1111011-0000001=1111010$, where for some reason $(01,01)\rightarrow01$ in the 5th and 6th columns.

OK, that was a silly rule. More work needed!

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  • $\begingroup$ Looks O.K. so far! $\endgroup$ – Conor O'Brien Jun 7 '15 at 23:33
  • $\begingroup$ @CᴏɴᴏʀO'Bʀɪᴇɴ I've added a rule that works, though it has quite a nasty exception case ... is there a simpler way to describe it? $\endgroup$ – Rand al'Thor Jun 7 '15 at 23:55
  • $\begingroup$ Much much much simpler. But, oh! Look at the time! I must be off (I'll be here all day...) $\endgroup$ – Conor O'Brien Jun 7 '15 at 23:59
1
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I am leaving my original partial answer, and adding a full answer with no changes in my thinking. The answer that is currently marked correct, while close, I did not think was correct. The question, through edits, I thought it became more incorrect.

I have to strip off the leading letter, otherwise things do not line up. Preformatting does not help.

$\mathfrak{C}^+(X,Y)$ is a binary OR operation.
$X \setminus Y$ is a binary XOR operation.

The original problem statement says
$\mathfrak{C}^+(4,7) = 9 \setminus \{\bot,\bot,\bot,\bot,\bot,\bot,\top\}$
0111010 is 4
1010010 is 7
1111010 is 4 OR 7 = $\mathfrak{C}^+(4,7)$
1111011 is 9
1111010 is 9 XOR the original set = $\mathfrak{C}^+(4,7)$
1111110 is 9 XOR the edited set, which does not match $\mathfrak{C}^+(4,7)$


The original post did not have $\sim$ as an operator listed. Thus it was...
$\mathfrak{C}^+(1,2) = 6$
0010010 is 1
1011101 is 2
1011111 is 1 OR 2 = $\mathfrak{C}^+(1,2)$
1101111 is 6
1111011 is $\sim$6 (according to the accepted answer) so that is farther from $\mathfrak{C}^+(1,2)$
Thus, I made the statement that there was a typo in 6.
The typo in 6 does not have an effect on $\mathfrak{C}^+(2,6)=8$

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  • $\begingroup$ Haha, yes, the answer checked is indeed correct--it is the very way I created the puzzle. I am fascinated, however, at the new answer you have formed for yourself. Very innovative! $\endgroup$ – Conor O'Brien Jun 14 '15 at 19:14
  • $\begingroup$ Could you please elaborate on the accepted answer then? There are still things that are not clear to me. $\endgroup$ – DanielKWinsor Jun 17 '15 at 10:21
0
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I am unable to comment, so this also is a partial answer. If you say this is correct I will post a full answer.

I believe that

s1101111 is a typo and should be s1011111

I understand the

letters before the string of 0s and 1s

and although not necessary, they sped up my ability to solve the puzzle.

You say that there is sufficient information, and there is, but

only if you read the spoiler hint. At first, I was avoiding this, but I think one needs to see the actual bits... oops, 0s and 1s.

The sheet's first line is interesting because

it's not strictly necessary to determine the function, and, in some ways follows naturally from the actual function.

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