[Guest]

The Encoder and Decoder both take each character of the two inputs, covert them into binary using the table below, then do 'something' that results in one output. This output is converted back using the same table and comes out the other end.

Table:

Bin     Chr     Bin     Chr     Bin     Chr     Bin     Chr

00000  <spc>	01000 	H	10000	P	11000	X

00010	A	01001	I	10001	Q	11001	Y

00010	B	01010	J	10010	R	11010	Z

00011	C	01011	K	10011	S	11011	.

00100	D	01100	L	10100	T	11100	,

00101	E	01101	M	10101	U	11101	?

00110	F	01110	N	10110	V	11110	!

00111	G	01111	O	10111	W	11111	-

[/codebox]

[b]

Code: BRAINDEN ROCKS

Key: N?F IDL? SCJ?V

CAN YOU THE DECODED TEXT?[/b]

Edited June 15, 2009 by adiace

[Guest]

The Encoder and Decoder both take each character of the two inputs, covert them into binary using the table below, then do 'something' that results in one output. This output is converted back using the same table and comes out the other end.

Table:

Bin     Chr     Bin     Chr     Bin     Chr     Bin     Chr

00000  <spc>	01000 	H	10000	P	11000	X

00010	A	01001	I	10001	Q	11001	Y

00010	B	01010	J	10010	R	11010	Z

00011	C	01011	K	10011	S	11011	.

00100	D	01100	L	10100	T	11100	,

00101	E	01101	M	10101	U	11101	?

00110	F	01110	N	10110	V	11110	!

00111	G	01111	O	10111	W	11111	-

[/codebox]

[b]

Code: BRAINDEN ROCKS

Key: N?F IDL? SCJ?V

CAN YOU THE DECODED TEXT?[/b]

Is A really supposed to be equal to 00010? It seems like it should be 00001.

[Guest]

Assuming that A is really 00001 and that the KEY = N?F MDL? SCJ?V

The answer is "LOGIC IS ALIVE".

[Guest]

Is A really supposed to be equal to 00010? It seems like it should be 00001.

Your right

[Prof. Templeton]

Assuming that A is really 00001 and that the KEY = N?F MDL? SCJ?V

The answer is "LOGIC IS ALIVE".

Ahhh, I see it.

[Guest]

Since 'A' is really 00001, if you perform an XOR operation on the code and the key, then you get 'LOGIG IS ALIVE'. If you change the 'I' in the Key to an 'M' (01101), then you get 'LOGIC IS ALIVE', which would make more sense.

The Encoder and Decoder both take each character of the two inputs, covert them into binary using the table below, then do 'something' that results in one output. This output is converted back using the same table and comes out the other end.

Table:

Bin     Chr     Bin     Chr     Bin     Chr     Bin     Chr

00000  <spc>	01000 	H	10000	P	11000	X

00010	A	01001	I	10001	Q	11001	Y

00010	B	01010	J	10010	R	11010	Z

00011	C	01011	K	10011	S	11011	.

00100	D	01100	L	10100	T	11100	,

00101	E	01101	M	10101	U	11101	?

00110	F	01110	N	10110	V	11110	!

00111	G	01111	O	10111	W	11111	-

[/codebox]

[b]

Code: BRAINDEN ROCKS

Key: N?F IDL? SCJ?V

CAN YOU THE DECODED TEXT?[/b]

[Guest]

The code is "LOGIG IS ALIVE"

[Guest]

The code is "LOGIG IS ALIVE"

The puzzle does not specify that the same operation is used to encode and decode, but only that the same conversion table be used. That means that any operation which can be reversed (from a complicated lossless compression, to a simple NOP) is satisfactory. If the problem statement truly meant the same operation, it is unnecessary to represent the black-box as two distinct entities (encoder vs. decoder).

[Guest]

yah but the only real processes that are reverseable are xor and not for example

A|B=010101

a could equal anything as long as a one does not appear where there is a zero.

Same for "and" except zero's cant appear where the answer has a one.

"not" is reversible but does not use any key or other input.

Therefore unlesss you gonna go wacky and say xor key -86 or something like that i think xor is the only way. but i might be forgeting an operation i guess. (for clarification obviously there are other encoding processes but i dont think they apply here)

havent taken circuits in a while tho

[Guest]

yah but the only real processes that are reverseable are xor and not for example

A|B=010101

a could equal anything as long as a one does not appear where there is a zero.

Same for "and" except zero's cant appear where the answer has a one.

"not" is reversible but does not use any key or other input.

Therefore unlesss you gonna go wacky and say xor key -86 or something like that i think xor is the only way. but i might be forgeting an operation i guess. (for clarification obviously there are other encoding processes but i dont think they apply here)

havent taken circuits in a while tho

I'm not speaking in terms of circuits but simply in terms of the problem statement. If you want the simplest solution, it's a no-op. Doing nothing to the sequence satisfies the requirements. If you want something a little more interesting, how about reversing the input (011011 -> 110110)? How about swapping every pair of bits (011011 -> 100111)? These operations require no XOR operations (the first is just a LIFO).

I'll grant that the OP probably expected the answer already given. However, this is brainden, so I thought I'd play devil's advocate and point out the open-endedness of the problem statement.

[Guest]

fair enough i hadnt thought about just switching bits that is possible in a circuit. so you would have to use the key xored before or after but in between whatever. good thought