We're given the following image,

and we are given the caption "I feel pretty", and we are questioned, "What's next?"

Does anyone have a guess at what the answer might be?

Looking at how the xors are connected, the output of the leftmost xor will output a 0.

I tried to take the resulting binary string, (i.e. 0101111101110111) and convert it to text via ASCII.

The text it returns is "_w" which is incorrect.

Any ideas?

EDIT: You are required to input the next consequtive shifts.

 

Keep going... shift one bit right at a time and record the results after every shift.

 

I didn't work it out, but I bet the answer is "OH so pretty."

 

I tried but didn't get much sense.

What was the answer?

I simulated two steps as shown in pdf's

 

isn't 1011 1110 1110 1111 BEEF in hex?

 

isn't 1011 1110 1110 1111 BEEF in hex?

Yes and the next step is 0101 1111 0111 0111 which is 5F77 in hex. But it doesn't make any sense.
Only if we know the relationship between 1011111011101111 and "I feel pretty", we can decode the next step.

 

At least I can now make "pretty pictures" even if it is all meaningless.

 

At least I can now make "pretty pictures" even if it is all meaningless.

Is that picture of yours some random step since it doesn't match with either of the 2 values(current and next)
Anyway nice picture

 

hmmm...
1011 1110 1110 1111 in decimal gives
11___ 14___ 14___ 15 which when converted to the corresponding number of the alphabet makes
k ____ n ____ n ___ o

then
0101 1111 0111 0111
5 15 7 7
EOOG

I don't think this is going anywhere...

 

Is that picture of yours some random step since it doesn't match with either of the 2 values(current and next)
Anyway nice picture

I thought the values were OK.

I've attached the screens I get after the first and second right shifts.

 

Perhaps it's a trick. Looks like an XOR encrypter of sorts.

Who was "pretty"? - The Mona Lisa.

It's the Da Vinci Code!

Where's Jegues? Has he stolen the encoder?

Insanity slowly overtakes me! I'm becoming paranoid.

Loosewire might be able to help.

 

It's an LFSR; Linear Feedback Shift Register.

 

1 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1
0 1 0 1 1 1 1 1 0 1 1 1 0 1 1 1
1 0 1 0 1 1 1 1 1 0 1 1 1 0 1 1
1 1 0 1 0 1 1 1 1 1 0 1 1 1 0 1
1 1 1 0 1 0 1 1 1 1 1 0 1 1 1 0
1 1 1 1 0 1 0 1 1 1 1 1 0 1 1 1
1 1 1 1 1 0 1 0 1 1 1 1 1 0 1 1
0 1 1 1 1 1 0 1 0 1 1 1 1 1 0 1
1 0 1 1 1 1 1 0 1 0 1 1 1 1 1 0
1 1 0 1 1 1 1 1 0 1 0 1 1 1 1 1
0 1 1 0 1 1 1 1 1 0 1 0 1 1 1 1
0 0 1 1 0 1 1 1 1 1 0 1 0 1 1 1
1 0 0 1 1 0 1 1 1 1 1 0 1 0 1 1
1 1 0 0 1 1 0 1 1 1 1 1 0 1 0 1
0 1 1 0 0 1 1 0 1 1 1 1 1 0 1 0
0 0 1 1 0 0 1 1 0 1 1 1 1 1 0 1
1 0 0 1 1 0 0 1 1 0 1 1 1 1 1 0
1 1 0 0 1 1 0 0 1 1 0 1 1 1 1 1
0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 1
1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1
0 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1
1 0 1 0 1 1 0 0 1 1 0 0 1 1 0 1
0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 0
0 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1
0 0 0 1 0 1 0 1 1 0 0 1 1 0 0 1
0 0 0 0 1 0 1 0 1 1 0 0 1 1 0 0
0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 0
0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1
0 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1
1 0 0 0 0 0 0 0 1 0 1 0 1 1 0 0
1 1 0 0 0 0 0 0 0 1 0 1 0 1 1 0
1 1 1 0 0 0 0 0 0 0 1 0 1 0 1 1
0 1 1 1 0 0 0 0 0 0 0 1 0 1 0 1
1 0 1 1 1 0 0 0 0 0 0 0 1 0 1 0
0 1 0 1 1 1 0 0 0 0 0 0 0 1 0 1
0 0 1 0 1 1 1 0 0 0 0 0 0 0 1 0
1 0 0 1 0 1 1 1 0 0 0 0 0 0 0 1
0 1 0 0 1 0 1 1 1 0 0 0 0 0 0 0
0 0 1 0 0 1 0 1 1 1 0 0 0 0 0 0
1 0 0 1 0 0 1 0 1 1 1 0 0 0 0 0
1 1 0 0 1 0 0 1 0 1 1 1 0 0 0 0
0 1 1 0 0 1 0 0 1 0 1 1 1 0 0 0
0 0 1 1 0 0 1 0 0 1 0 1 1 1 0 0
1 0 0 1 1 0 0 1 0 0 1 0 1 1 1 0
0 1 0 0 1 1 0 0 1 0 0 1 0 1 1 1
0 0 1 0 0 1 1 0 0 1 0 0 1 0 1 1
1 0 0 1 0 0 1 1 0 0 1 0 0 1 0 1
0 1 0 0 1 0 0 1 1 0 0 1 0 0 1 0
0 0 1 0 0 1 0 0 1 1 0 0 1 0 0 1
0 0 0 1 0 0 1 0 0 1 1 0 0 1 0 0
1 0 0 0 1 0 0 1 0 0 1 1 0 0 1 0
...

 

As SgtWookie correctly notes this is an example of an LFSR.

This configuration corresponds to a so-called Fibonacci LFSR. According to Wikipedia, this case has the property of being able to cycle through all of the possible (65535) 16-bit states other than the all zeros condition - without repetition. So as long as the starting value is non-zero it should work.

Have no idea concerning the relevance of the caption "I feel pretty".

 

perhaps the golden ratio plays in somewhere?

 

I figure that the answer would be, how many iterations does the LFSR have to go through in order to get to the start of the next verse of the West Side Story song, which would be "Oh", or 4F68 in hex, or 0100 1111 0110 1000 in binary.

It would take you a very long time to determine that answer by hand, as it will take you more than 33,000 iterations to arrive at the correct answer (very big hint).

There are three other possibilities with mixing of upper and lower case "O"s and "H"'s.

 

This challenge has been completed.

The answer was to submit the next 6 or so shifts that the circuit would do.

Here's the link to the next level which I'm currently stuck on,

http://forum.allaboutcircuits.com/showthread.php?t=57407

I'd love some more help!

 

This challenge has been completed.

The answer was to submit the next 6 or so shifts that the circuit would do.

So this two page discussion regarding that "I feel pretty" is a waste huh? I'm a bit disappointed.

 

This challenge has been completed.

The answer was to submit the next 6 or so shifts that the circuit would do.

Here's the link to the next level which I'm currently stuck on,

http://forum.allaboutcircuits.com/showthread.php?t=57407

I'd love some more help!

Since you are probably not winning any popularity contests around here at the moment I'd give the caption challenge idea a rest for a while.

 

Ah, well - here's what I came up with...

"oh" was at iteration 14,975
"OH" "   "  "         33,689
"Oh" "   "  "         46,001
"oH" "   "  "         63,614

Believe it or not, I used an Excel spreadsheet.