You are most likely busy creating a game for the game jam and you probably know that an announcement came 2 weeks before the jam as a secret message using steganography I was one of those chaps who actually solved the secret, but didn't do anything. The challenge can be found here. There was a solution, but that was not very explanatory.This post will explain how ~I~ solved the announcement. This post assumes basic knowledge of the binary system.

The first thing I did after seeing the newsletter was to read up on steganography.I found this on wikipedia:

Concealing messages within the lowest bits of noisy images or sound files

Interesting...

Reverse Engineering:

Then, I looked at the code.Apart from imports and loops to go through all pixels,this was the crux of the code:

>> and & are binary operators.shift_amt was 7 and all_except_LSB_mask was 254 What does the >> operator do? It shifts the bits towards towards right.For example: 11100010101 >> 6 outputs 11100 i.e. the last 6 bits are removed. In our case, all the 8 bit data(because 255 is the maximum in RGB color format and 255 is 8 bits) is shifted by 7 bits so that would leave only the highest bit as one or zero. This would also imply that anything greater than 127.5 (255 / 2) would leave 1 while anything lesser would leave zero. 10101010 >> 7 == 1 01010000 >> 7 == 0 So that's our secret pixel Then, we see the & operator with all_except_LSB_mask as 254.The & operator is called bitwise and. Basically, it returns one only if both expressions are one.Like 1&1 == 1 1&0 == 0 0&0 == 0 An important property of the & operator is that something & all 1s = something another one is that something & 0 = 0 10100100&11111111==10100100 254 in binary is 11111110 So,all the bits except the last one would remain intact 10011011 & 11111110 ==10011010 So,this would turn all public pixels into even numbers as the last bit is 2**0=1 and if it is absent, the number must be even. The final pixel is secret_pix + public_pix.We know that public pix is either 0 or 1.Since the last bit of public pix is always zero,The last bit of final pixel must reflect the secret image. So here's my code:

from PIL import Image
public=Image.open('out.png').convert('RGB')
secret=Image.new('RGB',(600,600))
for x in range(600):
for y in range(600):
secret.putpixel((x,y),tuple((i%2)*255 for i in public.getpixel((x,y))))
secret.save('secret.png')

All the juicy stuff is in one line:(i%2)*255 for i in public.getpixel((x,y) What this does is:

take the public pixel

compute modulo two (returns 1 if odd and 0 if even)

multiply that by 255 because 00000001 or 00000000 show no visible difference but 11111111 and 00000000 do

put this in the secret image, pixel by pixel And Heres my output:

## A Detailed Solution to Secret Announcement

## Background:

You are most likely busy creating a game for the game jam and you probably know that an announcement came 2 weeks before the jam as a secret message using steganography I was one of those chaps who actually solved the secret, but didn't do anything.

The challenge can be found here. There was a solution, but that was not very explanatory.This post will explain how ~I~ solved the announcement. This post assumes basic knowledge of the binary system.

The first thing I did after seeing the newsletter was to read up on steganography.I found this on wikipedia:

Interesting...

## Reverse Engineering:

Then, I looked at the code.Apart from imports and loops to go through all pixels,this was the crux of the code:

`>>`

and`&`

are binary operators.shift_amt was 7 and all_except_LSB_mask was 254What does the >> operator do? It shifts the bits towards towards right.For example:

`11100010101 >> 6`

outputs`11100`

i.e. the last 6 bits are removed.In our case, all the 8 bit data(because 255 is the maximum in RGB color format and 255 is 8 bits) is shifted by 7 bits so that would leave only the highest bit as one or zero. This would also imply that anything greater than 127.5 (255 / 2) would leave 1 while anything lesser would leave zero.

`10101010 >> 7 == 1`

`01010000 >> 7 == 0`

So that's our secret pixel

Then, we see the

`&`

operator with all_except_LSB_mask as 254.The & operator is called bitwise and. Basically, it returns one only if both expressions are one.Like`1&1 == 1`

`1&0 == 0`

`0&0 == 0`

An important property of the & operator is that

`something & all 1s = something`

another one is that`something & 0 = 0`

`10100100`

&`11111111`

==`10100100`

254 in binary is

`11111110`

So,all the bits except the last one would remain intact

`10011011 & 11111110 ==10011010`

So,this would turn all public pixels into even numbers as the last bit is 2**0=1 and if it is absent, the number must be even. The final pixel is secret_pix + public_pix.We know that public pix is either 0 or 1.Since the last bit of public pix is always zero,The last bit of final pixel must reflect the secret image.

So here's my code:

All the juicy stuff is in one line:

`(i%2)*255 for i in public.getpixel((x,y)`

What this does is:And Heres my output:

waow indeed

thanks for this! i didnt have any time so i didnt even try to solve the puzzle

@Nanowrimoijk your welcome! It wasn't time taking anyway