Day 16: Advent Calendar

[Kathis]

Hello Jellybeans, Day 16 of puzzles. Todays lesson is about Colour~ and a bit of bitlogic.

 

We'll start with the colour.

 

Red, Green, Blue, Magenta, Cyan, Yellow, Black, and White. Have a what I like to call a pure hex value. 

 

Black #000000

Red #FF0000

Green #00FF00

Blue #0000FF

Yellow #FFFF00

Cyan #00FFFF

Magenta #FF00FF

White #FFFFFF

 

 

The first two values are Red values 0-255  (00-FF)

the second values are Green values 0-255 (00-FF)

and the last values are Blue values 0-255 (00-FF)

#RRGGBB

Now I know you have paint, and you can use the eye dropper. To get the decimal values.  and check it in the edit colour. 

 

----

Now bitlogic. 

 

Math in bit logic is a bit... weird.  

bitwise Or ( | ) is simple.  Take these two bytes for example. 

 

1000 1011

0111  0101

 

we compare the 1 and the 0, if there is a 1, then the final answer would have a 1 in that slot. 

 

So since every slot has a 1 at some point, it's 1111 1111 or FF

 

bitwise And (&) however is a bit more... picky. Like Or it will compare the 0 and the 1, but they both need to have a 1, otherwise it's a 0.

With our pervious bytes the answer would be 0000 0001  since only that slot had a 1 in both bytes.

 

bitwise Not (~) 

Just flips the bit.  0's become 1, and 1's become zero. For this puzzle you will not need to worry about it. But it is worth mentioning

 

-----

 

That's the crash course.

So let us begin. The easy exercise / puzzle.

 

(I made the | bolder since it was harder to see...)

 

Provide the answer in decimal with our respective letters.

 

 

Good luck everyone. 

[tachy]

Blue (a) = 2B8BEC

Indigo (b) = 626AB5

Lime (c) = B9E730
 

Each hexadecimal number can be represented by 4 bits, from 0000 (0) to 1111 (F). 

So, each color would have 6 x 4 = 24 bits. 

Then, you can break it down and convert by dividing each number by 2 until the number is zero.

The remainder for each division is the binary bit starting from the "ones" place: 

2 = 0010 (2/2 = 1 R0, 1/2 = 0 R1)

B = 1011 (11/2 = 5 R1, 5/2 = 2 R1, 2/2 = 1 R0, 1/2 = 0 R1)

8 = 1000 (8 is just 2^3 so it's 1000, because 2, 4, 8, 16, and so on is a 1 followed by X 0s using X from 2^X)

B = 1011 (B was done)

E = 1110 (minus 1 from F 1111)

C = 1100 (add 1 to B)

and combining them still gives the same number 001010111000101111101100 = 2B8BEC

 

Or you can just convert it somewhere. 

In fact, you could just do: ████████ & ████████ | ███████
 

Logically, the order would be (a & b) | c = 12mXXXkXX6 but the answer doesn't seem right?

 

[Kathis]

Your hex values are correct for the colours. But somewhere the math didn't math. Your last combination number  is off by a few bits flips, and I am trying to figure out where that happened. What do you get when you do a & b ? 

[tachy]

a&b: ██████████████████████

[Kathis]

Yep, that is correct. Saw the issue, upped the reward it was a fault of mine. Redacted the obvious. I'll leave the others since it still is part of the community solving if they need a hint or help.