'Removing gray from RGB. I am using coral to learn programming + its apart of a course
Problem :
Summary: Given integer values for red, green, and blue, subtract the gray from each value.
Computers represent color by combining the sub-colors red, green, and blue (rgb). Each sub-color's value can range from 0 to 255. Thus (255, 0, 0) is bright red, (130, 0, 130) is a medium purple, (0, 0, 0) is black, (255, 255, 255) is white, and (40, 40, 40) is a dark gray. (130, 50, 130) is a faded purple, due to the (50, 50, 50) gray part. (In other words, equal amounts of red, green, blue yield gray).
Given values for red, green, and blue, remove the gray part.
Ex: If the input is 130 50 130, the output is:
80 0 80
My code :
integer red
integer green
integer blue
integer gray
red = Get next input
green = Get next input
blue = Get next input
gray = 50
if red >= 50 and red <= 255
red = red - gray
Put red to output
elseif green >= 50 and green <= 255
green = green - gray
Put green to output
elseif blue >= 50 and blue <= 255
blue = blue - gray
Put blue to output
I need help trying to substract the smallest value from an input of three numbers rather than just 50 and output those three integers rather than just the first integer.
Note : I wanted to use the 'min' function but there is not much in the built in library for coral. I think this challenge wants to focus on branching.
(I can fairly read and understand python and java code as well if you wanted to help me using them)
Solution 1:[1]
integer red
integer green
integer blue
integer grey
red = Get next input
green = Get next input
blue = Get next input
if red < green and red < blue
grey = red
elseif green < red and green < blue
grey = green
else
grey = blue
red = red - grey
green = green - grey
blue = blue - grey
Put red to output
Put " " to output
Put green to output
Put " " to output
Put blue to output
Put " " to output
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Sid Hartman |
