[SOLUTION] Mexican Blanket (#127)

Hello,

Here is my solution to the quiz. Looking at the pattern for a moment,
each
line of the pattern is really just the previous line, shifted over one
pixel. Thus to draw the pattern we can first calculate what all of the
pixels will be. Then we simply iterate over this list, drawing one line
at a
time. With that in mind, here is the code I used to create the gradient:

def create_gradient(colors, width=5)
pattern = []

for i in 0…(width)
(width-i).times { pattern.push(colors[0]) }
(i+1).times { pattern.push(colors[1]) }
end

for i in 0…(width)
(i+1).times { pattern.push(colors[2]) }
(width-i-1).times { pattern.push(colors[1]) }
end

pattern
end

And this function creates solid bands of color:

def create_solid(colors, width)
pattern = []
for color in colors
width.times { pattern.push(color) }
end
pattern
end

To draw ASCII we simply print each line in turn.

def draw_ascii(pattern, width)
for i in 0…(pattern.size-width+1)
puts pattern.slice(i, width).join
end
end

And to draw a JPEG we get a bit fancier, but its the same basic idea.
There
must be a better way to code this, because it takes much too long to
run.
But anyway, here it is:

require ‘RMagick’
include Magick

def draw(filename, pattern, width, height)
canvas = Magick::ImageList.new
canvas.new_image(width, height, Magick::HatchFill.new(‘white’,
‘white’))
pts = Magick::Draw.new

for y in 0… height
line = pattern.slice(y, width)

x = 0
for color in line
  pts.fill(color)
  pts.point(x, y)
  x = x + 1
end

end

pts.draw(canvas)
canvas.write(filename)
end

Finally, we draw the ASCII from the Quiz and a picture similar to the
one
that was posted.

1) Draw the ASCII representation

draw_ascii(create_gradient([‘R’, ‘B’, ‘Y’]), 28)

2) Draw a picture of the blanket

mex_flag = create_solid([‘rgb(0, 64, 0)’, ‘white’, ‘red’], 5)
border = create_solid([‘rgb(0, 64, 0)’], 25)

pattern = create_gradient([‘red’, ‘blue’, ‘yellow’])
pattern = pattern + mex_flag
pattern = pattern + border
pattern = pattern + create_gradient([‘black’, ‘red’, ‘orange’])
pattern = pattern + border
pattern = pattern + mex_flag.reverse
pattern = pattern + create_gradient([‘red’, ‘purple’, ‘black’], 8)
draw(“mexican_blanket.jpg”, pattern, 100, 200)

Here is a picture of the final blanket, if anyone is interested:
http://justin.ethier.googlepages.com/mexican_blanket.jpg

Any thoughts? The code could probably be more clever, but I tried to
make
everything as easy to follow as possible.
Thanks,

Justin

Justin E. wrote:

And to draw a JPEG we get a bit fancier, but its the same basic idea.
There
must be a better way to code this, because it takes much too long to run.
Nice solution! Regarding the speed, see my earlier reply to Jesse
Merriman. Instead of drawing a pixel at a time, create an array of Pixel
objects and then use import_pixels or store_pixels to “draw” a row at a
time. For that matter, both of these methods can handle arbitrary
rectangles so you can create as many rows as you want and stow them in
the image all at once. You could build the whole image in memory if
that’s what you wanted.

On Sunday 10 June 2007 16:54, Tim H. wrote:

For that matter, both of these methods can handle arbitrary rectangles so you can create as many rows as you want and stow them in the image all at once. You could build the whole image in memory if that's what you wanted.

Though that does give a speed boost, its not much. Here’s a mod of the
last version of draw_blanket I posted:

def draw_blanket blanket
pixels = []
blanket.each_row_with_index do |row, y|
row.split(//).each do |color_char|
pixels << Magick::Pixel.from_color(StringToColor[color_char])
end
end
store_pixels 0, 0, blanket.width, blanket.height, pixels
end

For a 800x800 image, on my box, this shaves about 1/20 second off a
total
run time of about 7 seconds.

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