how to do the following?
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Elysi Cochin
el 5 de Nov. de 2016
Editada: Elysi Cochin
el 5 de Nov. de 2016
If the color of v(x,y) is black and the colors of q(x,y) and r(x,y) are one white and one black, the white pixel is selected and set to black with the probability of 1/2. What does probability of 1/2 mean?
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Guillaume
el 5 de Nov. de 2016
This is really not complicated. So far you've only implemented one half of any statement. Maybe spend more time understanding the statement?
E.g., for the first one: If the color of v(x,y) is black and the colors of q(x,y) and r(x,y) are both white, one of q(x,y) and r(x,y) is selected with equal probability, and the selected pixel is set to black.
if V(x, y) == 0 && q(x, y) == 1 && r(x, y) == 1 %Implements: If the color of v(x,y) is black and the colors of q(x,y) and r(x,y)
if rand >0.5 %implements: one of q(x,y) and r(x,y) is selected with equal probability and set to black
q(x, y) = 0;
else
r(x, y) = 0;
end
end
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Walter Roberson
el 5 de Nov. de 2016
The chance that rand() exactly equals 0.5 is one in (2^53-2)
You should be following the same structure that Guillaume and I gave: when you have the test against rand to decide which to set, then you should have an else that sets the other one.
Más respuestas (2)
Walter Roberson
el 5 de Nov. de 2016
if rand() <= 0.5 %probability 1/2
set white pixel to black at this point
end
1 comentario
Walter Roberson
el 5 de Nov. de 2016
if V(x,y) == 0 && q(x,y) == 1 && r(x,y) == 1
if rand() <= 0.5
r(x,y) = 0;
else
q(x,y) = 0;
end
end
KSSV
el 5 de Nov. de 2016
You have two blacks v(x,y) and r(x,y); you have 50 - 50 chances to pick either v or r and set it to white. So the probability is 1/2.
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