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how to do the following?

조회 수: 1 (최근 30일)
Elysi Cochin
Elysi Cochin 2016년 11월 5일
편집: Elysi Cochin 2016년 11월 5일
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?

채택된 답변

Guillaume
Guillaume 2016년 11월 5일
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
  댓글 수: 3
Elysi Cochin
Elysi Cochin 2016년 11월 5일
편집: Elysi Cochin 2016년 11월 5일
So what about in the other case when color of v(x,y) is black and the colors of q(x,y) is white
Walter Roberson
Walter Roberson 2016년 11월 5일
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.

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추가 답변 (2개)

Walter Roberson
Walter Roberson 2016년 11월 5일
if rand() <= 0.5 %probability 1/2
set white pixel to black at this point
end
  댓글 수: 1
Walter Roberson
Walter Roberson 2016년 11월 5일
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

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KSSV
KSSV 2016년 11월 5일
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.
  댓글 수: 1
Elysi Cochin
Elysi Cochin 2016년 11월 5일
편집: Elysi Cochin 2016년 11월 5일
but how will it come in implementation side
if V(x,y) == 0 && q(x,y) == 1 && r(x,y) == 0
q(x,y) = 0;
end
i did like the above one, but what about the probability...

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