UGCNET-Dec2012-III: 71

1.3k views

You are given four images to represent as

$I_1 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, I_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $I_3 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, I_4 = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$

The value of entropy is maximum for image

1. $I_1$
2. $I_2$
3. $I_3$
4. $I_4$

recategorized

ans C

entropy of image represents the randomness in pixels. in I1 all pixels are same. entropy is 0.
in I3 2 pixels are same other 2 diff.
in i2 i4, 3 are same.
so most random is in I3

Related questions

1
1.5k views
The Z-buffer algorithm is used for Hidden surface removal of objects. The maximum number of objects that can be handled by this algorithm shall depend on the application be arbitrary no. of objects depend on the memory availability depend on the processor
The Mandelbrot set used for the construction of beautiful images is based on the following transformation. $x_{n+1}=x^2_n +z$. Here, Both $x$ and $z$ are real numbers Both $x$ and $z$ are complex numbers $x$ is real and $z$ is complex $x$ is complex and $z$ is real
Let $V_1 = 2I-J+k$ and $V_2 = I+J-K$, then the angle between $V_1$ and $V_2$ and a vector perpendicular to both $V_1$ and $V_2$ shall be $90^o \text{ and } (-2I+J-3K)$ $60^o \text{ and } (2I+J+3K)$ $90^o \text{ and } (2I+J-3K)$ $90^o \text{ and } (-2I-J+3K)$