Clipping Algorithms Questions and Answers

Q: Binary region codes assigned to line end points according to relative position with respect to the clipping rectangle in the row order are :

In row order the codes are 1001, 1000, 1010, 0001, 0000, 0010, 0101, 0100, 0110.Cohen-Sutherland line clipping gives each region a 4-bit code whose bits, read from left to right, stand for above, below, right and left of the window (TBRL); a bit is 1 when the point lies on that side. Reading the nine regions row by row:100110001010000100000010010101000110The top row is above-left, above, above-right; the middle row is left, inside, right; the bottom row is below-left, below, below-right. The other options put 1010 (above-right) in the first position of the second row, or swap codes within the second and third rows.

Q: The binary region code assigned to the line end point located at the bottom-right side of the clipping rectangle in the Cohen-Sutherland line clipping algorithm is ............. .

The bottom-right region code is 0110.Cohen-Sutherland gives each end point a 4-bit code, conventionally ordered top, bottom, right, left. A bit is 1 when the point lies beyond that edge of the clipping window.A point below and to the right of the window is beyond the bottom and right edges, so it gets 0 1 1 0.1010 would be top-right, 0101 bottom-left and 1001 top-left.

Q: Consider a window bounded by the lines : \(x = 0\); \(y = 0\); \(x = 5\) and \(y = 3\). The line segment joining \((-1, 0)\) and \((4, 5)\), if clipped against this window will connect the points

The clipped segment connects \((0, 1)\) and \((2, 3)\).The line through \((-1, 0)\) and \((4, 5)\) has slope 1, so it is \(y = x + 1\). It enters the window at \(x = 0\), giving \(y = 1\), which is the point \((0, 1)\). Inside the window \(y\) cannot exceed 3, so it leaves at \(y = 3\), where \(x = 2\), which is the point \((2, 3)\).The other pairs have a second point that does not lie on \(y = x + 1\), for example \((3, 3)\), \((4, 3)\) and \((3, 2)\).

Which of the given options best describes the truthfulness of the following statements? i) Clipping process determines each element into the visible and invisible portion. The visible portion is selected. An invisible portion is discarded. ii) Clipping can be applied through hardware as well as software. iii) Point clipping, area clipping and text clipping are the types of clipping.

A.i-True, ii-True, iii-False
B.i-True, ii-True, iii-True
C.i-True, ii-False, iii-True
D.i-False, ii-False, iii-False

Exam Year(s)

DSSSB TGT - 2023,

Answer: B

All three statements are true. Clipping decides which part of each element is in the visible region, keeps the visible portion and discards the invisible portion. It can be done in hardware as well as in software, and point clipping, area clipping and text clipping are among its types. So the second option is correct and the options that mark any statement false are wrong. Text clipping can be done by discarding whole strings, characters or parts of characters that fall outside the clip window.

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Binary region codes assigned to line end points according to relative position with respect to the clipping rectangle in the row order are :

A.1001, 1000, 0001, 1010, 0000, 0010, 0101, 0100, 0110
B.1001, 1000, 1010, 0001, 0000, 0010, 0101, 0100, 0110
C.1001, 1000, 0001, 1010, 0000, 0101, 0010, 0100, 0110
D.1001, 1000, 0001, 1010, 0000, 0010, 0100, 0101, 0110

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

In row order the codes are 1001, 1000, 1010, 0001, 0000, 0010, 0101, 0100, 0110.

Cohen-Sutherland line clipping gives each region a 4-bit code whose bits, read from left to right, stand for above, below, right and left of the window (TBRL); a bit is 1 when the point lies on that side. Reading the nine regions row by row:

1001	1000	1010
0001	0000	0010
0101	0100	0110

The top row is above-left, above, above-right; the middle row is left, inside, right; the bottom row is below-left, below, below-right. The other options put 1010 (above-right) in the first position of the second row, or swap codes within the second and third rows.

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The binary region code assigned to the line end point located at the bottom-right side of the clipping rectangle in the Cohen-Sutherland line clipping algorithm is ............. .

A.0110
B.1010
C.0101
D.1001

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The bottom-right region code is 0110.

Cohen-Sutherland gives each end point a 4-bit code, conventionally ordered top, bottom, right, left. A bit is 1 when the point lies beyond that edge of the clipping window.

A point below and to the right of the window is beyond the bottom and right edges, so it gets 0 1 1 0.

1010 would be top-right, 0101 bottom-left and 1001 top-left.

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If a and b are the end points of a line, then which one of the following is true ?

A.If both end points are left, right, above or below the window, the line is invisible.
B.If both end points are left, right, above or below the window, the line is completely visible.
C.If both end points are left, right, above or below the window, the line is trivially visible.
D.If both end points are left, right, above or below the window, the line is trivially invisible.

Exam Year(s)

UGC NET - 2012,

Answer: D

In line clipping with a window, if both end points of a line lie on the same outer side of the window, left, right, above or below, then the whole line is outside it, and it can be rejected at once without any further calculation; that is called trivially invisible, as in the Cohen-Sutherland method. Such a line is not visible, so the options saying it is visible, completely or trivially, are wrong, and 'invisible' alone is not the term the key uses. The answer is option (D).

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Consider a window bounded by the lines : \(x = 0\); \(y = 0\); \(x = 5\) and \(y = 3\). The line segment joining \((-1, 0)\) and \((4, 5)\), if clipped against this window will connect the points

A.(0, 1) and (2, 3)
B.(0, 1) and (3, 3)
C.(0, 1) and (4, 3)
D.(0, 1) and (3, 2)

Exam Year(s)

UGC NET - 2014,

Answer: A

The clipped segment connects \((0, 1)\) and \((2, 3)\).

The line through \((-1, 0)\) and \((4, 5)\) has slope 1, so it is \(y = x + 1\). It enters the window at \(x = 0\), giving \(y = 1\), which is the point \((0, 1)\). Inside the window \(y\) cannot exceed 3, so it leaves at \(y = 3\), where \(x = 2\), which is the point \((2, 3)\).

The other pairs have a second point that does not lie on \(y = x + 1\), for example \((3, 3)\), \((4, 3)\) and \((3, 2)\).

Clipping Algorithms Questions and Answers

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