A in between dda digital differential analyzer and mpl...

a. In between DDA (Digital Differential Analyzer) and MPL (Midpoint Line Algorithm), which one would you choose as an ideal algorithm to draw a line? Please state your reasons with sufficient explanation. b. Suppose the line y = -2.5x + 10 intersects the y-axis at A and the x-axis at B. Identify the zone of the straight line AB and derive essential derivatives for plotting the lines using the midpoint line drawing algorithm. Convert the endpoints of the line AB to zone-0. c. Using the Midpoint Line Algorithm and 8-way Symmetry, compute the first 6 pixels of the line that goes from A to B. You must show the values of d and the direction in which the pixels are being selected at each stage, along with the values of the pixels (in zone 0 & the actual zone in which the line belongs).

Transcript

00:02 We are given a data table of x and y coordinate pairs, and we're tasked with using a graphing utility to make a scatter plot, find the line of best fit, and also the r value.

00:16 This particular graph here was done using excel, and excel can automatically plot your data.

00:25 It can determine the r value if you use the correlation function.

00:29 And this guy's r value was negative 0 .7 -1495.

00:33 So it's an okay fit.

00:35 It's not fantastic.

00:36 And you can see from the data, there's clearly a downward trend of a line, but it's pretty widely spread.

00:43 Excel can also tell you the equation for the line of best fit, and that's part of formatting the trend line.

00:52 The second order of business was to determine the value of f of x at several x values.

00:59 And of course, first we had to decide if the data was appropriate for that.

01:03 So the first thing we're trying to find out is what is f of zero? and that is definitely a reasonable data point to try to find, given the data that we have.

01:15 It's inside the spread of x.

01:18 So when we look at the graph, we can see we expect the answer to be approximately three.

01:26 And of course, we would just by looking at the equation, right? f of 0 is going to end up just equaling the y intercept.

01:37 So it's negative 0 .4077 times 0 plus 2 .9457.

01:53 And when we calculate that out of the 0 .407 times 0 is 0, and we end up with the y intercept, probably obvious 9457.

02:05 All right, next data point.

02:08 The negative six, negative six is definitely also inside the data range, if you will.

02:18 And so if we look here just on the graph, we're going to expect the y value or the f of x, f of negative six, to be in the neighborhood of maybe five and a half.

02:31 That's our expected.

02:32 If we get too far off from that, then we'll assume a calculator fumbled.

02:36 So f of negative 6 equals negative 0 .477 times negative 6 plus 2 .9457.

02:55 And the calculator says 5 .3919.

03:04 Yeah, and that's awfully close to 5 .5 .5.

03:07 So we're going to say we probably did that right.

03:10 The next one, 12.

03:12 This one's debatable.

03:13 12 is on the outside of the x range, if you will.

03:19 But given how wide the x range is, i think it's reasonable to extrapolate to 12.

03:26 So that would, if we extended the line, that would take us right about here, which is approximately negative 2.

03:33 So let's go ahead and calculate that.