ECE 2030
Introduction to Computer Engineering
SPIM Programming Assignment 2
Arrays and Iteration
Purpose:
To learn how to implement iterative programming constructs at the assembly language level, utilize
branches and re-emphasize the 1-dimensional nature of memory.
Assignment:
Your assignment involves the processing of an array. A graphic image (such as a photograph) is usually
stored in a computer as an array of integers. For a black-and-white image, 0 represents black, 255
represents white, and the numbers in between are shades of gray. Since graphic files tend to be large,
graphics compression is a common task for a computer. One common step in image compression is called
Differential Pulse Code Modulation (DPCM). Compression tends to work best with smaller numbers than
with larger ones, and DPCM tries to change some of the large numbers in the array into small ones without
losing information. That way, you can restore the image to its original value when you want to look at it.
Here’s how it works: instead of storing the value of each element in the array, store the difference between
each element and the one just to its left. Here’s an example.
Original Image Second Image
[146 147 208 210] [146 1 61 2]
[148 130 200 202] è DPCM è [148 -18 70 2]
[148 148 197 295] [148 0 49 98]
[140 143 203 210] [140 3 60 7]
Here’s how we got the values in the first row:
146 (no change)
147 – 146 = 1
208 – 147 = 61
210 – 208 = 2
Overall, the numbers in the array are smaller then before and so hopefully they would compress better.
Notice that the first element in each row is always the same as in the original and notice that negative
numbers are possible.
Unfortunately, since there is nothing to the left of the elements in the leftmost column, those numbers are
still just as large as before. We can fix that by doing a DPCM again, this time vertically. Each element in
the result array will now be the difference between that element and the element above it in the array.
Here’s an example using the previous result image:
Second image Final Image
[146 1 61 2] [146 1 61 2]
[148 -18 70 2] è DPCM è [ 2 -19 9 0]
[148 0 49 98] [ 0 18 -21 96]
[140 3 60 7] [ -8 3 11 -91]
Here’s how we got the values in the first column:
146 (no change again)
148 – 146 = 2
148 – 148 = 0
140 – 148 = -8
