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Unformatted text preview: Chapter 11 Solutions: Case Study:
The Data Clustering
(coauthored by Nargess Memarsadeghi) CHALLENGE 11.1.
The original image takes mpb bits, while the clustered
image takes kb bits to store the cluster centers and mp log2 k bits to store the
cluster indices for all mp pixels. For jpeg images with RGB (red, green, blue)
values ranging between 0 and 255, we need 8 bits for each of the q = 3 values (red,
green, and blue). Therefore, an RGB image with 250,000 pixels takes 24 ∗ 250, 000 =
6, 000, 000 bits, while the clustered image takes about 250, 000 log2 4 = 500, 000 bits
if we have 3 or 4 clusters and 250,000 bits if we have 2 clusters. These numbers can
be further reduced by compression techniques such as run-length encoding. CHALLENGE 11.2.
(a) Neither D nor R is convex everywhere. Figure 11.2 plots these functions for a
particular choice of points as one of the cluster centers is moved. We ﬁx the data
points at 0 and 1 and one of the centers at 1.2, and plot D and R as a function of
the second center c. For c < −1.2 and c > 1.2, the function D is constant, since the
second cluster is empty, while for −1.2 < c < 1.2, the function is quadratic. Since
each function is above some of its secants (the line connecting two points on the
graph), each function fails to be convex.
(b) Neither D nor R is diﬀerentiable everywhere. Again see Figure 11.1. The
function D fails to be diﬀerentiable at c = −1.2 and c = 1.2. Trouble occurs at the
points where a data value moves from one cluster to another.
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- Fall '11