Scalar and Vector Quantization

Lloyd-Max Algorithm (3/3)
If the initial guess of y 1 does not fulfills the termination
condition:
y M
ˆ
/ 2
− y M / 2
≤ ε,
where
yˆ = M 2
2bM
/ 2 1
yM
/ 2 1,
y
+
/ − −
b
∫
M / 2
b
/ 2
= ( ) ∫
M
xf
X
x dx
b
b
M / 2−1
( x)
dx.
we must pick a different y 1 and repeat the process.
M
/ 2
M / 2−1
f
X
40/55