2
$\begingroup$

The Lloyd-Max quantizer is a scalar quantizer which can be seen as a special case of a vector quantizer (VQ) designed with the Linde Buzo Gray (LBG) algorithm.

In k-means clustering, we are given a set of n data points in d-dimensional space and an integer k and the problem is to determine a set of k points in $R^d$, called centers, so as to minimize the mean squared distance from each data point to its nearest center. A popular heuristic for k-means clustering is Lloyd's (1982) algorithm.

Confusion :

Is the Lloyd-Max algorithm alternatively in short called the Lloyd's algorithm ? Or are they completely different?

I am very confused, please help. Thank you.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ You seem to be asking four or five different questions. I'd suggest maybe seeing if you can break this up into several, more focused questions. Regarding what Matlab uses, those questions are probably off-topic and unanswerable except by MathWorks engineers; I'd suggest submitting them dirctly to their technical support. $\endgroup$
    – MBaz
    Commented Mar 24, 2017 at 0:16
  • $\begingroup$ @MBaz: Thank you for your feedback, I have broken the question in different subparts. $\endgroup$
    – Ria George
    Commented Mar 24, 2017 at 2:54

1 Answer 1

Reset to default
3
$\begingroup$

Different.

  • Lloyd-Max is a special type of scalar quantizer design which is optimized (in terms of MSE) to source pdf. Hence the quantizer is generally non-uniform.

  • Lloyd's algorithm (and the more generalized LBG algorithm) is a scheme to design vector quantization. More information here and here.

That said, one should note that Lloyd's algorithm for VQ is essentially an extension of the scalar Lloyd-Max quantization to higher dimensions (vectorized sources).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.