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Lloyd can stop at a strictly suboptimal finite-data fixed point

Statement refuted

Every fixed point of the exact finite-data Lloyd iteration is a global minimizer of empirical squared-error distortion.

Facts & Assumptions

Given: Five equally weighted data points 0,1,4,5,9∈R, two ordered centers, and the exact iteration of Exact finite-data Lloyd quantizer.

[F1]

Exact finite-data Lloyd quantizer: the iteration uses least-index nearest-center assignments, updates nonempty clusters to their weighted centroids, leaves empty centers unchanged, and stops when centers are unchanged.

[F2]

The weighted centroid minimizes finite-cluster squared error: each nonempty cluster's weighted squared error is minimized at its weighted centroid.

[F3]

Let X5:={0,1,4,5,9} and (x1,x2,x3,x4,x5):=(0,1,4,5,9). Put μ5:=15∑i=15δxi on X5, where δxi is unit point mass at xi. This is a probability measure on the relative Borel sigma algebra of the finite Borel set X5, with finite second moment 15∑i=15∣xi∣2<∞. For a center pair c=(c1,c2), let Qc:X5→{1,2} be its least-index nearest-center rule. Each fiber is Borel because X5 is finite, so Finite measurable squared-error quantizer and nearest-center rule gives

Dμ5(Qc;c1,c2)=15∑i=15min⁡1≤j≤2∣xi−cj∣2,

so the sample-average costs below are exactly the distortions for this explicitly specified empirical law.

Counterexample

1.1givenF1F3algebra

Give each point weight 1/5 and initialize centers at (5/2,9). The squared distances to these centers, in point order 0,1,4,5,9, are (25/4,81), (9/4,64), (9/4,25), (25/4,16), and (169/4,0); each comparison is strict, so the clusters are {0,1,4,5} and {9}.

1.2givenF3algebra

The alternative centers (1/2,6) assign 0,1 to 1/2 and 4,5,9 to 6, with strict comparisons: the respective squared-distance pairs are (1/4,36), (1/4,25), (49/4,4), (81/4,1), and (289/4,9). Their total squared error is (1/2)+(4+1+9)=29/2, hence their average distortion is 29/10.

2.1F1F2F3step 1.1algebra

The two weighted centroids are (0+1+4+5)/4=5/2 and 9, so the center update leaves (5/2,9) unchanged by [F2]. The nearest assignment is unchanged as well by step 1.1, so [F1] stops at this tuple. Its distortion is (25/4+9/4+9/4+25/4)/5=17/5 by [F3].

3.1step 2.1step 1.2algebra∎

Since 29/10<17/5, the alternative two-center quantizer has strictly lower distortion than the stopped Lloyd output in step 2.1. Thus that exact fixed point is not globally optimal, refuting the statement.

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