advanced29 min studyLesson 755 in Mathematics

Covering numbers and chaining

Approximate a class at several resolutions and account for the geometry hidden by a single net.

On this page

What you’ll understand

  • Define an empirical covering number using a stated normalized metric.
  • Compute exact covers for a one-parameter constant-function class.
  • Interpret a truncated Dudley bound and its residual term without claiming it is always sharp.

Counting functions treats two nearly identical prediction vectors as different objects. Covering numbers instead count representatives at a chosen resolution. A single coarse cover loses precision, while an extremely fine cover may be enormous. Chaining balances these effects across multiple scales. The metric and its normalization determine what a resolution means, so they belong in the theorem rather than in an unstated implementation detail.

Specify the geometry seen by the sample

For a fixed sample of n inputs, use the root mean square distance between prediction vectors. It is a pseudometric on functions: two different functions can have distance zero when they agree on the sample. An epsilon cover is a finite collection of representatives within epsilon of every prediction vector, and its covering number is the smallest such collection size. Here centers lie in the restricted function class and closed balls are allowed. Other conventions can change boundary counts or constants, so identify the one being used.

For constant functions with values theta between zero and one, the empirical root mean square distance is simply the absolute difference of their constants, independent of n. A radius-one-quarter cover needs only two centers, at one quarter and three quarters. One center cannot cover the entire interval because its ball has length at most one half. This small example separates the geometry of the class from the number of observed coordinates: duplicating the same constant value across more inputs does not create additional distinct parameter directions.

dS(f,g)=(1n∑i=1n(f(xi)−g(xi))2)1/2,N(ε,F,dS)=min⁡{∣V∣:V⊆F, ∀f∈F ∃v∈V dS(f,v)≤ε}d_S(f,g)=\left(\frac1n\sum_{i=1}^{n}(f(x_i)-g(x_i))^2\right)^{1/2},\qquad N(\varepsilon,F,d_S)=\min\{|V|:V\subseteq F,\ \forall f\in F\ \exists v\in V\ d_S(f,v)\le\varepsilon\}
Functions are identified through their sample restrictions; the closed-cover convention uses centers in that restricted class.↔ Scroll wide equations horizontally.

Successive approximations expose smaller increments

Choose covers at dyadically decreasing radii and associate every function with a representative at each scale. Its prediction vector can be written as a coarse representative plus successive differences, followed by a residual. Adjacent approximations are close, so their signed-sum fluctuations have smaller amplitude than the original functions even though finer covers contain more representatives. A finite-class maximal inequality applied separately to these increments yields a sum of radius-weighted square roots of log covering numbers.

Passing from that multiscale sum to an integral gives a Dudley bound. For function values in[−1,1], the displayed version bounds our conditional Rademacher average by a residual term four alpha plus twelve divided by the square root of n times an entropy integral from alpha to one. It applies under measurable or finite sample-restriction conditions. Taking alpha positive can avoid divergence at very fine scales; discarding the residual after truncation would omit the approximation error that makes the finite-resolution argument valid.

R^S(F)≤inf⁡0≤α≤1{4α+12n∫α1log⁡N(ε,F,dS) dε},F⊆[−1,1]X\widehat{\mathfrak R}_S(F)\le\inf_{0\le\alpha\le1}\left\{4\alpha+\frac{12}{\sqrt n}\int_\alpha^1\sqrt{\log N(\varepsilon,F,d_S)}\,d\varepsilon\right\},\qquad F\subseteq[-1,1]^{\mathcal X}
This no-absolute-value empirical bound uses normalized L2 covering numbers and an explicit truncation residual; infinite integrals are allowed as uninformative bounds.↔ Scroll wide equations horizontally.

Geometry can improve a bound without predicting an exact law

A single-scale argument often balances its cover radius against a square root of log cover size divided by n. Chaining uses all scales, which can remove an artificial logarithmic factor for classes with sufficiently mild entropy growth. Conversely, a diverging entropy integral does not prove that the class is unlearnable: truncation or a different complexity method may still yield control. The displayed constants are sufficient and can be conservative, particularly on tiny samples or very simple classes.

Covering numbers are also metric dependent. A population L2 cover and an empirical L2 cover answer different questions, and replacing the normalized distance by an unnormalized Euclidean norm changes all radii by a square-root sample-size factor. A numerical cover needs a proof that every function is represented; a scatterplot of selected parameters is not enough. The examples use constant functions precisely because the metric balls and the minimal cover count can be verified directly rather than inferred visually.

Two exact resolutions of the constant class

For F={f_theta:f_theta(x)=theta,0≤theta≤1}, find minimal empirical L2 covers at radii 1/4 and 1/8, for any nonempty input sample.

  1. The distance is|theta−theta′|. A closed radius-epsilon ball covers a parameter interval of length at most 2epsilon, so covering length one needs at least the ceiling of 1/(2epsilon) centers.

  2. At radius 1/4, choose centers 1/4 and 3/4. Their closed balls cover[0,1], reaching the lower bound of two centers.

  3. At radius 1/8, choose centers 1/8,3/8,5/8,7/8. Their four closed balls tile[0,1], again reaching the lower bound. The counts do not depend on n because the metric normalization cancels identical coordinates.

The result

The minimal cover sizes are two and four. The result concerns these exact radii and the stated closed-ball, within-class convention.

Explore the model

Interactive lab

Interactive lab · function

The squared distance before taking a square root

Pause and Reset, then set coefficient a = 1 and position x = 0.25. Interpret x as the difference of two constant predictions: the exact y=0.0625 is their squared empirical L2 distance (the readout rounds to 0.063), so the distance is 0.25. Reset and set both a = 1 and x = 0.50: squared distance becomes 0.25 and distance 0.50. The quadratic instrument displays only this metric ingredient, not covers, entropy integrals or an arbitrary function-class geometry.

The squared distance before taking a square rootThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA cover radius is a distance, and confusing it with squared distance changes the required number of representatives.

Check your understanding

If a Dudley integral is truncated at alpha>0, what is the role of the 4alpha term in the displayed bound?

Try it yourself

For the same constant class, find the minimal cover size at radius 0.10 and provide explicit centers. If the distance were the unnormalized Euclidean distance on one hundred inputs, what cover size would radius 0.10 require?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

First cover[0,1] with intervals of radius 0.10. Then determine how the unnormalized distance scales relative to the root mean square distance.

Compare with the solution

For normalized empirical distance, centers 0.10,0.30,0.50,0.70,0.90 cover the interval with five radius 0.10 balls, and the total-length lower bound proves five is minimal. On one hundred inputs, the unnormalized distance is 100∣θ−θ′∣=10∣θ−θ′∣\sqrt{100}|\theta-\theta'|=10|\theta-\theta'|. Radius 0.10 then corresponds to parameter radius 0.01, requiring fifty centers at 0.01,0.03,...,0.99. The change comes entirely from metric normalization, not from an expanded set of constant functions. This is why n must not be silently moved between a covering radius and the prefactor of a complexity theorem.

Take it with you

  • Empirical covers identify functions through their prediction vectors and a specified metric.
  • Chaining sums controlled increments over several resolutions.
  • Truncation leaves an explicit residual, and metric normalization changes cover radii.

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