advanced28 min studyLesson 636 in Mathematics

Alteration and randomized rounding

Repair a random candidate and account jointly for retained value and repair cost.

On this page

What you’ll understand

  • Derive an independent-set bound by sampling vertices and deleting edge obstructions.
  • Optimize a sampling probability within its allowed interval.
  • Separate expected rounding cost, failure probability, and feasibility after repair.

A random candidate need not be valid immediately. Sometimes it is easier to generate something valuable with a manageable number of defects, then repair those defects at a controlled cost. Alteration turns this principle into an existence proof. Randomized rounding uses a related principle to convert fractional decisions into discrete ones. In both settings, the accounting must follow one outcome through its repairs rather than combine unrelated favorable expectations.

Sample first and delete the obstructions

Take a finite simple graph with n vertices and m edges. Retain each vertex independently with probability p. The retained set has expected size np, while the number of edges inside it has expectation mp2mp^2, because both endpoints of each edge must survive. Process the surviving edges in a fixed order, deleting one endpoint whenever both endpoints are still present. No edge remains at the end, so the result is independent. The total deletions are at most the initial number of surviving edges.

For every outcome, final size is at least retained size minus initial surviving-edge count. Taking expectations of this pointwise inequality gives a lower bound on expected final size and hence an outcome at least that large. This proof never asks that a large retained set and a small edge count occur separately with positive probability. They are coupled through one score. Removing a vertex can repair several defects at once, which only improves the lower bound; counting every original edge as a potential deletion is safe overcounting.

α(G)≥E[∣S∣−e(G[S])]=np−mp2,0≤p≤1\alpha(G)\ge\mathbb E\bigl[|S|-e(G[S])\bigr]=np-mp^2,\qquad0\le p\le1
The graph is finite and simple, vertices are retained independently, and deletion produces an independent subset of the retained vertices.↔ Scroll wide equations horizontally.

Tune the experiment without leaving its domain

The quadratic score is concave when m is positive. Its unconstrained maximum occurs at p=n/(2m), but this is a probability only when it lies in the unit interval. If the proposed value exceeds one, use the boundary p=1 instead. If there are no edges, simply keep every vertex. A real-valued bound on the maximum independent-set size can be rounded up because that size is integer valued, provided the underlying inequality has been established exactly rather than from an unreliable rounded decimal.

The same repair logic applies to forbidden configurations in hypergraphs: sample vertices, count each unwanted configuration that survives, and remove one vertex from each surviving obstruction when needed. The survival probability depends on the number of distinct sampled vertices in the configuration. A proof must show that the repair operation destroys every forbidden configuration and does not introduce new ones. Vertex deletion is well suited to monotone forbidden-subgraph properties, whereas an operation that adds edges could create fresh defects and require a different analysis.

p∗=min⁡ ⁣(1,n2m)(m>0),np∗−mp∗2 is a certified retained-size boundp_*=\min\!\left(1,\frac{n}{2m}\right)\quad(m>0),\qquad np_*-mp_*^2\text{ is a certified retained-size bound}
Optimization is over the allowed probability interval; m=0 is handled separately.↔ Scroll wide equations horizontally.

Rounding a fractional cover needs two accounts

For a covering problem, suppose nonnegative fractional choices x have total mass at least one on each constraint. Select each item independently with probability the smaller of one and t times its fractional choice, for a chosen real t≥0. If a constraint contains an item selected with probability one, it is certainly covered. Otherwise independence bounds its failure probability by the exponential of minus t times its fractional mass, hence at most e−te^{-t}. With M constraints the probability of any uncovered constraint is at most Me−tMe^{-t}.

For nonnegative item costs, expected selected cost is at most t times fractional cost. Neither expected cost nor positive feasibility probability alone guarantees a feasible outcome at that expected cost. One option is to combine a cost-tail estimate with the feasibility union bound. Another is to repair each uncovered constraint by adding a known covering item, charging the expected repair cost explicitly. Such a repair needs those items and costs to exist, and overlapping repairs must not invalidate earlier constraints. This distinction prevents an expectation calculation from being mislabeled a complete approximation guarantee.

A guaranteed independent set from a random subset

A simple graph has sixty vertices and one hundred fifty edges. Optimize the sample-and-delete bound and give a guaranteed independent-set size.

  1. For retention probability p, the expected score is 60p−150p260p-150p^2. Its derivative is 60−300p, which vanishes at p=0.2 inside the probability interval.

  2. At this setting the expected retained count is twelve and expected surviving-edge count is six. The expected score is therefore six.

  3. Delete at most one vertex for each initial surviving edge whenever necessary. The final independent-set size dominates the score outcome by outcome, so some outcome has an independent set of at least six vertices.

The result

Every such graph has an independent set of size at least six by this particular alteration certificate; the actual maximum may be larger.

Explore the model

Interactive lab

Interactive lab · function

Inspect the quadratic penalty in alteration

Pause and Reset, then set coefficient a = 1 and position x = 0.20. Interpret x as p and multiply the displayed y=p² by 150; predict six expected surviving edges. Compute 60p−150y=6 alongside the instrument. Reset and set both a = 1 and x = 0.40: predict a penalty of 24 and score zero. The plotted function is only the quadratic penalty, not a graph sampler or an independent-set solver.

Inspect the quadratic penalty in alterationThis 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 forThe sampling gain is linear and the obstruction penalty is quadratic, so more sampling can worsen the certified result.

Check your understanding

Why does sample-and-delete prove an independent-set bound of np−mp²?

Try it yourself

A simple graph has eighty vertices and one hundred sixty edges. Find the optimizing p and independent-set guarantee. Separately, if M=20 covering constraints use t=log(80), bound their total rounding-failure probability.

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

Give me a hint

Optimize the same concave score. In the covering calculation, sum twenty bounds of e to the power minus t.

Compare with the solution

The optimizing retention is p=80/(2⋅160)=1/4p=80/(2\cdot160)=1/4. Expected retained size is twenty and expected surviving edges are 160/16=10160/16=10, giving an independent set of size at least ten after deletion. In the separate covering experiment, each constraint fails with probability at most e−log⁡80=1/80e^{-\log80}=1/80, so the union bound is at most 20/80=1/420/80=1/4. Thus feasibility has probability at least three quarters. This does not by itself promise a feasible rounded solution whose cost is at most its expectation; a cost-tail or repair analysis remains necessary for that combined guarantee.

Take it with you

  • Repair inequalities should hold outcome by outcome before averaging.
  • Sampling parameters must remain valid probabilities when optimized.
  • Expected rounding cost and feasibility probability need a joint argument for a combined approximation claim.

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