Method of conditional probabilities

The probabilistic method is used (primarily in combinatorics) to prove the existence of mathematical objects with desired properties. The proofs are probabilistic -- they work by showing that a random object chosen from some probability distribution has the desired properties with positive probability. Consequently, they are nonconstructive -- they don't explicitly describe an efficient method for computing the desired objects.

The method of conditional probabilities (Erdös & Selfridge 1973), (Raghavan 1988) converts such a proof, in a "very precise sense", into an efficient deterministic algorithm, one that is guaranteed to compute an object with the desired properties. It is a particular kind of derandomization.

Overview

(Raghavan 1988) gives this description:

We first show the existence of a provably good approximate solution using the probabilistic method... [We then] show that the probabilistic existence proof can be converted, in a very precise sense, into a deterministic approximation algorithm.

(Raghavan is discussing the method in the context of randomized rounding, but it works with the probabilistic method in general.)

The method of conditional probabilities

To apply the method to a probabilistic proof, the randomly chosen object in the proof must be choosable by a random experiment that consists of a sequence of "small" random choices. The diagram to the right shows a very simple example that illustrates the principle.

In this example the random experiment consists of rolling three fair three-sided die. There are twenty-seven outcomes, each corresponding to a leaf in the tree. A trial of the random experiment corresponds to taking a random walk from the root (the top node in the tree, where no die have been rolled) to a leaf. There are two successful outcomes:

• the first roll is 1, the second and third rolls are 3;
• the first roll is 3, the second and third rolls are 1.

(Thus, the probability of failure is 25/27 ≈ 0.93.) The interior nodes in the tree correspond to partially determined outcomes, where only 0, 1, or 2 of the die have been rolled so far.

To apply the method of conditional probabilities, one focuses on the conditional probability of failure, given the choices so far as the experiment proceeds. In the diagram, each node is labeled with this conditional probability. (For example, if only the first die has been rolled, and it comes up three, that corresponds to the third child of the root. Conditioned on that partial state, the probability of failure is 8/9 ≈ 0.89.)

By definition, the conditional probability at each leaf is either 0 or 1, and the conditional probability at any interior node is the average of the conditional probabilities of its children. The latter property is important because it implies that any interior node whose conditional probability is less then 1 has at least one child whose conditional probability is less than 1.

The conditional probability at the root is just the (unconditioned) probability that the experiment fails, which (by the original proof) must be less than 1. The method of conditional probabilities replaces the random root-to-leaf walk in the random experiment by a deterministic root-to-leaf walk, where each step is chosen to inductively maintain the following invariant:

the conditional probability of failure, given the current state, is less than 1.

In this way, it is guaranteed to arrive at a leaf with label 0, that is, a successful outcome.

Efficiency

In a typical application of the method, the goal is to be able to implement the resulting deterministic process by a reasonably efficient algorithm (formally, one taking time polynomial in the output size), even though typically the number of possible outcomes is huge (exponentially large). (E.g., consider the example above, but extended to some large number of die rolls.)

In the ideal case, given a partial state (a node in the tree), the conditional probability of failure (the label on the node) can be efficiently and exactly computed. If this is so, then the algorithm can select the next node to go to by computing the conditional probabilities at each of the children of the current node, then moving to any child whose conditional probability is less than 1. (As discussed above, there is guaranteed to be such a node.)

Unfortunately, in most applications, the conditional probability of failure is not easy to compute exactly. There are two standard and related techniques for dealing with this:

• Using a conditional expectation: Many probabilistic proofs work as follows: they implicitly define a random variable $Q$, show that (i) the expectation of $Q$ is at most (or at least) some threshold value, and (ii) in any outcome where $Q$ is at most (at least) this threshold, the outcome is a success. Then (i) implies that there exists an outcome where $Q$ is at most the threshold, and this and (ii) imply that there is an outcome that is a success. (For example, $Q$ may be the number of "bad" events (not necessarily disjoint) that occur in a given outcome, where each bad event corresponds to one way the experiment can fail, and the expected number of bad events that occur is less than 1.)

In this case, to keep the conditional probability of failure below 1, it suffices to keep the conditional expectation of $Q$ below (or above) the threshold. To do this, instead of computing the conditional probability of failure, the algorithm computes the conditional expectation of $Q$ and proceeds accordingly: at each interior node, there is some child whose conditional expectation is at most (at least) the node's conditional expectation; the algorithm moves from the current node to such a child, thus keeping the conditional expectation below (above) the threshold.

• Using a pessimistic estimator: In some cases, as a proxy for the exact conditional expectation of the quantity $Q$, one uses an appropriately tight bound called a pessimistic estimator. The pessimistic estimator is a function of the current state. It should upper (or lower) bound the conditional expectation of $Q$ given the current state, and it should be non-increasing (or non-decreasing) in expectation with each random step of the experiment. Typically, a good pessimistic estimator can be computed by precisely deconstructing the logic of the original proof.

Example using conditional expectations

This example demonstrates the use of conditional expectations.

Max-Cut Lemma

Given any undirected graph $G=(V,E)$, the Max cut problem is to color each vertex of the graph with one of two colors (say black or white) so as to maximum the number of edges whose endpoints have different colors. (Say such an edge is cut.)

Lemma:In any graph $G=(V,E)$, at least $|E|/2$ edges can be cut.

Probabilistic proof

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Color each vertex black or white by flipping a fair coin.

By calculation, for any edge e in $E$, the probability that it is cut is 1/2.

Thus, by linearity of expectation, the expected number of edges cut is $|E|/2$.

Thus, there exists a coloring that cuts at least $|E|/2$ edges.

Applying the method of conditional probabilities with conditional expectations

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To apply the method of conditional probabilities, first model the random experiment as a sequence of small random steps. In this case it is natural to consider each step to be the choice of color for a particular vertex (so there are $|V|$ steps).

Next, replace the random choice at each step by a deterministic choice, so as to keep the conditional probability of failure, given the vertices colored so far, below 1. (Here failure means that finally fewer than $|E|/2$ edges are cut.)

In this case, the conditional probability of failure is not easy to calculate (indeed the original proof did not calculate the probability of failure directly).

The proof worked by showing that the expected number of cut edges was at least $|E|/2$. Let random variable $Q$be the number of edges cut. To keep the conditional probability of failure below 1, it suffices to keep the conditional expectation of $Q$ at or above the threshold $|E|/2$. (This is because as long as the conditional expectation of $Q$ is at least $|E|/2$, there must be some still-reachable outcome where $Q$is at least $|E|/2$, so the conditional probability of reaching such an outcome is positive.) To keep the conditional expectation of $Q$ at $|E|/2$ or above, the algorithm will, at each step, color the vertex under consideration so as to maximize the resulting conditional expectation of $Q$.

Given that some of the vertices are colored already, what is this conditional expectation? Following the logic of the original proof, the conditional expectation of the number of cut edges is

the number of edges whose endpoints are colored differently so far

+ (1/2)*(the number of edges with at least one endpoint not yet colored).

Algorithm

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The algorithm colors each vertex to maximize the resulting value of the above conditional expectation. This is guaranteed to keep the conditional expectation at $|E|/2$ or above, and so is guaranteed to keep the conditional probability of failure below 1, which in turn guarantees a successful outcome. By calculation, the algorithm simplifies to the following:

 1. For each vertex $u$ in $V$ (in any order):
 2.   Consider the already-colored neighboring vertices of $u$.
 3.   If more of these are black then white, then color $u$ white.
 4.   Otherwise, color $u$ black.

Because of its derivation, this deterministic algorithm is guaranteed to cut at least half the edges of the given graph. (This makes it a 0.5-approximation algorithm for Max-cut.)

Example demonstrating the use of pessimistic estimators

The next example demonstrates the use of pessimistic estimators.

Turan's theorem

One way of stating Turan's theorem is the following:

Any graph $G=(V,E)$ contains an independent set of size at least $|V|/(D+1)$, where $D=2|E|/|V|$ is the average degree of the graph.

Probabilistic proof

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Consider the following random process for constructing an independent set $S$:

 1. Initialize $S$  to be the empty set.
 2. For each vertex $u$ in $V$ in random order:
 3.    If no neighbors of $u$ are in $S$, add $u$ to $S$
 4. Return $S$.

Clearly the process computes an independent set. Any vertex $u$ that is considered before all of its neighbors will be added to $S$. Thus, letting $d(u)$ denote the degree of $u$, the probability that $u$ is added to $S$ is at least $1/(d(u)+1)$. By linearity of expectation, the expected size of $S$ is at least

$\sum _{u\in V}{\frac {1}{d(u)+1}}~\geq ~{\frac {|V|}{D+1}}.$

(The inequality above follows because $1/(x+1)$ is convex in $x$, so the left-hand side is minimized, subject to the sum of the degrees being fixed at $2|E|/|V|$, that is, when each $d(u)=D=2|E|/|V|$.)

The method of conditional probabilities using pessimistic estimators

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In this case, the random process has $|V|$ steps. Each step considers some not-yet considered vertex $u$ and adds $u$ to $S$ if none of its neighbors have yet been added. Let random variable $Q$ be the number of vertices added to $S$. The proof shows that $E[Q]$ ≥ $|V|/(D+1)$.

We will replace each random step by a deterministic step that keeps the conditional expectation of $Q$ at or above $|V|/(D+1)$. This will ensure a successful outcome, that is, one in which the independent set $S$ has size at least $|V|/(D+1)$, realizing the bound in Turan's theorem.

Given that the first t steps have been taken, let $S^{(t)}$ denote the vertices added so far. Let $R^{(t)}$ denote those vertices that have not yet been considered, and that have no neighbors in $S^{(t)}$. Given the first t steps, following the reasoning in the original proof, any given vertex $w$ in $R^{(t)}$ has conditional probability at least $1/(d(w)+1)$ of being added to $S$, so the conditional expectation of $Q$ is at least

$|S^{(t)}|~+~\sum _{w\in R^{(t)}}{\frac {1}{d(w)+1}}.$

Let $Q^{(t)}$ denote the above quantity, which is called a pessimistic estimator for the conditional expectation.

The proof showed that the pessimistic estimator is initially at least $|V|/(D+1)$. (That is, $Q^{(0)}$ ≥ $|V|/(D+1)$.) The algorithm will make each choice to keep the pessimistic estimator from decreasing, that is, so that $Q^{(t+1)}$ ≥ $Q^{(t)}$ for each $t$. Since the pessimistic estimator is a lower bound on the conditional expectation, this will ensure that the conditional expectation stays above $|V|/(D+1)$, which in turn will ensure that the conditional probability of failure stays below 1.

Let $u$ be the vertex added by the algorithm in the next ($t+1$st) step.

If $u$ already has a neighbor in $S$, then $u$ is not added to $S$ and (by inspection of $Q^{(t)}$), the pessimistic estimator is unchanged.

If $u$ does not have a neighbor in $S$, then $u$ is added to $S$. By calculation, if $u$ is chosen randomly from the remaining vertices, the expected increase in the pessimistic estimator is non-negative. [The calculation: Conditioned on choosing a vertex in $R^{(t)}$, the probability that a given term $1/(d(w)+1)$ is dropped from the sum in the pessimistic estimator is at most $(d(w)+1)/|R^{(t)}|$, so the expected decrease in each term in the sum is at most $1/|R^{(t)}|$. There are $|R^{(t)}|$ terms in the sum. Thus, the expected decrease in the sum is at most 1. Meanwhile, the size of $S$ increases by 1.] Thus, there must exist some choice of $u$ that keeps the pessimistic estimator from decreasing.

Algorithm maximizing the pessimistic estimator

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The algorithm below chooses each vertex $u$ to maximize the resulting pessimistic estimator. By the previous considerations, this keeps the pessimistic estimator from decreasing and guarantees a successful outcome.

Below, $N^{(t)}$ denotes the neighbors of $u$ in $R^{(t)}$ (that is, neighbors of $u$ that are neither in $S$ nor have a neighbor in $S$).

1. Initialize $S$ to be the empty set.
2. While there exists a not-yet-considered vertex $u$ with no neighbor in $S$:
3.    Add such a vertex $u$ to $S$ where $u$ minimizes $\sum _{w\in N^{(t)}\cup \{u\}}{\frac {1}{d(w)+1}}$.
4. Return $S$.

Algorithms that keep the pessimistic estimator from decreasing

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For the method of conditional probabilities to work, it suffices if the algorithm keeps the pessimistic estimator from decreasing (or increasing, as appropriate). This gives some flexibility in deriving the algorithm. The next two algorithms illustrate this.

1. Initialize $S$ to be the empty set.
2. While there exists a vertex $u$ in the graph with no neighbor in $S$:
3.    Add such a vertex $u$ to $S$, where $u$ minimizes $d(u)$ (the initial degree of $u$).
4. Return $S$.

1. Initialize $S$ to be the empty set.
2. While the remaining graph is not empty:
3.    Add a vertex $u$ to $S$, where $u$ has minimum degree in the remaining graph.
4.    Delete $u$ and all of $u$'s neighbors from the graph.
5. Return $S$.

Each algorithm is analyzed with the same pessimistic estimator as before. With each step of either algorithm, the net increase in the pessimistic estimator is

$1-\sum _{w\in N^{(t)}\cup \{u\}}{\frac {1}{d(w)+1}},$

where $N^{(t)}(u)$ denotes the neighbors of $u$ in the remaining graph (that is, in $R^{(t)}$).

For the first algorithm, the net increase is non-negative because, by the choice of $u$,

$\sum _{w\in N^{(t)}\cup \{u\}}{\frac {1}{d(w)+1}}\leq (d(u)+1){\frac {1}{d(u)+1}}=1$,

where $d(u)$ is the degree of $u$ in the original graph.

For the second algorithm, the net increase is non-negative because, by the choice of $u$,

$\sum _{w\in N^{(t)}\cup \{u\}}{\frac {1}{d(w)+1}}\leq (d'(u)+1){\frac {1}{d'(u)+1}}=1$,

where $d'(u)$ is the degree of $u$ in the remaining graph.

See also

• Probabilistic method
• Randomized rounding

Original sources

• Erdös, Paul; Selfridge, J. L. (1973), "On a combinatorial game", Journal of Combinatorial Theory, Series A, 14 (3): 298–301, doi:10.1016/0097-3165(73)90005-8.
• Raghavan, Prabhakar (1988), "Probabilistic construction of deterministic algorithms: approximating packing integer programs", Journal of Computer and System Sciences, 37 (2): 130–143, doi:10.1016/0022-0000(88)90003-7.

Textbooks

• Alon, Noga; Spencer, Joel (2000), The probabilistic method, John Wiley and Sons, pp. 250–, ISBN 9780471370468

• Motwani, Rajeev; Raghavan, Prabhakar, Randomized algorithms, Cambridge University Press, ISBN 9780521474658

• Vazirani, Vijay, Approximation algorithms, Springer Verlag, ISBN 9783540653677