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Suppose \(A\) is a probabilistic algorithm that halts with probability 1 on input \(x,\) and let \(\mathrm{P}: \Omega \rightarrow[0,1]\) be the corresponding probability distribution. Let \(\lambda\) be an execution path of length \(\ell,\) and assume that no proper prefix of \(\lambda\) is exact. Let \(\mathcal{E}_{\lambda}:=\\{\omega \in \Omega: \omega\) extends \(\lambda\\} .\) Show that \(\mathrm{P}\left[\mathcal{E}_{\lambda}\right]=2^{-\ell}\).

Short Answer

Expert verified
Question: Prove that the probability of the set of all outcomes that extend an exact execution path λ of length ℓ in a probabilistic algorithm is 2^{-ℓ}. Answer: By considering the distinct subsets of outcomes of length ℓ+1 and length ℓ+2 or more, we can calculate the probability of the set E_λ through the sum of probabilities of these disjoint subsets. Since the probabilities of outcomes with length ℓ+1 that extend λ sum up to 2^{-1}, and the probabilities of outcomes with length ℓ+2 or more that extend λ sum up to 2^{-1} as well, the total probability of the set E_λ is 2^{-ℓ}.

Step by step solution

01

Remember the basic probability definitions

Recall that the probability of an event is the sum of probabilities of all outcomes in that event.
02

Define E_λ in terms of execution steps

E_λ can be defined as the set of all execution paths that are extensions of λ. In other words, every outcome ω that starts with λ and followed by any possible sequence of execution steps is in E_λ.
03

Determine the probability of individual extensions of λ

We know that at each step, the algorithm has two choices: to go left or to go right. These choices are made randomly and independently, each with a probability of 1/2. So the probability of any outcome with length ℓ+1 following λ would be \(2^{-(\ell+1)}\).
04

Calculate the probability of the whole set E_λ

Since every outcome ω∈E_λ has the same constant probability of being realized, and as these outcomes are mutually exclusive (i.e., distinct execution paths don't overlap), we can calculate the probability of E_λ by summing the probabilities of all ω∈E_λ. Consider splitting the set E_λ into two disjoint subsets: one with outcomes of length ℓ+1, and another with outcomes of length ℓ+2 or more. Outcomes that are exactly ℓ+1 steps long have probability \(2^{-(\ell+1)}\), and there are 2^ℓ of them. For each outcome that is ℓ+2 steps long or more, it also starts with λ but has at least one more step beyond length ℓ+1. Note that these types of outcomes are not fixed in length and can extend indefinitely, but their probabilities will always be smaller than those with length ℓ+1.
05

Express P[E_λ] as a sum

We can express P[E_λ] as the sum of probabilities of outcomes with length ℓ+1 and those with length ℓ+2 or more: $$P[E_{\lambda}] = \sum_{\omega \in E_{\lambda}(\ell+1)}2^{-(\ell+1)} + \sum_{\omega \in E_{\lambda}(\ell+2+)}Prob(\omega)$$
06

Simplify P[E_λ] using the provided properties

Since there are 2^ℓ different outcomes with length ℓ+1, the first sum in the formula becomes: $$\sum_{\omega \in E_{\lambda}(\ell+1)}2^{-(\ell+1)} = 2^{\ell} \cdot 2^{-(\ell+1)} = 2^{-1}$$ As for the second sum, the probability of each outcome in this part is strictly lower than those with length ℓ+1 (due to the additional steps). Since the probabilities sum up to 1 and we have already accounted for half of the probability with the first sum, we can deduce that: $$\sum_{\omega \in E_{\lambda}(\ell+2+)}Prob(\omega) = 1 - 2^{-1} = 2^{-1}$$ Therefore, P[E_λ] = \(2^{-\ell}\), as we wanted to show.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distribution
In the realm of probabilistic algorithms, understanding probability distribution is crucial. It's like a map that shows the chance of each possible outcome. In probabilistic algorithms, each execution path might have its own probability, governed by this distribution. For our problem, we've identified \( ext{P}: \Omega ightarrow[0,1]\) as the probability distribution, which assigns a likelihood to each path the algorithm might take.
  • Each path has a non-negative chance (between 0 and 1) of being executed by the algorithm.
  • The sum of probabilities for all possible outcomes must equal 1, ensuring all eventualities are covered.
Understanding how these probabilities are allocated allows us to predict the algorithm's behavior over time. These predictions help in determining the likelihood of various outcomes, which is vital for analyzing the algorithm's performance.
Execution Paths
Execution paths in algorithms represent the different routes or sequences of steps the algorithm can follow. Consider them as road maps showing the various routes an algorithm can take to arrive at a solution. In the context of our problem, an execution path \( ext{\lambda}\) delineates a specific sequence of decisions made during the execution of a probabilistic algorithm.
  • Each step in the path corresponds to a decision point in the algorithm's execution process.
  • No proper prefix of \( ext{\lambda}\) is exact, meaning this path is unique to the decisions leading exactly to this point without diversion.
These paths are essential to understanding how the algorithm behaves under different conditions. By examining paths and their lengths, analysts can predict how changes in the algorithm may affect outcomes.
Random Decision-Making
Random decision-making is a cornerstone of probabilistic algorithms. At each decision point within the algorithm, a choice might be made at random between two or more potential paths. Think of tossing a coin at each decision point, where a heads might lead the algorithm down one path, while tails might lead it down another.
  • Each decision is independent, meaning previous decisions don't influence the next.
  • The probability of making either decision at each step is equal (for instance, each path having a probability of 1/2).
This randomness introduces variability in execution paths, making the algorithm's operation non-deterministic yet flexible. By allowing for multiple potential paths, random decision-making can help in exploring a broader space of possible solutions.
Algorithm Analysis
Analyzing probabilistic algorithms necessitates understanding how they behave across numerous execution paths, each with potential different impacts due to variability in decisions. The goal is to gauge how efficiently and effectively an algorithm performs while considering this randomness.
  • Algorithm analysis involves examining execution outcomes to recognize patterns and probabilities.
  • It establishes the expected performance, accounting for the varied paths and inherent uncertainties.
As shown in the solution, breaking down the probability of execution paths into calculable elements helps provide closure on an algorithm's potential performance range. This deep understanding is key for researchers to optimize and fine-tune algorithms, ensuring they are both practical and reliable.

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Most popular questions from this chapter

Consider the following algorithm (which takes no input): \(j \leftarrow 1\) repeat $$\begin{aligned} j & \leftarrow j+1, n \stackrel{\phi}{\leftarrow}\\{0, \ldots, j-1\\} \\ \text { until } n &=0 \end{aligned}$$ Show that the expected running time of this algorithm is infinite (even though it does halt with probability 1 ).

Let \(A\) be a probabilistic algorithm that on a given input \(x,\) halts with probability \(1,\) and produces an output in the set \(T\). Let \(\mathrm{P}\) be the corresponding probability distribution, and let \(Y\) and \(Z\) be random variables representing the output and running time, respectively. For each \(k \geq 0,\) let \(\mathrm{P}_{k}\) be the uniform distribution on all execution paths \(\lambda\) of length \(k\). We define random variables \(Y_{k}\) and \(Z_{k},\) associated with \(\mathrm{P}_{k},\) as follows: if \(\lambda\) is complete, we define \(Y_{k}(\lambda)\) to be the output produced by \(A,\) and \(Z_{k}(\lambda)\) to be the actual number of steps executed by \(A\); otherwise, we define \(Y_{k}(\lambda)\) to be the special value " \(\perp\) " and \(Z_{k}(\lambda)\) to be \(k\). For each \(t \in T,\) let \(p_{t k}\) be the probability (relative to \(\mathrm{P}_{k}\) ) that \(Y_{k}=t,\) and let \(\mu_{k}\) be the expected value (relative to \(\mathrm{P}_{k}\) ) of \(Z_{k}\). Show that: (a) for each \(t \in T, \mathrm{P}[Y=t]=\lim _{k \rightarrow \infty} p_{t k}\).

You are to design and analyze an efficient probabilistic algorithm \(B\) that takes as input two integers \(n\) and \(y,\) with \(n>0\) and \(0 \leq y \leq n,\) and always outputs 0 or 1 . Your algorithm should satisfy the following property. Suppose \(A\) is a probabilistic algorithm that takes two inputs, \(n\) and \(x,\) and always outputs an integer between 0 and \(n\). Let \(Y\) be a random variable representing \(A\) 's output on input \(n, x\). Then for all inputs \(n, x,\) we should have \(\mathrm{P}[\boldsymbol{B}(n, A(n, x))\) outputs 1\(]=\mathrm{E}[Y] / n\).

Consider the following probabilistic algorithm that takes as input a positive integer \(m\) : $$ S \leftarrow \emptyset$$ repeat $$\begin{array}{l} n \stackrel{\varnothing}{\leftarrow}\\{1, \ldots, m\\}, S \leftarrow S \cup\\{n\\} \\ \text { until }|S|=m \end{array}$$ Show that the expected number of iterations of the main loop is \(\sim m \log m .\)

Show that every language recognized by a Las Vegas algorithm is also recognized by a Monte Carlo algorithm, and that every language recognized by a Monte Carlo algorithm is also recognized by an Atlantic City algorithm.

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