ইন্টোর মন্টি কার্লো অ্যালগরিদমে মিনিম্যাক্স নীতি

উদযাপিত ইয়াওর মিনিম্যাক্স নীতিটি বিতরণ জটিলতা এবং এলোমেলো জটিলতার মধ্যে সম্পর্ককে বর্ণনা করে। যাক $P$ একটি নির্দিষ্ট সেট নিয়ে কোনও সমস্যা হয়ে $\mathcal{X}$ ইনপুট এবং একটি সসীম সেট $\mathcal{A}$ সমাধানের জন্য নির্ণায়ক আলগোরিদিম $P$ । এছাড়াও দিন $\mathcal{D}$ ইনপুট বন্টন বোঝাতে দিন $\mathcal{R}$ উপর সম্ভাব্যতা বিতরণের বোঝাতে $\mathcal{A}$ । তারপরে নীতিটি

min A \in A E c o s t (A, D) \leq max x \in X E c o s t (R, x) for all D and R .

$\min_{A\in\mathcal{A}}\quad\mathbb{E} cost(A,\mathcal{D}) \leq \max_{x\in\mathcal{X}}\quad\mathbb{E} cost(\mathcal{R},x) \quad\quad\text{for all $\mathcal{D}$ and $\mathcal{R}$}.$ এই প্রমাণটি সরাসরি শূন্য-সমীকরণের খেলাগুলির জন্য ভন নিউমানের মিনিম্যাক্স উপপাদ্য থেকে অনুসরণ করে।

বেশিরভাগ ইয়াও নীতি কেবল লাস ভেগাস অ্যালগরিদমগুলির সাথে সম্পর্কিত , তবে এটি মন্টি কার্লো অ্যালগরিদমকে সাধারণভাবে অনুসরণ করা যেতে পারে ।

1 2 min A \in A E c o s t 2 ϵ (A, D) \leq max x \in X E c o s t ϵ (R, x) for all D, R and ϵ \in [0, 1 / 2]

$\frac12 \min_{A\in\mathcal{A}}\quad\mathbb{E} cost_{2\epsilon}(A,\mathcal{D}) \leq \max_{x\in\mathcal{X}}\quad\mathbb{E} cost_{\epsilon}(\mathcal{R},x)\quad\quad\text{for all $\mathcal{D}$, $\mathcal{R}$ and $\epsilon\in [0,1/2]$}$ যেখানে

costϵ(⋅,⋅) $cost_\epsilon(\cdot,\cdot)$ মন্টে কার্লো আলগোরিদিম খরচ দিতে হবে যা সর্বোচ্চ সম্ভাবনা ভুল উল্লেখ করে

ϵ $\epsilon$ ।

ইন ইয়াও মূল কাগজ , মন্টে কার্লো আলগোরিদিম জন্য সম্পর্ক প্রমানহীন উপপাদ্য 3 দেওয়া হয়। এটি প্রমাণ করার জন্য কোনও ইঙ্গিত?

randomized-algorithms

— ফেডেরিকো ম্যাগালানেজ
সূত্র

উত্তর:

এটি মার্কোসের উত্তরের উপর তাঁর স্বরলিপিটি ব্যবহার করে কেবল একটি বর্ধিত মন্তব্য। আমি তার যুক্তির বিবরণগুলি অনুসরণ করতে যথেষ্ট সক্ষম নই এবং নীচের একটিটি খুব ছোট এবং সহজ।

গড়, কসম

\sum A q (A) \sum x d (x) ϵ (A, x) = \sum x d (x) \sum A q (A) ϵ (A, x) \leq λ .

$\sum_A{q(A)\sum_x{d(x)\epsilon(A, x)}} = \sum_x{d(x)\sum_A{q(A)\epsilon(A, x)}} \leq \lambda.$

আসলে উপরে এবং মার্কভ এর বৈষম্য পরোক্ষভাবে । $\sum_{A \in \beta(2\lambda)}{q(A)} \geq 1/2$

সুতরাং আমরা পেতে:

max x \sum A q (A) r (A, x) \geq \sum x d (x) \sum A q (A) r (A, x) = \sum A q (A) \sum x d (x) r (A, x) \geq \sum A \in β (2 λ) q (A) \sum x d (x) r (A, x) \geq ⎛ ⎝ \sum A \in β (2 λ) q (A) ⎞ ⎠ min A \in β (2 λ) \sum x d (x) r (A, x) \geq 1 2 min A \in β (2 λ) \sum x d (x) r (A, x)

$\begin{align*} \max_x \sum_A{q(A)r(A,x)} &\geq \sum_x{d(x)\sum_A{q(A)r(A, x)}}\\ &= \sum_A{q(A)\sum_x{d(x)r(A, x)}}\\ &\geq \sum_{A \in \beta(2\lambda)}{q(A)\sum_x{d(x)r(A, x)}}\\ &\geq \left(\sum_{A \in \beta(2\lambda)}{q(A)}\right) \min_{A \in \beta(2\lambda)}{\sum_x{d(x)r(A, x)}}\\ &\geq \frac{1}{2}\min_{A \in \beta(2\lambda)}{\sum_x{d(x)r(A, x)}} \end{align*}$

— Sasho Nikolov
সূত্র

I'll give it a try on this. I'm going to use Yao's original notation. This way it will be easier to contrast with his paper and his definitions.

Let $\mathcal{I}$ be a finite set of inputs, and let $\mathcal{A}_0$ be a finite set of deterministic algorithms that may fail to give a correct answer for some inputs. Let also $\epsilon(A,x)=0$ if $A$ gives the correct answer for $x$ , and $\epsilon(A,x)=1$ otherwise. Also denote by $r(A,x)$ the number of queries made by $A$ on input $x$ , or equivalently, the depth of $A$ 's decision tree.

Average Cost: Given a probability distribution $d$ on $\mathcal{I}$ , the average cost of an algorithm $A\in \mathcal{A}_0$ is $C(A,d)=\sum_{x\in\mathcal{I}} d(x)\cdot r(A,x)$ .

Distributional Complexity: Let $\lambda\in[0,1]$ . For any distribution $d$ on the inputs, let $\beta(\lambda)$ be the subset of $\mathcal{A}_0$ given by $\beta(\lambda)=\{A : A\in \mathcal{A}_0, \sum_{x\in\mathcal{I}} d(x)\cdot \epsilon(A,x)\leq \lambda\}$ . The distributional complexity with error $\lambda$ for a computational problem $P$ is defined as $F_{1,\lambda}(P)=\max_{d} \min_{A\in \beta(\lambda)} C(A,d)$ .

$\lambda$ -tolerance: A distribution $q$ on the family $\mathcal{A}_0$ is $\lambda$ -tolerant if $\max_{x\in \mathcal{I}} \sum_{A\in\mathcal{A}_0} q(A)\cdot \epsilon(A,x)\leq \lambda$ .

Expected Cost: For a randomized algorithm $R$ , let $q$ be a probability distribution that is $\lambda$ -tolerant on $\mathcal{A}_0$ . The expected cost of $R$ for a given input $x$ is $E(R,x)=\sum_{A\in \mathcal{A}_0} q(A)\cdot r(A,x)$ .

Randomized Complexity: Let $\lambda\in[0,1]$ . The randomized complexity with error $\lambda$ is $F_{2,\lambda}=\min_R \max_{x\in\mathcal{I}} E(R,x)$ .

Now we are ready to go into business. What we want to prove is given a distribution $d$ on the inputs and a randomized algorithm $R$ (i.e., a distribution $q$ on $\mathcal{A}_0$ )

Yao's Minimax Principle for Montecarlo Algorithms
$max x \in I E (R, x) \geq 1 2 min A \in β (2 λ) C (A, d)$ $\begin{equation}\max_{x\in\mathcal{I}} E(R,x)\geq \frac{1}{2}\min_{A\in \beta(2\lambda)} C(A,d) \end{equation}$ for $\lambda\in[0,1/2]$ .

I will follow an approach given by Fich, Meyer auf der Heide, Ragde and Wigderson (see Lemma 4). Their approach does not yield a characterization for Las Vegas algorithms (only the lower bound), but it is sufficient for our purposes. From their proof, it easy to see that for any $\mathcal{A}_0$ and $\mathcal{I}$

Claim 1. $\max_{x\in \mathcal{I}} E(R,x)\geq \min_{A\in \mathcal{A}_0} C(A,d)$ .

To get the correct numbers there, we'll do something similar. Given that the probability distribution $q$ given by the randomized algorithm $R$ is $\lambda$ -tolerant on $\mathcal{A}_0$ we have that

λ \geq max x \in I {\sum A \in A 0 q (A) \cdot ϵ (A, x)} \geq \sum x \in I d (x) \sum A \in A 0 q (a) \cdot ϵ (A, x) = \sum A \in A 0 q (a) \sum x \in I d (x) \cdot ϵ (A, x) \geq min A \in A 0 {\sum x \in I d (x) \cdot ϵ (A, x)} .

$\begin{align*} \lambda &\geq \max_{x\in \mathcal{I}}\left\{ \sum_{A\in\mathcal{A}_0} q(A)\cdot \epsilon(A,x) \right\}\\ &\geq \sum_{x\in\mathcal{I}} d(x) \sum_{A\in \mathcal{A}_0} q(a)\cdot \epsilon(A,x)\\ &= \sum_{A\in \mathcal{A}_0} q(a)\sum_{x\in\mathcal{I}} d(x) \cdot \epsilon(A,x)\\ &\geq \min_{A\in \mathcal{A}_0}\left\{ \sum_{x\in\mathcal{I}} d(x) \cdot \epsilon(A,x) \right\}. \end{align*}$ If we replace the family

A0 $\mathcal{A}_0$ with

β(2λ) $\beta(2\lambda)$ we see that

λ \geq max x \in I {\sum A \in A 0 q (A) \cdot ϵ (A, x)} \geq max x \in I ⎧ ⎩ ⎨ \sum A \in β (2 λ) q (A) \cdot ϵ (A, x) ⎫ ⎭ ⎬ \geq \sum x \in I d (x) \sum A \in β (2 λ) q (a) \cdot ϵ (A, x) = \sum A \in β (2 λ) q (a) \sum x \in I d (x) \cdot ϵ (A, x) \geq min A \in β (2 λ) {1 2 \sum x \in I d (x) \cdot ϵ (A, x)},

$\begin{align*} \lambda &\geq \max_{x\in \mathcal{I}}\left\{ \sum_{A\in\mathcal{A}_0} q(A)\cdot \epsilon(A,x) \right\}\\ &\geq \max_{x\in \mathcal{I}}\left\{ \sum_{A\in\beta(2\lambda)} q(A)\cdot \epsilon(A,x) \right\}\\ &\geq \sum_{x\in\mathcal{I}} d(x) \sum_{A\in \beta(2\lambda)} q(a)\cdot \epsilon(A,x)\\ &= \sum_{A\in \beta(2\lambda)} q(a)\sum_{x\in\mathcal{I}} d(x) \cdot \epsilon(A,x)\\ &\geq \min_{A\in \beta(2\lambda)}\left\{ \frac{1}{2}\sum_{x\in\mathcal{I}} d(x) \cdot \epsilon(A,x) \right\}, \end{align*}$

where the second inequality follows because $\beta(2\lambda) \subseteq \mathcal{A}_0$ , and the last inequality is given by the definition of $\beta(2\lambda)$ where the summation divided by 2 cannot be greater than $\lambda$ . Hence,

max x \in I {\sum A \in A 0 q (A) \cdot ϵ (A, x)} \geq 1 2 min A \in β (2 λ) {\sum x \in I d (x) \cdot ϵ (A, x)} .

$\begin{equation}\max_{x\in \mathcal{I}}\left\{ \sum_{A\in\mathcal{A}_0} q(A)\cdot \epsilon(A,x) \right\}\geq\frac{1}{2} \min_{A\in \beta(2\lambda)}\left\{ \sum_{x\in\mathcal{I}} d(x) \cdot \epsilon(A,x) \right\}. \end{equation}$

By noting that $\epsilon$ maps to $\{0,1\}$ and $r$ maps to $\mathbb{N}$ and Claim 1 above, now we can safely replace the function $\epsilon$ in the inequality above by $r(A,x)$ to obtain the desired inequality.

— Marcos Villagra
সূত্র

Is there a short explanation for where the factor of 2 comes from?

— Robin Kothari

in short, it comes from the definition of

β(2λ) $\beta(2\lambda)$ . The summation in the definition divided by 2 is at most

λ $\lambda$ .

— Marcos Villagra

something seems strange to me. by definition,

maxA∈β(2λ)){12∑x∈Id(x),ϵ(A,x)}≤λ $\max_{A \in \beta(2\lambda))} \left\{\frac{1}{2} \sum_{x \in \mathcal{I}}{d(x), \epsilon(A,x)}\right\} \leq \lambda$ so why the min?

— Sasho Nikolov

and i don't understand the last sentence. how did you make an entire argument about

ϵ $\epsilon$ and then replaced it with

r $r$ ?

— Sasho Nikolov

regarding your first question, I added more details.

— Marcos Villagra