Merge branch 'main' of https://github.com/Trance-0/NoteNextra
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Zheyuan Wu
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# Lecture 11
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## More randomized algorithms
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> Caching problem: You have a cache with $k$ blocks and a sequence of accesses, called $\sigma$. The cost of a randomized caching algorithm is the expected number of cache misses on $\sigma$.
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### Randomized Marking Algorithm
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> A phase $i$ has $n_i$ new pages.
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Our goal is to optimize $m^*(\sigma)\geq \frac{1}{2}\sum_{i=1}^{n} n_j$ where $n_j$ is the number of new pages in phase $j$.
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Marking algorithm:
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- at a cache miss, evict an unmarked page uniformly at random
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- at the beginning of the algorithm, all the entries are unmarked
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- after $k$ unique accesses and one miss, all the entries are unmarked
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- old pages: pages in cache at the end of the previous phase
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- new pages: pages accessed in this phase that are not old.
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- new pages always cause a miss.
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- old pages can cause a miss if a new page was accessed and replaced that old page and then the old page was accessed again. This can also be caused by old pages replacing other old pages and creating this cascading effect.
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Reminder: Competitive ratio for our randomized algorithm is
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$$
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max_\sigma \{\frac{E[m(\sigma)]}{m^*(\sigma)}\}
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$$
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```python
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def randomized_marking_algorithm(sigma, k):
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cache = set()
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marked = set()
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misses = 0
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for page in sigma:
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if page not in cache:
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# once all the blocks are marked, unmark all the blocks
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if len(marked) == k:
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marked.clear()
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# if the cache is full, randomly remove a page that is not marked
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if len(cache) == k:
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for page in cache:
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if page not in marked:
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cache.remove(page)
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misses += 1
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# add the new page to the cache and mark it
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cache.add(page)
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marked.add(page)
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return misses
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```
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Example:
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A cache on phase $i$ has $k$ blocks and miss on page $x$:
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[$n_i$ new pages] [$o_i$ old pages] [$x$] [$\ldots$]
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$P[x \text{ causes a miss}] = P[x\text{ was evicted earlier}] \leq \frac{n_j}{k-o_i}$
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Proof:
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**Warning: the first few line of the equation might be wrong.**
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$$
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\begin{aligned}
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P\left[x \text{ was evicted earlier}\bigg\vert\begin{array}{c} n_j\text{ new pages}, \\ o_i\text{ old pages}, \\ k \text{ unmarked blocks} \end{array}\right] &=P[x\text{ was unmarked}]+P[x\text{ was marked}] \\
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&=P[x\text{ was unmarked (new page)}]+P[x\text{ was old page}]+P[x\text{ was in the remaining cache blocks}] \\
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&= \frac{1}{k}+\frac{o_i}{k} P\left[x \text{ was evicted earlier}\bigg\vert\begin{array}{c} n_j-1\text{ new pages}, \\ o_i-1\text{ old pages}, \\ k-1 \text{ unmarked blocks} \end{array}\right] +\frac{k-1-o_i}{k} P\left[x \text{ was evicted earlier}\bigg\vert\begin{array}{c} n_j-1\text{ new pages}, \\ o_i\text{ old pages}, \\ k-1 \text{ unmarked blocks} \end{array}\right] \\
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\end{aligned}
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$$
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Let $P(n_j, o_i, k)$ be the probability that page $x$ causes a miss when the cache has $n_j$ new pages, $o_i$ old pages, and $k$ unmarked blocks.
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Using $P(n_j, o_i, k)\leq \frac{n_j}{k-o_i}$, we have
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$$
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\begin{aligned}
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P(n_j, o_i, k) &= \frac{1}{k}+\frac{o_i}{k} P(n_j-1, o_i-1, k-1)+\frac{k-1-o_i}{k} P(n_j-1, o_i, k-1) \\
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&\leq \frac{1}{k}+\frac{o_i}{k} \frac{n_j-1}{k-1-o_i-1}+\frac{k-1-o_i}{k} \frac{n_j-1}{k-1-o_i} \\
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&= \frac{1}{k}+\left(1+\frac{o_in}{k-o_i}+\frac{n_j-1}{k-o_i}\right)\\
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&=\frac{1}{k}\left(\frac{k-o_i+o_in+(n_j-1)(k-o_i)}{k-o_i}\right)\\
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&= \frac{n_j}{k-o_i}
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\end{aligned}
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$$
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Fix a phase $j$, let $x_i$ be an indicator random variable
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$$
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x_i=\begin{cases}
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1 & \text{if page } i \text{th old page causes a miss} \\
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0 & \text{otherwise}
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\end{cases}
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$$
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$$
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\begin{aligned}
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P[x_i=1]&=P[i\text{th old page causes a miss}]\\
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&\leq \frac{n_j}{k-(i-1)}\\
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\end{aligned}
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$$
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$$
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\begin{aligned}
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E[x_i]&=E[\sum_{i=1}^{o_i} P[x_i=1]]\\
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&= E[n_j+\sum_{i=1}^{k-n_j}x_i]\\
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&=n_j+\sum_{i=1}^{k-n_j} E[x_i]\\
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&\leq n_j+\sum_{i=1}^{k-n_j} \frac{n_j}{k-(i-1)}\\
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&=n_j+\left(1+\frac{1}{k}+\frac{1}{k-1}+\cdots+\frac{1}{n_j}\right)\\
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&\leq n_j H_k\\
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\end{aligned}
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$$
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Let $N$ be the total number of phases.
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So the expected total number of misses is
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$$
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E[\sum_{i=1}^{N} x_i]\leq \sum_{i=1}^{N} E[x_i]\leq\sum_{j=1}^{N} n_j H_k
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$$
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So the competitive ratio is
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$$
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\frac{E[\sum_{i=1}^{N} x_i]}{\frac{1}{2}\sum_{j=1}^{N} n_j}\leq 2H_k=O(\log k)
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$$
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## Probabilistic boosting for decision problems
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Assume that you have a randomized algorithm that gives you the correct answer with probability $\frac{1}{2}+\epsilon$. for some $\epsilon>0$.
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I want to boost the probability of the correct decision to be $\geq 1-\delta$.
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What we can do is to run the algorithm $x$ times independently with probability $\frac{1}{2}+\epsilon$ and take the majority vote.
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The probability of the wrong decision is
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$$
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\binom{x}{\lfloor x/2\rfloor} \left(\frac{1}{2}-\epsilon\right)^{\lfloor x/2\rfloor}
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$$
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I want to choose $x$ such that this is $\leq \delta$.
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> $$(1-p)^{\frac{1}{p}}\leq e^{-1}$$
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So
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$$
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\begin{aligned}
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\binom{x}{\lfloor x/2\rfloor}\left(\frac{1}{2}-\epsilon\right)^{\lfloor x/2\rfloor}&\leq \left(\frac{xe}{x/2}\right)^{\lfloor x/2\rfloor}\left(\frac{1}{2}-\epsilon\right)^{-\lfloor x/2\rfloor\epsilon}
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\end{aligned}
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$$
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We use this to solve for $x$.
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@@ -46,6 +46,15 @@ List of known NP-hard problems:
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## Approximation Algorithms
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- Consider optimization problems whose decision problem variant is NP-hard. Unless P=NP, finding an optimal solution to these problems cannot be done in polynomial time.
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- In approximation algorithms, we make a trade-o↵: we’re willing to accept sub-optimal solutions in exchange for polynomial runtime.
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- The Approximation Ratio of our algorithm is the worst-case ratio of our solution to the optimal solution.
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- For minimization problems, this ratio is Ours/OPT, since our solution will be larger than OPT.
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- For maximization problems, this ratio is OPT/Ours, since our solution will be smaller than OPT.
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- If given an algorithm, and you need to show it has some desired approximation ratio, there are a few approaches.
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- In recitation, we saw Max-Subset Sum. We found upper bounds on the optimal solution and showed that the given algorithm would always give a solution with value at least half of the upper bound, giving our approximation ratio of 2.
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- In lecture, you saw the Vertex Cover 2-approximation. Here, you would select any uncovered edge $(u, v)$ and add both u and v to the cover. We argued that at least one of u or v must be in the optimal cover, as the edge must be covered, so at every step we added at least one vertex from an optimal solution, and potentially one extra. So, the size of our cover could not be any larger than twice the optimal.
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## Randomized Algorithms
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## Online Algorithms
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