তাত্ত্বিক কম্পিউটার বিজ্ঞানে একটি দ্রুত অ্যালগরিদম বলতে কী বোঝায়?

Question

যদি কোনও সমস্যা A এর জন্য সময় $O(f(n))$ এ চলমান অ্যালগরিদম হয় এবং সময় মতো ছুটে চলে আসা এলগরিদম নিয়ে আসে কেউ, $O(f(n)/g(n))$ , যেখানে $g(n) = o(f(n))$ , এটি কি পূর্ববর্তী অ্যালগোরিদমের তুলনায় উন্নতি হিসাবে বিবেচনা করা হয়?

তাত্ত্বিক কম্পিউটার বিজ্ঞানের প্রসঙ্গে কি এ জাতীয় অ্যালগরিদম নিয়ে আসা উচিত?

Comments

"দ্রুত অ্যালগরিদম" দ্বারা, আমরা "asympototically দ্রুত অ্যালগরিদম" এর অর্থ।

— যুবাল ফিল্মাস 14 ই

@YuvalFilmus কি সঙ্গে "এসিম্পটোটিকভাবে" মানে কী

— undefined

সময় চলমান । $o(f(n))$

— যুবাল ফিল্মাস

Answer 1

না, সময় চলছে এমন একটি অ্যালগরিদম , যেখানে g ( n ) = o ( f ( n ) ) , অগত্যা উন্নতি হিসাবে বিবেচিত হবে না। উদাহরণস্বরূপ, ধরুন যে f ( n ) = n এবং g ( n ) = 1 / n । তারপরে ও ( এফ ( এন ) / জি $O(f(n)/g(n))$$g(n) = o(f(n))$$f(n) = n$$g(n) = 1/n$ও ( এফ ( এন ) ) = ও ( এন ) এর চেয়ে খারাপ সময় বেঁধে দেওয়া।$O(f(n)/g(n)) = O(n^2)$$O(f(n)) = O(n)$

সময় চলমান অ্যালগরিদমকে উন্নত করার জন্য আপনাকে সময় ও ( এফ ( এন ) ) এ চলমান অ্যালগরিদম নিয়ে আসতে হবে , অর্থাৎ সময় জি ( $f(n)$$o(f(n))$$g(n)$ for some function $g(n) = o(f(n))$.

যদি আপনারা সকলেই জানেন যে একটি অ্যালগরিদম সময়ে চলমান থাকে তবে এটি O ( g ( n ) ) সময়ে চলমান একটি অ্যালগরিদম কোনও উন্নতি কিনা তা স্পষ্ট নয় , f ( n ) , g ( n ) ) হয়। এটি কারণ চলমান সময়ে বড় ও কেবলমাত্র একটি উপরের আবদ্ধ। পরিবর্তে, সবচেয়ে খারাপ সময়ের জটিলতা বিবেচনা করা এবং এটি একটি বড় as হিসাবে অনুমান করা সাধারণ $O(f(n))$$O(g(n))$$f(n),g(n)$$\Theta$ বরং একটি বড় হিসাবে চেয়ে ।$O$

Comments

আপনার প্রথম অনুচ্ছেদে নেওয়া ভাল । একটি ক্রমহ্রাসমান ফাংশনটি ব্যবহার করে কিছুটা ঠকানো feels $g(n)=1$

— ডেভিড রিচারবি

@ ডেভিডরিচার্বি: সম্ভবত কিছুটা হলেও ওপি কখনও বলেনি যে এ তাদের একটি অ্যালগরিদম চলছে তাই একঘেয়েমি অনুমান করা যায় না। $O(g(n))$

— কেভিন

@Kevin Sure but the context is computer science and, in computer science, big-O notation is usually used for nondecreasing functions. Probably the asker was thinking in those terms.

— David Richerby

Answer 2

Remember that $O(...)$ notation is meant for analyzing how the task grows for different sizes of input, and specifically leaves out multiplicative factors, lower-order term, and constants.

Suppose you have an $O(n^2)$ algorithm whose actual runtime is $1n^2+2n+1$ (assuming you can actually count the instructions and know the exact timings and so on, which is admittedly a huge assumption in modern systems). Then suppose you come up with a new algorithm that happens to be $O(n)$, but the actual runtime is $1000n + 5000$. Also suppose you know the software to use this algorithm will never see a problem size of $n>10$.

So, which would you chose - the $O(n)$ algorithm that's going to take 15000 units of time, or the $O(n^2)$ one that's only going to take 121 units? Now if your software evolves to handling problem sizes of $n>100000$, which one would you pick? What would you do if your problem size varies greatly?

Comments

"never see a problem size of n>10" - then we'd not use the O notation at all, would we...

— AnoE

@AnoE Simple numbers for the sake of argument. The same logic applies whether you're analyzing for a problem size of 10 vs 1e5 or analyzing for 1e6 vs 1e9.

— twalberg

@AnoE Most computer programs do not try to handle an infinitely growing problem size. So there will be a trade-off. That's why big-O is for theoretical computer science, and the concepts can be applied to improve actual programs.

— mbomb007

Exactly, @mbomb007. The question title is "What does a faster algorithm mean in theoretical computer science?" and he has this in the body: "Does it make sense, in the context of theoretical computer science...".

— AnoE

@AnoE From experience, O notation is used when n<10 all the time! Not that it's a good idea... but it's totally something that's done!

— Cort Ammon - Reinstate Monica

Answer 3

$g(n) \in o\bigl(f\left(n\right)\bigr)$$g$$f$ the time complexity of the old.

$\Theta(n^2)$$\Theta(n \log n)$ time, but it’s widely-used in practice because of its good average-case running time. It can additionally be tweaked to run very quickly in the cases that are most frequent in the wild, such as arrays that are mostly in the right order.

And sometimes, even theoretical computer scientists use “faster” the same way normal people do. For example, most implementations of String classes have Short String Optimization (also called Small String Optimization), even though it only speeds things up for short strings and is pure overhead for longer ones. As the input size gets larger and larger, the running time of a String operation with SSO is going to be higher by a small constant term, so by the definition I gave in the first paragraph, removing SSO from a String class makes it “faster.” In practice, though, most strings are small, so SSO makes most programs that use them faster, and most computer-science professors know better than to go around demanding that people only talk about orders of asymptotic time complexity.

Answer 4

There is not one unified definition of what a "faster algorithm" is. There is not a governing body which decides whether an algorithm is faster than another.

To point out why this is, I'd like to offer up two different scenarios which demonstrate this murky concept.

The first example is an algorithm which searches a linked list of unordered data. If I can do the same operation with an array, I have no change on the big Oh measure of performance. Both searches are O(n). If I just look at the big Oh values, I might say that I made no improvement at all. However, it is known that array lookups are faster than walking a linked list in the majority of cases, so one may decide that that made an algorithm "faster," even though the big Oh did not change.

If I may use the traditional example of programming a robot to make a PBJ sandwich, I can show what I mean another way. Consider just the point where one is opening the jar of peanut butter.

Pick up the jar
Grab the lid
Unscrew the lid

Versus

Pick up the jar
Put the jar back down
Pick up the jar
Put the jar back down
Pick up the jar
Put the jar back down
Pick up the jar
Put the jar back down
Pick up the jar
Put the jar back down
Pick up the jar
Grab the lid
Unscrew the lid

Even in the most academic theoretical setting I can think of, you'll find that people accept that the first algorithm is faster than the second, even though the big Oh notation results are the same.

By contrast, we can consider an algorithm to break RSA encryption. At the moment, it is perceived that this process is probably O(2^n), where n is the number of bits. Consider a new algorithm which runs n^100 faster This means my new process runs in O(2^n/n^100). However, in the world of cryptography, a polynomial speedup to an exponential algorithm is traditionally not thought of as a theoretical speed up at all. When doing security proofs, it's assumed that an attacker may discover one of these speed ups, and that it will have no effect.

So in one circumstance, we can change a O(n) to O(n), and call it faster. In a different circumstance, we can change a O(2^n) to O(2^n/n^100), and claim there was no meaningful speed up at all. This is why I say there is no one unified definition for a "faster algorithm." It is always contextually dependent.

Answer 5

আমি এখনও মন্তব্য করতে পারি না, তবে আমি বর্তমানের উত্তরগুলির মতো অনুভব করি, সঠিক এবং তথ্যবহুল হলেও এই প্রশ্নের অংশটি চিহ্নিত করিনা। প্রথমে এর সমতুল্য একটি অভিব্যক্তি লিখি$A(n) \in O(f(n))$।

$$\exists \ 0 \le c_f \lt \infty \mid \ \limsup_{n \to \infty} \frac{A(n)}{f(n)} = c_f$$

Now, let us assume we are talking about an arbitrarily increasing function $g(n)$ where $\limsup_{n\to\infty} g(n) = \infty$ and let us create the function $h(n) = \frac{f(n)}{g(n)}$.

We are given that the run-time of the "improved" algorithm $A'(n)$ is in $O(h(n))$. Suppose that the run-time of the original algorithm $A(n)$ is also in $O(h(n))$. This can be written as follows.

$$\exists \ 0 \le c_h \lt \infty \mid \ \limsup_{n \to \infty} \frac{A(n)}{h(n)} = c_h$$

Using the rules of limits, we can also write:

$$c_h = \limsup_{n \to \infty} \frac{A(n)}{h(n)} = \limsup_{n \to \infty} \frac{A(n)g(n)}{f(n)} = c_f \cdot \limsup_{n \to \infty} g(n)$$

Since $c_h \lt \infty$, this can only be true if $c_f = 0$.

The contrapositive statement is: If $c_f \neq 0$, then $A(n) \notin O(h(n))$.

In words, $A'(n)$ is an "improvement" on $A(n)$ under the additional conditions that $A(n) \in \Theta(f(n))$ and $g(n)$ is arbitrarily increasing.

Additionally, this should show why the statement that $A(n) \in O(f(n))$ is not strong enough to draw a conclusion about whether $A'(n)$ is an "improvement." In short, $A(n)$ could already be in $O(h(n))$.

Comments

Your limit should be limit superior.

— Yuval Filmus

@YuvalFilmus Updated

— Jared Goguen