Measuring Performance of a java algorithm - java

How can i measure the performance of my java algorithm effectively ? is there any accurate way of doing it?
i read other questions of same kind but not satisfied.
any help would be appreciated.

long reference=System.nanoTime();
your_funct();
long finishm=System.nanoTime();
System.out.println( ( (double)(finishm-reference) )/1000000000.0); //in seconds
Has a meaningful level of ~0.003 seconds in my machine. I mean, you measure in nano-seconds but smallest step is around 3000000 nanoseconds in my machine.

You ask for performance which indicates some sort of timing. But then what would you compare against?
A general way of measuring an algorithm is using Big O, which takes a simplified mathematical approach.
To explain this at a very basic level a simple linear search of a list of integers has a linear (n) worst case big o. Eg:
for(int i = 0; i < sizeofarray; ++i)
if(array[i] == to_find)
return i;
At the worst case this would take i iterations (often number is referred to as n in big o) - so we call it n or linear complexity algorithm.
Something like a bubblesort algorithm is a loop within a loop so we have n * n complexity = n^2 or quadratic complexity.
Comparing like with like, if we just consider sorting, quicksort is more efficient than quadratic complexity (it is n log n complexity) so you can consider quicksort being 'better' than bubblesort.
So when evaluating your algorithm think about it in terms of n. Is there a loop? how many? The fewer the better. No loops even better - constant big o.

You can use some profilers. Many IDEs (such as Netbeans) have one.

If can make it practical or theoretical. If practical then put a timer before the algorithm starts and stop it when it ends. If theoretical then use Big O notation (isn't that hard) and you'll get an estimate of its time or space complexity.

The best way to do it is still java.lang.System.currentTimeMillis() as it will work regardless the IDE you are using.

Related

Confusion in the concept of constant time/space complexity

I am confused with the concept of constant time/space complexity.
For example:
public void recurse(int x) {
if(x==0) return;
else recurse(x/10);
}
where, 1<x<=2147483647
If we want to express the space complexity for this function in terms of big O notation and count the stack space for recursion, what will be the space complexity?
I am confused between:
O(1) - The maximum value of int in java is 2147483647, so at max it will recurse 10 times.
O(log x) - Number of recursions is really dependent on the number of digits in x, so at max we will have ~log10x recursion.
If we say it is O(1), then wouldn't any algorithm which has some finite input can have its time/space complexity bounded by some number? So let's take case of insertion sort in an array of numbers in java. The largest array you can have in java is of size 2147483647, so does that mean T(n) = O(21474836472) = O(1)?
Or should I just look it as, O(1) is a loose bound, while O(log x) is a tighter bound?
Here is the definition I found on wikipedia:
An algorithm is said to be constant time (also written as O(1) time) if the value of T(n) is bounded by a value that does not depend on the size of the input. For example, accessing any single element in an array takes constant time as only one operation has to be performed to locate it. In a similar manner, finding the minimal value in an array sorted in ascending order; it is the first element. However, finding the minimal value in an unordered array is not a constant time operation as scanning over each element in the array is needed in order to determine the minimal value. Hence it is a linear time operation, taking O(n) time. If the number of elements is known in advance and does not change, however, such an algorithm can still be said to run in constant time.
When analysing the time and space complexity of algorithms, we have to ignore some limitations of physical computers; the complexity is a function of the "input size" n, which in big O notation is an asymptotic upper bound as n tends to infinity, but of course a physical computer cannot run the algorithm for arbitrarily large n because it has a finite amount of memory and other storage.
So to do the analysis in a meaningful way, we analyse the algorithm on an imaginary kind of computer where there is no limit on an array's length, where integers can be "sufficiently large" for the algorithm to work, and so on. Your Java code is a concrete implementation of the algorithm, but the algorithm exists as an abstract idea beyond the boundary of what is possible in Java on a real computer. So running this abstract algorithm on an imaginary computer with no such limits, the space complexity is O(log n).
This kind of "imaginary computer" might sound a bit vague, but it is something that can be mathematically formalised in order to do the analysis rigorously; it is called a model of computation. In practice, unless you are doing academic research then you don't need to analyse an algorithm that rigorously, so it's more useful to get comfortable with the vaguer notion that you should ignore any limits which would prevent the algorithm running on an arbitrarily large input.
It really depends on why you are using the big-O notation.
You are correct in saying that, technically, any algorithm is O(1) if it only works for a finite number of possible inputs. For example, this would be an O(1) sorting algorithm: "Read the first 10^6 bits of input. If there are more bits left in the input, output "error". Otherwise, bubblesort."
But the benefit of the notation lies in the fact that it usually approximates the actual running time of a program well. While an O(n) algorithm might as well do 10^100 * n operations, this is usually not the case, and this is why we use the big-O notation at all. Exceptions from this rule are known as galactic algorithms, the most famous one being the Coppersmith–Winograd matrix multiplication algorithm.
To sum up, if you want to be technical and win an argument with a friend, you could say that your algorithm is O(1). If you want to actually use the bound to approximate how fast it is, what you should do is imagine it works for arbitrarily large numbers and just call it O(log(n)).
Side note: Calling this algorithm O(log(n)) is a bit informal, as, technically, the complexity would need to be expressed in terms of size of input, not its magnitude, thus making it O(n). The rule of thumb is: if you're working with small numbers, express the complexity in terms of the magnitude - everyone will understand. If you're working with numbers with potentially millions of digits, then express the complexity in terms of the length. In this case, the cost of "simple" operations such as multiplication (which, for small numbers, is often considered to be O(1)) also needs to be factored in.
Constant time or space means that the time and space used by the algorithm don't depend on the size of the input.
A constant time (hence O(1)) algorithm would be
public int square(int x){
return x * x;
}
because for any input, it does the same multiplication and it's over.
On the other hand, to sum all elements of an array
public int sum(int[] array){
int sum = 0;
for(int i : array) sum += i;
return sum;
}
takes O(n) time, where n is the size of the array. It depends directly on the size of the input.
The space complexity behaves equally.
Any thing that doesn't rely on the size of any input is considered constant.
Applying asymptotic complexity to the real world is tricky as you have discovered.
Asymptotic complexity deals with the abstract situation where input size N has no upper limit, and you're only interested in what will happen with arbitrarily large input size.
In the real world, in the practical applications you're interested, the input size often has an upper limit. The upper limit may come from the fact that you don't have infinite resources (time/money) to collect data. Or it may be imposed by technical limitations, like the fixed size of int datatype in Java.
Since asymptotic complexity analysis does not account for real world limitations, the asymptotic complexity of recurse(x) is O(log x). Even though we know that x can only grow up to 2^31.
When your algo doesnt depend on size of input, it is said to have constant time complexity. For eg:
function print(int input) {
// 10 lines of printing here
}
Here, no matter what you pass in as 'input', function body statements will always run 10 times. If you pass 'input' as 10, 10 statements are run. If you pass 'input' as 20, still 10 statements are run.
Now on other hand, consider this:
function print(int input) {
// This loop will run 'input' times
for(int i=0;i<input;i++){
System.out.println(i);
}
}
This algo will run depending on the size of input. If you pass 'input' as 10, for loop will run 10 times, If you pass 'input' as 20, for loop will run 20 times. So, algo grows with the same pace as 'input' grows. So, in this case time complexity is said to be O(n)

knapsack with dynamic programming complexity doesn't match with the real result [duplicate]

I'd prefer as little formal definition as possible and simple mathematics.
Quick note, my answer is almost certainly confusing Big Oh notation (which is an upper bound) with Big Theta notation "Θ" (which is a two-side bound). But in my experience, this is actually typical of discussions in non-academic settings. Apologies for any confusion caused.
BigOh complexity can be visualized with this graph:
The simplest definition I can give for Big Oh notation is this:
Big Oh notation is a relative representation of the complexity of an algorithm.
There are some important and deliberately chosen words in that sentence:
relative: you can only compare apples to apples. You can't compare an algorithm that does arithmetic multiplication to an algorithm that sorts a list of integers. But a comparison of two algorithms to do arithmetic operations (one multiplication, one addition) will tell you something meaningful;
representation: BigOh (in its simplest form) reduces the comparison between algorithms to a single variable. That variable is chosen based on observations or assumptions. For example, sorting algorithms are typically compared based on comparison operations (comparing two nodes to determine their relative ordering). This assumes that comparison is expensive. But what if the comparison is cheap but swapping is expensive? It changes the comparison; and
complexity: if it takes me one second to sort 10,000 elements, how long will it take me to sort one million? Complexity in this instance is a relative measure to something else.
Come back and reread the above when you've read the rest.
The best example of BigOh I can think of is doing arithmetic. Take two numbers (123456 and 789012). The basic arithmetic operations we learned in school were:
addition;
subtraction;
multiplication; and
division.
Each of these is an operation or a problem. A method of solving these is called an algorithm.
The addition is the simplest. You line the numbers up (to the right) and add the digits in a column writing the last number of that addition in the result. The 'tens' part of that number is carried over to the next column.
Let's assume that the addition of these numbers is the most expensive operation in this algorithm. It stands to reason that to add these two numbers together we have to add together 6 digits (and possibly carry a 7th). If we add two 100 digit numbers together we have to do 100 additions. If we add two 10,000 digit numbers we have to do 10,000 additions.
See the pattern? The complexity (being the number of operations) is directly proportional to the number of digits n in the larger number. We call this O(n) or linear complexity.
Subtraction is similar (except you may need to borrow instead of carry).
Multiplication is different. You line the numbers up, take the first digit in the bottom number and multiply it in turn against each digit in the top number and so on through each digit. So to multiply our two 6 digit numbers we must do 36 multiplications. We may need to do as many as 10 or 11 column adds to get the end result too.
If we have two 100-digit numbers we need to do 10,000 multiplications and 200 adds. For two one million digit numbers we need to do one trillion (1012) multiplications and two million adds.
As the algorithm scales with n-squared, this is O(n2) or quadratic complexity. This is a good time to introduce another important concept:
We only care about the most significant portion of complexity.
The astute may have realized that we could express the number of operations as: n2 + 2n. But as you saw from our example with two numbers of a million digits apiece, the second term (2n) becomes insignificant (accounting for 0.0002% of the total operations by that stage).
One can notice that we've assumed the worst case scenario here. While multiplying 6 digit numbers, if one of them has 4 digits and the other one has 6 digits, then we only have 24 multiplications. Still, we calculate the worst case scenario for that 'n', i.e when both are 6 digit numbers. Hence Big Oh notation is about the Worst-case scenario of an algorithm.
The Telephone Book
The next best example I can think of is the telephone book, normally called the White Pages or similar but it varies from country to country. But I'm talking about the one that lists people by surname and then initials or first name, possibly address and then telephone numbers.
Now if you were instructing a computer to look up the phone number for "John Smith" in a telephone book that contains 1,000,000 names, what would you do? Ignoring the fact that you could guess how far in the S's started (let's assume you can't), what would you do?
A typical implementation might be to open up to the middle, take the 500,000th and compare it to "Smith". If it happens to be "Smith, John", we just got really lucky. Far more likely is that "John Smith" will be before or after that name. If it's after we then divide the last half of the phone book in half and repeat. If it's before then we divide the first half of the phone book in half and repeat. And so on.
This is called a binary search and is used every day in programming whether you realize it or not.
So if you want to find a name in a phone book of a million names you can actually find any name by doing this at most 20 times. In comparing search algorithms we decide that this comparison is our 'n'.
For a phone book of 3 names it takes 2 comparisons (at most).
For 7 it takes at most 3.
For 15 it takes 4.
…
For 1,000,000 it takes 20.
That is staggeringly good, isn't it?
In BigOh terms this is O(log n) or logarithmic complexity. Now the logarithm in question could be ln (base e), log10, log2 or some other base. It doesn't matter it's still O(log n) just like O(2n2) and O(100n2) are still both O(n2).
It's worthwhile at this point to explain that BigOh can be used to determine three cases with an algorithm:
Best Case: In the telephone book search, the best case is that we find the name in one comparison. This is O(1) or constant complexity;
Expected Case: As discussed above this is O(log n); and
Worst Case: This is also O(log n).
Normally we don't care about the best case. We're interested in the expected and worst case. Sometimes one or the other of these will be more important.
Back to the telephone book.
What if you have a phone number and want to find a name? The police have a reverse phone book but such look-ups are denied to the general public. Or are they? Technically you can reverse look-up a number in an ordinary phone book. How?
You start at the first name and compare the number. If it's a match, great, if not, you move on to the next. You have to do it this way because the phone book is unordered (by phone number anyway).
So to find a name given the phone number (reverse lookup):
Best Case: O(1);
Expected Case: O(n) (for 500,000); and
Worst Case: O(n) (for 1,000,000).
The Traveling Salesman
This is quite a famous problem in computer science and deserves a mention. In this problem, you have N towns. Each of those towns is linked to 1 or more other towns by a road of a certain distance. The Traveling Salesman problem is to find the shortest tour that visits every town.
Sounds simple? Think again.
If you have 3 towns A, B, and C with roads between all pairs then you could go:
A → B → C
A → C → B
B → C → A
B → A → C
C → A → B
C → B → A
Well, actually there's less than that because some of these are equivalent (A → B → C and C → B → A are equivalent, for example, because they use the same roads, just in reverse).
In actuality, there are 3 possibilities.
Take this to 4 towns and you have (iirc) 12 possibilities.
With 5 it's 60.
6 becomes 360.
This is a function of a mathematical operation called a factorial. Basically:
5! = 5 × 4 × 3 × 2 × 1 = 120
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
…
25! = 25 × 24 × … × 2 × 1 = 15,511,210,043,330,985,984,000,000
…
50! = 50 × 49 × … × 2 × 1 = 3.04140932 × 1064
So the BigOh of the Traveling Salesman problem is O(n!) or factorial or combinatorial complexity.
By the time you get to 200 towns there isn't enough time left in the universe to solve the problem with traditional computers.
Something to think about.
Polynomial Time
Another point I wanted to make a quick mention of is that any algorithm that has a complexity of O(na) is said to have polynomial complexity or is solvable in polynomial time.
O(n), O(n2) etc. are all polynomial time. Some problems cannot be solved in polynomial time. Certain things are used in the world because of this. Public Key Cryptography is a prime example. It is computationally hard to find two prime factors of a very large number. If it wasn't, we couldn't use the public key systems we use.
Anyway, that's it for my (hopefully plain English) explanation of BigOh (revised).
It shows how an algorithm scales based on input size.
O(n2): known as Quadratic complexity
1 item: 1 operations
10 items: 100 operations
100 items: 10,000 operations
Notice that the number of items increases by a factor of 10, but the time increases by a factor of 102. Basically, n=10 and so O(n2) gives us the scaling factor n2 which is 102.
O(n): known as Linear complexity
1 item: 1 operation
10 items: 10 operations
100 items: 100 operations
This time the number of items increases by a factor of 10, and so does the time. n=10 and so O(n)'s scaling factor is 10.
O(1): known as Constant complexity
1 item: 1 operations
10 items: 1 operations
100 items: 1 operations
The number of items is still increasing by a factor of 10, but the scaling factor of O(1) is always 1.
O(log n): known as Logarithmic complexity
1 item: 1 operations
10 items: 2 operations
100 items: 3 operations
1000 items: 4 operations
10,000 items: 5 operations
The number of computations is only increased by a log of the input value. So in this case, assuming each computation takes 1 second, the log of the input n is the time required, hence log n.
That's the gist of it. They reduce the maths down so it might not be exactly n2 or whatever they say it is, but that'll be the dominating factor in the scaling.
Big-O notation (also called "asymptotic growth" notation) is what functions "look like" when you ignore constant factors and stuff near the origin. We use it to talk about how thing scale.
Basics
for "sufficiently" large inputs...
f(x) ∈ O(upperbound) means f "grows no faster than" upperbound
f(x) ∈ Ɵ(justlikethis) mean f "grows exactly like" justlikethis
f(x) ∈ Ω(lowerbound) means f "grows no slower than" lowerbound
big-O notation doesn't care about constant factors: the function 9x² is said to "grow exactly like" 10x². Neither does big-O asymptotic notation care about non-asymptotic stuff ("stuff near the origin" or "what happens when the problem size is small"): the function 10x² is said to "grow exactly like" 10x² - x + 2.
Why would you want to ignore the smaller parts of the equation? Because they become completely dwarfed by the big parts of the equation as you consider larger and larger scales; their contribution becomes dwarfed and irrelevant. (See example section.)
Put another way, it's all about the ratio as you go to infinity. If you divide the actual time it takes by the O(...), you will get a constant factor in the limit of large inputs. Intuitively this makes sense: functions "scale like" one another if you can multiply one to get the other. That is when we say...
actualAlgorithmTime(N) ∈ O(bound(N))
e.g. "time to mergesort N elements
is O(N log(N))"
... this means that for "large enough" problem sizes N (if we ignore stuff near the origin), there exists some constant (e.g. 2.5, completely made up) such that:
actualAlgorithmTime(N) e.g. "mergesort_duration(N) "
────────────────────── < constant ───────────────────── < 2.5
bound(N) N log(N)
There are many choices of constant; often the "best" choice is known as the "constant factor" of the algorithm... but we often ignore it like we ignore non-largest terms (see Constant Factors section for why they don't usually matter). You can also think of the above equation as a bound, saying "In the worst-case scenario, the time it takes will never be worse than roughly N*log(N), within a factor of 2.5 (a constant factor we don't care much about)".
In general, O(...) is the most useful one because we often care about worst-case behavior. If f(x) represents something "bad" like the processor or memory usage, then "f(x) ∈ O(upperbound)" means "upperbound is the worst-case scenario of processor/memory usage".
Applications
As a purely mathematical construct, big-O notation is not limited to talking about processing time and memory. You can use it to discuss the asymptotics of anything where scaling is meaningful, such as:
the number of possible handshakes among N people at a party (Ɵ(N²), specifically N(N-1)/2, but what matters is that it "scales like" N²)
probabilistic expected number of people who have seen some viral marketing as a function of time
how website latency scales with the number of processing units in a CPU or GPU or computer cluster
how heat output scales on CPU dies as a function of transistor count, voltage, etc.
how much time an algorithm needs to run, as a function of input size
how much space an algorithm needs to run, as a function of input size
Example
For the handshake example above, everyone in a room shakes everyone else's hand. In that example, #handshakes ∈ Ɵ(N²). Why?
Back up a bit: the number of handshakes is exactly n-choose-2 or N*(N-1)/2 (each of N people shakes the hands of N-1 other people, but this double-counts handshakes so divide by 2):
However, for very large numbers of people, the linear term N is dwarfed and effectively contributes 0 to the ratio (in the chart: the fraction of empty boxes on the diagonal over total boxes gets smaller as the number of participants becomes larger). Therefore the scaling behavior is order N², or the number of handshakes "grows like N²".
#handshakes(N)
────────────── ≈ 1/2
N²
It's as if the empty boxes on the diagonal of the chart (N*(N-1)/2 checkmarks) weren't even there (N2 checkmarks asymptotically).
(temporary digression from "plain English":) If you wanted to prove this to yourself, you could perform some simple algebra on the ratio to split it up into multiple terms (lim means "considered in the limit of", just ignore it if you haven't seen it, it's just notation for "and N is really really big"):
N²/2 - N/2 (N²)/2 N/2 1/2
lim ────────── = lim ( ────── - ─── ) = lim ─── = 1/2
N→∞ N² N→∞ N² N² N→∞ 1
┕━━━┙
this is 0 in the limit of N→∞:
graph it, or plug in a really large number for N
tl;dr: The number of handshakes 'looks like' x² so much for large values, that if we were to write down the ratio #handshakes/x², the fact that we don't need exactly x² handshakes wouldn't even show up in the decimal for an arbitrarily large while.
e.g. for x=1million, ratio #handshakes/x²: 0.499999...
Building Intuition
This lets us make statements like...
"For large enough inputsize=N, no matter what the constant factor is, if I double the input size...
... I double the time an O(N) ("linear time") algorithm takes."
N → (2N) = 2(N)
... I double-squared (quadruple) the time an O(N²) ("quadratic time") algorithm takes." (e.g. a problem 100x as big takes 100²=10000x as long... possibly unsustainable)
N² → (2N)² = 4(N²)
... I double-cubed (octuple) the time an O(N³) ("cubic time") algorithm takes." (e.g. a problem 100x as big takes 100³=1000000x as long... very unsustainable)
cN³ → c(2N)³ = 8(cN³)
... I add a fixed amount to the time an O(log(N)) ("logarithmic time") algorithm takes." (cheap!)
c log(N) → c log(2N) = (c log(2))+(c log(N)) = (fixed amount)+(c log(N))
... I don't change the time an O(1) ("constant time") algorithm takes." (the cheapest!)
c*1 → c*1
... I "(basically) double" the time an O(N log(N)) algorithm takes." (fairly common)
c 2N log(2N) / c N log(N) (here we divide f(2n)/f(n), but we could have as above massaged the expression and factored out cNlogN as above)
→ 2 log(2N)/log(N)
→ 2 (log(2) + log(N))/log(N)
→ 2*(1+(log2N)-1) (basically 2 for large N; eventually less than 2.000001)
(alternatively, say log(N) will always be below like 17 for your data so it's O(17 N) which is linear; that is not rigorous nor sensical though)
... I ridiculously increase the time a O(2N) ("exponential time") algorithm takes." (you'd double (or triple, etc.) the time just by increasing the problem by a single unit)
2N → 22N = (4N)............put another way...... 2N → 2N+1 = 2N21 = 2 2N
[for the mathematically inclined, you can mouse over the spoilers for minor sidenotes]
(with credit to https://stackoverflow.com/a/487292/711085 )
(technically the constant factor could maybe matter in some more esoteric examples, but I've phrased things above (e.g. in log(N)) such that it doesn't)
These are the bread-and-butter orders of growth that programmers and applied computer scientists use as reference points. They see these all the time. (So while you could technically think "Doubling the input makes an O(√N) algorithm 1.414 times slower," it's better to think of it as "this is worse than logarithmic but better than linear".)
Constant factors
Usually, we don't care what the specific constant factors are, because they don't affect the way the function grows. For example, two algorithms may both take O(N) time to complete, but one may be twice as slow as the other. We usually don't care too much unless the factor is very large since optimizing is tricky business ( When is optimisation premature? ); also the mere act of picking an algorithm with a better big-O will often improve performance by orders of magnitude.
Some asymptotically superior algorithms (e.g. a non-comparison O(N log(log(N))) sort) can have so large a constant factor (e.g. 100000*N log(log(N))), or overhead that is relatively large like O(N log(log(N))) with a hidden + 100*N, that they are rarely worth using even on "big data".
Why O(N) is sometimes the best you can do, i.e. why we need datastructures
O(N) algorithms are in some sense the "best" algorithms if you need to read all your data. The very act of reading a bunch of data is an O(N) operation. Loading it into memory is usually O(N) (or faster if you have hardware support, or no time at all if you've already read the data). However, if you touch or even look at every piece of data (or even every other piece of data), your algorithm will take O(N) time to perform this looking. No matter how long your actual algorithm takes, it will be at least O(N) because it spent that time looking at all the data.
The same can be said for the very act of writing. All algorithms which print out N things will take N time because the output is at least that long (e.g. printing out all permutations (ways to rearrange) a set of N playing cards is factorial: O(N!) (which is why in those cases, good programs will ensure an iteration uses O(1) memory and doesn't print or store every intermediate step)).
This motivates the use of data structures: a data structure requires reading the data only once (usually O(N) time), plus some arbitrary amount of preprocessing (e.g. O(N) or O(N log(N)) or O(N²)) which we try to keep small. Thereafter, modifying the data structure (insertions/deletions/ etc.) and making queries on the data take very little time, such as O(1) or O(log(N)). You then proceed to make a large number of queries! In general, the more work you're willing to do ahead of time, the less work you'll have to do later on.
For example, say you had the latitude and longitude coordinates of millions of road segments and wanted to find all street intersections.
Naive method: If you had the coordinates of a street intersection, and wanted to examine nearby streets, you would have to go through the millions of segments each time, and check each one for adjacency.
If you only needed to do this once, it would not be a problem to have to do the naive method of O(N) work only once, but if you want to do it many times (in this case, N times, once for each segment), we'd have to do O(N²) work, or 1000000²=1000000000000 operations. Not good (a modern computer can perform about a billion operations per second).
If we use a simple structure called a hash table (an instant-speed lookup table, also known as a hashmap or dictionary), we pay a small cost by preprocessing everything in O(N) time. Thereafter, it only takes constant time on average to look up something by its key (in this case, our key is the latitude and longitude coordinates, rounded into a grid; we search the adjacent gridspaces of which there are only 9, which is a constant).
Our task went from an infeasible O(N²) to a manageable O(N), and all we had to do was pay a minor cost to make a hash table.
analogy: The analogy in this particular case is a jigsaw puzzle: We created a data structure that exploits some property of the data. If our road segments are like puzzle pieces, we group them by matching color and pattern. We then exploit this to avoid doing extra work later (comparing puzzle pieces of like color to each other, not to every other single puzzle piece).
The moral of the story: a data structure lets us speed up operations. Even more, advanced data structures can let you combine, delay, or even ignore operations in incredibly clever ways. Different problems would have different analogies, but they'd all involve organizing the data in a way that exploits some structure we care about, or which we've artificially imposed on it for bookkeeping. We do work ahead of time (basically planning and organizing), and now repeated tasks are much much easier!
Practical example: visualizing orders of growth while coding
Asymptotic notation is, at its core, quite separate from programming. Asymptotic notation is a mathematical framework for thinking about how things scale and can be used in many different fields. That said... this is how you apply asymptotic notation to coding.
The basics: Whenever we interact with every element in a collection of size A (such as an array, a set, all keys of a map, etc.), or perform A iterations of a loop, that is a multiplicative factor of size A. Why do I say "a multiplicative factor"?--because loops and functions (almost by definition) have multiplicative running time: the number of iterations, times work done in the loop (or for functions: the number of times you call the function, times work done in the function). (This holds if we don't do anything fancy, like skip loops or exit the loop early, or change control flow in the function based on arguments, which is very common.) Here are some examples of visualization techniques, with accompanying pseudocode.
(here, the xs represent constant-time units of work, processor instructions, interpreter opcodes, whatever)
for(i=0; i<A; i++) // A * ...
some O(1) operation // 1
--> A*1 --> O(A) time
visualization:
|<------ A ------->|
1 2 3 4 5 x x ... x
other languages, multiplying orders of growth:
javascript, O(A) time and space
someListOfSizeA.map((x,i) => [x,i])
python, O(rows*cols) time and space
[[r*c for c in range(cols)] for r in range(rows)]
Example 2:
for every x in listOfSizeA: // A * (...
some O(1) operation // 1
some O(B) operation // B
for every y in listOfSizeC: // C * (...
some O(1) operation // 1))
--> O(A*(1 + B + C))
O(A*(B+C)) (1 is dwarfed)
visualization:
|<------ A ------->|
1 x x x x x x ... x
2 x x x x x x ... x ^
3 x x x x x x ... x |
4 x x x x x x ... x |
5 x x x x x x ... x B <-- A*B
x x x x x x x ... x |
................... |
x x x x x x x ... x v
x x x x x x x ... x ^
x x x x x x x ... x |
x x x x x x x ... x |
x x x x x x x ... x C <-- A*C
x x x x x x x ... x |
................... |
x x x x x x x ... x v
Example 3:
function nSquaredFunction(n) {
total = 0
for i in 1..n: // N *
for j in 1..n: // N *
total += i*k // 1
return total
}
// O(n^2)
function nCubedFunction(a) {
for i in 1..n: // A *
print(nSquaredFunction(a)) // A^2
}
// O(a^3)
If we do something slightly complicated, you might still be able to imagine visually what's going on:
for x in range(A):
for y in range(1..x):
simpleOperation(x*y)
x x x x x x x x x x |
x x x x x x x x x |
x x x x x x x x |
x x x x x x x |
x x x x x x |
x x x x x |
x x x x |
x x x |
x x |
x___________________|
Here, the smallest recognizable outline you can draw is what matters; a triangle is a two dimensional shape (0.5 A^2), just like a square is a two-dimensional shape (A^2); the constant factor of two here remains in the asymptotic ratio between the two, however, we ignore it like all factors... (There are some unfortunate nuances to this technique I don't go into here; it can mislead you.)
Of course this does not mean that loops and functions are bad; on the contrary, they are the building blocks of modern programming languages, and we love them. However, we can see that the way we weave loops and functions and conditionals together with our data (control flow, etc.) mimics the time and space usage of our program! If time and space usage becomes an issue, that is when we resort to cleverness and find an easy algorithm or data structure we hadn't considered, to reduce the order of growth somehow. Nevertheless, these visualization techniques (though they don't always work) can give you a naive guess at a worst-case running time.
Here is another thing we can recognize visually:
<----------------------------- N ----------------------------->
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x
x x x x x x x x
x x x x
x x
x
We can just rearrange this and see it's O(N):
<----------------------------- N ----------------------------->
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x|x x x x x x x x|x x x x|x x|x
Or maybe you do log(N) passes of the data, for O(N*log(N)) total time:
<----------------------------- N ----------------------------->
^ x x x x x x x x x x x x x x x x|x x x x x x x x x x x x x x x x
| x x x x x x x x|x x x x x x x x|x x x x x x x x|x x x x x x x x
lgN x x x x|x x x x|x x x x|x x x x|x x x x|x x x x|x x x x|x x x x
| x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x
v x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x
Unrelatedly but worth mentioning again: If we perform a hash (e.g. a dictionary/hashtable lookup), that is a factor of O(1). That's pretty fast.
[myDictionary.has(x) for x in listOfSizeA]
\----- O(1) ------/
--> A*1 --> O(A)
If we do something very complicated, such as with a recursive function or divide-and-conquer algorithm, you can use the Master Theorem (usually works), or in ridiculous cases the Akra-Bazzi Theorem (almost always works) you look up the running time of your algorithm on Wikipedia.
But, programmers don't think like this because eventually, algorithm intuition just becomes second nature. You will start to code something inefficient and immediately think "am I doing something grossly inefficient?". If the answer is "yes" AND you foresee it actually mattering, then you can take a step back and think of various tricks to make things run faster (the answer is almost always "use a hashtable", rarely "use a tree", and very rarely something a bit more complicated).
Amortized and average-case complexity
There is also the concept of "amortized" and/or "average case" (note that these are different).
Average Case: This is no more than using big-O notation for the expected value of a function, rather than the function itself. In the usual case where you consider all inputs to be equally likely, the average case is just the average of the running time. For example with quicksort, even though the worst-case is O(N^2) for some really bad inputs, the average case is the usual O(N log(N)) (the really bad inputs are very small in number, so few that we don't notice them in the average case).
Amortized Worst-Case: Some data structures may have a worst-case complexity that is large, but guarantee that if you do many of these operations, the average amount of work you do will be better than worst-case. For example, you may have a data structure that normally takes constant O(1) time. However, occasionally it will 'hiccup' and take O(N) time for one random operation, because maybe it needs to do some bookkeeping or garbage collection or something... but it promises you that if it does hiccup, it won't hiccup again for N more operations. The worst-case cost is still O(N) per operation, but the amortized cost over many runs is O(N)/N = O(1) per operation. Because the big operations are sufficiently rare, the massive amount of occasional work can be considered to blend in with the rest of the work as a constant factor. We say the work is "amortized" over a sufficiently large number of calls that it disappears asymptotically.
The analogy for amortized analysis:
You drive a car. Occasionally, you need to spend 10 minutes going to
the gas station and then spend 1 minute refilling the tank with gas.
If you did this every time you went anywhere with your car (spend 10
minutes driving to the gas station, spend a few seconds filling up a
fraction of a gallon), it would be very inefficient. But if you fill
up the tank once every few days, the 11 minutes spent driving to the
gas station is "amortized" over a sufficiently large number of trips,
that you can ignore it and pretend all your trips were maybe 5% longer.
Comparison between average-case and amortized worst-case:
Average-case: We make some assumptions about our inputs; i.e. if our inputs have different probabilities, then our outputs/runtimes will have different probabilities (which we take the average of). Usually, we assume that our inputs are all equally likely (uniform probability), but if the real-world inputs don't fit our assumptions of "average input", the average output/runtime calculations may be meaningless. If you anticipate uniformly random inputs though, this is useful to think about!
Amortized worst-case: If you use an amortized worst-case data structure, the performance is guaranteed to be within the amortized worst-case... eventually (even if the inputs are chosen by an evil demon who knows everything and is trying to screw you over). Usually, we use this to analyze algorithms that may be very 'choppy' in performance with unexpected large hiccups, but over time perform just as well as other algorithms. (However unless your data structure has upper limits for much outstanding work it is willing to procrastinate on, an evil attacker could perhaps force you to catch up on the maximum amount of procrastinated work all-at-once.
Though, if you're reasonably worried about an attacker, there are many other algorithmic attack vectors to worry about besides amortization and average-case.)
Both average-case and amortization are incredibly useful tools for thinking about and designing with scaling in mind.
(See Difference between average case and amortized analysis if interested in this subtopic.)
Multidimensional big-O
Most of the time, people don't realize that there's more than one variable at work. For example, in a string-search algorithm, your algorithm may take time O([length of text] + [length of query]), i.e. it is linear in two variables like O(N+M). Other more naive algorithms may be O([length of text]*[length of query]) or O(N*M). Ignoring multiple variables is one of the most common oversights I see in algorithm analysis, and can handicap you when designing an algorithm.
The whole story
Keep in mind that big-O is not the whole story. You can drastically speed up some algorithms by using caching, making them cache-oblivious, avoiding bottlenecks by working with RAM instead of disk, using parallelization, or doing work ahead of time -- these techniques are often independent of the order-of-growth "big-O" notation, though you will often see the number of cores in the big-O notation of parallel algorithms.
Also keep in mind that due to hidden constraints of your program, you might not really care about asymptotic behavior. You may be working with a bounded number of values, for example:
If you're sorting something like 5 elements, you don't want to use the speedy O(N log(N)) quicksort; you want to use insertion sort, which happens to perform well on small inputs. These situations often come up in divide-and-conquer algorithms, where you split up the problem into smaller and smaller subproblems, such as recursive sorting, fast Fourier transforms, or matrix multiplication.
If some values are effectively bounded due to some hidden fact (e.g. the average human name is softly bounded at perhaps 40 letters, and human age is softly bounded at around 150). You can also impose bounds on your input to effectively make terms constant.
In practice, even among algorithms which have the same or similar asymptotic performance, their relative merit may actually be driven by other things, such as: other performance factors (quicksort and mergesort are both O(N log(N)), but quicksort takes advantage of CPU caches); non-performance considerations, like ease of implementation; whether a library is available, and how reputable and maintained the library is.
Programs will also run slower on a 500MHz computer vs 2GHz computer. We don't really consider this as part of the resource bounds, because we think of the scaling in terms of machine resources (e.g. per clock cycle), not per real second. However, there are similar things which can 'secretly' affect performance, such as whether you are running under emulation, or whether the compiler optimized code or not. These might make some basic operations take longer (even relative to each other), or even speed up or slow down some operations asymptotically (even relative to each other). The effect may be small or large between different implementation and/or environment. Do you switch languages or machines to eke out that little extra work? That depends on a hundred other reasons (necessity, skills, coworkers, programmer productivity, the monetary value of your time, familiarity, workarounds, why not assembly or GPU, etc...), which may be more important than performance.
The above issues, like the effect of the choice of which programming language is used, are almost never considered as part of the constant factor (nor should they be); yet one should be aware of them because sometimes (though rarely) they may affect things. For example in cpython, the native priority queue implementation is asymptotically non-optimal (O(log(N)) rather than O(1) for your choice of insertion or find-min); do you use another implementation? Probably not, since the C implementation is probably faster, and there are probably other similar issues elsewhere. There are tradeoffs; sometimes they matter and sometimes they don't.
(edit: The "plain English" explanation ends here.)
Math addenda
For completeness, the precise definition of big-O notation is as follows: f(x) ∈ O(g(x)) means that "f is asymptotically upper-bounded by const*g": ignoring everything below some finite value of x, there exists a constant such that |f(x)| ≤ const * |g(x)|. (The other symbols are as follows: just like O means ≤, Ω means ≥. There are lowercase variants: o means <, and ω means >.) f(x) ∈ Ɵ(g(x)) means both f(x) ∈ O(g(x)) and f(x) ∈ Ω(g(x)) (upper- and lower-bounded by g): there exists some constants such that f will always lie in the "band" between const1*g(x) and const2*g(x). It is the strongest asymptotic statement you can make and roughly equivalent to ==. (Sorry, I elected to delay the mention of the absolute-value symbols until now, for clarity's sake; especially because I have never seen negative values come up in a computer science context.)
People will often use = O(...), which is perhaps the more correct 'comp-sci' notation, and entirely legitimate to use; "f = O(...)" is read "f is order ... / f is xxx-bounded by ..." and is thought of as "f is some expression whose asymptotics are ...". I was taught to use the more rigorous ∈ O(...). ∈ means "is an element of" (still read as before). In this particular case, O(N²) contains elements like {2 N², 3 N², 1/2 N², 2 N² + log(N), - N² + N^1.9, ...} and is infinitely large, but it's still a set.
O and Ω are not symmetric (n = O(n²), but n² is not O(n)), but Ɵ is symmetric, and (since these relations are all transitive and reflexive) Ɵ, therefore, is symmetric and transitive and reflexive, and therefore partitions the set of all functions into equivalence classes. An equivalence class is a set of things that we consider to be the same. That is to say, given any function you can think of, you can find a canonical/unique 'asymptotic representative' of the class (by generally taking the limit... I think); just like you can group all integers into odds or evens, you can group all functions with Ɵ into x-ish, log(x)^2-ish, etc... by basically ignoring smaller terms (but sometimes you might be stuck with more complicated functions which are separate classes unto themselves).
The = notation might be the more common one and is even used in papers by world-renowned computer scientists. Additionally, it is often the case that in a casual setting, people will say O(...) when they mean Ɵ(...); this is technically true since the set of things Ɵ(exactlyThis) is a subset of O(noGreaterThanThis)... and it's easier to type. ;-)
EDIT: Quick note, this is almost certainly confusing Big O notation (which is an upper bound) with Theta notation (which is both an upper and lower bound). In my experience this is actually typical of discussions in non-academic settings. Apologies for any confusion caused.
In one sentence: As the size of your job goes up, how much longer does it take to complete it?
Obviously that's only using "size" as the input and "time taken" as the output — the same idea applies if you want to talk about memory usage etc.
Here's an example where we have N T-shirts which we want to dry. We'll assume it's incredibly quick to get them in the drying position (i.e. the human interaction is negligible). That's not the case in real life, of course...
Using a washing line outside: assuming you have an infinitely large back yard, washing dries in O(1) time. However much you have of it, it'll get the same sun and fresh air, so the size doesn't affect the drying time.
Using a tumble dryer: you put 10 shirts in each load, and then they're done an hour later. (Ignore the actual numbers here — they're irrelevant.) So drying 50 shirts takes about 5 times as long as drying 10 shirts.
Putting everything in an airing cupboard: If we put everything in one big pile and just let general warmth do it, it will take a long time for the middle shirts to get dry. I wouldn't like to guess at the detail, but I suspect this is at least O(N^2) — as you increase the wash load, the drying time increases faster.
One important aspect of "big O" notation is that it doesn't say which algorithm will be faster for a given size. Take a hashtable (string key, integer value) vs an array of pairs (string, integer). Is it faster to find a key in the hashtable or an element in the array, based on a string? (i.e. for the array, "find the first element where the string part matches the given key.") Hashtables are generally amortised (~= "on average") O(1) — once they're set up, it should take about the same time to find an entry in a 100 entry table as in a 1,000,000 entry table. Finding an element in an array (based on content rather than index) is linear, i.e. O(N) — on average, you're going to have to look at half the entries.
Does this make a hashtable faster than an array for lookups? Not necessarily. If you've got a very small collection of entries, an array may well be faster — you may be able to check all the strings in the time that it takes to just calculate the hashcode of the one you're looking at. As the data set grows larger, however, the hashtable will eventually beat the array.
Big O describes an upper limit on the growth behaviour of a function, for example the runtime of a program, when inputs become large.
Examples:
O(n): If I double the input size the runtime doubles
O(n2): If the input size doubles the runtime quadruples
O(log n): If the input size doubles the runtime increases by one
O(2n): If the input size increases by one, the runtime doubles
The input size is usually the space in bits needed to represent the input.
Big O notation is most commonly used by programmers as an approximate measure of how long a computation (algorithm) will take to complete expressed as a function of the size of the input set.
Big O is useful to compare how well two algorithms will scale up as the number of inputs is increased.
More precisely Big O notation is used to express the asymptotic behavior of a function. That means how the function behaves as it approaches infinity.
In many cases the "O" of an algorithm will fall into one of the following cases:
O(1) - Time to complete is the same regardless of the size of input set. An example is accessing an array element by index.
O(Log N) - Time to complete increases roughly in line with the log2(n). For example 1024 items takes roughly twice as long as 32 items, because Log2(1024) = 10 and Log2(32) = 5. An example is finding an item in a binary search tree (BST).
O(N) - Time to complete that scales linearly with the size of the input set. In other words if you double the number of items in the input set, the algorithm takes roughly twice as long. An example is counting the number of items in a linked list.
O(N Log N) - Time to complete increases by the number of items times the result of Log2(N). An example of this is heap sort and quick sort.
O(N^2) - Time to complete is roughly equal to the square of the number of items. An example of this is bubble sort.
O(N!) - Time to complete is the factorial of the input set. An example of this is the traveling salesman problem brute-force solution.
Big O ignores factors that do not contribute in a meaningful way to the growth curve of a function as the input size increases towards infinity. This means that constants that are added to or multiplied by the function are simply ignored.
Big O is just a way to "Express" yourself in a common way, "How much time / space does it take to run my code?".
You may often see O(n), O(n2), O(nlogn) and so forth, all these are just ways to show; How does an algorithm change?
O(n) means Big O is n, and now you might think, "What is n!?" Well "n" is the amount of elements. Imaging you want to search for an Item in an Array. You would have to look on Each element and as "Are you the correct element/item?" in the worst case, the item is at the last index, which means that it took as much time as there are items in the list, so to be generic, we say "oh hey, n is a fair given amount of values!".
So then you might understand what "n2" means, but to be even more specific, play with the thought you have a simple, the simpliest of the sorting algorithms; bubblesort. This algorithm needs to look through the whole list, for each item.
My list
1
6
3
The flow here would be:
Compare 1 and 6, which is biggest? Ok 6 is in the right position, moving forward!
Compare 6 and 3, oh, 3 is less! Let's move that, Ok the list changed, we need to start from the begining now!
This is O n2 because, you need to look at all items in the list there are "n" items. For each item, you look at all items once more, for comparing, this is also "n", so for every item, you look "n" times meaning n*n = n2
I hope this is as simple as you want it.
But remember, Big O is just a way to experss yourself in the manner of time and space.
Big O describes the fundamental scaling nature of an algorithm.
There is a lot of information that Big O does not tell you about a given algorithm. It cuts to the bone and gives only information about the scaling nature of an algorithm, specifically how the resource use (think time or memory) of an algorithm scales in response to the "input size".
Consider the difference between a steam engine and a rocket. They are not merely different varieties of the same thing (as, say, a Prius engine vs. a Lamborghini engine) but they are dramatically different kinds of propulsion systems, at their core. A steam engine may be faster than a toy rocket, but no steam piston engine will be able to achieve the speeds of an orbital launch vehicle. This is because these systems have different scaling characteristics with regards to the relation of fuel required ("resource usage") to reach a given speed ("input size").
Why is this so important? Because software deals with problems that may differ in size by factors up to a trillion. Consider that for a moment. The ratio between the speed necessary to travel to the Moon and human walking speed is less than 10,000:1, and that is absolutely tiny compared to the range in input sizes software may face. And because software may face an astronomical range in input sizes there is the potential for the Big O complexity of an algorithm, it's fundamental scaling nature, to trump any implementation details.
Consider the canonical sorting example. Bubble-sort is O(n2) while merge-sort is O(n log n). Let's say you have two sorting applications, application A which uses bubble-sort and application B which uses merge-sort, and let's say that for input sizes of around 30 elements application A is 1,000x faster than application B at sorting. If you never have to sort much more than 30 elements then it's obvious that you should prefer application A, as it is much faster at these input sizes. However, if you find that you may have to sort ten million items then what you'd expect is that application B actually ends up being thousands of times faster than application A in this case, entirely due to the way each algorithm scales.
Here is the plain English bestiary I tend to use when explaining the common varieties of Big-O
In all cases, prefer algorithms higher up on the list to those lower on the list. However, the cost of moving to a more expensive complexity class varies significantly.
O(1):
No growth. Regardless of how big as the problem is, you can solve it in the same amount of time. This is somewhat analogous to broadcasting where it takes the same amount of energy to broadcast over a given distance, regardless of the number of people that lie within the broadcast range.
O(log n):
This complexity is the same as O(1) except that it's just a little bit worse. For all practical purposes, you can consider this as a very large constant scaling. The difference in work between processing 1 thousand and 1 billion items is only a factor six.
O(n):
The cost of solving the problem is proportional to the size of the problem. If your problem doubles in size, then the cost of the solution doubles. Since most problems have to be scanned into the computer in some way, as data entry, disk reads, or network traffic, this is generally an affordable scaling factor.
O(n log n):
This complexity is very similar to O(n). For all practical purposes, the two are equivalent. This level of complexity would generally still be considered scalable. By tweaking assumptions some O(n log n) algorithms can be transformed into O(n) algorithms. For example, bounding the size of keys reduces sorting from O(n log n) to O(n).
O(n2):
Grows as a square, where n is the length of the side of a square. This is the same growth rate as the "network effect", where everyone in a network might know everyone else in the network. Growth is expensive. Most scalable solutions cannot use algorithms with this level of complexity without doing significant gymnastics. This generally applies to all other polynomial complexities - O(nk) - as well.
O(2n):
Does not scale. You have no hope of solving any non-trivially sized problem. Useful for knowing what to avoid, and for experts to find approximate algorithms which are in O(nk).
Big O is a measure of how much time/space an algorithm uses relative to the size of its input.
If an algorithm is O(n) then the time/space will increase at the same rate as its input.
If an algorithm is O(n2) then the time/space increase at the rate of its input squared.
and so on.
It is very difficult to measure the speed of software programs, and when we try, the answers can be very complex and filled with exceptions and special cases. This is a big problem, because all those exceptions and special cases are distracting and unhelpful when we want to compare two different programs with one another to find out which is "fastest".
As a result of all this unhelpful complexity, people try to describe the speed of software programs using the smallest and least complex (mathematical) expressions possible. These expressions are very very crude approximations: Although, with a bit of luck, they will capture the "essence" of whether a piece of software is fast or slow.
Because they are approximations, we use the letter "O" (Big Oh) in the expression, as a convention to signal to the reader that we are making a gross oversimplification. (And to make sure that nobody mistakenly thinks that the expression is in any way accurate).
If you read the "Oh" as meaning "on the order of" or "approximately" you will not go too far wrong. (I think the choice of the Big-Oh might have been an attempt at humour).
The only thing that these "Big-Oh" expressions try to do is to describe how much the software slows down as we increase the amount of data that the software has to process. If we double the amount of data that needs to be processed, does the software need twice as long to finish it's work? Ten times as long? In practice, there are a very limited number of big-Oh expressions that you will encounter and need to worry about:
The good:
O(1) Constant: The program takes the same time to run no matter how big the input is.
O(log n) Logarithmic: The program run-time increases only slowly, even with big increases in the size of the input.
The bad:
O(n) Linear: The program run-time increases proportionally to the size of the input.
O(n^k) Polynomial: - Processing time grows faster and faster - as a polynomial function - as the size of the input increases.
... and the ugly:
O(k^n) Exponential The program run-time increases very quickly with even moderate increases in the size of the problem - it is only practical to process small data sets with exponential algorithms.
O(n!) Factorial The program run-time will be longer than you can afford to wait for anything but the very smallest and most trivial-seeming datasets.
What is a plain English explanation of Big O? With as little formal definition as possible and simple mathematics.
A Plain English Explanation of the Need for Big-O Notation:
When we program, we are trying to solve a problem. What we code is called an algorithm. Big O notation allows us to compare the worse case performance of our algorithms in a standardized way. Hardware specs vary over time and improvements in hardware can reduce the time it takes an algorithms to run. But replacing the hardware does not mean our algorithm is any better or improved over time, as our algorithm is still the same. So in order to allow us to compare different algorithms, to determine if one is better or not, we use Big O notation.
A Plain English Explanation of What Big O Notation is:
Not all algorithms run in the same amount of time, and can vary based on the number of items in the input, which we'll call n. Based on this, we consider the worse case analysis, or an upper-bound of the run-time as n get larger and larger. We must be aware of what n is, because many of the Big O notations reference it.
Ok, my 2cents.
Big-O, is rate of increase of resource consumed by program, w.r.t. problem-instance-size
Resource : Could be total-CPU time, could be maximum RAM space. By default refers to CPU time.
Say the problem is "Find the sum",
int Sum(int*arr,int size){
int sum=0;
while(size-->0)
sum+=arr[size];
return sum;
}
problem-instance= {5,10,15} ==> problem-instance-size = 3, iterations-in-loop= 3
problem-instance= {5,10,15,20,25} ==> problem-instance-size = 5 iterations-in-loop = 5
For input of size "n" the program is growing at speed of "n" iterations in array. Hence Big-O is N expressed as O(n)
Say the problem is "Find the Combination",
void Combination(int*arr,int size)
{ int outer=size,inner=size;
while(outer -->0) {
inner=size;
while(inner -->0)
cout<<arr[outer]<<"-"<<arr[inner]<<endl;
}
}
problem-instance= {5,10,15} ==> problem-instance-size = 3, total-iterations = 3*3 = 9
problem-instance= {5,10,15,20,25} ==> problem-instance-size = 5, total-iterations= 5*5 =25
For input of size "n" the program is growing at speed of "n*n" iterations in array. Hence Big-O is N2 expressed as O(n2)
A simple straightforward answer can be:
Big O represents the worst possible time/space for that algorithm. The algorithm will never take more space/time above that limit. Big O represents time/space complexity in the extreme case.
Big O notation is a way of describing the upper bound of an algorithm in terms of space or running time. The n is the number of elements in the the problem (i.e size of an array, number of nodes in a tree, etc.) We are interested in describing the running time as n gets big.
When we say some algorithm is O(f(n)) we are saying that the running time (or space required) by that algorithm is always lower than some constant times f(n).
To say that binary search has a running time of O(logn) is to say that there exists some constant c which you can multiply log(n) by that will always be larger than the running time of binary search. In this case you will always have some constant factor of log(n) comparisons.
In other words where g(n) is the running time of your algorithm, we say that g(n) = O(f(n)) when g(n) <= c*f(n) when n > k, where c and k are some constants.
"What is a plain English explanation of Big O? With as little formal
definition as possible and simple mathematics."
Such a beautifully simple and short question seems at least to deserve an equally short answer, like a student might receive during tutoring.
Big O notation simply tells how much time* an algorithm can run within,
in terms of only the amount of input data**.
( *in a wonderful, unit-free sense of time!)
(**which is what matters, because people will always want more, whether they live today or tomorrow)
Well, what's so wonderful about Big O notation if that's what it does?
Practically speaking, Big O analysis is so useful and important because Big O puts the focus squarely on the algorithm's own complexity and completely ignores anything that is merely a proportionality constant—like a JavaScript engine, the speed of a CPU, your Internet connection, and all those things which become quickly become as laughably outdated as a Model T. Big O focuses on performance only in the way that matters equally as much to people living in the present or in the future.
Big O notation also shines a spotlight directly on the most important principle of computer programming/engineering, the fact which inspires all good programmers to keep thinking and dreaming: the only way to achieve results beyond the slow forward march of technology is to invent a better algorithm.
Algorithm example (Java):
public boolean search(/* for */Integer K,/* in */List</* of */Integer> L)
{
for(/* each */Integer i:/* in */L)
{
if(i == K)
{
return true;
}
}
return false;
}
Algorithm description:
This algorithm searches a list, item by item, looking for a key,
Iterating on each item in the list, if it's the key then return True,
If the loop has finished without finding the key, return False.
Big-O notation represents the upper-bound on the Complexity (Time, Space, ..)
To find The Big-O on Time Complexity:
Calculate how much time (regarding input size) the worst case takes:
Worst-Case: the key doesn't exist in the list.
Time(Worst-Case) = 4n+1
Time: O(4n+1) = O(n) | in Big-O, constants are neglected
O(n) ~ Linear
There's also Big-Omega, which represent the complexity of the Best-Case:
Best-Case: the key is the first item.
Time(Best-Case) = 4
Time: Ω(4) = O(1) ~ Instant\Constant
Big O notation is a way of describing how quickly an algorithm will run given an arbitrary number of input parameters, which we'll call "n". It is useful in computer science because different machines operate at different speeds, and simply saying that an algorithm takes 5 seconds doesn't tell you much because while you may be running a system with a 4.5 Ghz octo-core processor, I may be running a 15 year old, 800 Mhz system, which could take longer regardless of the algorithm. So instead of specifying how fast an algorithm runs in terms of time, we say how fast it runs in terms of number of input parameters, or "n". By describing algorithms in this way, we are able to compare the speeds of algorithms without having to take into account the speed of the computer itself.
Big O
f(x) = O(g(x)) when x goes to a (for example, a = +∞) means that there is a function k such that:
f(x) = k(x)g(x)
k is bounded in some neighborhood of a (if a = +∞, this means that there are numbers N and M such that for every x > N, |k(x)| < M).
In other words, in plain English: f(x) = O(g(x)), x → a, means that in a neighborhood of a, f decomposes into the product of g and some bounded function.
Small o
By the way, here is for comparison the definition of small o.
f(x) = o(g(x)) when x goes to a means that there is a function k such that:
f(x) = k(x)g(x)
k(x) goes to 0 when x goes to a.
Examples
sin x = O(x) when x → 0.
sin x = O(1) when x → +∞,
x2 + x = O(x) when x → 0,
x2 + x = O(x2) when x → +∞,
ln(x) = o(x) = O(x) when x → +∞.
Attention! The notation with the equal sign "=" uses a "fake equality": it is true that o(g(x)) = O(g(x)), but false that O(g(x)) = o(g(x)). Similarly, it is ok to write "ln(x) = o(x) when x → +∞", but the formula "o(x) = ln(x)" would make no sense.
More examples
O(1) = O(n) = O(n2) when n → +∞ (but not the other way around, the equality is "fake"),
O(n) + O(n2) = O(n2) when n → +∞
O(O(n2)) = O(n2) when n → +∞
O(n2)O(n3) = O(n5) when n → +∞
Here is the Wikipedia article: https://en.wikipedia.org/wiki/Big_O_notation
You want to know all there is to know of big O? So do I.
So to talk of big O, I will use words that have just one beat in them. One sound per word. Small words are quick. You know these words, and so do I. We will use words with one sound. They are small. I am sure you will know all of the words we will use!
Now, let’s you and me talk of work. Most of the time, I do not like work. Do you like work? It may be the case that you do, but I am sure I do not.
I do not like to go to work. I do not like to spend time at work. If I had my way, I would like just to play, and do fun things. Do you feel the same as I do?
Now at times, I do have to go to work. It is sad, but true. So, when I am at work, I have a rule: I try to do less work. As near to no work as I can. Then I go play!
So here is the big news: the big O can help me not to do work! I can play more of the time, if I know big O. Less work, more play! That is what big O helps me do.
Now I have some work. I have this list: one, two, three, four, five, six. I must add all things in this list.
Wow, I hate work. But oh well, I have to do this. So here I go.
One plus two is three… plus three is six... and four is... I don’t know. I got lost. It is too hard for me to do in my head. I don’t much care for this kind of work.
So let's not do the work. Let's you and me just think how hard it is. How much work would I have to do, to add six numbers?
Well, let’s see. I must add one and two, and then add that to three, and then add that to four… All in all, I count six adds. I have to do six adds to solve this.
Here comes big O, to tell us just how hard this math is.
Big O says: we must do six adds to solve this. One add, for each thing from one to six. Six small bits of work... each bit of work is one add.
Well, I will not do the work to add them now. But I know how hard it would be. It would be six adds.
Oh no, now I have more work. Sheesh. Who makes this kind of stuff?!
Now they ask me to add from one to ten! Why would I do that? I did not want to add one to six. To add from one to ten… well… that would be even more hard!
How much more hard would it be? How much more work would I have to do? Do I need more or less steps?
Well, I guess I would have to do ten adds… one for each thing from one to ten. Ten is more than six. I would have to work that much more to add from one to ten, than one to six!
I do not want to add right now. I just want to think on how hard it might be to add that much. And, I hope, to play as soon as I can.
To add from one to six, that is some work. But do you see, to add from one to ten, that is more work?
Big O is your friend and mine. Big O helps us think on how much work we have to do, so we can plan. And, if we are friends with big O, he can help us choose work that is not so hard!
Now we must do new work. Oh, no. I don’t like this work thing at all.
The new work is: add all things from one to n.
Wait! What is n? Did I miss that? How can I add from one to n if you don’t tell me what n is?
Well, I don’t know what n is. I was not told. Were you? No? Oh well. So we can’t do the work. Whew.
But though we will not do the work now, we can guess how hard it would be, if we knew n. We would have to add up n things, right? Of course!
Now here comes big O, and he will tell us how hard this work is. He says: to add all things from one to N, one by one, is O(n). To add all these things, [I know I must add n times.][1] That is big O! He tells us how hard it is to do some type of work.
To me, I think of big O like a big, slow, boss man. He thinks on work, but he does not do it. He might say, "That work is quick." Or, he might say, "That work is so slow and hard!" But he does not do the work. He just looks at the work, and then he tells us how much time it might take.
I care lots for big O. Why? I do not like to work! No one likes to work. That is why we all love big O! He tells us how fast we can work. He helps us think of how hard work is.
Uh oh, more work. Now, let’s not do the work. But, let’s make a plan to do it, step by step.
They gave us a deck of ten cards. They are all mixed up: seven, four, two, six… not straight at all. And now... our job is to sort them.
Ergh. That sounds like a lot of work!
How can we sort this deck? I have a plan.
I will look at each pair of cards, pair by pair, through the deck, from first to last. If the first card in one pair is big and the next card in that pair is small, I swap them. Else, I go to the next pair, and so on and so on... and soon, the deck is done.
When the deck is done, I ask: did I swap cards in that pass? If so, I must do it all once more, from the top.
At some point, at some time, there will be no swaps, and our sort of the deck would be done. So much work!
Well, how much work would that be, to sort the cards with those rules?
I have ten cards. And, most of the time -- that is, if I don’t have lots of luck -- I must go through the whole deck up to ten times, with up to ten card swaps each time through the deck.
Big O, help me!
Big O comes in and says: for a deck of n cards, to sort it this way will be done in O(N squared) time.
Why does he say n squared?
Well, you know n squared is n times n. Now, I get it: n cards checked, up to what might be n times through the deck. That is two loops, each with n steps. That is n squared much work to be done. A lot of work, for sure!
Now when big O says it will take O(n squared) work, he does not mean n squared adds, on the nose. It might be some small bit less, for some case. But in the worst case, it will be near n squared steps of work to sort the deck.
Now here is where big O is our friend.
Big O points out this: as n gets big, when we sort cards, the job gets MUCH MUCH MORE HARD than the old just-add-these-things job. How do we know this?
Well, if n gets real big, we do not care what we might add to n or n squared.
For big n, n squared is more large than n.
Big O tells us that to sort things is more hard than to add things. O(n squared) is more than O(n) for big n. That means: if n gets real big, to sort a mixed deck of n things MUST take more time, than to just add n mixed things.
Big O does not solve the work for us. Big O tells us how hard the work is.
I have a deck of cards. I did sort them. You helped. Thanks.
Is there a more fast way to sort the cards? Can big O help us?
Yes, there is a more fast way! It takes some time to learn, but it works... and it works quite fast. You can try it too, but take your time with each step and do not lose your place.
In this new way to sort a deck, we do not check pairs of cards the way we did a while ago. Here are your new rules to sort this deck:
One: I choose one card in the part of the deck we work on now. You can choose one for me if you like. (The first time we do this, “the part of the deck we work on now” is the whole deck, of course.)
Two: I splay the deck on that card you chose. What is this splay; how do I splay? Well, I go from the start card down, one by one, and I look for a card that is more high than the splay card.
Three: I go from the end card up, and I look for a card that is more low than the splay card.
Once I have found these two cards, I swap them, and go on to look for more cards to swap. That is, I go back to step Two, and splay on the card you chose some more.
At some point, this loop (from Two to Three) will end. It ends when both halves of this search meet at the splay card. Then, we have just splayed the deck with the card you chose in step One. Now, all the cards near the start are more low than the splay card; and the cards near the end are more high than the splay card. Cool trick!
Four (and this is the fun part): I have two small decks now, one more low than the splay card, and one more high. Now I go to step one, on each small deck! That is to say, I start from step One on the first small deck, and when that work is done, I start from step One on the next small deck.
I break up the deck in parts, and sort each part, more small and more small, and at some time I have no more work to do. Now this may seem slow, with all the rules. But trust me, it is not slow at all. It is much less work than the first way to sort things!
What is this sort called? It is called Quick Sort! That sort was made by a man called C. A. R. Hoare and he called it Quick Sort. Now, Quick Sort gets used all the time!
Quick Sort breaks up big decks in small ones. That is to say, it breaks up big tasks in small ones.
Hmmm. There may be a rule in there, I think. To make big tasks small, break them up.
This sort is quite quick. How quick? Big O tells us: this sort needs O(n log n) work to be done, in the mean case.
Is it more or less fast than the first sort? Big O, please help!
The first sort was O(n squared). But Quick Sort is O(n log n). You know that n log n is less than n squared, for big n, right? Well, that is how we know that Quick Sort is fast!
If you have to sort a deck, what is the best way? Well, you can do what you want, but I would choose Quick Sort.
Why do I choose Quick Sort? I do not like to work, of course! I want work done as soon as I can get it done.
How do I know Quick Sort is less work? I know that O(n log n) is less than O(n squared). The O's are more small, so Quick Sort is less work!
Now you know my friend, Big O. He helps us do less work. And if you know big O, you can do less work too!
You learned all that with me! You are so smart! Thank you so much!
Now that work is done, let’s go play!
[1]: There is a way to cheat and add all the things from one to n, all at one time. Some kid named Gauss found this out when he was eight. I am not that smart though, so don't ask me how he did it.
Not sure I'm further contributing to the subject but still thought I'd share: I once found this blog post to have some quite helpful (though very basic) explanations & examples on Big O:
Via examples, this helped get the bare basics into my tortoiseshell-like skull, so I think it's a pretty descent 10-minute read to get you headed in the right direction.
I've more simpler way to understand the time complexity
he most common metric for calculating time complexity is Big O notation. This removes all constant factors so that the running time can be estimated in relation to N as N approaches infinity. In general you can think of it like this:
statement;
Is constant. The running time of the statement will not change in relation to N
for ( i = 0; i < N; i++ )
statement;
Is linear. The running time of the loop is directly proportional to N. When N doubles, so does the running time.
for ( i = 0; i < N; i++ )
{
for ( j = 0; j < N; j++ )
statement;
}
Is quadratic. The running time of the two loops is proportional to the square of N. When N doubles, the running time increases by N * N.
while ( low <= high )
{
mid = ( low + high ) / 2;
if ( target < list[mid] )
high = mid - 1;
else if ( target > list[mid] )
low = mid + 1;
else break;
}
Is logarithmic. The running time of the algorithm is proportional to the number of times N can be divided by 2. This is because the algorithm divides the working area in half with each iteration.
void quicksort ( int list[], int left, int right )
{
int pivot = partition ( list, left, right );
quicksort ( list, left, pivot - 1 );
quicksort ( list, pivot + 1, right );
}
Is N * log ( N ). The running time consists of N loops (iterative or recursive) that are logarithmic, thus the algorithm is a combination of linear and logarithmic.
In general, doing something with every item in one dimension is linear, doing something with every item in two dimensions is quadratic, and dividing the working area in half is logarithmic. There are other Big O measures such as cubic, exponential, and square root, but they're not nearly as common. Big O notation is described as O ( ) where is the measure. The quicksort algorithm would be described as O ( N * log ( N ) ).
Note: None of this has taken into account best, average, and worst case measures. Each would have its own Big O notation. Also note that this is a VERY simplistic explanation. Big O is the most common, but it's also more complex that I've shown. There are also other notations such as big omega, little o, and big theta. You probably won't encounter them outside of an algorithm analysis course.
See more at: Here
What is a plain English explanation of “Big O” notation?
Very Quick Note:
The O in "Big O" refers to as "Order"(or precisely "order of")
so you could get its idea literally that it's used to order something to compare them.
"Big O" does two things:
Estimates how many steps of the method your computer applies to accomplish a task.
Facilitate the process to compare with others in order to determine whether it's good or not?
"Big O' achieves the above two with standardized Notations.
There are seven most used notations
O(1), means your computer gets a task done with 1 step, it's excellent, Ordered No.1
O(logN), means your computer complete a task with logN steps, its good, Ordered No.2
O(N), finish a task with N steps, its fair, Order No.3
O(NlogN), ends a task with O(NlogN) steps, it's not good, Order No.4
O(N^2), get a task done with N^2 steps, it's bad, Order No.5
O(2^N), get a task done with 2^N steps, it's horrible, Order No.6
O(N!), get a task done with N! steps, it's terrible, Order No.7
Suppose you get notation O(N^2), not only you are clear the method takes N*N steps to accomplish a task, also you see that it's not good as O(NlogN) from its ranking.
Please note the order at line end, just for your better understanding.There's more than 7 notations if all possibilities considered.
In CS, the set of steps to accomplish a task is called algorithms.
In Terminology, Big O notation is used to describe the performance or complexity of an algorithm.
In addition, Big O establishes the worst-case or measure the Upper-Bound steps.
You could refer to Big-Ω (Big-Omega) for best case.
Big-Ω (Big-Omega) notation (article) | Khan Academy
Summary
"Big O" describes the algorithm's performance and evaluates it.
or address it formally, "Big O" classifies the algorithms and standardize the comparison process.
Assume we're talking about an algorithm A, which should do something with a dataset of size n.
Then O( <some expression X involving n> ) means, in simple English:
If you're unlucky when executing A, it might take as much as X(n) operations to
complete.
As it happens, there are certain functions (think of them as implementations of X(n)) that tend to occur quite often. These are well known and easily compared (Examples: 1, Log N, N, N^2, N!, etc..)
By comparing these when talking about A and other algorithms, it is easy to rank the algorithms according to the number of operations they may (worst-case) require to complete.
In general, our goal will be to find or structure an algorithm A in such a way that it will have a function X(n) that returns as low a number as possible.
If you have a suitable notion of infinity in your head, then there is a very brief description:
Big O notation tells you the cost of solving an infinitely large problem.
And furthermore
Constant factors are negligible
If you upgrade to a computer that can run your algorithm twice as fast, big O notation won't notice that. Constant factor improvements are too small to even be noticed in the scale that big O notation works with. Note that this is an intentional part of the design of big O notation.
Although anything "larger" than a constant factor can be detected, however.
When interested in doing computations whose size is "large" enough to be considered as approximately infinity, then big O notation is approximately the cost of solving your problem.
If the above doesn't make sense, then you don't have a compatible intuitive notion of infinity in your head, and you should probably disregard all of the above; the only way I know to make these ideas rigorous, or to explain them if they aren't already intuitively useful, is to first teach you big O notation or something similar. (although, once you well understand big O notation in the future, it may be worthwhile to revisit these ideas)
Say you order Harry Potter: Complete 8-Film Collection [Blu-ray] from Amazon and download the same film collection online at the same time. You want to test which method is faster. The delivery takes almost a day to arrive and the download completed about 30 minutes earlier. Great! So it’s a tight race.
What if I order several Blu-ray movies like The Lord of the Rings, Twilight, The Dark Knight Trilogy, etc. and download all the movies online at the same time? This time, the delivery still take a day to complete, but the online download takes 3 days to finish.
For online shopping, the number of purchased item (input) doesn’t affect the delivery time. The output is constant. We call this O(1).
For online downloading, the download time is directly proportional to the movie file sizes (input). We call this O(n).
From the experiments, we know that online shopping scales better than online downloading. It is very important to understand big O notation because it helps you to analyze the scalability and efficiency of algorithms.
Note: Big O notation represents the worst-case scenario of an algorithm. Let’s assume that O(1) and O(n) are the worst-case scenarios of the example above.
Reference : http://carlcheo.com/compsci
Definition :- Big O notation is a notation which says how a algorithm performance will perform if the data input increases.
When we talk about algorithms there are 3 important pillars Input , Output and Processing of algorithm. Big O is symbolic notation which says if the data input is increased in what rate will the performance vary of the algorithm processing.
I would encourage you to see this youtube video which explains Big O Notation in depth with code examples.
So for example assume that a algorithm takes 5 records and the time required for processing the same is 27 seconds. Now if we increase the records to 10 the algorithm takes 105 seconds.
In simple words the time taken is square of the number of records. We can denote this by O(n ^ 2). This symbolic representation is termed as Big O notation.
Now please note the units can be anything in inputs it can be bytes , bits number of records , the performance can be measured in any unit like second , minutes , days and so on. So its not the exact unit but rather the relationship.
For example look at the below function "Function1" which takes a collection and does processing on the first record. Now for this function the performance will be same irrespective you put 1000 , 10000 or 100000 records. So we can denote it by O(1).
void Function1(List<string> data)
{
string str = data[0];
}
Now see the below function "Function2()". In this case the processing time will increase with number of records. We can denote this algorithm performance using O(n).
void Function2(List<string> data)
{
foreach(string str in data)
{
if (str == "shiv")
{
return;
}
}
}
When we see a Big O notation for any algorithm we can classify them in to three categories of performance :-
Log and constant category :- Any developer would love to see their algorithm performance in this category.
Linear :- Developer will not want to see algorithms in this category , until its the last option or the only option left.
Exponential :- This is where we do not want to see our algorithms and a rework is needed.
So by looking at Big O notation we categorize good and bad zones for algorithms.
I would recommend you to watch this 10 minutes video which discusses Big O with sample code
https://www.youtube.com/watch?v=k6kxtzICG_g
Simplest way to look at it (in plain English)
We are trying to see how the number of input parameters, affects the running time of an algorithm. If the running time of your application is proportional to the number of input parameters, then it is said to be in Big O of n.
The above statement is a good start but not completely true.
A more accurate explanation (mathematical)
Suppose
n=number of input parameters
T(n)= The actual function that expresses the running time of the algorithm as a function of n
c= a constant
f(n)= An approximate function that expresses the running time of the algorithm as a function of n
Then as far as Big O is concerned, the approximation f(n) is considered good enough as long as the below condition is true.
lim T(n) ≤ c×f(n)
n→∞
The equation is read as
As n approaches infinity, T of n, is less than or equal to c times f of n.
In big O notation this is written as
T(n)∈O(n)
This is read as T of n is in big O of n.
Back to English
Based on the mathematical definition above, if you say your algorithm is a Big O of n, it means it is a function of n (number of input parameters) or faster. If your algorithm is Big O of n, then it is also automatically the Big O of n square.
Big O of n means my algorithm runs at least as fast as this. You cannot look at Big O notation of your algorithm and say its slow. You can only say its fast.
Check this out for a video tutorial on Big O from UC Berkley. It is actually a simple concept. If you hear professor Shewchuck (aka God level teacher) explaining it, you will say "Oh that's all it is!".
Preface
algorithm: procedure/formula for solving a problem
How do analyze algorithms and how can we compare algorithms against each other?
example: you and a friend are asked to create a function to sum the numbers from 0 to N. You come up with f(x) and your friend comes up with g(x). Both functions have the same result, but a different algorithm. In order to objectively compare the efficiency of the algorithms we use Big-O notation.
Big-O notation: describes how quickly runtime will grow relative to the input as the input get arbitrarily large.
3 key takeaways:
Compare how quickly runtime grows NOT compare exact runtimes (depends on hardware)
Only concerned with runtime grow relative to the input (n)
As n gets arbitrarily large, focus on the terms that will grow the fastest as n gets large (think infinity) AKA asymptotic analysis
Space complexity: aside from time complexity, we also care about space complexity (how much memory/space an algorithm uses). Instead of checking the time of operations, we check the size of the allocation of memory.
I found a really great explanation about big O notation especially for a someone who's not much into mathematics.
https://rob-bell.net/2009/06/a-beginners-guide-to-big-o-notation/
Big O notation is used in Computer Science to describe the performance
or complexity of an algorithm. Big O specifically describes the
worst-case scenario, and can be used to describe the execution time
required or the space used (e.g. in memory or on disk) by an
algorithm.
Anyone who's read Programming Pearls or any other Computer Science
books and doesn’t have a grounding in Mathematics will have hit a wall
when they reached chapters that mention O(N log N) or other seemingly
crazy syntax. Hopefully this article will help you gain an
understanding of the basics of Big O and Logarithms.
As a programmer first and a mathematician second (or maybe third or
fourth) I found the best way to understand Big O thoroughly was to
produce some examples in code. So, below are some common orders of
growth along with descriptions and examples where possible.
O(1)
O(1) describes an algorithm that will always execute in the same time
(or space) regardless of the size of the input data set.
bool IsFirstElementNull(IList<string> elements) {
return elements[0] == null;
}
O(N)
O(N) describes an algorithm whose performance will grow linearly and
in direct proportion to the size of the input data set. The example
below also demonstrates how Big O favours the worst-case performance
scenario; a matching string could be found during any iteration of the
for loop and the function would return early, but Big O notation will
always assume the upper limit where the algorithm will perform the
maximum number of iterations.
bool ContainsValue(IList<string> elements, string value) {
foreach (var element in elements)
{
if (element == value) return true;
}
return false;
}
O(N2)
O(N2) represents an algorithm whose performance is directly
proportional to the square of the size of the input data set. This is
common with algorithms that involve nested iterations over the data
set. Deeper nested iterations will result in O(N3), O(N4) etc.
bool ContainsDuplicates(IList<string> elements) {
for (var outer = 0; outer < elements.Count; outer++)
{
for (var inner = 0; inner < elements.Count; inner++)
{
// Don't compare with self
if (outer == inner) continue;
if (elements[outer] == elements[inner]) return true;
}
}
return false;
}
O(2N)
O(2N) denotes an algorithm whose growth doubles with each additon to
the input data set. The growth curve of an O(2N) function is
exponential - starting off very shallow, then rising meteorically. An
example of an O(2N) function is the recursive calculation of Fibonacci
numbers:
int Fibonacci(int number) {
if (number <= 1) return number;
return Fibonacci(number - 2) + Fibonacci(number - 1);
}
Logarithms
Logarithms are slightly trickier to explain so I'll use a common
example:
Binary search is a technique used to search sorted data sets. It works
by selecting the middle element of the data set, essentially the
median, and compares it against a target value. If the values match it
will return success. If the target value is higher than the value of
the probe element it will take the upper half of the data set and
perform the same operation against it. Likewise, if the target value
is lower than the value of the probe element it will perform the
operation against the lower half. It will continue to halve the data
set with each iteration until the value has been found or until it can
no longer split the data set.
This type of algorithm is described as O(log N). The iterative halving
of data sets described in the binary search example produces a growth
curve that peaks at the beginning and slowly flattens out as the size
of the data sets increase e.g. an input data set containing 10 items
takes one second to complete, a data set containing 100 items takes
two seconds, and a data set containing 1000 items will take three
seconds. Doubling the size of the input data set has little effect on
its growth as after a single iteration of the algorithm the data set
will be halved and therefore on a par with an input data set half the
size. This makes algorithms like binary search extremely efficient
when dealing with large data sets.

How to know if method is N or N^2

I often see you guys talking about N methods and N^2 methods, which, correct me if I'm wrong, indicate how fast a method is. My question is: how do you guys know which methods are N and which are N^2? And also: are there other speed indications of methods then just N and N^2?
This talks abnout the complexity of an algorithm (which is an indicator of how fast it will be, yes)
In short, it tells how many "operations" (with operations being a very vague and abstract term) will be needed for a input to the method of size "N".
e.g. if your input is an List-type object, and you must iterate over all items in the list, the complexity is "N". (often expressed O(N) ).
if your input is an list-type object, and you need only to look at the first (or last), and the list gurantees to you that such a look at the item is O(1); your method will be O(1) - independent from the input size.
If your input is a list, and you need to compare every item to every other item the complexity will be O(N²) or O(N*log(n))
correct me if I'm wrong, indicate how fast a method is.
Its says how an algorithm will scale on an ideal machine. It deliberately ignores the factor involved which can mean that an O(1) could be slower than an O(N) which could be slower than an O(N^2) for your use-case. e.g. Arrays.sort() will use insertion sort O(N^2) for small collections (length < 47 in Java 7) in preference to quick sort O(N ln N)
In general, using lower order algorithms are a safer choice because they are less likely to break in extreme cases which you may not get a chance to test thoroughly.
The way to guesstimate the big-O complexity of a program is based on experience with dry-running code (running it in your mind). Some cases are dead obvious, but the most interesting ones aren't: for example, calling library methods known to be O(n) in an O(n) loop results in O(n2) total complexity; writing to a TreeMap in an O(n) loop results in O(nlogn) total, and so on.

Time Complexity of an algorithm : How to decide which algorithm after calculated the time

Today i'm come across with the blog in msdn and i noticed how to calculate the time complexity of an algorithm. I perfectly understand how to calculate the time complexity of an algorithm but in the last the author mentioned the below lines
Adding everything up I get
(N+4)+(5N+2)+(4N+2) = 10N+8
So the asymptotic time complexity for the above code is O(N), which
means that the above algorithm is a liner time complexity algorithm.
So how come the author is said it's based on the liner time complexity algorithm. The link for the blog
http://blogs.msdn.com/b/nmallick/archive/2010/03/30/how-to-calculate-time-complexity-for-a-given-algorithm.aspx.
He said that because 10N + 8 is a linear equation. If you plot that equation you get a straight line. Try typing 10 * x + 8 on this website (function graphs) and see for yourself.
Ascending order of time complexities(the common one)
O(1) - Constant
O(log n) - logarithmic
O(n) - linear
O(n log n) - loglinear
O(n^2) - quadratic
Note: N increases without bounds
For complexity theory you definitely should read some background theory. It's usually about asymptotic complexity, which is why you can drop the smaller parts, and only keep the complexity class.
The key idea is that the difference between N and N+5 becomes neglibile once N is really big.
For more details, start reading here:
http://en.wikipedia.org/wiki/Big_O_notation
The author just based on his experience in choosing most appropriate . You should know, that counting algorithm complexity almost always means to find a Big-Oh function, which, in turn, is just an upper bound to given function (10N+8 in your case).
There is only a few well-known complexity types: linear complexity, quadratic complexity, etc. So, the final step of counting a time complexity consists of choosing what is the less complex type (i mean, linear is less complex than a quadratic, and quadratic is less complex that exponential, and so on) can be used for the given function, which correctly describe its complexity.
In your case, O(n) and O(n^2) and even O(2^n) are the right answers indeed. But the less complex function, which suits perfectly in Big-Oh notation definition is O(n), which is an answer here.
Here is a real good article, fully explained Big-Oh notation.
A very pragmatic rule is:
when the complexity of an algorithm si represented by a poly like A*n^2+B*n+C then the order of complexity ( that is to say the O(something) ) is equal to the highest order of the variable n.
In the A*n^2+B*n+C poly the order is O(n^2).
Like josnidhin explained, if the poly has
order 1 (i.e. n)- it is called linear
order 2 (i.e. n^2) - it is called quadratic
... and so on.

How can I prove that one algorithm is faster than another in Java

Is there anything in Java that would allow me to take a code snippit and allow me to see exactly how many "ticks" it takes to execute. I want to prove that an algorithm I wrote is faster than another.
"Ticks"? No. I'd recommend that you run them several times each and compare the average results:
public class AlgorithmDriver {
public static void main(String [] args) {
int numTries = 1000000;
long begTime = System.currentTimeMillis();
for (int i = 0; i < numTries; ++i) {
Algorithm.someMethodCall();
}
long endTime = System.currentTimeMillis();
System.out.printf("Total time for %10d tries: %d ms\n", numTries, (endTime-begTime));
}
}
You probably are asking two different questions:
How can you measure the run time of a java implementation (Benchmarking)
How can you prove the asymptotic run time of an algorithm
For the first of these I wouldn't use the solutions posted here. They are mostly not quite right. Forst, its probably better to use System.nanoTime than System.currentTimeMillis. Second, you need to use a try catch block. Third, take statistic of running times of your code running many times outside of the metric, so that you can have a more complete picture.
Run code that looks vaguely like this many times:
long totalTime = 0;
long startTime = System.nanoTime();
try{
//method to test
} finally {
totalTime = System.nanoTime() - startTime;
}
Getting benchmarking correct is hard. For example, you must let your code "warmup"" for a few minutes before testing it. Benchmark early and often, but dont over believe your benchmarks. Particularly small micro benchmarks almost always lie in one way or another.
The second way to interpret your question is about asymptotic run times. The truth is this has almost nothing to do with Java, it is general computer science. Here the question we want to ask is: what curves describe the behavior of the run time of our algorithm in terms of the input size.
The first thing is to understand Big-Oh notation. I'll do my best, but SO doesn't support math notation. O(f(n)) denotes a set of algorithms such that in the limit as n goes to infinity f(n) is within a constant factor of an upper bound on the algorithm run time. Formally, T(n) is in O(f(n)) iff there exists some constant n0 and some constant c such that for all n > n0 c*f(n) >= n. Big Omega is the same thing, except for upper bounds, and big Theta f(n) just means its both big Oh f(n) and big Omega f(n). This is not two hard.
Well, it gets a little more complicated because we can talk about different kinds of run time, ie "average case", best case, and worst case. For example, normall quicksort is O(n^2) in the worst case, but O(n log n) for random lists.
So I skipped over what T(n) means. Basically it is the number of "ticks." Some machine instructions (like reading from memory) take much longer than others (like adding). But, so long as they are only a constant factor apart from each other, we can treat them all as the same for the purposes of big Oh, since it will just change the value of c.
Proving asymptotic bounds isn't that hard. For simple structured programming problems you just count
public int square(int n){
int sum = 0
for(int i = 0, i < n, i++){
sum += n
}
return sum
}
In this example we have one instruction each for: initializing sum, initializing i, and returning the value. The loop happens n times and on each time we do a comparison, and addition, and an increment. So we have O(square(n)) = O(3 + 3n) using n0 of 2 and c of 4 we can easily prove this is in O(n). You can always safely simplify big Oh expressions by removing excess constant terms, and by dividing by constant multiples.
When you are faced with a recursive function you have to solve a recurrence relation. If you have a function like T(n) = 2*T(n/2) + O(1) you want to find a closed form solution. You sometimes have to do this by hand, or with a computer algebra system. For this example, using forward substitution, we can see the pattern (in an abuse of notation) T(1) = O(1), T(2) = O(3), T(4) = O(7), T(8) = (15) this looks alot like O(2n - 1), to prove this is the right value:
T(n) = 2*T(n/2) + 1
T(n) = 2*(2(n/2) - 1) + 1
T(n) = 2*(n-1) + 1
T(n) = 2n - 2 + 1
T(n) = 2n - 1
As we saw earlier you can simplify O(2n -1) to O(n)
More often though you can use the master theorem which is a mathematical tool for saving you time on this kind of problem. If you check wikipedia you can find the master theorem, which if you plug and play the example above you get the same answer.
For more, check out an algorithms text book like Levitin's "The Design & Analysis of Algorithms"
You could use System.currentTimeMillis() to get start and end times.
long start = System.currentTimeMillis();
// your code
long end = System.currentTimeMillis();
System.out.println( "time: " + (end - start) );
You can measure wall time with System.currentTimeMillis() or System.nanoTime() (which have different characteristics). This is relatively easy as you just have to print out the differences at the end.
If you need to count specific operations (which is common in algorithms), the easiest is to simply increment a counter when the operations are being done , and then print it when you are done. long is well suited for this. For multiple operations use multiple counters.
I had to do this algorithm efficiency proofs mostly on my Data Structures lesson this year.
First,I measured the time like they mentioned upper.
Then I increased the method's input number with squaring each time(10,100,1000,...)
Lastly,I put the time measurements in an Excel file and drawed graphics for these time values.
By this way,you can check if one algorithm is faster than other or not,slightly.
I would:
Come up with a few data sets for the current algorithm: a set where it performs well, a set where it performs ok, and a data set where it performs poorly. You want to show that your new algorithm outperforms the current one for each scenario.
Run and measure the performance of each algorithm multiple times for increasing input sizes of each of the three data sets, then take average, standard deviation etc. Standard deviation will show a crude measure of the consistency of the algorithm performance.
Finally look at the numbers and decide in your case what is important: which algorithm's performance is more suitable for the type of input you will have most of the time, and how does it degrade when the inputs are not optimal.
Timing the algorithm is not necessarily everything - would memory footprint be important as well? One algorithm might be better computationally but it might create more objects while it runs.. etc. Just trying to point out there is more to consider than purely timing!
I wouldn't use the current time in ms as some of the others have suggested. The methods provided by ThreadMXBeans are more accurate (I dare not say 100% accurate).
They actually measure the cpu time taken by the thread, rather then elapsed system time, which may be skewed due to context switches performed by the underlying OS.
Java Performance Testing
I am not too familiar with the Java Framework but i would do it the following way:
Define a set of test cases (mostly example data) that can be used for both algorithms
Implement a timing method to measure the amount of time that a specific function takes
Create a for loop and execute method A (repeatedly, e.g. 1000 times with the whole test data). Measure the timing of the loop, not the sum of the single functions since timing functions can bias your result when called a lot)
Do the same for method B
Compare your result and choose a winner
If both algorithms have the same definition of a macro-level "tick" (e.g. walking one node in a tree) and your goal is to prove that your algorithm accomplishes its goal in a lower number of those macro-level ticks than the other, then by far the best way is to just instrument each implementation to count those ticks. That approach is ideal because it doesn't reward low-level implementation tricks that can make code execute faster but are not algorithm-related.
If you don't have that luxury, but you are trying to calculate which approach solves the problem using the least amount of CPU resources, contrary to the approaches listed here involving System.currentTimeMillis etc, I would use an external approach: the linux time command would be ideal. You have each program run on the same set of (large) inputs, preferably that take on the order of minutes or hours to process, and just run time java algo1 vs time java algo2.
If your purpose is compare the performances between two pieces of code, the best way to do is using JMH. You can import via maven and is now official in openjdk 12.
https://openjdk.java.net/projects/code-tools/jmh/

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