This is a question popped into my mind while reading the halting problem, collatz conjecture and Kolmogorov complexity. I have tried to search for something similar but I was unable to find a particular topic maybe because it is not of great value or it could just be a trivial question.
For the sake of simplicity I will give three examples of programs/functions.
function one(s):
return s
function two(s):
while (True):
print s
function three(s):
for i from 0 to 10^10:
print(s)
So my questions is, if there is a way to formalize the length of a program (like the bits used to describe it) and also the internal memory used by the program, to determine the minimum/maximum number of time/steps needed to decide whether the program will terminate or run forever.
For example, in the first function the program doesn't alter its internal memory and halts after some time steps.
In the second example, the program runs forever but the program also doesn't alter its internal memory. For example, if we considered all the programs with the same length as with the program two that do not alter their state, couldn't we determine an upper bound of steps, which if surpassed we could conclude that this program will never terminate ? (If not why ?)
On the last example, the program alters its state (variable i). So, at each step the upper bound may change.
[In short]
Kolmogorov complexity suggests a way of finding the (descriptive) complexity of an object such as a piece of text. I would like to know, given a formal way of describing the memory-space used by a program (computed in runtime), if we could compute a maximum number of steps, which if surpassed would allow us to know whether this program will terminate or run forever.
Finally, I would like to suggest me any source that I might find useful and help me figure out what I am exactly looking for.
Thank you. (sorry for my English, not my native language. I hope I was clear)
If a deterministic Turing machine enters precisely the same configuration twice (which we can detect b keeping a trace of configurations seen so far), then we immediately know the TM will loop forever.
If it known in advance that a deterministic Turing machine cannot possibly use more than some fixed constant amount of its input tape, then the TM must explicitly halt or eventually enter some configuration it has already visited. Suppose the TM can use at most k tape cells, the tape alphabet is T and the set of states is Q. Then there are (|T|+1)^k * |Q| unique configurations (the number of strings over (T union blank) of length k times the number of states) and by the pigeonhole principle we know that a TM that takes that many steps must enter some configuration it has already been to before.
one: because we are given that this function does not use internal memory, we know that it either halts or loops forever.
two: because we are given that this function does not use internal memory, we know that it either halts or loops forever.
three: because we are given that this function only uses a fixed amount of internal memory (like 34 bits) we can tell in fewer than 2^34 iterations of the loop whether the TM will halt or not for any given input s, guaranteed.
Now, knowing how much tape a TM is going to use, or how much memory a program is going to use, is not a problem a TM can solve. But if you have an oracle (like a person who was able to do a proof) that tells you a correct fixed upper bound on memory, then the halting problem is solvable.
I do not know if the language I am using in the title is correct, but here is an example that illustrates what I am asking.
What would the time complexity for this non-optimal algorithm that removes character pairs from a string?
The function will loop through a string. When it finds two identical characters next to each other it will return a string without the found pair. It then recursively calls itself until no pair is found.
Example (each line is the return string from one recurisive function call):
iabccba
iabba
iaa
i
Would it be fair to describe the time complexity as O(|Characters| * |Pairs|)?
What about O(|Characters|^2) Can pairs be used to describe the time complexity even though the number of pairs is not knowable at the initial function call?
It was argued to me that this algorithm was O(n^2)because the number of pairs is not known.
You're right that this is strictly speaking O(|Characters| * |Pairs|)
However, in the worst case, number of pairs can be same as number of charachters (or same order of magnitude), for example in the string 'abcdeedcba'
So it also makes sense to describe it as O(n^2) worst-case.
I think this largely depends on the problem you mean to solve and and it's definition.
For graph algorithms for example, everyone is comfortable with a writing as complexity O(|V| + |E|), although in the worst case of a dense graph |E| = |V|^2. In other problems we just look at the worst possible case and write O(n^2), without breaking it into more specific variables.
I'd say that if there's no special convention, or no special data in the problem regarding number of pairs, O(...) implies worst-case performance and hence O(n^2) would be more appropriate.
I'm thinking more about how much system memory my programs will use nowadays. I'm currently doing A level Computing at college and I know that in most programs the difference will be negligible but I'm wondering if the following actually makes any difference, in any language.
Say I wanted to output "True" or "False" depending on whether a condition is true. Personally, I prefer to do something like this:
Dim result As String
If condition Then
Result = "True"
Else
Result = "False"
EndIf
Console.WriteLine(result)
However, I'm wondering if the following would consume less memory, etc.:
If condition Then
Console.WriteLine("True")
Else
Console.WriteLine("False")
EndIf
Obviously this is a very much simplified example and in most of my cases there is much more to be outputted, and I realise that in most commercial programs these kind of statements are rare, but hopefully you get the principle.
I'm focusing on VB.NET here because that is the language used for the course, but really I would be interested to know how this differs in different programming languages.
The main issue making if's fast or slow is predictability.
Modern CPU's (anything after 2000) use a mechanism called branch prediction.
Read the above link first, then read on below...
Which is faster?
The if statement constitutes a branch, because the CPU needs to decide whether to follow or skip the if part.
If it guesses the branch correctly the jump will execute in 0 or 1 cycle (1 nanosecond on a 1Ghz computer).
If it does not guess the branch correctly the jump will take 50 cycles (give or take) (1/200th of a microsecord).
Therefore to even feel these differences as a human, you'd need to execute the if statement many millions of times.
The two statements above are likely to execute in exactly the same amount of time, because:
assigning a value to a variable takes negligible time; on average less than a single cpu cycle on a multiscalar CPU*.
calling a function with a constant parameter requires the use of an invisible temporary variable; so in all likelihood code A compiles to almost the exact same object code as code B.
*) All current CPU's are multiscalar.
Which consumes less memory
As stated above, both versions need to put the boolean into a variable.
Version A uses an explicit one, declared by you; version B uses an implicit one declared by the compiler.
However version A is guaranteed to only have one call to the function WriteLine.
Whilst version B may (or may not) have two calls to the function WriteLine.
If the optimizer in the compiler is good, code B will be transformed into code A, if it's not it will remain with the redundant calls.
How bad is the waste
The call takes about 10 bytes for the assignment of the string (Unicode 2 bytes per char).
But so does the other version, so that's the same.
That leaves 5 bytes for a call. Plus maybe a few extra bytes to set up a stackframe.
So lets say due to your totally horrible coding you have now wasted 10 bytes.
Not much to worry about.
From a maintainability point of view
Computer code is written for humans, not machines.
So from that point of view code A is clearly superior.
Imagine not choosing between 2 options -true or false- but 20.
You only call the function once.
If you decide to change the WriteLine for another function you only have to change it in one place, not two or 20.
How to speed this up?
With 2 values it's pretty much impossible, but if you had 20 values you could use a lookup table.
Obviously that optimization is not worth it unless code gets executed many times.
If you need to know the precise amount of memory the instructions are going to take, you can use ildasm on your code, and see for yourself. However, the amount of memory consumed by your code is much less relevant today, when the memory is so cheap and abundant, and compilers are smart enough to see common patterns and reduce the amount of code that they generate.
A much greater concern is readability of your code: if a complex chain of conditions always leads to printing a conditionally set result, your first code block expresses this idea in a cleaner way than the second one does. Everything else being equal, you should prefer whatever form of code that you find the most readable, and let the compiler worry about optimization.
P.S. It goes without saying that Console.WriteLine(condition) would produce the same result, but that is of course not the point of your question.
I got a theoretical question, will appreciate if you advise me here.
Say, we have these two pieces of code.
First one:
For Each cell In rng1
collectionOfValues.Add (cell.Value)
Next
For Each cell In rng2
collectionOfAddresses.Add (cell.Address)
Next
For i = 1 To collectionOfAddresses.Count
Range(collectionOfAddresses.Item(i)) = collectionOfValues.Item(i)
Next i
Here we add addresses from one range to a certain collection, and values from another range to a second collection, and then fill cells on these addresses with the values.
Here is the second code, which makes the same:
For i = 1 To rng1.Rows.Count
For j = 1 To rng1.Columns.Count
rng2.Cells(i, j) = rng1.Cells(i, j)
Next j
Next i
So, the question is - what is the time of execution in both cases? I mean, it's clear that the second case is O(n^2) (to make it easier we assume the range is square).
What about the first one? Is For Each considered a nested loop?
And if so, does it mean that the time of the first code is O(n^2) + O(n^2) + O(n^2) = 3*O(n^2) which makes pretty the same as the second code time?
In general, do these two codes differ apart from the fact that the first one takes additional memory when creating collections?
Thanks a lot in advance.
Actually, your first example is O(n^4)!
That might sound surprising, but this is because indexing into a VBA Collection has linear, not constant, complexity. The VBA Collection essentially has the performance characteristics of a list - to get element N by index takes a time proportional to N. To iterate the whole thing by index takes a time proportional to N^2. (I switched cases on you to distinguish N, the number of elements in the data structure, from your n, the number of cells on the side of a square block of cells. So here N = n^2.)
That is one reason why VBA has the For...Each notation for iterating Collections. When you use For...Each, VBA uses an iterator behind the scenes so walking through the entire Collection is O(N) not O(N^2).
So, switching back to your n, your first two loops are using For...Each over a Range with n^2 cells, so they are each O(n^2). Your third loop is using For...Next over a Collection with n^2 elements, so that is O(n^4).
I actually don't know for sure about your last loop because I don't know exactly how the Cells property of a Range works - there could be some extra hidden complexity there. But I think Cells will have the performance characteristics of an array, so O(1) for random access by index, and that would make the last loop O(n^2).
This is a good example of what Joel Spolsky called "Shlemiel the painter's algorithm":
There must be a Shlemiel the Painter's
Algorithm in there somewhere. Whenever
something seems like it should have
linear performance but it seems to
have n-squared performance, look for
hidden Shlemiels. They are often
hidden by your libraries.
(See this article from way before stackoverflow was founded: http://www.joelonsoftware.com/articles/fog0000000319.html)
More about VBA performance can be found at Doug Jenkins's webstie:
http://newtonexcelbach.wordpress.com/2010/03/07/the-speed-of-loops/
http://newtonexcelbach.wordpress.com/2010/01/15/good-practice-best-practice-or-just-practice/
(I will also second what cyberkiwi said about not looping through Ranges just to copy cell contents if this were a "real" program and not just a learning excercise.)
You are right that the first is 3 x O(n^2), but remember that O-notation does not care about constants, so in terms of complexity, it is still an O(n^2) algorithm.
The first one is not considered a nested loop, even if it is working on the same size as the loop in the second. It is just a straight iteration over an N-item range in Excel. What makes it N^2 is the fact that you are defining N as the length of a side, i.e. number of rows/columns (being square).
Just an Excel VBA note, you shouldn't be looping through cells nor storing addresses anyway. Neither of the approaches is optimal. But I think they serve to illustrate your question to understand O-notation.
rng1.Copy
rng2.Cells(1).PasteSpecial xlValues
Application.CutCopyMode = False
Remember not to confuse the complexity of YOUR code with the complexity of background Excel functions. Over all the amount of work done is N^2 in both cases. However, in your first example - YOUR code is actually only 3N (N for each of the three loops). The fact that a single statement in Excel can fill in multiple values does not change the complexity of your written code. A foreach loop is the same as a for loop - N complexity by itself. You only get N^2 when you nest loops.
To answer your question about which is better - generally it is preferable to use built in functions where you can. The assumption should be that internally Excel will run more efficiently than you could write yourself. However (knowing MS) - make sure you always check that assumption if performance is a priority.
This question already has answers here:
What is a plain English explanation of "Big O" notation?
(43 answers)
Closed 9 years ago.
What is Big O notation? Do you use it?
I missed this university class I guess :D
Does anyone use it and give some real life examples of where they used it?
See also:
Big-O for Eight Year Olds?
Big O, how do you calculate/approximate it?
Did you apply computational complexity theory in real life?
One important thing most people forget when talking about Big-O, thus I feel the need to mention that:
You cannot use Big-O to compare the speed of two algorithms. Big-O only says how much slower an algorithm will get (approximately) if you double the number of items processed, or how much faster it will get if you cut the number in half.
However, if you have two entirely different algorithms and one (A) is O(n^2) and the other one (B) is O(log n), it is not said that A is slower than B. Actually, with 100 items, A might be ten times faster than B. It only says that with 200 items, A will grow slower by the factor n^2 and B will grow slower by the factor log n. So, if you benchmark both and you know how much time A takes to process 100 items, and how much time B needs for the same 100 items, and A is faster than B, you can calculate at what amount of items B will overtake A in speed (as the speed of B decreases much slower than the one of A, it will overtake A sooner or later—this is for sure).
Big O notation denotes the limiting factor of an algorithm. Its a simplified expression of how run time of an algorithm scales with relation to the input.
For example (in Java):
/** Takes an array of strings and concatenates them
* This is a silly way of doing things but it gets the
* point across hopefully
* #param strings the array of strings to concatenate
* #returns a string that is a result of the concatenation of all the strings
* in the array
*/
public static String badConcat(String[] Strings){
String totalString = "";
for(String s : strings) {
for(int i = 0; i < s.length(); i++){
totalString += s.charAt(i);
}
}
return totalString;
}
Now think about what this is actually doing. It is going through every character of input and adding them together. This seems straightforward. The problem is that String is immutable. So every time you add a letter onto the string you have to create a new String. To do this you have to copy the values from the old string into the new string and add the new character.
This means you will be copying the first letter n times where n is the number of characters in the input. You will be copying the character n-1 times, so in total there will be (n-1)(n/2) copies.
This is (n^2-n)/2 and for Big O notation we use only the highest magnitude factor (usually) and drop any constants that are multiplied by it and we end up with O(n^2).
Using something like a StringBuilder will be along the lines of O(nLog(n)). If you calculate the number of characters at the beginning and set the capacity of the StringBuilder you can get it to be O(n).
So if we had 1000 characters of input, the first example would perform roughly a million operations, StringBuilder would perform 10,000, and the StringBuilder with setCapacity would perform 1000 operations to do the same thing. This is rough estimate, but O(n) notation is about orders of magnitudes, not exact runtime.
It's not something I use per say on a regular basis. It is, however, constantly in the back of my mind when trying to figure out the best algorithm for doing something.
A very similar question has already been asked at Big-O for Eight Year Olds?. Hopefully the answers there will answer your question although the question asker there did have a bit of mathematical knowledge about it all which you may not have so clarify if you need a fuller explanation.
Every programmer should be aware of what Big O notation is, how it applies for actions with common data structures and algorithms (and thus pick the correct DS and algorithm for the problem they are solving), and how to calculate it for their own algorithms.
1) It's an order of measurement of the efficiency of an algorithm when working on a data structure.
2) Actions like 'add' / 'sort' / 'remove' can take different amounts of time with different data structures (and algorithms), for example 'add' and 'find' are O(1) for a hashmap, but O(log n) for a binary tree. Sort is O(nlog n) for QuickSort, but O(n^2) for BubbleSort, when dealing with a plain array.
3) Calculations can be done by looking at the loop depth of your algorithm generally. No loops, O(1), loops iterating over all the set (even if they break out at some point) O(n). If the loop halves the search space on each iteration? O(log n). Take the highest O() for a sequence of loops, and multiply the O() when you nest loops.
Yeah, it's more complex than that. If you're really interested get a textbook.
'Big-O' notation is used to compare the growth rates of two functions of a variable (say n) as n gets very large. If function f grows much more quickly than function g we say that g = O(f) to imply that for large enough n, f will always be larger than g up to a scaling factor.
It turns out that this is a very useful idea in computer science and particularly in the analysis of algorithms, because we are often precisely concerned with the growth rates of functions which represent, for example, the time taken by two different algorithms. Very coarsely, we can determine that an algorithm with run-time t1(n) is more efficient than an algorithm with run-time t2(n) if t1 = O(t2) for large enough n which is typically the 'size' of the problem - like the length of the array or number of nodes in the graph or whatever.
This stipulation, that n gets large enough, allows us to pull a lot of useful tricks. Perhaps the most often used one is that you can simplify functions down to their fastest growing terms. For example n^2 + n = O(n^2) because as n gets large enough, the n^2 term gets so much larger than n that the n term is practically insignificant. So we can drop it from consideration.
However, it does mean that big-O notation is less useful for small n, because the slower growing terms that we've forgotten about are still significant enough to affect the run-time.
What we now have is a tool for comparing the costs of two different algorithms, and a shorthand for saying that one is quicker or slower than the other. Big-O notation can be abused which is a shame as it is imprecise enough already! There are equivalent terms for saying that a function grows less quickly than another, and that two functions grow at the same rate.
Oh, and do I use it? Yes, all the time - when I'm figuring out how efficient my code is it gives a great 'back-of-the-envelope- approximation to the cost.
The "Intuitition" behind Big-O
Imagine a "competition" between two functions over x, as x approaches infinity: f(x) and g(x).
Now, if from some point on (some x) one function always has a higher value then the other, then let's call this function "faster" than the other.
So, for example, if for every x > 100 you see that f(x) > g(x), then f(x) is "faster" than g(x).
In this case we would say g(x) = O(f(x)). f(x) poses a sort of "speed limit" of sorts for g(x), since eventually it passes it and leaves it behind for good.
This isn't exactly the definition of big-O notation, which also states that f(x) only has to be larger than C*g(x) for some constant C (which is just another way of saying that you can't help g(x) win the competition by multiplying it by a constant factor - f(x) will always win in the end). The formal definition also uses absolute values. But I hope I managed to make it intuitive.
It may also be worth considering that the complexity of many algorithms is based on more than one variable, particularly in multi-dimensional problems. For example, I recently had to write an algorithm for the following. Given a set of n points, and m polygons, extract all the points that lie in any of the polygons. The complexity is based around two known variables, n and m, and the unknown of how many points are in each polygon. The big O notation here is quite a bit more involved than O(f(n)) or even O(f(n) + g(m)).
Big O is good when you are dealing with large numbers of homogenous items, but don't expect this to always be the case.
It is also worth noting that the actual number of iterations over the data is often dependent on the data. Quicksort is usually quick, but give it presorted data and it slows down. My points and polygons alogorithm ended up quite fast, close to O(n + (m log(m)), based on prior knowledge of how the data was likely to be organised and the relative sizes of n and m. It would fall down badly on randomly organised data of different relative sizes.
A final thing to consider is that there is often a direct trade off between the speed of an algorithm and the amount of space it uses. Pigeon hole sorting is a pretty good example of this. Going back to my points and polygons, lets say that all my polygons were simple and quick to draw, and I could draw them filled on screen, say in blue, in a fixed amount of time each. So if I draw my m polygons on a black screen it would take O(m) time. To check if any of my n points was in a polygon, I simply check whether the pixel at that point is green or black. So the check is O(n), and the total analysis is O(m + n). Downside of course is that I need near infinite storage if I'm dealing with real world coordinates to millimeter accuracy.... ...ho hum.
It may also be worth considering amortized time, rather than just worst case. This means, for example, that if you run the algorithm n times, it will be O(1) on average, but it might be worse sometimes.
A good example is a dynamic table, which is basically an array that expands as you add elements to it. A naïve implementation would increase the array's size by 1 for each element added, meaning that all the elements need to be copied every time a new one is added. This would result in a O(n2) algorithm if you were concatenating a series of arrays using this method. An alternative is to double the capacity of the array every time you need more storage. Even though appending is an O(n) operation sometimes, you will only need to copy O(n) elements for every n elements added, so the operation is O(1) on average. This is how things like StringBuilder or std::vector are implemented.
What is Big O notation?
Big O notation is a method of expressing the relationship between many steps an algorithm will require related to the size of the input data. This is referred to as the algorithmic complexity. For example sorting a list of size N using Bubble Sort takes O(N^2) steps.
Do I use Big O notation?
I do use Big O notation on occasion to convey algorithmic complexity to fellow programmers. I use the underlying theory (e.g. Big O analysis techniques) all of the time when I think about what algorithms to use.
Concrete Examples?
I have used the theory of complexity analysis to create algorithms for efficient stack data structures which require no memory reallocation, and which support average time of O(N) for indexing. I have used Big O notation to explain the algorithm to other people. I have also used complexity analysis to understand when linear time sorting O(N) is possible.
From Wikipedia.....
Big O notation is useful when analyzing algorithms for efficiency. For example, the time (or the number of steps) it takes to complete a problem of size n might be found to be T(n) = 4n² − 2n + 2.
As n grows large, the n² term will come to dominate, so that all other terms can be neglected — for instance when n = 500, the term 4n² is 1000 times as large as the 2n term. Ignoring the latter would have negligible effect on the expression's value for most purposes.
Obviously I have never used it..
You should be able to evaluate an algorithm's complexity. This combined with a knowledge of how many elements it will take can help you to determine if it is ill suited for its task.
It says how many iterations an algorithm has in the worst case.
to search for an item in an list, you can traverse the list until you got the item. In the worst case, the item is in the last place.
Lets say there are n items in the list. In the worst case you take n iterations. In the Big O notiation it is O(n).
It says factualy how efficient an algorithm is.