I thought it meant it takes a constant amount of time to run. Is that different than one step?
O(1) is a class of functions. Namely, it includes functions bounded by a constant.
We say that an algorithm has the complexity of O(1) iff the amount of steps it takes, as a function of the size of the input, is bounded by a(n arbirtary) constant. This function can be a constant, or it can grow, or behave chaotically, or undulate as a sine wave. As long as it never exceeds some predefined constant, it's O(1).
For more information, see Big O notation.
It means that even if you increase the size of whatever the algorithm is operating on, the number of calculations required to run remains the same.
More specifically it means that the number of calculations doesn't get larger than some constant no matter how big the input gets.
In contrast, O(N) means that if the size of the input is N, the number of steps required is at most a constant times N, no matter how big N gets.
So for example (in python code since that's probably easy for most to interpret):
def f(L, index): #L a list, index an integer
x = L[index]
y=2*L[index]
return x + y
then even though f has several calculations within it, the time taken to run is the same regardless of how long the list L is. However,
def g(L): #L a list
return sum(L)
This will be O(N) where N is the length of list L. Even though there is only a single calculation written, the system has to add all N entries together. So it has to do at least one step for each entry. So as N increases, the number of steps increases proportional to N.
As everyone has already tried to answer it, it simply means..
No matter how many mangoes you've got in a box, it'll always take you the same amount of time to eat 1 mango. How you plan on eating it is irrelevant, there maybe a single step or you might go through multiple steps and slice it nicely to consume it.
Related
Consider the following algorithm (just as an example as the implementation is obviously inefficient):
def add(n):
for i in range(n):
n += 1
return n
The program adds one number with itself and returns it. Now the efficiency of an algorithm is sometimes modelled as a function between the size of the input and the number of primitive steps the algorithm has to compute. In this case, the input is an integer, n, and as n gets increased the number of steps necessary to complete the algorithm also increase (in this case linearly). But is it true that the size of the input increases? Let's assume that the machine where the program is running is representing integers in 8 bits. So if I increase the hypthetical input 3 for example to 7, the number of bits involved remains the same: 00000011 -> 00000111. However, the steps necessary to compute the algorithm increase. So it seems like that it's not always true that algorithmic efficiency can be modelled as a relation between input size and steps to compute. Could somebody explain to me where I go wrong or if I don't go wrong, why it still makes sense to model the efficiency of an algorithm as a function between the size of the input and the number of primitive steps to be computed?
Let S be the size of the input n. (Normally we'd use n for this size, but since the argument is also called n, that's confusing). For positive n, there's a relation between S and n, namely S = ceil(ln(n)). The program loops n times, and since n < 2^S, it loops at most 2^S times. You can also show it loops at least 1/2 * 2^S times, so the runtime (measured in loop iterations) is Theta(2^S).
This shows there's a way to model the runtime as a function of the size, even if it's not exact.
Whether it makes sense. In your example it doesn't much, but if your input is an array for sorting, taking size as the number of elements in the array does makes sense. (And it's typically what's used for example to model the number of comparisons done by different sort algorithms).
Wikipedia defines time complexity as
In computer science, the time complexity of an algorithm quantifies
the amount of time taken by an algorithm to run as a function of the
length of the string representing the input.
What's mean of the strong part?
I know algorithm may be treated as a function but why its input must be "the length of the string representing"?
The function in the bolder part means the time complexity of the algorithm, not the algorithm itself. An algorithm may be implemented in a programming language that has a function keyword, but that's something else.
Algorithm MergeSort has as input a list of 32m bits (assuming m 32-bit values). It's time complexity T(n) is a function of n = 32m, the input size, and in the worst case is bound from above by O(n log n). MergeSort could be implemented as a function in C or JavaScript.
The defition is derived from the context of Turing machines where you define different states. Every function which you can compute with a computer is also computeable with turing machine.(i would say that computer computes a function on the basis of turing machine)
Every fucntion is just a mapping from one domain to other domain or same domain.
Before going to Turing machines look at the concept of finite-automata.It has finite states.If your input is of length n then it's possible that it only needs two stats but it has to visit those states n times where n is legnth of the string.
Not a good sketch but look at the image below ,our final state is C , means if
end with a string in c our string will be accepted.
We use unary numberal system. We want to check if this this string gets accepted by our automata: string is 010101010
When we read 0 from A we move to B and if we read again a 0 we will move to C and if we end with 0 our string gets accepted otherwise we move to A again.
In computer you represent numbers as strings with length n and in order to compute it you have to visit each character of the string.
Turing machines work on the same way but finite-automata only is limited to regular languages. How this is a big theory
Did you ever try to think how computer computes a function 2*x where x is your input.
It's fun :D . Suppose i want to compute 20*2 and i represent this number with unary numeral system because it's easy. so we represetn 0 with , 1 with 11 , 2 with 111 and etc so if we convert 20 to unary sytem we get 1111. You can think of a turing machine or computer(not advanced) , a system with linear memory.
Suppose empty spots in your memory are presented with # .
With you input you have something like this: ###1111#### where # means empty slot in memory, with your input head of your turing machine is at first 1 so you keep moving forward until you find first # once you find this you just replace it with * which is just a helping symbol and change the right side of # with 1 now move back and change one more 1 to * and write one 1 on the right side when you find a # keep doing this and you will be left with all * on the left hand side and all 1s on the right side, Now change all *s back to 1 and you have 2*x. Here is the trace and you have 2*x where x was your input.
The point is that the only thing these machines remember is the state.
#####1111#######
#####111*1######
#####11**11#####
#####1***111####
#####****1111###
#####11111111###
Summary
If there is some input, it is expressed as a string. So you have a length of that string. Then you have a function F that maps the length of the input (as a string) to the time needed by A to compute this input (in the worst case).
We call this F time complexity.
Say we have an algorithm A. What is its time complexity?
The very easy case
If A has constant complexity, the input doesn't matter. The input could be a single value, or a list, or a map from strings to lists of lists. The algorithm will run for the same amount of time. 3 seconds or 1000 ticks or a million years or whatever. Constant time values not depending on the input.
Not much complexity at all to be honest.
Adding complexity
Now let's say for example A is an algorithm for sorting list of integer numbers. It's clear that the time needed by A now depends on the length of the list. A list of length 0 is literally sorted in no time (but checking the length of the list) but this changes if the length of the input list grows.
You could say there exists a function F that maps the list length to the seconds needed by A to sort a list of that length. But wait! What if the list is already sorted? So for simplicity let's always assume a worst case scenario: F maps list length to the maximum of seconds needed by A to sort a list of that length.
You could measure in seconds, CPU cycles, ticks, or whatever. It doesn't depend on the units.
Generalizing a bit
What with all the other algorithms? How to measure time complexity for an algorithm that cooks me a nice meal?
If you cannot define any input parameter then we're back in the easy case: constant time. If there is some input it is expressed as a string. So you have a length of that string. And - similar to what has been said above - then you have a function F that maps the length of the input (as a string) to the time needed by A to compute this input (in the worst case).
We call this F time complexity.
That's too simple
Yeah, I know. There is the average case and the best case, there is the big O notation and asymptotic complexity. But for explaining the bold part in the original question this is sufficient, I think.
Usually I am lazy and calculate modular inverses as follows:
def inv(a,m):
return 1 if a==1 else ((a-inv(m%a,a))*m+1)//a
print(inv(100**7000,99**7001))
But I was curious to know whether the method of passing more information back, namely the solution to Bezout's theorem (instead of just one of the pair), yields a faster or slower algorithm in practice:
def bez(a,b):
# returns [x,y,d] where a*x+b*y = d = gcd(a,b) since (b%a)*y+a*(x+(b//a)*y)=d
if a==0: return [0,1,b] if b>0 else [0,-1,-b]
r=bez(b%a,a)
return [r[1]-(b//a)*r[0],r[0],r[2]]
def inv(a,m):
r=bez(a,m)
if r[2]!=1: return None
return r[0] if r[0]>=0 else r[0]+abs(m)
print(inv(100**7000,99**7001))
I was amazed to find that the latter ran more than 50 times faster than the former! But both use nearly the same number of recursive calls and 1 integer division and 1 modulo operation per call, and the bit-length of the operands is roughly twice in the former on average (because the arguments involved are identical), so I only expect its time complexity to be roughly 4 times that of the latter, not 50 times.
So why am I observing this strange speedup?
Note that I am using Python 3, and I observed this when running online, with the following code added at the top to stop the Python interpreter complaining about exceeding maximum recursion depth or stack space:
import resource,sys
sys.setrecursionlimit(1000000)
resource.setrlimit(resource.RLIMIT_STACK,[0x10000000,resource.RLIM_INFINITY])
I figured it out finally. Suppose the initial inputs are n-bit integers. In typical cases, besides the first call to inv or inv2, the recursive calls have parameters whose sizes differ by just O(1) on average, and there are O(n) recursive calls on average, both due to some number-theoretic phenomenon. This O(1) size difference implies that the two methods actually have drastically different average time complexities. The implicit assumption in my question was that multiplication or division of two n-bit integers takes roughly O(n^2) time; this holds in the worst-case (for school-book multiplication), but not in the average case for the fast method!
For the fast method, it can be proven that when p,q > 0, the triple [x,y,d] returned by bez(p,q) satisfies abs(x) ≤ q and abs(y) ≤ p. Thus when each call bez(a,b) performs b%a and (b//a)*r[0], the modulo and division and multiplication each takes O(size(a)) time on average, since size(b)-size(a) ∈ O(1) on average and abs(r[0]) ≤ a. Total time is therefore roughly O(n^2).
For the slow method, however, when each call inv(a,m) performs ((a-inv(m%a,a))*m+1)//a, we have a-inv(m%a,a) roughly the same size as a, and so this division is of two integers where the first is roughly twice the size of the second, which would take O(size(a)^2) time! Total time is therefore roughly O(n^3).
Indeed, simulation yields very close agreement with the above analysis. For reference, using the fast method for inv(100**n,99**(n+1)) where n = 1400·k for k∈[1..6] took time/ms 17,57,107,182,281,366, while using the slow method where n = 500·k for k∈[1..6] took time/ms 22,141,426,981,1827,3101. These are very well fit by 9·k^2+9·k and 13·k^3+8·k^2 respectively, with mean fractional error ≈3.3% and ≈1.6% respectively. (We include the first two highest-order terms because they contribute a small but significant amount.)
Here's something I've been thinking about: suppose you have a number, x, that can be infinitely large, and you have to find out what it is. All you know is if another number, y, is larger or smaller than x. What would be the fastest/best way to find x?
An evil adversary chooses a really large number somehow ... say:
int x = 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9
and provides isX, isBiggerThanX, and isSmallerThanx functions. Example code might look something like this:
int c = 2
int y = 2
while(true)
if isX(y) return true
if(isBiggerThanX(y)) fn()
else y = y^c
where fn() is a function that, once a number y has been found (that's bigger than x) does something to determine x (like divide the number in half and compare that, then repeat). The thing is, since x is arbitrarily large, it seems like a bad idea to me to use a constant to increase y.
This is just something that I've been wondering about for a while now, I'd like to hear what other people think
Use a binary search as in the usual "try to guess my number" game. But since there is no finite upper end point, we do a first phase to find a suitable one:
Initially set the upper end point arbitrarily (e.g. 1000000, though 1 or 1^100 would also work -- given the infinite space to work in, all finite values are equally disproportionate).
Compare the mystery number X with the upper end point.
If it's not big enough, double it, and try again.
Once the upper end point is bigger than the mystery number, proceed with a normal binary search.
The first phase is itself similar to a binary search. The difference is that instead of halving the search space with each step, it's doubling it! The cost for each phase is O(log X). A small improvement would be to set the lower end point at each doubling step: we know X is at least as high as the previous upper end point, so we can reuse it as the lower end point. The size of the search space still doubles at each step, but in the end it will be half as large as would have been. The cost of the binary search will be reduced by only 1 step, so its overall complexity remains the same.
Some notes
A couple of notes in response to other comments:
It's an interesting question, and computer science is not just about what can be done on physical machines. As long as the question can be defined properly, it's worth asking and thinking about.
The range of numbers is infinite, but any possible mystery number is finite. So the above method will eventually find it. Eventually is defined such as that, for any possible finite input, the algorithm will terminate within a finite number of steps. However since the input is unbounded, the number of steps is also unbounded (it's just that, in every particular case, it will "eventually" terminate.)
If I understand your question correctly (advise if I do not), you're asking about how to solve "pick a number from 1 to 10", except that instead of 10, the upper bound is infinity.
If your number space is truly infinite, the following are true:
The value will never be held in an int (or any other data type) on any physical hardware
You will NEVER find your number
If the space is immensely large but bound, I think the best you can do is a binary search. Start at the middle of the number range. If the desired number turns out to be higher or lower, divide that half of the number space, and repeat until the desired number is found.
In your suggested implementation you raise y ^ c. However, no matter how large c is chosen to be, it will not even move the needle in infinite space.
Infinity isn't a number. Thus you can't find it, even with a computer.
That's funny. I've wondered the same thing for years, though I've never heard anyone else ask the question.
As simple as your scenario is, it still seems to provide insufficient information to allow the choice of an optimal strategy. All one can choose is a suitable heuristic. My heuristic had been to double y, but I think that I like yours better. Yours doubles log(y).
The beauty of your heuristic is that, so long as the integer fits in the computer's memory, it finds a suitable y in logarithmic time.
Counter-question. Once you find y, how do you proceed?
I agree with using binary search, though I believe that a ONE-SIDED binary search would be more suitable, since here the complexity would NOT be O( log n ) [ Where n is the range of allowable numbers ], but O( log k ) - where k is the number selected by your adversary.
This would work as follows : ( Pseudocode )
k = 1;
while( isSmallerThanX( k ) )
{
k = k*2;
}
// At this point, once the loop is exited, k is bigger than x
// Now do normal binary search for the range [ k/2, k ] to find your number :)
So even if the allowable range is infinity, as long as your number is finite, you should be able to find it :)
Your method of tetration is guaranteed to take longer than the age of the universe to find an answer, if the opponent merely uses a paradigm which is better (for example, pentation). This is how you should do it:
You can only do this with symbolic representations of numbers, because it is trivial to name a number your computer cannot store in floating-point representation, even if it used arbitrary-precision arithmetic and all its memory.
Required reading: http://www.scottaaronson.com/writings/bignumbers.html - that pretty much sums it up
How do you represent a number then? You represent it by a program which will, if run to completion, print out that number. Even then, your computer is incapable of computing BusyBeaver(10^100) (if you dictated a program 1 terabyte in size, this well over the maximum number of finite clock cycles it could run without looping forever). You can see that we could easily have the computer print out 1 0 0... each clock cycle, making the maximum number it could say (if we waited nearly an eternity) would be 10^BusyBeaver(10^100). If you allowed it to say more complicated expressions like eval(someprogram), power-towers, Ackermann's function, whatever-- then I believe that would be no better than increasing the original 10^100 by some constant proportional to the complexity of what you described (plus some logarithmic interpreter factor, see Kolmogorov complexity).
So let's frame this another way:
Your opponent picks a finite computable number, and gives you a function tells you if the number is smaller/larger/equal by computing it. He also gives you a representation for the output (in a sane world this would be "you can only print numbers like 99999", but he can make it more complicated; it actually doesn't matter). Proceed to measure the size of this function in bits.
Now, answer with your own function, which is twice the size of his function (in bits), and prints out the largest number it can while keeping the code to less than 2N bits in length. (You use the same representation he chose: In a world where you can only print out numbers like "99999", that's what you do. If you can define functions, it gets slightly more complicated.)
I do not understand the purpose here, but I this is what I thought of:
Reading your comments, I suppose you aren't looking for infinitely large number, but a "super large number" instead. And whatever be the number, it will have a large no. of digits. How you got them, isn't the concern. Keeping this in mind:
No complex computation is required. Just type random keys on your numeric keyboard to have a super large number, and then have a program randomly add/remove/modify digits of that number. You get a list of very large numbers - select any one out of them.
e.g: 3672036025039629036790672927305060260103610831569252706723680972067397267209
and keep modifying/adding digits to get more numbers
PS: If you state the purpose in your question clearly, we might be able to give better answers.
In an array with integers between 1 and 1,000,000 or say some very larger value ,if a single value is occurring twice twice. How do you determine which one?
I think we can use a bitmap to mark the elements , and then traverse allover again to find out the repeated element . But , i think it is a process with high complexity.Is there any better way ?
This sounds like homework or an interview question ... so rather than giving away the answer, here's a hint.
What calculations can you do on a range of integers whose answer you can determine ahead of time?
Once you realize the answer to this, you should be able to figure it out .... if you still can't figure it out ... (and it's not homework) I'll post the solution :)
EDIT: Ok. So here's the elegant solution ... if the list contains ALL of the integers within the range.
We know that all of the values between 1 and N must exist in the list. Using Guass' formula we can quickly compute the expected value of a range of integers:
Sum(1..N) = 1/2 * (1 + N) * Count(1..N).
Since we know the expected sum, all we have to do is loop through all the values and sum their values. The different between this sum and the expected sum is the duplicate value.
EDIT: As other's have commented, the question doesn't state that the range contains all of the integers ... in this case, you have to decide whether you want to optimize for memory or time.
If you want to perform the operation using O(1) storage, you can perform an in-place sort of the list. As you're sorting you have to check adjacent elements. Once you see a duplicate, you know you can stop. Optimal sorting is an O(n log n) operation on average - which establishes an upper bound for find the duplicate in this manner.
If you want to optimize for speed, you can use an additional O(n) storage. Using a HashSet (or similar structure), insert values from your list until you determine you are inserting a duplicate into the HashSet. Inserting n items into a HashSet is an O(n) operation on average, which establishes that as an upper bound for this method.
you may try to use bits as hashmap:
1 at position k means that number k occured before
0 at position k means that number k did not occured before
pseudocode:
0. assume that your array is A
1. initialize bitarray(there is nice class in c# for this) of 1000000 length filled with zeros
2. for each num in A:
if bitarray[num]
return num
else
bitarray[num] = 1
end
The time complexity of the bitmap solution is O(n) and it doesn't seem like you could do better than that. However it will take up a lot of memory for a generic list of numbers. Sorting the numbers is an obvious way to detect duplicates and doesn't require extra space if you don't mind the current order changing.
Assuming the array is of length n < N (i.e. not ALL integers are present -- in this case LBushkin's trick is the answer to this homework problem), there is no way to solve this problem using less than O(n) memory using an algorithm that just takes a single pass through the array. This is by reduction to the set disjointness problem.
Suppose I made the problem easier, and I promised you that the duplicate elements were in the array such that the first one was in the first n/2 elements, and the second one was in the last n/2 elements. Now we can think of playing a game in which two people each hold a string of n/2 elements, and want to know how many messages they have to send to be sure that none of their elements are the same. Since the first player could simulate the run of any algorithm that takes a pass through the array, and send the contents of its memory to the second player, a lower bound on the number of messages they need to send implies a lower bound on the memory requirements of any algorithm.
But its easy to see in this simple game that they need to send n/2 messages to be sure that they don't hold any of the same elements, which yields the lower bound.
Edit: This generalizes to show that for algorithms that make k passes through the array and use memory m, that m*k = Omega(n). And it is easy to see that you can in fact trade off memory for time in this way.
Of course, if you are willing to use algorithms that don't simply take passes through the array, you can do better as suggested already: sort the array, then take 1 pass through. This takes time O(nlogn) and space O(1). But note curiously that this proves that any sorting algorithm that just makes passes through the array must take time Omega(n^2)! Sorting algorithms that break the n^2 bound must make random accesses.