Now, fortunately when we analyze algorithms, actually not too many different functions arise and actually that property allows us to really classify algorithms according to their performance as the problem size grows.
So that’s what we’ll talk about next. So the good news is there’s only these few functions turn up about the algorithms that we are interested in. We can craft things that have other functions and there are counter examples to this. But really a great number of the algorithms that we consider are described by these few functions and that are plotted here. And [cough] the when we are talking about the order of growth, we are not talking about the leading constant. Normally we’ll say the running time of the algorithm is proportional to N log N. That means we that we think that our hypothesis is that the running time is tilde C lg N, N lg N, where C is some constant. And in these plots, these are lg, lg plots that not really give a good idea of what’s going on. If a order of growth is logarithmic or constant, doesn’t matter how big the thing is. It’s going to be fast of the running time for is T for say a thousand, and for half a million it will be pretty close to T.
If it’s linear, if it’s auto growth is proportional to N then as the running time, as the size increases the running time increases correspondingly. And the same is true, almost, if it’s N log N. So those are the algorithms that we strive for. They scale with the input size. As the input grows, so grows the running time. And that’s, a reasonable situation to be in. As we talked about when we talked about Union-find. If it’s quadratic, the running time grows much faster than the input size. And it’s not feasible to use such an algorithm for large inputs. And qubic is even worse.
So what we find is for many algorithms our first task is really, simply, make sure it’s not quadratic or qubit. And these order of growth classifications actually come from kind of simple patterns in terms of the code that we write. So if our code has no loops in it, then the order of growth is going to be constant. If our code has some kind of loop where the input’s divided in half, and so binary search algorithm is an example of that. Then our order growth will be logarithmic and we’ll take a look at that analysis and but if you do the doubling test, it grows almost linearly, if you have a huge input and you double the size it’s, it’s still going to be I’m sorry, not linearly, constant just like if it’s constant. You’ll hardly notice that lg N. If you have a loop where you touch everything in your input. Than the running time is linear, proportional to end so a typical example of that would be find the maximum, or to count the number of zeros. Our one some problem. A very interesting category is a so-called N lg N algorithms or linear rhythmic algorithms. And those are the ones that arise from a particular algorithms design technique called the divide and conquer. And the Mergesort algorithm, which we’ll talk about in a couple of weeks, is a prime example of that. And then if you have double four loops like our two sum algorithm, that’s going to be time proportional to N^2. As we saw, that’s quadratic, or triple four loop like our 3-sum algorithm, that’s going to be cubic or time proportional to N^3.
For a quadratic algorithm or a cubic algorithm, the doubling factor is four or eight as the input size double for cubic algorithm, the running time goes up by a factor of eight, and that’s the kind of calculation that you can do in your head while waiting for a program to finish.
There’s also a category of algorithms who’s running time is exponential and in those algorithms n doesn’t get very large at and we’ll talk about those at the end part two of the course.
So these are some practical implications of, of the order growth. And we really dwell on this too much, except to come back to the point that the algorithms we are really interested in, that can solve huge problems, are the linear and N lg N algorithms. Because even now a quadr atic algorithm on a typical fast computer could only solve problems and saying that tens of thousands in a cubic algorithm only in the size of thousands. And nowadays those are just not useful because the amount of data that we have is more like the millions or billions or trillions. That fact is becoming more and more evident as time wears on the ancient times would have some discussion about whether quadratic algorithm might be useful but the situation gets worse as the time goes on, so we need better algorithms.
To illustrate the process of developing a mathematical model for describing a performance through an algorithm, we’ll look at a familiar algorithm called binary search. It’s, the goal is that you have a sorted array of integers, say and you’re given a key. And you want to know, is that key in the array? And if it is, what, what’s its index? And a fast algorithm for doing this is known as binary search, where we compare the key against the middle entry. In this case, if we’re looking for 33, we compare it against 53. If its smaller we know its in the left half of the array, if it’s larger we know it’s in the right half of the array, if it’s equal, we found it. And then we apply the same algorithm recursively.
So let’s quickly look at a demo. So we’re looking for 33 in this array, compare it against the middle entry in the array. 53 and it’s less so we go left, so now we can concentrate just on the left half of the array, now we look in the middle of this half, that’s 25, 33 is bigger so we go right.
And now we concentrate on the right half or the left half and we have a smaller sub array. Look at the middle, 33 is less so we go left and now we have only the one element to look at and we found our key 33 in the array and we return that index four.
If we’re looking for something that’s not in the array, we do the same process. So, say, we’re looking for 34. It’s going to be the same. Look in the left half, look in the right half. Look to the left of the 43. Now, there’s only one key to look at. And it’s not 34, so we say, it’s not there.
So that’s binary search. So here’s the code for binary search.
Actually, Binary Search although it’s a simple algorithm, its notoriously tricky to get every detail right. In fact one paper claimed, that the first bug free binary search wasn’t published until 1962, and even in 2006, a bug was found in Java’s implementation of binary search, just an indication of the care that we have to take in developing algorithms especially for libraries that are going to be used by millions of people. So here’s an implementation. It’s not recursive although often we can implement this recursively. And it’s just reflexing code, what I described in words, we have to find. A key, whether a key’s in an array. And we use two pointers, low and high, to, indicate the part of the array we are interested in, as long as low is less and equal to high, we compute the middle. And then we compare our key against the middle, actually its a three way compare, see its less or greater or if its equal, we, we return that mid index. If its less we reset the high pointer, if its greater, we reset the low pointer, and we keep on going until the pointers are equal. If they are equal and we haven’t found it then we return -one.
And it’s easy to persuade ourselves that this program works as advertised by thinking about this invariant, if the keys in the array, then it’s between low and high in the array.
Alright, so that’s a program that, you are probably familiar with. Lets look at the mathematical analysis of that program. And this a, a theorem that we are going to prove easily. We want to a lot of proofs but this is one worth doing. So its say that binary search uses at most one + lg base two event compares, to complete a search, in a sorted array of size f. So we do that, to setup the problem by defining, a variable T(N), which is the number of compares that binary search needed for its array size and. And then we write down a recurrence relation that is reflex the code. And what the code does is, it divides the problem size in half so that. If the event is less or equal to the event over two plus depending on how you count what the compare is think of it as a two way compare so divided in half by doing one compare and that’s true as long as N is bigger than one.
If it’s equal to one the solution is one. So it’s a recurrent relation describing the computation. And so we, we can go ahead and, solve this recurrence by applying the recurrence itself, to the first term on the right. Now that’s called telescoping. So if this is true and we can apply the same thing to T(N/2). And throw out another one and if that’s, this is true, apply the same thing to N over four, and throw out another one and so forth until we get down to just one. In which case we have lg N ones left. Now this is a true sketch you might have noticed that, that this proof actually only holds if N is a power of two.
Because we nearly specify in this recurrence what we mean if N is odd. But it’s possible to go ahead and sorry, possible to go ahead and take care of that detail as well and show that binary search running time is logarithmic always.
All right, so given that fact we can develop a faster algorithm for a threesome. It’s a sorting based algorithm. And so what we’re going to do is we’re going to take the numbers that we have as input and sort them. We’ll talk about sorting algorithms next week. And we get that time in time proportional to N lg N but that’s not the main part of the computation. The main part of the computation is to after the numbers are sorted, we’ll go through and for each pair of numbers ai and aj. We’ll do a binary search for -ai + ij. If we find it then we’ll have three numbers that sum to zero. So if we [cough] sort our numbers and then go through for each pair do a binary search to see if it’s there, so -40, zero. Minus that is 40, we do a binary search that’s in there so we have one solution to the 3-sum problem. And do that for all pairs of numbers.
Then a quick analysis says the order of growth of running time is going to be N^2 lg N. Then you need a good sort, well, you could use the elementary insertion sort the first one we talk about but the running time of the binary search for each of the pairs, each of the N^2 pairs or N^2/2 pairs we’re going to do the binary search, so we get a N^2 lg N running time. So, a quick example of how we could improve the performance, we could find an imroved algorithm to solve a problem. N^2 lg N is much less than N^3 for large N.
And so, we’re implicitly making the hypothesis that if we do this, do the sort base thing and use binary search, we’re going to have a faster program. And, sure enough we can go ahead and run some experiments and find that whereas it took us 50 seconds to solve the problem for 8,000 numbers before. It’s taking less than a second now. In 50 seconds we can solve up to 64,000. So typically we expect that better order of growth means. Faster in practice and but when it comes to examining the algorithms in detail we can, we can go ahead and do the tests and figure out which algorithm is faster. And certainly going from N^3 to N^2 lg N we’re going to expect that we’re going to have a much better algorithm.