The idea of divide and conquer is that we solve a large problem by solving a number of smaller problems and then we apply this idea recursively.
From the description above we see that the complexity of a divide and conquer solution has three parts.
Let T(N) be the time required to solve an instance of the problem having time N. The time required to split the problem and combine the subproblems is typically a function of N, say f(N).
More interesting is the time required to solve the subproblems. If the problem has been split in half then the time required for each subproblem is T(N/2).
Since the total time required includes splitting, solving both subproblems, and combining we get.
T(N) = 2T(N/2)+f(N)
Very often the splitting and combining are fast, specifically linear in N. Then we get
T(N) = 2T(N/2)+bNfor some constant b. (We probably should say ≤ rather than =, but we will soon be using big-Oh and friends so we can afford to be a little sloppy.
What if N is not divisible by 2? We should be using floor or ceiling or something, but we won't. Instead we will be assuming for recurrences like this one that N is a power of two so that we can keep dividing by 2 and get an integer. (The general case is not more difficult; but is more tedious)
But that is crazy! There is no integer that can be divided by 2 forever and still give an integer! At some point we will get to 1, the so called base case. But when N=1, the problem is almost always trivial and has a O(1) solution. So we write either
b if N = 1 T(N) = 2T(N/2)+bN if N > 1or
T(1) = b T(N) = 2T(N/2)+bN if N > 1
No we will now see three techniques that, when cleverly applied, can solve a number of problems. We will also see a theorem that, when its conditions are met, gives a solution without our being clever.
Also called ``plug and chug'' since we plug the equation into itself and chug along.
T(N) = 2T(N/2)+bN = 2[ ]+bN = 2[2T((N/2)/2)+b(N/2)]+bN = 4T(N/4)+2bN now do it again = 8T(N/8)+3bN
A flash of inspiration is now needed. When the smoke clears we get
T(N) = 2^{i}T(N/2^{i})+ibN
When i=log(N), N/2^{i}=1 and we have the base case. This gives the final result
T(N) = 2^{log(N)}T(N/2^{log(N)})+ibN = N T(1) +log(N)bN = Nb +log(N)bNHence T(N) is O(Nlog(N))
The idea is similar but we use a visual approach.
We already studied the recursion tree when we analyzed merge-sort. Let's look at it again.
The diagram shows the various subproblems that are executed, with their sizes. (It is not always true that we can divide the problem so evenly as shown.) Then we show that the splitting and combining that occur at each node (plus calling the recursive routine) only take time linear in the number of elements. For each level of the tree the number of elements is N so the time for all the nodes on that level is Θ(N) and we need to find the height of the tree. When the tree is split so evenly the sizes of all the nodes on each level go down by a factor of two so we reach a node with size 1 in logN levels (assuming N is a power of 2). Thus T(N) is Θ(Nlog(N)).
This method is really only useful after you have practice in recurrences. The idea is that, when confronted with a new problem, you recognize that it is similar to a problem you have seen the solution to previously and you guess that the solution to the new problem is similar to the old.
But we don't have enough experience for this to be very useful.
In this section, we apply the heavy artillery. The following theorem, which we will not prove, enables us to solve some problems by just plugging in.
We will only be considering complexities of the form
T(1) = c T(N) = aT(N/b)+f(N) if N > 1
The idea is that we have done some sort of divide and conquer where there are a subproblems of size at most N/b. As mentioned earlier f(N) accounts for the time to divide the problem into subproblems and to combine the subproblem solutions.
Theorem [The Master Theorem]: Let f(N) and T(N) be as above.
Proof: Not given.
Remarks:
Now we can solve some problems easily and will do two serious problems after that.
Example: T(N) = 4T(N/2)+N.
Example: T(N) = 2T(N/2) = Nlog(N)
Example: T(N) = T(N/4) + 2N
Example: T(N) = 9T(N/3) + N^{2.5}
Homework: R-5.4
We want to multiply big integers.
When we multiply without machine help, we use as knowledge the times tables for all 1 digit numbers and then do the standard 5th-grade algorithm that requires Θ(N^{2}) steps for N digit numbers. On a computer the known tables are for all 32-bit numbers (or 64-bit on some machines). If we want to multiply two 32N-bit numbers on the computer we could write the 5-th grade algorithm in software (using 32-bit numbers as ``digits'') and again compute the result in Θ(N^{2}) time.
To make the wording easier lets assume we only know how to multiply 1-bit numbers and we want to multiply two N-bit numbers X and Y. Using the 5th grade algorithm we can do this in time Θ(N^{2}). We want to go faster.
If we simply divide the bits in half (the high order and the low order), we see that to multiply X and Y we need to compute 4 sub-products Xhi*Yhi+Xhi*Ylo+Xlo*Yhi+Xlo*Ylo. (Note that when I write the products above I don't include the powers of 2 needed. This corresponds to writing the numbers in the correct column with the 5th grade algorithm.)
Since addition of K-bit numbers is Θ(K) and our multiplications are only of N/2-bit values we get
T(N) = 4T(N/2)+cnBut when we apply the master theorem (case 1) we just get T(N)=N^{2}, which is no improvement. We try again and with cleverness are able to get by with three instead of four multiplications.
We compute Xhi*Yhi, Xlo*Ylo, and the miraculous (Xhi-Xlo)*(Ylo-Yhi). The miracle multiplication produces the two remaining terms we need but has added to this two other terms. But those are just the (negative of) the two terms we did compute so we can add them back and they cancel.
The summary is that T(N) = 3T(N/2)+cN, which by the master theorem (case 1) gives T(N) is Θ(N^{log23}N). But log_{2}3<1.585 so we get the surprising
Theorem: We can multiple two N-bit numbers in O(N^{1.585}) time.
If you thought the integer multiplication involved pulling a rabbit out of our hat, get ready.
This algorithm was a sensation when it was discovered by Strassen. The standard algorithm for multiply two NxN matrices is Θ(N^{3}). We want to do better
First try. Assume N is a power of 2 and break each matrix into 4 parts as shown on the right. Then X = AE+BG and similarly for Y, Z, and W. This gives 8 multiplications of half size matrices plus Θ(N^{2}) scalar addition so
T(N) = 8T(N/2) + bN^{2}We apply the master theorem and get T(N) = Θ(N^{3}), i.e., no improvement.
But strassen found the way! He (somehow, I don't know how) decided to consider the following 7 (not 8) multiplications of half size matrices.
S1 = A(F-H) S2 = (A+B)H S3 = (C+D)E S4 = D(G-E) S5 = (A+D)(E+H) S6 = (B-D)(G+H) S7 = (A-C)(E+F)
Now we can compute X, Y, Z, and W from the S's
X = S5+S6+S4-S2 Y = S1+S2 Z = S3+S4 W = S1-S7-S3+S5
This computation shows that
T(N) = 7T(N/2) + bN^{2}
Thus the master theorem now gives
Theorem(Strassen): We can multiply two NxN matrices in time O(n^{log7}).
Remarks: