Divide and Conquer. Maximum/minimum. Integer Multiplication. CS125 Lecture 4 Fall 2015


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1 CS125 Lecture 4 Fall 2015 Divide ad Coquer We have see oe geeral paradigm for fidig algorithms: the greedy approach. We ow cosider aother geeral paradigm, kow as divide ad coquer. We have already see a example of divide ad coquer algorithms: mergesort. The idea behid mergesort is to take a list, divide it ito two smaller sublists, coquer each sublist by sortig it, ad the combie the two solutios for the subproblems ito a sigle solutio. These three basic steps divide, coquer, ad combie lie behid most divide ad coquer algorithms. With mergesort, we kept dividig the list ito halves util there was just oe elemet left. I geeral, we may divide the problem ito smaller problems i ay coveiet fashio. Also, i practice it may ot be best to keep dividig util the istaces are completely trivial. Istead, it may be wise to divide util the istaces are reasoably small, ad the apply a algorithm that is fast o small istaces. For example, with mergesort, it might be best to divide lists util there are oly four elemets, ad the sort these small lists quickly by isertio sort. Maximum/miimum Suppose we wish to fid the miimum ad maximum items i a list of umbers. How may comparisos does it take? A atural approach is to try a divide ad coquer algorithm. Split the list ito two sublists of equal size. (Assume that the iitial list size is a power of two.) Fid the maxima ad miima of the sublists. Two more comparisos the suffice to fid the maximum ad miimum of the list. Hece, if T () is the umber of comparisos, the T () = 2T (/2) + 2. (The 2T (/2) term comes from coquerig the two problems ito which we divide the origial; the 2 term comes from combiig these solutios.) Also, clearly T (2) = 1. By iductio we fid T () = (3/2) 2, for a power of 2. Iteger Multiplicatio The stadard multiplicatio algorithm takes time Θ( 2 ) to multiply together two digit umbers. This algorithm is so atural that we may thik that o algorithm could be better. Here, we will show that better algorithms 41
2 Lecture exist (at least i terms of asymptotic behavior). Imagie splittig each umber x ad y ito two parts: x = 10 /2 a + b,y = 10 /2 c + d. The xy = 10 ac + 10 /2 (ad + bc) + bd. The additios ad the multiplicatios by powers of 10 (which are just shifts!) ca all be doe i liear time. We have therefore reduced our multiplicatio problem ito four smaller multiplicatios problems, so the recurrece for the time T () to multiply two digit umbers becomes T () = 4T (/2) + O(). The 4T (/2) term arises from coquerig the smaller problems; the O() is the time to combie these problems ito the fial solutio (usig additios ad shifts). Ufortuately, whe we solve this recurrece, the ruig time is still Θ( 2 ), so it seems that we have ot gaied aythig. The key thig to otice here is that four multiplicatios is too may. Ca we somehow reduce it to three? It may ot look like it is possible, but it is usig a simple trick. The trick is that we do ot eed to compute ad ad bc separately; we oly eed their sum ad + bc. Now ote that (a + b)(c + d) = (ad + bc) + (ac + bd). So if we calculate ac, bd, ad (a + b)(c + d), we ca compute ad + bc by the subtractig the first two terms from the third! Of course, we have to do a bit more additio, but sice the bottleeck to speedig up this multiplicatio algorithm is the umber of smaller multiplicatios required, that does ot matter. The recurrece for T () is ow T () = 3T (/2) + O(), ad we fid that T () = log , improvig o the quadratic algorithm. If oe were to implemet this algorithm, it would probably be best ot to divide the umbers dow to oe digit. The covetioal algorithm, because it uses fewer additios, is probably more efficiet for small values of. Moreover, o a computer, there would be o reaso to cotiue dividig oce the legth is so small that the multiplicatio ca be doe i oe stadard machie multiplicatio operatio! It also turs out that usig a more complicated algorithm (based o a similar idea) the asymptotic time for multiplicatio ca be made arbitrarily close to liear that is, for ay ε > 0 there is a algorithm that rus i time O( 1+ε ).
3 Lecture Strasse s algorithm Divide ad coquer algorithms ca similarly improve the speed of matrix multiplicatio. Recall that whe multiplyig two matrices, A = a i j ad B = b jk, the resultig matrix C = c ik is give by c ik = a i j b jk. j I the case of multiplyig together two by matrices, this gives us a Θ( 3 ) algorithm; computig each c ik takes Θ() time, ad there are 2 etries to compute. Let us agai try to divide up the problem. We ca break each matrix ito four submatrices, each of size /2 by /2. Multiplyig the origial matrices ca be broke dow ito eight multiplicatios of the submatrices, with some additios. A C B D E G F H = AE + BG CE + DG AF + BH CF + DH Lettig T () be the time to multiply together two by matrices by this algorithm, we have T () = 8T (/2)+ Θ( 2 ). Ufortuately, this does ot improve the ruig time; it is still Θ( 3 ). As i the case of multiplyig itegers, we have to be a little tricky to speed up matrix multiplicatio. (Strasse deserves a great deal of credit for comig up with this trick!) We compute the followig seve products: P 1 = A(F H) P 2 = (A + B)H P 3 = (C + D)E P 4 = D(G E) P 5 = (A + D)(E + H) P 6 = (B D)(G + H) P 7 = (A C)(E + F) The we ca fid the appropriate terms of the product by additio: AE + BG = P 5 + P 4 P 2 + P 6
4 Lecture AF + BH = P 1 + P 2 CE + DG = P 3 + P 4 CF + DH = P 5 + P 1 P 3 P 7 Now we have T () = 7T (/2) + Θ( 2 ), which give a ruig time of T () = Θ( log7 ). Faster algorithms requirig more complex splits exist; however, they are geerally too slow to be useful i practice. Strasse s algorithm, however, ca improve the stadard matrix multiplicatio algorithm for reasoably sized matrices, as we will see i our secod programmig assigmet. FFT We will ow cosider aother divide ad coquer algorithm, the Fast Fourier trasform (FFT). The Fast Fourier trasform is a classic algorithm that has prove icredibly useful across a wide spectrum of applicatios, ad as such remais i widespread use today. The algorithm is based o a clever idea that is worth emphasizig before we begi. The key poit is that it is very importat to cosider how to represet your data. I this case, he key to the Fast Fourier trasform is a exceedigly clever ad oituitive represetatio of polyomials. To motivate the FFT, suppose we are keepig track of traffic low alog a highway. We have a hardware device that samples the traffic every five miutes. That is, it outputs the umber of cars that have passed i the last five miutes at five miute itervals. Let us call these samples a 0,a 1,a 2,... We may wat to keep a ruig average of the traffic over the last half a hour. I that case, we would wat to compute +5 1 i= 6 a i. Alteratively, we may wat to keep a weighted average of the traffic over the last half a hour. If we are judgig traffic flow, it makes sese that the last five miutes are perhaps more importat tha a whole half hour ago. At the same time, it would be a mistake to discout the previous iformatio etirely; if we look oly at five miute itervals, we might overreact to radom traffic fluctuatios. Suppose we use weightig coefficiets b 0,b 1,...,b 5. The most recet time iterval will be weight by b 0, ad so o. The after a is output we wish to compute the weighted average c give by c = 5 b 0 a i.
5 Lecture This sum of products should look familiar. It is the coefficiet of x 5 i the product of ( 5 )( 5 a i x i b i x ). i Ideed, more geerally, if we look at the various coefficiets of we ca obtai several values of c at oce. ( i)( k 5 i x b i x ), a i This sort of processig occurs ofte, especially i digital sigal processig. With this motivatio, we see that it is of great iterest to compute the product of two give polyomials extremely quickly. We therefore focus o determiig a efficiet algorithm for multiplyig two polyomials. That is, give A(x) ad B(x), we wish to compute the coefficiets of C(x) = A(x)B(x). Without loss of geerality, we will assume that C has degree 1, where is a power of 2. Also, we will thik of A ad B of also beig degree 1; the coefficiets of some terms of A ad B will just be 0. It is clear that we ca compute the coefficiets of C i time Θ( 2 ) usig the stadard algorithm. But ca we do better? Thus far, we have bee thikig of a polyomial as beig represeted by its coefficiets. We thik of the polyomial A(x) as A(x) = 1 t=0 a tx t. But there are other ways to represet a polyomial. I particular, a polyomial of degree 1 is completely defied by poits. For example, a lie (a polyomial of degree oe) is defied by ay two poits it passes through; a quadratic is defied by ay three poits it passes through. Thus, istead of represetig A(x) by the sequece of coefficiets (a 0,a 1,...,a 1 ) (call this the coefficiet represetatio), we could represet it by the sequece of values at poits x 0,x 1,x 2,...,x 1, amely (A(x 0 ),A(x 1 ),A(x 2 ),...,A(x 1 )) (call this the poit represetatio). Now multiplyig is easy i the poit represetatio: we ca compute C(x i ) = A(x i ) B(x i ). Sice to represet the polyomial C uiquely we will eed to evaluate C at poits, we must have A ad B evaluated at poits. But assumig that we use this represetatio, we ca multiply A(x) ad B(x) with just O() multiplicatios! The problem is that we may ot wat to keep the polyomial i the poit represetatio. For example, it is hard to evaluate C at ay ew poits whe we represet it i this way, while this is a straightforward operatio i the coefficiet represetatio. Also, as we saw i our motivatio sectio, we actually wat as output the coefficiets of C. Agai, we see that differet represetatios are useful i differet situatios! What we really wat ow is a quick way to move back ad forth betwee these two represetatios. I geeral, goig back ad forth betwee the two represetatios is ot apparetly quicker tha Ω( 2 ). To do
6 Lecture better, we eed to be eve trickier. We must carefully choose which poits to use i the poit represetatio. By choosig these poits cleverly, we will be able to move back ad forth betwee represetatios i time O(log). The right poits to evaluate A ad B at tur out to be the solutios to x = 1. These solutios are the called the th roots of uity. Now, i the real umbers, of course, there are oly 2 solutios to this equatio, amely 1 ad 1. I the complex umbers, however, there are 2 differet solutios. I fact, these solutios are all powers of a geerator w, so the solutios ca be writte as w,w 2,w 3,...,w 1,w = 1. Note also that w /2 = 1! Fially, a very importat fact is that for all roots w i other tha 1, whe we add its powers 1 + w i + w 2i +...w ( 1)i, we must get 0. This is because (1 + w i + w 2i +...w ( 1)i ) (1 w i ) = 1 w i = 1 (w ) i = 0. Sice the first term i the product is ot 0 (as log as w i 1), the first term is. (Iterestigly, we do ot eve have to work over complex umbers we ca work over other mathematical structures, such as fiite fields. The poit is we just eed a geerator a w with w = 1, such that w k 1 for ay k with 1 k <.) We ow wat to fid a fast way to compute the values A(w j 1 ) = a i w i j, j = 0,..., 1. It turs out we ca use a divide ad coquer scheme. We divide the coefficiets of A ito two subsequeces: the eve subsequece a 0,a 2,... ad the odd subsequece a 1,a 3,... The A(w j ) = = a 2i w 2i j + a 2i (w 2 ) i j + w j 2i j+ j a 2i+1 w = E((w 2 ) j ) + w j O((w 2 ) j ) a 2i+1 (w 2 ) i j Now we have reduced the problem of computig A(w j ) to computig two smaller problems of size /2 of the same kid amely computig E(w j 2 ) ad O(w j 2 ), where w 2 = w 2! Notice that this recursio will work because 1,w 2,w 4,..., which are the poits where we will compute O ad E, are /2d roots of uity. (Agai, ote that there are /2 poits; we have ideed reduced to the problem to two subproblems of half the size.) To obtai A(w j ) from the subproblems, we do a multiply (by w j ) ad a additio. This is stadard divide ad coquer, with the recurrece: T () = 2T ( 2 ) + c,
7 Lecture where the work liear i comes from puttig the parts back together via multiplicatios ad additios. Of course we immediately have T () = O(log). This process is kow as the Fast Fourier Trasform, or FFT. (Aside: Aother way of expressig this is we wat to compute A(x) at the roots of uity. Note that A(x) = A 0 (x 2 ) + xa 1 (x 2 ) where A 0 is the polyomial with the eve coefficiets ad A 1 is the polyomial with the odd coefficiets. We eed to fid A 0 ad A 1 at the poits 1,w 2,w 4,... = 1,v,v 2,... where v is a (/2)d root of uity. This gives us the recursio above. ) We ve show how to go quickly from the coefficiet represetatio to a poit represetatio. How do we go back? Amazigly, this ca be doe with aother FFT! We start with a poit represetatio (C(1),C(w),C(w 2 ),...,C(w 1 )). To ivert, we thik of these values as coefficiets! (This is a tricky thought!) The we do a FFT o these coefficiets, usig w 1 istead of w i the FFT. The jth value we obtai is 1 C(w i )w i j = = 1 1 ( k=0 1 k=0 1 c k ( c k w ik )w i j w i(k j) ). Now, cosider the ier sum i the last lie. If j ad k are the same, the ier sum is just. Otherwise, we get the sum of all the powers of w but we kow this sum equals 0! Hece we have 1 C(w i )w i j = c k, so doig this iverse FFT returs to us the proper coefficiets (scaled up by a factor of ). To summarize, to multiply two polyomials A ad B i O(log) time, we do the followig: We first compute the FFT o the coefficiets (a 0,a 1,...,a 1 ) for A, ad similarly for B. This puts us i the poit represetatio. We compute C(w i ) = A(w i ) B(w i ) for i = 0,..., 1. Note that these are multiplicatos o complex umbers! We treat (C(w 0 ),C(w 1 ),...,C(w 1 )) as coefficiets, ad by computig the iverse FFT (usig w 1!) ad dividig the results by, we obtai the coefficiet represetatio (c 0,c 1,...,c 1 ).
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