$OJRULWKPV. (Feodor F. Dragan) Department of Computer Science Kent State University

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1 $GYDQF $OJRULWKPV (Feodor F. Dragan) Department of Computer Scence Kent State Unversty Advanced Algorthms, Feodor F. Dragan, Kent State Unversty Textbook: Thomas Cormen, Charles Lesterson, Ronald Rvest, and Clff Sten, Introducton to Algorthms, McGraw Hll Publshng Company and MIT Press, (nd Edton). Gradng: Homework: 4% Mdterm Exam: 3% Fnal Exam: 3% All the course sldes wll be avalable at Advanced Algorthms, Feodor F. Dragan, Kent State Unversty

2 Course Outlne We wll cover the followng topcs (the topcs and order lsted are tentatve and subect to change; some topcs may only be quckly surveyed to add breadth, whle others wll be covered n reasonable depth). Dynamc Programmng Optmal greedy algorthms Amortzed analyss Parallel and crcut algorthms Network flow algorthms Randomzed algorthms Number theoretc & cryptographc algorthms Strng matchng algorthms Computatonal geometry algorthms Algorthms for NP-hard and NP-complete problems Approxmaton algorthms Onlne algorthms Lnear programmng algorthms Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 3 What s a computer program exactly? Input Some mysterous processng Output Programs = Data Structures + Algorthms Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 4

3 CHAPTER 5 Dynamc Programmng We begn dscusson of an mportant algorthm desgn technque, called dynamc programmng (or DP for short). It s not a partcular algorthm, t s a metatechnque. Programmng = tableau method, not wrtng a code. The technque s among the most powerful for desgnng algorthms for optmzaton problems. Ths s true for two reasons. Dynamc programmng solutons are based on a few common elements. Dynamc programmng problems are typcal optmzaton problems (fnd the mnmum or maxmum cost soluton, subect to varous constrants. The technque s related to dvde-and-conquer, n the sense that t breaks problems down nto smaller problems that t solves recursvely. However, because of the somewhat dfferent nature of dynamc programmng problems, standard dvde-and-conquer solutons are not usually effcent. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 5 The basc elements that characterze a dynamc programmng algorthm are: Substructure: Decompose your problem nto smaller (and hopefully smpler) subproblems. Express the soluton of the orgnal problem n terms of solutons for smaller problems. (Unlke dvde-and-conquer problems, t s not usually suffcent to consder one decomposton, but many dfferent ones.) Table-structure: Store the answers to the subproblems n a table. Ths s done because (typcally) subproblem solutons are reused many tmes, and we do not want to repeatedly solve the same problem. Bottom-up computaton: Combne solutons on smaller subproblems to solve larger subproblems, and eventually to arrve at a soluton to the complete problem. (Our text also dscusses a top-down alternatve, called memozaton.) Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 6 3

4 The most mportant queston n desgnng a DP soluton to a problem s how to set up the subproblem structure. Ths s called the formulaton of the problem. Dynamc programmng s not applcable to all optmzaton problems. There are two mportant elements that a problem must have n order for DP to be applcable. Optmal substructure: Optmal soluton to the problem contans wthn t optmal solutons to subproblems. Ths s sometmes called the prncple of optmalty. It states that for the global problem to be solved optmally, each subproblem should be solved optmally. Polynomally many subproblems: An mportant aspect to the effcency of DP s that the total number of dstnct subproblems to be solved should be at most a polynomal number. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 7 Strngs One mportant area of algorthm desgn s the study of algorthms for character strngs. There are a number of mportant problems here. Among the most mportant has to do wth effcently searchng for a substrng or generally a pattern n large pece of text. (Ths s what text edtors do when you perform a search.) In many nstances you do not want to fnd a pece of text exactly, but rather somethng that s smlar. Ths arses for example n genetcs research. Genetc codes are stored as long DNA molecules. The DNA strands conssts of a strng of molecules of four basc types: C, G, T, A. Exact matches rarely occur n bology because of small changes n DNA replcaton. For ths reason, t s of nterest to compute smlartes between strngs that do not match exactly. One common method of measurng the degree of smlarty between two strngs s to compute ther longest common subsequence. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 8 4

5 Longest Common Subsequence Gven two sequences of characters X = x, x,..., x m and Z = z, z,..., zk, we say that Z s a subsequence of X f there s a strctly ncreasng sequence of k ndces,..., ( < <... < ) such that Z = x, x,..., x. In X = ABRACADABRA, Z = AADAA, T = ADAD Z s a subsequence of X, T s not. Gven two strngs X and Y, the longest common subsequence (LCS) of X and Y s a longest sequence Z whch s both a subsequence of X and Y. The LCS of s Z =, k m k X = ABRACADABRA and Y = YABBADABBADOO ABADABA. k The longest common subsequence problem s the followng. Gven two sequences X = x, x,..., x m and Y = y, y,..., yn, determne a LCS of X and Y. It s not always unque. LCS of (ABC) and (BAC) s ether (AC) or (BC). Brute-force algorthm: (very neffcent) for every subsequence of X check f t s n Y. O( nm) tme ( m subseqences of X, O(n) for scannng Y wth X-subseq.) Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 9 Dynamc Programmng Approach In typcal DP fashon, we need to break the problem nto smaller peces. There are many ways to do ths for strngs. But t turns out for ths problem that consderng all pars of prefxes wll suffce for us. A prefx of a sequence s ust an ntal strng of values, X = x, x,..., x. s the empty sequence. X The dea wll be to compute the longest common subsequence for every possble par of prefxes ( polynomal (how many?) number of subproblems). Let c[] denote the length of the longest common subsequence of X and Y. Eventually we are nterested n c[m, n] snce ths wll be the LCS of the two entre strngs (where s the prncple of optmalty?). The dea s to compute c[ ] assumng that we already know the values of c[, ] for and (but not both equal). We begn wth some observatons. Bass: c[ ] = c[,] =. If ether sequence s empty, then the longest common subsequence s empty. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 5

6 Dynamc Programmng Approach (cont.) Last characters match: Suppose x = y. Snce both end n x, we clam that the LCS must also end n x. (We wll leave the proof as an exercse.) Snce the s part of the LCS we may fnd the overall LCS by removng x from both sequences and takng the LCS of X and Y, and then addng to the x end. Thus f x = y then c[ ] = c[, ] +. Last characters do not match: Suppose x y. In ths case x and y cannot both be n LCS (snce they would have to be the last character of the LCS). Thus ether x s not part of the LCS, or y s not part of the LCS (and possbly both are not part of the LCS). In the frst case the LCS of X and Y s the LCS of X and Y, whch s c[-,]. In the second case the LCS s the LCS of and Y, whch s c[-]. We do not know whch s the case, so we try both and take the one that gves us the longer LCS. Thus, f x y then c[ ] = max( c[, ], c[ ]). X Advanced Algorthms, Feodor F. Dragan, Kent State Unversty x Implementng the Rule Thus, c[ ] = max( c[, ], c[ ]) c[, ] + = or =, > and x y, > and x = y The task now s to smply mplement ths rule. We concentrate only on computng the maxmum length of the LCS. Later we wll see how to extract the actual sequence. We wll store some helpful ponters n a parallel array, b[..m,..n]. The code s shown below. LCS (char x[..m], char y[..n] ) { // compute LCS table nt c[..m,..n]; char b[..m,..n]; for = to m do {c[]=; b[]=skipx;} // ntalze column for = to n do {c[,]=; b[,]=skipy;} // ntalze row for = to m do { // fll rest of table for = to n do { f (x[]==y[]) {c[]=c[-,-]+; b[]=addxy; } // take x[] (=y[]) for LCS else f (c[-,]>=c[-]) { // x[] s not n LCS c[]=c[-,]; b[]=skipx;} else {c[]=c[-]; b[]=skipy;} } } // y[] s not n LCS return c[m,n]; } // return length of LCS Advanced Algorthms, Feodor F. Dragan, Kent State Unversty f f f. 6

7 Extractng the Actual Sequence Extractng the fnal LCS s done by usng the back ponters stored n b[..m,..n]. Intutvely b[ ] = ADDXY means that x[] and y [] together form the last character of the LCS. So, we take ths common character, and contnue wth entry b[-,-] to the northwest ( ). If b[]=skipx, then we know that x[] s not n the LCS, and we skp t and go to b[-,] ( ). Smlarly, f b[ ] = SKIPY, then we know that y[] s not n the LCS, and so we skp t and go to b[ ] to the left ( ). Followng these back ponters, and outputtng a character wth each dagonal move gves the fnal subsequence. getlcs (char x[..m], char y[..n], char b[..m,..n]) { // extract the LCS LCS=empty strng; =m; =n; // start at lower rght whle (!= &&!=) { // go untll upper left swtch b[] { case ADDXY: add x[] (or equvalently y[]) to front of LCS --; --; break case SKIPX: --; break // skp x[] case SKIPY: --; break } } // skp y[] return LCS } Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 3 Tme and Space Bounds and an Example The runnng tme of the algorthm LCS s clearly O(mn) snce there are two nested loops wth m and n teratons, respectvely. The runnng tme of the algorthm getlcs s O(m+n). Both algorthms use O(mn) space. X: 3 4 Y: 3 4 = n B D C B B A C D X=BACDB Y=BDCB LCS=BCB m= 5 B 3 m= 5 B Y: 3 4 = n B A C D B D C B LCS Length Table wth back ponters ncluded Longest common subsequence example. READ Ch. 5.4 n CLRS. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 4 7

8 Chan Matrx Multplcaton Ths problem nvolves the queston of determnng the optmal sequence for performng a seres of operatons. Ths general class of problem s mportant n compler desgn for code optmzaton and n databases for query optmzaton. We wll study the problem n a very restrcted nstance, where the dynamc programmng ssues are easest to see. Suppose that we wsh to multply a seres of matrces A A... A n Matrx multplcaton s an assocatve but not a commutatve operaton. Ths means that we are free to parenthesze the above multplcaton however we lke, but we are not free to rearrange the order of the matrces. Also recall that when two (non-square) matrces are beng multpled, there are restrctons on the dmensons. A matrx has p rows and q columns. You can multply a p q p matrx A tmes a q q r matrx B, and the result wll be a p r matrx C. The number of columns of A must equal the number of rows of B. In partcular, for p and r, A * B = C [, ] = q q C A[ k] B[ k, ]. p p k= r There are p r total entres n C and each r takes O(q) tme to compute, thus the total tme q (e.g. number of multplcatons) to multply these two matrces s p q r. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 5 Note that although any legal parentheszaton wll lead to a vald result, not all nvolve the same number of operatons. Consder the case of 3 matrces: Then Problem Formulaton A be 4, A be 4 6, A be multcost[((a A)A3)] = (5 4 6) + (5 6 ) = 8; multcost[(a ( A A ))] = (4 6 ) + (5 4 ) = Even for ths small example, consderable savngs can be acheved by reorderng the evaluaton sequence.. Chan Matrx Multplcaton Problem: Gven a sequence of matrces A A... A and dmensons p, p, p,..., p n n where A s of dmenson p p, determne the order of multplcaton that mnmzes the number of operatons. Important note: Ths algorthm does not perform the multplcatons, t ust fgures out the best order n whch to perform the multplcatons. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 6 8

9 Nave Algorthm We could wrte a procedure whch tres all possble parentheszatons. Unfortunately, the number of ways of parentheszng an expresson s very large. If you have ust one tem, then there s only one way to parenthesze. If you have n tems, then there are n- places where you could break the lst wth the outermost par of parentheses, namely ust after the frst tem, ust after the second tem, etc., and ust after the (n-)st tem. When we splt ust after the kth tem, we create two sublsts to be parentheszed, one wth k tems, and the other wth n-k tems. Then we could consder all the ways of parentheszng these. Snce these are ndependent choces, f there are L ways to parenthesze the left sublst and R ways to parenthesze the rght sublst, then the total s L tmes R. Ths suggests the followng recurrence for P(n), the number of dfferent ways of parentheszng n tems: f n =, P( n) = n P( k) P( n k) f n. k= Ths s related to famous functon n combnatorcs called the Catalan numbers (whch n turn s related to the number of dfferent bnary trees on n nodes). In partcular P (n) = C(n-), where C(n) s the nth Catalan number; ( ) n C n =. n+ n Applyng Strlng s formula, we fnd that C( n) Ω(4n / n 3/ ). Snce 4n s exponental and n 3/ s ust polynomal, the exponental wll domnate, mplyng that functon grows very fast. Thus, ths wll not be practcal except for very small n. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 7 Dynamc Programmng Approach Ths problem, lke other dynamc programmng problems nvolves determnng a structure (n ths case, a parentheszaton). We want to break the problem nto subproblems, whose solutons can be combned to solve the global problem. For convenence we can wrte A.. to be the result of multplyng matrces through. It s easy to see that A s a p p.. matrx. In parentheszng the expresson, we can consder the hghest level of parentheszaton. At ths level we are smply multplyng two matrces together. That s, for any k, k n, A.. n = A.. k Ak +.. n. Thus the problem of determnng the optmal sequence of multplcatons s broken up nto two questons: how do we decde where to splt the chan (what s k?) and how do we parenthesze the subchans A..k and A k+.. n? The subchan problems can be solved by recursvely applyng the same scheme. So, let us thnk about the problem of determnng the best value of k. At ths pont, you may be tempted to consder some clever deas. For example: snce we want matrces wth small dmensons, pck the value of k that mnmzes p k. Although ths s not a bad dea, t turns out that t does not work n ths case. Instead, (as n almost all dynamc programmng solutons), we wll do the dumb thng. We wll consder all possble values of k, and take the best of them. Notce that ths problem satsfes the prncple of optmalty, because once we decde to break the sequence nto the product A.. k A k +.. n, we should compute each subsequence optmally. That s, for the global problem to be solved optmally, the subproblems must be solved optmally as well. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 8 9

10 Dynamc Programmng Formulaton We wll store the solutons to the subproblems n a table, and buld the table n a bottom-up manner. For n, let m[ ] denote the mnmum number of multplcatons needed to compute A... The optmum cost can be descrbed by the followng recursve formulaton. Bass: Observe that f = then the sequence contans only one matrx, and so the cost s. (There s nothng to multply.) Thus, m[ ] =. Step: If, then we are askng about the product. Ths can be splt by consderng each k, k n, as A. A.. A.. k k+.. The optmum tme to compute k s m[ k], and the optmum tme to compute k + s m[k +, ]. We may assume that these values have been computed prevously and stored n our array. Snce A.. s a p p k matrx, and A k k +.. s a p k p matrx, the tme to multply them s p p p. Ths suggests the followng recursve rule for computng m[ ]. k m[ ] =, k< m[ ] = mn ( m[ k] + m[ k +, ] + p p p ) k for <. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 9 A.. A.. Implementng the Rule It s not hard to convert ths rule nto a procedure, whch s gven below. The only trcky part s arrangng the order n whch to compute the values. In the process of computng m[ ] we wll need to access values m[ k] and m[k +, ] for k lyng between and. Ths suggests that we should organze our computaton accordng to the number of matrces n the subchan. Let L = -+ denote the length of the subchan beng multpled. The subchans of length (m[ ]) are trval. Then we buld up by computng the subchans of length, 3,, n. The fnal answer s m[,n]. We need to be a lttle careful n settng up the loops. If a subchan of length L starts at poston then = + L-. Snce we want n, ths means that +L- n, or n other words, n-l+. So our loop for runs from to n-l+ (to keep n bounds). Matrx-Chan(array p[..n], nt n) { array s[..n-,..n]; for = to n do m[]=; // ntalze for L= to n do { // L=length of subchan for = to n-l+ do { =+L-; m[]=nfnty; for k= to - do { // check all splts q=m[k]+m[k+,]+p[-]*p[k]*p[]; f (q<m[]) {m[]=q; s[]=k; } } } } return m[,n](fnal cost) and s (splttng markers); } The runnng tme of ths procedure s (3 nested loops, each terates n tmes.) Θ( n 3 ). Advanced Algorthms, Feodor F. Dragan, Kent State Unversty

11 Extractng Optmum Sequence The array s[ ] can be used to extract the actual sequence. To extract t s a farly easy extenson. The basc dea s to keep n s[] a splt marker ndcatng what the best splt s, that s, what value of k leads to the mnmum value of m[ ]. s[ ] = k tells us that the best way to multply the subchan A s to frst multply the subchan A.... k and then multply the subchan A, and fnally multply these k +.. together. Intutvely, s[ ] tells us what multplcaton to perform last. Note that we only need to store s[ ] when we have at least two matrces, that s, f >. The actual multplcaton algorthm uses the s[ ] value to determne how to splt the current sequence. Assume that the matrces are stored n an array of matrces A[..n], and that s[ ] s global to ths recursve procedure. The procedure returns a matrx. Mult() { f ( ==) return A[]; // basc case else { k=s[]; X=Mult(k]; // X=A[] A[k] Y=Mult(k+,]; // Y=A[k+] A[] return X*Y; // multply matrces X and Y } } Advanced Algorthms, Feodor F. Dragan, Kent State Unversty Chan Matrx Multplcaton Example Ths algorthm s trcky, so t would be a good dea to trace through ths example (and the one gven n the text). The ntal set of dmensons are 5; 4; 6; ; 7 meanng that we are multplyng A (5 4) tmes A (4 6) tmes A tmes 3 (6 ) A 4 ( 7) The optmal sequence s ((A(AA3))A4) A p A A A3 4 p p p 3 p4 m[] 3 s[] 3 A A A3 A4 READ Ch. 5. n CLRS. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty

12 Recursve Implementaton We have descrbed dynamc programmng as a method that nvolves the ``bottom-up'' computaton of a table. However, the recursve formulatons that we have derved have been set up n a ``top-down'' manner. Must the computaton proceed bottom-up? Consder the followng recursve mplementaton of the chan-matrx multplcaton algorthm. The call Rec-Matrx-Chan(p, ) computes and returns the value of m[ ]. The ntal call s Rec-Matrx-Chan(p,, n). We only consder the cost here. Rec-Matrx-Chan(array p, nt nt ) { f (==) m[]=; // basc case else { m[]=nfnty; // ntalze for k= to - do { // try all splts cost=rec-matrx-chan(p,k)+ Rec-Matrx-Chan(p,k+,)+p[-]*p[k]*p[]; f (cost<m[]) m[]=cost; } } // update f better return m[]; } // return fnal cost (Note that the table m[..n,..n] s not really needed. We show t ust to make the connecton wth the earler verson clearer.) Ths verson of the procedure certanly looks much smpler, and more closely resembles the recursve formulaton that we gave prevously for ths problem. So, what s wrong wth ths? Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 3 Clam: Recursve Implementaton (cont.) The answer s the runnng tme s much hgher than the Θ( n 3 ) algorthm that we gave before. In fact, we wll see that ts runnng tme s exponental n n. Ths s unacceptably slow. Let T(n) denote the runnng tme of ths algorthm on a sequence of matrces of length n. (That s, n = +.) If =, then we have a sequence of length, and the tme s Θ(). Otherwse, we do Θ() work and then consder all possble ways of splttng the sequence of length n nto two sequences, one of length k and the other of length n-k, and nvoke the procedure recursvely on each one. So, we get the followng recurrence, defned for n. (We have replaced the Θ() s wth the constant.) f n =, T( n) = n + ( T( k) + T ( n k)) f n. k = T( n) n Proof: By nducton on n. Clearly, ths s true for n=. The nducton hypothess s that for all m<n. Usng ths we have ( m) m n n - gnore the term T(n-k), T( n) = + ( T ( k) + T ( n k)) + ( ) = T k k k= - apply the nducton, and n k n k n n + = + = + ( ) =. - apply the formula for the geometrc seres. k= k= Why s ths so much worse than the dynamc programmng verson? If you ``unravel'' the recursve calls on a reasonably long example, you wll see that the procedure s called repeatedly wth the same arguments. The bottom-up verson evaluates each entry exactly once. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 4

13 Memozaton Is t possble to retan the nce top-down structure of the recursve soluton, whle keepng the same O( n 3 ) effcency of the bottom-up verson? The answer s yes, through a technque called memozaton. Here s the dea. Let's reconsder the functon Rec-Matrx-Chan() gven above. It's ob s to compute m[ ], and return ts value. As noted above, the man problem wth the procedure s that t recomputes the same entres over and over. So, we wll fx ths by allowng the procedure to compute each entry exactly once. One way to do ths s to ntalze every entry to some specal value (e.g. UNDEFINED). Once an entres value has been computed, t s never recomputed. Mem-Matrx-Chan(array p, nt nt ) { f (m[]!= UNDEFINED) return m[]; // already defned else f (==) m[]=; // basc case else { m[]=nfnty; // ntalze for k= to - do { // try all splts cost=mem-matrx-chan(p,k)+ Mem-Matrx-Chan(p,k+,)+p[-]*p[k]*p[]; f (cost<m[]) m[]=cost; } } // update f better return m[]; } // return fnal cost Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 5 Memozaton (cont.) Ths verson runs n O ( n 3 ) tme. Intutvely, ths s because each of the O( n ) table entres s only computed once, and the work needed to compute one table entry (most of t n the for-loop) s at most O(n).... Memozaton s not usually used n practce, snce t s generally slower than the bottom-up method. However, n some DP problems, many of the table entres are smply not needed, and so bottom-up computaton may compute entres that are never needed. In these cases memozaton may be a good dea. If you know that most of the table wll not be needed, here s a way to save space. Rather than storng the whole table explctly as an array, you can store the ``defned'' entres of the table n a hash table, usng the ndex par ( ) as the hash key. See Chapter n CLRS for more nformaton on hashng. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 6 3

14 Polygons and Trangulatons Let's consder a geometrc problem that outwardly appears to be qute dfferent from chan-matrx multplcaton, but actually has remarkable smlartes. We begn wth a number of defntons. Defne a polygon to be a pecewse lnear closed curve n the plane. In other words, we form a cycle by onng lne segments end to end. The lne segments are called the sdes of the polygon and the endponts are called the vertces. A polygon s smple f t does not cross tself, that s, f the sdes do not ntersect one another except for two consecutve sdes sharng a common vertex. A smple polygon subdvdes the plane nto ts nteror, ts boundary and ts exteror. A smple polygon s sad to be convex f every nteror angle s at most 8 degrees. Vertces wth nteror angle equal to 8 degrees are normally allowed, but for ths problem we wll assume that no such vertces exst. Polygon Smple Polygon Convex Polygon Gven a convex polygon, we assume that ts vertces are labeled n counterclockwse order P = v, v, v,..., vn. We assume that ndexng s done modulo n, so v = vn. Ths polygon has n sdes, v v. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 7 Trangulatons (cont.) Gven two non-adacent vertces v, v ( < ), the lne segment s a chord. (If the polygon s smple but not convex, we nclude the addtonal requrement v v that the nteror of the segment les entrely n the nteror of P. Any chord subdvdes the polygon nto two polygons: v v,..., v and v, v +,..., v A trangulaton of a convex polygon s a maxmal set T of parwse non-crossng chords. In other words, every chord that s not n T ntersects the nteror of some chord n T. It s easy to see that such a set of chords subdvdes the nteror of the polygon nto a collecton of trangles wth parwse dsont nterors (and hence the name trangulaton). It s not hard to prove (by nducton) that every trangulaton of an n-sded polygon conssts of n-3 chords and n- trangles. Trangulatons are of nterest for a number of reasons. Many geometrc algorthms operate by frst decomposng a complex polygonal shape nto rectangles. Then an algorthm can be appled trangle by trangle. Defne the dual graph of the trangulaton to be a graph whose vertces are the trangles, and n whch two vertces share a sde (are adacent) f the two trangles share a common chord. Observe that the dual graph s a tree. Hence algorthms for traversng trees can be used for traversng the trangles of a trangulaton. +. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 8 4

15 Mnmum-Weght Convex Polygon Trangulaton In general, gven a convex polygon, there are many possble trangulatons. In fact, the number s exponental n n, the number of sdes. Whch trangulaton s the ``best''? There are many crtera that are used dependng on the applcaton. One crteron s to magne that you must ``pay'' for the nk you use n drawng the trangulaton, and you want to mnmze the amount of nk you use. Ths suggests the followng nterestng problem. Mnmum-weght convex polygon trangulaton: Gven a convex polygon, determne the trangulaton that mnmzes the sum of the permeters of ts trangles. A trangulaton Another trangulaton The dual tree Gven three dstnct vertces v, we defne the weght of the assocated v, vk trangle by the weght functon w v, v, v ) = v v + v v + v v, ( k k k where v v denotes the length of the lne segment v v. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 9 Correspondence to Bnary Trees One explanaton behnd the smlarty of trangulatons and the chan matrx multplcaton algorthm s to observe that both are fundamentally related to bnary trees. In the case of the chan matrx multplcaton, the assocated bnary tree s the evaluaton tree for the multplcaton, where the leaves of the tree correspond to the matrces, and each node of the tree s assocated wth a product of a sequence of two or more matrces. To see that there s a smlar correspondence here, consder an (n + )-sded convex polygon P = v, v, v,..., v n, v n, and fx one sde of the polygon (say v v n ). Now consder a rooted bnary tree whose root node s the trangle contanng sde v v n, whose nternal nodes are the nodes of the dual tree, and whose leaves correspond to the remanng sdes of the polygon. Observe that parttonng the polygon nto trangles s equvalent to a bnary tree wth n leaves, and vce versa. Ths s llustrated n the fgure below. Note that every trangle s assocated wth an nternal node of the tree and every edge of the orgnal polygon, except for the dstngushed startng sde v v n, s assocated wth a leaf node of the tree. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 3 5

16 A Correspondence to Bnary Trees (cont.) v A v root v A root A 3 v v A v 3 A 4 v 9 v 4 v 8 A 9 A A A3 A4 A5 A6 A7 A8 A9 AA A 5 v 5 A 6 v 6 A 7 v 7 A 8 Once you see ths connecton. Then the followng two observatons follow easly. Observe that the assocated bnary tree has n leaves, and hence (by standard results on bnary trees) n- nternal nodes. Snce each nternal node other that the root has one edge enterng t, there are n- edges between the nternal nodes. Each nternal node corresponds to one trangle, and each edge between nternal nodes corresponds to one chord of the trangulaton. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 3 Dynamc Programmng Soluton Let us consder an (n+)-sded polygon P = v v, v,..., v n, v, n To derve a DP formulaton we need to defne a set of subproblems from whch we can derve the optmum soluton. For n, defne t[ ] to be the weght of the mnmum weght trangulaton for the subpolygon v, v..., v, assumng that the chord v v s already present n the trangulaton. (The reason that we start wth v rather than v s to keep the structure as smlar as possble to the chan-matrx multplcaton problem. We wll see ths below.) Observe that f we can compute ths quantty for all such and, then the weght of the mnmum weght trangulaton of the entre polygon wll be t[, n]. As a bass case, we defne the weght of the trval ``-sded polygon'' to be zero, mplyng that t[ ] (the lne segment ) to be. v v In general, to compute t[], consder the subpolygon, where <. One of the chords of ths polygon s the sde v. We may v, splt v,..., ths v subpolygon v by ntroducng a trangle whose base s ths chord, and whose thrd vertex s any vertex v k, where k.. Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 3 6

17 Dynamc Programmng Soluton (cont.) Ths subdvdes the polygon nto the subpolygons v, v..., vk and v k, vk +,..., v whose mnmum weght are already known to us as t[k] and t[k+,]. In addton we should consder the weght of newly added trangle v vkv. Thus, we have the followng recursve rule: t[ ] = mn f =, ( t[ k] + t[ k+, ] + w( v v v )) f <. k k Note that ths has exactly the same structure as the recursve defnton used n the chan-matrx multplcaton algorthms. The same Θ( n 3 ) algorthm can be appled wth only mnor changes. READ Ch. 5.3 n CLRS and agan all sldes about trangulatons (ths part was taken from the frst edton of the book (Ch. 6.4), you wll not fnd ths n the second edton). Homework : Problems 5.-(p.338), 5.-5(p.339), 5.3-3(p.349), 5.3-5(p.35), 5.4-(p.355), 5.4-5(p.356). Extra credt problem: 5-7(p.369). Advanced Algorthms, Feodor F. Dragan, Kent State Unversty 33 7