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# Analysis Of Algorithms CS 3343

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GPA 3.55

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This 61 page Class Notes was uploaded by Mireya Heidenreich on Thursday October 29, 2015. The Class Notes belongs to CS 3343 at University of Texas at San Antonio taught by Carola Wenk in Fall. Since its upload, it has received 7 views. For similar materials see /class/231376/cs-3343-university-of-texas-at-san-antonio in ComputerScienence at University of Texas at San Antonio.

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Date Created: 10/29/15

CS 3343 Fall 2007 ALGORlTHMS Sorting Carola Wenk Slides courtesy of Charles Leiserson with small anges by Carola Wenk mm CS diAmlymaAlgmm 1 How fast can we sort l a All the sorting algorithms we have seen so far are camparisan sumquot only use comparisons to determine the relative order of elements I Eg insertion sort merge sort quicksort heapsort The best worstcase running time that we ve seen for comparison sorting is On log n Is 0n log n the best we can do Decision trees can help us answer this question mm c liAmlym army th 2 liq Decisiontree example 1 V Sort lta1 a2 Each intemal node is labeled apaj for if e 1 n IThe le subtree shows subsequent comparisons if ax a I The right subtree shows subsequent comparisons if a 2 L17 3 mm c diAmlym 17 ng Decisiontree example 1 Sort a1 a2 a3gt gt Each internal node is labeled apal for if e 1 n I The le subtree shows subsequent comparisons if ax L17 I The right subtree shows subsequent comparisons if a 2 L17 3 mm c liAmlym army th 2001 by Charles E Leiserson small changes by Carola S01tlta1 a2 a3gt lt 9 4 6 gt1 Each intelnal node is labeled 1111 for ij e 1 n I The le subtree shoWs subsequent comparisons if ax I The light subtree shoWs subsequent comparisons if a 2 a 105407 assummtym mtg th s Decisiontree example is S01tlta1 a2 a3gt lt 9 4 6 gt1 Each intelnal node is labeled upaj fol ij e 1 n I The le subtlee shows subsequent colnpalisons if ax I The light subtree shoWs subsequent colnpalisons if a 2 a mm minimum 11234197 th 5 u 5 a Decisiontree example S01tlta1 a2 a3gt lt 9 4 6 gt Each intelnal node is labeled 1111 for ij e 1 n I The le subtree shoWs subsequent colnpalisons if ax a I The light subtree shoWs subsequent compalisons if a 2 a 7 105407 assummtym mtg th Decisiontree example La S01tlta1 a2 a3gt 7 9 4 6 4 S 6 S Each leafcontains a pellnutation lt71 7r2 7rngt to indicate that the oldeling aw aw am has been established 105407 minimum 11234197 th z 2001 by Charles E Leiserson small changes by Carola L Dec1s10ntree model A decision tree can model the execution of any comparison sorting algorithm I One tree for each input size n I The tree contains all possible comparisons ifbmnches that could be executed for any input of size n I The tree contains all comparisons along all possible instruction traces control ows for all inputs of size n I For one input only one path to a leaf is executed unning time length of the path taken I Worstcase running time height of tree 105407 mummy mtg th 9 m Lower bound for 39 t r w comparlson sorting Theorem Any decision tree that can sort n elements must have height Qnlogn Proof The tree must contain 2 n leaves since there are n possible permutations A heighth binary tree has S 2h leaves Thus n S 2h h 2 logn log is mono increasing 2 log nequot Stirling s formula n og n 7 n og e Qn log n D mm summon qulgrmhm m Lower bound for comparison sortlng Corollary Heapsort and merge sort are asymptotically optimal comparison sorting algorithms D 105407 mummy mtg th u PM Sorting in linear time Counting sort No comparisons between elements IInputAl n whereAUEl 2 k I Output Bl n sorted IAuxiliury storage Cl k mm summon qulgrmhm 12 2001 by Charles E Leiserson small changes by Carola 2forj lt 1 to n do 0mm e 0mm 1 gt cm key m 3for i lt 2 to k do Ci lt Ci Ci71 lgt Ci ery S il 4forj lt n downto 1 doBCAU e Am cA U E own1 1 105407 mmmtm mtg th 13 Countingsort example 2 3 4 5 1 2 3 4 l E mm mmmtm 11234197 th m 1forilt1tok do Cilt0 105407 mmmtm mtg th IS 2forj lt 1 to n d0 CAj e CAj 1 Igt Ci ery iH mm mmmtm 11234197 th 15 2001 by Charles E Leiserson small changes by Carola 2forj lt 1 to n do CAj lt CAj 1 lgt Ci ery il 105407 mmmtm mtg th 17 Loop 2 2 4 5 1 2 3 4 n lo ll 2forj lt 1 to n do CAj lt CAj 1 lgt Ci ery il mm mmmtm 11234197 th 12 2forj lt 1 to n d0 CAj e CAj 1 Igt Ci ery il 105407 mmmtm mtg th 19 2forj lt 1 to n d0 CAj e CAj 1 Igt Ci ery il mm mmmtm 11234197 th 20 2001 by Charles E Leiserson small changes by Carola rm L00p3 2 3 4 5 1 2 3 4 r l C I 3forilt2tok 3forilt2tok do Ci lt Ci Ci71 lgt Ci ery S il do Ci lt Ci Ci71 lgt Ci ery S il mm mumwmmmm 21 mm magnum2mm 22 1 2 3 4 1 2 3 4 5 1 2 3 C 4forj lt n downto 1 3for z lt 2 to k 7 do BCAU 9 AU do C1 lt Cz Cll Igt C1 7 ke S 1 y cum e cam 71 mm mmmmmgmm 2 mm maimmmgm 24 2001 by Charles E Leiserson small changes by Carola 4forj lt n downto 1 doBCAU 9 AU CAU e CAU 1 105407 mmmtm mtg th 4forj lt n downto 1 d0 BCAU 9 AU CAU e CAj 1 mm mmmtm 11234197 th 25 4forj lt n downto 1 doBCAU 9 AU CAU e CAU 1 105407 mmmtm mtg th 4forj lt n downto 1 d0 BCAU 9 AU CAU e CAj 1 mm mmmtm 11234197 th 22 2001 by Charles E Leiserson small changes by Carola 4forj lt n downto 1 doBCAU 9 AU CAU e CAU 1 105407 mmmtm mtg th 2 l 3 4 cunnn 4forj lt n downto 1 d0 BCAU 9 AU CAU e CAj 1 mm mmmtm 11234197 th 4forj lt n downto 1 doBCAU 9 AU CAU e CAU 1 105407 mmmtm mtg th 2001 by Charles E Leiserson small changes by Carola 4forj lt n downto 1 d0 BCAU 9 AU CAU e CAj 1 mm mmmtm 11234197 th S 2mm cs 3343 Spring 2009 AiooniTHMs More Divide amp Conquer Carola Wenk lides courtesy of Charles Leiserson with small changes by Carola Wenk 54mm 17 ng l a Powering a number Problem Compute aquot where n E N Naive algorithm n Divideandconquer algorithm recursive squaring n vimam ifnis even a a HyZ ultquot 1 a ifn is odd rm rot2 1 rm log n 25mg CS lfAnalym army th ilu mm m Fibonacci numbers Recursive definition 0 ifn039 F 1 ifn139 FHFH ifn22 0112358132134 Naive recursive algorithm QM exponential time where 1 43 2 is the golden ratio a ammo 17 ng Computing Fibonacci numbers Naive recursive squaring F MAB rounded to the nearest integer Recursive squaring log n time This method is unreliable since oatingpoint arithmetic is prone to roundoff errors Bottomup onerdimensional dynamic programming Compute F0 F1 F2 Fn in order forming each number by summing the two previous Running time n 25mg CS lfAnalym army th Convex Hull n Convex Hull Divide amp Conquer Given a set ofpins on apinhoard com zgfsmg 5 quot 1quot mints by X39 And ambb band mm 111quotquot Divide the set ofpoints into two How does the rubber band look sets A and B when it snaps tight A contains the le Lnzj points B contains the right Fn points A B We represent convex hull as the Recursiver compute the convex sequence ofpoints on the convex ull of A hu11 polygon in counterclockwise Recursively compute me convex order hu11 ofB Merge the two convex hu11s 2st summon qulgamfmi s 2st minimums 12214197 th 5 1quot Merging Find upper and lower mngenl with those tangents the convex hull ofAUB can be computed from the convex hu11s ofA and the convex hull whilemnot 1owertangent to both convex hu11s ofA and B do of B In 001 linear time while T not 1owertangent to c hu11 ofA do right turn or le tum swamm 12214197 th quot 2st summon qulgamfmi 7 2st 5 Convex Hull Runtime Preprocessing son the points by x coordinate On log n Just once Divide the set ofpoints into two setsAandB 01 A contains the le Lnzj points B contains the right Fnz l points Recursiver compute the convex Tn2 hu11 ofA Recursiver compute the convex hull ofB Tn2 Merge the two convex hu11s On 2529 unwanmmgmm 9 inn Convex Hull Runtime Ruttime Recurrence Tn 2 Tn2 on Solves to Tn n log n 2st minimumx 11234197 th m Matrix multiplication Input A ayB IL1 I I Output C Cy AB 1 1 2 n F511 512 Cm ran an am l bu biz bin all an 52 7 a21 aZZ an I bu bzz bZnJ Cm an Cm am anZ am bu an bm 7L cl Zalk bg 161 2st CX AIzlym momm u 1 Standard algorithm forielton doforjelton do CVEO forkelton do cyecya kbkj Running time n3 2st minimumx 11234197 th i2 quot Divideandconquer algorithm IDEA nxn matrix 2x2 matrix of n2xn2 submatrices alu b CA39B warmm afAlgamPmr Strassen s idea Multiply 2x2 matrices with onl P1alfih r P5P4PZP6 P2ablh s P1PZ P3cde t P3P4 P4dlg e P5P1 P3 P7 P5adleh P6 b 7 d39g h 7Nrriult1s18 aiddssubs P7aic39ef oe ore anceon commutativity of mult 2st warmm afAlgamPmr Analysis of DampC algorithm TM submatrices Work adding submatrices submatrix size 141 n1 g28 n3 5 CASE 1 5 Tn n3 N0 better than the ordinary algorithm 2st minimum 11234197 th Strassen s idea Multiply 2x2 matrices with only 7 recursive mults Pl afih r P5P4 P2P6 P2abh adeh P3 0de dgeabh P4dg7e bidgh P5adeh aeahdedh P6bidgh dgideiahibh P7a7cef bgbhidgidh ae 2st minimum 11234197 th CS 3343 Spring 2009 ALGDRlTHMS Minimum Spanning Trees Carola Wenk Slides courtesy of Charles Leiserson with changes and additions by Carola Wenk NQDQ c iidemlym 17 ng NE EJ Minimum spanning trees Input A connected undirected graph G V E with weight function w E gt For simplicity assume that all edge weights are distinct CLRS covers the general case Output A spanning tree T7 a tree that connects all vertices i ofminimum weig t wT Zwuv mar Mme mammal army th 2 0 Example of MST NQDQ c iidemlym 17 ng Hallmark for greedy algorithms Greedychoice property A locally optimal choice is globally optimal Theorem Let The the MST ofG V E andletA g V Suppose that u v E E is the leastweight edge connectingA to VA Then u v E T Mme mammal army th a 6 Proof of theorem Proof Suppose u v E T Cut and paste T o e A u v leastWeight edge E VA connecting1 to VA 45409 Cx AmlymafAlgamhm s Proof of theorem Proof Suppose u v e T Cut and paste T o E A u v leastWeight edge E VA connecting110 VA Consider the unique simple path from u to v in T AVE03 minimum 11234197 th 5 Proof of theorem Proof Suppose u v e T Cut and paste T o e A u v leastWeight edge E VA connecting110 VA Consider the unique simple path from u to v in T Swap u v with the rst edge on this path that connects a vertex inA to a vertex in VA 4mg assummtym mtg th 7 Proof of theorem Proof Suppose u v e T Cut and paste T o e A u v leastWeight edge E VA connecting110 VA Consider the unique simple path from u to v in T Swap u v with the rst edge on this path that connects a vertex inA to a vertex in VA A lighterweight spanning tree than T results I AVE03 minimum 11234197 th A A Ea Prim s algorithm IDEA Maintain VIA as a priority queue Q Key each vertex in Q with the weight of the least weight edge connecting it to a vertex in ll 339quot 4quot hr r k2yv ewforallv a V dc k2ys e 0 for some arbitrary s a V while Q do u e EXTRACTMNQ for each v a Adju do in a Q and w14v lt k2yv then k2yv e wu v gt DECREASEKEY At the end v 7rv forms the MST f Amlym mtg th 9 u e EXTRACTVMIMQ fur each v e Ariu dn ifv s Q andwu v lt 1wyv 1wyv e wu wgt DECREASEKEY e u we 7 V minimum 12224197 th m u e EXTRACTVMIMQ fur each v e Ariu dnil39v s Q andwhr v lt1wyv than kzyv e wu vgt DECREASEKEY we WM e u eaacmmm u Example of Prim s algorithm SEA oeVA u e EXTRACTVMIMQ fur each v e Ariu dn ifv s Q andwu v lt 1wyv 1wyv e wu vgt DECREASEKEY was quotM e quot minimum 12224197 th 12 a Example of Prim s algorithm h e EXTRACTVMIMQ fur each v e Adh an il39v s Q ahdwh v lt 1wyv 1wyv e wh vgt DECREASEKEY m quotM a quot CS AmlymaAlgamhm 13 h e EXTRACTVMIMQ fur each v e Adh an ifv s Q ahdwh v lt 1wyv 1wyv e wh wgt DECREASEKEY h e EXTRACTVMIMQ fur each v e Adh dnil39v s Q ahdwgh v own then 1wyv e wh vgt DECREASEKEY W WM a u mammhgm u he NIH mamame h 5 Example of Prim s algorithm SEA oeVA h e EXTRACTVMIMQ fur each v e Adh dn ifv s Q ahdwh v lt 1wyv 1wyve wh vgt DECREASEKEY was quotM e quot minimum 11234197 th 15 fa Example of Prim s algorithm 6 12 h e EXTRACTVMIMQ fur each v e Adh an il39v s Q ahdwh v lt 1wyv 1wyv e wh vgt DECREASEKEY m quotM a quot CS AmlymaAlgamhm 17 h e EXTRACTVMIMQ fur each v e Adh dn ifv s Q ahdwh v lt 1wyv 1wyve wh wgt DECREASEKEY was 7 V a quot magnumAte th u h e EXTRACTVMIMQ fur each v e Adh 39 dnil39v s Q ahdwgh v own then 1wyv e wh vgt DECREASEKEY W WM a u mammhgm w Example of Prim s algorithm SEA 6 6 oeVA 9Q i 5 a 7 14 739 15 a 8 or a 3 10 u e EXTRACTVMIMQ 8 d hr each v e A Ah flu in e Q andwh v lt 1wyv kzyv lt wu vgt DECREASEKEY was 1 e quot minimum 11234197 th 20 SEquot Example of Prim s algorithm u e EXTRACTVMIMQ fur each v e Ariu an il39v s Q andwu v lt 12yv 12yv e W01 vgt DECREASEKEY m quotM E quot mmmm 112341517 th 1 06A EVA AVE03 u e EXTRACTVMIMQ In each v e Ariu an in s Q andwu v lt1wyv 1wyv e Wei 01gt DECREASEKEY quotM E quot CS AmlymaAlgrmhm w Example of Prim s algorithm Analysis of Prim Q e V kgyv e w for all v e V total keyv e 0 for some arbitrary r e V while Q Q do u e EXTRACTMINQ for eachv e Adju do if v e Q and wu v lt keyv then keyv e wu v 7rv e u V times 013373301 times Handshaking Lemma gt LE implicit DECREASEKEYS Time garD39TEXTRACTVMIN iEDITDECREASEVKEY Analysis of Prim continued Time iVD39TEXTRACT7MIN iEiITDECREASEVKEY AVE03 4mg assummtym mtg th minimum 11234197 th Q TEXTRACTVMIN TDECREASE39KEY Tomi arr y OOVD 01 OOVV bin 1162 ang M ang M Fibonacci OlogVD 01 OQE V logWD heap am rtize rtize worst case LL Kruskal s algorithm IDEA again greedy Repeatedly pick edge With smallest Weight as long as it does not form a cycle I The algorithm creates a set oftrees a forest I During the algorithm the added edges merge the trees together such that in the end only one tree remains I The correctness of this greedy strategy is not obvious and needs to be proven Proof skipped here mmmtm mtg th 25 m Example of Kr39uskal s algorithm s eh9d2 12 mngn MST edges 1 set repr Every node is a single tree AVE03 mmmtm 11234197 th Example of Kr39uskal s algorithm s eh2df 12 g 211 MST edges 1 set repr Edge 3 merged two singleton trees mmmtm mtg th Example of Kr39uskal s algorithm S 2df g 12 g h E 0 MST edges 1 set repr AVE03 mmmtm 11234197 th 5 0 Exalnple of Kruskal s algorithm ktmmsxg in MEkd MST edges 1 set repr 4mg assummtym mtg th 29 twin Exalnple of Kruskal s algorithm kugxg 12QMJQEQQ MST edges 1 set repr AVE03 minimum 11234197 th 30 Exalnple of Kruskal s algorithm Sdg 12 ghabcf MST edges 1 set repr Edge 8 merged the two bigger trees 4mg assummtym mtg th 1 Exalnple of Kruskal s algorithm 1 S g 12 ghabcfd 9 MST edges 1 set repr AVE03 minimum 11234197 th 2 Exalnple of Kruskal s algorithm S g 2habcfd 9 M Exalnple of Kruskal s algorithm S g 1 g h a b c f d MST edges 12 1 set repr MST edges 1 set repr Skip edge 10 as it would cause a cycle Skip edge 12 as it would cause a cycle m9 Cx AmlqufAlgamhm 33 Ala09 magnumMm 4 Exalnple of Kruskal s algorithm S g 2 h a b c f d MST edges 12 1 set repr Exalnple of Kruskal s algorithm p Se h a b c f d g MST edges 1 set repr Skip edge 14 as it would cause a cycle 4mg assummtym 112341517 th as AVE03 mmmtm 11234197 th 35 AM OK i r ri39M39s CS 3343 Fall 2007 L quot pv ALGORITHMS Redblack trees Carola Wenk Slides courtesy of Charles Leiserson with small changes by Carola Wenk 101107 CS 334Analysis ofAlgorithms Search Trees A binary search tree is a binary tree Each node stores a key The tree fulfills the binary search tree property For every node x holds 0 leftpcjs x if x s left child eftx exists 0 x S rightx if x s right child rightx exists 101107 CS 334 Analysis ofAlgorilhms AlGORi39IHMs Search Trees Vr 12 Different variants of search trees Balanced search trees guarantee height of log n for n elements k ary search trees such as Btrees 23trees Search trees that store the keys only in the leaves and store additional splitvalues in the internal nodes 101107 CS 334 Analysis ofAlgorilhms 3 Vr 12 101107 CS 334 Analysis ofAlgorilhms 39 ADT Dictionary Dynamic Set Abstract data type ADT Dictionary also called Dynamic Set A data structure which supports operations Insert Delete Find Using balanced binary search we can implement a dictionary data structure such that each operation takes 0log n time 7 Balanced search trees Balanced search tree A searchtree data structure for which a height of 010g n is guaranteed when implementing a dynamic set of n items AVL trees 23 trees Examples 234 trees Btrees 101107 CS 334 Analysis ofAlgorilhms N Redblack trees T 33939 This data structure requires an extra one bit color eld in each node Redblack properties Every node is either red or black The root is black The leaves NIL s are black If a node is red then both its children are black All simple paths from any node x excluding x to a descendant leaf have the same number of black nodes blackheightx 101107 CS 334 Analysis ofAlgorilhms 6 01ngi 101107 CS 334 Analysis ofAlgorilhms NIL NIL NIL NIL NIL NIL 1 Every node is either red or black 101107 CS 334 Analysis ofAlgorilhms NIL NIL NIL NIL NIL NIL 2 3 The root and leaves NIL S are black 101107 CS 334 Analysis ofAlgorilhms NIL IL NIL IL NIL IL 4 If a node is red then both its children are black 101107 CS 334 Analysis ofAlgorilhms 10 bh 0 NIL NIL NIL NIL NIL NIL 5 All simple paths from any node x excluding x to a descendant leaf have the same number of black nodes blackheightx 101107 CS 334 Analysis ofAlgorilhms ll Height of a redblack tree Theorem A redblack tree with n keys has height h s 210gn 1 Proof The book uses induction Read carefully INTUITION Merge red nodes into their black parents 101107 CS 334 Analysis ofAlgorilhms 12 Height of a redblack tree Theorem A redblack tree with n keys has height h s 210gn 1 Proof The book uses induction Read carefully INTUITION Merge red nodes into their black parents 101107 CS 334 Analysis ofAlgorilhms 13 Height of a redblack tree Theorem A redblack tree with n keys has height h s 210gn 1 Proof The book uses induction Read carefully INTUITION Merge red nodes into their black parents 101107 CS 334 Analysis ofAlgorilhms 14 Height of a redblack tree Theorem A redblack tree with n keys has height h s 210gn 1 Proof The book uses induction Read carefully INTUITION Merge red nodes into their black parents 101107 CS 334 Analysis ofAlgorilhms 15 Height of a redblack tree Theorem A redblack tree with n keys has height h s 210gn 1 Proof The book uses induction Read carefully INTUITION Merge red nodes into their black parents 101107 CS 334 Analysis ofAlgorilhms 16 Height of a redblack tree Theorem A redblack tree with n keys has height h s 2 logn 1 Proof The book uses induction Read carefully INTUITION I Merge red nodes into their black parents This process produces a tree in which each node has 2 3 or 4 children The 234 tree has uniform depth h39 of leaves 101107 CS 334 Analysis ofAlgorilhms l7 i s Proof continued We have h39 2 h2 since at most half the leaves on any path are red The number of leaves in each tree is n 1 gt n 1 2 2 gt logn 1 2 h39 2 h2 2h3210gn1 101107 CS 334 Analysis ofAlgorilhms 18 101107 e Query operations Corollary The queries SEARCH MIN MAX SUCCESSOR and PREDECESSOR all run in 0log n time on a redblack tree with n nodes NIL NlL NIL NIL NIL NIL CS 334 Analysis ofAlgorilhms 19 101107 Modifying operations The operations INSERT and DELETE cause modi cations to the redblack tree 1 the operation itself 2 3 color changes restructuring the links of the tree Via rotations CS 334 Analysis ofAlgorilhms 20 ALGo Iii rH Ms Rotations maintain the inorder ordering of keys ae ocbe Bcey gt aSASbSBSc Rotations maintain the binary search tree property 39 A rotation can be performed in 01 time 101107 CS 334 Analysis ofAlgorilhms 21 N Redblack trees T 33939 This data structure requires an extra one bit color eld in each node Redblack properties Every node is either red or black The root is black The leaves NIL s are black If a node is red then both its children are black All simple paths from any node x excluding x to a descendant leaf have the same number of black nodes blackheightx 101107 CS 334 Analysis ofAlgorilhms 22 01ngi i sv N Insertion into a redblack tree Vr 33939 IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the Violation up the tree by recoloring until it can be xed with rotations and recoloring 101107 CS 334 Analysis ofAlgorilhms 23 439 N Insertion into a redblack tree Vr 12 IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the Violation up the tree by recoloring until it can be xed with rotations and recoloring Example Insert x 15 72 3H 24 101107 CS 334 Analysis ofAlgorilhms 24 439 N Insertion into a redblack tree Vr 12 IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the Violation up the tree by recoloring until it can be xed with rotations and recoloring Example Insert x 15 72 3H 24 101107 CS 334 Analysis ofAlgorilhms 25 439 N Insertion into a redblack tree Vr 12 IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the violation up the tree by recoloring until it can be xed with rotations and recoloring Example Insertx15 Recolor moving th violation u the tree 101107 CS 334 Analysis ofAlgorilhms 26 439 N Insertion into a redblack tree Vr 12 IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the violation up the tree by recoloring until it can be xed with rotations and recoloring Example Insertx15 Recolor moving th violation u the tree 101107 CS 334 Analysis ofAlgorilhms 27 V V 12 Insertion into a redblack tree IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the violation up the tree by recoloring until it can be xed with rotations and recoloring Example k Insert x 15 Recolor moving th violation up the tree RIGHTROTATE 18 r H3 quot l l 101107 CS 334 Analysis ofAlgorilhms 28 V V 12 Insertion into a redblack tree IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the violation up the tree by recoloring until it can be xed with rotations and recoloring Example Insert x 15 Recolor moving the violation up the tre RIGHTROTATE 18 l lquot quot Hes 101107 CS 334 Analysis ofAlgorilhms 29 V V 12 101107 Insertion into a redblack tree IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the violation up the tree by recoloring until it can be xed with rotations and recoloring Example Insert x 15 Recolor moving the violation up the tree RIGHTROTATE18 LEFTROTATE7 CS 334 Analysis ofAlgorilhms 30 V V 12 Insertion into a redblack tree IDEA Insert x in tree Color x red Only red black property 4 might be violated Move the violation up the tree by recoloring until it can be xed with rotations and recoloring Example Insert x 15 Recolor moving the violation up the tree R1GHTR0TATE18 LEFTROTATE7 101107 CS 334 Analysis ofAlgorilhms 31 A1 0mm HMS Pse d de V RBINSERTT x TREEINSERTT x colorx lt RED D only RB property 4 can be violated while x 7 r00ZT and colorpx RED do mm leftppx then y lt righlppx D y auntuncle of x if colory RED then Case 1 else if x rightpx then Case 2 D Case 2 falls into Case 3 Case 3 else then clause with left and right swapped colorr00l lt BLACK 101107 CS 334 Analysis ofAlgorilhms 32 1 N Graphical notation Vr a2 Let denote a subtree with a black root A11 s have the same blackheight 101107 CS 334 Analysis ofAlgorilhms 33 A1 GORi39I HMS Or A s children are swapped Push C s black onto A and D and recurse pm le mmxu since C s parent may be y righlmmx red colory RED 101107 CS 334 Analysis ofAlgorilhms 34 Px 16 PPx y righlmmx colory BLACK x righlpx 101107 CS 334 Analysis ofAlgorilhms 35 Transform to Case 3 PM le PPx y righlmmx colory BLACK x leflPx 101107 J and recolor R1GHTR0TATEC Done No more Violations of RB property 4 are possible CS 334 Analysis ofAlgorilhms 36 6 N Analysis Vr 12 Go up the tree performing Case 1 which only recolors nodes If Case 2 or Case 3 occurs perform 1 or 2 rotations and terminate Running time 0log n with 01 rotations RBDELETE same asymptotic running time and number of rotations as RBINSERT see textbook 101107 CS 334 Analysis ofAlgorilhms 37 A1 0mm HMS Pseudocode part 11 else then clause with left and right swapped gtpx righrmmxn then y lt leftppx D y auntuncle of x if colory RED then Case 1 else ifx leftpx then Case 2 D Case 2 falls into Case 3 Case 3 colorr00l7 lt BLACK 101107 CS 334 Analysis ofAlgorilhms 38 A1 GORi39I HMS Or A s children are swapped Push C s black onto A and D and recurse pm righlwmxn since C s parent may be y 1 PPx red colory RED 101107 CS 334 Analysis ofAlgorilhms 39 PM W39th PM y 16fZ PPx colory BLACK x leflPx 101107 CS 334 Analysis ofAlgorilhms 40 Transform to Case 3

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