1 (12 points) Red-Black trees and Red-Purple trees

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1 CS6 Hoework 3 Due: 29 April 206, 2 oo Subit o Gradescope Haded out: 22 April 206 Istructios: Please aswer the followig questios to the best of your ability. If you are asked to desig a algorith, please describe it clearly ad cocisely, prove its correctess, ad aalyze its ruig tie. If you are asked to show your work, please iclude relevat calculatios for derivig your aswer. If you are asked to explai your aswer, give a short setece) ituitive descriptio of your aswer. If you are asked to prove a result, please write a coplete proof at the level of detail ad rigor expected i prior CS Theory classes i.e. 03). Whe writig proofs, please strive for clarity ad brevity i that order). Cite ay sources you referece. 2 poits) Red-Black trees ad Red-Purple trees It s Barr the Bear s birthday! To celebrate, he decides to save the plaet... by platig trees! I particular, he wats to plat Red-Black trees RB trees) i his garde. Ufortuately, Barr the Bear s birthday coicides with Prak Day celebrated by the peguis, ad Poe the Pegui has bee up to soe ischief. He praks the RB trees ito Red-Purple trees RP trees) by alterig the structure of the trees i soe way. a) 3 pts) If you believe a tree is a valid RB tree, provide a valid colorig for the tree to covice Barr to keep it. Otherwise, cocisely prove why there is o valid colorig to covice Barr to destroy it. b) 3 pts) If you believe a tree is a valid RB tree, provide a valid colorig for the tree to covice Barr to keep it. Otherwise, cocisely prove why there is o valid colorig to covice Barr to destroy it.

2 c) After reovig the RP trees, Barr ow wats to upgrade his RB tree show below by isertig the uber 6 ito it. Show how this ca be doe such that after the isertio it reais to be a valid RB tree. For full-credit, you should show iterediate steps i.e., by showig how the tree is beig trasfored ito the fial oe) by drawig a few trees poits) The followig sequece of operatios are perfored whe we isert 6. ) Isert a ew red ode with 6 as key as a left child of the ode with 200 as key. 2) Right-rotate aroud the ode with 200 as key. 3) Re-color the ode with 6 50, resp.) as key to black red, res.). 4) Left-rotate aroud the ode with 50 as key. 5) This results i a valid RB-tree. To receive a full credit, you eed to show all iterediate steps by drawig the trees with proper keys ad colors; steps 3-4 ca be erged ito oe tree, though). The trees are show o the last page of this solutio due to space). 2 4 poits) Terary Search Trees o 2D Plae I this proble, we will be talkig about terary search trees TST). Terary search trees are siilar to biary search trees but rather tha havig just 2 poiters left ad right), they have 3 poiters left, iddle, ad right). A TST T is a terary tree i which each ode cotais a poit of the for x, y) for soe x, y Z. Moreover, o two poits i T ca share the sae x value ad o two poits ca share the sae y value. The the followig properties hold for every ode u with key x, y) i ay TST: Ay poit x L, y L ) i the left-subtree of u has x L < x ad y L < y. Ay poit x M, y M ) i the iddle-subtree of u has x M > x ad y M < y. Ay poit x R, y R ) i the right-subtree of u has x R > x ad y R > y. Clarificatio: You ca assue that for ay ode v i TST T, you ca copute the uber of odes that belog to the subtree of v i tie O) e.g., it is stored as part of TST). a) Prove or disprove: For ay set of distict poits where o two poits share the sae x-coordiate or y-coordiate), there exists a terary search tree T o these poits with ht ) = Of)): i) whe f) =, 2

3 ii) whe f) = log. To prove the clai, provide a cocise proof. To disprove the clai, provide a couterexaple. 3 poits) Stateet i) is true ad stateet ii) is false. 2 poits) To prove stateet i): Let S be ay set of poits. Let x, y ) be the poit i S with sallest x-coordiate. We ake the root ode cotaiig x, y ), ad let S R = {x, y) S : y > y } ad S M = {x, y) S : y < y }. Notice that {x, y )} S L S R = S ad they partitio S. We recursively build a subtree o S R ad ake it the right-subtree of the root ad also build a subtree o S M ad ake it the iddle-subtree of the root. Sice S =, the height of this tree is trivially O). To see this is a valid TST, by our costructio of S M of S R, resp.) all poits i the iddle-subtree right-subtree, resp.) satisfy the properties of TST at the root ode. I fact this stateet reais true at every ode i TST we build oe ca prove this by iductio o the uber of odes; we oit the details). Note that provig this clai for soe exaple tree would ot receive full credit. Also, provig by iductio o that oe ca isert a ew poit to a arbitrary TST is wrog, as the exaple for 2-a-ii shows.) poit) To disprove stateet ii): Cosider poits give by {i, i) : i [, ]}; that is, x-coordiates are icreasig i i ) ad y-coordiates are decreasig. Due to the properties of TST, it is easy to see that, ) ust be the root of ay valid TST that cotais these poits otherwise, if soe other poit i, i) is the root, the, ) caot belog to ay of its subtrees). All the other poits ow ust belog to the iddle subtree of, ). Repeatig this arguet, we ed up with a TST of height i which each ode that cotais i, i) has i, i )) as its paret uless i = ) ad has i +, i + )) as its iddle-child uless i = ). Note that disprovig this clai for soe exaple tree would ot receive full credit.) b) Suppose that you are give a poit x, y ) ad a TST T which cotais poits ad has height h. You wish to deterie whether x, y ) is cotaied i T or ot. Desig a efficiet algorith to solve this proble. Prove the correctess of the algorith. Aalyze its ruig tie i ters of ad also i ters of h. 4 poits) poit) Algorith: The followig recursive algorith works, give a ode of TST we would call SearchT.root)). v L, v M, v R deote the childre of v if they do ot exist, they are NILs). Algorith Searchv) : If [v = NIL] the retur FALSE 2: If [x, y) = x, y )] the retur TRUE 3: If [x < x) y < y)] the retur Searchv L ) 4: If [x > x) y < y)] the retur Searchv M ) 5: If [x > x) y > y)] the retur Searchv R ) 6: retur FALSE At this poit, x = x or y = y or x < x) y > y). Let 2 poits) Correctess: First, it is clear that if this algorith fids x, y ), the it does exist i TST due to Lie 2). We ow prove that if x, y ) exists i TST, the this algorith fids it. Together, they prove a if-ad-oly-if stateet). 3

4 We prove by iductio o the uber of poits i TST, assuig that x, y ) exists i T. If TST cotais oly oe ode, the the root ode ust cotai x, y ), ad thus Lie 2 will be correctly executed. Now assue that TST cotais poits. If the root ode of TST cotais x, y ), the Lie 2 will be executed correctly. Otherwise, oe of the descedats of the root ode ust cotai x, y ). Sice all poits i TST have distict x-coordiates, oly oe of the sub-trees of the root ode ca ad ust) cotai x, y ). Due to the properties of TST, x, y ) ust satisfy oe of the coditios listed i Lies 3-5; the three cases are pairwise exclusive, ad if x, y ) satisfies oe of the, the it caot belog to ay subtrees of the root, which is a cotradictio. Hece by iductive hypothesis whe we call our recursive algorith o oe of the child of the root, it will correctly fid x, y ) sice it subtree cotais at ost poits). Note: There are other ways to prove correctess.) poit) Ruig tie: I the worst case this algorith visits every ode, ad its ruig tie is O) a exaple of such TST is the oe fro previous part). I ters of h, this algorith traverses a path fro the root ode to a leaf ode i the worst case, ad thus we have Oh). c) Now, suppose that give a TST T o odes ad height h ad a poit x, y ), you wish to deterie the uber of poits x, y) i the tree such that x x ad y y. Give a recursive algorith that searches for these poits, startig fro the root of T. At each recursive step at a ode u of T, how ay childre of u does your algorith eed to recurse o i the worst case? Aalyze the worst case) rutie Uh) of your algorith i ters of h. What is the worst case) ruig tie i ters of whe h ca be arbitrary)? What would the ruig tie be i ters of if the tree is copletely balaced ad h = log 3? 7 poits) 3 poits) Algorith: The followig recursive algorith works, give a ode of TST we would call CoutT.root)). Let v L, v M, v R deote the childre of v if they do ot exist, they are NILs). Let cv) be the uber of odes belogig to the subtree that is rooted at v. Algorith 2 Coutv) : If [v = NIL] the retur 0 2: x, y) poit i v 3: If [x, y) = x, y )] the z 4: Else z 0 5: If [x x) y y)] the retur Coutv L ) + Coutv R ) + cv M ) + z 6: If [x x) y < y)] the retur Coutv L ) + Coutv M ) + z 7: If [x > x) y y)] the retur Coutv M ) + Coutv R ) + z 8: If [x > x) y < y)] the retur Coutv M ) + z poit) I the worst case the algorith recurses o at ost 2 childre of ay ode. Note: If your algorith recurses o at ost 3 childre i the worst case, you lose 3 poits for the algorith ad poit for this questio. You also receive 0 poits for part d).) poit) Rutie Uh) is O2 h ) because the brachig factor at each ode is 2 i.e., we recurse o at ost two odes fro each ode), ad the height of tree is h. Usig a recurrece relatio, we ca write it as Uh) 2Uh ) + O) i the worst case, both subtrees ca have height h ), which leads to Uh) = O2 h ).) poit) Rutie is O) if h ca be arbitrary because i the worst case we ust visit every ode). 4

5 poit) If the tree is copletely balaced, the we have h = log 3, ad the worst-case rutie is O2 h ) = O2 log 3 ) = O log 3 2 ) which is asyptotically better tha O), ad gives us a better upper boud i ters of ). d) Bous: 4 poits) Show that your ruig tie Uh) above is tight by providig a exaple tree ad iput poit that cause the algorith to use at least ΩUh)) tie. To show the tight boud, you ust show that for arbitrary there exists a valid TST o poits ad a poit x, y ) you ca choose your TST ad its poits as well as x, y )) such that the algorith i Part c) rus i ΩUh)) = Ω2 h ) tie. Ituitively we wat to have a TST such that at each o-leaf) ode our algorith recurses o two odes. There are three cases out of four) i which the algorith akes two recursive calls i our solutio, Lies 4-6). You ca pick ay, but choosig Lie 4 would be easiest choosig Lie 5 or Lie 6 ay lead to a uch ore coplicated way of costructig a exaple). Let us create a TST o odes as follows. Let x i, y i ) = i, i) for all i []. Notice that o two poits share the sae x or y coordiates. Let x /2, y /2 ) be the poit cotaied i the root ode of our TST if is ot eve, roudig up or dow does ot atter). Clearly, the poits with saller idex tha /2 ust all belog to the left-subtree of the root, ad the others to the right-subtree of the root; they are of equal sizes sice we picked the iddle oe, they differ by at ost ), ad we recursively choose the iddle oe as the root of each subtree, ad so o. This defies our TST. Note that it satisfies all three coditios o x, y-coordiates of poits i subtrees of each ode ad the coditio that o two poits ca share the sae x or y coordiate. The height of the TST is h = log. For the poit beig queried, we choose x, y ) = 0, + ) ay x < ad y > works, too). This esures that Lie 4 is executed at every o-leaf ode. Thus the tie that the algorith takes o this poit is Ω) Ω2 h ). Note: If your exaple does NOT lead to Ω2 h ) rutie for the correct algorith for Part c), the you get 0 poits for this part i particular, if your algorith for Part c) is wrog, the you are likely to receive o credit for this part). The rubric icluded i the solutio we posted earlier is out of date. Please refer to Gradescope s rubric istead.) 3 2 poits) Hash Table Cosider a hash table with buckets ad eleets hashed ito it. We will show that with high probability, the size of the largest bucket is goig to be very sall. We will assue that the hash fuctio beig cosidered satisfies the siple uifor hashig property. That is, the probability that a eleet is iserted ito ay of the buckets is / ad idepedet of the eleets already there. a) Let b i be the uber of eleets i the i-th bucket. The b i follows a bioial distributio. What are the paraeters of this distributio? What is E[b i ]? 4 poits) b i Bio, ), E[b i] =. b) Show that li Prax i b i l ) = 0. I other words, for sufficietly large, we ca be assured that with high probability, oe of the buckets will have ore tha l eleets hashed to the. You ay fid the followig iequalities helpful: The Cheroff Boud: If X is a bioially ) distributed rado variable with E[X] = µ, the for ay µ. e δ > 0, PrX + δ)µ) < δ +δ) +δ 5

6 The Uio Boud: Give evets A, A 2,... A, we have Pr i= A i) i= PrA i). I aother words, the probability of at least oe of the evets occurrig is at ost the total su of the idividual evet probabilities. 8 poits) Substitutig δ = l, ad µ = ito the Cheroff boud, we get Prb i l ) < e δ + δ) +δ = e e +δ + δ) +δ = e e l )l = e l ) l = e expl l ) l = e = e = e el l l l l l l ) Usig the uio boud, we see that ax i Prb i l ) Prb i l ) e 2 l l, which goes to 0 as goes to. Note: Oe coo error is to apply the Uio boud o the quatities li Prb i l )). 4 6 poits) Hash Table: Siple Uifor Hashig Poe the Pegui uderstads ad likes hash tables. Barr the Bear uderstads hash tables but does ot like chaiig. Help Barr choose the size of his hash table such that he would ot eed to worry about chaiig. Let be the uber of ites that Barr has ad be the size of his hash table to be deteried. We assue is sufficietly large ad is uch larger tha. Assue Barr s hash fuctio satisfies the siple uifor hashig property. a) What is the probability that a particular slot is epty after hashig ites? Derive a lower boud i the for of c 0 + c where c 0 ad c are costats. Hit: Take the two ost sigificat ters i the bioial expasio x + y) = j=0 j) x j y j. 3 poits) We choose c 0 = ad c =. The probability that a particular slot is epty is exactly ). We ca lower boud this as follows: ) = i=0 i ) = > ) i 2 ) ) ) ) sice the last ter i previous lie is positive) 6

7 ) ) i To obtai the iequality above, we eed to show that i=2 2 i > 0. It is sufficiet to prove the i 2 iequality ) i > i+) for i [2, ] because each positive ter would be followed by a egative ter saller tha it). ) ) i i + = = = ) i ) i i + ) ) i + ) i) i i + ) ) i + i)) > 0. i i + ) Note that >, ad therefore i) = + i > 0.) b) What is the probability that a particular slot has exactly oe ite? Derive a lower boud i the for of f) + f2) where f 2 ad f 2 are polyoial fuctios i. 3 poits) We choose f ) = ad f 2 ) = ). The probability that a particular slot has oe ite is exactly. ) Fro Part a), we kow that ) > whe > ). Hece, we ca lower boud this probability as follows: ) > ) = ) 2. c) Derive a upper boud o the probability that there exists a slot with at least two ites. Use the bouds fro Parts a) - b). 5 poits) The probability that a particular slot has at least two ites is ius the probability that it has either zero or oe ite. By Parts a) ad b), this is at ost ) ) ) ) 2 = 2. By the uio boud, the probability that there exists a slot out of slots) with at least two ites is at ost ). d) Let ɛ > 0 be give. Usig Part c), deterie i ters of ad ɛ, such that the probability that all slots have at ost oe ite o chaiig!) is at least / ɛ. Note: should be i the for gɛ) where g is a liear fuctio i ɛ. 5 poits) We choose = 2+ɛ or, gɛ) = 2 + ɛ). The, the probability that there exists a slot with at least two ites is at ost ) = 2 < 2 2+ɛ = / ɛ. The probability that all slots have at ost oe ite is 2+ɛ the at least / ɛ. 7

8 Trees for Q-c