66 Lecture 3 Random Search Tree i unique. Lemma 3. Let X and Y be totally ordered et, and let be a function aigning a ditinct riority in Y to each ele

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1 Lecture 3 Random Search Tree In thi lecture we will decribe a very imle robabilitic data tructure that allow inert, delete, and memberhi tet (among other oeration) in exected logarithmic time. Thee reult were rt obtained by Pugh in 988 (ee [88]), who called hi robabilitic data tructure ki lit. We will follow the reentation of Aragon and Seidel [7], whoe data tructure i omewhat dierent and more cloely related to the elf-adjuting tree reented in the lat lecture, and whoe robabilitic analyi i articularly elegant. 3. Trea Conider a binary tree, not necearily balanced, with node drawn from a totally ordered et, ordered in inorder; that i, if i i in the left ubtree of k and j i in the right ubtree of k, then i < k < j. Recall that the rotate oeration dicued in the reviou lecture reerve thi order. Now uoe that each element k ha a unique riority (k) drawn from ome other totally ordered et, and that the element are ordered in hea order according to riority; that i, an element of maximum riority in any ubtree i found at the root of that ubtree. A tree in which the data value k are ordered in inorder and the rioritie (k) are ordered in hea order i called a trea (for tree-hea, one uoe). It may not be obviou at rt that trea alway exit for every riority aignment. They do! Moreover, if the rioritie are ditinct, then the trea 65

2 66 Lecture 3 Random Search Tree i unique. Lemma 3. Let X and Y be totally ordered et, and let be a function aigning a ditinct riority in Y to each element of X. Then there exit a unique trea with node X and rioritie. Proof. Let k be the unique element of X of maximum riority; thi mut be the root. Partition the remaining element into two et fi 2 X j i<kg; fi 2 X j i>kg : Inductively build the unique trea out of thee two et and make them the left and right ubtree of k, reectively Random Trea A random trea i a trea in which the rioritie have been aigned randomly. Thi i bet done in ractice by calling a random number generator each time a new element m i reented for inertion into the trea to aign a random riority tom. Under ome highly idealized but reaonable aumtion about the random number generator 3, two element receive the ame riority with robability zero, and if all element in the trea are orted by riority, then every ermutation i equally likely. When a new element m i reented for inertion or to tet memberhi, we tart at the root and work our way down ome ath in the trea, comaring m to element along the ath to ee which way to go to nd m' aroriate inorder oition. If we ee m on the ath on the way down, we can anwer the memberhi query armatively. If we make it all the way down without eeing m, we can anwer the memberhi query negatively. If m i to be inerted, we attach m a a new leaf in it aroriate inorder oition. At that oint we call the random number generator to aign a random riority (m), which by Lemma 3. ecie a unique oition in the trea. We then rotate m uward a long a it riority i greater than that of it arent, or until m become the root. At that oint the tree i in hea order with reect to the rioritie and in inorder with reect to the data value. To delete m, we rt nd m by earching down from the root a decribed above, then rotate m down until it i a leaf, taking care to chooe the direction 3 A call to the random number generator give a uniformly ditributed random real number in the interval [0; ), and ucceive call are tatitically indeendent; i.e. if x ;:::;xn are the reult of n ucceive call, then ^ Pr( xi 2 Ai) = in Y Pr(xi 2 Ai) : in

3 Lecture 3 Random Search Tree 67 of rotation o a to maintain hea order. For examle, if the children of m are j and k and (j) > (k), then we rotate m down in the direction of j, ince the rotate oeration will make j an ancetor of k. When m become a leaf, we rune it o. The beauty of thi aroach i that the oition of any element in the trea i determined once and for all at the time it i inerted, and it tay ut at that level until it i deleted; there i not a lot of retructuring going on a with lay tree. Moreover, a we will how below, the exected number of rotation for an inertion or deletion i at mot two. 3.3 Analyi We now how that, averaged over all random riority aignment, the exected time for any inert, memberhi tet, or delete i O(log n). We will do the analyi for delete only; it i not hard to ee that the time bound for memberhi tet and inert i roortionally no wore than for delete. Suoe that at the moment, the trea contain n data item (without lo of generality, ay f; 2;:::;ng), and we wih to delete m. The rioritie have been choen randomly, o that if the et f; 2;:::;ng i orted in decreaing order by riority to obtain a ermutation of f; 2;:::;ng, every i equally likely. In order to locate m in the trea, we follow the ath from the root down to m. The amount of time to do thi i roortional to the length of the ath. Let u calculate the exected length of thi ath, averaged over all oible random ermutation. Let m = f; 2;:::;mg m = fm; m +;:::;ng : Let A be the et of ancetor of m, including m itelf. The denition of m and m do not deend on, but the denition of A doe. Let X be the random variable X = length of the ath from the root down to m = jm \ Aj + jm \ Aj,2 : The 2 i ubtracted becaue m i counted in both m and m. We are intereted in EX, the exected value of X; by linearity of exectation, we have EX = Ejm \ Aj + Ejm \ Aj,2 : By ymmetry, it will uce to calculate Ejm \ Aj. Note that if the element of m are orted in decending order by riority, then

4 68 Lecture 3 Random Search Tree every ermutation of m i equally likely; an element of m i in A if and only if it i larger than all reviou element of m in orted order. In other word, ermute m randomly, then can the reulting lit from left to right, checking o thoe element k that are larger than anything to the left of k; the quantity Ejm \ Aj i the exected number of check. Examle 3.2 Let n = 0 and m = 8. Suoe that when rioritie are aigned randomly to f; 2;:::; 0g and thee element are orted in decreaing order by riority, we get the ermutation Thi reult in the following trea: = (4; 5; 9; 2; ; 7; 3; 0; 8; 6) : 4 5,,@ 9, 0 8 Then m = f; 2; 3; 4; 5; 6; 7; 8g. If we retrict the random ermutation to thi et, we obtain the ermutation (4; 5; 2; ; 7; 3; 8; 6). Scanning from left to right and checking only thoe element k that are greater than all element to the left of k, we get the equence (4; 5; 7; 8). Thi i exactly the equence of element in m aearing on the ath from the root down to m in the trea. A ymmetric argument uing m give the equence (9; 8), which i the equence of element in m aearing on the ath from the root down to m. The length of the ath i then the um of the two length of thee equence le 2. 2 We are thu left with the roblem of determining the exected value of the random variable H m, the number of check obtained when canning a random ermutation of f; 2;:::; mg from left to right and checking every element that i greater than anything to it left. We claim that thi number i exactly EH m = mx k= k : (23) We will obtain thi by olving a imle recurrence, uing the linearity of exectation.

5 Lecture 3 Random Search Tree 69 Suoe we ermute f;:::;mg randomly to get the random ermutation. Deleting from, we get a random ermutation 0 of f2; 3;:::;mg. Note that an element other than i checked when canning if and only if it i checked when canning 0 ; thu the reence or abence of doe not aect whether 2 i checked (however, the reence or abence of 2 might very well aect whether i checked). Thu the exected number ofcheck on element other than i the ame in a in 0,orEH m,. The element i checked if and only if it occur rt in, and thi occur with robability. Thu the m exected number of check on the element,averaged over all ermutation, i. By linearity of exectation, m EH m = EH m, + m : The unique olution to thi recurrence with EH = i (23). The quantity (23) i O(log m). Thi can be veried by aroximating the um above and below with denite integral involving the function and, x x+ and recalling from calculu that Z m dx x = ln m = ln 2 log 2 m : 3.4 Exected time for deletion A imilar analyi allow u to calculate the exected number of rotation neceary to delete m from it oition in the trea. The number of rotation needed i the um of the length of the rightmot ath in the left ubtree of m and the length of the leftmot ath in the right ubtree of m. To ee thi, try rotating m down; if you rotate to the left (right), the length of the rightmot (leftmot) ath in the left (right) ubtree decreae by one and the length of the leftmot (rightmot) ath in the right (left) ubtree tay the ame. Let u calculate the exected value of G m, the length of the rightmot ath of the left ubtree of m. By ymmetry, the exected length of the leftmot ath of the right ubtree of m i EG n,m+, and by the linearity of exectation, the exected number of rotation to remove m i EG m + EG n,m+. We will how below that thi number i le than 2! An analyi imilar to the analyi for EH m above reveal that EG m i the exected number of check obtained when canning a random ermutation of the et f; 2;:::;mg from left to right, where we check an element k rovided that k occur trictly to the right of m; k i greater than all element of f; 2;:::;m, g occurring to the left of k and either to the left or to the right ofm.

6 70 Lecture 3 Random Search Tree Thi i the ame a the exected number of check obtained when canning a random ermutation of the et f; 2;:::;m, g from left to right, where we check element k if it i greater than all element to it left, then lace m randomly in the lit and erae thoe check occurring to the left of m. Examle 3.3 For m = 3, we have the following ix ituation, all occurring with equal robability: The exected number of check i = It i eay to ee that the exected value of G m i at mot that of H m,, which wewould get if the check to the left of m were not eraed; thu EG m EH m, = O(log m), and thi uce for our comlexity bound. In fact, it turn out that EG m <. A above, the exected number of check on element other than i EG m,, and the robability that i checked i, ince i checked if and only if m occur leftmot, followed m(m,) immediately by. Again, by linearity of exectation, EG m i the exected number of check on element other than lu the exected number of check on : EG m = EG m, + and EG =0. The olution to thi recurrence i EG m = m, m : m(m, )