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# アルゴリズムイントロダクション 第5.4章

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アルゴリズムイントロダクション 第5.4章 勉強会資料

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### アルゴリズムイントロダクション 第5.4章

1. 1. 2010 CS 3  2010/06/12 niky1   h:p://www.tniky1.com/ Copyright©  2010  tniky1    All  rights  reserved.   Page  1
2. 2. 2010 CS  1                2              3         n  i   best Copyright©  2010  tniky1    All  rights  reserved.   Page  2
3. 3. 2010 CS 5.4.4   (^^;) Copyright©  2010  tniky1    All  rights  reserved.   Page  3
4. 4. 2010 CS •    –    –  ( )   –    •    –  k ( )   –  k+1     p109   Copyright©  2010  tniky1    All  rights  reserved.   Page  4
5. 5. 2010 CS 5.4 Probabilistic analysis and further uses of indicator r the ﬁrst k applicants, and hiring th then rejecting a higher score than all preceding applicants. If applicant was among the ﬁrst k interviewed, th This strategy is formalized in the procedure O   appears below. Procedure O N -L INE -M AXIMUM  1              2                                    k k+1        k+2                                              n we wish to hire. O N -L INE -M AXIMUM (k, n) 1 bestscore ← −∞ 2 for i ← 1 to k 3 do if score(i) > bestscore 4 then bestscore ← score(i) 5 for i ← k + 1 to n  I    I   bestscore 6 do if score(i) > bestscore (i≤k)   bestscore (i>k)   7 then return i 8 return n     We wish to determine, for each possible va 1 k     hire the most qualiﬁed applicant. We will the     n   k   implement the strategy with that value. For th Copyright©  2010  tniky1    All  rights  reserved.   Let M( j ) = max1≤i≤ j {score(i)} denote the m Page  5 through j . Let S be the event that we succe
6. 6. 7 then return i n8 i return n 2010 CS We wish to determine, for each possible value Chapter 5 Probabilistic Analysis and Randomized Algorithms of k, the probability that Chapter Probabilistic Analysis e, hire the most qualiﬁed applicant. We will thenthat we the best possible k, a for each5 possible value of k, and Randomized Algorithms the probability choose applicant.•  We will 5 Probabilisticthe bestand Randomized Algorithms implementChapter then positions Analysis possiblewhereas Biassume that k on ﬁxe 116 thevalues in choose 1value. For the moment, depends only is strategy with that through i S   1, k, and ordering of the −with that value.max1≤i≤ j moment, assume that maximum score among applicants Let M( the = For the {score(i)} ) k denote the k is ﬁxed. whether • ) Svalue in position i is greater than the values in all other positions. The j (k ( ordering of. of values inthe event 1 throughsucceed − not affect whether thescore(i)} denotetheSvalues in positionswe through iin choosing the Bi depends ordering k) through j the Let be positions that 1   i − 1 does 1, the maximum score among applicants 1 whereas best-qualiﬁ event positionletis in position is greater than value in − in all does positi evalue in thatand valueiof the valuesi in positions 1 through ithe 1, whereas not app applicant, iorderingSgreater than all of them, best-qualiﬁed position i other Bi dep whether the isucceed the choosing the and the the values best-qualiﬁed we be in event that we succeed when event • thatith theofthe value in the best-qualiﬁed ithanare does not we other p the ordering whether values when position 1 1throughappli- 1Thus we can all have th cant the orderingsucceed in positions i is greater Si the values in apply we various − affect is the of one Sinterviewed. positionsthethrough i − 1. disjoint, affect wh the values in Since i     viewed. =(C.15) theobtain the Si are disjoint, wesucceed when− 1 best-qualiﬁed a Pr {S} Since to {Si }. Noting that we k of them, and the valuedoes not affec ordering of n Pr various values in positions have that i the in position i equation in position i is greater than all never 1 through value value in,2,•••,k       i is greater than all of them, and the value in posit i=1 =  1 –   never position when the best-qualiﬁed ap-Noting that one of the succeedwe have that Pr {S } = 0 for i = 1, 2, . . . , k. Thus, plicant is we ﬁrst the values in positions 1 through i − 1. Thus we c k, affect the{B ∩ Oi } = Pr {Bi i {Oi } 2, . . . k. Thus, we } = Praffect {S of 0 for Pr the . ordering = Pr {Sihave thati Pr thei }ordering} of= 1,values,ini positions 1 through i − 1. Thus k, we obtain equation equationto obtain obtain (C.15)+1,•••,n    to The probability Prk i(C.15) –    =   {B } is clearly 1/n, since the maximum is equally likely to be n inPr {Sone ofPr {B{Si } O{B= Pr {Bi } PrOi to}Pr ){O i:  the maximum value in po- (5.1 Pr any i= = thei }ni= Pr. i ii: ∩For }event i {Bi } occur,i }k+1 i-­‐1 {S} } Pr {S positions. Oi = Pr {Oi . (O . Pr ∩ }(B ) sitions 1 through i − 1 must be in one of the ﬁrst k (5.13) i=k+1 positions, and it is equally n likely to be The probability− 1 clearly Consequently, Pr {Oi } maximum isand lik The probability Pr {Bii} is {Bi } is clearly 1/n, the maximum is − 1) equall in any of these Pr positions. 1/n, since since the = k/(i equally compute Pr {Si }. n order to succeed Pr {Sany k/(n(i − onenUsing equation (5.13),For Oi when ithe best-qualiﬁed applica We nowin any 1)). of the In positions. we event O to occur, the maximum i } = one of the positions. For event haveSi }.in p109-­‐110 to succeed when the best-qualiﬁedthe to occur, the maximum valu is the ith one, two1things must− 1 mustFirst, applicant ﬁrst k positions, and In order i happen. be of must the best-qualiﬁed applicant must sitions the best-qualiﬁedin one in one ofbe k positions, and it i n First, throughmust be applicant the ﬁrsts must happen. an eventll  − 1eserved.  we denote by B . Second, the algorithm must n in {S} =1likely010  tniky1  i  A in any of these i − 1 positions. Consequently, Pr {O } = sitions Copyright©  2 position throughbe} rights  r Pr we denote by {Si . which the algorithm imust notwhich i, Pr to B Second, Page  6 likely to be in any iof these i − 1 positions. Consequently, Pr {O } = k/(i i
7. 7. 7 then returnPr {Si } = Pr {Bi }∩ Oi } {BiPr {Bii{Si } {O{Bi.{Bi ∩ O}} .= Pr {Bi } Pr { i {S = Pr = ∩ Pr } Pr Pr iPr } Pr {Oi i O = = }8 return n 2010 CS The probability Pr {Bi } is Pr The} probability1/n, maximummaxim The probability clearly 1/n, since Pr {Bi } is clearly equ {Bi is clearly the since the is 1/ 116 any one of theonepositions. positions. ofOthe noccur,i the occur, e in inChapter 5 the n For event For to and Randomize any n of Probabilistic Analysis positions.maxim in any one i event O to ForChapter 5 Probabilistic Analysissitions each possible value ofmust the in one of mustﬁrst in p We wish to determine, and Randomized Algorithms isitions k,through i ﬁrst1k the that for 1 through 1i through − 1in 1 be the − positions, sitions − 1 must be one of probability be k be We to be in any − 1 positions. Consequently,− 1{Oi a likely will then of these ibe 1 any of possible k, } likely toandin any of these Algorithms − in positions. Consequentlhire the most qualiﬁed applicant. Randomized i likely to the best these i Pr posi Chapter 5 Probabilistic Analysis chooseimplementChapter 5 Probabilistic= k/(n(i} andi RandomizediUsing(5.13),− 1)).only wethro116ordering of the strategy with that1value. = 1)). of the {S } = Biassume (5.13), on ﬁxe Pr {Si } Pr {Si thevalues in positions Analysis −For 1,−Prequationk/(n(ipositionsk1ishave k/(n(i 1)). Using moment, equation that equati ordering−the whereas in through values depends Algorithms we have Using n n nwhether the = max1≤i≤ j {score(i)} Prwhether ithe value all=position {Sis applicantsLet M( j ) value in position i {S}greater than the }maximum score among} greater th denote= {S Pr is = {S} Prthe values{Si } in other positions. The Pr in Pr {S} Pr i i ordering Let be positions that 1 i i=k+1 the − not whereas best-qualiﬁordering of. of theSvalues in positionswe throughdoes choosing positions the throthrough j the values inthe event 1 throughsucceed iinvalues i=k+1 the Bi depends ordering − 1 i=k+1 of 1, affect whether 1 in −applicant, orderingSgreater      values  i ofk+1    weink  position      i  values      n  best-qualiﬁedi app whether the valueiof      the than      all invalue+2  succeed      when    i    the 1,n  whereas not dep and letis  in    position     k is positions    n1        through   greater     k allBof th 1    the      event that    k and    the      the      is             in all other positi be 2             greater    than value  in position than nvalue in position i them,          k             i doescant the ordering values in in position 1 1through thanofThus we can1) positio ordering whether the value positions in(i − 1) n(i Si1) = does n(i inaffect whp ith one the values in =Since =the ordering 1.1 i=k+1 not all other is greater various − thedisjoint,− inaffect is the of theof interviewed. positionsthethrough ii−−arethe valueswe have th affect i=k+1 through i − 1 does not affec values apply {S} =(C.15) to obtain the values inwe nnever(C.15) 1 whenvalue in1position ia ordering of Noting that positionssucceed 1 n i=k+1equation in position ii }. greater than all of them, and obtain n best-qualiﬁedPrvalue i=1 Pr {S is equation1 n k k to the ktheplicant is one of in i-­‐1ﬁrst thei values  n positions 10throughn1, 2, 1. −,1k.in posit value theposition we have thatthan all = them,i and the . . k, is greater Pri {Si1 of i − 1 = i − value Thus, c affect the{Bi ∩ the} ordering} of = i }in = in i=k+1 for 1 = i=k+1.ii − 1. Thus ordering= Pr {Bi Pr {O values − }Pr {Si } = Praffect Oi of thePri=k+1} = Pr {Bi ∩ Oi } = Pr {BiThus we . .{S n positions through } Pr {Oi }obtain equation equationto obtain obtain k n−1 1 k n−1 1 (C.15) (C.15) to i 1 2 k The probability Prk+1,•••,n     k n−1 likely to be –    =   {Bi } is clearly 1/n, since= . maximum is equally . the 1 n = The probability Pr {B } is clearly 1/n, sin . = any i } = the } i= Pr. i }(Bi: ∩For }event O } occur,i }k+1 i-­‐1 i n i=k iinPr {Sone ofPr {B{Si } O{B= Pr {Bi }n ii=ki ito}Pr )i=k(Oi:  the maximum value in po- (5.1 Pr {B n. of .Pr {S} = Pr {Si n positions. Oi =in anyi ione{Oi the n positions. For event Pr ∩ i Pr {O ) sitions 1 through i − 1 must be in one of the ﬁrst k positions, and it is equally i=k+1 n sitions since since −i 1 summation from lik The probability Pr {Bii}We clearly approximate bytheapproximate k/(i besummationo in any of these is {Bi }We clearly 1/n,We Pr {Oto maximum in equall is 1/n, 1 through i the} = by this is one must equally likely to be The probability− 1approximate Consequently,bound thisboundis − 1) and abov Pr positions. by integrals to maximum integrals to boun integrals compute Pr {Sithe In positions. For event O of best-qualiﬁed applica }. n order(5.13), we have we to occur, the maximum We nowin any 1)). of the inequalities (A.12), we in any thethese i − positions to succeed when havePr {Sany k/(n(i − onenUsing equationlikely to be(A.12),inequalities (A.12),1we have valu in i } = one of the the inequalities have the positions. For event Oi to occur, the maximum iis the p110 one, two1things must− 1 mustFirst,n−111 best-qualiﬁed1k positions, and ith sitions throughn i1 happen. 1 ibe= k/(n(i 1− 1)).n−1 applicant must n−1 n inthe ﬁrst k positions, and it(5 Pr1{S } n−1 one nofn−1 1 ﬁrst the Using equation i n−1 sitions 1 through i  All  − 1eserved.  we ≤ indx≤ of .the . dx ≤dx . algorithm 1must nin {S} = Copyright©  2010  tniky1   }in any x dx be x − 1 positions.x Consequently, Pr {Odx . positionlikely Pr {S rights  r must denote by Bi i x≤n i, an to be whichof these i i one n event ≤ Second, the i ≤ dx x }= Page  7 Pr x k positions. Consequently, Pr {O } = k/(i likely to be in any iof these i − 1k i=k k−1 i=k k k−1 i=k k−1 i
8. 8. nLet Pr {S}j )== max{Si } {Bji }{score(i)} denotePr {Smaximum score among appli M( probability1≤i≤ is clearly 1/n, since thethe } =isPr {B likely to be Pr {B } The Pr Pr maximum equally 2010 CS any one of the n positions. For event Oi to occur,i the maximumi value in}po- in ∩ Oi = ithrough j . 1Let i=k+1 S be the event that we succeed in choosing the best-q sitions through i − 1 must be in one of the ﬁrst k positions, and it is equally napplicant,=and let k iofbe the−event thatConsequently, Pr {Oi } = k/(i − 1) iandis clearly likely to be in any 1) these i 1 positions. we succeed when Pr {B } S The probability the best-qualiﬁedcant is the n(i − = k/(n(i interviewed. Since in any i=k+1 Pr {Si } ith one− 1)). Using equation (5.13), thehave one of the n positions. ha we various Si are disjoint, we F k n n 1 nPr {S} = = i=1 Pri {S1 }. Noting i sitions 1 through − best-qualiﬁ that we never succeed wheni the 1 must be Pr {S} n i=k+1 −Pr {Si } =plicant is one n−1i=k+1 ﬁrst k, we have that likely}to be forany of these. i, − 1 of the =  1,2,•••,k   Pr {Si = 0 in i = 1, 2, . . k. Th k 1obtain = n i=k i n. k =  k+1,•••,n   Pr {Si } = k/(n(i − 1)). Using eq = n n(i − 1) i=k+1 nPr {S} approximatePr {Si } .1to bound this summation from{S} and below. By Pr {S } We = by integrals k n Pr above = = the inequalities (A.12), we have i i=k+1 n i=k+1 i − 1 n n−1 i=k+1 1/k  +  1/(k+1)  +  1/(k+2)  +  •••  +  1/(n-­‐1) 1 1 n−1 n−1 1 We now≤compute 1 .x{Si .}. In order to succeed when the best-qualiﬁed k x dx = i=k i k≤ Pr dx k−1 n k apis the ith one, two things mustus the bounds First, the= n i=k i Evaluating these deﬁnite integrals gives happen. best-qualiﬁed applicant m n(i − 1) i=k+1the algorithm min position i, an event which bounddenote by BfromSecond, below. By k We approximate by integrals to we this summation i . above and k (ln n − ln k) ≤ Pr {S} ≤ (ln(n − 1) − ln(k − 1)) , nselect anyinequalities (A.12), we have positions k + 1 through ik− 1, which happens n the of the applicants in n 1 which provide a that k + 1 ≤ j ≤ i Because we wish to =for each nj1such rather1tight bound1 for Pr {S}.− 1, we ﬁnd that score( j ) < bestscore in maximize our probability of success, let us focus on choosing the value of k that maximizesi=k+1 i − 1n the n−1 n−1 dx ≤ are unique, we. can ignore the possibility of score( j ) = bes ≤ dx  (Because scores {S}.i (Besides, the lower-bound expression is easier to   maximize lowerkbound on Pr k Pr{S} x k−1 x than the upper-bound must be the case that all of the n k n−1 1 i=kIn other words, it expression.) Differentiating the expression (k/n)(ln values score(k + 1) t −ln k)score(i − 1) are less than M(k); if any are greater = n p110-­‐111 Evaluatingk, we obtain with respect to these deﬁnite integrals gives us the bounds . than M(k) we will instead i k k 1 Copyright©  2010  tniky1    All  rights  reserved.   i=k Page  8 the index of−the .ﬁrst one (ln(n 1) − ln(k We use Oi to denote the event th (ln n(ln ln k − 1) ≤ Pr − n ln k) {S} ≤ that is−greater. − 1)) ,
9. 9. Pr {S} = Pr {Si } i=k+1We approximate by integrals to bound this summation from above and below. By 2010 CS n the inequalities (A.12), we have k = 1068 Appendix A Summations i=k+1 n(i − 1) n n−1 n−1 1 1 1 k x dx ≤ i=k i = ≤ k n dx . k−1 x 1 n i=k+1 i − 1 f (x)Evaluating these deﬁnite integrals gives us the bounds k n−1 1k = k . (ln n − ln k) ≤ Pr {S} i=k i (ln(n − 1) − ln(k − 1)) , n ≤n nwhich provide a rather tight bound for Pr {S}. Because we wish to maximize our We approximate by integrals to bound this summation from above … below. By and f (m+1) f (m+2) f (n–2) f (n–1) f (m) f (n) …probability of success, let us focus on choosing the value of k that maximizes the the inequalities (A.12), we havelower bound on Pr {S}. (Besides, the lower-bound expression is easier to maximize x … …than the upper-bound n−1 1 n 1 expression.) Differentiating themexpression (k/n)(ln n −ln k) n–1 n n−1 1 –1 m m+1 m+2 n–2 n+1with respect to dx we obtain≤ k, ≤ dx . (a) k x i=k i k−1 x1 (ln n −Evaluating. these deﬁnite integrals gives us the bounds ln k − 1) f (x)n k k (ln n − ln k) ≤ Pr {S} ≤ (ln(n − 1) − ln(k − 1)) , n n which provide a rather tight bound for Pr {S}. Because we wish to maximize our probability of success, let us focus on choosing the value of k that maximizes the f (m+1) f (m+2) f (n–2) f (n–1) lower bound on Pr {S}. (Besides, the lower-bound expression is … easier to maximize f (m) f (n) … than the upper-bound expression.) Differentiating the expression (k/n)(ln n −ln k) x with respect to k, we obtain m –1 m m+1 m+2 … … n–2 n–1 n n+1 (b) p325 1 (ln n − ln k − 1) . n Copyright©  2010  tniky1    All  rights  reserved.   Figure A.1 Approximation of n Page  9 k=m f (k) by integrals. The area of each rectangle is shown within the rectangle, and the total rectangle area represents the value of the summation. The in- tegral is represented by the shaded area under the curve. By comparing areas in (a), we get
10. 10. nLet Pr {S}j )== max{Si } {Bji }{score(i)} denotePr {Smaximum score among appli M( probability1≤i≤ is clearly 1/n, since thethe } =isPr {B likely to be Pr {B } The Pr Pr maximum equally 2010 CS any one of the n positions. For event Oi to occur,i the maximumi value in}po- in ∩ Oi = ithrough j . 1Let i=k+1 S be the event that we succeed in choosing the best-q sitions through i − 1 must be in one of the ﬁrst k positions, and it is equally napplicant,=and let k iofbe the−event thatConsequently, Pr {Oi } = k/(i − 1) iandis clearly likely to be in any 1) these i 1 positions. we succeed when Pr {B } S The probability the best-qualiﬁedcant is the n(i − = k/(n(i interviewed. Since in any i=k+1 Pr {Si } ith one− 1)). Using equation (5.13), thehave one of the n positions. ha we various Si are disjoint, we F k n n 1 nPr {S} = = i=1 Pri {S1 }. Noting i sitions 1 through − best-qualiﬁ that we never succeed wheni the 1 must be Pr {S} n i=k+1 −Pr {Si } =plicant is one n−1i=k+1 ﬁrst k, we have that likely}to be forany of these. i, − 1 of the =  1,2,•••,k   Pr {Si = 0 in i = 1, 2, . . k. Th k 1obtain = n i=k i n. k =  k+1,•••,n   Pr {Si } = k/(n(i − 1)). Using eq = n n(i − 1) i=k+1 nPr {S} approximatePr {Si } .1to bound this summation from{S} and below. By Pr {S } We = by integrals k n Pr above = = the inequalities (A.12), we have i i=k+1 n i=k+1 i − 1 n n−1 i=k+1 1/k  +  1/(k+1)  +  1/(k+2)  +  •••  +  1/(n-­‐1) 1 1 n−1 n−1 1 We now≤compute 1 .x{Si .}. In order to succeed when the best-qualiﬁed k x dx = i=k i k≤ Pr dx k−1 n k apis the ith one, two things mustus the bounds First, the= n i=k i Evaluating these deﬁnite integrals gives happen. best-qualiﬁed applicant m n(i − 1) i=k+1the algorithm min position i, an event which bounddenote by BfromSecond, below. By k We approximate by integrals to we this summation i . above and k (ln n − ln k) ≤ Pr {S} ≤ (ln(n − 1) − ln(k − 1)) , nselect anyinequalities (A.12), we have positions k + 1 through ik− 1, which happens n the of the applicants in n 1 which provide a that k + 1 ≤ j ≤ i Because we wish to =for each nj1such rather1tight bound1 for Pr {S}.− 1, we ﬁnd that score( j ) < bestscore in maximize our probability of success, let us focus on choosing the value of k that maximizesi=k+1 i − 1 n the n−1 n−1 dx ≤ are unique, we. can ignore the possibility of score( j ) = bes ≤ dx  (Because scores {S}.i (Besides, the lower-bound expression is easier to   maximize lowerkbound on Pr k Pr{S} x k−1 x than the upper-bound must be the case that all of the n k n−1 1 i=kIn other words, it expression.) Differentiating the expression (k/n)(ln values score(k + 1) t −ln k)score(i − 1) are less than M(k); if any are greater = n p111 Evaluatingk, we obtain with respect to these deﬁnite integrals gives us the bounds . than M(k) we will 1instead i k k 1 Copyright©  2010  tniky1    All  rights  reserved.   i=k Page   0 the index of−the .ﬁrst one (ln(n 1) − ln(k We use Oi to denote the event th (ln n(ln ln k − 1) ≤ Pr − n ln k) {S} ≤ that is−greater. − 1)) ,
11. 11. k We i=k i k−1 approximate by integrals to bound this summation from above and below. B in any onethe inequalities (A.12), we For event Oi the bounds the maximum valu of the these deﬁnite integrals gives us to occur, Evaluating nn positions. have 2010 CS Pr {S} = Pr {Si } i −in ordering 1kthrough i=k+1 11positions in through the ﬁrstwhereas Bi depends is sitions of the valuesn−1 mustn−1 1 one of i − 1, k positions, and it o n be 1 k1 likely the value ink) ≤ Pr ≤ − (ln(n. − 1) − ln(k values in Pr {Oi } position whetherto n (ln nx− ln≤positioni ik−1 1greater thanConsequently,all other = k/(i be in any of these ≤ positions. the − 1)) , dx n {S} is dx i k nx k = − 1)). Using equation (5.13), we have Pr {Si } of the values n(i − 1) ordering = k/(n(i i=k+1in positions 1 through{S}.− 1 does not affect whet i=k Pr i Because we wish to maximiz which provide a rather tight bound for us the bounds Evaluating these deﬁnite integrals gives value in positionn i is greater than all of them, and the value of k position i do probability of success, let us focus on choosing the value in that maximize k n 1 k Pr {S} = k bound k) ≤ {S}. ≤ in(ln(n − lower-bound expression easier to can lower = ln the values k   affect the≥ (ln n −ofon Pr}Pr− 1(Besides, the 1) − ln(k − 1)) , i − 1. isThus wemax {Si ordering Pri=k+1 i {S} n positions 1 through n i=k+1 n upper-bound expression.) Differentiating the expression (k/n)(ln n − than theto obtain equation (C.15) provide a rather tight bound for Pr {S}. Because we wish to maximize ou n−1 which n with respect 1 k to k, we obtain k . let random variables = of indicator us focus on choosing the value of k that maximizes th istic analysis and further usesn of =i Pr {B } Pr {O } . 117 Pr {Si } = Prprobabilityi } success, i = {Bi ∩ O i=k i 1 lower bound on Pr1) (Besides, the lower-bound expression is easier to maximiz 117− ln k − 1) {S}. (ln i=k+1 n(i − . n The probability Prn{Bi }by integrals to1/n, since the maximum is and below.−ln k n than the upper-bound expression.) Differentiating the expression equallynlikel We approximate is clearly bound this summation from above (k/n)(ln By of k n 0 1 with respect to k, we obtain k in any one to the we (A.12), we have event Oi bound on the probability derivative equal the inequalities see thatFor lower to occur, the maximum value = 1 0, positions. the sitions 1 through − ln ki1− 1. n−1 1 in one of the ﬁrst k positions, and it is e n i=k+1ed on the ln k = lnn ndx ≤ − 1 ≤ln(n/e) or, equivalently, when k = n/e. n (ln n i n−1 − must be 1 − 1 = 1)d whenlikely to be in any of these i − 1 positions. Consequently, Pr {O } = k/(i − probability dx . k xk n−1 i 1 i k−1 xtly, when{Siour k/(n(i − 1)).. Using= n/e, we will succeed in hiring our implementk } = = = strategy with k equation (5.13), we have n/e. i=k Pr k!! d applicant Evaluating these deﬁnite integrals gives us the boundsucceed in hiring probability at least 1/e. with our i=k i n n k [ k]   Pr {S} ≥ n = (ln n − ln k){Si Pr {S} ≤ (ln(n − 1) − ln(k − 1)) , Pr ≤} k=n/e   n   We approximate by integrals to bound this summation from above and be i=k+1 which provide a rather tight bound for Pr {S}. Because we wish to maximize our the inequalities (A.12), welet us focus on choosing the value of k that maximizes the p111 probability of success, have n k  1/e     =lowern−1 on Pr Copyright©  2010  tniky1    All  rights  reserved.  {S}. (Besides, the lower-bound expression is easier to maximize bound n 1 than the upper-bound 1 i=k+1 1 n(i − 1) expression.) Differentiating the expression (k/n)(ln n1−ln k) n−1 Page   1
12. 12. 116 Chapter 5 Probabilistic Analysis and Randomized Algorithms Pr {Si } = k/(n(i − 1)). Using equation (5.13), we have2010 CS n Pr {S} = ordering of the values in positions Pr {Si } ordering of the values in positions 1 through i − in position i Bi gre i=k+1 whether the value 1, whereas is de (whether the value in position i is greater than the valuesin positions )   TVn k ordering of the values in all other = ordering of the values in positions 1 in position −is greater than al i=k+1 n(i − 1) value through i i 1 does not affe value in position i is greater than all of them, and thethe values in value in posi •  k n 1 affect the ordering of affect the ordering of the values in positions 1 through i − 1. Thus = equation   (C.15) to obtain i − 1 equation (C.15) to obtain n i=k+1 –  k n−1 1 10 Pr {Si } = Pr {Bi ∩ Oi } = Pr {Bi } Pr   = . Pr {Si } = Pr {Bi ∩ On}i=k iPr {Bi } Pr {Oi } . i = –  (The probability Pr {B )   is clearly 1/ i} The probability Pr {Bi } by integrals to bound this summation from aboveequal We approximate is clearly 1/n, one ofthe maximum is and b in any since the n positions. For in any onethe inequalities (A.12), we For event through  0.1 1the maximum of the n positions. sitions 1 Oi1/10  occur, must be in have to = i − sitions 1 through i − 1 mustn−1 likely to be in any of k positions,pos n n−1 be in one of the ﬁrst these i − 1 an 1 1 1 dx of ≤ positions. Consequently, Pr {Oi } = (n=10)   dx . likely to be in any ≤ these i − 1Pr {S } = k/(n(i − 1)). Using equat k   k x i k−1 x   i Pr {Si } = k/(n(i − 1)). Using equation (5.13),k=3   have i=k !(^^;) we Evaluating these deﬁnite integrals gives us the  nbounds  0.3611   Pr{S}   n = k k {S} = Pr Pr {S} ≥ (ln n − ln k) ≤ }Pr {S} ≤ (ln(n − 1) − ln(k  Pr{S}   ,  0.3665     Pr {Si k=4   Pr {Si } − i=k+11)) n i=k+1 n n which provide a rather/e   10/(2.7182 Pr {S}. .679 kwe wish to maxi n k  =  n tight bound for )    3 Because =   probability of success, let us focus on= k choosing the value of Page  12 maxim n(i − 1)k that = Copyright©  2010  tniky1    All  rights  reserved.   i=k+1 lower bound on Pr {S}. (Besides, the lower-bound expression is easier to m
13. 13. 2010 CS n k k/n lgn lgk logn-logk Pr{S} 10 1 0.1 2.302585093 0 2.302585093 0.230258509 10 2 0.2 2.302585093 0.693147181 1.609437912 0.321887582 10 3 0.3 2.302585093 1.098612289 1.203972804 0.361191841 10 4 0.4 2.302585093 1.386294361 0.916290732 0.366516293 10 5 0.5 2.302585093 1.609437912 0.693147181 0.34657359 10 6 0.6 2.302585093 1.791759469 0.510825624 0.306495374 10 7 0.7 2.302585093 1.945910149 0.356674944 0.249672461 10 8 0.8 2.302585093 2.079441542 0.223143551 0.178514841 10 9 0.9 2.302585093 2.197224577 0.105360516 0.094824464 10 10 1 2.302585093 2.302585093 0 0 Copyright©  2010  tniky1    All  rights  reserved.   Page  13