ln 2 = 1 + max{c m,n/2 2 t 1, t + C m 1,n/2 1} + m 107

Transcription

1 Errata to The Analyss of Algorthms, Second Prntng Usually just the corrected segment of text s gven. Negatve lne numbers ndcate the number of lnes from the bottom. p, l 3. (Only for n > does n 0.1 become notcably larger than lg n. p 18, ex. 4. t( ln(1 p ln (1+ ln ln(1 p ( ln /( ln(1 p ln + t( ln(1 p. ln 1 p 0, ex.. Show that the number of tmes you go around the loop n the Eucldean Algorthm (Algorthm 1.5 s no more than lg n. p 1, l 1 (m n our example p 7, l. A trangular matrx s one n whch all the elements above p 7, ex.. prevous exercse can be wrtten as p 36, l 1, Step 5 s done the same number of or one less tme than Step 4 each tme Step s done. p 49, l 7. order. The frst element n each fle has subscrpt p 49, l 9. once agan the frst remanng element has subscrpt p 49, Step 3. Steps 3 5 and 6 8 are p 5, eq. ( max{c m,n/ t 1 + m, t + C m 1,n/ + m 1}. p 53, eq. ( C m,n 1 + max{c m,n/ t 1 + m, t + C m 1,n/ + m 1} C m,n 1 + max{c m,n/ t 1 + m, t + C m 1,n/ + m 1} max{c m,n/ t 1, t + C m 1,n/ 1} + m 107 provded n s even, m, and n m +. Now comes a clever porton of the proof, a crucal observaton: by eq. ( max{c m,n/ t 1, t + C m 1,n/ 1} C m,n/, so we have p 53, l 9. left sde of eq. p 55, ex. 3. C m,m+1 p 55, l 0. A predcate P s called -complete f, whenever P (x s true for all x n δ + (y, then P (y s also true. (In partcular, P (y s true when δ + (y s empty. p 55, l 10. A relaton s confluent f x y mples x y. Confluence s mportant because, n a Noetheran confluent relaton, p 61 l 1. For example, the formula 0 n a s not n closed form, because the summaton sgn stands for the addton of n + 1 elements. p 65, l 10. The second sum p 67, No perod at the end of equaton 31. p 7 l 19. n Steps 3 and 6 of Splt. p 74, TABLE.1

2 Poston Input Intermedate Output 0 A A A A A A 1 G B A A A A C C B B B B 3 H A C C C C 4 I F F F F F 5 F G G G G G 6 A I I I H H 7 J J J I I I 8 B H H H J J 9 J J J J J J p 74, l 9. ; ths tme wth l 6, r 8. Splt selects the element n poston 6 (I as the splttng element and produces the results shown n the fourth ntermedate column of Table.1. Qucksort (level calls Qucksort (level 3 wth l 6, r 6 (Qucksort (level 3 returns from Step 1, and then wth l 8, r 8 (Qucksort (level 3 returns from Step 1. Then (level and (level 1 fnsh to gve the results n the output column of Table.1. p 75 ex. 4. What s the worst-case tme for the varaton of Qucksort? p 77, l 18. would look n locaton of x p 79 l 5. the sequence {0, c 0, c 0 + c 1, c 0 + c 1 + c,... } p 81, eqs A n a, 77 where a 0 j< p j. Now defne 1 n q Prob(the step s done or more tmes j p j. 78 Wth ths defnton q 1 a. Thus A 1 n q. 79 p 81 l 18. If the q are easer to compute than the p (as they were n Secton p 85 l 17 [the degree of R( s less than that of D(]. For example, p 88 l 11. The method n ths secton p 98 ex. 3. m ( ( e t t x 1 dt Γ m (x e t t x 1 dt + e t 1 t m t x 1 dt m 0 m p 104, l 11. Suppose, for example, that you need to smplfy ( n+ x, where x < 1. p 110. The smudges on ths page appear to be on the plate. If so, a new plate should be shot. p 16 l 10. ω n/ (ω 1/ + ω 1/ n ω n/ ( cos(π/3 n. Lkewse, 0

3 p 130, ex.. ( ( m m j. + j m p 133, l and eq In general, the Strlng number [ n k] s the sum of all the products of (n k dfferent ntegers taken from 1 to n 1; that s, [ ] n 1 n k. k 1 1< < < n k n 1 p 134, l 3 and eq In general, the Strlng number { n k} s the sum of all the products of n k ntegers taken from 1 to k; that s, { } n 1 k. k 1 1 n k k p 135, l 1. One of these ways, however, has one part empty. p 144, eq. 3. ( ( r s + ( 1 s!(s + 1 ( r n!n!(s n!(s n + 0 p 145, l 5. Many results on hypergeometrc functons can be generalzed to basc hypergeometrc functons, whch are obtaned p 146, eq. 43. N( N + ( 1 N(A j1 A j... A j. p 146, eq. 44. p 147 eq. 45. p 147 eq. 46. p 147, l 6. pgeonhole p 150, l k N(A 1 A... A N φ(n n + 1 k ( 1 1 j 1<j < <j k N(A 1N(A N(A N, 1 j 1<j < <j k n p j1 p j p j, φ(n n (1 (1 1p1 1p (1 1pk. mn {g L (x} y max {g U (x}. x 0 x x 1 x 0 x x 1 p 150, l. where x wll be requred to be n some range x 0 x x 1 and U (alternately L s an upper (lower bound on f (n+1 (x/(n + 1! n the range of x. p 151, eq. 15. e x 1 + x for x 0 and e x 1 1 x 3 for 0 x < 1. 15

4 4 p 151, l 11. for some c n the range 0 c x. Usng Prncple 9 (wth n 0 gves p 151, eq (1 1/ p 153, l. n N, testng 1 + (n 1 N,..., n N 1. (Actually, there s no need to test aganst N because we know the value s less than or equal to N. p 154, l 7. The trees for ( ( d 0 and d d each contan a sngle node. p 154, l 15. For testng values up to N we need a tree wth at least N leaves. p 154, l 1. tree has at least N leaves. p 154, l 10. A bnomal tree wll have at least N leaves p 154, l 6. h tems by dong at most (h!n 1/h + h 1 tests. p 159, l 19 a m+1 > 0, we have p 159, l 8 to 6 U a m+ + a m+3 r + (f(r 0 m+1 a r /r m+, whch s fnte. Ths establshes eq. (45 wth the C for eq. (3 equal to a m+1 / p 163, eq. 71. ( [ ( ] 1 e 1/ O p 164, ex. 7. Show that (n s the varable p 165, l 4. at x 0 f lm n [f(x 0 s n (x 0 ] 0. p 169, l 15. Backtrackng s a method of organzng a search through p 173, l. The probablty that a clause does not contan the lteral p 175 l 18. for another example. p 175, l 10. Let s use r k for the probablty that P ( k 1 (false,..., false true. p 175, 1. only lterals for the frst k varables and all the lterals are postve, where k s p 176, l. that a clause does not have ths form. p 177, l 9. Now eq. (10 s a decreasng functon of [because (1 p p 179, eq p 179, eq t( ln(1 p ln (1+ ln ln(1 p ( ln ln + t( ln(1 p ln /( ln(1 p ( t+ln /( ln(1 p N v+1 ln. (1 ln + t( ln(1 p p 183, l 5. sum s smaller than the ntegral, but t s larger than the ntegral that s obtaned f the curve s shfted rght one unt. p 186, The Bernoull numbers B m, whch occur as coeffcents n the Bernoull polynomals, are defned by the recurrence B 0 1, B n 1 n n 1 ( n + 1 B for n Except for B 1 all the Bernoull numbers of odd ndex are zero. The frst few Bernoull numbers are gven n Table 4.1.

5 Hgher-degree approxmatons can be obtaned by ntegratng eq. (153 by parts. To do ths we wll need to ntegrate B 1 (x. As the ntegraton by parts proceeds, t wll gve rse to a sequence of polynomals called Bernoull polynomals. The Bernoull polynomal B m (x s defned recursvely as mb m 1 (x dx wth the constant of ntegraton equal to B m. An equvalent defnton s p 186, ex. Show that 1 <n p 188, ex. 1. f( 1 <m f( + n m f(x dx 1 n (f(n f(m + B 1 ({x}f (x dx. m 5 EXERCISES 1. Show that 1 <n 1/ 3 n3/ 1 n1/ n 1/ n 5/ and that 1/ 3 n3/ 1 n1/ n 1/. Hnt: Use the generalzed Euler 1 <n summaton formula wth m 3. p 190, l 3. approaches B 1 m ({x} dx/x m+1. p 190, eq B m ({x} 1 x m+1 dx R mn n B m ({x} x m+1 dx 173 p 191, ex. 4 and 5. Every lower lmt of 0 should be changed to 1 on the summatons. p 19, eq A 1 + k ( ( N 1 k (N k + O N N + O k(n k. p 196, ex. 3 p 06 eq j n 1 j(j ( 1 1 j n 3(3(3T n/ 3 + n/ + n/ + n p 09, l 8. wth boundary condton T p 09 eq. 70. T k k k 1 1 j(j 1. p 10 ex. 6. Solve T (n / r nt (n + bn. (Ths exercse fts wthout resettng any addtonal pages. p 16 eq. 10. p 17 l 19. T n a(n + 1 lg(n b 3a + c n + c a b.

6 6 The classcal algorthm for matrx multplcaton (Algorthm 1.9 [modfed to save one addton n the nner loop] p 17 EXERCISES 1. Consder a sequence of k 1 numbers wth the mddle number frst, then the smaller numbers, and then the larger numbers. The small part (and also the large part obey the same rule. For k 3 the bnary sequence 100, 001, 011, 010, 110, 101, 111 s such a sequence. Show that the sequence formed by these rules causes the Splt algorthm to produce an equal dvson each tme.. Show that the sequence of the last exercse results n Splt extng the frst tme t gets to Step Show that f Splt exts the frst tme t gets to Step 7 and f the tme to go around the loop n Step 3 s equal to the tme to go around the loop n Step 6, then the tme for Qucksort obeys the recurrence T n an + b + T n1 + T n when the splttng element s such that Splt dvdes the data nto sets of sze n 1 and n (where n n 1 + n + 1. Remove the prevous exercse 1 and renumber the remanng ones. The changes on p. 17 result n pages beng retypeset. Some thought should be gven to ways to reduce the number of reset pages. p 0, l 9. Snce s less than 3, p 3, l. The materal p 6 l. F n+1 (1/ 5(φ n+1 ˆφ n+1. p 35, eq z n b n z n. 0 n<k 0 n a T n + n k p 36, l 6. By replacng wth k the left sde of eq. (190 can be rewrtten as p 37, l 7 to 8. T p (z (1 λ p z ( 1 ( 1βp+1 c p,q λ βp β p 1 pz q, (0 0 q max{0,q β p+1} q so usng d q (p for the coeffcent of z q n T p (z/(1 λ p z βp, we have ( 1 d q (p ( 1 βp+1 c p,q λ p (03 β p 1 max{0,q β p+1} q ( 1 ( 1 βp+1 λ q p c p,q λ q+ p. (04 β p 1 max{0,q β p+1} q Now replace q by ( q + and drop the prme to obtan ( q 1 d q (p ( 1 βp+1 λ q p c p, λ p β p 1. (05 max{ q, β p+1} 0

7 Fnally replace by to obtan d q (p ( 1 βp+1 λ q p 0 mn{q,β p 1} 7 ( q 1 c p, λ p β p 1. (06 Thus the form of d q (p s an exponental n q (.e., λ q p tmes a polynomal n q of degree β p 1 [.e., the rest of the rght sde of eq. (06]. To obtan the coeffcent of z q n the generatng functon G(z for the homogenous case, t s necessary to sum the contrbutons from each T p (z to obtan d q (p. (07 p 39, eq p j f 1,0 1 + f 1, , p 39, l 5. so the soluton that matches p 40, ex.. Fnd the general soluton of the recurrence F n+ + F n+1 + F n n. Hnt: Frst try to fnd a partcular soluton of the form c 1 n + c 0. p 40, ex. 4. T n 4T n 1 5T n p 47, l 3. the grand medan and the remanng elements. The remanng p 48, l. For eq. (53 p 49, eq. (55. Solvng eq. (56 gves β 1 9 β + α + β β + 4α + 4β ( β. (56 β (57 p 49, l 8. β p 49, l 10 gve p 49, eq. (59. p 56, ex n 7830 J n+1 (z n z J n(z + J n 1 (z 0, p 56, ex. 4. Show that the F (n F 1 (n, (b + a/(c + 1; b; c + 1 s a soluton to the recurrence ncf n+1 + [(1 cn + a]f n + (b nf n 1 0. Fnd the general soluton to the recurrence. ncf n+1 + [(1 cn + a]f n + (b nf n 1 0. Fnd the general soluton to the recurrence. p 59, 46. r +1 (n[b(n B(n 1] r 0 (nb(n r k+1 (nb(n k 1. 0 <k+1 p 61, ex. 4. Show that ths equaton s satsfed by n r

8 8 p 63, eq. 83. ( A ( 1! A 1 + Y 0 1 j< 1. j! For large the sum s close to e 1. p 66. Exercses move to page 68. p 66, eq a 0 (nb j (n 1 k a (nb j (n. p 68. New secton called Annhlaton of the Nonhomogeneous Part. These two changes cause the rest of the chapter (pp to be reset. p 7, l 6. wth the boundary condtons p 1 δ 0 p 73, eq p 73, eq G n (z ( 1n n!z G n (z ( 1n z [ n ( 1 n ( z. n p 73, l 10 The rght sde of eq. (143 s the product p 74, l 17. P 1 and P p 75 eq ] ( z 1 n! [ ] n z 1, q q 1 + q 11 + q 10 (q 13 ( q 1 ( 1 + q 11 ( 1 + q 10 +, p 75 eq p 76 eq. 170 q q 1 + q 11 + q 10 q q q 13 q 13 + q 1 3 q 1 + q 1 + q 11 q 11 + q q 13 q 1 + q 11 q 10 p 78, l 10. The boundary condton s C 0 0. p 78, eq. 7. C n n C for n 1, 7 n p 79, eq <n nc n (n 1C n 1 n (n 1 + n (n 1 + C C 9 0 <n 0 <n 1 n + C n 1 for n, 10

9 p 79, eq a frst order lnear equaton. The soluton for the boundary condton C 1 s C n j j j j p 79, eq. 14. <j n 1 n n <j n n (n + 1(H n C n (n + 1(H n+1 1 n ln n + O(n. 14 p 87, l 8-9. To smplfy ths equaton, we can use the fact that C(x obeys eq. (159. We can rewrte eq. (159 as p 87, l 8. We can apply eq. (37 to the last term of p 87, l 3 and apply eq. (37. Thus we have p 87, eq. 54. C C n C n+1 C n. p 88, eq. 55. K n 0 n 1 0 n 1 p, ex. 3. t n n n 1 t p 95, eq p 98, eq C n 1 K + C n+1 C n + δ n0. aze z + e pz [ a(1 pze (1 pz + e p(1 pz( a(1 p ze (1 p z. az 0 <v (1 p e (1 p z 0 j< e p(1 pj z + v G 0 ((1 p v z 0 l 3 + l 4 0 j<v e p(1 pj z 9 p 98, eq C n l 1 + (l 1 n + H n + nh n. 113 p 98, l 5. we consdered the recurrence p 304, 11. If we say that the heght of a tree s the number of nodes on the longest path to the root, then the number of such trees wth heght no more than n s gven by p 304, 6. The frst term n the recurrence allows for the empty tree. p 308, ex. 4. φ n (1 p(φ 1 φ n + φ φ n φ n φ 1 for n,

10 10 p 314, ex. 1. f n (x mn 0 y x {y ln y + f n 1(x y} p 315, 9 label any of the leaves n ts left subtree. p 3, eq. 04. t n n 1 + (t + t n, 04 1 n 1 p 331, Step 4. If > F t+1 and Y true, then set M Ft+1 true. p 339, eq c 11 + c 1 + c 13, p 339, eq. 73 p 339, eq. 77 p 340, eq. 85 p 340, eq. 91 x 3 (l c 11 λ λ 3 + λ λ 3 (λ 1 λ (λ 1 λ 3. c 11 λ 1 (3λ 1 + 1(λ 1 1. (λ + 1λ 3λ + 1 λn + O(0.738 n. λ l+1 (3λ + 1(λ 1 + O(0.738l 0.336λ l. p 341, l 16. where I s the dentty matrx and A s the matrx p 34, eqs T Polyphase Merge Lλ (3λ + 1(λ 1 + O(0.401l 1 j n nlλ (3λ + 1(λ 1 + O(1 λ L lg N + O( L lg N. (3λ + 1(λ 1 lg λ p 34, l 14. Merge about 80 percent faster than Smple Merge p 343. Check wth Q. Stout to see f there are any more errors on ths page. p 344, eq p 345, eq d P k (t dt µp k 1 (t (kλ + µp k (t + (k + 1λP k+1 (t, A(t, v at + ( t p (1 p t A(t, v 1. p 346, l 1. (usng the boundary condtons of the orgnal problem, whch reduces to p 346,. The general soluton of eq. (16 s the functon

11 p 348, eq ( sn q x (n, x ps qr ( sn q y (n, y ps qr p 351, eq z (n, z rn + p, ps qr, rn + p ps qr ( sn q rn + p, ps qr ps qr ( n+ a n x(c + a[n ], d + b[n ], y(c + a[n ], d + b[n ], z(c + a[n ], d + b[n ]. ( n+ 1 + p 355, 1. When n 1, H n s zero except for 0, and H 10 1, so F obeys p 356, ex. 8 n T n (n T n 1, 1 + T n 1, for n 3. p 356, ex. 8 The number of k-dmensonal cubes (k-cubes n an n-cube obeys the recurrence p 360, l 5. The materal n ths secton s adapted from Cohen [81], [Carefully check that the above correcton s made correctly!] p 370. Parent Class Parent Class Parent Class Parent Class 3 4, ,1 1,3 9 30, , ,16 3 4,5 3 33,34 6, ,18 6 7, ,40 8 9, ,1 8 9, , ,13 TABLE 9.1. The nontrval equvalence classes for the flow graph n Fgure 9.5. Two nodes are equvalent f there exsts a node wth arcs to both of them. The equvalence classes wth just one node are not gven. p 377, l 17. Input: A text strng T whch s an array p 378, l. next j equals the largest such that p 380, l 6. calls to Unon and Fnd. p 380, eq. 3. nt 0 + t 1 f(a, S p 387, l 3. Ths s a much better p 388, Step 4. If y t, then set z y, y parent(y, parent(z t, and repeat ths step. p 389, Fg A dot s needed n the mddle of the three long paths. That s n the rght path of the thrd fgure, the second from the left path of the fourth fgure and the lower part of the rghtmost path of the fourth fgure. p 393, TABLE 9.9. Ax 1 p 393, l. to evaluatng the polynomal 0 k<n x ky k at the ponts y j ω j for 0 j < n. 11

12 1 p 394, l 4 and many other p 395, l 1. so eq. (37 p 397, l one for each of the two subpart, p 398. Level Calculaton y 0 y 0 + y 1 x j + x j+4 y 1 y 0 y 1 x j x j+4 1 y 0 y 0 + y x j + x j+4 + x j+ + x j+6 y y 0 y x j + x j+4 x j+ x j+6 y 1 y 1 + ω y 3 x j x j+4 + ω x j+ ω x j+6 y 3 y 1 ω y 3 x j x j+4 ω x j+ + ω x j+6 0 y 0 y 0 + y 4 x 0 + x 4 + x + x 6 y 4 y 0 y 4 x 0 + x 4 + x + x 6 +x 1 + x 5 + x 3 + x 7 x 1 x 5 x 3 x 7 y 1 y 1 + ωy 5 x 0 x 4 + ω x ω x 6 +ωx 1 ωx 5 + ω 3 x 3 ω 3 x 7 y 5 y 1 ωy 5 x 0 x 4 + ω x ω x 6 ωx 1 + ωx 5 ω 3 x 3 + ω 3 x 7 y y + ω y 6 x 0 + x 4 x x 6 +ω x 1 + ω x 5 ω x 3 ω x 7 y 6 y ω y 6 x 0 + x 4 x x 6 ω x 1 ω x 5 + ω x 3 + ω x 7 y 3 y 3 + ω3 y 7 x 0 x 4 ω x + ω x 6 +ω 3 x 1 ω 3 x 5 ω 5 x 3 + ω 5 x 7 y 7 y 3 ω3 y 7 x 0 x 4 ω x + ω x 6 ω 3 x 1 + ω 3 x 5 + ω 5 x 3 ω 5 x 7 p 399, l. dfferent ways and then does not make any addtonal use of the orgnal p 399, l 1. the varable q used n Step has a leadng zero, followed by the j hgher-order bts of the n reversed order, followed by k j 1 tralng zeros. p 399, Step. For 0 < n do the rest of ths step. If k j 1 0, then set q 0 k j k j+1... k , odd ω q x + k j 1, p 401, l 8. We need (see exercse p 405, l 13. (for specal values of such numbers are called Mersenne prmes n the frst case and Fermat prmes n the second case. p 40, l 4. Arrays T 1, T, T 3, T 4, and T 5. p 454 eq. 3 F (x 1 x (y µ exp ( π σ dy.

13 p 454 eq. 33 p 463 eq. 70. Prob(X 70 x f(x 1 (x µ exp ( π σ. ( ( p 463 eq. 73. Prob( X ( 400 p 463 l 1. reduce the lmt to p 464 eq. 83. ( n p (1 p n exp[ ax + n ln(1 p + pe a ] p 478, l 17. rather than a j. p 479 eq p 479 eq p 488, eq. 7 p 494, eq. 5 Γ dz z z [y a bx ] 0, [y a bx ]x y a 1 3 y x a ( ( f(x x m n 1 1 x 1 /x 1 b 1 3 x b 1 3 x 0, x 0, 13 1 e πθ d(e πθ π dθ π. ( x /x ( 1 1 x n /x x m n + (x 1 + x + + x n x m n 1 +, (5 p 496, ref. 16 Mlton Abramowtz and Irene A. Stegun (eds. p 501, ref Ene neue Art von Zahlen, hre Egenschaften und Anwendung n der mathematschen Statstk p 505, eq..79 A q. p 531, Delete Euler, number 353. p 531, Euleran number n (3