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Sahni s  handbook of data structures and applications

Sahni s handbook of data structures and applications

... 100 0 Now when n = 100 , it would take 3.17 years to execute n 10 instructions and ∗ 101 0 years to execute 2n instructions n 10 20 30 40 50 100 103 104 105 106 n 01 µs 02 µs 03 µs 04 µs 05 µs 10 ... the adds) + 200 0 ∗ (for the stores) = 118 ,00 0 cycles If the second algorithm has 100 percent L2 misses, it will take 200 0 ∗ 100 (L2 misses) + 100 0 ∗ (adds) + 100 0 ∗ (stores) = 202 ,00 0 cycles So ... the computer would need second when n = 100 0, 1 10. 67 minutes when n = 10, 000 , and 11.57 days when n = 100 ,00 0 Figure 1.13 gives the time that a 1 ,00 0 ,00 0 ,00 0 instructions per second computer needs...

Ngày tải lên: 13/09/2015, 09:45

1,3K 6,8K 0
Đề tài " Classical and modular approaches to exponential Diophantine equations I. Fibonacci and Lucas perfect powers " pdf

Đề tài " Classical and modular approaches to exponential Diophantine equations I. Fibonacci and Lucas perfect powers " pdf

... APPROACHES 999 where a = 100 704 59885442777 702 417941827394441148299 900 2799, b = 100 704 59885442777 702 417941827394441148299 900 2 801 , c = 201 409 197 708 85555 404 835883654788882296599 800 5599, and the value ... K K0 = + + − − − + (K0 + 1)2 K0 + K0 + 3 2 (K0 − 1)I I K K0 + = − − (K0 + 1) K0 + 12 I (K0 − 1)(K0 + 1) K0 (K0 + 2)(K0 + 1)2 − 1+ K0 + I 12I which proves the lemma = , 100 4 YANN BUGEAUD, MAURICE ... 100 3 CLASSICAL AND MODULAR APPROACHES Lemma 12.2 Let K0 , L and I be positive integers with K0 ≥ 3, L ≥ and I ≥ K0 (K0 + 1)/2 Then I2 2(K0 + 1) Θ(K0 , I) ≥ 1+ (K0 − 1)(K0 + 1) K0 (K0 + 2)(K0...

Ngày tải lên: 06/03/2014, 08:21

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candlesticks, fibonacci and chart pattern trading tools - fischer 2003

candlesticks, fibonacci and chart pattern trading tools - fischer 2003

... the ratios 0. 618, 1 .00 0, and 1.618 of the PHI series: FIBONACCI ANALYSIS • 17 • 38.2 percent is the result of the division 0. 618 ÷ 1.618 • 50. 0 percent is the transformed ratio 1 .00 0 • 61.8 percent ... ISBN 0- 471-44861-3 (hard : CD-ROM) Investments Securities Investment analysis I Fischer, Jens II Title HG4521.F584 200 3 332.63′ 204 2—dc21 200 300 6623 Printed in the United States of America 10 This ... Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07 0 30, 201 -748- 601 1, fax 201 -748- 600 8, e-mail: permcoordinator@wiley.com Limit of Liability/ Disclaimer of Warranty:...

Ngày tải lên: 03/05/2014, 12:58

273 1,4K 0
Báo cáo toán học: "The q-Binomial Theorem and two Symmetric q-Identities" doc

Báo cáo toán học: "The q-Binomial Theorem and two Symmetric q-Identities" doc

... certain e L-functions, Duke Math J., 108 ( 200 1), 395–419 [5] R Chapman, Combinatorial proofs of q-series identities, J Combin Theory Ser A, 99 ( 200 2), 1 16 [6] G Gasper, Elementary derivations ... Publishers, Dordrecht, 1999, pp 101 –113 [3] G E Andrews, The Theory of Partitions, Cambridge University Press, Cambridge, 1998 the electronic journal of combinatorics 10 ( 200 3), #R34 [4] G E Andrews, ... σ(λ) gives the 2-modular diagram 2 2 2 2 2 2 2 2 2 2 the electronic journal of combinatorics 10 ( 200 3), #R34 Namely, σ(λ) = (19, 13, 6, 5, 3) Clearly, the number of rows in the diagram of λ is...

Ngày tải lên: 07/08/2014, 07:21

8 313 0
Báo cáo toán học: "Hybrid Proofs of the q-Binomial Theorem and other identities" ppsx

Báo cáo toán học: "Hybrid Proofs of the q-Binomial Theorem and other identities" ppsx

... J Number Theory 80 ( 200 0), no 2, 273–2 90 [6] G E Andrews, A Knopfmacher and P Paule, An infinite family of Engel expansions of Rogers-Ramanujan type, Adv in Appl Math 25 ( 200 0), no 1, 2–11 [7] ... that ∞ 2 (−q ; q )∞ n =0 ∞ (−q; q )∞ n =0 qn = (q ; q )n n2 +n q = (q ; q )n (−q; q )n the electronic journal of combinatorics 18 ( 201 1), #P 60 ∞ C(n)q n n =0 ∞ D(n)q n n =0 16 As in the proof of ... of 10 counted by B( 10) , C( 10) and D( 10) Those counted by B( 10) are {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {4, 1, 1, 1, 1, 1, 1}, {4, 4, 1, 1}, {6, 1, 1, 1, 1}, {6, 4}, {9, 1}, those counted by C( 10) ...

Ngày tải lên: 08/08/2014, 14:22

21 190 0
Data Structures and Algorithms - Chapter 8: Heaps pptx

Data Structures and Algorithms - Chapter 8: Heaps pptx

... is empty) return underflow else MinData = Data [0] Data [0] = Data[count -1] count = count - ReheapDown (0, count -1) return success End DeleteHeap 16 DeleteHeap (ref MinData ) ... function ReheapDown position = count / -1 loop (position > =0) ReheapDown(position, count-1) position = position - return End BuildHeap2 20 Complexity of Binary Heap Operations 21 d-heaps  d-heap ... (position>=count ) OR (position

Ngày tải lên: 15/03/2014, 17:20

41 619 3
Báo cáo khoa học: Roles of adenine anchoring and ion pairing at the coenzyme B12-binding site in diol dehydratase catalysis pptx

Báo cáo khoa học: Roles of adenine anchoring and ion pairing at the coenzyme B12-binding site in diol dehydratase catalysis pptx

... 10) 4 Wild-type Sa224A Sa224N Kb135R Kb135A Kb135Q Kb135E 336 64 17 254 196 211 7.7 0. 15 0. 15 1. 90 0.12 0. 39 0. 39 0. 40 2.2 0. 43 0. 009 2.1 0. 50 0.54 0. 019 0. 027 0. 46 0. 0 70 0 .02 7 0. 059 0. 068 0. 008 5 ... 0. 008 5 75 0. 8 1.5 56 20 19 5.4 ( 100 ) (19) (5) (76) (58) (63) (2) ± ± ± ± ± ± ± 0. 02 0. 01 0. 01 0. 01 0. 02 0. 01 0. 06 FEBS Journal 275 ( 200 8) 6 204 –6 216 ª 200 8 The Authors Journal compilation ª 200 8 FEBS ... during 6 208 0. 94 0. 36 3.4 0. 25 2.2 5.4 127 kcat ⁄ Km (AdoCbl) · 10) 8 (s)1ÆM)1) Km for AdoCbl (lM) ± ± ± ± ± ± ± 0. 11 0. 05 0. 2 0. 04 0. 4 0. 6 11 Ki for CN-Cbl (lM) 3.6 1.8 0. 05 10. 2 0. 89 0. 39 0. 000 61...

Ngày tải lên: 23/03/2014, 06:20

13 380 0
Báo cáo hóa học: "EXTENDING GENERALIZED FIBONACCI SEQUENCES AND THEIR BINET-TYPE FORMULA" pot

Báo cáo hóa học: "EXTENDING GENERALIZED FIBONACCI SEQUENCES AND THEIR BINET-TYPE FORMULA" pot

... variables 0 ,α−1 , ,α−r+1 : V1 = a0 0 + a1 α−1 + · · · + ar −1 α−r+1 , V2 = a0 V1 + a1 0 + a2 α−1 + · · · + ar −1 α−r+2 + a0 α−r+1 , V3 = a0 V2 + a1 V1 + a2 0 + a3 α−1 + · · · + ar −1 α−r+3 + a0 α−r+2 ... have m1 −1 lim i→∞ t =0 α1,t (−i) t = ⎧ ⎨+∞ ⎩ α 1 ,0 (= 0) if t > 0, if t = 0, (3.15) Extending generalized Fibonacci sequences and hence we have limi→∞ |B−i | = +∞ or |α1 ,0 |, since |λ1 | ≤ This ... i→∞ i→∞ t =0 j =1 α j,t (−i)t λj −i = 0, (3 .16) we have i→∞ (−i)m −1 m j −1 = lim −1 m j −1 = lim i→∞ j =1 t =0 j =1 t =0 i λ λj α j,t (−i)t (3.17) λ λj α j,t (−i)t−m +1 m −1 i + t =0 α ,t (−i)t−m...

Ngày tải lên: 22/06/2014, 22:20

11 371 0
Báo cáo toán học: " On the Symmetric and Rees Algebras of Some Binomial Ideals" pdf

Báo cáo toán học: " On the Symmetric and Rees Algebras of Some Binomial Ideals" pdf

... specialization, J Algebra 81 (1983) 202 –224 Ph Gimenez, M Morales, and A Simis, The analytical spread of the ideal of codimension mononial varieties, Result Math 35 (1999) 2 50 259 M L Ha and M Morales, ... 21 (1993) 647–664 G Valla, On the symmetric and Rees algebras of an ideal, Manuscripta Math 30 (19 80) 239–255 ... −β1 xμ− xr ⎠ s −x xt Hence, we have the following relations β2 xν− xr fu + α1 xμ+ fv − xs fu+v = 0, α2 xν+ fu + β1 xμ− xr fv − xt fu+v = Let us remark that if either xs = or xt = then I is a complete...

Ngày tải lên: 06/08/2014, 05:20

8 431 0
Báo cáo toán học: "Meta-Fibonacci Sequences, Binary Trees and Extremal Compact Codes" pptx

Báo cáo toán học: "Meta-Fibonacci Sequences, Binary Trees and Extremal Compact Codes" pptx

... a0 a1 a2 d0 d1 d2 p0 p1 p2 1 1 1 1 1 2 1 0 2 1 2 1 2 0 11 14 12 15 4 11 14 17 1 12 15 18 4 1 16 20 24 10 4 0 17 21 25 11 1 19 23 27 12 1 20 24 28 13 0 23 27 31 14 1 24 28 32 15 8 1 26 30 34 16 ... = 110r1 110r2 · · · 110r2n−1 , which will finish the proof of the first equality since D0 = E∞ By induction En+1 = En En = 110r1 110r2 · · · 110r2n−1 110r1 110r2 · · · 110r2n−1 = 110r1 110r2 · ... known as the “ruler function” (A 001 511) If the alternating 0 s are removed from the sequence r1 − 1, r2 − 1, r3 − 1, r4 − 1, = 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, then the ruler function...

Ngày tải lên: 07/08/2014, 13:21

15 197 0
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