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Analyzing Series

Analyzing Series. CSC2110 Tutorial 9 Darek Yung. Outline. Continue on Cryptography Review on Series Examples Class Work 3. Repeated Squaring. Calculate 5795 73 (mod 21) by repeated squaring 72 = 64 + 8 + 1 5795  20 (mod 21) 5795 2  1 (mod 21)

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Analyzing Series

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  1. Analyzing Series CSC2110 Tutorial 9 Darek Yung

  2. Outline • Continue on Cryptography • Review on Series • Examples • Class Work 3

  3. Repeated Squaring • Calculate 579573 (mod 21) by repeated squaring • 72 = 64 + 8 + 1 • 5795  20 (mod 21) • 57952 1 (mod 21)  57954 57958 579516 1 (mod 21) 579532 579564 1 (mod 21) • 579572 579564 * 57958 * 57951  20 (mod 21) • Can we apply Fermat’s Little Theorem i.e. 579520 1 (mod 21) and thus 579573 579513(mod 21)??? • NO!!!! 21 is not prime!!!

  4. RSA • Under RSA, given p = 13, q = 37: • Calculate n = pq, T = (p-1)(q-1) • Given e =35, calculate d = e-1 (mod T) • Encrypt the message m = 37 into m’ • Decrypt the message m’ into m

  5. RSA • p = 13, q = 37: • Calculate n = pq, T = (p-1)(q-1) • N = (13)(37) = 481 • T = (p-1)(q-1) = 432

  6. RSA • p = 13, q = 37, n = 481, T = 432 • Given e =35, calculate d = e-1 (mod T) • 432 = 12(35) + 12  12 = 432 – 12(35) • 35 = 2(12) + 11  11 = 35 – 2(12) = -2(432) + 25(35) • 12 = 1(11) + 1  1 = 12 – 1(11) = 3(432) – 37(35) • 11 = 11(1) + 0 • d = -37 = 432 – 37 = 395

  7. RSA • p = 13, q = 37, n = 481, T = 432, e = 35 • Encrypt the message m = 37 into m’ • 371 37 (mod n) • 372 372 407(mod n) • 374 4072 185(mod n) • 378 1852 74(mod n) • 3716 742 185(mod n) • 3732 1852 74(mod n) • m’  3735 (3732) (372) (371)  (74) (407) (37)  370 (mod n)

  8. RSA • p = 13, q = 37, n = 481, T = 432, d = 395, m’ = 370 • Decrypt the message m’ into m • 3701 370 (mod n) • 3702 3702 296(mod n) • 3704 2962 74(mod n) • 3708 742 185(mod n) • 37016 1852 74(mod n) • 37032 742 185(mod n) • 37064 1852 74(mod n) • 370128 742 185(mod n) • 370256 1852 74(mod n) • m’  370395 (370256) (370128) (3708) (3702) (3701)  37 (mod n) • NO WORRY! If repeated squaring appears in next Wednesday’s class work, the numbers will be small~

  9. Another RSA • Under RSA, given p = 17, q = 41: • Calculate n = pq, T = (p-1)(q-1) • Given e =49, calculate d = e-1 (mod T) • Encrypt the message m = 21 into m’ • Decrypt the message m’ into m

  10. Another RSA • p = 17, q = 41: • Calculate n = pq, T = (p-1)(q-1) • N = (13)(37) = 697 • T = (p-1)(q-1) = 680

  11. Another RSA • p = 17, q = 41, n = 697, T = 680 • Given e =49, calculate d = e-1 (mod T) • 680= 13(49) + 43  43 = 680 – 13(49) • 49 = 1(43) + 6  6 = 49 – 43 = -680 + 14(49) • 43 = 7(6) + 1  1 = 43 – 7(6) = 8(680) – 111(49) • 6 = 6(1) + 0 • d = -111 = 680 – 111 = 569

  12. Another RSA • p = 17, q = 41, n = 697, T = 680, e = 49 • Encrypt the message m = 21 into m’ • 211 21 (mod n) • 212 212 441(mod n) • 214 4412 18(mod n) • 218 182 324(mod n) • 2116 3242 426(mod n) • 2132 4262 256(mod n) • m’  2149 (2132) (2116) (211)  (256) (426) (21)  531 (mod n)

  13. Another RSA • p = 17, q = 41, n = 697, T = 680, d = 569, m’ = 531 • Decrypt the message m’ into m • 5311 531 (mod n) • 5312 5312 373(mod n) • 5314 3732 426(mod n) • 5318 4262 256(mod n) • 53116 2562 18(mod n) • 53132 182 324(mod n) • 53164 3242 426(mod n) • 531128 4262 256(mod n) • 531256 2562 18(mod n) • 531512 182 324(mod n) • m’  531569 (531512) (53132) (53116) (5318) (5311)  21 (mod n)

  14. Review on Series • Geometric Series • Future and Current Value • Annuity • Related Sums (talk after Rental Problem) • Integral Method (talk after Rental Problem)

  15. Geometric Series • Gn: 1 + x + x2 +x3 + … + xn • Gn = (1 – xn+1) / (1 – x) • Gn = 1 / (1 – x) for n  infinity, |x| < 1

  16. Future and Current Value • With annual interest rate r • $m now = $m(1+r)n after n years Future Value • $m after n years = $m / (1+r)n now  Current / Present Value • E.g. With $1 next year • Current Value = $1 / (1+r) • Future Value = $1(1+r) for 2 years from now

  17. Annuity • Borrow $V from bank with annual interest rate r, and return $A annually for n years • Current Value of ith payment = A / (1+r)i  V = A / (1+r) * [1 – (1+r)-n] / [1 – (1+r)-1] (By Geometric sum, 1st term = A / (1+r) )  V = A * [(1+r)n – 1] / [r *(1+r)n]  A = V * [r * (1+r)n] / [(1+r)n – 1]

  18. Flat Rental Problem • Given • price of a flat is $1,000,000 • Interest rate is 10% • Inflation rate is 5% • Loan period is 20 years • Suppose John wants to borrow 100% of loan to buy the flat • What’s the annual payment? • What’s the value of the flat after 20 years? • Suppose he can rent the flat for $80,000 and invest the money saved to get 10% annual return, how much money will he have after 20 years?

  19. Flat Rental Problem • What’s the annual payment? • Recall • V = A / (1+r) * [1 – (1+r)-n] / [1 – (1+r)-1] • V = A * [(1+r)n – 1] / [r *(1+r)n] • A = V * [r * (1+r)n] / [(1+r)n – 1] • Substitute V = 1,000,000, r = 10%, n = 20, • A = $117,460

  20. Flat Rental Problem • What’s the value of the flat after 20 years? • Value after 20 years • = 1,000,000 * (1+5%)20 • = $2,653,298

  21. Flat Rental Problem • How much money will John have if he rent the flat? • Money saved annually m • = 117,460 – 80,000 = $37,460 • After 20 years, Investment make at ith year will contribute $ [ m * (1+r)20-i ] • Therefore, total money after 20 years M is

  22. Flat Rental Problem • How much money will John have if he rent the flat? • By closed form of Geometric Sum, • Substitute m = $37,460, r = 10%, M = $2,145,521

  23. Another Flat Rental Problem • Given • price of a flat is $2,000,000 • Interest rate is 12% • Inflation rate is 5% • Loan period is 20 years • Suppose John wants to borrow 70% of loan to buy the flat • What’s the annual payment? • What’s the value of the flat after 20 years? • Suppose he can rent the flat for $180,000 and invest the money saved to get 12% annual return, how much money will he have after 20 years?

  24. Another Flat Rental Problem • What’s the annual payment? • Again, recall • V = A / (1+r) * [1 – (1+r)-n] / [1 – (1+r)-1] • V = A * [(1+r)n – 1] / [r *(1+r)n] • A = V * [r * (1+r)n] / [(1+r)n – 1] • Substitute V = 1,400,000, r = 12%, n = 20, • A = $187,430

  25. Another Flat Rental Problem • What’s the value of the flat after 20 years? • Value after 20 years • = 2,000,000 * (1+5%)20 • = $5,306,595

  26. Another Flat Rental Problem • How much money will John have if he rent the flat? • Money saved annually m • = 187430 – 180,000 = $7,430 • After 20 years, Investment make at ith year will contribute $ [ m * (1+r)20-i ] • Therefore, total money after 20 years M is

  27. Another Flat Rental Problem • How much money will John have if he rent the flat? • By closed form of Geometric Sum, • Substitute m = $7,430, r = 12%, M = $535,350

  28. Another Flat Rental Problem • How much money will John have if he rent the flat? • M = $535,350 • On another hand, the 30% down payment contribute 2,000,000 * 30% * (1+12%)20 • = $5,787,776 after 20 years • Therefore, total amount of money = $6,323,126

  29. y = f(x+1) f(6) f(5) y = f(x) f(4) f(3) f(2) f(1) f(0) 0 1 2 3 4 5 6 Integral Method

  30. Integral Method • When f(x) is increasing, summation of 1 to n terms have • Upper bound = integration on f(x+1) from 0 to n • Lower bound = integration on f(x) from 0 to n

  31. f(0) y = f(x) f(1) f(2) y = f(x+1) f(3) f(4) f(5) f(6) 0 1 2 3 4 5 6 Integral Method

  32. Integral Method • When f(x) is decreasing, summation of 1 to n terms have • Upper bound = integration on f(x) from 0 to n • Lower bound = integration on f(x+1) from 0 to n

  33. Example on Integral Method • Use Integral Method to find upper and lower bounds for Summation of i3 from i = 1 to n • i3 is increasing function • Upper bounded by integration on (i+1)3 • Lower bounded by integration on i3

  34. Related Sums • Sn: x + 2x2 + 3x3 + … + nxn • Sn = [x – (n+1)xn+1 + nxn+2] / (1 – x)2 • Sn = x / (1 – x)2 for n  infinity, |x| < 1

  35. Example on Closed form • Express 3 + 3(33) + 5(35) + … + (2n-1)(32n-1) in closed form • 3 + 3(33) + 5(35) + … + (2n-1)(32n-1) • = [3 + 2(32) + 3(33) + … + (2n)(32n)] - 2 [(32) + 2((32)2) + 3((32)3) + … + (n)((32)n)] • = [3 – (2n+1)32n+1 + (2n)32n+2]/(1-3)2 - 2 * [32 – (n+1)32n+1 + (n)32n+4]/(1-32)2 (By Sn = [x – (n+1)xn+1 + nxn+2] / (1 – x)2, take x = 3, and 32) • = [-(n)32n+4 – (17n+1)32n+2 + (16n+8)32n+1 + 15] / 4

  36. Class Work 3 • On 14 Nov, Wednesday • For more than 80% of the class work • Straight Forward, not tricky • No proof is needed (but may include express summation in closed form) • Numbers are selected that calculations (like GCD Algorithm) will not take too many iterations • Make sure understand sample class work

  37. Check List for Number Theory • Include but may not limit to: • Concepts • Applications • Interesting Examples

  38. Concept • Properties on Divisibility • Properties on GCD • Linear Combination & GCD = SPC

  39. Concept • Properties of Prime • Properties of Modular Arithmetic • Multiplicative Inverse & Fermat’s Little Theorem

  40. Concept • Closed Form of Sums • Future and Current Value • Annuity

  41. Application • (Extended) GCD Algorithm (e.g. Q43) • Die Hard Problem (e.g. Q44)

  42. Application • Calculating Date (e.g. Q48) • Evaluating divisibility of 9 and 11 (e.g. Q61) • Calculating Multiplicative Inverse • SPC (GCD) Method (e.g. Q52) • Repeated Squaring (e.g. Q51) • Turing’s Code (Version 1 & 2) (e.g. Q58,59) • RSA (e.g. Q60)

  43. Application • Express in closed form (e.g. Q53) • Annuity (e.g. Q63) • Integral Method (e.g. Q57) • Calculating Future / Current Value (e.g. Q64)

  44. Interesting Example • Examples Discussed during Lectures~

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