analyzing series

Analyzing Series

CSC2110 Tutorial 9

Darek Yung

Outline

Continue on Cryptography Review on Series Examples Class Work 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!!!

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

RSA

p = 13, q = 37: Calculate n = pq, T = (p-1)(q-1)

N = (13)(37) = 481 T = (p-1)(q-1) = 432

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

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)

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 nu

mbers will be small~

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

Another RSA

p = 17, q = 41: Calculate n = pq, T = (p-1)(q-1)

N = (13)(37) = 697 T = (p-1)(q-1) = 680

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

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)

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)

Review on Series

Geometric Series Future and Current Value Annuity Related Sums (talk after Rental Problem) Integral Method (talk after Rental Problem)

Geometric Series

Gn: 1 + x + x2 +x3 + … + xn

Gn = (1 – xn+1) / (1 – x)

Gn = 1 / (1 – x) for n infinity, |x| < 1

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

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]

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?

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

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

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 con

tribute $ [ m * (1+r)20-i ] Therefore, total money after 20 years M is

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

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?

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

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

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 con

tribute $ [ m * (1+r)20-i ] Therefore, total money after 20 years M is

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

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 contribu

te 2,000,000 * 30% * (1+12%)20

= $5,787,776 after 20 years Therefore, total amount of money = $6,323,126

Integral Method

f(1)

f(2)

f(3)f(4)

f(5) f(6)

0 1 2 3 4 5 6

f(0)

y = f(x)

y = f(x+1)

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

f(5)f(4)f(3)

f(2)f(1)

f(0)

0 1 2 3 4 5 6

f(6)

Integral Method

y = f(x)

y = f(x+1)

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

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

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

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

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

Check List for Number Theory

Include but may not limit to: Concepts Applications Interesting Examples

Concept

Properties on Divisibility Properties on GCD Linear Combination & GCD = SPC

Concept

Properties of Prime Properties of Modular Arithmetic Multiplicative Inverse & Fermat’s Little

Theorem

Concept

Closed Form of Sums Future and Current Value Annuity

Application

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

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)

Application

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

Q64)

Interesting Example

Examples Discussed during Lectures~