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Assignment 3

CS-GY 6003 INET Spring 2022

Due date:

11:55pm on April 15th 2022 on Gradescope.

Instructions:

Write or type out your solutions and hand in on Gradescope before the deadline. Justify your answers!

Question 1: Counting

.

(a) 5 points Use a combinatorial argument to prove each of:

(4n)!

22n · n! · n!

=

(

4n

2n

)

(2n− 1)2(2n− 3)2(2n− XXXXXXXXXX)

(b) 4 points Use pascal’s triangle to prove(

n + 1

+ 1

)

=

n∑

k=

(

k

)

(c) 3 points Use Pascal’s identify to prove that:

22n = 2

[(

2n

0

)

+

(

2n

1

)

+

(

2n

2

)

+

(

2n

3

)

+ . . . +

(

2n

n− 1

)]

+

(

2n

n

)

Question 2: Number Theory

.

(a) 4 points Let p = 4567 and q = 113. Pick a public key e where 86 ≤ e ≤ 94. Encode the message

m = 100 and show how it is successfully decoded.

(b) 3 points Suppose that in mod m we determine that the inverse of x is y. Explain why xk has an

inverse in mod m for any k ≥ 1. What is the inverse of x3?

(c) 5 points Solve the following for x in mod 397. Show your work without using a calculator.

XXXXXXXXXXx ≡ 2402 mod 397

(d) 5 points Use CRT to find a value of 0 ≤ x < XXXXXXXXXXthat satisfies the following congruences:

2x ≡ 15 mod 29

19x ≡ 100 mod 1000

x ≡ 52 mod 117

(e) 5 points Use CRT to solve the following for 0 ≤ x < 899 . You must find all 4 distinct solutions

in that range.

x2 ≡ 33 mod 29

x2 ≡ 66 mod 31

1

(f) 5 points Determine integers s and t such that

5744s = 3260t mod 4

(g) 6 points For each of the following, solve for x or prove that there is no solution:

• 6x + 5 ≡ 20 mod 91

• 13x + 2 ≡ 20 mod 91

• 7x + 2 ≡ 23 mod 91

(h) 6 points

• Update the solution to problem 14 in set 11 so that Power(x,a,n) returns the value of xa mod n,

where 0 ≤ a < n is an integer.

• Write a procedure FastExp(x,a,p) which determines the value of xa mod p where p ≥ 2 is a prime

number. Your procedure can assume that a ≥ 0 and n ≥ 1, which are both integers. Your solution

must use Fermat’s Last theorem, and may call Power(x,a,n) from above.

Question 3: Recu

ence Relations:

. (a) 6 points

• Let F (n), n ≥ 1 be the Fibonacci sequence defined in class. Used the closed-form solution of the

Fibonacci sequence to find F (15). Does this give you the exact same value as if you compute F (15)

using the recu

ence? Explain why or why not.

• Define a sequence g(n) , n ≥ 1 as follows: Let g(1) = 5 and g(2) = 8, and g(n) = g(n− 1) + g(n− 2)

for n ≥ 3. What is the closed-form solution for g(n)? Justify your answer.

.

(b) 6 points A child is building a tower out of a total of n blocks. Let r of those blocks be red, and

the remaining blocks are blue. The child builds a tower using all of the blocks. Suppose that the tower is

uilt with no adjacent red blocks. Let B(n, r) be the number of ways to build such a tower. Write a

ecursive definition for B(n, r), where n ≥ 0, and 0 ≤ r ≤ n. Include your base cases. What is the solution

to this recu

ence? Use your recu

ence to solve for B(6, 2) and B(7, 4), and verify they are co

ect using

your solution to the recu

ence.

(c) 5 points Solve the following recu

ence and prove your result is co

ect using induction:

a0 = 1

a1 = 1

an = 4an−2 + 1 for n ≥ 2

(d) 4 points Solve the following recu

ence:

an = 5an−2 − 4an−4 for n ≥ 4

a0 = 0, a1 = 1, a2 = 1, a3 = 1

(e) 5 points Suppose we are given a rod of length n. We can cut rod into pieces in such a way that

each piece length is a natural number of size at least 1. There are many ways to cut the rod. For example,

if n = 5, we could cut it into pieces of size 1, 2, 2 or 4, 1 etc. Write a recursive definition for P (n, k) which

is the number of ways to cut a rod of length n where the maximum size of the pieces is k. Your definition

2

is for n ≥ 1 and k ≥ 1. You do not need to solve the recu

ence. Use your recu

ence to solve for P (7, 7)

and P (7, 2).

(f) 5 points A child is building a tower out of red, black, and green blocks. The red and black blocks

have size 1 and the green block has size 2. A tower of height n is built using a sequence of the colored

locks, with the condition that no red blocks are adjacent. Write a recursive definition T (n), n ≥ 1, fo

the number of possible towers of height n. You do not need to solve the recu

ence.

Question 4: Recursive Procedures

.

(a) 6 points Recall that in week 6 we considered the problem of distributing n identical items into k

unlabelled groups. In this problem, you must write a recursive procedure called NumberPartitions(n, k)

that returns the number of ways of distributing n identical pennies into k piles, where each pile has at

least 1 penny. The piles are not labelled, so the only thing that matters is the number of coins in each

pile. Your procedure must function properly for n ≥ 1 and 1 ≤ k ≤ n. Use your procedure to verify the

output for n = 8 and k = 4.

(b) 6 points Consider a making a postage of n cents using stamps of size 2, 3 and 4. Suppose we must

use at least one of each stamp denomination. Write a recursive procedure called PrintPostage(n) that

prints out a set of stamps that create a postage of size n. You must determine the co

ect base case and

ensure that your recursive procedure handles that co

ectly. If no postage is possible, your procedure must

print “No postage”. Your procedure must handle all inputs of n ≥ 0.

For example: FindPostage(25) outputs stamps 4, 4, 4, 4, 4, 3, 2. There are other options possible, but just

one co

ect option must be printed.

(c) 6 points Recall problem 4(c) from assignment 2. Consider a new version of this problem where the

ills have denominations 1, 5, 52, 53, XXXXXXXXXXAs in the previous assignment, you owe your friend a debt of n

dollars. You must transfer a set of bills between you and your friend in such a way that the debt is repaid.

In this case, a bill denomination may be used at most two times.

Write a recursive procedure called PrintBills(n) that returns a list of the necessary bills, for n ≥ 0.

The list should include positive numbers for bills you give your friend, and negative numbers for bills you

friend gives you. For example, if n = 21, the output list should be [52,−5, 1]. For n = 32, the output list

should be [25, 5, 1, 1].

*Note: This is not a course on data-structures. So you may assume that you can create an empty list

with the pseudo-code “x = list()” and use the following operations:

-add elements to the list using x.add(25)

-update/access element at index i using x[i]

-determine the length using x.length().

3

Answered 2 days AfterApr 07, 2022

Assignment 3 CS-GY 6003 INET Spring 2022

The above is equivalent to proving

=*

Right side first factor is selecting 2n from 4n objects.

Left side denominator is (by using one 2 to multiply with one number)

= (1.2.3.4….(2n))^2 (because all odd numbers upto to 2n are on right side and all even numbers in first term of right side

=

Hence right side is nothing but

Thus proved

Q no2

Here p and q are prime

We write n =pq =516071

The element e belongs to {86,87….94}

E=89 since this is prime we have gcd of (89, 4566, 112) =1

So public message is e =89 and n = 516071

The private key will be the inverse of 89 in Z_515958

There decoding they factorise 516071 into 113 and 4567 (i.e. two big primes)

The receiver determines the inverse of e modulo (p-1)(q-1)

(p-1) (q-1) = 4566*112 =511392

Set up a division problem where a is larger than b.

a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.

511392 ÷ 89 = 5745 R 87 (511392 = 5745 × 89 + 87)

89 ÷ 87 = 1 R 2 (89 = 1 × 87 + 2)

87 ÷ 2 = 43 R 1 (87 = 43 × 2 + 1)

2 ÷ 1 = 2 R 0 (2 = 2 × 1 + 0)

When remainder R = 0, the GCF is the divisor, b, in the last equation. GCF = 1

Taking reverse

89(-252823)+ 44*511392 =1

Inverse is coefficient of 89 i.e. -252923

Hence private key is d = -252923

S= 100^89 = 100^89 mod 516071 =(483929)^33*100mod 516071 = 123673

The receiver would do s^d and decode it.

Given that inverse of x is y. i.e. x*y = 1 mod m

Or m-1 is divisible by xy. This is possible only if gcd (x,m)=1

For any k>1, x^k will never have any common factor with m. Since (x^k, m) have gcd 1, we get x^k also definitely has an inverse

For x^3, gcd (x^3,m) =1

There exists integers s and t such that s*x^3+t*m=1

We already have x*s1+t1*m =1

Subtracting s1(x^3-x) +m(t-t1) =0

Or s1(x^3-x) = 0 mod m.

Or x is the inverse of x^3

We have 2^9 = 512 and 2^9 mod 397 = 115 mod 297

Hence

X=285-20^3963

2x = 15 mod 29

i.e 2x = {….-14, 15, 44, 73, 102,….}

For x to be an integer we have to select only even numbers on right

i.e. x ={…-7, 22, 51,…}

Or x = 22 mod 29 = (22,51,80, …3392993)

19x = 100 mod 1000 = {-900, 100, 1100, 2100,…….17100, …36100,…,55100,…}

Only 17100 is the least integer divisible by 19

Hence x = {…,900, 1000, 1100,1200,…}

X=0 mod 100

Since x square cannot be...

The above is equivalent to proving

=*

Right side first factor is selecting 2n from 4n objects.

Left side denominator is (by using one 2 to multiply with one number)

= (1.2.3.4….(2n))^2 (because all odd numbers upto to 2n are on right side and all even numbers in first term of right side

=

Hence right side is nothing but

Thus proved

Q no2

Here p and q are prime

We write n =pq =516071

The element e belongs to {86,87….94}

E=89 since this is prime we have gcd of (89, 4566, 112) =1

So public message is e =89 and n = 516071

The private key will be the inverse of 89 in Z_515958

There decoding they factorise 516071 into 113 and 4567 (i.e. two big primes)

The receiver determines the inverse of e modulo (p-1)(q-1)

(p-1) (q-1) = 4566*112 =511392

Set up a division problem where a is larger than b.

a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.

511392 ÷ 89 = 5745 R 87 (511392 = 5745 × 89 + 87)

89 ÷ 87 = 1 R 2 (89 = 1 × 87 + 2)

87 ÷ 2 = 43 R 1 (87 = 43 × 2 + 1)

2 ÷ 1 = 2 R 0 (2 = 2 × 1 + 0)

When remainder R = 0, the GCF is the divisor, b, in the last equation. GCF = 1

Taking reverse

89(-252823)+ 44*511392 =1

Inverse is coefficient of 89 i.e. -252923

Hence private key is d = -252923

S= 100^89 = 100^89 mod 516071 =(483929)^33*100mod 516071 = 123673

The receiver would do s^d and decode it.

Given that inverse of x is y. i.e. x*y = 1 mod m

Or m-1 is divisible by xy. This is possible only if gcd (x,m)=1

For any k>1, x^k will never have any common factor with m. Since (x^k, m) have gcd 1, we get x^k also definitely has an inverse

For x^3, gcd (x^3,m) =1

There exists integers s and t such that s*x^3+t*m=1

We already have x*s1+t1*m =1

Subtracting s1(x^3-x) +m(t-t1) =0

Or s1(x^3-x) = 0 mod m.

Or x is the inverse of x^3

We have 2^9 = 512 and 2^9 mod 397 = 115 mod 297

Hence

X=285-20^3963

2x = 15 mod 29

i.e 2x = {….-14, 15, 44, 73, 102,….}

For x to be an integer we have to select only even numbers on right

i.e. x ={…-7, 22, 51,…}

Or x = 22 mod 29 = (22,51,80, …3392993)

19x = 100 mod 1000 = {-900, 100, 1100, 2100,…….17100, …36100,…,55100,…}

Only 17100 is the least integer divisible by 19

Hence x = {…,900, 1000, 1100,1200,…}

X=0 mod 100

Since x square cannot be...

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