1. Randomized Algorithm & Public Key Cryptography
Binhai Zhu
Computer Science Department, Montana State University
Frequently, presenters must deliver material of a technical nature to an audience unfamiliar with the topic or vocabulary. The material may be complex or heavy with detail. To present technical material effectively, use the following guidelines from Dale Carnegie Training®.
Consider the amount of time available and prepare to organize your material. Narrow your topic. Divide your presentation into clear segments. Follow a logical progression. Maintain your focus throughout. Close the presentation with a summary, repetition of the key steps, or a logical conclusion.
Keep your audience in mind at all times. For example, be sure data is clear and information is relevant. Keep the level of detail and vocabulary appropriate for the audience. Use visuals to support key points or steps. Keep alert to the needs of your listeners, and you will have a more receptive audience.
5/21/2019

2. Randomized Algorithm
How to flip a coin?
5/21/2019

3. Randomized Algorithm How to flip a coin?
Easy. But the interesting part is that this kind of simple idea can help us designing algorithms.
5/21/2019

4. Randomized Algorithm How to flip a coin?
Easy. But the interesting part is that this kind of simple idea can help us designing algorithms.
Example. Quicksort.
5/21/2019

5. Randomized Algorithm How to flip a coin?
Easy. But the interesting part is that this kind of simple idea can help us designing algorithms.
Example. Quicksort.
In designing randomized algorithms, you can assume that we will toss a dice which can have many faces. (With a coin, it is a dice with 2 faces.)
5/21/2019

6. Assistant Hiring Problem
Suppose that you are assigned the job to interview and hire an office assistant, the rule is that the current best candidate will be hired.
Assume that the cost to interview a candidate is Ci and the cost to hire a candidate is Ch (typically Ch >> Ci).
What would you do?
5/21/2019

7. Assistant Hiring Problem
Assume that the cost to interview a candidate is Ci and the cost to hire a candidate is Ch (typically Ch >> Ci).
What would you do? Easy!
Hire-Assistant(n) //1 to n are the candidates
1. best ← 0
2. For i =1 to n
interview candidate i
If i is better than best
then best ← i
hire candidate i
5/21/2019

8. Assistant Hiring Problem
Assume that the cost to interview a candidate is Ci and the cost to hire a candidate is Ch (typically Ch >> Ci).
What would you do? Easy!
Hire-Assistant(n) //1 to n are the candidates
1. best ← 0
2. For i =1 to n
interview candidate i
If i is better than best
then best ← i
hire candidate i
What is the total cost of this algorithm?
5/21/2019

9. Assistant Hiring Problem
Assume that the cost to interview a candidate is Ci and the cost to hire a candidate is Ch (typically Ch >> Ci).
What would you do? Easy!
Hire-Assistant(n) //1 to n are the candidates
1. best ← 0
2. For i =1 to n
interview candidate i
If i is better than best
then best ← i
hire candidate i
What is the total cost of this algorithm?
If m candidates are hired, the total cost is
O(nCi+mCh).
As n, Ci, Ch are all constants, m is the crucial parameter!
5/21/2019

10. Assistant Hiring Problem
As m = n in the worst case, the worst case running time of Hire-Assistant(-) is O(n(Ci+Ch)). But intuitively this can’t be true in reality — it would be insane to hire all the candidates!
What can we do?
5/21/2019

11. Assistant Hiring Problem
As m = n in the worst case, the worst case running time of Hire-Assistant(-) is O(n(Ci+Ch)). But intuitively this can’t be true in reality — it would be insane to hire all the candidates!
What can we do?
Analyze the behavior of m, with a randomized algorithm!
5/21/2019

12. Assistant Hiring Problem
An algorithm is randomized if its behavior is determined not only by input but also by random numbers.
5/21/2019

13. Assistant Hiring Problem
An algorithm is randomized if its behavior is determined not only by input but also by random numbers.
Randomized-Hire-Assistant(n)
0. Randomly permute the list of candidates
1. best ← 0
2. For i =1 to n
3. interview candidate i
4. If i is better than best
then best ← i
hire candidate i
5/21/2019

14. Indicator Random Variable I{A}
I{A} = 1, if event A occurs.
I{A} = 0, if event A does not occur.
//let A- be the complement of A
Lemma. Given a sample space S and an event A in the sample space, let XA=I{A}. Then E{XA}=Pr{A}.
5/21/2019

15. Indicator Random Variable I{A}
I{A} = 1, if event A occurs.
I{A} = 0, if event A does not occur.
//let A- be the complement of A
Lemma. Given a sample space S and an event A in the sample space, let XA=I{A}. Then E{XA}=Pr{A}.
Proof. E{XA} = E[I{A}]
= 1 x Pr{A} + 0 x Pr{A-}
= Pr{A}.
5/21/2019

16. Indicator Random Variable I{A}
I{A} = 1, if event A occurs.
I{A} = 0, if event A does not occur.
//let A- be the complement of A
Lemma. Given a sample space S and an event A in the sample space, let XA=I{A}. Then E{XA}=Pr{A}.
Example. Let’s look at flipping n coins.
Yi — random variable denoting the outcome of the i-th flip
Xi — I{Yi=H}
X — random variable denoting the total number of HEADs in n coin flipping
What is E[X]?
5/21/2019

17. Indicator Random Variable I{A}
Lemma. Given a sample space S and an event A in the sample space, let XA=I{A}. Then E{XA}=Pr{A}.
Example. Let’s look at flipping n coins.
Yi — random variable denoting the outcome of the i-th flip
Xi — I{Yi=H}
X — random variable denoting the total number of HEADs in n coin flipping
What is E[X]?
X = Σ i=1 to n Xi, so
E[X] = E [Σ i=1 to n Xi ] = Σ i=1 to n E[Xi]
= Σ i=1 to n Pr[event Xi occurs] = Σ i=1 to n ½ = n/2
5/21/2019

18. Analysis of Randomized-Hire-Assistant
Xi = I{candidate i is hired}; i.e., Xi=1 if and only if i is hired.
X = Σ i=1 to n Xi
What is E[X]?
E[X] = E [Σ i=1 to n Xi ] = Σ i=1 to n E[Xi]
= Σ i=1 to n Pr[candidate i is hired occurs] //is the best
= Σ i=1 to n 1/i = 1 + ½ + 1/3 + …+ 1/n = O(log n).
The answer is natural, as log n << n.
5/21/2019

19. Public Key Cryptography
Private Key Cryptography is easy: everybody keeps a secret key for your account.
But this is not efficient when you (e.g., a bank) need to communicate with many people.
5/21/2019

20. Public Key Cryptography
Private Key Cryptography is easy: everybody keeps a secret key for your account.
But this is not efficient when you (e.g., a bank) need to communicate with many people.
The idea is that everybody maintains two keys, a private one and a public one. When Alice sends a message M to Bob, Alice encrypts the message using Bob’s public key PB, sends PB(M) to Bob and Bob decrypts it using his own private key SB, i.e., SB(PB(M))=M. (The tricky part is that you can’t decrypt the message using PB.)
5/21/2019

21. Public Key Cryptography
Another application is called Digital Signatures, you want to know the message you have received is from the true party from whom you are expecting a message.
The idea is again that everybody maintains two keys, a private one and a public one. When Alice sends a digital signature to Bob, Alice encrypts the signature M’ using her secret key SA, i.e., δ=SA(M’), sends (δ,M’) to Bob and Bob decrypts δ using Alice’s public key PA and verify whether PA(δ)=M’ or not. If not, then Bob can claim that the Alice who sends the signature is a fake.
5/21/2019

22. Public Key Cryptography
Another application is called Digital Signatures, you want to know the message you have received is from the true party from whom you are expecting a message.
The idea is again that everybody maintains two keys, a private one and a public one. When Alice sends a digital signature to Bob, Alice encrypts the signature M’ using her secret key SA, i.e., δ=SA(M’), sends (δ,M’) to Bob and Bob decrypts δ using Alice’s public key PA and verify whether PA(δ)=M’ or not. If not, then Bob can claim that the Alice who sends the signature is a fake.
PA(SA(M))=SA(PA(M))=M is the foundation for the whole public key system!
5/21/2019

23. Public Key Cryptography
Q: All these sound like fairy tales, how can it work?
A: Mathematics!
5/21/2019

24. Public Key Cryptography
Q: All these sound like fairy tales, how can it work?
A: Mathematics!
Of course, most of you are probably not at the level to understand the proofs yet. But let’s at least see a working algorithm.
5/21/2019

25. RSA Public Key Cryptography
Select two large random prime numbers p,q (>1000 bits).
n = pq.
Select a small odd integer e that is relative prime to (p-1)(q-1).
Find d such that de = 1 (mod (p-1)(q-1)).
Create keys, public key: (e,n),
secret key: (d,n)
P(M) = Me(mod n)
S(C) = Cd (mod n)
5/21/2019

26. RSA Public Key Cryptography
Select two large random prime numbers p,q (>1000 bits).
n = pq.
Select a small odd integer e that is relative prime to (p-1)(q-1).
Find d such that de = 1 (mod (p-1)(q-1)).
Create keys, public key: (e,n),
secret key: (d,n)
P(M) = Me(mod n)
S(C) = Cd (mod n)
Example:
p=11,q=29,n=319
(p-1)(q-1)=280
e=3
5/21/2019

27. RSA Public Key Cryptography
Select two large random prime numbers p,q (>1000 bits).
n = pq.
Select a small odd integer e that is relative prime to (p-1)(q-1).
Find d such that de = 1 (mod (p-1)(q-1)).
Create keys, public key: (e,n),
secret key: (d,n)
P(M) = Me(mod n)
S(C) = Cd (mod n)
Example:
p=11,q=29,n=319
(p-1)(q-1)=280
e=3
d=187
5/21/2019

28. RSA Public Key Cryptography
Select two large random prime numbers p,q (>1000 bits).
n = pq.
Select a small odd integer e that is relative prime to (p-1)(q-1).
Find d such that de = 1 (mod (p-1)(q-1)).
Create keys, public key: (e,n),
secret key: (d,n)
P(M) = Me(mod n)
S(C) = Cd (mod n)
Mde = Med = M (mod n) is the core of the proof.
5/21/2019