Numb3rs 513: Trouble in Chinatown

In this episode, Charlie and the team are trying to find the men responsible for the ghost bride service, which provides brides to families of Chinese men who died bachelors, to be buried alive with them. An undercover agent is selected as a bride. While being dragged into a van, she is able to make a phone call and state the license plate number. Now the team has to clear the tape of this call from all the background noise, to make her words intelligible. They use forensic audiology to filter out the unwanted sounds, such as car alarms and fireworks.

Audio Forensics

Audio forensics is a widely used science that helps clear up recordings, such as cassette tapes, or as in this episode a voicemail tape. Do you remember how in movies the undercover detectives wear a wire while talking to a suspect in a park? Well, since the wire is usually well hidden under layers of clothing and since the actual suspect is quite far from the microphone, the recording would be unintelligible. That is where the audio forensics comes in. It can be used to diminish the noise that is made by the clothes rubbing against the microphone, the children screaming in the background, the cars honking, etc. The audio experts also amplify the voices of the detective and the suspect in the recording. What we end up with, is a clear conversation between the two.

Audio forensics examiners are also the ones who can authenticate a recording for a trial. They can use the frequency signatures of local electrical power sources to pinpoint where and when recordings were made since the digital recorders that are plugged into electrical sockets capture the frequency signature of the local power supply -- a signature that varies over time. Now there are many software which can detect irregularity of a recording (undetectable by naked ear), hinting on evidence tampering.

RSA chryptosystem

A multi-layer encryption is designed to prevent people from accessing information that they are not authorized to see. The underlying algorithm used for multi-layer encryption is RSA. RSA is based on the idea that multiplying out two prime numbers is easy, while factoring a number into its prime factors is computationally hard.

Say we have two friends, Jane and Bob, who would like to exchange secret messages via postal service. If Bob just sends a message to Jane, then any postal worker could potentially read the mail. Even if Bob thinks of a secret method to encode his message, he would still need to let Jane know his method in order for her to decode the letter (and of course, if Bob sends the method over the mail, it could be used by a stranger to read his original letter). This is why Bob would like to use RSA encryption for his message. The process of using RSA is as follows:

1. Jane creates "public key" and "private key". She sends Bob the "public key", which consists of two fairly large numbers. She keeps the "private key" secret.
2. Bob uses this "public key" to encode his message and sends the encoded letter to Jane.
3. Jane uses her "private key" to decode Bob's message.

Notice here that the "private key" is never sent over the mail and it's the key to decoding the message. Hence, no one except for Jane who created the "private key" can decode Bob's message.

Now we are going to show, in more detail, what happens during the three steps above.

Step 1: Creation of the "public key" and "private key".

Note that this step will be done by Jane.

1. First Jane has to pick randomly two large prime numbers, we'll call them p and q.
2. Jane needs to compute n = pq (n is called modulus, since we will be moding out large numbers by n).
3. Jane will compute f(n)= (p-1)(q-1) (which would actually give the number of positive integers less than or equal to n that are coprime to n. Try to show this!)
4. Now Jane will choose an integer e, strictly between 1 and f(n), such that gcd(e, f(n)) = 1.
5. Jane computes the secret exponent d, also strictly between 1 and f(n), such that ed =1 (mod f(n)) (For this one might need to use Euclidean Algorithm.)

Here, the numbers n and e are part of the "public key" and d is part of the "private key", thus is kept secret. So for step 1, Jane sends a letter to Bob consisting of the values for n and e. (All other values such as p, q and f(n) are kept secret as well, since d could be calculated from them).

Step 2: Encoding of the message.

This step will be done by Bob after he receives the "public key".

1. Bob turns his message, M, into some integer strictly between zero and n using padding scheme (which we are not going to discuss here)
2. Bob computes C by taking m^e (mod n).

Bob sends the ciphertext, C, to Jane.

Step 3: Decryption.

This step is performed by Jane after she receives C from Bob.

1. All Jane has to do to recover m, is to take c to the power d and then mod out the result by n, m=c^d(mod n). (Note here that Jane is using her "private key", d, that nobody else knows).
2. Now by reversing the padding scheme for m, Jane can easily recover the original message, M.

Let us try this out on an easy example. Note that in reality the primes that are picked should be much larger.

Step 1: Creation of the "public key" and "private key".

1. Say Jane picked prime numbers p=23 and q=41. Say Jane picked prime numbers p=23 and q=41.
2. Jane computes n=pq=(23)(41)=943.
3. Jane also computes f(n)= (p-1)(q-1)=(22)(40)=880.
4. Say she picks e=7 (she could've just as well picked e=3, 9, 13, etc; but not 2, 5, 8, etc).
5. Jane computes the secret exponent in this case to be d=503 (Check that it works!)

Jane sends numbers n=943 and e=7 to Bob.

Step 2: Encoding of the message.

This step will be done by Bob after he receives the "public key".

1. Say Bob wants to encode a message M, for which the padding scheme gives m=3.
2. Bob computes C=m^e(mod n)=3^7(mod 943)=2187(mod 943)=301.

Bob sends the ciphertext, C=301, to Jane.

Step 3: Decryption.

This step is performed by Jane after she receives C from Bob.

1. To decode the message, Jane calculates m=c^d(mod n)=301^503(mod 943)=3 (this requires some number theory to do efficiently).
2. Now by reversing the padding scheme for m, Jane can easily recover the original message, M.

Reference:

http://www.wired.com/science/discoveries/news/2007/10/audio_forensics

http://www.wisegeek.com/what-is-audio-forensics.htm