The homework problem(s)

Each person individually or in a group should try to "break RSA". The major part of the assignment, however, follows.

Each working group has a public key in the table below. This public key should be used to send messages to that group. I will send every member of each group the private key for that group. The group members should use the group's private key to decrypt encrypted messages sent to them by other groups. You will need to be very careful! Words are changed to numbers as in previous assignments: A --> 01, B --> 02, etc. There's a table below which you can use. Each group should do the following:

• Step 0 Read the group's entries in the columns "Sending instructions" and "The message". Both of these were written using RSA with the group's own public key. The group should decrypt each of these entries with their private key. The sending instructions will ask the group to send "The message" to another group.
• Step 1 Encrypt the message (using the receiving group's public key), and send the encrypted message to every member of the receiving group. Also send me a message (with decrypted content!) telling me that you have accomplished this step. I'll change the Step 1 column entry from (red) to (blue) to record your achievement.
• Step 2 When you get a message from another group, decrypt it, and send me a copy of the decrypted message. I'll change the Step 2 column entry from (red) to (blue) to record your achievement.

Comments and cautions and information

Please note that although RSA and close relatives are widely used now for highly secure applications, what I've done here should not be viewed as secure or useful in the "real world". It is a class exercise and isn't really safe for industrial secrets, love letters, or espionage (!). The N used here isn't big enough and "good" key selection has not been practiced: keys should be selected using somewhat more constraints than I have suggested. Finally, consquences could result not only from government reaction (as the policy discussion might suggest) but also from corporations asserting copyright or patent protection (see, for example, rsa.com or pgp.com).

I am interested in furthering your education at every opportunity. I therefore have made each message the title of one of the Sherlock Holmes short stories written by Arthur Conan Doyle. There are fifty-six of them. In fact, each of these short stories cited is in public domain and can be read on the web: see http://www.citsoft.com/holmes3.html for some really nice fiction. I was going to use the sonnets of Shakespeare, but I am not literate enough.

Breaking RSA

This might be the last contest. The prize budget is running low ... In an effort to spread around the joy, previous winners will earn my respect and congratulations but only previous non-winners are eligible for the new valuable prize. Previous non-winners may work with previous winners, however. Every student should enter the contest either individually or as a member of a group -- it is part of the homework assignment.

What's known

The public sees the following information:

    Bob (to Alice)   53397684690326489272244
    Note: this is a message to Alice, so it uses Alice's public key.

    Alice (to Bob)   26563411027343523303980
    Note: this is a message to Bob, so it uses Bob's public key.

The Contest

Break the encryption. Find the messages, and send me the content. You'll certainly need help from some tool like Maple. Try to find and use the appropriate Maple instruction which will make reading these messages very easy.

Using Maple for RSA

I don't know the internal workings of Maple, but I guess that when Maple encounters A^B modC;, it computes A^B in a rather direct fashion, multiplying A by itself B times, divides the result by C, and then prints the remainder. If B is large, straightforward computation of A^B takes a long time and may result in a number too large for Maple to handle. Several people had this problem doing the Diffie-Hellman homework assignment.

The instruction 3^12345 mod 2; just took about a quarter of a second on one of the Math Department computers (not a particularly fast machine). 12345^12345 mod 2; then took over 17 seconds, and asking for 123456789^123456789 mod 2; resulted in the message Error, object too large with no answer provided. By the way, the answer in all cases should have been 1 since the numbers involved were powers of odd integers. We need something better if we want to use Maple to do RSA encryption and decryption. Here we will need to compute A^B (mod C) when A, B, and C are each 20 digits or more long.

The computations can be done efficiently by taking advantage of repeated squaring to do exponentiation, as discussed in class. I've written a small Maple procedure to exponentiate using this idea. Here is the procedure. You do not need to know or worry about what happens "inside" it. I'll describe how to use it in the text following.

ES:=proc(base,power,modulus)
local A,B,C; 
A:=1; B:=power;C:=base; 
while type(B,positive) do 
	if type(B,odd) then A:=A*C mod modulus; B:=B-1; fi;
        B:=B/2; C:=C^2 mod modulus 
        od;
RETURN(A);
end;

Maybe that's enough chatter. To use ES, you must copy the procedure into your Maple session. You should be able to use your mouse to point and grab the material above (take exactly what's in boldface type). I'll show and comment on a Maple session below using RSA. Please read this material. Questions are welcome.

First the RSA setup: I will assume that there are two users, Charlie and Deborah. What do we need to know?

For your information, N is the product of the primes P=19669087 and Q=90693287. This is unknown to the "public". Otherwise the security of the encryption would be effectively compromised.

Charlie would normally not know Deborah's private key, nor would Deborah know Charlie's private key. Only the column of public keys would be known generally.

Charlie will send the message TOAD to Deborah. Deborah will reply with the message FROG. We'll need to translate these alphabetic messages into numbers. Let's continue using our simple-minded technique, which is helped by using the table below. In reality, people could use the ASCII code which is described in the text. You may use this table for the other parts of the homework.

Using this scheme, T   O   A   D becomes the number 20   15   01   04 -- just 20150104 without the spaces, and FROG becomes the number 06181507 . Note that the initial 0 in the number will not be reported by Maple. For example, the number 0123 is just 123 to Maple. So Charlie will send the number to Deborah encrypted with Deborah's public key, and Deborah will decrypt it with her private key. She then will reply to Charlie, encrypted with his public key. In turn, Charlie will decrypt this message with his private key.

Here is a record of an actual Maple session which will accomplish the computations needed by Charlie and Deborah. The Maple commands and replies will be in this typeface and I will add comments in this typeface, indented. The comments were not part of the Maple session.

lagrange>maple     

      Instead of "xmaple" I typed maple at the prompt.
      This is a different way to use the program, with 
      no snazzy icons, colors, etc. Here we're just doing
      a few computations. The buttons and windows may not
      be useful and could indeed be distracting.

    |\^/|     Maple V Release 4 (Rutgers University)
._|\|   |/|_. Copyright (c) 1981-1996 by Waterloo Maple Inc. All rights
 \  MAPLE  /  reserved. Maple and Maple V are registered trademarks of
 <____ ____>  Waterloo Maple Inc.
      |       Type ? for help.
> ES:=proc(base,power,modulus)
> local A,B,C; 
> A:=1; B:=power;C:=base; 
> while type(B,positive) do 
>     if type(B,odd) then A:=A*C mod modulus; B:=B-1; fi;
>         B:=B/2; C:=C^2 mod modulus 
>         od;
> RETURN(A);
> end;
ES := proc(base, power, modulus)
local A, B, C;
    A := 1;
    B := power;
    C := base;
    while type(B, positive) do
        if type(B, odd) then A := A*C mod modulus; B := B - 1 fi;
        B := 1/2*B;
        C := C^2 mod modulus
    od;
    RETURN(A)
end

      All that I've done here is exactly copy the text
      of the ES procedure to an input line. Maple then
      indicated it understood the procedure by writing 
      it back to me. It only remembers this procedure
      during this session. If I want to use ES during
      another Maple session, I must copy it again.
  
> 231^456 mod 789;
                                      186

      Now I'll check that things are working the way
      they're supposed to. I asked Maple to compute
      231456 (mod 789) and got the answer 186.

> ES(231,456,789);
                                      186

      This command returned the same answer, so I now do
      believe things are working well. I check things like
      this almost always. The checks don't use much time or 
      effort. They help me find typing errors or any other
      problems.

> N:=1783854152318969;
                             N := 1783854152318969

      I told Maple N's value. I used the mouse to
      grab this and other long numbers but I check
      the first and last digits of numbers that I 
      copy. My mouse technique isn't perfect.

> C_pub_key:=604305623921;
                           C_pub_key := 604305623921

      And I now told it Charlie's public key. I tend
      to use long identifying names for variables in 
      situations like this. Confusion is too easy.

> C_priv_key:=155833484920913;
                         C_priv_key := 155833484920913

      Charlie's private key.

> D_pub_key:= 745580047409;
                           D_pub_key := 745580047409

> D_priv_key:=889041818372597;
                         D_priv_key := 889041818372597

      Deborah's public and private keys.

> ES(20150104,D_pub_key,N);
                               1654115086409835

      This is Charlie's encryption of TOAD using
      Deborah's public key. With this instruction, 
      Charlie has Maple compute 20150104(D_pub_key) 
      (mod N). This number is 20150104745580047409 
      (mod 1783854152318969). Wow! I don't think 
      anyone would want or could do computations
      like this by hand! Charlie sends this number 
      to Deborah.
 
> ES(1654115086409835,D_priv_key,N);
                                   20150104

      Deborah decodes the message by exponentiating it
      with her private key. We get back 20150104, TOAD
      again. We have accomplished meaningful secret 
      communication (if you like toads). You can
      practice with various messages until you feel
      comfortable with this method.

> ES(06181507,C_pub_key,N);
                                199543635933342

      Now it is Deborah's turn. She sends a message to
      Charlie, encrypting it with Charlie's public key.
      The long number is what is sent to him in public.
      The word FROG seems well disguised!

> ES(199543635933342,C_priv_key,N);
                                    6181507

      Charlie decodes the message by exponentiating it
      with his private key. He should read the digits
      in pairs from right to left. The missing "0" in
      "F" will be apparent. The word FROG appears to
      Charlie.

> quit;

Maintained by greenfie@math.rutgers.edu and last modified 3/5/2000.