Cracking the Code – The Mathematics of Cryptanalysis
Lesson 7 – Review
The Zimmerman Telegram
The message which was the answer to Cypher #3 in Lesson 4 has a very
interesting history.
By 1917 World War I was raging in Europe. The United States remained
neutral. President Woodrow Wilson believed that a negotiated settlement
was possible, and that the United States, by remaining neutral, could broker
it. The British strongly desired America’s entrance into the war on behalf of
the Allies who were fighting against Germany.
In 1915, a German U-boat had sunk the civilian ocean liner Lusitania,
resulting in great loss of life, including 128 Americans. Germany had
persuaded the United States not to declare war at that time by promising that
in the future German U-boats would surface before they attacked, thereby
lessening the chance of an attack on a non-military ship. By 1917, the
Germans were ready to renege on their promise and begin unrestricted
submarine warfare in hopes of prompting an Allied surrender. However,
they were afraid that such a course of action would draw the United States
into the war on the side of the Allies.
The German foreign minister, Arthur Zimmerman, proposed that Germany
make alliances with Mexico and Japan, encouraging these countries to attack
the United States. The idea was that the United States would be so busy
defending itself at home that it would not have the resources to wage war in
Europe. Zimmerman sent a telegram to the German ambassador in
Washington outlining his plan and instructing the ambassador to deliver the
proposal to Mexico. The text of this telegram is the answer to Cypher #3.
The telegram was intercepted by the British and deciphered. The British
delivered the telegram to the Americans. By this time, the Germans had
actually begun unrestricted submarine warfare, but President Wilson had
maintained his position of neutrality. Upon receipt of the telegram,
however, he changed his position and on 2 April 1917 he requested that
Congress declare war on Germany. Thus began the United States’ military
involvement in WWI.
For more details about the Zimmerman telegram, the reader is referred to
The Zimmerman Telegram, Barbara W. Tuchman, Ballantine, 1994.
The telegram was actually encrypted with a code which was much more
sophisticated than a multiplicative cipher. The encrypted telegram appears
below. The image is from Decimal File 862.20212/82A (1910-1929),
General Records of the Department of State, Record Group 59.
(http://www.archives.gov/digital_classroom/lessons/zimmermann_telegram/
zimmermann_telegram.html)
Review Questions
1. Explain what is meant by the term “frequency analysis,” and explain
how frequency analysis is helpful in breaking certain types of ciphers.
2. Make an addition table for addition modulo 3.
3. (5+10) (mod 12)
4. 53 (mod 26)
5. 410 (mod 12)
6. (22+9) (mod 26)
7. What is the additive inverse of 2 mod 5?
8. What is the additive inverse of 19 mod 26?
9. Encrypt the word “Monday” with a Caesar cipher with shift 3.
10. The following word was encrypted using a Caesar cipher with shift 20.
Decrypt this word: FIN
11. What is the multiplicative inverse of 5 mod 26?
12. Encrypt the word “Monday” with a multiplicative cipher with key 3.
13. The following word was encrypted using a multiplicative cipher with
shift 5. Decrypt this word: OEL
14. What is the multiplicative inverse of 2 mod 5?
15. What does it mean for two numbers to be relatively prime? Give an
example of two whole numbers which are relatively prime. Give an
example of two whole numbers which are not relatively prime.
16. What does it mean for a whole number to be prime? Give an example
of a prime. Give an example of a whole number (other than 1) which is
not prime.
17. List all the whole numbers between 2 and 8 which have multiplicative
inverses mod 9.
18. Encrypt the word “bus” with an affine cipher with multiplicative key 5
and additive key 3. (Apply the multiplicative key first.)
19. The following word was encrypted with an affine cipher with
multiplicative key 5 and additive key 3. Decipher it. QZPB
20. Solve for m: 5m = 7 (mod 26)
21. (optional) Suppose we have an affine cipher in which cyphertext “J”
corresponds to plaintext “b” and cyphertext “R” corresponds to
plaintext “s.” Set up the system of equations which one would need to
solve in order to find the additive and multiplicative keys.
22. (optional) Solve this system of equations:
7m + a = 21 (mod 26)
3m + a = 11 (mod 26)