The Lumenary · Grades 3–5

Secret codes

Name: ______________________

Date: ______________________

You will learn to

Encode and decode messages with a Caesar shift using an alphabet wheel
Explain why shifting past Z wraps round to A, and why a shift forward by $k$ is undone by a shift forward by $26 - k$
Break a Caesar cipher by trying all 25 shifts that change the message
Break a simple substitution cipher by counting letter frequencies
Explain why a message is not safe just because it looks scrambled, and why the key must be secret

1.
This message was made with a Caesar shift of 3 (each letter moved 3 places forward): KHOOR. What does it say?
- A.HELLO
- B.NKRRU
- C.JGNNQ
2.
Encode ZEBRA with a Caesar shift of 3.
- A.CHEUD
- B.ZHEUD
- C.WBYOX
3.
A message was encoded by shifting every letter 7 places forward. Ngozi's alphabet wheel only turns forward. By how many places forward must she shift to decode it? (Give a number from 1 to 25.)
4.
Tobi wants to break a Caesar cipher by trying every possible shift. How many different shifts does he need to try that actually change the message?
5.
Chidi shifts a message 10 places forward, and then shifts the result 20 places forward. Which single shift (a number from 0 to 25) would have done the same?
6.
In a long message made with a Caesar shift, the most common letter is K. If K stands for E, how many places forward was each letter shifted?
7.
A long English message was coded with a substitution cipher (a scrambled alphabet). In the coded message the most common letter is Q. Which plain letter is Q most likely to stand for?
- A.E
- B.Q
- C.A
- D.Z
8.
Kai sends KHOOR ZRUOG using a Caesar shift. "Nobody can read it," he says. "It looks like complete nonsense!" Is his message safe?
- A.Yes: it looks like nonsense, so nobody can read it.
- B.No: there are only 25 shifts to try, so anyone can break it quickly.
- C.Yes, as long as nobody knows he used a Caesar shift.
9.
In Lean, caesar k message encodes a message with a Caesar shift by k. Which statement says that shifting by 26 − k decodes every message encoded with key k (for any key k from 0 to 26)?
- A.caesar (26 - k) (caesar k message) = message
- B.caesar k (caesar k message) = message
- C.caesar 23 "PHHW DW QRRQ" = "MEET AT NOON"
10.
Write a function shift_letter(letter, k) that shifts one capital letter k places forward, wrapping round after Z. For example shift_letter("A", 3) is "D" and shift_letter("Y", 3) is "B". Use the string ALPHABET: ALPHABET.index(letter) gives the letter's number (A = 0), and ALPHABET[n] gives the letter with number n. In Python, n % 26 is the remainder when n is divided by 26.
Computational — Practice

Answer key — Secret codes

1.
Move each letter 3 places back: K → H, H → E, O → L, O → L, R → O. The message is HELLO.
2.
Move each letter 3 places forward, wrapping round after Z: Z → C, E → H, B → E, R → U, A → D. The answer is CHEUD. With numbers: Z = 25, and $25 + 3 = 28$ , $28 - 26 = 2$ , which is C.
3.
Going 26 places is once round the wheel, which changes nothing. The message has already gone 7 places, so it needs $26 - 7 =$ 19 more.
4.
Every shift does the same as one of the 26 shifts 0, 1, …, 25, and these are all different. The shift by 0 changes nothing, so there are 25 shifts that change the message — few enough to try them all by hand.
5.
The two shifts move every letter $10 + 20 = 30$ places. Once round the wheel is 26 places and changes nothing, so 30 places does the same as $30 - 26 =$ 4 places.
6.
From E to K is $10 - 4 =$ 6 places forward (E → F → G → H → I → J → K). So the key is a shift of 6, and the message is decoded by shifting 6 back (or $26 - 6 = 20$ forward).
7.
A substitution cipher changes the letters but keeps their counts. In long English texts the most common letter is E (roughly one letter in eight), so Q most likely stands for E. It is a good guess, not a certainty — a code breaker checks it by trying to read the message.
8.
No. A Caesar shift has only 25 keys that change the message, so anyone can try them all. With a shift of 3 back, KHOOR ZRUOG becomes HELLO WORLD. A message is safe only if it is hard to decode without the key, even for someone who knows the method.
9.
caesar (26 - k) (caesar k message) = message says: encode with key k, then with key 26 - k, and the original message comes back — for any message. Lean proves it (caesar_undo) from the fact that $k + (26 - k) = 26$ is once round the wheel. The other options are a false claim (encoding twice with the same key) and a single example.
10.
```
ALPHABET = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

def shift_letter(letter, k):
    n = ALPHABET.index(letter)
    return ALPHABET[(n + k) % 26]
```
The remainder % 26 does the wrapping round: for Y with a shift of 3, n + k is $24 + 3 = 27$ , and $27 \bmod 26 = 1$ , which is B. Then "".join(shift_letter(c, k) for c in message) encodes a whole word.