Caesar Cipher Part 2

Last time we encrypted the following statement:

“walk for five minutes, turn left, then walk for three minutes, the club house is up in the brown tree.”

As:

“zdon iru ilyh plqxwhv, wxuq ohiw, wkhq zdon iru wkuhh plqxwhv, wkh foxe krxvh lv xs lq wkh eurzq wuhh.”

Unfortunately, for our friend, there is an easy way to break this encryption. English is an interesting language. Some letters, like “e” and “t” get used quite often while other letters, such as “q” and “z” are almost never used (below is a table of the letter frequency in English). We can use this fact to our advantage.

letter frequency

Let’s count how many times each letter occurs in the encrypted text.

H

W

Q

U

X

K

L

I

O

R

V

Z

D

E

N

P

F

S

Y

12

9

6

6

6

5

5

4

4

4

4

3

2

2

2

2

1

1

1

Since H occurs most often, we can make the reasonable guess that H must be representing E, since E is usually the letter that appears most often. To get from E to H, our clever friend must have counted up 3 letters. Hence, if we count backwards by 3 letters, we should be able to decrypt our message.

Now let’s take it one step further. We can encrypt a message by counting up as many letters as we want. We could count up 5 letters, for example. Then the word “SECRET” would become “XJHWJY.” Or maybe we could count up 10 letters. Then the word “SECRET” would become “COMBOD.”

Ok, enough talk. Here is some encrypted text you intercepted. Try to decrypt it and leave your response in the comments below.

H GZUD MDUDQ KDS LX RBGNNKHMF HMSDQEDQD VHSG LX DCTBZSHNM. LZQJ SVZHM

If you are having trouble with this, leave a comment and I’ll give you a hint.

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1 Comment

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One response to “Caesar Cipher Part 2

  1. “I have never let my schooling interfere with my education.” ~Mark Twain
    😀

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