Encode the following message using a shift 4 cipher, as Francis Bacon supposedly did. Note: You may use the encryption tool described here or you may do this by hand.

Francis Bacon

shift_encrypt('francis bacon',4)
'jvergmw fegsr'

Now encode it using a shift 15 cipher.

shift_encrypt('Francis Bacon',15)
'Fgpcrxh Bprdc'
 >shift_encrypt('francis bacon',15)
'ugpcrxh qprdc'

The following messages have been encoding using a shift cipher with key 8. Decode.

1. emtkwum bw bpm eiksg ewztl wn kzgxbwozixpg >shift_encrypt('emtkwum bw bpm eiksg ewztl wn kzgxbwozixpg',18) 'welcome to the wacky world of cryptography'
2. tivociom qa zmlcvlivb shift_encrypt('tivociom qa zmlcvlivb',18) 'language is redundant'

The following message has been encoded using a shift cipher. What is the key?

n puvyq funyy yrnq gurz

Answer 13

The brute force method:

Cipher text:  n puvyq funyy yrnq gurz
---------------------------
    Shift: 0: n puvyq funyy yrnq gurz
    Shift: 1: o qvwzr gvozz zsor hvsa
    Shift: 2: p rwxas hwpaa atps iwtb
    Shift: 3: q sxybt ixqbb buqt jxuc
    Shift: 4: r tyzcu jyrcc cvru kyvd
    Shift: 5: s uzadv kzsdd dwsv lzwe
    Shift: 6: t vabew latee extw maxf
    Shift: 7: u wbcfx mbuff fyux nbyg
    Shift: 8: v xcdgy ncvgg gzvy oczh
    Shift: 9: w ydehz odwhh hawz pdai
    Shift: 10: x zefia pexii ibxa qebj
    Shift: 11: y afgjb qfyjj jcyb rfck
    Shift: 12: z bghkc rgzkk kdzc sgdl
    Shift: 13: a child shall lead them
    Shift: 14: b dijme tibmm mfbe uifn
    Shift: 15: c ejknf ujcnn ngcf vjgo
    Shift: 16: d fklog vkdoo ohdg wkhp
    Shift: 17: e glmph wlepp pieh xliq
    Shift: 18: f hmnqi xmfqq qjfi ymjr
    Shift: 19: g inorj yngrr rkgj znks
    Shift: 20: h jopsk zohss slhk aolt
    Shift: 21: i kpqtl apitt tmil bpmu
    Shift: 22: j lqrum bqjuu unjm cqnv
    Shift: 23: k mrsvn crkvv vokn drow
    Shift: 24: l nstwo dslww wplo espx
    Shift: 25: m otuxp etmxx xqmp ftqy

Cipher text:  n puvyq funyy yrnq gurz
---------------------------
    Shift: 0: n
    Shift: 1: o
    Shift: 2: p
    Shift: 3: q
    Shift: 4: r
    Shift: 5: s
    Shift: 6: t
    Shift: 7: u
    Shift: 8: v
    Shift: 9: w
    Shift: 10: x
    Shift: 11: y
    Shift: 12: z
    Shift: 13: a
    Shift: 14: b
    Shift: 15: c
    Shift: 16: d
    Shift: 17: e
    Shift: 18: f
    Shift: 19: g
    Shift: 20: h
    Shift: 21: i
    Shift: 22: j
    Shift: 23: k
    Shift: 24: l
    Shift: 25: m

Cipher text:  n puvyq funyy yrnq gurz
---------------------------
    Shift: 13: a child
    Shift: 21: i kpqtl

 >affine_encrypt('attack at dawn',3,2)
'chhcig ch lcqp'

Decrypt the following message assuming it is an affine cipher with a slope of 3 and a shift of 2:

goov wskb ixap kv

If (m,b) is the encryption key, then (m^-1, - (m^-1 * b)) is the decryption key.

We need to find the inverse of 3. The multiplication table for 3 is:

Multiplication table for 3 modulus 26

	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25
3	0	3	6	9	12	15	18	21	24	1	4	7	10	13	16	19	22	25	2	5	8	11	14	17	20	23

So therefore the decryption key should be:

(9, -(9*2)) = (9,-18) = (9,8)

 >affine_encrypt('goov wskb ixap kv',9,8)
'keep your chin up'

'g' => 'k'  (9,8)
encoding 'g' = 6
6 * 9 = 2 mod 26
2 + 8 = 10
decoding 10 = 'k'

Verifying:

 >affine_encrypt('keep your chin up',3,2)
'goov wskb ixap kv'

Determine the shift and slope for the following affine encryption scheme:

Plain Text	a	b	c	d	e	f	g	h	i	j	k	l	m	n	o	p	q	r	s	t	u	v	w	x	y	z
Cipher Text	f	m	t	a	h	o	v	c	j	q	x	e	l	s	z	g	n	u	b	i	p	w	d	k	r	y

Compute encodings:

Plain Text	a	b	c	d	e	f	g	h	i	j	k	l	m	n	o	p	q	r	s	t	u	v	w	x	y	z
Encoding	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25
Cipher Codes	5	12	19	0	7	14	21	2	9	16	23	4	11	18	25	6	13	20	1	8	15	22	3	10	17	24
Cipher	f	m	t	a	h	o	v	c	j	q	x	e	l	s	z	g	n	u	b	i	p	w	d	k	r	y

the simplest way to do this:

1. The encoding of 'a'(=0) tells us the shift (b):

   (m * 0) + b = 5
   b = 5
   

2. Now we plug this info into the encoding for 'b'(=1):

   (m * 1) + 5 = 12
   m + 5 =12
   m = 7
   

3. So the key is (7,5).
4. Verifying with another example 'd'(=3):

   (7 * 3) + 5 = 26 = 0 
   decoding 0 = 'a'
   

   Yes.

Suppose we change the ALPHABET we're encoding to include " "(space). That's one more character. That means we're now doing arithmetic mod 27 instead of mod 26.

Previously (in mod 26) our valid slopes were all the odd numbers minus 13. What is our new set of valid slopes (in mod 27)?

Valid slopes must be relatively prime to 27. The only factors of 27 are powers of 3. Therefore any multiple of 3 is excluded, leaving:

1 2 4 5 6 7 8 10 11 13 14 16 17 19 20 22 23 25 26