This is a cryptography class . I need to create a python programs atbash.py , SpartanScytale.py and kidkrypto.py can anyone help please? these are the answered just the code I need

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Solution: ZMW ZUGVI GSVN GSV PRMT LU HSVHSZXS HSZOO WIRMP

Solution : THIS WHOLE LAND SHALL BECOME A RUIN AND A WASTE

DECIPHER

Solution : JUSTA SWATE RREFL ECTST

HEFAC ESOON EHUMA NHEAR

TREFL ECTSA NOTHE R

The message was written on the Scytale that allowed for 6 lines that means we can separate the message into chunks that are six characters long:

EIEPAA -

DSNERD

UATREV

CNIFE

AONTUR

TRPYGS

INRAEI

OAONIT

NMSDNY

Reading columns we get Education is an ornament in prosperity and a refuge in adversity.

THESCYTALE

The scytale was an early example o a transposition cypher

7- Ijmc&miclTzvvgxgzthAfmaneokoLy&qfkcydIak

Printed out copy of the wheels from the book, using K as the pointer . mention J drop , polyalphabetic

9- First write the plain text as it is. For the key start with the primary key corresponding to the first letter of the plain text. The next corresponding keys just follow from the main text. Then use vigenere square to get the cypher text.

Plane Text: WALTER RALEIGH BRINGS TOBACCO TO ENGLAND FROM AMERICA

key: BWALTE RRALEIG HBRING STOBACC OT OENGLAN DFRO MAMERIC

cipher text: XWAEXV IRAPMON ISZVTY LHPBAEQ HH SRTVLAQ IWFA MAQVZKC

14- The pigpen cipher relies on its users memorizing the diagrams shown in the book on page 35 in figure 1.12. Decipher the message : Well got wealth may meet disaster but ill got wealth destroys its master

19- Well be writing them in a column 6 characters wide to encode the message Army moves 6 December using the ADFGVX table . Subtitution

The next step is using the keyword Berlin we order the columns 126435 and the solution is : AFGVFF GFVAAG VAXGVG AVXDXX VVAXVA GVGDGG

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M = ab -1

e= AM +a

d= BM +b

10 digit no.

N = (ed-1)/

Y= e.x mod n

N = 1722629

E= 41996

Y= 1305808

X= ?

X = (y/e)mod n e^-1 =a

= ye^-1 modn

Multiplication inverse

Another way of doing it

D = d.y mod n

= d.(e.x) mod n

ed = Mn+1 mod n

ed = 16 + 1 mod 16

ed1 = 32 +1 mod 16 potential answer

ed2 = 48 +1 mod 16 potential answer

ed = Mn +1

2.5 = 10 +1 mod 10

10 = 20 + 1 mod 10

M:1033, e: 11410, d:5187, n:57293

Y:53803, undone: 1958

You can find Walters pin easily. Since you know how to find the ciphered text and theres only 9999 or so pins just loop through each pin and encrypt it and find which ever matches the enciphered pin.

ATBASH.py

Import string

Cipher =

Qfhgz hdzgv iivuo vxghg

Svuzx vhllm vsfnz msvzi

Gvuo vxghz mlgsv i

Alpha = abcdefghegklmnopqrstuvwxyz

Cipher = cipher.lower()

# basically tr/abc./cba/

Deciphered = cipher.translate ( string.maketrans9alpha, alpha[ ::-11]))

Print Deciphered/Ciphered: + str(deciphered)

SpartanScytale.py