Decryption Problem help:)

thelibertyboy

Heres the message im trying to decypher in really lost on this:(

DNSAGIVZG ST PIQWG ST CIBCABIDKSV JWXWIBWZ DNW GWCJWD ST DNW PBIVWDG SJFKD

It is known that the letter h is encrypted as n and the letter m is encrypted as y in the modular encryption scheme which enciphered the above message. Use this information to find the equations satisfied by the numbers a and b in the encryption map y ≡ (ax+b) mod 26, solve these equations to find the values of a and b, construct the decryption map x ≡ (cy+d) mod 26 and, hence, decipher the message. Fill in all of your answers in the cells provided below on this sheet

1. Equations satisfied by a and b

H encrypted as N implies ()a +b ≡()mod26
M encrypted as y implies ()a +b ≡()mod26

2.Solutions for a and b

The value of a is () (give least residue, mod 26, for a)
The value of b is () (give least residue, mod 26, for b)

3. Decryption map x ≡ (cy+d) mod 26

The value of c is () (give least residue, mod 26, for c)
The value of d is () (give least residue, mod 26, for d)

Ficheall

thelibertyboy wrote: »

DNSAGIVZG ST PIQWG ST CIBCABIDKSV JWXWIBWZ DNW GWCJWD ST DNW PBIVWDG SJFKD

I've no background in this, and so can't answer any of your questions, but I've worked out what the code says, if that's of any use to you?

thelibertyboy

yea tht wud be great howd u figure it out ?:)

Ficheall

"Thousands of hours of calculation revealed the planets orbit."

I guessed DNW was THE and just guessed at what might fill in the gaps left after inserting those letters...

tricky D

Ficheall wrote: »

"Thousands of hours of calculation revealed the planets orbit."

I guessed DNW was THE and just guessed at what might fill in the gaps left after inserting those letters...

pages not hours

thelibertyboy

how did ya figurev tht out tricky??:)

Ficheall

"Pages" indeed. I neglected to go back and look at where I'd written it out. Apologies.

I'd love to know how one is supposed to work these things out - when I had time to attend a cryptography course I did not have the inclination, and now that I have the inclination I do not have the time...

tricky D

thelibertyboy wrote: »

how did ya figurev tht out tricky??:)

Just like Fichell, DNW is obviously 'the', then the rest is educated guessing. It helps if you do the number to word newspaper puzzles* or also in my case used to work for a crypto company.

* can't remember the name but they're like crosswords except have numbers filling all the boxes instead. They give you one or two to get you started then you find the common ones like vowels, s and d, and so on.

edit:
Found one:

Ficheall wrote: »

"Pages" indeed. I neglected to go back and look at where I'd written it out. Apologies.

Now write it out a hundred times:p

Pygmalion

The way I'd approach this is to see that it's a simple substitution cypher.
That is the same letter always encrypts to the same letter no matter whereabouts it is, and the same for words.

As the above said it's then a matter of looking at the likelihood.
DNW appears twice, and "THE" is a good bet for a 3 letter word, so assume that T encrypts to D, H encrypts to N (which you're given in the question, confirming this somewhat) and E encrypts to W and replace them, and you get:

THSAGIVZG ST PIQEG ST CIBCABITKSV JEXEIBEZ THE GECJET ST THE PBIVETG SJFKT

With the letters in bold being the original, unencryted letters and the rest of the letters currently unknown.
Then you can see what fits.
You could try and guess what "ST" is here, since there's only a limited number of 2 letter words, or you could notice that the letter "G" appears a good bit, usually at the end of words, so you could guess that it's an 'S'.
The word "Secret" is half-revealed at the moment too.

Really it's just a matter of seeing what fits and what letters are likely to appear.

Thing is this approach doesn't really have much to do with the hints your given in OP or the questions you're asked, so maybe you're expected to do it a different way, but this way seems easy enough if you're willing to give a few guesses :P.

hivizman

Ficheall wrote: »

I'd love to know how one is supposed to work these things out - when I had time to attend a cryptography course I did not have the inclination, and now that I have the inclination I do not have the time...

This is a simple substitution cipher, where the same letter is substituted in the encrypted text for every occurrence of a given letter in the plain text. The "modular encryption scheme" simply means that each letter from A to Z is numbered in sequence (A=1, B=2, etc.). The letter in the encrypted text that represents a given letter in the plain text is determined by multiplying the numerical value of the letter in the plain text by a given number (a) and adding another given number (b), then working out what letter the answer represents. This is done by repeatedly subtracting 26 from the answer until it falls between 1 and 26, and then taking the letter represented by that number. More formally, the "residue modulo 26" is determined.

The question basically asks you to work out a and b. You are given that the plain letter H is encrypted as N and the plain letter M is encrypted as Y, so you have:

8a + b = 14 (mod 26)
13a + b = 25 (mod 26)

Subtract the top equation from the bottom equation to get:

5a = 11 (mod 26)

This is enough information to calculate a, and b is found by substitution into either of the original equations.

The final part of the question asks you for the decryption formula. This is found from the following equations:

14c + d = 8 (mod 26)
25c + d = 13 (mod 26)

I suspect that the point of the question was to test your knowledge of solving simultaneous equations in modular arithmetic, and being able to decrypt the text was a way of allowing you to check that you'd got the correct mathematical answer. For simple substitution cyphers, the quickest approach, as Pygmalion has just pointed out, is often a combination of guessing short words (what Ficheall and tricKy D did by guessing that DNW represented THE) and looking at letter frequencies - the most common letter in the encrypted text is likely to be E in the plain text.