Cracking the Playfair cipher

Question

I have the ciphertext and an encrypting program (with the key hardcoded in). How would I go about finding the key? Surely the availability of the encryptor must open up possibilities beyond brute-forcing it.

Answer 1

Yes knowing the algorithm may help in decoding the cypher text, but only if there is a flaw in the algorithm that may be exploited. (the good news is Playfair has some flaws that can be exploited)

Here are a few good starting points to read.

1. Wikipedia (read it all - particularly Cryptanalysis)
2. Basic Cryptanalysis (look at chapter 7)

The second one is not what I would call a light read, but interesting if you're into cyphers.

Answer 2

I've found a way in five lines (obviously re-evaluated a bit, and admittedly very long lines):

(a,b,c)="".join((input("CODE: ")).split()),input("Polybius Square: "),"" 
for i in a:
 c+=str(int(((b.find(i))-((b.find(i))%5))/5))+str((b.find(i))%5)
for j in range(0,(int(len(c)/2))):
 print((b[((5*(int((c[:(int(len(c)/2))])[j])))+(int((c[(int(len(c)/2)):])[j])))]).lower(),end="")

NB: When prompted to enter the polybius square, first enter row 1, then row 2 etc, no spaces

Then you just have to remove unnecessary 'x's and voila!

Answer 3

Try:

(a,b,f,g,c)="".join(input("CODE: ").split()),input("Polybius S: "),"","",1
for(i)in(a):
 if(c%2)==0:
 g+=i
 else:
 f+=i
 c+=1
for(j)in(range(0,len(f))):
 if(b.find(f[j])%5)!=(b.find(g[j])%5)and(int(((b.find(f[j]))-(b.find(f[j])%5))/5))!=(int(((b.find(g[j]))-(b.find(g[j])%5))/5)):
 print(b[((int(((b.find(f[j]))-(b.find(f[j])%5))/5))*5)+(b.find(g[j])%5)],end="")
 print(b[((int(((b.find(g[j]))-(b.find(g[j])%5))/5))*5)+(b.find(f[j])%5)],end="")
 elif(b.find(f[j])%5)==(b.find(g[j])%5)and(int(((b.find(f[j]))-(b.find(f[j])%5))/5))!=(int(((b.find(g[j]))-(b.find(g[j])%5))/5)):
 print(b[((((int(((b.find(f[j]))-(b.find(f[j])%5))/5))-1)%5)*5)+b.find(f[j])%5],end="")
 print(b[((((int(((b.find(g[j]))-(b.find(g[j])%5))/5))-1)%5)*5)+b.find(g[j])%5],end="")
 elif(b.find(f[j])%5)!=(b.find(g[j])%5)and(int(((b.find(f[j]))-(b.find(f[j])%5))/5))==(int(((b.find(g[j]))-(b.find(g[j])%5))/5)):
 print(b[((int(((b.find(f[j]))-(b.find(f[j])%5))/5))*5)+((b.find(f[j])%5)-1)%5],end="")
 print(b[((int(((b.find(g[j]))-(b.find(g[j])%5))/5))*5)+((b.find(g[j])%5)-1)%5],end="")

It's not pretty, but it works