[Baypiggies] Python in the Movies | Decoding an ISBN?

[Guido van Rossum]

Did you Google it? Others have solved it...

--Guido (on mobile)
On Oct 3, 2015 10:04 AM, "Glen Jarvis" <glen at glenjarvis.com> wrote:

> I have been watching "Ex Machina" (http://www.imdb.com/title/tt0470752/).
> [It's worth watching, by the way).
>
> When I see a television program and/or movie that has "code," I always
> pause and look. It's amazing how often a thin veneer of "code" is put up on
> the screen as the actor hammer at the keyboard and the "code" just flows
> down the page.
>
> I admit, usually I do this exercise only for my family to groan because I
> usually piss and moan how badly it is done. For example, Criminal Minds
> once had an IP address similar to: 76.21.58.572. They even had the actors
> read it in the dialogue.. Arghhh
>
> "Ex Machina," at roughly index 1:09:37 has two "windows." One looks like
> C. And, the second, the foreground window Python. It's not PEP8 compliant,
> has bad spacing and doesn't win the efficiency award, but it's genuine code
> and syntactically (but not semantically) correct :)  woot!
>
> Here is the code below. What grabbed my interest is the "puzzle" included:
> (An obfuscated way to write "ISBN = "):
>
> sys.stdout.write("".join(chr(i) for i in (73,83,66,78,32,61,32)))
>
> I expect the rest is an ISBN number. I'm curious what book was intended --
> even if their "code" is broken.
>
> I believe their "sieve of Eratosthenes" is incorrect. Reference video:
> https://www.youtube.com/watch?v=V08g_lkKj6Q
>
>
> Can anyone work out which ISBN number they were trying to reference?  1206
> isn't prime. And, if I just assume they completely meant to take code[i] -
> key[i] ignoring prime's completely, I get this `12053003823` -- which
> isn't an ISBN reference.
>
>
> Hmm... I'm still mildly curious...
>
>
> # BlueBook code decryption
>
>
> import sys
>
>
> def sieve(n):
>
> # Compute primes using sieve of Eratosthenes
>
>     x = [1]*n
>
>     x[1] = 0
>
>     for i in range(2,n/2):
>
>         j = 2*i
>
>         while j < n:
>
>             x[j] = 0
>
>             j = j+1
>
>     return x
>
>
> def prime(n,x):
>
> # Find the nth prime
>
>
>     i = 1
>
>     j = 1
>
>     while j <= n:
>
>         if x[i] == 1:
>
>             j = j+1
>
>         i = i+1
>
>     return i-1
>
>
>
> x = sieve(1000)
>
>
> code = [1206,301,384,5]
>
> key= [1,1,2,2]
>
>
> sys.stdout.write("".join(chr(i) for i in (73,83,66,78,32,61,32)))
>
>
> for i in range(0,4):
>
>     sys.stdout.write(str(prime(code[i],x)-key[i]))
>
> print
>
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