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Challenge "Beaver Code"

Challenge "Beaver Code"  

  By: admin on Aug. 13, 2012, 8:15 p.m.

The Beaver Code is a transposition cipher. Apart from the solution of this challenge, can you find a formula for any given ciphertext length, which generates the permutation needed to decrypt the ciphertext?
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 Last edited by: admin on Oct. 31, 2021, 2:55 a.m., edited 1 time in total.

Re: Challenge  

  By: Veselovský on Aug. 13, 2012, 11:54 p.m.

We would like to know whether there is an algorithm that can
generate the permutation which is needed to decrypt the ciphertext
for messages of any given length.

Why it shouldn't?

To code a message we need s steps
[img]http://www.mysterytwisterc3.org/phpbb3/download/file.php?mode=view&id=10&sid=6e6a9f8992f29f127adda84f9c245276[/img]
(n is the length of message, || is the ceiling function)

2^(k-1) divisions are needed in each k-th step, so then

Total number of divisions for the whole coding process is exactly:
[img]http://www.mysterytwisterc3.org/phpbb3/download/file.php?mode=view&id=11&sid=6e6a9f8992f29f127adda84f9c245276[/img]

Same number of steps and divisions is needed for the decoding process.

This is of course the most straightforward algorithm so a better question would be if there is a more efficient/better/faster algorithm.

Re: Challenge  

  By: Killer on Aug. 15, 2012, 2:06 a.m.

[quote:24wzjcoh]Uns würde interessieren, ob es eine Formel gibt, die für alle
Text-Längen direkt die Permutation erzeugen kann, mit der man
aus einem Geheimtext den entsprechenden Klartext erzeugen kann.[/quote:24wzjcoh]
The following is a more or less handy description of the permutation:

If the length L of the text is a power of two, say L=2^n, then the encryption permutation P is given by bit reversal P=BR_n, in particular it is an involution then. For general L choose n with 2^n >= L. Then P(i)=BR_n(k_i) where k_i is the (i+1)-th element in 0,1,…,2^n-1 such that BR_n(k_i) < L.

[ Encryption]

Re: Challenge  

  By: Veselovský on Aug. 15, 2012, 3:07 a.m.

:-)
It seems that the English version of the challenge is not an exact translation of the German one.
In the appendix of English version they asked for an algorithm that can generate the permutation and in German one they asked for a formula.

Re: Challenge  

  By: DarkFibre on Aug. 15, 2012, 5:24 a.m.

We would like to know whether there is an algorithm that can
generate the permutation which is needed to decrypt the ciphertext
for messages of any given length.

Yes.

Re: Challenge  

  By: aurelie on Aug. 22, 2012, 6:45 a.m.

:-)
It seems that the English version of the challenge is not an exact translation of the German one.
In the appendix of English version they asked for an algorithm that can generate the permutation and in German one they asked for a formula.

Indeed, the English translation was imprecise. We are interested in a formula to solve this transpostion cipher.
A new version of the pdf file is going to be uploaded within the next days.

Re: Challenge  

  By: barr306 on Oct. 21, 2012, 5:01 p.m.

Im pretty sure ive decoded the cipher but everytime i enter the answer it doesnt work. what should i do?

Re: Challenge  

  By: be on Oct. 21, 2012, 11:56 p.m.

Im pretty sure ive decoded the cipher but everytime i enter the answer it doesnt work. what should i do?

Hello, According to the log you solved this challenge already (18:22). Could you please add a comment, how you was able to circumvent the problem? Thanks, be

Re: Challenge  

  By: i2gh0st on Nov. 14, 2012, 7:23 p.m.

hello,
little help please :)
so if i got 10 letters of the alphabet

a b c d e f g h i j
it will go like this ?
acegi | bdfhj
aei | cg | bfj dh
ai e |cg |bj f |dh

==> aiecgbjfdh ?

Re: Challenge  

  By: aurelie on Nov. 15, 2012, 11:57 a.m.

Hello i2gh0st,

a b c d e f g h i j
it will go like this ?
acegi | bdfhj
aei | cg | bfj dh
ai e |cg |bj f |dh
==> aiecgbjfdh ?

yes, you're right. There is also an example on slide 3/6.

Good luck!

Re: Challenge  

  By: i2gh0st on Nov. 15, 2012, 2:37 p.m.

thank you :)

i was on the good track last night but probably to tired.
easy … workaround discovered :)

Challenge "The Book Code: A Challenge for Bookworms"  

  By: Sansibar on Feb. 15, 2013, 3:32 p.m.

Hallo

Habe die Authorin / den Author gefunden.

Um sicher zu gehen:

15.00 bedeutet Zeile 15 und Spalte 00
1.41 bedeutet Zeile 1 und Spalte 41
6.01 bedeutet Zeile 6 und Spalte 01

Ist das wohl richtig?

Danke für die tollen Rätsel!

Gruss
Sansibar

Re: Challenge "The Book Code: A Challenge for Bookworms"  

  By: aurelie on Feb. 17, 2013, 10:18 p.m.

15.00 bedeutet Zeile 15 und Spalte 00
1.41 bedeutet Zeile 1 und Spalte 41
6.01 bedeutet Zeile 6 und Spalte 01

Ja, das ist richtig: Die Zeilen sind beginnend mit 1 durchnummeriert, bei den Spalten beginnt man mit 00, 01, 02, …

Re: Challenge  

  By: Anita on Nov. 28, 2016, 4:37 p.m.

Halli Hallo

Viellicht kann mir jemand helfen bei dieser Challenge. Habe die Buchstaben genau wie im Beispiel geordnet, raus kommt aber nichts sinnvolles. [HTML_REMOVED]

Grüsse Anita

Hi

Can anybody help me with this challenge? I rearranged the letters like in the example but there is only nonsense [HTML_REMOVED]

Greetings Anita

Re: Challenge  

  By: grigorios on Nov. 30, 2016, 10:56 a.m.

Halli Hallo

Viellicht kann mir jemand helfen bei dieser Challenge. Habe die Buchstaben genau wie im Beispiel geordnet, raus kommt aber nichts sinnvolles. [HTML_REMOVED]

Grüsse Anita

Hi

Can anybody help me with this challenge? I rearranged the letters like in the example but there is only nonsense [HTML_REMOVED]

Greetings Anita

Hi there Anita.

To help you understand how this challenge works try encoding and decoding a sentence of your own.

I hope that will help you understand how the cipher works if not PM me for an extra hint.


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