Another free giveaway: 9 books on cryptology

Apparently, you, the readers of this blog sphere, are not into “historical” guitar magazines. No one showed any interest in my recent offer of giving away my collection for free.

Could it be that you are unaware of the fact that music is science? At least, music was considered science until not too long ago. Well, until the 17th century, at least. For me, math and music share the same space in my brain. In case you never came across this, listen to any of Johann Sebastian Bach’s pieces. Other good choices are music from  Schönberg or Webern. Or — even better — read Douglas Hostadter’s “Gödel, Escher, Bach: An Eternal Golden Braid”, a must-read for computer science students.

Today, I would like to offer 9 of my books on cryptology as a giveaway. The giveaway will not be for free, however. In order to get the books, you need to give me a useful answer  to the following question which touches the connection between math and music:

In music, an octave means double frequency, i.e. the frequency relation 2:1. Likewise, a fifth means a frequency relation of 3:2. It is common to think of stacking fifths. This is called the “circle of fifths”. If one puts 2 fifths over the note C, one gets the note D. This corresponds to squaring 3:2, i.e. 9:4.

How many fifths do you need to stack to get back to some note C again? And why do you need to cheat here? Who became known for coming up with a “hack” to this problem.

I am offering:

  • Menezes et al.: Handbook of Applied Cryptography
  • Stinson: Cryptography Theory and Practice
  • Schneier: Applied Cryptography
  • Rosing: Implementing Elliptic Curve Cryptography
  • Knudsen: Java Cryptography
  • Pincock et al.: Geheime Codes
  • Levy: Crypto
  • Singh: Geheime Botschaften
  • Rijmen: The Design of Rijndael

(The winner needs to pick up the books in my office.)

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3 thoughts on “Another free giveaway: 9 books on cryptology

  1. Manfred

    One would need 12 fifths, regarding to what the “Quintenzirkel” says, right? But why is it 12 and not 14 (which would be somewhat logical – 7 times 2 fifths).
    Maybe this is because of the half tones between E/F and H/C? Because if you add 2 fifths to E, you will get the note F#, not F, and analogous H + 2 fifths gets you a C#. So you will “reach the C 2 quints earlier” than one would think in the first place.
    I assume this is the “hack” you mentioned?

    Nonetheless, I am interested in some of the books as well as the answer to your question.

    Reply
    1. kcposch Post author

      > One would need 12 fifths, regarding to what the “Quintenzirkel” says, right?

      Yes.

      >But why is it 12 and not 14 (which would be somewhat logical – 7 times 2 fifths).
      Maybe this is because of the half tones between E/F and H/C? Because if you add 2 fifths to E, you will get the note F#, not F, and analogous H + 2 fifths gets you a C#. So you will “reach the C 2 quints earlier” than one would think in the first place.

      No. This reasoning is wrong.

      >I assume this is the “hack” you mentioned?

      No.

      >Nonetheless, I am interested in some of the books as well as the answer to your question.

      I am sorry, but all the books are gone by now. But I thank you anyway for your try.

      Reply

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