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A comment on The Catch 22 of Base64: Attacker Dilemma from a Defender Point of View.

Re:The Catch 22 of Base64: Attacker Dilemma from a Defender ... by boneitis on Tuesday May 01, 2018 @01:27PM (#16969174)
To really crank up the pedant-O-meter, wouldn't it be accurate to instead describe the output growth in point #1 as polynomial, even sub-quadratic?
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Re:The Catch 22 of Base64: Attacker Dilemma from a Defender ... by boneitis on Wednesday May 02, 2018 @04:24AM (#16975020)
Perhaps i can blame the booze, as i an admittedly very drunk but still unconvinced and/or confused.

I asked my question coming from the angle of big-oh analysis, and all three (as of the time of this reply) responses in the thread seem to describe a growth by a constant factor of 1.3bar, bar over 3, otherwise known as polynomial growth, yes?

Absolutely, the growth can be mathematically expressed with an exponent, but we verbally characterize the actual growth to be polynomial if the scalar stays constant (a.k.a. a literal number in the exponents place) rather than increasing as the amouny of input increases (independent variable lying within the actual exponent), no?

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Re:The Catch 22 of Base64: Attacker Dilemma from a Defender ... by boneitis on Wednesday May 02, 2018 @12:20PM (#16978504)
Man, I'm dyslexic.
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Re:The Catch 22 of Base64: Attacker Dilemma from a Defender ... by benchaney on Tuesday May 01, 2018 @01:56PM (#16969488)
No. It is exponential, not polynomial. That is, it is (4/3) ^ x rather than x^(4/3).
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Re:The Catch 22 of Base64: Attacker Dilemma from a Defender ... by tempodox on Tuesday May 01, 2018 @01:47PM (#16969397)
The least one could say is: The growth factor is not 1.3333, but 4/3.
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Re:The Catch 22 of Base64: Attacker Dilemma from a Defender ... by matharmin on Tuesday May 01, 2018 @01:56PM (#16969496)
The output size is O(4/3^n), which is exponential.
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