Re: Will Ada-95 Programs Written in MS Windows Run in MacOS and Linux Without Some Tweaking.

[Austin Obyrne <austin.obyrne@hotmail.com>]

On Monday, December 16, 2013 9:19:15 PM UTC, Simon Wright wrote:
> Austin Obyrne <austin.obyrne@hotmail.com> writes: > I am the first to admit that this cryptography has very high > ciphertext expansion but I cannot understand it being 870 Megabytes > for the example you gave. I have no doubt you are right. Each character is written as 3 integers, each occupying 11 characters. I've put a new line between every group of 3 integers, whereas you put a new line where there's a new line in the original (I think; anyway, not a good idea when the input isn't text), => 34 characters per byte (35 on Windows). 34 * 27_014_786 = 918_502_724 (875 MB). > Most of what you say is technically beyond my ken also but its is very > encouraging that you have not decided to condemn it out of hand as a > lost cause like for instance the current complexity-theoretic crypto > that is headed for the scrap heap. My interest is in helping people to get code working right, not in cryptography; I'd probably use GPG if I needed to. It seems to me that yours is a symmetric-key scheme (I can't read the documentation on your site because you've packaged up multiple PDF files in one, which is a feature that's not supported by Mac Preview, and I'm not going to install Acrobat).

< Each character is written as 3 integers, each occupying 11 characters

An integer occupying 11 characters like say 21474836470 would not be possible on a 32-bit Windows computer where the largest positive integer is 2147483647 and it would cause a "Constraint Error" to appear  due to integer overflow.

Am I reading you right?

Austin.