Time for a throwback!  This year, I posted my new article “The Art of the Compensating Control” over a three week period back in April.  A reader recently contacted me about a claim I make in Part 4 of the posting.  He says:

In your April 2009 blog The Art of the Compensating Control (Part 4) Tax day special, you stated that using the random function in COBOL to generate your key was in a sense, “a really bad idea”.  I have no knowledge of encryption so I don’t see the fault with the process.  How would this be equivalent to only 53 bits of encryption?

Excellent question!  The basis of this post relies on a tool by Mandylion Labs called the Brute Force Calculator.

True 128-bit encryption means that the key is 128 “bits” of data long.  A bit of data can either be a 1 or a 0.  8 bits make a byte, 1024 bytes make a kilobyte, and so on.  An example 128-bit key might look like…

0010100010101011001010001010101100101000101010110010
1111101010110010100010101011001010001010101100101000
101010110010100010101011

Save, by rpongsaj

Theoretically, this would be truly random, and you can use the entire “key space” offered by the key size and the algorithm you wish to use because you are placing a 1 or a 0 in each of the 128 bits.

Now let’s look at how ASCII-based numbers (or if you want, EBCDIC) are represented in binary.  Each ASCII character (or number in our case) is equivalent to 8 bits of data, or 1 byte of data (yes, technically 7 with 1 bit of parity).  For example, the binary value for the ASCII character ‘1’ is ‘00110001’.  So now, 16 digits (128-bits / 8 bits [because each number takes 8 bits to be represented in binary] = 16 positions), or characters, would be all you can use to make up your key.  If you know that your key is based on binary representation of ASCII numerical values, then you effectively know that each 8 bits of data can only be one of the following values:

• 0:    00110000
• 1:    00110001
• 2:    00110010
• 3:    00110011
• 4:    00110100
• 5:    00110101
• 6:    00110110
• 7:    00110111
• 8:    00111000
• 9:    00111001

So based on this, you know that at least every 4 bits of data will ALWAYS be 0011, with no exception, because you based your binary key on the ASCII numerical values.

Using the calculator referenced above, you can calculate the “effective” key strength by understanding that 16 digits gets you a total of 10 quadrillion (10 with 15 zeros behind it) possible permutations.  By taking that number and performing a logarithm using a base of 2 (because remember, binary numbers can be either 0 or 1, two different values) you find that 10 quadrillion effectively equals just over 2^53, or 53-bits of effective encryption.  For comparison, if you are able to use the entire 128-bit key space, the total number of possible keys would be 340,282,366,920,938,000,000,000,000,000,000,000,000.  That’s a pretty big number!

Granted that these examples are painting one particular picture.  Yes you could get creative and just pull the last two digits of the binary values and use MORE digits, but isn’t that the point?  Create a better, more random key?  If you are going down that road, why not just ask your user to type randomly on a keyboard while generating your key, then measure the time between keystrokes, combine it with the random number generator on your machine, and get about the closest you will get to true randomness with 128 0 or 1 rounded values?

So using the assumptions in the calculator (which may be outdated by way of computing power, especially in light of the recent improvements in GPU processing), 1,000 machines could crack it in less than 300 hours.

Give the calculator a try!  There are some other great things you can do with this sheet like calculating how effective your enterprise password scheme would be against a (non-rainbow table based) brute force attack.

In your April 2009 blog The Art of the Compensating Control (Part 4) Tax day special, you stated that using the random function in COBOL to generate your key was in a sense, “a really bad idea”.  I have no knowledge of encryption so I don’t see the fault with the process.  How would this be equivalent to only 53 bits of encryption?

This post originally appeared on BrandenWilliams.com.

Possibly Related Posts:
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}