Re: Boolean math: Fill in the blanks.

Very interesting topic!

I see an overall pattern, which I would like to share :-)

First, a few basic observations:

There seems to be a certain symmetry evident from how R & R results in 8 bits set on average (halfway between 1 and 16) versus R | R resulting in 24 bits set on average (halfway between 16 and 32). I'll be sure to make my hypothesis symmetrical, in that regard.
If we apply AND, the average bits set goes down due to higher likelihood of 0 bits swallowing the 1 bits. Conversely, if we apply OR, the average bits set goes up due to higher likelihood of 1 bits overriding the 0 bits.
Non-associative rules apply for AND and OR:
So evaluating the expression from left-to-right is important! Update: Perl gives precedence to the & operator over |

Hypothesis:

Each AND/OR operation is evaluated in the following way:

( <left_gained_avg_bits_set> <op> <right_modifier> ) = <resulting_avg_bits_set>

Whenever an AND/OR <op> is applied, it induces a change to the already gained average bits on the left-side. That change is a multiplicative factor determined by the average bits set on the right-side:

right_mod_factor = right_avg_set_bits / total_bits

Cases:

For <op> = AND, we would take right_mod_factor and multiply it with left_avg_bits_set to get <resulting_avg_bits_set>. Examples:

R & (R & R) => 16 * (8 / 32) = 4 bits set on average

(R | R) & (R & R & R) => 24 * (4 / 32) = 3 bits set on average

That is, if there are more bits set on average on the AND right-side, there will be less chance of swallowing, so more bits are set on average in the result.

Conversely, for <op> = OR, we would take right_mod_factor and multiply it with left_avg_unset_bits. Then, take that result and add it to left_avg_set_bits to get <resulting_avg_bits_set>. Examples:

R | (R | R) => 16 + 16 * (24 / 32) = 28 bits set on average [second 16 is # of avg unset bits]

(R | R) | (R & R & R) => 24 + 8 * (4 / 32) = 25 bits set on average

R & R & R | R | R = (R & R & R) | (R | R) => 4 + 28 * (24 / 32) = 25 bits set on average

R & R & R | R | R = (R & R & R) | R | R => 4 + 28 * (16 / 32) + 14 * (16 / 32) = 25 bits set on average [got 14 from 32 - ( 4 + 28 * (16 / 32) )]

That is, if there are more bits set on average on the OR right-side, there will be more chance of overriding, so more bits are set on average in the result.

Conclusion:

So, following my pattern hypothesis, I see some inconsistent results in your "full set". ~~For instance, 7 of ( R & R | R & R ) & R evaluates to 5 bits set on average.~~ Attention has to be paid to precedence of AND/OR evaluations. Also, 22 & 26 are the same, when 22 of R & R | R | R evaluates to 26 bits set on average.

There may be some fallacy in my hypothesis, so feedback is welcomed :)

So, given the following legend:

n  = total # bits per sample
L1 = # bits   set on average on  left-side of AND/OR
L0 = # bits unset on average on  left-side of AND/OR
   = n - L1
R1 = # bits   set on average on right-side of AND/OR
[download]

we can derive a set of equations to calculate the resulting average bits set for each AND/OR operation:

AND : L1 * R1 / n

OR  : L1 + L0 * R1 / n = L1 + (n - L1) * R1 / n
                       = L1 + R1 - L1  * R1 / n
                       = L1 * ( 1 - R1 / n ) + R1
[download]

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Re^2: Boolean math: Fill in the blanks. by BrowserUk (Patriarch) on Oct 12, 2008 at 08:25 UTC
Conclusion: ... I see some inconsistent results ... 7 of `( R & R \| R & R ) & R` evaluates to 5 bits set on average. If you plug `( R & R \| R & R ) & R` into the OP code and run it a few times, you'll see that the average does approximate 7: `C:\test>booleanBuk -N=1e2 7.02 C:\test>booleanBuk -N=1e2 6.88 C:\test>booleanBuk -N=1e2 7.06 C:\test>booleanBuk -N=1e2 7.06 C:\test>booleanBuk -N=1e2 6.92 C:\test>booleanBuk -N=1e2 7.08` [download] And as you increase the number of iterations, the more closely the observations coincide with the theoretical value: `C:\test>booleanBuk -N=1e5 7.00138 C:\test>booleanBuk -N=1e5 6.98986 C:\test>booleanBuk -N=1e5 6.99812 C:\test>booleanBuk -N=1e5 7.00729 C:\test>booleanBuk -N=1e6 6.994402 C:\test>booleanBuk -N=1e6 7.002038` [download] That's how I like my proofs. Tangible :) Sorry if I have misunderstood your hypothesis. Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice. "Too many [] have been sedated by an oppressive environment of political correctness and risk aversion."	[reply] [d/l] [select]
Re^3: Boolean math: Fill in the blanks. by repellent (Priest) on Oct 12, 2008 at 16:54 UTC
But plug in `(( R & R ) \| R ) & R & R` into the OP code and it tends towards 5, which was how I was interpreting the expression: `( R & R \| R & R ) & R` My mistake. Perl evaluates `( R & R \| R & R ) & R` as `( (R & R) \| (R & R) ) & R` and that, according to my hypothesis, results in 7 bits set on average.	[reply] [d/l] [select]