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QuestionThe Bessel-Overhauser Spline interpolation - suitable values for the weight function Pin
Kenneth Haugland4-Apr-14 0:34
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QuestionFactoring algorithm Pin
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Peter,

I was waiting for someone to figure that out. Check with Ravi, I proposed this to him in Email in March. The following is the second half of the original Email to him. Too bad this would not work, I think that the actual algo would work except for the fact that the file requirements are insane:

<pre>
------------------------------------------------------------------------------------------

The following discussion is why this algorithm (the part above this point) should be
published in Algorithms on April 1st without this lower section - let the wrecking crew of
experts in CP determine what is so wrong with the solution (the algorithm is correct but
just will not work).

------------------------------------------------------------------------------------------

Now, the number of primes with 300 digits is:

((10^300) / ln(10^300)) = (3.33 * (10^297))

The max size of a prime number in that range is 52 DWORDS or 26 QWORDS which would convert
to 232 characters so the total size of all path names describing all of these prime
numbers will be:

(232 * 3.33 * (10^297)) = (7.7 * (10^299)) BYTES

The number of terabytes to contain this data would be:

(7.7 * (10^299) / (10^12)) = (7.7 * (10^287)) TB

The number of 4 TB drives to contain this total size would thus be:

((1/4) * 7.7 * (10^287)) = (1.9 * (10^287)) drives

I don't think that Seagate has enough material to create that many drives. Penrose in his
book "The Emperor's New Mind" estimated that the number of particles in the visible
universe was 10^80, and after that problem has been solved, you still have another
problem about how much time it would take to write out these prime number directories to
the drives. So much for having the prime numbers available to use as trial multipliers to
see if the product matches the semi prime.

Lest you think it would make a difference (in the count and total size of the prime
numbers) if you only saved the prime numbers that lie between 10^200 and 10^300:

((10^300) / ln(10^300) - (10^200) / ln(10^200)) =
(1.447 * (10^297) - 2.171 * (10^197)) =
(1.447 * (10^297))

or

(1.447 * (10^297)) = (1.447 * (10^297))

Saving the number of prime numbers in a range of digits (from 200 to 300) would not even
make a visible dent in the count because the actual calculation is subtracting a 197 digit
number from a 297 digit number, leaving a 297 digit number with a diminished lowest value
(the last 196 digits if a borrow was needed else 197 digits), but the most significant 100
(or at least 99) digits would be unchanged. Only if you displayed the result in its actual
297 digit decimal form (instead of in truncated scientific notation) would you be able to
see the difference.

Another way to visualize this is to see how big a block of (1.9 * (10^287)) 4 TB drives
would be. Now my drives are (in inches) about 1.75 * 4.75 * 7. The diagonal from the
center of the drive to any corner is:

((((1.75 / 2)^2 + (4.75 / 2)^2 + (7 / 2)^2))^(1 / 2))) = 4.32 inches

A slightly smaller version of the block of drives would be to block the drives as a 4 wide
and 2 deep block, almost making them a cube. This would group the drives in blocks of 8
drives giving a total number of blocks as:

((1.9 / 8 * (10^287))) = (2.375 * (10^286)) blocks

The diagonal from the center of the block to any corner is:

((((1.75 * 4 / 2)^2 + (4.75 * 2 / 2)^2 + (7 / 2)^2))^(1 / 2))) = 6.86 inches

The dimensions (in the number of blocks) of a block with the same number of blocks in each
row, column, and layer would be:

((2.375 * (10^286))^(1/3)) = (2.874 * (10^95))

The diagonal from the center of this block to any corner would be

(6.86 * 2.874 * (10^95)) = (1.97 * (10^96)) inches
((6.86 * 2.874 / 12) * (10^95)) = (1.64 * (10^95)) feet
((6.86 * 2.874 / 5280) * (10^95)) = (3.11 * (10^91)) miles

The radius of a sphere that would enclose this block of drives would then be:

(3.11 * (10^91)) miles.

The radius of the the sphere that would enclose our solar system would be equal to the
orbit size of the planetoid Pluto (I guess I should be politically correct and not call
it a planet anymore) which is:

4.583 billion miles

or

(4.583 * (10^9)) miles

How much bigger would the radius of the sphere enclosing the block of drives be relative
to the radius of the sphere enclosing the solar system? To compare two spheres for "which
one is bigger, and by how much", consider comparing a baseball with a basketball. You
compare their height or width differences and not their circumference or volumes
differences (whether you are comparing diameters or radii), so try:

((radius of drives enclosing sphere) / (radius of solar system enclosing sphere))
((3.11 * (10^91)) / (4.583 * (10^9))) = (6.79 * (10^81))

It would be (6.79 * (10^81)) times as big as the radius of our solar system, and,
furthermore, you would be filling that space with 4 TB drives instead of the almost
complete vacuum that exists there today.

Where would you get that much material to build the drives? Maybe we could be seeing some
quantum sharing of particles between (10^287) different drives, all at the same time,
without dropping any recorded bits??? Let's not even consider the power requirements, or
the write time to populate the directory entries before the factoring algorithm starts.
</pre>

Dave.
QuestionTeam Contract Algorithm Pin
Laurence Senna27-Mar-14 21:23
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Richard MacCutchan28-Mar-14 1:22
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QuestionDetecting File Changes Pin
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QuestionHow much is my encryption algorithm worth? Pin
Daniel Mullarkey22-Mar-14 21:47
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Richard MacCutchan23-Mar-14 5:15
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GeneralRe: How much is my encryption algorithm worth? Pin
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