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Topic: Find the perfect square closest to n(x), i just want the perfect square above or below no decimals. Can it be solved using geometry?
Replies: 7   Last Post: Nov 1, 2013 8:53 PM

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Graham Cooper

Posts: 4,320
Registered: 5/20/10
Re: Find the perfect square closest to n(x), i just want the perfect
square above or below no decimals. Can it be solved using geometry?

Posted: Nov 1, 2013 8:53 PM
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On Friday, November 1, 2013 4:13:39 AM UTC-7, scattered wrote:
> On Friday, November 1, 2013 12:22:07 AM UTC-4, graham...@gmail.com wrote:
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> > On Thursday, October 31, 2013 8:01:53 PM UTC-7, Ben Bacarisse wrote:
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> > > jonas.thornvall@gmail.com writes:
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> > > <snip>
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> > > > Actually Ben i have a similar that may be easier for you to follow,
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> > > > any square can be divided into 4 sub squares. And if we have a number
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> > > > we can find the 10^x above it and 10^x-1 below it.
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> > > > So 10^x is 1 now we can chose if we want 0 at real zero or zero at
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> > > > square 10^x-1 If we choose the later we close in faster. Now the area
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> > > > between the lesser and bigger square or if we use zero, can be
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> > > > described as a percentage ratio of the height.
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> > > Sorry, I can't make head nor tail of this.
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> > > <snip>
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> > > > The perfect square we find is subtracted from our number and now we
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> > > > work same approach for this smaller square. This is repeated until the
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> > > > full number is encoded to a series of squares + a small integer less
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> > > > then 4.
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> > > Yes, this bit I've understood, but why? What's the point of doing this?
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> > I'd like a SQRT(X) algorithm that goes to 10,000 digits.
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> > useful for 1 time cypher pads in this age of eavesdropping!
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> > In Javascript, so no more than 100,000 maths operations..
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> > Binary is fine too!
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> > I understand Ben's point... if you are using repeated approximations
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> > then Newtons method is very fast.
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> > Herc
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> > --
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> > www.PrologDatabase.com
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> How would *any* mathematical operation be useful in the
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> creation of 1-time pads? Running a random bit-pattern through
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> a function is at best harmless and at worst will introduce
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> statistical correlations between the bits which could only
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> hurt security. On the other hand -- if the input to the
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> function has not been generated by a genuinely random
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> process then it isn't a 1-time pad. It is true that the output
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> of a pseudo-random number generator *might* be improved (from the point
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> of view of passing various statistical tests for randomness) by running it
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> through a function, but that has nothing to do with 1-time pads.




You can INDEX the start of the CYPHER string anywhere using a PRIVATE KEY.

E.g. SQRT( 94847473475474838374572943754743842392373748484745 )

would give you and your communications staff a practically un-crackable
1 time pad, by using 94847473475474838374572943754743842392373748484745
as a SCRAMBLE/DESCRAMBLE password that isn't amenable to brute force
password hacking.



MY Algorithm is to use SQRT(day of year) or some such, changes once per day
for the BASE 10,000 random digits.

Once you have 10,000 random digits you have massive, massive random space
to encode things.



TAKE the 1st 50 pairs of digits as the SCRAMBLE SET.

23
54
01
04
99
33
...
49


Use these as SAWTOOTH HARMONICS to extend the CYPHER to 2,000,000 digits.

Break the 1,000,000 long CYPHER into 1000 segments of 1000 digit pairs.


TO send 1 sentence, generate 3 Random Segment Blocks.

START OF SENTENCE
343
666
999




TO encode each character, generate 3 random offsets 1 in each segment.

343 : 234 = 44
666 : 653 = 58 44+58+77 = 'C'
999 : 111 = 77

343 : 031 = 22
666 : 343 = 77 22+77+21 = 'A'
999 : 043 = 21

343 : 232 = 87
666 : 543 = 13
999 : 001 = 13



So to send a sentence would be:

343-666-999-234-653-111-031-343-043-232-543-001
^^^^^^^^^^^
SENTENCE BLOCKS



You Generate the 2,000,000 Cypher Pad on the receivers side
using the Starting Key... which is a bit more complicated
to fetch the set of 50 harmonic values from the 10,000 digits
by using a private password entry...(rather than the 1st 50 digit pairs..)

You subtract the harmonics from the message,
subtract the CYPHER from the message,
this gives you the value at each SEGMENT:OFFSET
by adding the 3 random values... which is the
ASCII Value of the letters...

BAS BAS BAS - OFF OFF OFF - OFF OFF OFF - OFF OFF OFF
C A T

There are 100s of parallel equations to decript the message
so much that it is easier to crack the scrambler password by brute force
(which could be up to 100 digits long).

Even with numerous repetitions of each of the 1,000,000 cypher values
(i.e. years of communications) you would only get 1 of the 3
cypher digits at a time using frequency analysis in 1,000,000 possible
and each digit is a triple of 3 random cypher positions!



Herc
--
www.ChatZombie.com






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