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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?
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7
Last Post:
Nov 1, 2013 8:53 PM



JT
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4/7/12


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 12:20 PM


Den fredagen den 1:e november 2013 kl. 15:03:39 UTC+1 skrev jonas.t...@gmail.com: > Den fredagen den 1:e november 2013 kl. 04:01:53 UTC+1 skrev Ben Bacarisse: > > > jonas.thornvall@gmail.com writes: > > > > > > <snip> > > > > > > > Actually Ben i have a similar that may be easier for you to follow, > > > > > > > any square can be divided into 4 sub squares. And if we have a number > > > > > > > we can find the 10^x above it and 10^x1 below it. > > > > > > > > > > > > > > So 10^x is 1 now we can chose if we want 0 at real zero or zero at > > > > > > > square 10^x1 If we choose the later we close in faster. Now the area > > > > > > > between the lesser and bigger square or if we use zero, can be > > > > > > > described as a percentage ratio of the height. > > > > > > > > > Sorry, I can't make head nor tail of this. > > > > > I find it hard to beleive that a skille mathematician can not understand this principle. > > HERE ENDS BIG SQUARE BETWEEN SMALL BIG IS RANGE > > Sqrt=Height=number*1,0 (10*10)/100=1.0 of area > > Sqrt=Height=number*0,9 (9*9)/100=0.81 of area > > Sqrt=Height=number*0,8 (8*8)/100=0.64 ... > > Sqrt=Height=number*0,7 (7*7)/100=0.49 ... > > Sqrt=Height=number*0,6 (6*6)/100=0.36 ... > > Sqrt=Height=number*0,5 (5*5)/100=0.25 ... > > Sqrt=Height=number*0,4 (4*4)/100=0.16 ... > > Sqrt=Height=number*0,3 (3*3)/100=0.09 ... > > Sqrt=Height=number*0,2 (2*2)/100=0.04 ... > > Sqrt=Height=number*0,1 (1*1)/100=0.01 ... > > HERE START SMALL SQUARE BETWEEN SMALL AND BIG IS RANGE
I must try start doing something about my spelling and slopiness, it should of course be the small square ending, thus we have a range between where the small start and big end. And as i show we either can chose to close in on the number using a precalculated of power 10^x of ten 1/10, 2/10, 3/10... 1/100, 2/100 and creating the size of precalculated list we chose, now each term in the list is actually an area reflecting a percentage of our GREATER SQUARE do you agree?
And for each entry in list there is also a precalculated SQRT, anyone knowing that if we take 50% of height =h * 0,5 of our MAX square, we also do know that if we square. It will represent 1/4 AREA and as you know 5*5/100, and from this we can make the list that i presented above, to any precision of our like. 10,100,1000 entrys, well you pick.
Let us say I chose number 5763331 the closest bigger 10^x is 10 millions. 5 763 331/10 000 000=0,5763331 i have only 10 item in my precomuted in list. We chose the closest fit and begin our above below game. And when an item reoccur in list we know it is the correct square and we can finish.
An alternative way to do this is working just with squares chosing high or low.
NOW TO SOMETHING ***VERY*** INTERESTING When you realise this one understand it is a way to compress a number. It can be encoded, like MAX + OUR CHOICES ENCODED BINARY ABOVE=0 BELOW=1
So our number will end up look something like this... MAX 6=110 and OUR CHOICES 11011011 UNTIL FOUND.
SOMEHTING EVEN MORE INTERESTING, HOW WE BREAK DOWN NUMBER MAY BE DIGIT DEPENDENT, WE CAN FIND REFINED METHODS/FUNCTIONS COMPRESSING DEPENDENT OF SIZE AND RELATION OF DIGITS.
Basicly what i say is that random noice is recursively comrpessible down to a limit, information how it is done are the rules formed in our program.
> > > > > > <snip> > > > > > > > The perfect square we find is subtracted from our number and now we > > > > > > > work same approach for this smaller square. This is repeated until the > > > > > > > full number is encoded to a series of squares + a small integer less > > > > > > > then 4. > > > > > > > > > > > > Yes, this bit I've understood, but why? What's the point of doing this? > > > > > > > > > > > > <snip> > > > > > >  > > > > > > Ben.



