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Re: Encoding very big numbers, estimation challenge.
Posted:
Nov 1, 2013 11:29 PM


On Friday, November 1, 2013 1:56:37 PM UTC7, jonas.t...@gmail.com wrote: > Estimation challenge > > > > Given we do know our algorithm always work with a 512 byte, 4096 bit, number we do not need to encode numbers of digit in it correct? > > > > Now we look at first two digit, turns out to be 77. > > > > Thus we set MAX to 77 and MIN to 76 any objection? > > > > Now we start an automatised guessing game from a pregenerated table using an extended list of below instead of jumps of 10% use 1000 values so each jump is 0.1percent. > > > > ARRAY WITH INITIAL LIST TO COMPARE TO our number. > > Sqrt=Height=number*1,0 (10*10)/100=1.0 of area This is max square > > 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 ... > > > > OUR GAME FOLLOW THE SIMPLE ROWS BELOW (small dummy example) > > Number 222222 range MAX=10^x MIN^10^(x1) > > UNTIL floor.TRIED=floor.CURRENT > > 0,4*1000=400 400^2=160000 BELOW > > 0,45*1000=450 450^2=202500 BELOW > > 0,45+0,025*1000=475 475^2=225625 ABOVE > > 0,45+0,0125*1000=462,5 462,5^2=213906,25 BELOW > > 0,45+0,0125+0,00625*1000=468,75 468,75^2=219726,5625 BELOW > > 0,45+0,0125+0,00625+0,003125*1000=471,875 471,875^2=222666,015625 ABOVE > > 0,45+0,0125+0,00625+0,0015625*1000=470,3125 470,3125^2=221193,84765625 BELOW > > 0,45+0,0125+0,00625+0,0015625+0,00078125=471,09375 471,09375^2=221929,3212890625 IS IN LIST OF TRIED BREAK > > > > Each above below is encoded to either 0 or 1. > > > > Now we must start using two reserved decimal digits encoding MAX becuase our algorithm choose a value between 10256 to zoom in on number unfortunatly 77 came a bit between so our encoding starts... > > > > 0000000001001101 > > > > Now to the question > > > > 1.How many above/below guesses does our algorithm need to find our 2048 bit number. (Every above/below represent one bit) > > > > I am not mathematician so do not know, but i will implement this to check up your answers.
Just add them up each to add up those rules, for the definition of your language. Can you pick up a sequence and always find the start? These are sequences in binary with code in them thus information and data. You add up the time terms from for how much time it takes to decode the sequence in constant bit rate access to the sequence. Similarly storage is along the lines here of compression, for organization, in for example Huffman codes, for that the serial access is the network access or other I/O resource where it's generally tedious to read out the codes instead of for example reading words, (or updating the value) those basically contain the information there you convey. They convey it so exactly from information as quantitied into a value exactly with signal analysis and information theory as to Shannon, Huffman, modern analysis around the continuous and discrete, sampling theory, all this is built into the expectation of the routine, of what takes longer: looking up the code, or waiting for the code on the wire, here as its value. Compression then is generally to write the shortest code, for the information, on shared encoder and decoder state on data with value in data, and expectations on usual data. Then the encoder and decoder share the usual in terms of data, value, and transport. That is usually a time resource to add up, and space resource.
So, how does it?



