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the wwii enigma machine and criptology


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i have in my mobil phone a free enigma machine and im wacthing videos on turing

 

i think the flaw of turing machines he use to uncrypt enigma its they get in a loop too easy

 

i like uncryptology :) im currently on factoring the base of modern cryptolgy and a brute force aproach to test all posible programs that for example to find the one that produces primes or pi for example

 

i think the key for always halting turing machines is ban the backwards jum and substitute it by loops of fixed size however much i love the goto

 

this is something spiritual im dealing with after all information is eternal and immaterial

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On 2/24/2021 at 2:03 PM, raaaid said:

 

this is something spiritual im dealing with after all information is eternal and immaterial

 

Makes me think of that conservation of information and Black Hole paradox.

 

Dat chit bends my little brain.

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its not posible to destroy information not even by a blackhole for it can be restored by the algorythm a=a+1 which contains all posible information

 

another thing is how to index those info elements unless with the element itself

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im onto a factorizing method that i think compromises rsa encryption security, do you think i could get in trouble if i was right?

 

ill elaborate on my method:

 

imagine you want to factorize the number 121, by its termination you know one of its factos its gonna end in 1,3,7 or 9, the rest of terminations are just not posible so im starting to siege

i make a for so i test all numbers from zero to nine like 01,11,21,31,41...,03,13,23,33,...

and i go on comparing terminations 11*01=11, termination dont match eliminated, 11*11=121, teminations match so 11 its a candidtae

 

on next iteration ill try 011,111,211,311...

011 is a winner

 

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I doubt that anyone will be bothered by your factorising method, Raaaid. Assuming it works, it looks like taking rather a long time to arrive at the answer, for the size of numbers actually used in encryption. The Sun will probably have gone into into its red-giant stage and swallowed up the Earth by the time it's finished. 

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my factorizing algorythm is looking good, so far im getting 13 digits prime factors in 10 seconds:

 

 

here the program if youre interested is pretty simple you can run it in this ide:

 

https://smallbasic-publicwebsite.azurewebsites.net/

 

its my favourite ide, its designed for kids, kids get away with load of sith

 


main()

Sub main
  objetivo=1234567913*1234567927
  number=objetivo
  TextWindow.WriteLine(number)
  iter=1
  comienzo()
  digitos=Math.Floor(Text.GetLength(number)/2)
  While flag<>1
    iter=iter+1
    factor()
    TextWindow.WriteLine(posibles)
    digitos=digitos-1
  EndWhile
  TextWindow.WriteLine("the factor are: "+fac1+"*"+fac2)
EndSub

sub factor 
ar=1
ar2=1
For times=1 To Array.GetItemCount(posibles)
  TextWindow.WriteLine(Array.GetItemCount(posibles)-times)
    for i=0 To 9
      posfac=Text.Append(i,posibles[times])  
      posibles2[ar]=posfac
      ar=ar+1
    EndFor
EndFor
posibles=posibles2
posibles2=""
ar=1
For times=1 To Array.GetItemCount(posibles)
  For times2=1 To Array.GetItemCount(posibles)
    If times>times2  Then
      Goto break2
    EndIf
    TextWindow.WriteLine(digitos+"*"+(Array.GetItemCount(posibles)-times)+"*"+iter)
    a=posibles[times]
    b=posibles[times2]
    prod=a*b
    If prod=number then
       fac1=a
       fac2=b
       flag=1
       Goto fin2
       EndIf
     While Text.GetLength(prod)<10
       prod=Text.Append(0,prod)
     EndWhile  
     l=Text.GetLength(prod) 
     prod=Text.GetSubText(prod,l-iter,l)
     If Text.EndsWith(number,prod)  Then
        while Text.GetLength(a)<iter
          a=Text.Append(0,a)
        EndWhile
        while Text.GetLength(b)<iter
          b=Text.Append(0,b)
          EndWhile
          posibles2[ar2]=a
          ar2=ar2+1
          posibles2[ar2]=b
          ar2=ar2+1
      EndIf
      break2:
    EndFor
  EndFor
  posibles=posibles2
  posibles2=""
  iguales()
  sort()
  fin2:
EndSub

Sub comienzo
for i=1 to 9
  for j=1 to 9
  
        mult=(i*j)
        a= text.GetSubTextToEnd(mult,Text.GetLength(mult)) 
        b= text.GetSubTextToEnd(objetivo,Text.GetLength(objetivo)) 
         if a=b Then
              ar=ar+1  
              posibles[ar]=i
             ar=ar+1    
            posibles[ar]=j
        
        EndIf
    EndFor
  EndFor
  iguales()
EndSub

Sub iguales
  
   y=0
  ye=0
  For ye=1 To Array.GetItemCount(posibles)
    temp=posibles[ye]
    If Array.ContainsValue(posiblesbis,temp) Then
    Else
      y=y+1
      posiblesbis[y]=temp
    EndIf
  EndFor
  posibles=posiblesbis
EndSub 
  
Sub sort
  var=posibles
  'bubble sort variables - must be in array called var
  For ii = 1 To Array.GetItemCount(var)-1
    For jj = ii+1 To Array.GetItemCount(var)
      If (var[jj] < var[ii]) Then
        temp = var[ii]
        var[ii] = var[jj]
        var[jj] = temp
      EndIf
    EndFor
  EndFor
  posibles=var
EndSub

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Yay! A bubble sort! Haven't seen one of them in years.

 

Seriously though, you should probably look into Big O notation, and why it isn't wise to guess how long a large task is going to take, based on how long it takes to do a smaller one. Is the scaling linear? Quadratic? Exponential? I'm not going to try to figure it out for your algorithm, since that sort of maths is a bit beyond me, but I can tell you from experience that what looks like a good algorithm at first can turn out to be useless later. Try it on the size of number you are ultimately going to use it on, and then come back when its finished. If it does at all, in any reasonable timeframe, which I suspect it won't.

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after being in the likely not posible compresion so long factorization is a joy, dont they say aim for the moon and youll touch the sky?

 

currently im factorizing into primes of 12 digits in tenths of seconds but im not sure why it only works with some numbers

 

trying to account for the reminder has improved eficiency, also using instead a true natural numbers who account for the zero to the left seems to be solving the zeros problem

 

the last years when im about to wake up i dream im trying to debug reality and i use assembler :) 

 

also i noticed with my method the iterations you have to make for each new digits dont grow linearly, either they dont grow at all or they grow arithmetically

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8 minutes ago, raaaid said:

the last years when im about to wake up i dream im trying to debug reality and i use assembler :) 

 

At least you know you were dreaming. Unless 'being awake now' is a dream and the 'dream' was real. In which case, good luck with the debugging. Is there anywhere I can send bug reports? 😃

 

 

 

Edited by AndyJWest
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havent you ever been called a zero to the left?

 

why neglecting the zeros the left?

 

this comes form natural numbers coming from the first manual computers, abacus due to the use of fixed rows for digits

 

but if they had used numbered cards instead we would be have accounted indeed for the zero to the left and our numbers would be more natural

 

the number 123 acounting for the zero fto the left is 012 and the number 1000  889

 

so if the number of the beast was expressed in true natural numbers that number in our numbers is 777

 

neglecting the zero to the left in the unnatural numbers means you can not do basic operations as reversing a number what you can perfectly do in the true natural numbers as when a number ends in 0

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  • 2 weeks later...

ok here it goes a mathematical question that has bugged me for years:

 

you count to one in the first second

 

count to two in next half second

 

count to four in the next quarter of second

 

count to eight in the next eighth of second

 

count to sixteen in the next sixteenth of second

 

and so on

 

what number will you be counting in second two?

 

and what is worse what will you be counting in second three?

 

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