Information security and cryptography: Secure random number

Some theory

Well, the next step is to verify the keys of the One-Time-Pad randomly generated by Python's random method

There are many types of methods for checking if the numbers are random.

Characteristics:

Uniformly distributed
No numbers repeated on a specified length
Numbers must be independent of each other
The series should be reproducible
The numbers must be generated quickly

Part of the previous program had to generate the random numbers

Code

	def crearlibretas(): ## aqui creamos las libretas para la encriptacion
	x = 0
	y = 0
	pepe = open("libreta1.txt","w") ##libreta el que envia
	ana = open("libreria2.txt","w")
	while y<=20:
	while x<= 20:
	numeror = random.randint(0,2) ##valor random entre: 0<= x <2
	#float(numeror)
	pepe.write(str(numeror)) ## escribimos el caracter
	ana.write(str(numeror))
	# maria.write(str(numeror)) ## escribimos el mismo caracter
	x = x + 1
	#print "y = ",y,""
	pepe.write("\n") ##salto de linea en el archivo
	ana.write("\n")

view raw Random hosted with ❤ by GitHub

In this code I generate random numbers between 0 and 1 to create different keys, now is necesary to verify that this numbers are really random

Test used "Run Test"

In this Test is verified if the hypothesis of a number is really random between two elements, in this case 0 and 1.

Parameters

amount of the first element in this case "0"
amount of the second element in this case "1"
amount of elements
amount of blocks

Regarding the blocks, a block is when the same element is repeated until meet the other element and it starts over with a new block:

For example:
In this case, we have 5 Blocks

Having this elements clearly defined, we can begin seeing how the test works

First, we get the arithmetic mean:

After we got the arithmetic mean, the variance:

Finally we got the probability of Z

Code:

	import math
	entrada = open("libreta1.txt")
	nn = 0
	nm = 0

	y = 0
	totalb = 0
	bloquea = 0
	#bloqueb = 0
	x = 0
	#def bloques():
	u = 0

	##leemos la libreta que contienelos numeros
	for i in entrada.xreadlines():
	tam = len(i)
	######guardamos la cantidad de 1 y 0######
	while x < tam:
	if i[x] == "0":
	nm = nm + 1
	else:
	nn = nn + 1
	#x = x + 1
	###chekamos cauntos bloques tiene###########
	if y != 0:
	if i[x] == bloquea:
	print "si es del bloque"
	else:
	totalb = totalb + 1
	bloquea = i[x]
	print "otro numero diferente"

	else:
	totalb = totalb + 1
	bloquea = i[x]
	y = y + 1
	x = x + 1

	x = 0



	#u = float((2(nn*nm)/(nn + nm)) + 1)
	##proceso a realizar##########################
	#####primero sacar la media aritmetica#######
	u = (2nnnm)/(nn + nm) + 1
	#desviacion estandar recordando que esta al cuadrado ##########
	desviacion = ((u - 1)*(u - 2))/(nn + nm - 1)
	#####sacamos la raiz#############################
	des = math.sqrt(desviacion)
	####sacamos la probabilidad o resultado para comprobarlo con la probabilidad de Z0.005 = 1.96##
	z = (totalb - u)/des

	print "total de bloques = ",totalb,""
	print "total de 1 = ",nn,"total de 0= ",nm,""
	print u
	print desviacion

	print z
	####comparamos si s aleatorio o no 1.96 se saca en las tablas de distribucion
	if z < 1.960:
	print "Alta probabilidad que si sean random"
	else:
	print "No hay probabilidad"

	entrada.close()

view raw Comprobación hosted with ❤ by GitHub

Screen Printing

In this case is randomly, but in mayor of the case arenot randomly

Is randomly = 20%

Not randomly = 80 %

Test was performed 10 times

Sorry, here is more information, the value 1.96 is found in the critical value, it depends of the level of reliability, in this case 95 percent.

Graphics and table: