Enpass Discussion Forum

# Backdoor in password generator ?

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Hi !

Enpass looks nice. It's full of features. Of course, it makes me willing to use it.

As I noticed for some  software companies that spending time in developing features is a natural marketing constraint to catch more customers, it pushes security concerns to second priority.

Then, I'd like to make my mind about

and try to assess how good behaves your password generator ( that is one of the secure component that everybody can understand )

A simple test: measure the occurence of consecutive characters.

with a simple setup (only numbers) : probability of having 2 consecutive number is below 1% for length=10

but statistics (I have tried ~50 iterations ) shows 50% !

Having a long password with funny characters will not help if the sequence is generated by an algorithm that limits the statistics...

More than a bug in the randomizer, it seems to me a backdoor algorithm runs behind.

Make me change my mind please

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Is there new information or a solution to the problem?

Edited by Area
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Hi @Sam,

Welcome to Enpass.

Do you mean to say "Measure of occurrence of consecutive number pairs in a random password of length 10, using only digits"?

Regards,
Vinod

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Hi Vinod,

the problem is : the password generator does not behave randomly, at all

the question is  : Can you explain why ?

the matter of "occurrence of consecutive number pairs in a random password of length 10, using only digits" is just an example of the problem observation.

I reported this example beause the probability is easy to calculate and statistitictsis is easy to measure.

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Hi @Sam,

Let me have this opportunity to assure you that there is such no backdoor in the algorithm. There is clearly some misunderstanding here.

Let us find out the probability of having at least two consecutive digits in a number of various length using a computer program

P(length 2) = 10 / 100 = 0.1
P(length 3) = 190 / 1000 = 0.19
P(length 4) = 2710 / 10000 ≈ 0.27
P(length 5) = 34390 / 100000 ≈ 0.34
P(length 6) = 409510 / 1000000 ≈ 0.40
P(length 7) = 4685590 / 10000000 ≈ 0.46

So, as length increases probability increases. and is approximately

```P ≈ (∑k - ∑(k-n+1))/k^2 where k >= n

k is sample space of single digit

In our case k=10 and n=10
P(length 10) = (55 - 1)/100 = 0.54

So, Probability of occurrence of at least two consecutive digits in a random password of length 10, using only digits is 0.54 i.e. 54%.
And it is very close to your statistical calculation.

Cheers:)

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Hi Vinod

Thanks for demonstration that show my mistake and exonerate the generator .

thanks 