+1. One suggestion: can you always use "secret phrase" in the wording, instead of one time using "secret phrase" and another time using "password"? imho.

On "What happens to the static key if you lose your Yubikey?"  You are only storing one part of your NXT Passphrase in the pseudo 2 factor authentication use case described.  If lost, it cannot be used to gain access to your Nxt account without ALSO knowing the first part of the Nxt Passphrase (which user would memorize)

I realize that Lastpass signs and encrypts locally before transmitting encrypted data.  STILL, some security paranoid users may not feel comfortable with any option but LOCAL backup of private keys.

I agree the Yubikey standard looks like an interesting option. 

(emphasis mine)

Yubikey offers a Key Registration service that allows multiple keys to be remotely wiped at https://admin.yubico.com/yubirevoke/login.php

A Yubikey costs $30 and is worth far more than that to protect valuable digital assets.

Lastpass uses a third party server verification and for the most security paranoid this is not acceptable.

Wesleyh, can you code your login so that users can enter their OWN STRONG password so that the described Yubikey pseudo 2 factor authentication method can be used?

To quote Eadeqa, "Huh? I never mentioned yubikey. I think that's for 2-factor authentication. It won't even work with Nxt as Nxt is local login to NRS. "

Yubikey has a second slot for a user programmed static password.  The second slot is not involved with 2-factor authentication by server.

But newbie don't even need to know there is a wallet.dat when they first launch the client. They directly have an account and can send NXT on it.
It's an easier solution.

This good enough for a start page? Shown ONLY ONCE to the user, afterwards it's back to the secret phrase input box being shown by default.

I have not the math skills to do even this.

1600 * 10 = 16,000  

Yes, and now calculate the combinations and come back with the number

I have no idea how a yubikey works, sorry.

Even worse. It's 126-bit strong.

In the future will able to bruteforce the password (most computers are strong, they will do many calculations)

hm, I'm not sure if that is perfect...
35 random characters (let's say from 50 different characters, numbers and signs) gives 10^59 possibilities.
12 words (or characters) out of 1626 are "only" 10^38 possibilites...

There is a physical argument that a 128-bit symmetric key is computationally secure against brute-force attack. The so-called Landauer limit implied by the laws of physics sets a lower limit on the energy required to perform a computation of kT  · ln 2 per bit erased in a computation, where T is the temperature of the computing device in kelvins, k is the Boltzmann constant, and the natural logarithm of 2 is about 0.693. No irreversible computing device can use less energy than this, even in principle.[2] Thus, in order to simply flip through the possible values for a 128-bit symmetric key (ignoring doing the actual computing to check it) would theoretically require 2128 − 1 bit flips on a conventional processor. If it is assumed that the calculation occurs near room temperature (~300 K) the Von Neumann-Landauer Limit can be applied to estimate the energy required as ~1018 joules, which is equivalent to consuming 30 gigawatts of power for one year. This is equal to 30×109 W×365×24×3600 s = 9.46×1017 J or 262.7 TWh (more than 1/100th of the world energy production).[citation needed] The full actual computation – checking each key to see if you have found a solution – would consume many times this amount. Furthermore, this is simply the energy requirement for cycling through the key space; the actual time it takes to flip each bit is not considered, which is certainly greater than 0.

Here's my new logic for my client http://nxtra.org/nxt-client (to be available later today, not yet uploaded)

Start page:

- didn't you read this
when you create your account?

But if we partially save the password and the users just have to remember a Pin code. There will be no problems.

Wesleyh, I understand your client is going to be the official client that will replace NRS at nxtcrypto.org this week.
About passwords: It might be included already, but it is obvious we need something like a series of prompts to make sure new users use a long password.

Thats how I did it already. Your approach intrigues me because I don't know how to do that, I dont like that  

In this example, the number is 3,481,269,321. The table of words contains 1626 words, which are indexed from 0 to 1625. To encode this number, the ME system divides the number by the radix, 1626, which yields an integer quotient of 2,141,002 and a remainder of 69 (line 102). The ME system then uses the remainder as an index into the table and retrieves the indexed word, which is "BUS." The system sets the mnemonic encoding to that indexed word. The ME system then divides the integer quotient by the radix, 1626, which yields a new integer quotient of 1316 and a remainder of 1186 (line 103). The ME system then uses the remainder 1186 as an index into the table and retrieves the indexed word, which is "ART." The ME system then adds the word as the left-most word of the mnemonic encoding, which is now "ART BUS." The ME system then divides the integer quotient by the radix 1626, which yields a new integer quotient of 0 and a remainder of 1316 (line 104). The ME system uses the remainder 1316 as an index into the table and retrieves the indexed word, which is "DRUM." The ME system then adds the word "DRUM" as the left-most word of the mnemonic encoding, which is now "DRUM ART BUS." Since the integer quotient is zero, the encoding is complete. To decode the mnemonic encoding of "DRUM ART BUS," the ME system initializes the number that is represented by the mnemonic encoding to zero (line 105). The ME system then removes the left-most word from the encoding, which is "DRUM." The ME system then determines the index for that word in the table. Since the index of that word is 1316, the ME system sets the resultant number to the value 1316 (line 106). The ME system then again removes the left-most word of the current encoding, which is "ART." The ME system then determines the index for that word in the table, which is 1186. The ME system then sets the value of the resultant number to 2,141,002, which is the sum of the index (1186) and of the resultant number (1316) times the radix (1626) (line 107). The ME system then removes the left-most word from the encoding, which is "BUS." The ME system then determines the index of that word in the table, which is 69 (line 108). The ME system then sets the resultant number to 3,481,269,321, which is the sum of the index (69) and of the resultant number (2,141,002) times the radix (1626). Since the encoding is now empty, the current value of the resultant number represents the number for the mnemonic encoding. Each word in the mnemonic encoding corresponds to a radix position in the numbering system. For example, the mnemonic encoding "DRUM ART BUS," which has indices 1316, 1186, and 69, respectively, represents the number that is the result of 1316 * 1626.sup.2 +1186 * 1626.sup.1 +69 * 1626.sup.0.

Just out of curiosity. Could you give a brief example, I don't immediately see how to implement this.

I don't have a reference, but say you want to map a random value R between 0 and 15 to a value P between 0 and 9 and use P=(R modulo 10):


As you see, having the input value R completely random, doesn't mean that P is as random, since you will get values 0 to 5 twice as often as values 6 to 9.

I picked the example to show the problem. With the very large ranges for R (e.g. integer) and very small ranges for P (e.g. 0 to 8191), the problem might just be a very theoretical one.