Terrible advice! Do not do this! It will vastly decrease the security of your wallet!

They do not have the same security profile. Your method reveals half the information of your seed, the OP's method does not. In fact it can be shown that seed A and seed B as described by OP are both random numbers and reveal absolutely no information at all.

To see this consider a simpler case where we have a three index dictionary {0, 1, 2} and our seed is only one number.

Let S be our seed chosen randomly, then:

S = 0 with 1/3 probability; S = 1 with 1/3 probability and S = 2 with 1/3 probability.

We chose A the same way:

A = 0 with 1/3 probability; A = 1 with 1/3 probability and A = 2 with 1/3 probability.

Note that since A is chosen randomly it reveals zero information about S.

B is calculated from S and A as per the scheme: B = (S - A) % 3

So consider the three cases for S:

Case: S = 0 then:

B = (S - A) % 3  = (0 - 0) % 3 = 0 with 1/3 probability (A = 0 1/3 of the time)
B = (S - A) % 3  = (0 - 1) % 3 = 2 with 1/3 probability (A = 1 1/3 of the time)
B = (S - A) % 3  = (0 - 2) % 3 = 1 with 1/3 probability (A = 2 1/3 of the time)

Case: S = 1 then:

B = (S - A) % 3  = (1 - 0) % 3 = 1 with 1/3 probability
B = (S - A) % 3  = (1 - 1) % 3 = 0 with 1/3 probability
B = (S - A) % 3  = (1 - 2) % 3 = 2 with 1/3 probability

Case: S = 2 then:

B = (S - A) % 3  = (2 - 0) % 3 = 2 with 1/3 probability
B = (S - A) % 3  = (2 - 1) % 3 = 1 with 1/3 probability
B = (S - A) % 3  = (2 - 2) % 3 = 0 with 1/3 probability


So we see no matter the value for S; B = 0 with 1/3 probability; B = 1 with 1/3 probability and B = 2 with 1/3 probability. Thus with no knowledge of A, B is indistinguishable from a random number and like A reveals zero information about S.


Your method of dividing the secret in two halves does reveal information about the seed. Consider a similar example of a four digit pin:

If I want to brute force crack the pin I have to try all combinations which is 10*10*10*10 = 10^4 = 10,000 tries. However if I discover half, I only need to crack the remaining two digits, thus I need only 10^2 = 100 tries. A factor of 100 speed up from the brute force method!