Changes of string in 30214: A better approach would be to authenticate Alice without storing her password in clear on Bob's computer. For this, Alice computes a `hash chain` as proposed by Lamport in [Lamport1981]_. A hash chain is a sequence of applications of a hash function (`H`) on an input string. If Alice's password is `P`, then her 10 steps hash chain is : :math:`H(H(H(H(H(H(H(H(H(H(P))))))))))`. The result of this hash chain will be stored on Bob's computer together with the value `10`. This number is the maximum number of remaining authentications for Alice on Bob's computer. To authenticate Alice, Bob sends the remaining number of authentications, i.e. `10` in this example. Since Alice knows her password, `P`, she can compute :math:`H^9(P)=H(H(H(H(H(H(H(H(H(P)))))))))` and send this information to Bob. Bob computes the hash of the value received from Alice (:math:`H(H^9(P))`) and verifies that this value is equal to the value stored in his database. It then decrements the number of authorized authentications and stores :math:`H^9(P)` in his database. Bob is now ready for the next authentication of Alice. When the number of authorized authentications reaches zero, the hash chain needs to be reinitialized. If Eve captures :math:`(H^n(P))`, she cannot use it to authenticate herself as Alice on Bob's computer because Bob will have decremented its number of authorized authentications. Furthermore, given that hash functions are not invertible, Eve cannot compute :math:`H^{n-1}(P)` from :math:`H^{n}(P)`. @ Weblate

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A better approach would be to authenticate Alice without storing her password in clear on Bob's computer. For this, Alice computes a `hash chain` as proposed by Lamport in [Lamport1981]_. A hash chain is a sequence of applications of a hash function (`H`) on an input string. If Alice's password is `P`, then her 10 steps hash chain is : :math:`H(H(H(H(H(H(H(H(H(H(P))))))))))`. The result of this hash chain will be stored on Bob's computer together with the value `10`. This number is the maximum number of remaining authentications for Alice on Bob's computer. To authenticate Alice, Bob sends the remaining number of authentications, i.e. `10` in this example. Since Alice knows her password, `P`, she can compute :math:`H^9(P)=H(H(H(H(H(H(H(H(H(P)))))))))` and send this information to Bob. Bob computes the hash of the value received from Alice (:math:`H(H^9(P))`) and verifies that this value is equal to the value stored in his database. It then decrements the number of authorized authentications and stores :math:`H^9(P)` in his database. Bob is now ready for the next authentication of Alice. When the number of authorized authentications reaches zero, the hash chain needs to be reinitialized. If Eve captures :math:`(H^n(P))`, she cannot use it to authenticate herself as Alice on Bob's computer because Bob will have decremented its number of authorized authentications. Furthermore, given that hash functions are not invertible, Eve cannot compute :math:`H^{n-1}(P)` from :math:`H^{n}(P)`.

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