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65 ../../principles/security.rst:775
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Now consider that Mallory wants to be authenticated as Alice. The above protocol has a subtle flaw that could be exploited by Mallory. This flaw can be exploited if Alice and Bob can act as both client and server. Knowing this, Mallory could operate as follows. Mallory starts an authentication with Bob faking himself as Alice. He sends a first message to Bob including Alice's identity. 388/10000 · 388
66 ../../principles/security.rst:799
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — In this exchange, Bob authenticates himself by signing the :math:`RA` nonce that was sent by Mallory. Now, to authenticate as Alice, Mallory needs to compute the signature of nonce :math:`RB` with Alice's private key. Mallory does not know Alice's key, but he could exploit the protocol to force Alice to perform the required computation. For this, Mallory can start an authentication to Alice as shown below. 409/10000 · 409
67 ../../principles/security.rst:822
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — In this example, Mallory has forced Alice to compute :math:`E_p(Alice_{priv},RB)` which is the information required to finalize the first exchange and be authenticated as Alice. This illustrates a common problem with authentication schemes when the same information can be used for different purposes. The problem comes from the fact that Alice agrees to compute her signature on a nonce chosen by Bob (and relayed by Mallory). This problem occurs if the nonce is a simple integer without any structure. If the nonce includes some structure such as some information about Alice and Bob's identities or even a single bit indicating whether the nonce was chosen by a user acting as a client (i.e. starting the authentication) or as a server, then the protocol is not vulnerable anymore. 784/10000 · 784
68 ../../principles/security.rst:837
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — To cope with some of the above mentioned problems, public-key cryptography is usually combined with certificates. A `certificate` is a data structure that includes a signature from a trusted third party. A simple explanation of the utilization of certificates is to consider that Alice and Bob both know Ted. Ted is trusted by these two users and both have stored Ted's public key : :math:`Ted_{pub}`. Since they both know Ted's key, he can issue certificates. A certificate is mainly a cryptographic link between the identity of a user and his/her public key. Such a certificate can be computed in different ways. A simple solution is for Ted to generate a file that contains the following information for each certified user : 728/10000 · 728
69 ../../principles/security.rst:849
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — his/her identity 16/10000 · 16
70 ../../principles/security.rst:850
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — his/her public key 18/10000 · 18
71 ../../principles/security.rst:851
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — a hash of the entire file signed with Ted's private key 55/10000 · 55
72 ../../principles/security.rst:853
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Then, knowing Ted's public key, anyone can verify the validity of a certificate. When a user sends his/her public key, he/she must also attach the certificate to prove the link between his/her identity and the public key. In practice, certificates are more complex than this. Certificates will often be used to authenticate the server and sometimes to authenticate the client. 376/10000 · 376
73 ../../principles/security.rst:860
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — A possible protocol could then be the following. Alice sends :math:`Cert(Alice_{pub},Ted)`. Bob replies with a random nonce. 124/10000 · 124
74 ../../principles/security.rst:885
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Until now, we have only discussed the authentication problem. This is an important but not sufficient step to have a secure communication between two users through an insecure network. To securely exchange information, Alice and Bob need to both : 247/10000 · 247
75 ../../principles/security.rst:890
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — mutually authenticate each other 32/10000 · 32
76 ../../principles/security.rst:891
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — agree on a way to encrypt the messages that they will exchange 62/10000 · 62
77 ../../principles/security.rst:893
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Let us first explore how this could be realized by using public-key cryptography. We assume that Alice and Bob have both a public-private key pair and the corresponding certificates signed by a trusted third party : Ted. 220/10000 · 220
78 ../../principles/security.rst:898
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — A possible protocol would be the following. Alice sends :math:`Cert(Alice_{pub},Ted)`. This certificate provides Alice's identity and her public key. Bob replies with the certificate containing his own public key : :math:`Cert(Bob_{pub},Ted)`. At this point, they both know the other public key and could use it to send encrypted messages. Alice would send :math:`E_p(Bob_{pub},M1)` and Bob would send :math:`E_p(Alice_{pub},M2)`. In practice, using public key encryption techniques to encrypt a large number of messages is inefficient because these cryptosystems require a large number of computations. It is more efficient to use secret key cryptosystems for most of the data and only use a public key cryptosystem to encrypt the random secret keys that will be used by the secret key encryption scheme. 805/10000 · 805
79 ../../principles/security.rst:913
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Key exchange 12/10000 · 12
80 ../../principles/security.rst:920
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — When users want to communicate securely through a network, they need to exchange information such as the keys that will be used by an encryption algorithm even in the presence of an eavesdropper. The most widely used algorithm that allows two users to safely exchange an integer in the presence of an eavesdropper is the one proposed by Diffie and Hellman [DH1976]_. It operates with (large) integers. Two of them are public, the modulus, p, which is prime and the base, g, which must be a primitive root of p. The communicating users select a random integer, :math:`a` for Alice and :math:`b` for Bob. The exchange starts as : 627/10000 · 627
81 ../../principles/security.rst:931
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Alice selects a random integer, :math:`a` and sends :math:`A=g^{a} \mod p` to Bob 81/10000 · 81
82 ../../principles/security.rst:933
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Bob selects a random integer, :math:`b` and sends :math:`B=g^{b} \mod p` to Alice 81/10000 · 81
83 ../../principles/security.rst:935
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — From her knowledge of :math:`a` and :math:`B`, Alice can compute :math:`Secret=B^{a} \mod p= (g^{b} \mod p) ^{a} \mod p=g^{a \times b} \mod p` 142/10000 · 142
84 ../../principles/security.rst:937
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — From is knowledge of :math:`b` and :math:`A`, Bob can compute :math:`Secret=A^{b} \mod p=(g^{a} \mod p) ^{b} \mod p=g^{a \times b} \mod p` 138/10000 · 138