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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
85 ../../principles/security.rst:940
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The security of this protocol relies on the difficulty of computing discrete logarithms, i.e. from the knowledge of :math:`A` (resp. :math:`B`), it is very difficult to extract :math:`\log(A)=\log(g^{a} \mod p)=a` (resp. :math:`\log(B)=\log(g^{b} \mod p)=b`). 259/10000 · 259
86 ../../principles/security.rst:945
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — An example of the utilization of the Diffie-Hellman key exchange is shown below. Before starting the exchange, Alice and Bob agree on a modulus (:math:`p=23`) and a base (:math:`g=5`). These two numbers are public. They are typically part of the standard that defines the protocol that uses the key exchange. 308/10000 · 308
87 ../../principles/security.rst:951
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Alice chooses a secret integer : :math:`a=8` and sends :math:`A= g^{a} \mod p= 5^{8} \mod 23=16` to Bob 103/10000 · 103
88 ../../principles/security.rst:953
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Bob chooses a secret integer : :math:`b=13` and sends :math:`B= g^{b} \mod p=5^{13} \mod 23=21` to Alice 104/10000 · 104
89 ../../principles/security.rst:955
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Alice computes :math:`S_{A}=B^{a} \mod p= 21^{8} \mod 23=3` 59/10000 · 59
90 ../../principles/security.rst:956
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Bob computes :math:`S_{B}=A^{b} \mod p= 16^{13} \mod 23=3` 58/10000 · 58
91 ../../principles/security.rst:958
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Alice and Bob have agreed on the secret information :math:`3` without having sent it explicitly through the network. If the integers used are large enough and have good properties, then even Eve who can capture all the messages sent by Alice and Bob cannot recover the secret key that they have exchanged. There is no formal proof of the security of the algorithm, but mathematicians have tried to solve similar problems with integers during centuries without finding an efficient algorithm. As long as the integers that are used are random and large enough, the only possible attack for Eve is to test all possible integers that could have been chosen by Alice and Bob. This is computationally very expensive. This algorithm is widely used in security protocols to agree on a secret key. 788/10000 · 788
92 ../../principles/security.rst:970
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Unfortunately, the Diffie-Hellman key exchange alone cannot cope with man-in-the middle attacks. Consider Mallory who sits in the middle between Alice and Bob and can easily capture and modify their messages. The modulus and the base are public. They are thus known by Mallory as well. He could then operate as follows : 320/10000 · 320
93 ../../principles/security.rst:976
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Alice chooses a secret integer and sends :math:`A= g^{a} \mod p` to Mallory 75/10000 · 75