Exercises 2013-02-28 Information Security Course Eric Laermans – Tom Dhaene.

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Exercises Information Security Course Eric Laermans – Tom Dhaene

Information Security Department of Information Technology – Internet Based Communication Networks and Services (IBCN) p. 2 Addition (large numbers) Addition given: wanted: sum classical solution (full adder):  s 0 = a 0  b 0 c 1 = a 0  b 0  s i = a i  b i  c i c i+1 = (a i  b i )  (a i  c i )  (b i  c i )  s k = c k  time complexity?  parallellisation?

Information Security Department of Information Technology – Internet Based Communication Networks and Services (IBCN) p. 3 Addition (large numbers) Addition given: wanted: sum A + B + D better solution?  carry-save-adder:  s i = a i  b i  d i c i+1 = (a i  b i )  (a i  d i )  (b i  d i )  S + C = A + B + D  time complexity?  parallellisation?  drawbacks?

Information Security Department of Information Technology – Internet Based Communication Networks and Services (IBCN) p. 4 Addition modulo N Addition mod N given:  A < N and B < N wanted: sum S = A + B mod N required adaptations to traditional solution? with three numbers ( A, B and D )  adaptation using carry-save-adders possible?  issues?

Information Security Department of Information Technology – Internet Based Communication Networks and Services (IBCN) p. 5 Multiplication given: A and B wanted: product P classical solution? improved solutions?  Karatsuba-Comba –(2 m r + s) (2 m t + u) = 2 2m rt + 2 m ((r + s)(t + u) – rt – su) + su –time complexity?

Information Security Department of Information Technology – Internet Based Communication Networks and Services (IBCN) p. 6 Multiplication modulo N Multiplication mod N given: A < N and B < N wanted: product P = A B mod N required adaptation to traditional solution? possible issues  with improved solution?  with Karatsuba-Comba?

Information Security Department of Information Technology – Internet Based Communication Networks and Services (IBCN) p. 7 Multiplication modulo N Multiplication mod N ( N odd) given: A < N and B < N wanted: product P = 2 –k A B mod N  Montgomery multiplication solution u := 0 ; for i := 0 to (k-1) do u := u + a i B ; if odd(u) then u := u+ N fi; u := u/2 ; od advantages? drawbacks? how to compute A B ? carry-save-adders possible?

Information Security Department of Information Technology – Internet Based Communication Networks and Services (IBCN) p. 8 Exponentiation modulo N Exponentiation mod N given: A < N and e <  (N) wanted: A e mod N efficient solution?