Hash Functions, the MD5 Algorithm and the Future (SHA-3)
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The document discusses hash functions and the MD5 algorithm. It explains that a hash function maps inputs of arbitrary size to outputs of a fixed size, and that it is virtually impossible to derive the input given only the hash output. The document then provides a detailed overview of how the MD5 algorithm works, including converting the input to binary, padding it to a multiple of 512 bits, breaking it into 512-bit blocks, assigning initialization values, and performing 64 rounds of logical operations on each block that combines it with the output of the previous block.
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Hash Functions, the MD5 Algorithm and the Future (SHA-3)
- 1. Hash Functions, the MD5 Algorithm and the Future (SHA-3) Dylan Field, Fall ’08 SSU Math Colloquium
- 2. What is a hash?
- 4. Humpty Dumpty sat on a wall.
- 5. Humpty Dumpty had a great fall.
- 6. All the king’s horses and all the king’s men
- 7. Couldn’t put Humpty together again.
- 9. X
- 11. h(x)
- 12. BUT h(x) is a one way function
- 13. ... so they can’t put Humpty together again.
- 14. x hash function h(x) Humpty falls
- 15. ‘ hello’ MD5 x hash function h(x) Humpty falls
- 16. 5d41402abc4b ‘ hello’ MD5 2a76b9719d91 1017c592 x hash function h(x) Humpty falls
- 17. - going backwards - - sdrawkcab gniog -
- 18. - going backwards - - sdrawkcab gniog -
- 19. - going backwards - NO!!! - sdrawkcab gniog -
- 20. - going backwards - 5d41402abc4b 2a76b9719d91 1017c592 - sdrawkcab gniog -
- 21. - going backwards - 5d41402abc4b 2a76b9719d91 1017c592 ‘ hello’ - sdrawkcab gniog -
- 22. Requirements h(x)
- 23. Requirements h(x) Given h(x) cannot find x 1
- 24. Requirements h(x) Given h(x) h(x) is cannot find x constant 1 2
- 25. Requirements h(x) Given h(x) h(x) is Can’t find x2 cannot find x constant so h(x2)=h(x1) 1 2 3
- 26. Requirement #3 - Humpty Dumpty Style
- 27. Requirement #3 - Humpty Dumpty Style ≠
- 28. Requirement #3 - Humpty Dumpty Style ≠ ≠ ≠ ≠ .........
- 29. so how does it work?
- 30. ‘ hello’
- 32. we’re going to focus on MD5
- 34. 1. Convert ‘x’ to binary
- 35. ‘ hello’ 0110100001100101011011000110110001101111
- 36. 1. Convert ‘x’ to binary 2. Pad ‘x’ so that size of x (mod 512) = 0
- 37. ‘hello’ in binary 0110100001100101011011000110110001101111 1 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00000 0000000000101000
- 38. ‘hello’ in binary 0110100001100101011011000110110001101111 1 add ‘1’ 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00000 0000000000101000
- 39. ‘hello’ in binary 0110100001100101011011000110110001101111 1 add ‘1’ 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0’s until 0000000000 0000000000 0000000000 0000000000 0000000000 x mod 512 = 496 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00000 0000000000101000
- 40. ‘hello’ in binary 0110100001100101011011000110110001101111 1 add ‘1’ 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0’s until 0000000000 0000000000 0000000000 0000000000 0000000000 x mod 512 = 496 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00000 add 16 bit binary 0000000000101000 representation of x
- 41. xpadded = 0110100001100101011011000110110001101111 1 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00000 0000000000101000
- 42. 1. Convert ‘x’ to binary 2. Pad ‘x’ so that size of x (mod 512) = 0 3. Break ‘x’ into 512 bit sub parts and 32 bit words
- 43. 0110100001100101011011000110110001101111 1 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00000 0000000000101000 W1 = 01101000011001010110110001101100
- 44. 1. Convert ‘x’ to binary 2. Pad ‘x’ so that size of x (mod 512) = 0 3. Break ‘x’ into 512 bit sub parts and 32 bit words 4. Assign values to k[i], r[i], w[g], h0, h1, h2 and h3.
- 45. k[i] = |sin(i+1)| x 232 where ‘i’ is in radians
- 46. k[i] = |sin(i+1)| x 232 where ‘i’ is in radians r[i] = Various round shift amounts
- 47. k[i] = |sin(i+1)| x 232 where ‘i’ is in radians r[i] = Various round shift amounts w[g] = Word number (0 – 15)
- 48. k[i] = |sin(i+1)| x 232 where ‘i’ is in radians r[i] = Various round shift amounts w[g] = Word number (0 – 15) h0 = a = 0x67452301 h1 = b = 0xEFCDAB89 h2 = c = 0x98BADCFE h3 = d = 0x10325476
- 49. 1. Convert ‘x’ to binary 2. Pad ‘x’ so that size of x (mod 512) = 0 3. Break ‘x’ into 512 bit sub parts and 32 bit words 4. Assign values to k[i], r[i], w[g], h0, h1, h2 and h3. 5. Perform 64 rounds on each sub part
- 52. But first... binary operations!
- 53. ∧
- 54. ∧ (AKA ‘AND’)
- 55. p q ∧ T T
- 56. p q ∧ T T T
- 57. p q ∧ T T T T F
- 58. p q ∧ T T T T F F
- 59. p q ∧ T T T T F F F T
- 60. p q ∧ T T T T F F F T F
- 61. p q ∧ T T T T F F F T F F F
- 62. p q ∧ T T T T F F F T F F F F
- 63. In binary: T=1 F=0
- 64. p q ∧ T T T T F F F T F F F F
- 65. p q ∧ bit 1 bit 2 ∧ T T T 1 1 1 T F F 1 0 0 F T F 0 1 0 F F F 0 0 0
- 66. ∨
- 67. ⊕
- 68. bit 1 bit 2 ∨ 1 1 1 1 0 1 0 1 1 0 0 0
- 69. ⊕ “XOR is a type of logical disjunction on two operands that results in a value of “true” if and only if exactly one of the operands has a value of ‘true’”
- 70. bit 1 bit 2 ∨ bit 1 bit 2 ⊕ 1 1 1 1 1 F 1 0 1 1 0 T 0 1 1 0 1 T 0 0 0 0 0 F
- 71. ¬
- 72. ¬ (not)
- 73. ¬1=0 ¬0=1
- 74. << (bit shift)
- 75. 1 0 1 0 1 0
- 76. 0 1 0 1 0 0 1 0 1 0 0 0
- 79. Operation A f = (b ∧ c) ∨ (¬ b ∧ d) g=i
- 80. Operation B f = (d ∧ b) ∨ ((¬ d) ∧ c) g = (5i + 1) mod 16
- 81. Operation C f=b⊕c⊕d g = (3i + 5) mod 16
- 82. Operation D f = c ⊕ (b ∨ (¬ d)) g = (7i) mod 16
- 84. A B C D
- 85. A B C D
- 86. A B C D
- 87. B b + {(a + f + k[i] + w[g]) << r[i]}
- 88. b + {(a + f + k[i] + w[g]) << r[i]} h1 h0 Calculated in The gth word Operations A-D (32 bit chunk) |sin(i+1)| x 232 ith pre-designated where ‘i’ is in radians shift
- 89. After all 64 rounds...
- 90. 1. Convert ‘x’ to binary 2. Pad ‘x’ so that size of x (mod 512) = 0 3. Break ‘x’ into 512 bit sub parts and 32 bit words 4. Assign values to k[i], r[i], w[g], h0, h1, h2 and h3. 5. Perform 64 rounds on each sub part 6. Add a, b, c and d to register values
- 91. h0 = h0 + a h1 = h1 + b h2 = h2 + c h3 = h3 + d
- 92. 1. Convert ‘x’ to binary 2. Pad ‘x’ so that size of x (mod 512) = 0 3. Break ‘x’ into 512 bit sub parts and 32 bit words 4. Assign values to k[i], r[i], w[g], h0, h1, h2 and h3. 5. Perform 64 rounds on each sub part 6. Add a, b, c and d to register values 7. Append the register values to create digest
- 93. 128 bit digest
- 94. ‘ hello’
- 96. So?
- 97. Applications
- 99. Message Integrity Applications Password Protection
- 100. Message Integrity Applications Digital Password Signatures Protection
- 101. Password Protection
- 103. When you registered... MD5 ‘password’ 5f4dcc3b5aa765d61d8327deb882cf99
- 104. When you registered... MD5 ‘password’ 5f4dcc3b5aa765d61d8327deb882cf99 Data Base
- 106. ‘password’
- 107. MD5 ‘password’
- 108. MD5 ‘password’ 5f4dcc3b5aa765d61d8327deb882cf99
- 109. 5f4dcc3b5aa765d61d8327deb882cf99 = stored, hashed password?
- 110. 5f4dcc3b5aa765d61d8327deb882cf99 = stored, hashed password? No. Give ‘incorrect password’ error
- 111. 5f4dcc3b5aa765d61d8327deb882cf99 = stored, hashed password? No. Yes. Give ‘incorrect Let user password’ error into website
- 114. Attacks
- 115. Rainbow Tables
- 117. omgyouarenever 1c9fee8bd70a5afb6 goingtocrackthis 30fc4f38e97123f 123
- 118. omgyouarenever 1c9fee8bd70a5afb6 goingtocrackthis 30fc4f38e97123f 123
- 119. and Brute Force Attacks
- 120. Message Integrity
- 122. digest
- 123. File Verification
- 124. File Verification Guarding against corruption
- 125. File Verification Guarding against corruption Proving you have something before you release it
- 126. Attacks
- 127. Nostradamus Attack
- 129. But on November 30th 2007...
- 130. “We have used a Sony Playstation 3 to correctly predict the outcome of the 2008 US presidential elections. In order not to influence the voters we keep our prediction secret, but commit to it by publishing its cryptographic hash on this website. The document with the correct prediction and matching hash will be revealed after the elections.” - Marc Stevens, Arjen Lenstra and Benne de Weger
- 132. But how could they have known!?!?
- 133. But how could they have known!?!? They didn’t.
- 135. Digital Signatures
- 137. MD5 hash
- 138. MD5 hash private key encrypted
- 139. MD5 hash private key hash encrypted public key
- 140. MD5 hash private MD5 key hash encrypted public key
- 141. MD5 hash private MD5 key hash ✔ encrypted public key
- 142. Attacks
- 143. Collision Attack
- 144. hash private MD5 key hash ✔ encrypted public key
- 145. Changed hash Message MD5 hash ✔ encrypted public key
- 146. Very Dangerous!
- 147. Birthday Attack
- 148. Relies on ‘Birthday Paradox’
- 149. Relies on ‘Birthday Paradox’ First we calculate the chance no one has the same birthday
- 150. p(1)=100%
- 151. p(2)=(1)(1 - 1/365)
- 152. p(3)=(1)(1 - 1/365)(1 - 2/365)
- 153. To Generalize...
- 154. P(n)= 365! . 365 n(365-n)!
- 155. 23 50% chance
- 156. 30 70.6% chance
- 157. 50 97% chance
- 158. We can use this property to find out how many hashes must be calculated to find a collision.
- 159. Current State of MD5
- 160. MD5 =
- 161. MD5 = Broken
- 162. The Future of Hashes
- 164. Submissions were due on October 30th
- 165. Currently Submitted
- 166. Skein Maraca BLAKE MD6 Keccak CubeHash Edon-R Ponic EnRUPT SHAMATA MCSSHA-3 Sgàil Blue Midnight Wish Grøstl ESSENCE WaMM Boole NaSHA NKS2D Waterfall
- 167. Skein BLAKE MD6 Maraca Keccak CubeHash Edon-R Ponic EnRUPT SHAMATA MCSSHA-3 Sgàil Blue Midnight Wish Grøstl ESSENCE WaMM Boole NaSHA NKS2D Waterfall
- 169. Thank you for coming!
- 170. Any Questions?