Lecture on 18 December 2018 Role of Cryptography in Blockchain RSA and SHA Blockchain for Beginners Elective course from the Faculty of Information Technology, Thai - Nichi Institute of Technology, Bangkok for undergraduate students. #BlockchainTNI2018

1. Blockchain Technology
Week 6
Unit III – Role of Cryptography in Blockchain
Ferdin Joe John Joseph, PhD
Faculty of Information Technology
Thai-Nichi Institute of Technology, Bangkok
Venue: D603

2. Week 6 – Unit III
Agenda
• Role of Cryptography in Blockchain
• RSA and SHA
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3. Blockchain’s Encryption
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4. Faculty of Information Technology, Thai-Nichi Institute of
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5. Faculty of Information Technology, Thai-Nichi Institute of
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6. Faculty of Information Technology, Thai-Nichi Institute of
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7. Bitcoin Hash
• The corresponding SHA-256 of the sentence “How to buy
Bitcoin?” looks like this:
156aedcfab1d49f73abddd89faf78d9930e4b523ab804026310c973bf
a707d37
• If we remove only one symbol – for example the question mark “?” –
the hash of “How to buy Bitcoin” looks like
this: 4314d903f04e90e4a5057685243c903fbcfa4f8ec75ec797e1780e
d5c891b1bf
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8. What is hashing
• Hashing means taking an input string of any length and giving out an
output of a fixed length.
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9. Properties of Hashing
• Deterministic
• Quick Computation
• Pre-image resistance
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10. Properties of hashing - Deterministic
• No matter how many times you parse a particular input through a
hash function you will always get the same result.
• This is critical because if you get different hashes every single time it
will be impossible to keep track of the input.
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11. Properties of hashing – Quick Computation
• The hash function should be capable of returning the hash of an input
quickly.
• If the process isn’t fast enough then the system simply won’t be
efficient.
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12. Properties of hashing – Pre-image resistance
• It is infeasible to determine A, where A is the input and H(A) is the
output hash.
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13. Faculty of Information Technology, Thai-Nichi Institute of
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14. The RSA Algorithm
• Based on the idea that factorization of integers into their prime
factors is hard.
★ n=p．q, where p and q are distinct primes
• Proposed by Rivest, Shamir, and Adleman
in 1977 and a paper was published in The Communications of ACM in
1978
• A public-key cryptosystem

15. RSA Algorithm
• Bob chooses two primes p,q and compute n=pq
• Bob chooses e with gcd(e,(p-1)(q-1))=
gcd(e, ψ(n))=1
• Bob solves de≡1 (mod ψ(n))
• Bob makes (e,n) public and (p,q,d) secret
• Alice encrypts M as C≡Me (mod n)
• Bob decrypts by computing M≡Cd (mod n)

16. Proof for the RSA Algorithm
• Cd ≡ (Me)d ≡ Med ≡ M1+kφ(n) ≡M (mod n) by Euler’s theorem and
Exercise 19 on p.192
• p=885320963, q=238855417,
• n=p．q=211463707796206571
• Let e=9007, ∴ d=116402471153538991
• M=“cat”=30120, C=113535859035722866

17. Another Example
• n=127x193=24511, φ(n)=24192
• e=1307, d=10643
• Encrypt “box” with M=21524, then
C=?
Encrypt the following message
Formosa means a beautiful island

18. Selected Problems from P.192-200
(1) n=11413=101x113, so p=101, q=113
ψ(n)=(p-1)x(q-1)=100x112=11200
Choose e=7467, then gcd(e, ψ(n))=1
Solve de≡1 (mod ψ(n)) to get d=3
If the ciphertext C=5859, then the plaintext
M≡Cd ≡58593 ≡1415 (mod 11413)

19. Fast Computation of xd (mod n)
• 1235 mod 511
• 1235 ≡ 28153056843 mod 511
• 1232 ≡ 310 (mod 511)
• 1234 ≡ 32 (mod 511)
• 1235 ≡ 123101b ≡1234 ×123
≡ 359 (mod 511)

20. Fast Computation for xd (mod n)
y=1;
while (d != 0) {
if ((d%2) != 0) { y=(y*x)%n; d--; }
d>>1;
x=(x*x)%n; /* x^(2k) */
}

21. Fast Computation for xd (mod n)
Let t be the number of bits for integer d, e.g.,
If d=5=1012 , then t=3
y=1;
for (i=t; i≧0; i--) {
y=(y*y)%n;
if (d[i]==1) y=(y*x)%n;
}

22. The Concept and Criteria
1. Ek(Dk(m))=m and Dk(Ek(m))=m for every message m in M, the set of
possible messages, every key k in K, the set of possible keys
2. For every m and every k, then values of Ek(m) and Dk(m) are easy to
compute
3. For every k, if someone knows only the function Ek, it is
computationally infeasible to find an algorithm to compute Dk
4. Given k, it’s easy to find the functions Ek and Dk

23. Secure Hash Algorithm (SHA)
• SHA-0 1993
• SHA-1 1995
• SHA-2 2002
• SHA-224, SHA-256, SHA-384, SHA-512
SHA-1
A message
composed of
b bits
160-bit
message
digest
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Algorithm
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24. Step 1 -- Padding
• Padding  the total length of a padded message is multiple of 512
• Every message is padded even if its length is already a multiple of 512
• Padding is done by appending to the input
• A single bit, 1
• Enough additional bits, all 0, to make the final 512 block exactly 448 bits long
• A 64-bit integer representing the length of the original message in bits
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25. Padding (cont.)
Message Message length1 0…0
64 bits
Multiple of 512
1 bit
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Algorithm
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26. Example
• M = 01100010 11001010 1001 (20 bits)
• Padding is done by appending to the input
• A single bit, 1
• 427 0s
• A 64-bit integer representing 20
• Pad(M) = 01100010 11001010 10011000 … 00010100

27. Example
• Length of M = 500 bits
• Padding is done by appending to the input:
• A single bit, 1
• 459 0s
• A 64-bit integer representing 500
• Length of Pad(M) = 1024 bits

28. Step 2 -- Dividing Pad(M)
• Pad (M) = B1, B2, B3, …, Bn
• Each Bi denote a 512-bit block
• Each Bi is divided into 16 32-bit words
• W0, W1, …, W15
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29. Step 3 – Compute W16 – W79
• To Compute word Wj (16<=j<=79)
• Wj-3, Wj-8, Wj-14 , Wj-16 are XORed
• The result is circularly left shifted one bit
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30. Step 4 – Initialize A,B,C,D,E
• A = H0
• B = H1
• C = H2
• D = H3
• E = H4
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31. Initialize 32-bit words
• H0 = 67452301
• H1 = EFCDAB89
• H2 = 98BADCFE
• H3 = 10325476
• H4 = C3D2E1F0
• K0 – K19 = 5A827999
• K20 – K39 = 6ED9EBA1
• K40 – K49 = 8F1BBCDC
• K60 – K79 = CA62C1D6
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Algorithm
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32. Step 5 – Loop
For j = 0 … 79
TEMP = CircLeShift_5 (A) + fj(B,C,D) + E + Wj + Kj
E = D; D = C;
C = CircLeShift_30(B);
B = A; A = TEMP
Done
+  addition (ignore overflow)
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Algorithm
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33. Four functions
• For j = 0 … 19
• fj(B,C,D) = (B AND C) OR ( B AND D) OR (C AND D)
• For j = 20 … 39
• fj(B,C,D) = (B XOR C XOR D)
• For j = 40 … 59
• fj(B,C,D) = (B AND C) OR ((NOT B) AND D)
• For j = 60 … 79
• fj(B,C,D) = (B XOR C XOR D)
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Algorithm
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34. Step 6 – Final
• H0 = H0 + A
• H1 = H1 + B
• H2 = H2 + C
• H3 = H3 + D
• H4 = H4 + E
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Algorithm
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35. Done
• Once these steps have been performed on each 512-bit block (B1, B2,
…, Bn) of the padded message,
• the 160-bit message digest is given by
H0 H1 H2 H3 H4
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Algorithm
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36. SHA
Output
size
(bits)
Internal
state size
(bits)
Block
size
(bits)
Max
message
size (bits)
Word
size
(bits)
Rounds Operations
Collisions
found
SHA-0 160 160 512 264 − 1 32 80
+, and, or,
xor, rot
Yes
SHA-1 160 160 512 264 − 1 32 80
+, and, or,
xor, rot
None
(252 attack)
SHA-2
256/224 256 512 264 − 1 32 64
+, and, or,
xor, shr, rot
None
512/384 512 1024 2128 − 1 64 80
+, and, or,
xor, shr, rot
None
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Algorithm
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37. Next Week
• Fraud and Intrusion Detection
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Faculty of Information Technology, Thai-Nichi Institute of
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