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Highlights of Computing: From theAncient Indian Vedic Literature to
Modern Practical Cryptography
Uwe Wystup
February 2010
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1. Numeral Systems
2. Vedic Mathematics
3. Zero Search Methods
4. Credit Cards
5. Cryptography
6. Denitions
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1. Numeral Systems
Fundamentals: Any integer has a unique represen-tation as a weighted sum of powers of a chosen base.
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1. Numeral Systems
Fundamentals: Any integer has a unique represen-tation as a weighted sum of powers of a chosen base.
Example: 245 = 2102 + 4 101 + 5 100
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1. Numeral Systems
Fundamentals: Any integer has a unique represen-tation as a weighted sum of powers of a chosen base.
Example: 245 = 2102 + 4 101 + 5 100
The Base 10 is in no way special except
that
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1. Numeral Systems
Fundamentals: Any integer has a unique represen-tation as a weighted sum of powers of a chosen base.
Example: 245 = 2102 + 4 101 + 5 100
The Base 10 is in no way special except
that
The Mayas also used the feet (base=20), seehttp://en.wikipedia.org/wiki/Maya_numerals .
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1. Numeral Systems
Fundamentals: Any integer has a unique represen-tation as a weighted sum of powers of a chosen base.
Example: 245 = 2102 + 4 101 + 5 100
The Base 10 is in no way special except
that
The Mayas also used the feet (base=20), seehttp://en.wikipedia.org/wiki/Maya_numerals .
Computers use a dual system (power off or on, base 2) or a hexadecimal system(0, 1, 2, . . . , 9,A,B,C,D,E,F ) for memory size:1 byte = F F .
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1. Numeral Systems
Fundamentals: Any integer has a unique represen-tation as a weighted sum of powers of a chosen base.
Example: 245 = 2102 + 4 101 + 5 100
The Base 10 is in no way special except
that
The Mayas also used the feet (base=20), seehttp://en.wikipedia.org/wiki/Maya_numerals .
Computers use a dual system (power off or on, base 2) or a hexadecimal system(0, 1, 2, . . . , 9,A,B,C,D,E,F ) for memory size:1 byte = F F .
More on http://en.wikipedia.org/wiki/Numeral_system
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1.1. ExerciseWrite the hexadecimal numberCAFE in the dual, dec-imal and Maya number system.
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2. Vedic Mathematics
1. Introduction2. Multiplication
3. Squares
4. Division
5. Divisibility
6. Square Roots
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2.1. Introductionveda (Sanskrit) means: knowledge
Veda Upaveda
Rigveda AyurvedaSamaveda Gandharvaveda
Yajurveda Dhanurveda
Atharvaveda Sthapatyaveda
Table 1: Vedas and Upavedas (supplementary vedas)
2 1 1 Th 16 S tras
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2.1.1. The 16 Sutras
are part of a Parisista (Appendix) of theAtharvaveda 1. By one more than the one before
2. All from 9 and the last from 103. Vertically and crosswise
4. Transpose and apply
5. If the Samuccaya is the same it is zero
6. If one is in ratio the other is zero
7. By addition and by subtraction
8. By the completion or non-completion
9. Differential calculus
10. By the deciency
11. Specic and general
12. The remainders by the last digit
13. The ultimate and twice the penultimate
14. By one less than the one before
15. The product of the sum
16. All the multipliers
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2.1.2. Jagadguru Swami Sri Bharati Krsna TirthajiMaharaja
Explained the sutras in his books.Jagadguru Swami SriBharati Krsna TirthajiMaharaja (March, 1884 -February 2, 1960) was theJagadguru (literally, teacherof the world; assigned toheads of Hindu mathas)of the Govardhana mathaof Puri during 1925-1960.
He was one of the mostsignicant spiritual guresin Hinduism during the 20thcentury. He is particularlyknown for his work on Vedic
mathematics.
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2.2. Multiplication
Example with working base 10:
9 - 1
7 - 36 / 3
= 63
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2.2. Multiplication
Example with working base 10:
9 - 1
7 - 36 / 3
= 63
7 - 3
6 - 43 / 1 2
= 42
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2.2. Multiplication
Example with working base 10:
9 - 1
7 - 36 / 3
= 63
7 - 3
6 - 43 / 1 2
= 42
13 + 3
12 + 215 / 6
= 156
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2.2. Multiplication
Example with working base 10:
9 - 1
7 - 36 / 3
= 63
7 - 3
6 - 43 / 1 2
= 42
13 + 3
12 + 215 / 6
= 156
12 + 2
8 - 210 / 4
= 96
Reason: (x + a)(x + b) = x(x + a + b) + ab
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2.2. Multiplication
Example with working base 10:
9 - 1
7 - 36 / 3
= 63
7 - 3
6 - 43 / 1 2
= 42
13 + 3
12 + 215 / 6
= 156
12 + 2
8 - 210 / 4
= 96
Reason: (x + a)(x + b) = x(x + a + b) + ab
Origin of the-sign comes from this method
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Example with working base 100:
91 - 9
96 - 4
87 / 36
= 8736
111 + 11
109 + 9
120 / 99
= 12099
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Example with working base 100:
91 - 9
96 - 4
87 / 36
= 8736
111 + 11
109 + 9
120 / 99
= 12099
108 + 8
97 - 3
105 / 24
= 10476
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Other working bases (division case)
100/2=50
49 - 1
49 - 1
2)48 / 01
24 / 01= 2401
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Other working bases (division case)
100/2=50
49 - 1
49 - 1
2)48 / 01
24 / 01= 2401
100/2=50
54 + 4
46 - 4
2)50 / 16
25 / 16= 2484
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Other working bases (multiplication case)
102=2019 - 1
19 - 1
2)18 / 136 / 1
= 361
106=6062 + 2
48 - 12
6)50 / 2 4300 / 2 4
= 2976
2.2.1. Exercises
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Multiply the following
a 9494b 9789c 8799d 8798e 87
95
f 9595g 7996h 9896i 9299 j 8888k 9756l 9763m 92196
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Multiply the following mentally
a 667998b 768997c 989998d 885997e 883
998
f 86g 891989h 88889996i 69999997 j 9090999994k 7898999997l 98769989
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Multiply the following mentally
a 133103b 107108c 171101d 102104e 132
102
f 1412g 1813h 12221003i 10511007 j 1511110003k 125105l 1060710008
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2.3. Squares
Using the sutra Ekadhikena Purvena (by one morethan the previous one) we get
152 = 1 2/ 25 = 2/ 25 = 225252 = 2 3/ 25 = 6/ 25 = 625352 = 3 4/ 25 = 12/ 25 = 1225...
1152 = 11 12/ 25 = 132/ 25 = 13225
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2.3.1. ExercisesMultiply the following mentally
a 652
b 852
c 0.52
d 7.52
e 0.02252
f 10502
g 1752
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2.4. Division
Lots of tricks are available. We do only some high-lights.
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2.4. Division
Lots of tricks are available. We do only some high-lights. Find the exact decimal representation of
119 .
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2.4. Division
Lots of tricks are available. We do only some high-lights. Find the exact decimal representation of
119 .
Standard methods are cumbersome.
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2.4. Division
Lots of tricks are available. We do only some high-lights. Find the exact decimal representation of
119 .
Standard methods are cumbersome. Using the Ekadhika Purva Sutra it is easy: Start with 1 and then work from right to left mul-tiplying by 2.
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2.4. Division
Lots of tricks are available. We do only some high-lights. Find the exact decimal representation of
119 .
Standard methods are cumbersome. Using the Ekadhika Purva Sutra it is easy: Start with 1 and then work from right to left mul-tiplying by 2.
. 0 5 2 6 3 1 5 7 8
1 1 1 1 1 1
/ 9 4 7 3 6 8 4 2 1
1 1 1
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2.4.1. Exercises
Compute the exact decimal number of a 129b 149
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2.5. Divisibility
Use Ekadhika as an osculator.
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2.5. Divisibility
Use Ekadhika as an osculator. For 9, 19, 29, 39 etc. the Ekadhikas are 1, 2, 3, 4,etc.
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2.5. Divisibility
Use Ekadhika as an osculator. For 9, 19, 29, 39 etc. the Ekadhikas are 1, 2, 3, 4,etc. For 3, 13, 23, 33 etc. multiply them by 3 and youget 1, 4, 7, 10, etc. as the Ekadhikas.
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2.5. Divisibility
Use Ekadhika as an osculator. For 9, 19, 29, 39 etc. the Ekadhikas are 1, 2, 3, 4,etc. For 3, 13, 23, 33 etc. multiply them by 3 and youget 1, 4, 7, 10, etc. as the Ekadhikas. For 7, 17, 27, 37 etc. multiply them by 7 and youget 5, 12, 19, 26, etc. as the Ekadhikas.
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2.5. Divisibility
Use Ekadhika as an osculator. For 9, 19, 29, 39 etc. the Ekadhikas are 1, 2, 3, 4,etc. For 3, 13, 23, 33 etc. multiply them by 3 and youget 1, 4, 7, 10, etc. as the Ekadhikas. For 7, 17, 27, 37 etc. multiply them by 7 and youget 5, 12, 19, 26, etc. as the Ekadhikas. For 1, 11, 21, 31 etc. multiply them by 9 and youget 1, 10, 19, 28, etc. as the Ekadhikas.
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2.5. Divisibility
Use Ekadhika as an osculator. For 9, 19, 29, 39 etc. the Ekadhikas are 1, 2, 3, 4,etc. For 3, 13, 23, 33 etc. multiply them by 3 and youget 1, 4, 7, 10, etc. as the Ekadhikas. For 7, 17, 27, 37 etc. multiply them by 7 and youget 5, 12, 19, 26, etc. as the Ekadhikas. For 1, 11, 21, 31 etc. multiply them by 9 and youget 1, 10, 19, 28, etc. as the Ekadhikas. Now test if 112 is divisible by 7 osculating by 5:2 5 + 11 = 21, which is divisible by 7. Therefore: yes
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2.5. Divisibility
Use Ekadhika as an osculator. For 9, 19, 29, 39 etc. the Ekadhikas are 1, 2, 3, 4,etc. For 3, 13, 23, 33 etc. multiply them by 3 and youget 1, 4, 7, 10, etc. as the Ekadhikas. For 7, 17, 27, 37 etc. multiply them by 7 and youget 5, 12, 19, 26, etc. as the Ekadhikas. For 1, 11, 21, 31 etc. multiply them by 9 and youget 1, 10, 19, 28, etc. as the Ekadhikas. Now test if 112 is divisible by 7 osculating by 5:2 5 + 11 = 21, which is divisible by 7. Therefore: yes
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Is 2774 divisible by 19? Osculate by 2:
2 7 7 4
+ 8
2 8 5+ 1 0
3 8
+ 1 6
1 9
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One more example: Is 5293240096 divisible by 139? The Ekadhika (osculator) is 14.
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One more example: Is 5293240096 divisible by 139? The Ekadhika (osculator) is 14.
5 2 9 3 2 4 0 0 9 6
139 89 36 131 29 131 19 51 93
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One more example: Is 5293240096 divisible by 139? The Ekadhika (osculator) is 14.
5 2 9 3 2 4 0 0 9 6
139 89 36 131 29 131 19 51 93
Answer: yes
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2.5.1. Exercises
Using the osculation method, check if
a 32896 is divisible by 29b 93148 is divisible by 29
c 4914 is divisible by 39
d 14061 is divisible by 43
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2.6. Square Roots (Vargamula)
1, 5, 6 and 0 at the end of a number reproducethemselves as the last digits in the square.
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2.6. Square Roots (Vargamula)
1, 5, 6 and 0 at the end of a number reproducethemselves as the last digits in the square. Squares of complements from 10 have the same lastdigit; thus 12 and 92 end in 1, 22 and 82 end in 4
etc.
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2.6. Square Roots (Vargamula)
1, 5, 6 and 0 at the end of a number reproducethemselves as the last digits in the square. Squares of complements from 10 have the same lastdigit; thus 12 and 92 end in 1, 22 and 82 end in 4
etc.
2, 3, 7 and 8 are out of court altogether.
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2.6. Square Roots (Vargamula)
1, 5, 6 and 0 at the end of a number reproducethemselves as the last digits in the square. Squares of complements from 10 have the same lastdigit; thus 12 and 92 end in 1, 22 and 82 end in 4
etc.
2, 3, 7 and 8 are out of court altogether. If the given number hasn digits, then the squareroot will contain n2 or n +12 digits.
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2.6. Square Roots (Vargamula)
1, 5, 6 and 0 at the end of a number reproducethemselves as the last digits in the square. Squares of complements from 10 have the same lastdigit; thus 12 and 92 end in 1, 22 and 82 end in 4
etc.
2, 3, 7 and 8 are out of court altogether. If the given number hasn digits, then the squareroot will contain n2 or n +12 digits. Systematic computation of an exact square root re-quires the Dvandvayoga (Duplex) process.
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2.6.2. Square Root of a Perfect Square
Find 1849.Group in pairs, taking a single extra digit on the left asextra digit.
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2.6.2. Square Root of a Perfect Square
Find 1849.Group in pairs, taking a single extra digit on the left asextra digit.
1 8 4 98) 2
4
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2.6.2. Square Root of a Perfect Square
Find 1849.Group in pairs, taking a single extra digit on the left asextra digit.
1 8 4 98) 2
4
4 is the largest integer whose square does not exceed 18.18/ 4 is 4 with remainder 2.The divisor 8 is two times 4.
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Next we divide 24 by the divisor 8. This gives 3 remain-der 0, placed as
1 8 4 9
8) 2 0
4 3
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Next we divide 24 by the divisor 8. This gives 3 remain-der 0, placed as
1 8 4 9
8) 2 0
4 3
Now we see 09 and we deduct from this the duplex of the last answer gure 3, i.e. 09D(3) = 09 32 =099 = 0. This means that the answer is exactly 43.
1 3 6 9
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6) 4
3
1 3 6 9
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6) 4
3
3 is the largest integer whose square does not exceed 13.13/ 3 is 3 with remainder 4.
The divisor 6 is two times 3.Next we divide 46 by the divisor 6. This gives 7 remain-der 4, placed as
1 3 6 9
6) 4 43 7
1 3 6 9
6) 4
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6) 4
3
3 is the largest integer whose square does not exceed 13.13/ 3 is 3 with remainder 4.
The divisor 6 is two times 3.Next we divide 46 by the divisor 6. This gives 7 remain-der 4, placed as
1 3 6 9
6) 4 43 7
49D(7) = 0, so 37 is the exact square root of 1369.
2.6.3. Exercises
Find the square root of the following.
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a 3136
b 3969c 5184
d 3721
e 6889
f 1296
2.6.3. Exercises
Find the square root of the following.
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a 3136
b 3969c 5184
d 3721
e 6889
f 1296
Find out how the method extends to 6-digit squares andnd the square roof of the numbers
a 119025
b 524176
c 519841
d 375769
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3. Zero Search Methods1. Newtons Method
2. Herons Square Root Finder
3. Exercise: Divisions without Dividing
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Sir Isaac Newton1643 - 1727
3.1. Newtons Method
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Given a differentiable functionf (x) we want to deter-
mine x such thatf (x) = 0. (6)
Starting with x0 we take the tangent to the curvethrough the point (x0 , f (x0)) and use its intersectionwith the x-axis x1 as a new starting point. We repeatthis method until no further changes occur. The recur-sive relation is
xn +1 = xn f (xn )f (xn )
(7)
and the result is
limn
xn = x. (8)
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Newtons Method - graphically
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3.1.1. Problems
Problems can occur due to
1. multiple solutions
2. non convexf , reection points3. solutions at extreme values
4. |f | = 5. pathological cases
3.1.2. Rate of Convergence
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g
Considering all these problems, why is Newtons methodstill so popular? The reason lies in the rate of conver-gence. Dene the error by
en= xn x (9)
From the denition of the Newton iteration, we haveen +1 = xn +1 x
= xn f (xn )f (xn ) x
= en f (xn )f (xn )
=en f (xn ) f (xn )
f (xn )(10)
By Taylors Theorem, we have
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0 = f (x) = f (xn en ) (11)= f (xn ) en f (xn ) + 12e2n f (n ) (12)
where n is a number betweenxn and x. A rearrange-ment of this equation yields
en f (xn ) f (xn ) = 12f (n )e2n (13)Putting this in (10) leads to
en +1 =1
2
f (n )
f (xn )e2n
1
2
f (x)
f (x)e2n = Ce2n (14)
This equation tells us that en +1 is roughly a con-stant times e2n . This desirable state of affairs is calledquadratic convergence .
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3.3. Exercise: Divisions without Dividing
Use Newtons method to determine an algorithmthat computes the reciprocal of a given number x,without ever performing any division.
apply your method to compute1
19 exact to 8 decimalplaces.
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4. Credit CardsWe follow An Introduction to the Mathematics of Money by Lovelock, Mendel and Wright [3].
9876
BIN5432 1987 654
cardholder ID3
checksum(16)
BIN: Bank Identication Number
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4. Credit CardsWe follow An Introduction to the Mathematics of Money by Lovelock, Mendel and Wright [3].
9876
BIN5432 1987 654
cardholder ID3
checksum(16)
BIN: Bank Identication Number More on http://en.wikipedia.org/wiki/Credit_card_numbers
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4. Credit CardsWe follow An Introduction to the Mathematics of Money by Lovelock, Mendel and Wright [3].
9876
BIN5432 1987 654
cardholder ID3
checksum(16)
BIN: Bank Identication Number More on http://en.wikipedia.org/wiki/Credit_card_numbers checksum: designed to protect against accidentalerrors, not malicious attacks. How?
d d
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4. Credit CardsWe follow An Introduction to the Mathematics of Money by Lovelock, Mendel and Wright [3].
9876
BIN5432 1987 654
cardholder ID3
checksum(16)
BIN: Bank Identication Number More on http://en.wikipedia.org/wiki/Credit_card_numbers checksum: designed to protect against accidentalerrors, not malicious attacks. How? Luhn algorithm
l
4.1. The Luhn Algorithm
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Hans Peter Luhn
The Luhn algorithm or Luhn formula , also known asthe modulus 10 or mod 10 algorithm , is a simple check-sum formula used to validate a variety of identicationnumbers, such as credit card numbers and Canadian
Social Insurance Numbers.
N l S t
4.1. The Luhn Algorithm
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Hans Peter Luhn
The Luhn algorithm or Luhn formula , also known asthe modulus 10 or mod 10 algorithm , is a simple check-sum formula used to validate a variety of identicationnumbers, such as credit card numbers and Canadian
Social Insurance Numbers.It was created by IBM scientist Hans Peter Luhn anddescribed in US Patent 2,950,048, led on January 6,1954, and granted on August 23, 1960.
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4.1. The Luhn Algorithm
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Hans Peter Luhn
The Luhn algorithm or Luhn formula , also known asthe modulus 10 or mod 10 algorithm , is a simple check-sum formula used to validate a variety of identicationnumbers, such as credit card numbers and Canadian
Social Insurance Numbers.It was created by IBM scientist Hans Peter Luhn anddescribed in US Patent 2,950,048, led on January 6,1954, and granted on August 23, 1960.
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9876BIN
5432 1987 654cardholder ID
3checksum
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BIN cardholder ID checksum9 8 7 6 5 4 3 2 1
2 2 2 2 2 18 8 14 6 10 4 6 2 2
9 8 7 6 5 4 add up
2 2 2 digits9 16 7 12 5 8 82
Table 2: Example of the Luhn Algorithm
Note: 18 counts as 1 + 8, i.e. count only the digits
Numeral Systems 98765432 1987 654 3
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9876
BIN
5432 1987 654
cardholder ID
3
checksum
Add the checksum 3 to the 82 and obtain 85.
Numeral Systems 98765432 1987 654 3
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BIN
cardholder ID
checksum
Add the checksum 3 to the 82 and obtain 85. If the new total is divisible by 10, then the creditcard number has passed the validation test.
Numeral Systems 98765432 1987 654 3
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BIN
cardholder ID
checksum
Add the checksum 3 to the 82 and obtain 85. If the new total is divisible by 10, then the creditcard number has passed the validation test.
our number fails, whereas9876
BIN5432 1987 654
cardholder ID8
checksumpasses.
Numeral Systems 98765432 1987 654 3
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BIN
cardholder ID
checksum
Add the checksum 3 to the 82 and obtain 85. If the new total is divisible by 10, then the creditcard number has passed the validation test.
our number fails, whereas
9876
BIN5432 1987 654
cardholder ID8
checksumpasses. Note: Amex has one digit less and starts with thesecond digit.
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4.2. Exercises
1. Implement the Luhn-Algorithm in Excel.
2. Apply the Luhn-Algorithm to your own credit cardnumber.
3. In your group think of a credit card number andask another group to verify it.
4. Does the Luhn Algorithm pick up any incorrect en-try of a single digit?
5. Will the Luhn Algorithm pick up any incorrecttransposition of adjacent digits?
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5. Cryptography1. The Inventors of the RSA-Algorithm
2. Method
3. Examples4. Exercise
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5. Cryptography1. The Inventors of the RSA-Algorithm
2. Method
3. Examples4. Exercise
5.1. The Inventors of the RSA-Algorithm (1978)
Ron Rivest, Adi Shamir, and Leonard Adleman [5]
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Ronald L. RivestAndrew and Erna Viterbi Professor of Electrical Engi-neering and Computer Science in MITs Department of Electrical Engineering and Computer Science
http://theory.lcs.mit.edu/ ~rivest/
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Adi ShamirPaul and Marlene Borman Professor of Applied Math-ematics at Weizmann Institute
http://www.weizmann.ac.il/math/profile04/scientists/shamir-prof04.html
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Professor Leonard AdlemanDepartment of Computer Science University of South-ern California
http://www.usc.edu/dept/molecular-science/fm-adleman.htm
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V di M h i
5.2. MethodWe follow Francis Litterio instructions onhttp://world.std.com/ ~franl/crypto/rsa-guts.html .
1 Find P and Q two large (e g 1024-bit) prime num-
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1. Find P and Q, two large (e.g., 1024-bit) prime num-
bers.2. Choose E ]1, P Q [, no prime factors in commonwith (P 1)(Q 1).3. Compute D (the multiplicative inverse) such that
DE = 1 mod (P 1)(Q 1). (Find an integer X which causesD = ( X (P 1)(Q 1)+1) /E to bean integer, then use that value of D .)4. The encryption function is C = T E mod P Q,
where C is the ciphertext (a positive integer), T
is the plaintext (a positive integer). The messagebeing encrypted,T , must be less than the modulus,P Q .
5. The decryption function is T = C D mod P Q,where C is the ciphertext (a positive integer), T is the plaintext (a positive integer).
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V di M th ti
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Your public key is the pair (PQ,E ). Your private keyis the number D (reveal it to no one). The productP Qis the modulus (often calledN in the literature). E isthe public exponent. D is the secret exponent.
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Your public key is the pair (PQ,E ). Your private keyis the number D (reveal it to no one). The productP Qis the modulus (often calledN in the literature). E isthe public exponent. D is the secret exponent.
You can publish your public key freely, because thereare no known easy methods of calculatingD, P , orQ given only (PQ,E ) (your public key). If P and Qare each 1024 bits long, the sun will burn out beforethe most powerful computers presently in existence can
factor your modulus intoP and Q.
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5.3. Examples
prime 1 P 5 11 37 61
prime 2 Q 7 7 41 53
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prime 2 Q 7 7 41 53
public modulus N = P Q 35 77 1517 3233
P 1 4 10 36 60Q 1 6 6 40 52(P 1)(Q 1) 24 60 1440 3120
public exponent E 5 7 7 17
inverse of E D 5 43 823 2753
check DE 25 301 5761 46801
check DE 1 24 300 5760 46800check [DE 1]/ [(P 1)(Q 1)] 0 0 0 0plaintext T 5 6 100 123encryption function C = T E mod P Q 10 41 1062 855
decryption function T = C D mod P Q 5 6 100 123
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5.4. Exercise1. Set up an RSA tool in Excel/VBA and verify the
examples.
2. Take two prime numbers P and Q between 100and 200, e.g. fromhttp://primes.utm.edu andcompute E and D.
3. Based on a plain text of your choice compute itsciphertext.
4. Give your public key and the ciphertext to the nextgroup and determine the plaintext of another group.
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6.1. Prime Numbers
a number that can only be divided evenly by 1 and thenumber itself.
6.2. ModulusFor integers K , R and N the equation
K = R mod N (17)
means that R is the remainder of the division of K byN or alternatively there exists an integerL such that
N L + R = K. (18)
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7. Contact InformationProfessor Dr Uwe Wystup
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Professor Dr. Uwe WystupAnsua Dutta-WystupManaging Directors
MathFinance AGMainluststrae 4
60329 Frankfurt am MainGermanyPhone +49-700-MATHFINANCEMore papers are available athttp://www.mathfinance.com/wystup/papers.
phpThese slides and handouts are available athttp://www.mathfinance.com/seminars/vedic.php
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References[1] Burden , R. L. and Faires , J. D. (1993). Numerical analysis . PWS
Publishing Company.
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[2] Datta , B. and Singh , A.N. (1962). History of Hindu Mathematics .Asia Publishing House, Calcutta.
[3] Lovelock , D., Mendel , M. and Wright , A.L. (2007). An Intro-duction to the Mathematics of Money , Springer , New York.
[4] Maharaja , Bharati Krsna Tirthaji (1992). Vedic Mathematics , Moti-lal Banarsidass Publishers Private Ltd, Delhi.
[5] Rivest , R.L., Shamir , A. and Adleman , L.M. (1978). A Method forObtaining Digital Signatures and Public-Key Cryptosystems. Commu-nications of the ACM 21,2 , 120126.
[6] Schonard , A. and Kokot , C. (2006). Der Mathekn uller . http://www.matheknueller.de .
[7] Williams , K.R. (2002). Vedic Mathematics - Teachers Manual . Ad-vanced Level. Motilal Banarsidass Publishers Private Limited, Delhi.http://www.mlbd.com
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Vedic Mathematics
Indexb k id ti ti b (BIN) 40 i b 53
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bank identication number (BIN), 40BIN, 40
division, 18, 21dual system, 3duplex, 26dvandvayoga, 26
ekadhika, 21
Herons Square Root Finder, 38hexadecimal system, 3
Luhn algorithm, 41
Maya numerals, 3modulus, 53multiplication, 9
Newtons Method, 33numeral systems, 3
osculator, 21
prime number, 53
quadratic convergence, 37
RSA-Algorithm, 45
square root, 25squares, 16sutras, 7
vedas, 6
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