- Fact:
- Decryption =
the inverse operation of
the encrypt operation
- In general:
X = encrypt( A ); Then: A = decrypt( X ) An therefore: decrypt( encrypt( A ) ) = A
- Decryption =
the inverse operation of
the encrypt operation
- Main idea in making
public key encryption possible:
- The decryption algorithm is not easy to figure out even if you know the encryption algorithm.
Let's look closer what this means...
- The encryption
algorithm consists of:
- A publicly available program
- A publicly available encryption key
- The decryption
algorithm consists of:
- A publicly available program (because all programs are made publicly available !!!)
- A private decryption key
- Therefore: the main idea in making
public key encryption possible:
- The decryption key is not easy to figure out even if you know the encryption key.
However:
- That is also the
difficulty to create a
public key encryption method:
- If you tell me how you "shuffle" (= encrypt) the plaintxt, in most cases, it is trivial to "unshuffle" (= decrypt) the cyphertext !!!
- Perfect shuffle:
- Perfect shuffle (of a deckof cards) = you cut the deck of cards exactly in half and perfectly interlace the cards
Example:
- If we use a deck of 8 cards:
1, 2, 3, 4, 5, 6, 7, 8
A perfect shuffle will re-arrange the deck as follows:
- Interesting fact:
- If you perform 8 perfect shuffles in a row, the deck of cards will return back to its original order
Behold:
Original deck: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 After shuffle #0: 1 27 2 28 3 29 4 30 5 31 6 32 7 33 8 34 9 35 10 36 11 37 12 38 13 39 14 40 15 41 16 42 17 43 18 44 19 45 20 46 21 47 22 48 23 49 24 50 25 51 26 52 After shuffle #1: 1 14 27 40 2 15 28 41 3 16 29 42 4 17 30 43 5 18 31 44 6 19 32 45 7 20 33 46 8 21 34 47 9 22 35 48 10 23 36 49 11 24 37 50 12 25 38 51 13 26 39 52 After shuffle #2: 1 33 14 46 27 8 40 21 2 34 15 47 28 9 41 22 3 35 16 48 29 10 42 23 4 36 17 49 30 11 43 24 5 37 18 50 31 12 44 25 6 38 19 51 32 13 45 26 7 39 20 52 After shuffle #3: 1 17 33 49 14 30 46 11 27 43 8 24 40 5 21 37 2 18 34 50 15 31 47 12 28 44 9 25 41 6 22 38 3 19 35 51 16 32 48 13 29 45 10 26 42 7 23 39 4 20 36 52 After shuffle #4: 1 9 17 25 33 41 49 6 14 22 30 38 46 3 11 19 27 35 43 51 8 16 24 32 40 48 5 13 21 29 37 45 2 10 18 26 34 42 50 7 15 23 31 39 47 4 12 20 28 36 44 52 After shuffle #5: 1 5 9 13 17 21 25 29 33 37 41 45 49 2 6 10 14 18 22 26 30 34 38 42 46 50 3 7 11 15 19 23 27 31 35 39 43 47 51 4 8 12 16 20 24 28 32 36 40 44 48 52 After shuffle #6: 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 After shuffle #7: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link(s) and
save in a scratch directory
- To compile: javac Shuffle1.java
- To run: java Shuffle1
- Interesting encryption/decryption method:
- A plaintext is encrypted
by using k perfect shuffles
- To decrypt the cyphertext, we perform another 8 − k perfect shuffles
- A plaintext is encrypted
by using k perfect shuffles
- Important fact from Mathematics:
- Let p = a
prime number
- Suppose you have the integers:
1 2 3 4 5 ... (p-1)
- If a number a is
no divisible by p, then:
(a)%p (2a)%p (3a)%p (4a)%p ... ((p-1)a)%pis a permutation of 1, 2, 3, ...., (p−1).
- Let p = a
prime number
- Examples:
p = 5 a = 8 ( a)%5 = 8%5 = 3 (2a)%5 = 16%5 = 1 (3a)%5 = 24%5 = 4 (4a)%5 = 32%5 = 2 is a permutation of 1, 2, 3, 4
p = 7 a = 12 ( a)%7 = 12%7 = 5 (2a)%7 = 24%7 = 3 (3a)%7 = 36%7 = 1 (4a)%7 = 48%7 = 6 (5a)%7 = 60%7 = 4 (6a)%7 = 72%7 = 2 is a permutation of 1, 2, 3, 4, 5, 6
- If we want to use this
general shuffle mechanism to
shuffle a
deck of cards, we need
to construct a permutation of
the numbers:
1, 2, 3, 4, ....., 52 (because we have 52 crads)From the Mathematical Fact above, we know that:
- If a is
not divisible by
53, then:
( a)%53 (2a)%53 (3a)%53 .... (52a)%53is a permutation of the numbers 1, 2, 3, 4, ..., 52.
(Which we can use to shuffle the deck of cards)
- If a is
not divisible by
53, then:
- Java code:
public static int [] shuffle( int[] x , int a ) { int[] out = new int[53]; int j; for ( int i = 1; i <= 52; i++ ) { a = a%53; // Prevent overflow j = (a*k)%53; // Computes: a, 2a, 3a, ... out[j] = x[i]; // Permute } return out; }
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link(s) and
save in a scratch directory
- To compile: javac Shuffle2.java
- To run: java Shuffle2
Sample outputs:
Enter a = 40 Shuffle i = 1 4 8 12 16 20 24 28 32 36 40 44 48 52 3 7 11 15 19 23 27 31 35 39 43 47 51 2 6 10 14 18 22 26 30 34 38 42 46 50 1 5 9 13 17 21 25 29 33 37 41 45 49 Shuffle i = 2 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 Shuffle i = 3 19 38 4 23 42 8 27 46 12 31 50 16 35 1 20 39 5 24 43 9 28 47 13 32 51 17 36 2 21 40 6 25 44 10 29 48 14 33 52 18 37 3 22 41 7 26 45 11 30 49 15 34 Shuffle i = 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
Enter a = 6 Shuffle i = 1 9 18 27 36 45 1 10 19 28 37 46 2 11 20 29 38 47 3 12 21 30 39 48 4 13 22 31 40 49 5 14 23 32 41 50 6 15 24 33 42 51 7 16 25 34 43 52 8 17 26 35 44 Shuffle i = 2 31 9 40 18 49 27 5 36 14 45 23 1 32 10 41 19 50 28 6 37 15 46 24 2 33 11 42 20 51 29 7 38 16 47 25 3 34 12 43 21 52 30 8 39 17 48 26 4 35 13 44 22 Shuffle i = 3 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 Shuffle i = 4 42 31 20 9 51 40 29 18 7 49 38 27 16 5 47 36 25 14 3 45 34 23 12 1 43 32 21 10 52 41 30 19 8 50 39 28 17 6 48 37 26 15 4 46 35 24 13 2 44 33 22 11 Shuffle i = 5 23 46 16 39 9 32 2 25 48 18 41 11 34 4 27 50 20 43 13 36 6 29 52 22 45 15 38 8 31 1 24 47 17 40 10 33 3 26 49 19 42 12 35 5 28 51 21 44 14 37 7 30 Shuffle i = 6 28 3 31 6 34 9 37 12 40 15 43 18 46 21 49 24 52 27 2 30 5 33 8 36 11 39 14 42 17 45 20 48 23 51 26 1 29 4 32 7 35 10 38 13 41 16 44 19 47 22 50 25 Shuffle i = 7 24 48 19 43 14 38 9 33 4 28 52 23 47 18 42 13 37 8 32 3 27 51 22 46 17 41 12 36 7 31 2 26 50 21 45 16 40 11 35 6 30 1 25 49 20 44 15 39 10 34 5 29 Shuffle i = 8 21 42 10 31 52 20 41 9 30 51 19 40 8 29 50 18 39 7 28 49 17 38 6 27 48 16 37 5 26 47 15 36 4 25 46 14 35 3 24 45 13 34 2 23 44 12 33 1 22 43 11 32 Shuffle i = 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52Note:
- The number of shuffles needed to return the deck of cards to its original state varies with the value of a !!!
- Focus on one single card
in the deck:
Enter a = 40 Original state: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 Shuffle i = 1 4 8 12 16 20 24 28 32 36 40 44 48 52 3 7 11 15 19 23 27 31 35 39 43 47 51 2 6 10 14 18 22 26 30 34 38 42 46 50 1 5 9 13 17 21 25 29 33 37 41 45 49 Shuffle i = 2 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 Shuffle i = 3 19 38 4 23 42 8 27 46 12 31 50 16 35 1 20 39 5 24 43 9 28 47 13 32 51 17 36 2 21 40 6 25 44 10 29 48 14 33 52 18 37 3 22 41 7 26 45 11 30 49 15 34 Shuffle i = 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
How do we get positions of this card:
Originally: pos = 8 Shuffle 1: pos = (8*40) %53 = 2 Shuffle 1: pos = (8*402) %53 = 2