Modular Arithmetic

(合同算術)

Discrete Mathematics I

13th lecture, January 10, 2020

https://www.sw.it.aoyama.ac.jp/2019/Math1/lecture13.html

Martin J. Dürst

AGU

© 2005-20 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

 

Remaining Schedule

About makeup classes: The material in the makeup class is part of the final exam. If you have another makeup class at the same time, please inform the teacher now.

補講について: 補講の内容は期末試験の対象。補講が別の補講とぶつかる場合には今すぐ申し出ること。

 

Questions about Final Exam

 

Homework Due January 6

By using formula manipulation, show that the Wolfram axiom of Boolean logic (((AB)⊼C)⊼(A⊼((AC)⊼A))=C) is a tautology. For each simplification step, indicate which law(s) you used.

Hints: Simplify the left side to obtain the right side. There should be between 15 and 20 steps.

(AB)C)(A((AC)A)) [definition of ⊼, six times]
= ¬(¬(¬(AB)∧C)¬(A∧¬(¬(AC)∧A))) [DeMorgan]
= ¬¬(¬(AB)∧C)∨¬¬(A∧¬(¬(AC)∧A)) [double negation, twice]
= ¬(AB)∧C ∨(A¬(¬(AC)∧A)) [DeMorgan, twice]
= (¬A∨¬B)∧C ∨(A∧(¬¬(AC)∨¬A)) [double negation]
= (¬A∨¬B)∧C ∨(A∧((AC)∨¬A)) [distributive law]
= (¬A∨¬B)∧C ∨(A∧((A∨¬A)∧(C∨¬A))) [law of excluded middle]
= (¬A∨¬B)∧C ∨(A∧(T∧(C∨¬A))) [property of true]
= (¬A∨¬B)∧C ∨(A(C∨¬A)) [distributive law]
= (¬A∨¬B)∧C ∨(ACA∧¬A) [law of noncontradiction]
= (¬A∨¬B)∧C ∨(AC∨F) [property of false]
= (¬A∨¬B)C ∨(AC) [distributive law (reverse)]
= (A∨¬B)∨A)∧C [associative/commutative laws]
= (¬AA∨¬B)∧C [law of excluded middle]
= (T∨¬B)∧C [property of true]
= T∧C [property of true]
= C

The left side of the Wolfram axiom can be transformed to the right side using only the laws of Boolean logic. This means that the Wolfram axiom is a tautology.

 

Comments on Homework Due January 9

 

Homework Due January 9

Draw the Hasse diagram of a Boolean algebra of order 4 (16 elements). There will be a deduction if you use the same elements of the group as another student.

Solution: For an example, see handout of last lecture

 

Summary of Last Lecture

 

Summary of Algebraic Structures

(hierarchy of objects)

 

Applications of Bitwise Operations

Operations work on 8, 16, 32, 64 or more bits concurrently

 

Other Bitwise Operations

 

More Applications of Bitwise Operations

For many more advanced examples, see Hacker's Delight, Henry S. Warren, Jr., Addison-Wesley, 2003

 

Addition in Different Numeral Systems

Works the same as in the decimal system:

Example (base 7):

Operand 1 3 6 5 1 2
Operand 2 6 0 3 3 4
carry 1 1 1
sum in base 10 1 10 7 8 4 6
sum in base 7 1 13 10 11 4 6
Result 1 3 0 1 4 6

 

Addition using Bitwise Operations

Single Digit Addition
0 1
0 0 1
1 1 10

 

Example of Addition Using Bitwise Operations

Example: a0 = 47 = 101111、b0 = 58 = 111010

x 0 1 2
ax (sx-1) 101111 010101 1000001
bx (cx-1<<1) 111010 1010100 0101000
sx (a0^b0) 010101 1000001 1101001 = 105 (result)
cx (a0&b0) 101010 010100 0000000 = 0 (finished)
cx<<1 1010100 0101000 00000000

 

Modular Arithmetic

 

Congruence Relation

 

Properties of Congruence Equations

 

Properties of the Modulo Operation

Reason: a(mod n) a mod n, and so on

Where to use: a and c may be very large numbers, but a mod n and c mod n may be much smaller, so calculation becomes easier.

 

Modulo Operation for Negative Operands

See English Wikipedia article on Modulo Operation

 

An Example of Using the Modulo Operation

Output some data, three items on a line.

A simple way:

int items = 0;
for (i=0; i<count; i++) {
    /* print out data[i] */
    items++;
    if (items==3) {
        printf("\n");
        items = 0;
    }
}

Using the modulo operation:

for (i=0; i<count; i++) {
    /* print out data[i] */
    if (i%3 == 2)
        printf("\n");
}

 

Congruence and Groups

 

Congruence and Division

Division and inverse for rationals: a/b = ca·1/b = a b-1 = c

Condition for (multiplicative) inverse b-1: b b-1 = 1

Condition for modular multiplicative inverse: bb-1 ≡ 1

n b 0 1 2 3 4 5 6 7 8
2 - 1
3 - 1 2
4 - 1 - 3
5 - 1 3 2 4
6 - 1 - - - 5
7 - 1 4 5 2 3 6
8 - 1 - 3 - 5 - 7
9 - 1 5 - 7 2 - 4 8

The modular multiplicative inverse is only defined if n and b are coprime (i.e. GCD(n, b) = 1)

Various methods to calculate, very time-consuming

a/b (mod n) is defined as a b-1 (mod n)

Example:
3/4 (mod 7) = 3 · 4-1 (mod 7) = 3 · 2 (mod 7) = 6
(Check: 6 · 4 (mod 7) = 24 (mod 7) = 3)

 

Application of Congruence: Simple Calculation of Remainder

Example: 216 mod 29 = ?

216 = 25 · 25 · 25 · 2

216 = 25 · 25 · 25 · 2 = 32 · 32 · 32 · 2(mod 29) 3 · 3 · 3 · 2 = 54(mod 29) 25

 

Remainder Calculation: More Examples

318 mod 7 = ?

318 = (33)6 = 276 (mod 7) (-1)6 = 1

4110 mod 37 = ?

4110 (mod 37) 410 = 220 = 324 (mod 37) (-5)4 = 252 (mod 37) (-12)2 = (6 · 2)2 = 36 · 4 (mod 37) (-1) · 4 = -4 (mod 37) 33

 

Homework

Prepare for final exam

 

Student Survey

(授業改善のための学生アンケート)

WEB Survey

お願い: 自由記述に必ず良かった点、問題点を具体的に書きましょう

(悪い例: 発音が分かりにくい; 良い例: さ行が濁っているかどうか分かりにくい)

 

Glossary

rotation
回転
reflection
反射
hierarchy
階層
concurrent
同時
shift
シフト
invert
逆転、反転
modular arithmetic
合同算術
congruence (equation)
合同式
modulus
residue
剰余
modulo operation
剰余演算
congruence relation
合同関係
congruence class
合同類
residue class
剰余類
cyclic group
巡回群
modular multiplicative inverse
モジュラー逆数
coprime
互いに素
operator
演算子
dividend
被除数
divisor
除数
implementation
実装