Resolution (7A) - upload.wikimedia.org (8B) Boolean Functions 4 Young Won Lim 3/22/18 Truth Table and minterms (1) 0 1 1 0 1 0 0 0 1 0 0 0 1 1 1 1 1 0 1 0 1 1 0 0 the case …

Young Won Lim3/19/18

Resolution (7A)


Copyright (c) 2015 – 2018 Young W. Lim.

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Logic (7A)Resolution 16 Young Won Lim

3/19/18

Example B – (1)

p → q∨r

p∨¬q

r∨q

¬p∨q∨r

p∨¬q

r∨q

Discrete Mathematics, Johnsonbough

q∨¬q∨r

r∨q

r∨q

q∨¬q∨r = T∨r = T

p → q∨r

p∨¬q

r∨q

¬p∨q∨r

p∨¬q

r∨q

p∨r


3/19/18

Example B – (2)

p → q∨r

p∨¬q

r∨q

p → q∨r

¬p → ¬q

q∨r

Discrete Mathematics, Johnsonbough

p → q∨r

q → p

q∨r

p∨r


3/19/18

p q r ¬p ¬p∨q∨r

T T T F T

T T F F T

T F T F T

T F F F F

F T T T T

F T F T T

F F T T T

F F F T T

p q r ¬q p∨¬q

T T T F T

T T F F T

T F T T T

T F F T T

F T T F F

F T F F F

F F T T T

F F F T T

Truth Table

p q r q∨r

T T T T

T T F T

T F T T

T F F F

F T T T

F T F T

F F T T

F F F F

p q r ¬p∨q∨r p∨¬q q∨r H 1∧H 2∧H 3 p∨rT T T T T T T T

T T F T T T T T

T F T T T T T T

T F F F T F F TF T T T F T F T

F T F T F T F F

F F T T T T T TF F F T T F F F

H 1 = ¬p∨q∨r

H 2 = p∨¬q

H 3 = q∨r

H 1∧H 2∧H 3 → H 3

H 1∧H 2∧H 3 → H 2

H 1∧H 2∧H 3 → H 1

H 1∧H 2∧H 3 → (p∨r )


3/19/18

p q r H 1∨H 2∨H 3

T T T T

T T F T

T F T T

T F F F

F T T F

F T F F

F F T T

F F F F

Truth Table and K-Map

p q r H 1∨H 2∨H 3

1 1 1 1

1 1 0 1

1 0 1 1

1 0 0 0

0 1 1 0

0 1 0 0

0 0 1 1

0 0 0 0

p q r H 1∨H 2∨H 3

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1

0 0 0 1 1 1 1 0

0 1 0 0

0 1 1 1

0

1

p q r


3/19/18

K-Map and Logic Minimization

0 0 0 1 1 1 1 0

0 1 0 0

0 1 1 1

0

1

p q r

0 0 0 1 1 1 1 0

0

1

p q r

p q r pq r pq r

0 0 0 1 1 1 1 0

0

1q r

p q

p q r + p q r = ( p + p)q r = q r

p q r + pq r = pq(r + r) = p q


3/19/18

K-Map : Verification

0 0 0 1 1 1 1 0

0

1q r

p qH 1∧H 2∧H 3 ≡ q r + pq

p q r q q r pq q r+ pq

0 0 0 1 0 0 0

0 0 1 1 1 0 1

0 1 0 0 0 0 0

0 1 1 0 0 0 0

1 0 0 1 0 0 0

1 0 1 1 1 0 1

1 1 0 0 0 1 1

1 1 1 0 0 1 1

p q r H 1∨H 2∨H 3

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1


3/19/18

Adding two don’t care conditions

0 0 0 1 1 1 1 0

0 1 X X

0 1 1 1

0

1

p q r

0 0 0 1 1 1 1 0

0

1r

0 0 0 1 1 1 1 0

0

1p

0 0 0 1 1 1 1 0

0 1 X X

0 1 1 1

0

1

p q r

p∨r


3/19/18

K-Map : Verification

H 1∧H 2∧H 3 → p∨r

p q r p∨r

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

p q r H 1∨H 2∨H 3 p∨q

0 0 0 0 0

0 0 1 1 1

0 1 0 0 0

0 1 1 0 1

1 0 0 0 1

1 0 1 1 1

1 1 0 1 1

1 1 1 1 1

0 0 0 1 1 1 1 0

0

1r

0 0 0 1 1 1 1 0

0

1p


References

[1] http://en.wikipedia.org/[2]


Boolean Algebra (8A)






Logic (8A)Boolean Algebra

3 Young Won Lim3/22/18

Argument



Boolean Algebra

https://en.wikipedia.org/wiki/Boolean_algebra



Operators


xy x+y x



Laws (1)


x+(y+z) = (x+y)+z

x(yz) = (xy)zx+y = y+x

xy = yz

x(y+z) = xy + xz

x+0=x

x*1=x

x*0=0



Laws (2)


x+1=1

x+x=xx*x=x

x(x+y)=x

x+xy=x

x+yz=(x+y)(x+z)

xx=0

x+x=1

x y = (x+y)

(x+y) = xy



Digital Logic Gates




NOT Gate

https://en.wikipedia.org/wiki/Logic_gate



AND, OR Gates




NAND, NOR Gates




XOR, XNOR Gates




CMOS Logic Gates

https://en.wikipedia.org/wiki/CMOS



Identity and Null Element Theorem

0 1

0 1

0 1

x

x

x

x

x

x

x

x

x

x

0

1

x⋅0 = 0 x⋅1 = x

x+ 0 = x x+ 1 = 1

x+ 0 = x x+ 1 = x

https://en.wikiversity.org/wiki/Discrete_Mathematics_in_plain_view#Algorithms



Distributive

x⋅( y + z) = x⋅ y + x⋅z

x + ( y⋅ z) = (x + y)⋅(x + z)

≠ x⋅ y + z

= x + y⋅ z

This parenthesis cannot be deleted

This parenthesis can be deleted

Operator precedence : ⋅ > +




Inclusion

x⋅(x + y ) = x

= x⋅(1 + y)

= x

x y

x + y

x + xy = x

x + xy = x⋅1 + x⋅ y

= x⋅(1 + y )

= x

x y

x y

x⋅(x + y ) = x⋅ x + x⋅ y

= x + x⋅ y




Inclusion

x⋅(x + y ) = x

= x⋅(1 + y)

= x

x y

x + y

x + xy = x

x + xy = x⋅1 + x⋅ y

= x⋅(1 + y )

= x

x y

x y

x⋅(x + y ) = x⋅ x + x⋅ y

= x + x⋅ y




`

Eliminate

x⋅(x + y ) = x y

= x⋅ y

x y

x + y

x + x y = x + y

= 1⋅(x + y )

= x + y

x y

x y

x⋅(x + y ) = x⋅ x + x⋅ y

= 0 + x⋅ y

x + x y = (x + x )⋅(x + y)



References



Boolean Functions (8B)






Logic (8B)Boolean Functions


Truth Table

x

y

z?

1

0

1

0

0

0

0

1

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

x y z F

F

I/O relationship

https://en.wikiversity.org/wiki/The_necessities_in_Digital_Design



Truth Table and minterms (1)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

the case when x=0 and y=1 and z=1








inputs

All possible combination of inputs

x y z

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

For the output of an and gate to be 1, all inputs must be 1

x=1

y=1

z=1

x=1

y=0

z=1

1




Truth Table and minterms (2)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

the case when the minterm








inputs


x y z

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

3

2

0

1

7

6

4

5

index

m0

=

m1

=

m2

=

m3

=

m4

=

m5

=

m6

=

m7

=

m5minterm index 5

binary 101

x y z = 1

0

bar

x=1

y=1

z=1

1




Truth Table and MAXterms (1)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0









inputs


x y z

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

For the output of an or gate to be 0, all inputs must be 0

x=0

y=0

z=0

x+ y+ z = 0

x=1

y=0

z=1

0




Truth Table and MAXterms (2)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

the case when the MAXterm








inputs


x y z

3

2

0

1

7

6

4

5

index

M0

=

M1

=

M2

=

M3

=

M4

=

M5

=

M6

=

M7

=

M5minterm index 5

binary 101 1

bar

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z

x=0

y=0

z=0

0




1

0

0

0

xy

Maxterm and minterm Conditions

1 1

1 0

0 1

0 0

x y

1

1

1

0

x+y

1 1

1 0

0 1

0 0

x y

0

0

0

1

xy

0 0

0 1

1 0

1 1

x y

0

1

1

1

x+y

0 0

0 1

1 0

1 1

x y

1

01

01

1

1

1

0

0

0

0




1

0

1

0

0

0

0

1

Boolean Function with minterms (1)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output


x y z

3

2

0

1

7

6

4

5

index

F The output F becomes 1, for one of the three following cases

(the case when x=0 and y=1 and z=1)



or

or

x y z = 1

x y z = 1

x y z = 1

m1

=

m3

=

m4

=

Domain Range

m1

m3

m4

m0

m2

m5

m6

m7

1

0




1

0

1

0

0

0

0

1


0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output


x y z

3

2

0

1

7

6

4

5

index

F

F = m1+ m

3+ m

4

The output F becomes 1, m

1=1either m

3=1 m

4=1or or

For the output of an or gate to be 1, at least one must be 1

m1+ m

3+ m

4=1 F = 1

m1

m3

m4

Fm

0

m2

m7

Fm

5

m6




1

0

1

0

0

0

0

1


0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output


x y z

3

2

0

1

7

6

4

5

index

F

F = m1+ m

3+ m

4

The output F becomes 1, m

1=1either m

3=1 m

4=1or or

For the output of an or gate to be 1, at least one must be 1

m1+ m

3+ m

4=1 F = 1

The output F becomes 0, either

F = 0

m0=1 m

2=1 m

5=1 m

6=1 m

7=1or or or or

m0+m

2+m

5+m

6+m

7=1

F = m0+m

2+m

5+m

6+m

7









1

0

1

0

0

0

0

1

Boolean Function with Maxterms (1)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output


x y z

3

2

0

1

7

6

4

5

index

F

The output F becomes 0, for one of the five following cases

or

or

Domain

Range

M1

M3

M4

M0

M2

M5

M6

M7

1

0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

x+ y+ z = 0

or

or

ororor




1

0

1

0

0

0

0

1


0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output


x y z

3

2

0

1

7

6

4

5

index

FThe output F becomes 0, M

0=0either M

2=0 M

5=0 M

6=0 M

7=0or or or or

For the output of an and gate to be 0, at least one input must be 0

F = 0M0⋅M

2⋅M

5⋅M

6⋅M

7=0

F = M0⋅M

2⋅M

5⋅M

6⋅M

7

M0

M2

M7

M5

M6

M1

M3

M4

F F




1

0

1

0

0

0

0

1


0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output


x y z

3

2

0

1

7

6

4

5

index

FThe output F becomes 0, M

0=0either M

2=0 M

5=0 M

6=0 M

7=0or or or or

For the output of an and gate to be 0, at least one input must be 0

F = 0

The output F becomes 1, either

F = 1

M0⋅M

2⋅M

5⋅M

6⋅M

7=0

F = M0⋅M

2⋅M

5⋅M

6⋅M

7

M1=0 M

3=0 M

4=0or or

M1⋅M

3⋅M

4=0

F = M1⋅M

3⋅M

4




SOP and POS

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

AND

AND

AND

AND

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

OR

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

AND

AND

AND

AND

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

OR

1

0

1

0

SOP

POS


sum+

product*

sum+

product*



1

0

1

0

0

0

0

1

Boolean Function with minterms

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F

1

11

11

0

1

0

0

0

0

1

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F1

11

1

1

1

F F




1

0

1

0

0

0

0

1

Boolean Function with Maxterms

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F

0

00

01

0

1

0

0

0

0

1

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F0

0

0

0

0

FF

0



References



K-Map (8C)






Logic (8C)K-Map




Logic (8C)K-Map


K-Map 3 variables (1)

2310

6754

y=1y=0

z=1z=0 z=0

x=0

x=1

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

3

2

0

1

7

6

4

5

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y z

2310

6754

10110100

0

1

x y zindex minterms

Logic (8C)K-Map



0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

3

2

0

1

7

6

4

5

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y

x y

x y

x y

x y z + x y z = x y ( z+ z) = x y

2310

6754

y=1y=0

z=1z=0 z=0

x=0

x=1

10110100

0

1

x y x y

x y x y

index minterms

x y z

21

a group of 2 minterms

Logic (8C)K-Map



0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

3

2

0

1

7

6

4

5

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x z

x z

x z

x z

x y z + x y z = x z( y+ y) = x z

2310

6754

y=1y=0

z=1z=0 z=0

x=0

x=1

10110100

0

1

x z x z

x z x z

index minterms

x y z

21


x z

x z

Logic (8C)K-Map



0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

3

2

0

1

7

6

4

5

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y z

y z

y z

y z

y z

x y z + x y z = y z ( x+x ) = y z

2310

6754

y=1y=0

z=1z=0 z=0

x=0

x=1

10110100

0

1

y z y z y z y z

index minterms

x y z

21


Logic (8C)K-Map



0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

3

2

0

1

7

6

4

5

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x

x

2310

6754

y=1y=0

z=1z=0 z=0

x=0

x=1

10110100

0

1 x

x

index minterms

x y z

22


Logic (8C)K-Map



0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

3

2

0

1

7

6

4

5

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y z

y

y

2310

6754

y=1y=0

z=1z=0 z=0

x=0

x=1

10110100

0

1

y y

index minterms

x y z

22


Logic (8C)K-Map



0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

3

2

0

1

7

6

4

5

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y z

z

z

2310

6754

y=1y=0

z=1z=0 z=0

x=0

x=1

10110100

0

1 z z

index minterms

x y z

22


Logic (8C)K-Map


K-Map, minterms, and Maxterms

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

3

2

0

1

7

6

4

5

x y z

x y z

x y z

x y z

x y z

x y z

x y z

x y z

2310

6754

10110100

0

1

x y zindex minterms

m0

m1 m

2m

3

m4

m5

m6

m7

2310

6754

10110100

0

1

M0

M1

M2

M3

M4

M5

M6

M7

x y z

Each rectangle is associated with a minterm or a maxterm which represents a particular input variable conditions.

In this table, output function value is overlaid

Logic (8C)K-Map


1

0

1

0

0

0

0

1

Boolean Function with minterms

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F

1

1

1

1

F

2310

6754

x

y

z

1 1

1F = x z + x y z

F=1 when xz = 1

F=1 when xyz = 1

2310

6754

10110100

0

1

m0

m1 m

2m

3

m4

m5

m6

m7

an overlaid table

a simplified function

Logic (8C)K-Map



1

0

1

0

0

0

0

1

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F0

0

0

0

0

F

0

2310

6754

0 0

0 0 0F = ( x+ z)( y+ z)(x+ z)

x

y

z

F=0 when x+z=0

F=0 when y+z=0F=0 when x+z=0

2310

6754

M0

M1

M2

M3

M4

M5

M6

M7

an overlaid table

a simplified function

Logic (8C)K-Map


Logic (8C)K-Map



0 0 1 1

0 0 1 0

0 0 0 1

0 0 0 0

0 1 1 1

0 1 1 0

0 1 0 1

0 1 0 0

1 0 1 1

1 0 1 0

1 0 0 1

1 0 0 0

1 1 1 1

1 1 1 0

1 1 0 1

1 1 0 0

3

2

0

1

7

6

4

5

11

10

8

9

15

14

12

131000 1001 1011 1010

1100 1101 1111 1110

0100 0101 0111 0110

0000 0001 0011 0010

2310

6754

101198

14151312

x y z w

x y z w

x y z w

x y zw

x y z w

x y z w

x y z w

x y zw

x y z w

x y z w

x y z w

x y zw

x y z w

x y z w

x y z w

x y zw

z=1z=0

w=1w=0 w=0

x=1

x=0

y=1

y=0

y=0

10110100

00

01

10

11

index minterms

Logic (8C)K-Map



0 0 1 1

0 0 1 0

0 0 0 1

0 0 0 0

0 1 1 1

0 1 1 0

0 1 0 1

0 1 0 0

1 0 1 1

1 0 1 0

1 0 0 1

1 0 0 0

1 1 1 1

1 1 1 0

1 1 0 1

1 1 0 0

3

2

0

1

7

6

4

5

11

10

8

9

15

14

12

131000 1001 1011 1010

1100 1101 1111 1110

0100 0101 0111 0110

0000 0001 0011 0010

2310

6754

101198

14151312

x y z w

x y z w

x y z w

x y zw

x y z w

x y z w

x y z w

x y zw

x y z w

x y z w

x y z w

x y zw

x y z w

x y z w

x y z w

x y zw

z=1z=0

w=1w=0 w=0

x=1

x=0

y=1

y=0

y=0

index minterms


References



Set Operations (1A)


Copyright (c) 2017 - 2 018 Young W. Lim.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License,Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled "GNU Free Documentation License".



Set Operations (1A) 14 Young Won Lim3/23/18

Power Set

https://en.wikipedia.org/wiki/Power_set

Set Operations (1A) 15 Young Won Lim3/23/18

Power Set Example

https://en.wikipedia.org/wiki/Power_set


References



Matrix (2A)


Copyright (c) 2015 - 2018 Young W. Lim.




Determinant (3A) 18Young Won Lim

03/22/2018

Gauss-Jordan Elimination

+ 2 + 1 −1

−3 −1 + 2

−2 + 1 + 2

x1

x2

x3

= + 8

= −11

= −3

x1

x2

x3

=

=

=

1 0 0

0 1 0

0 0 1

*

*

*

×c

×c+ + + +

https://en.wikiversity.org/wiki/Linear_Algebra_in_plain_view


03/22/2018

Gauss-Jordan Elimination – Step 1

+ 2 x1+ x

2− x

3= 8

−3 x1− x

2+ 2 x

3= −11

−2 x1+ x

2+ 2 x

3= −3

(L1)

(L2)

(L3)

+ 1 x1+ 1

2x

2− 1

2x

3= 4

−3 x1− x

2+ 2 x

3= −11

−2 x1+ x

2+ 2 x

3= −3

( 1

2× L

1)

(L2)

(L3)

+2

–3

–2

+1

–1

+1

–1

+2

+2

+8

–11

–3

+1

–3

–2

+1/2

–1

+1

–1/2

+2

+2

+4

–11

–3

+2/2 +1/2 –1/2 +8/2+ 1 x1+ 1

2x

2− 1

2x

3= 4 ( 1

2× L

1)



03/22/2018


(L1)

(L2)

(L3)

+ 1 x1+ 1

2x

2− 1

2x

3= + 4

−3 x1− x

2+ 2 x

3= −11

−2 x1+ x

2+ 2 x

3= −3

(L1)

(3 × L1+ L

2)

(2 × L1+ L

3)

+1

–3

–2

+1/2

–1

+1

–1/2

+2

+2

+4

–11

–3

+1

0

0

+1/2

+1/2

+2

–1/2

+1/2

+1

+4

+1

+5

+ 1 x1+ 1

2x

2− 1

2x

3= + 4

0 x1+ 1

2x

2+ 1

2x

3= + 1

0 x1+ 2 x

2+ 1 x

3= + 5

(3 × L1)+ 3 x

1+ 3

2x

2− 3

2x

3= + 12

(L2)−3 x

1− x

2+ 2 x

3= −11

(2 × L1)+ 2 x

1+ 2

2x

2− 2

2x

3= + 8

(L3)−2 x

1+ x

2+ 2 x

3= −3

+3 +3/2 –3/2 +12

–3 –1 +2 –11

+2 +2/2 –2/2 +8

–2 +1 +2 –3



03/22/2018


(L1)

(2 × L2)

(L3)

+1

0

0

+1/2

+1

+2

–1/2

+1

+1

+4

+2

+5

+ 1 x1+ 1

2x

2− 1

2x

3= + 4

0 x1+ 1 x

2+ 1 x

3= + 2

0 x1+ 2 x

2+ 1 x

3= + 5

+1

0

0

+1/2

+1/2

+2

–1/2

+1/2

+1

+4

+1

+5

+ 1 x1+ 1

2x

2− 1

2x

3= + 4

0 x1+ 1

2x

2+ 1

2x

3= + 1

0 x1+ 2 x

2+ 1 x

3= + 5

(L1)

(L2)

(L3)

0 +1 +1 +20 x1+ 1 x

2+ 1 x

3= + 2 (2 × L

2)



03/22/2018


(L2)

(−2 × L2+ L

3)

0

+1

0

+1/2

+1

0

–1/2

+1

–1

+4

+2

+1

0 x1+ 1 x

2+ 1 x

3= + 2

0 x1+ 0 x

2− 1 x

3= + 1

+1

0

0

+1/2

+1

+2

–1/2

+1

+1

+4

+2

+5

+ 1 x1+ 1

2x

2− 1

2x

3= + 4

0 x1+ 1 x

2+ 1 x

3= + 2

0 x1+ 2 x

2+ 1 x

3= + 5

(L1)

(L2)

(L3)

0 –2 –2 –40 x1− 2 x

2− 2 x

3= −4 (−2 × L

2)

0 +2 +1 +50 x1+ 2 x

2+ 1 x

3= + 5 (L

3)

0

+ 1 x1+ 1

2x

2− 1

2x

3= + 4 (L

1)



03/22/2018


0

0

+1

0

+1

–1

+2

+1

0 x1+ 1 x

2+ 1 x

3= + 2

0 x1+ 0 x

2− 1 x

3= + 1 0

(L2)

(L3)

0 0 +1 –10 x1− 0 x

2+ 1 x

3= −1 0(−1 × L

3)

0

0

+1

0

+1

+1

+2

–1

0 x1+ 1 x

2+ 1 x

3= + 2

0 x1+ 0 x

2+ 1 x

3= −1 0

(L2)

(−1 × L3)

+1 +1/2 –1/2 +4+ 1 x1+ 1

2x

2− 1

2x

3= + 4 (L

1)

+1 +1/2 –1/2 +4+ 1 x1+ 1

2x

2− 1

2x

3= + 4 (L

1)



03/22/2018

+1

Forward Phase

+2

–3

–2

+1

–1

+1

–1

+2

+2

+8

–11

–3

+1

–3

–2

+1/2

–1

+1

–1/2

+2

+2

+4

–11

–3

+1

0

0

+1/2

+1/2

+2

–1/2

+1/2

+1

+4

+1

+5

+1

0

0

+1/2

+1

+2

–1/2

+1

+1

+4

+2

+5

0

+1

0

+1/2

+1

0

–1/2

–1

+4

+2

+10

0

0

+1

0

+1

+1

+2

–10

+1 +1/2 –1/2 +4

Forward Phase - Gaussian Elimination



03/22/2018


0

+1

0

+1/2

+1

0

0

0

+1

+7/2

+3

–1

+1 x1+ 1

2x

2+ 0 x

3= + 7

2

0 x1+ 1 x

2+ 0 x

3= + 3

0 x1+ 0 x

2+ 1 x

3= −1 0

(+ 1

2× L

3+ L

1)

(−1 × L3+ L

2)

(L3)

0

0

+1

0

+1

+1

+2

–1

0 x1+ 1 x

2+ 1 x

3= + 2

0 x1+ 0 x

2+ 1 x

3= −1 0

(L2)

(L3)

0 0 +1/2 –1/20 x1+ 0 x

2+ 1

2x

3= − 1

20(+ 1

2× L

3)

+1 +1/2 –1/2 +4(L1)

0 0 –1 +10 x1+ 0 x

2− 1 x

3= + 1 0(−1 × L

3)

0 +1 +1 +20 x1+ 1 x

2+ 1 x

3= + 2 (L

2)

+1 +1/2 –1/2 +4+ 1 x1+ 1

2x

2− 1

2x

3= + 4 (L

1)

+ 1 x1+ 1

2x

2− 1

2x

3= + 4



03/22/2018


0

+1

0

0

+1

0

0

0

+1

+2

+3

–1

+ 1 x1+ 0 x

2− 0 x

3= + 2

0 x1+ 1 x

2+ 0 x

3= + 3

0 x1+ 0 x

2+ 1 x

3= −1 0

(− 1

2× L

2+ L

1)

(L2)

(L3)

0

+1

0

+1/2

+1

0

0

0

+1

+7/2

+3

–1

+1 x1+ 1

2x

2+ 0 x

3= + 7

2

0 x1+ 1 x

2+ 0 x

3= + 3

0 x1+ 0 x

2+ 1 x

3= −1 0

(L2)

(L3)

(L1)

0 –1/2 0 –3/20 x1− 1

2x

2+ 0 x

3= − 3

2(−1

2× L

2)

+1 +1/2 0 +7/2+ 1 x1+ 0 x

2− 0 x

3= + 2 (L

1)



03/22/2018

Backward Phase

0

+1

0

+1/2

+1

0

0

0

+1

+7/2

+3

–10

0

0

+1

0

+1

+1

+2

–10

+1 +1/2 –1/2 +4

0

+1

0

0

+1

0

0

0

+1

+2

+3

–10



03/22/2018

Gauss-Jordan Elimination

+1

+2

–3

–2

+1

–1

+1

–1

+2

+2

+8

–11–3

+1

–3

–2

+1/2

–1

+1

–1/2

+2

+2

+4

–11–3

+1

0

0

+1/2

+1/2

+2

–1/2

+1/2

+1

+4

+1

+5

+1

0

0

+1/2

+1

+2

–1/2

+1

+1

+4

+2

+5

0

+1

0

+1/2

+1

0

–1/2

–1

+4

+2

+10

0

0

+1

0

+1

+1

+2

–10

+1 +1/2 –1/2 +4

0

+1

0

+1/2

+1

0

0

0

+1

+7/2

+3

–10

0

0

+1

0

+1

+1

+2

–1

+1 +1/2 –1/2 +4

0

+1

0

0

+1

0

0

0

+1

+2

+3

–10

Backward Phase – Guass-Jordan Elimination

Forward Phase – Gaussian Elimination



References



Growth Functions (2A)


Copyright (c) 2015 - 2018 Young W. Lim.




Growth Functions (2A) 3 Young Won Lim3/22/18

Small Range

Zoom Out

All are distinguishable x2+2 x+1

x2

2 x

1

A1

x2+2 x+1

for x >−0.5 x2 < x

2+2 x+1


Medium Range

Zoom Out More

2 x

similarx

2+2 x+1x

2

A2

for x >−0.5 x2 < x

2+2 x+1


Large Range

Indistinguishable x

2+2 x+1x

2

A3

for x >−0.5 x2 < x

2+2 x+1


Small Range, C1=2

Zoom Out

distinguishable

x2+2 x+1

2 x2

2 x

1

B1

for x > 2.414 x2+2 x+1 < 2 x

2

for x > k f (n) < C g (n)


Medium Range, C1=2

Zoom Out

distinguishable

x2+2 x+1

2 x2

2 x

1

B2

for x > 2.414 x2+2 x+1 < 2 x

2



Large Range, C1=2

Zoom Out

distinguishable

x2+2 x+1

2 x2

2 x

1

B3

for x > 2.414 x2+2 x+1 < 2 x

2



Small Range, C1=10

Zoom Out

distinguishable

x2+2 x+1

10 x2

2 x1

C1

for x > 0.462 x2+2 x+1 < 10 x

2



Medium Range, C1=10

Zoom Out

distinguishable

x2+2 x+1

10 x2

C2

for x > 0.462 x2+2 x+1 < 10 x

2



Large Range, C1=10

distinguishable

x2+2 x+1

10 x2

C3

for x > 0.462 x2+2 x+1 < 10 x

2



Small Range, C1=10

Zoom Out

distinguishable

x2+2 x+1

x2

10 x

1

D1

for 0.127 < x < 7.873 x2+2 x+1 < 10 x


Medium Range, C1=10

Zoom Out

distinguishable x2+2 x+1

x2

D2

10 x

for x > 7.873 10 x < x2+2 x+1

for x > k C g(n) < f (n)


Large Range, C1=10

distinguishable

D3

x2+2 x+1

x2

10 x

for x > 7.873 10 x < x2+2 x+1

for x > k C g(n) < f (n)


Big Oh

Let f and g be functions (Z→R or R→R)from the set of integers or

the set of real numbersto the set of real numbers.

We say f(x) is O(g(x)) “f(x) is big-oh of g(x)”

If there are constants C and k such that

|f(x)| ≤ C|g(x)| whenever x > k.


Big Oh

f (n)≤ C|g(n)|

f (n) is O(g(n))

for k < n

k

f (n)

C g(n)


Big Omega

Let f and g be functions (Z→R or R→R)from the set of integers or

the set of real numbersto the set of real numbers.

We say f(x) is Ω(g(x)) “f(x) is big-omega of g(x)”

If there are constants C and k such that

C|g(x)| ≤ |f(x)| whenever x > k.


Big Oh

for k < n

k

f (n)C|g(n)|≤ f (n)

f (n) is Ω (g(n))C g(n)


Matrix Notation

C|g(n)|≤ f (n)

for k < n

f (n) is Ω (g(n))

f (n)≤ C|g(n)| f (n) is O(g(n))

C1|g(n)|≤ f (n)≤ C2|g(n)| f (n) is Θ (g(n))


Big-Oh, -Omega, -Theta Examples

for x >−0.5 1 x2 < x

2+2 x+1

for x > 2.414 x2+2 x+1 < 2 x

2

for x > 0.462 x2+2 x+1 < 10 x

2

for x > 7.873 10 x < x2+2 x+1

x2+2 x+1 is Ω (x2)

x2+2 x+1 is O (x2)

x2+2 x+1 is O (x2)

x2+2 x+1 is Ω (x )

x2+2 x+1 is Θ (x2)


Matrix Notation

for k < n

f (n) is Θ (g(n)) f (n) is O(g(n))

f (n) is Ω (g(n))f (n) is Θ (g(n))

f (n) is Θ (g(n)) f (n) is O(g(n))

f (n) is Ω (g(n))f (n) is Θ (g(n))


Big-O Examples

x2+2 x+1 is O (x2)

x2+2 x+1 is O (x3)

x2+2 x+1 is O (x4)

x2+2 x+1

2⋅x2x

3

x4


Big-Ω Examples

x2+2 x+1 is Ω (x )

x2+2 x+1 is Ω (√x )

x2+2 x+1 is Ω ( log x )

√x

x

x2+2 x+1

log x


Big-Θ Examples (1)

x2+2 x+1 is Θ (x2)

2⋅x2

1⋅x2

x2+2∗x+1

Zoom Out


Big-Theta Examples (2)

x2+2 x+1 is Θ (x2)

2⋅x2

1⋅x2x

2+2∗x+1


Common Growth Functions

x2 x log x x

√x

log x


References


Documents

Resolution (7A) - upload.wikimedia.org (8B) Boolean Functions 4 Young Won Lim 3/22/18 Truth Table and minterms (1) 0 1 1 0 1 0 0 0 1 0 0 0 1 1 1 1 1 0 1 0 1 1 0 0 the case …