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Inverses

Today: finding inverses quickly.

Euclid’s Algorithm.Runtime.

Euclid’s Extended Algorithm.

Inverses

Euclid’s Algorithm.

Runtime.Euclid’s Extended Algorithm.

Inverses

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Does 2 have an inverse mod 8?

Does 2 have an inverse mod 9? Yes. 52(5) = 10 = 1 mod 9.

Does 6 have an inverse mod 9? No.

x has an inverse modulo m if and only ifgcd(x ,m)> 1? No.gcd(x ,m) = 1? Yes.

Today:Compute gcd!Compute Inverse modulo m.

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Yes. 52(5) = 10 = 1 mod 9.

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Does 2 have an inverse mod 9? Yes.

52(5) = 10 = 1 mod 9.

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Does 2 have an inverse mod 9? Yes. 5

2(5) = 10 = 1 mod 9.

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Does 2 have an inverse mod 9? Yes. 5

2(5) = 10 = 1 mod 9.

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x has an inverse modulo m if and only if

gcd(x ,m)> 1? No.gcd(x ,m) = 1? Yes.

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x has an inverse modulo m if and only ifgcd(x ,m)> 1?

No.gcd(x ,m) = 1? Yes.

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x has an inverse modulo m if and only ifgcd(x ,m)> 1? No.gcd(x ,m) = 1?

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Today:Compute gcd!

Compute Inverse modulo m.

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Divisibility...

Notation: d |x means “d divides x” or

x = kd for some integer k .

x = `d and y = kd

=⇒ x−y = kd − `d = (k − `)d =⇒ d |(x−y)

Divisibility...

Notation: d |x means “d divides x” orx = kd for some integer k .

x = `d and y = kd

=⇒ x−y = kd − `d = (k − `)d =⇒ d |(x−y)

Divisibility...

Fact: If d |x and d |y then d |(x +y) and d |(x−y).

Proof: d |x and d |y orx = `d and y = kd

=⇒ x−y = kd − `d = (k − `)d =⇒ d |(x−y)

Divisibility...

x = `d and y = kd

=⇒ x−y = kd − `d = (k − `)d =⇒ d |(x−y)

Divisibility...

x = `d and y = kd

=⇒ x−y = kd − `d = (k − `)d =⇒ d |(x−y)

Divisibility...

x = `d and y = kd

=⇒ x−y = kd − `d

= (k − `)d =⇒ d |(x−y)

Divisibility...

x = `d and y = kd

=⇒ x−y = kd − `d = (k − `)d

=⇒ d |(x−y)

Divisibility...

x = `d and y = kd

=⇒ x−y = kd − `d = (k − `)d =⇒ d |(x−y)

Divisibility...

x = `d and y = kd

=⇒ x−y = kd − `d = (k − `)d =⇒ d |(x−y)

More divisibility

Lemma 1: If d |x and d |y then d |y and d | mod (x ,y).Proof: mod (x ,y) = x−bx/yc ·y

= x−s ·y for integer s= kd −s`d for integers k , `= (k −s`)d

Therefore d | mod (x ,y). And d |y since it is in condition.

Lemma 2: If d |y and d | mod (x ,y) then d |y and d |x .Proof...: Similar. Try this at home. .

GCD Mod Corollary: gcd(x ,y) = gcd(y , mod (x ,y)).Proof: x and y have same set of common divisors as x andmod (x ,y) by Lemma.Same common divisors =⇒ largest is the same.

More divisibility

Lemma 1: If d |x and d |y then d |y and d | mod (x ,y).

Proof: mod (x ,y) = x−bx/yc ·y= x−s ·y for integer s= kd −s`d for integers k , `= (k −s`)d

More divisibility

= x−s ·y for integer s

= kd −s`d for integers k , `= (k −s`)d

More divisibility

= x−s ·y for integer s= kd −s`d for integers k , `

= (k −s`)d

More divisibility

Therefore d | mod (x ,y).

And d |y since it is in condition.

More divisibility

Lemma 2: If d |y and d | mod (x ,y) then d |y and d |x .Proof...: Similar.

Try this at home. .

More divisibility

Lemma 2: If d |y and d | mod (x ,y) then d |y and d |x .Proof...: Similar. Try this at home.

More divisibility

GCD Mod Corollary: gcd(x ,y) = gcd(y , mod (x ,y)).

Proof: x and y have same set of common divisors as x andmod (x ,y) by Lemma.Same common divisors =⇒ largest is the same.

More divisibility

GCD Mod Corollary: gcd(x ,y) = gcd(y , mod (x ,y)).Proof: x and y have same set of common divisors as x andmod (x ,y) by Lemma.

Same common divisors =⇒ largest is the same.

More divisibility

Euclid’s algorithm.GCD Mod Corollary: gcd(x ,y) = gcd(y , mod (x ,y)).

gcd (x, y)if (y = 0) then

return xelse

return gcd(y, mod(x, y)) ***

Theorem: Euclid’s algorithm computes the greatest common divisorof x and y if x ≥ y .

Proof: Use Strong Induction.Base Case: y = 0, “x divides y and x”

=⇒ “x is common divisor and clearly largest.”Induction Step: mod (x ,y)< y ≤ x when x ≥ y

call in line (***) meets conditions plus arguments “smaller”and by strong induction hypothesiscomputes gcd(y , mod (x ,y))

which is gcd(x ,y) by GCD Mod Corollary.

return xelse

Proof: Use Strong Induction.

Base Case: y = 0, “x divides y and x”=⇒ “x is common divisor and clearly largest.”

Induction Step: mod (x ,y)< y ≤ x when x ≥ y

return xelse

=⇒ “x is common divisor and clearly largest.”

Induction Step: mod (x ,y)< y ≤ x when x ≥ y

return xelse

call in line (***) meets conditions plus arguments “smaller”

and by strong induction hypothesiscomputes gcd(y , mod (x ,y))

return xelse

call in line (***) meets conditions plus arguments “smaller”and by strong induction hypothesis

computes gcd(y , mod (x ,y))which is gcd(x ,y) by GCD Mod Corollary.

return xelse

Excursion: Value and Size.

Before discussing running time of gcd procedure...

What is the value of 1,000,000?

one million or 1,000,000!

What is the “size” of 1,000,000?

Number of digits: 7.

Number of bits: 21.

For a number x , what is its size in bits?

n = b(x)≈ log2 x

Number of bits: 21.

n = b(x)≈ log2 x

Number of bits: 21.

n = b(x)≈ log2 x

Number of bits: 21.

n = b(x)≈ log2 x

Number of bits: 21.

n = b(x)≈ log2 x

Number of bits: 21.

n = b(x)≈ log2 x

Number of bits: 21.

n = b(x)≈ log2 x

Number of bits: 21.

n = b(x)≈ log2 x

Number of bits: 21.

n = b(x)≈ log2 x

GCD procedure is fast.

Theorem: GCD uses 2n “divisions” where n is the number of bits.

Is this good? Better than trying all numbers in {2, . . .y/2}?Check 2, check 3, check 4, check 5 . . . , check y/2.

2n−1 divisions! Exponential dependence on size!

101 bit number. 2100 ≈ 1030 = “million, trillion, trillion” divisions!

2n is much faster! .. roughly 200 divisions.

Is this good?

Better than trying all numbers in {2, . . .y/2}?Check 2, check 3, check 4, check 5 . . . , check y/2.

Is this good? Better than trying all numbers in {2, . . .y/2}?

Check 2, check 3, check 4, check 5 . . . , check y/2.

Is this good? Better than trying all numbers in {2, . . .y/2}?Check 2,

check 3, check 4, check 5 . . . , check y/2.

Is this good? Better than trying all numbers in {2, . . .y/2}?Check 2, check 3,

check 4, check 5 . . . , check y/2.

Is this good? Better than trying all numbers in {2, . . .y/2}?Check 2, check 3, check 4,

check 5 . . . , check y/2.

101 bit number.

2100 ≈ 1030 = “million, trillion, trillion” divisions!

2n is much faster!

.. roughly 200 divisions.

Algorithms at work.

Trying everything

“gcd(x, y)” at work.

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)gcd(12, 4)

gcd(4, 0)4

Notice: The first argument decreases rapidly.At least a factor of 2 in two recursive calls.

(The second is less than the first.)

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)gcd(12, 4)

gcd(4, 0)4

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)gcd(12, 4)

gcd(4, 0)4

Algorithms at work.

Trying everything

gcd(700,568)

gcd(568, 132)gcd(132, 40)

gcd(40, 12)gcd(12, 4)

gcd(4, 0)4

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)gcd(12, 4)

gcd(4, 0)4

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)

gcd(40, 12)gcd(12, 4)

gcd(4, 0)4

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)

gcd(12, 4)gcd(4, 0)

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)

gcd(12, 4)

gcd(4, 0)4

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)

gcd(12, 4)gcd(4, 0)

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)

gcd(12, 4)gcd(4, 0)

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)

gcd(12, 4)gcd(4, 0)

Notice: The first argument decreases rapidly.

At least a factor of 2 in two recursive calls.

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)

gcd(12, 4)gcd(4, 0)

Algorithms at work.

Trying everything

gcd(700,568)gcd(568, 132)

gcd(132, 40)gcd(40, 12)

gcd(12, 4)gcd(4, 0)

Proof.gcd (x, y)

if (y = 0) thenreturn x

elsereturn gcd(y, mod(x, y))

Theorem: GCD uses O(n) ”divisions” where n is the number of bits.

Proof:

Fact:First arg decreases by at least factor of two in two recursive calls.

After 2 log2 x = O(n) recursive calls, argument x is 1 bit number.One more recursive call to finish.1 division per recursive call.O(n) divisions.

Proof of Fact: Recall that first argument decreases every call.

Case 1: y ≤ x/2, first argument is y=⇒ true in one recursive call;

Case 2: Will show “y > x/2” =⇒ “mod(x ,y)≤ x/2.”

mod (x ,y) is second argument in next recursive call,and becomes the first argument in the next one.

When y > x/2, then

bxyc= 1,

mod (x ,y) = x−ybxyc= x−y ≤x−x/2 = x/2

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

After 2 log2 x = O(n) recursive calls, argument x is 1 bit number.

One more recursive call to finish.1 division per recursive call.O(n) divisions.

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

After 2 log2 x = O(n) recursive calls, argument x is 1 bit number.One more recursive call to finish.

1 division per recursive call.O(n) divisions.

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

After 2 log2 x = O(n) recursive calls, argument x is 1 bit number.One more recursive call to finish.1 division per recursive call.

O(n) divisions.

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

mod (x ,y) is second argument in next recursive call,

and becomes the first argument in the next one.

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

mod (x ,y) = x−ybxyc=

x−y ≤x−x/2 = x/2

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

mod (x ,y) = x−ybxyc= x−y ≤x−x/2

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Proof.gcd (x, y)

Proof:

When y > x/2, then

bxyc= 1,

Finding an inverse?

We showed how to efficiently tell if there is an inverse.

Extend Euclid’s algo to find inverse.

Finding an inverse?

We showed how to efficiently tell if there is an inverse.

Extend Euclid’s algo to find inverse.

Euclid’s GCD algorithm.

return xelse

return gcd(y, mod(x, y))

Computes the gcd(x ,y) in O(n) divisions.

For x and m, if gcd(x ,m) = 1 then x has an inverse modulo m.

return xelse

Multiplicative Inverse.

GCD algorithm used to tell if there is a multiplicative inverse.

How do we find a multiplicative inverse?

Multiplicative Inverse.

GCD algorithm used to tell if there is a multiplicative inverse.

How do we find a multiplicative inverse?

Extended GCD

Euclid’s Extended GCD Theorem: For any x ,y there are integersa,b such that

ax +by = gcd(x ,y)

= d where d = gcd(x ,y).

“Make d out of sum of multiples of x and y .”

What is multiplicative inverse of x modulo m?

By extended GCD theorem, when gcd(x ,m) = 1.

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

So a multiplicative inverse of x if gcd(a,x) = 1!!Example: For x = 12 and y = 35 , gcd(12,35) = 1.

(3)12+(−1)35 = 1.

a = 3 and b =−1.The multiplicative inverse of 12 (mod 35) is 3.

Extended GCD

ax +by = gcd(x ,y) = d where d = gcd(x ,y).

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1

ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

So a multiplicative inverse of x if gcd(a,x) = 1!!

Example: For x = 12 and y = 35 , gcd(12,35) = 1.

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

a = 3 and b =−1.

The multiplicative inverse of 12 (mod 35) is 3.

Extended GCD

ax +bm = 1ax ≡ 1−bm ≡ 1 (mod m).

(3)12+(−1)35 = 1.

Make d out of x and y ..?

gcd(35,12)

gcd(12, 11) ;; gcd(12, 35%12)gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

How did gcd get 11 from 35 and 12?35−b35

12c12 = 35− (2)12 = 11

How does gcd get 1 from 12 and 11?12−b12

11c11 = 12− (1)11 = 1

Algorithm finally returns 1.

But we want 1 from sum of multiples of 35 and 12?

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

Get 11 from 35 and 12 and plugin.... Simplify. a = 3 and b =−1.

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)

gcd(1,0)1

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

How did gcd get 11 from 35 and 12?

35−b3512c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

How does gcd get 1 from 12 and 11?

12−b1211c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.

1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11

= 12− (1)(35− (2)12)

= (3)12+(−1)35

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11 = 12− (1)(35− (2)12)

= (3)12+(−1)35

Get 11 from 35 and 12 and plugin....

Simplify. a = 3 and b =−1.

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11 = 12− (1)(35− (2)12) = (3)12+(−1)35

Get 11 from 35 and 12 and plugin.... Simplify.

a = 3 and b =−1.

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11 = 12− (1)(35− (2)12) = (3)12+(−1)35

Get 11 from 35 and 12 and plugin.... Simplify.

a = 3 and b =−1.

gcd(35,12)gcd(12, 11) ;; gcd(12, 35%12)

gcd(11, 1) ;; gcd(11, 12%11)gcd(1,0)

12c12 = 35− (2)12 = 11

11c11 = 12− (1)11 = 1

Get 1 from 12 and 11.1 = 12− (1)11 = 12− (1)(35− (2)12) = (3)12+(−1)35

Extended GCD Algorithm.

ext-gcd(x,y)if y = 0 then return(x, 1, 0)

else(d, a, b) := ext-gcd(y, mod(x,y))return (d, b, a - floor(x/y) * b)

Claim: Returns (d ,a,b): d = gcd(a,b) and d = ax +by .Example:

a−bx/yc ·b =

1−b11/1c ·0 = 10−b12/11c ·1 =−11−b35/12c · (−1) = 3

ext-gcd(35,12)ext-gcd(12, 11)

ext-gcd(11, 1)ext-gcd(1,0)return (1,1,0) ;; 1 = (1)1 + (0) 0

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

Claim: Returns (d ,a,b): d = gcd(a,b) and d = ax +by .

Example:

a−bx/yc ·b =

1−b11/1c ·0 = 10−b12/11c ·1 =−11−b35/12c · (−1) = 3

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

a−bx/yc ·b =

1−b11/1c ·0 = 10−b12/11c ·1 =−11−b35/12c · (−1) = 3

ext-gcd(35,12)

ext-gcd(12, 11)ext-gcd(11, 1)

ext-gcd(1,0)return (1,1,0) ;; 1 = (1)1 + (0) 0

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

a−bx/yc ·b =

1−b11/1c ·0 = 10−b12/11c ·1 =−11−b35/12c · (−1) = 3

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

a−bx/yc ·b =

1−b11/1c ·0 = 10−b12/11c ·1 =−11−b35/12c · (−1) = 3

ext-gcd(11, 1)

ext-gcd(1,0)return (1,1,0) ;; 1 = (1)1 + (0) 0

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

a−bx/yc ·b =

1−b11/1c ·0 = 10−b12/11c ·1 =−11−b35/12c · (−1) = 3

ext-gcd(11, 1)ext-gcd(1,0)

return (1,1,0) ;; 1 = (1)1 + (0) 0return (1,0,1) ;; 1 = (0)11 + (1)1

return (1,1,-1) ;; 1 = (1)12 + (-1)11return (1,-1, 3) ;; 1 = (-1)35 +(3)12

Claim: Returns (d ,a,b): d = gcd(a,b) and d = ax +by .Example: a−bx/yc ·b =

1−b11/1c ·0 = 10−b12/11c ·1 =−11−b35/12c · (−1) = 3

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

Claim: Returns (d ,a,b): d = gcd(a,b) and d = ax +by .Example: a−bx/yc ·b =1−b11/1c ·0 = 1

0−b12/11c ·1 =−11−b35/12c · (−1) = 3

return (1,0,1) ;; 1 = (0)11 + (1)1

return (1,1,-1) ;; 1 = (1)12 + (-1)11return (1,-1, 3) ;; 1 = (-1)35 +(3)12

1−b11/1c ·0 = 1

0−b12/11c ·1 =−1

1−b35/12c · (−1) = 3

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

1−b11/1c ·0 = 10−b12/11c ·1 =−1

1−b35/12c · (−1) = 3

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

a−bx/yc ·b =

1−b11/1c ·0 = 10−b12/11c ·1 =−11−b35/12c · (−1) = 3

return (1,0,1) ;; 1 = (0)11 + (1)1return (1,1,-1) ;; 1 = (1)12 + (-1)11

return (1,-1, 3) ;; 1 = (-1)35 +(3)12

Theorem: Returns (d ,a,b), where d = gcd(a,b) and

d = ax +by .

Theorem: Returns (d ,a,b), where d = gcd(a,b) and

d = ax +by .

Correctness.

Proof: Strong Induction.1

Base: ext-gcd(x ,0) returns (d = x ,1,0) with x = (1)x +(0)y .

Induction Step: Returns (d ,A,B) with d = Ax +ByInd hyp: ext-gcd(y , mod (x ,y)) returns (d∗,a,b) with

d∗ = ay +b( mod (x ,y))

ext-gcd(x ,y) calls ext-gcd(y , mod (x ,y)) so

d = d∗ = ay +b · ( mod (x ,y))

= ay +b · (x−bxycy)

= bx +(a−bxyc ·b)y

And ext-gcd returns (d ,b,(a−b xy c ·b)) so theorem holds!

1Assume d is gcd(x ,y) by previous proof.

Correctness.

d∗ = ay +b( mod (x ,y))

d = d∗ = ay +b · ( mod (x ,y))

Correctness.

d∗ = ay +b( mod (x ,y))

d = d∗ = ay +b · ( mod (x ,y))

Correctness.

d∗ = ay +b( mod (x ,y))

d = d∗ = ay +b · ( mod (x ,y))

Correctness.

d∗ = ay +b( mod (x ,y))

d = d∗ = ay +b · ( mod (x ,y))

Correctness.

d∗ = ay +b( mod (x ,y))

d = d∗ = ay +b · ( mod (x ,y))

Correctness.

d∗ = ay +b( mod (x ,y))

d = d∗ = ay +b · ( mod (x ,y))

Correctness.

d∗ = ay +b( mod (x ,y))

d = d∗ = ay +b · ( mod (x ,y))

Correctness.

d∗ = ay +b( mod (x ,y))

d = d∗ = ay +b · ( mod (x ,y))

Review Proof: step.

Recursively: d = ay +b(x−b xy c ·y) =⇒ d = bx− (a−b x

y cb)y

Returns (d ,b,(a−b xy c ·b)).

Review Proof: step.

Recursively: d = ay +b(x−b xy c ·y)

=⇒ d = bx− (a−b xy cb)y

Review Proof: step.

y cb)y

Review Proof: step.

y cb)y

Wrap-up

Conclusion: Can find multiplicative inverses in O(n) time!

Very different from elementary school: try 1, try 2, try 3...

Inverse of 500,000,357 modulo 1,000,000,000,000? ≤ 80divisions.

versus 1,000,000

Internet Security.Public Key Cryptography: 512 digits.

512 divisions vs.(10000000000000000000000000000000000000000000)5 divisions.

Next lecture!

Wrap-up

versus 1,000,000

Next lecture!

Wrap-up

versus 1,000,000

Next lecture!

Wrap-up

Inverse of 500,000,357 modulo 1,000,000,000,000?

≤ 80divisions.

versus 1,000,000

Next lecture!

Wrap-up

versus 1,000,000

Next lecture!

Wrap-up

versus 1,000,000

Next lecture!

Wrap-up

versus 1,000,000

Next lecture!

Wrap-up

versus 1,000,000

Internet Security.

Public Key Cryptography: 512 digits.512 divisions vs.(10000000000000000000000000000000000000000000)5 divisions.

Next lecture!

Wrap-up

versus 1,000,000

Next lecture!

Wrap-up

versus 1,000,000

Internet Security.Public Key Cryptography: 512 digits.512 divisions vs.

(10000000000000000000000000000000000000000000)5 divisions.

Next lecture!

Wrap-up

versus 1,000,000

Internet Security.Public Key Cryptography: 512 digits.512 divisions vs.(10000000000000000000000000000000000000000000)5 divisions.

Next lecture!

Wrap-up

versus 1,000,000

Internet Security.Public Key Cryptography: 512 digits.512 divisions vs.(10000000000000000000000000000000000000000000)5 divisions.

Next lecture!

Inverses - EECS Instructional Support Group Home Pagecs70/fa16/static/slides/lec-9.pdfRefresh Does 2...

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