Addition and Subtraction

How fast do we add (or subtract) two numbers n and m?

How fast can we add (or subtract) two numbers n and m?

Definition. Let A(d) denote the number of steps required toadd two numbers with d bits.

Theorem. A(d) ⇣ d.

Addition and Subtraction

How fast do we add (or subtract) two numbers n and m?

How fast can we add (or subtract) two numbers n and m?

Definition. Let A(d) denote the number of steps required toadd two numbers with d bits.

Theorem. A(d) ⇣ d.

Addition and Subtraction

How fast do we add (or subtract) two numbers n and m?

How fast can we add (or subtract) two numbers n and m?

Definition. Let A(d) denote the number of steps required toadd two numbers with d bits.

Theorem. A(d) ⇣ d.

Addition and Subtraction

How fast do we add (or subtract) two numbers n and m?

How fast can we add (or subtract) two numbers n and m?

Definition. Let A(d) denote the number of steps required toadd two numbers with d bits.

Theorem. A(d) ⇣ d.

Addition and Subtraction

How fast do we add (or subtract) two numbers n and m?

How fast can we add (or subtract) two numbers n and m?

Definition. Let A(d) denote the number of steps required toadd two numbers with d bits.

Theorem. A(d) ⇣ d.

Theorem. S(d) ⇣ d.

Definition. Let A(d) denote the maximal number of stepsrequired to add two numbers with d bits.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Attempt 1

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Attempt 1

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Attempt 1

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Attempt 1

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Attempt 2

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = ??????????????????????????????????, deduceM(d) 3M(r + 1) + O(r) (3 + ")M(r + 1).

• Hence, M(d) (3 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (3 + ")log2 d = dlog(3+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Attempt 2

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From

nm = anam22r+�(an+bn)(am+bm)�anam�bnbm

�2r+bnbm,

deduce M(d) 3M(r + 2) + O(r) (3 + ")M(r + 2).

• Hence, M(d) (3 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (3 + ")log2 d = dlog(3+")/ log 2.

Attempt 2

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From

nm = anam22r+�(an+bn)(am+bm)�anam�bnbm

�2r+bnbm,

deduce M(d) 3M(r + 2) + O(r) (3 + ")M(r + 2).

• Hence, M(d) (3 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (3 + ")log2 d = dlog(3+")/ log 2.

Attempt 2

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From

nm = anam22r+�(an+bn)(am+bm)�anam�bnbm

�2r+bnbm,

deduce M(d) 3M(r + 2) + O(r) (3 + ")M(r + 2).

• Hence, M(d) (3 + ")sM

�(d + 2s+2 � 4)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (3 + ")log2 d = dlog(3+")/ log 2.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

log 3

log 2= 1.5849625

Theorem. M(d) ⌧ d1.585.

Attempt 2

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = ??????????????????????????????????, deduceM(d) 3M(r + 1) + O(r) (3 + ")M(r + 1).

• Hence, M(d) (3 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (3 + ")log2 d = dlog(3+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

log 3

log 2= 1.5849625

Theorem. M(d) ⌧ d1.585.

Multiplication

How fast do we multiply two numbers n and m?

How fast can we multiply two numbers n and m?

Definition. Let M(d) denote the number of steps required tomultiply two numbers with d bits.

Theorem. M(d) ⌧ d2.

Can we do better? Yes

How can we see “easily” that something better is possible?

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

HW: Due January 31 (Friday)

Page 3, Problems 1 and 2

Page 5, unnumbered homework (first set)

(you may use (log 5/ log 3) + " instead of log 5/ log 3)

September 7

Attempt 1

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From nm = anbn22r + (anbm + ambn)2r + bnbm, deduceM(d) 4M(r + 1) + O(r) (4 + ")M(r + 1).

• Hence, M(d) (4 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (4 + ")log2 d = dlog(4+")/ log 2.

Theorem. M(d) ⌧ d (log d) log log d.

Attempt 2

• Suppose M(d) � d1.5.

• Let d be large, and let " > 0.

• Let n and m have d bits, and write n = an ⇥ 2r + bn

and m = am ⇥ 2r + bm, where r = bd/2c and the aj andbj are integers with bj < 2r.

• From

nm = anam22r+�(an+bn)(am+bm)�anam�bnbm

�2r+bnbm,

deduce M(d) 3M(r + 2) + O(r) (3 + ")M(r + 2).

• Hence, M(d) (3 + ")sM

�(d + 2s+1 � 2)/2s

�.

• Take s = blog2 dc � C (with C big). Then 2s � d/2C+1.

• Conclude, M(d) ⌧ (3 + ")log2 d = dlog(3+")/ log 2.

log 3

log 2= 1.5849625

Theorem. M(d) ⌧ d1.585.

Idea for Doing Better

log 3

log 2= 1.5849625

Theorem. M(d) ⌧ d1.585.

Idea for Doing Better

Think in terms of writing

n = an22r + bn2

r + cn and m = am22r + bm2r + cm,

where r = bd/3c.

How many multiplications does it take to expand nm ?

log 3

log 2= 1.5849625

Theorem. M(d) ⌧ d1.585.

Idea for Doing Better

Think in terms of writing

n = an22r + bn2

r + cn and m = am22r + bm2r + cm,

where r = bd/3c.

How many multiplications does it take to expand nm ?

Homework

(3) The value of f(n) =nX

k=1

1/k can be estimated by com-

paring it’s value to an integral. For example, by comparingthe sum of the areas of the rectangles indicated in the graphbelow with the area under the graph of y = 1/x, one obtains

f(9) �Z 10

1

1

xdx = log 10.

(c) Prove that f(n) ⇠ log n.

Theorem. M(d) ⌧ d (log d) log log d.Theorem. For every " > 0, we have M(d) ⌧" d

1+".

Theorem. Given distinct numbers x0, x1, . . . , xk and num-bers y0, y1, . . . , yk, there is a unique polynomial f of degree k such that f(xj) = yj for all j.

Theorem. For every " > 0, we have M(d) ⌧" d1+".

Theorem. Given distinct numbers x0, x1, . . . , xk and num-bers y0, y1, . . . , yk, there is a unique polynomial f of degree k such that f(xj) = yj for all j.

Lagrange Interpolation:

f(x) =kX

i=0

✓ Y

0jk

j 6=i

x � xj

xi � xj

◆yi

Theorem. For every " > 0, we have M(d) ⌧" d1+".

Theorem. Given distinct numbers x0, x1, . . . , xk and num-bers y0, y1, . . . , yk, there is a unique polynomial f of degree k such that f(xj) = yj for all j.

Lagrange Interpolation:

f(x) =kX

i=0

✓ Y

0jk

j 6=i

x � xj

xi � xj

◆yi

Theorem. For every " > 0, we have M(d) ⌧" d1+".

Theorem. Given distinct numbers x0, x1, . . . , xk and num-bers y0, y1, . . . , yk, there is a unique polynomial f of degree k such that f(xj) = yj for all j.

Lagrange Interpolation:

f(x) =kX

i=0

✓ Y

0jk

j 6=i

x � xj

xi � xj

◆yi

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where

f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where

f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

Theorem. For every " > 0, we have M(d) ⌧" d1+".

Theorem. Given distinct numbers x0, x1, . . . , xk and num-bers y0, y1, . . . , yk, there is a unique polynomial f of degree k such that f(xj) = yj for all j.

Lagrange Interpolation:

f(x) =kX

i=0

✓ Y

0jk

j 6=i

x � xj

xi � xj

◆yi

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where

f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where

f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where

f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

Theorem. For every " > 0, we have M(d) ⌧" d1+".

Theorem. Given distinct numbers x0, x1, . . . , xk and num-bers y0, y1, . . . , yk, there is a unique polynomial f of degree k such that f(xj) = yj for all j.

Lagrange Interpolation:

f(x) =kX

i=0

✓ Y

0jk

j 6=i

x � xj

xi � xj

◆yi

Theorem. For every " > 0, we have M(d) ⌧" d1+".

Theorem. Given distinct numbers x0, x1, . . . , xk and num-bers y0, y1, . . . , yk, there is a unique polynomial f of degree k such that f(xj) = yj for all j.

Lagrange Interpolation:

f(x) =kX

i=0

✓ Y

0jk

j 6=i

x � xj

xi � xj

◆yi

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where

f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where

f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

• Suppose n and m have kr digits. Write

n =

k�1X

u=0

au2ur

and m =

k�1X

v=0

bv2vr

.

• Then nm = f(2r), where

f(x) =

✓ k�1X

u=0

auxu

◆✓ k�1X

v=0

bvxv

◆.

• Compute the 2k�1 numbers yj = f(j), for 0 j 2k�2,

using 2k�1 multiplications of two r+ck digit numbers.

• Compute the coe�cients of f(x) expanded, using Lagrange

interpolation, in Ok(r) steps.

• Deduce M(kr) (2k � 1)M(r + ck) + c0kr so that

M(d) ⌧ (2k � 1)logk d ⌧ d

log(2k�1)/ log k.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Division

Problem: Given two positive integers n and m, determine

the quotient q and the remainder r when n is divided by m.

These should be integers satisfying

n = mq + r and 0 r < m.

Definition. Let M0(d) denote an upper bound on the number

of steps required to multiply two numbers with d bits. Let

D0(d) denote an upper bound on the number of steps required

to obtain q and r given n and m each have d binary digits.

Theorem. Suppose M0(d) has the form df(d) where f(d) is

an increasing function of d. Then D0(d) ⌧ M

0(d).

We need only compute 1/m to su�cient accuracy.

Suppose n and m have s digits. If 1/m = 0.d1d2d3d4...

(base 2) with d1, . . . , ds known, then

n

m=

1

2s(n ⇥ d1d2 . . . ds) + ✓, where 0 ✓ 1.

Write this in the form

n

m=

1

2s(q

02

s+ q

00) + ✓,

so n = mq0+✓

0where 0 ✓

0< 2m. Try q = q

0and q = q

0+1.

Newton’s Method

Say we want to compute 1/m. Take a function f(x) which

has root 1/m. If x0

is an approximation to the root, then

how can we get a better approximation? Take f(x) = m �1/x. Starting with x

0= x0, this leads to the approximations

xn+1 = 2xn � mx2

n.

Note that if xn = (1 � ")/m, then xn+1 = (1 � "2)/m.

Newton’s Method

Say we want to compute 1/m. Take a function f(x) which

has root 1/m. If x0

is an approximation to the root, then

how can we get a better approximation? Take

f(x) = m � 1/x.

Starting with x0= x0, this leads to the approximations

xn+1 = 2xn � mx2

n.

Note that if xn = (1 � ")/m, then xn+1 = (1 � "2)/m.

Newton’s Method

Say we want to compute 1/m. Take a function f(x) which

has root 1/m. If x0

is an approximation to the root, then

how can we get a better approximation? Take f(x) = m �1/x. Starting with x

0= x0, this leads to the approximations

xn+1 = 2xn � mx2

n.

Note that if xn = (1 � ")/m, then xn+1 = (1 � "2)/m.

Newton’s Method

Say we want to compute 1/m. Take a function f(x) which

has root 1/m. If x0

is an approximation to the root, then

how can we get a better approximation? Take

f(x) = m � 1/x.

Starting with x0= x0, this leads to the approximations

xn+1 = 2xn � mx2

n.

Note that if xn = (1 � ")/m, then xn+1 = (1 � "2)/m.

Newton’s Method

Say we want to compute 1/m. Take a function f(x) which

has root 1/m. If x0

is an approximation to the root, then

how can we get a better approximation? Take

f(x) = m � 1/x.

Starting with x0= x0, this leads to the approximations

xn+1 = 2xn � mx2

n.

Note that if xn = (1 � ")/m, then xn+1 = (1 � "2)/m.

Algorithm from Knuth, Vol. 2, pp. 295-6

Algorithm R. Let v in binary be v = (0.v1v2v3 . . . )2, with

v1 = 1. The algorithm outputs z satisfying

|z � 1/v| 2�n

.

R1. [Initialize] Set z 1

4b32/(4v1 + 2v2 + v3)c and k 0.

R2. [Newton iteration] (At this point, z 2 has the binary

form (⇤⇤.⇤⇤ · · · ⇤)2 with 2k+1 places after the radix point.)

Calculate z2

exactly. Then calculate Vkz2

exactly, where

Vk = (0.v1v2 . . . v2k+1+3

)2. Then set z 2z � Vkz2

+ r,

where 0 r < 2�2

k+1�1is added if needed to “round up”

z so that it is a multiple of 2�2

k+1�1. Finally, set k k+1.

R3. [End Test] If 2k

< n, go back to step R2; otherwise the

algorithm terminates.

Algorithm from Knuth, Vol. 2, pp. 295-6

Algorithm R. Let v in binary be v = (0.v1v2v3 . . . )2, with

v1 = 1. The algorithm outputs z satisfying

|z � 1/v| 2�n

.

R1. [Initialize] Set z 1

4b32/(4v1 + 2v2 + v3)c and k 0.

R2. [Newton iteration] (At this point, z 2 has the binary

form (⇤⇤.⇤⇤ · · · ⇤)2 with 2k+1 places after the radix point.)

Calculate z2

exactly. Then calculate Vkz2

exactly, where

Vk = (0.v1v2 . . . v2k+1+3

)2. Then set z 2z � Vkz2

+ r,

where 0 r < 2�2

k+1�1is added if needed to “round up”

z so that it is a multiple of 2�2

k+1�1. Finally, set k k+1.

R3. [End Test] If 2k

< n, go back to step R2; otherwise the

algorithm terminates.

Algorithm from Knuth, Vol. 2, pp. 295-6

Algorithm R. Let v in binary be v = (0.v1v2v3 . . . )2, with

v1 = 1. The algorithm outputs z satisfying

|z � 1/v| 2�n

.

R1. [Initialize] Set z 1

4b32/(4v1 + 2v2 + v3)c and k 0.

R2. [Newton iteration] (At this point, z 2 has the binary

form (⇤⇤.⇤⇤ · · · ⇤)2 with 2k+1 places after the radix point.)

Calculate z2

exactly. Then calculate Vkz2

exactly, where

Vk = (0.v1v2 . . . v2k+1+3

)2. Then set z 2z � Vkz2

+ r,

where 0 r < 2�2

k+1�1is added if needed to “round up”

z so that it is a multiple of 2�2

k+1�1. Finally, set k k+1.

R3. [End Test] If 2k

< n, go back to step R2; otherwise the

algorithm terminates.

z 2 [0, 2]