unary, binary, and beyond
DESCRIPTION
Unary, Binary, and Beyond. X n – 1. 1 + X 1 + X 2 + X 3 + … + X n-2 + X n-1 =. X - 1. We are going to need this fundamental sum: The Geometric Series. A Frequently Arising Calculation. ( X -1 ) ( 1 + X 1 + X 2 + X 3 + … + X n-2 + X n-1 ) - PowerPoint PPT PresentationTRANSCRIPT
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Unary, Binary, and Beyond
Great Theoretical Ideas In Computer Science
Steven Rudich
CS 15-251 Spring 2004
Lecture 3 Jan 20, 2004 Carnegie Mellon University
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We are going to need this fundamental
sum:
The Geometric Series
1 + X1 + X11 + X + X22 + X + X 33 + … + X + … + Xn-2n-2 + X + Xn-1n-1 ==
XXn n – 1 – 1 XX - -
11
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A Frequently Arising Calculation
(X-1) ( 1 + X1 + X2 + X 3 + … + Xn-2 + Xn-1 )
= X1 + X2 + X 3 + … + Xn-1 + Xn
- 1 - X1 - X2 - X 3 - … - Xn-2 – Xn-1
= - 1 + Xn
= Xn - 1
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The Geometric Series
(X-1) ( 1 + X1 + X2 + X 3 + … + Xn-1 ) = Xn - 1
1 + X1 + X11 + X + X22 + X + X 33 + … + X + … + Xn-2n-2 + X + Xn-1n-1 ==
XXn n – 1 – 1 XX - -
11 when X1
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Last time we talked about unary
notation.
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nth Triangular Number
n = 1 + 2 + 3 + . . . + n-1 + n
= n(n+1)/2
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nth Square Number
n = 1 + 3 + … + 2n-1
= Sum of first n odd numbers
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nth Square Number
n = n + n-1
= n2
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(( nn))2 =2 = (( n-1n-1))2 2
++nn
(( nn))2 2 = + + . . . = + + . . . + +
nn
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We will define sequences of
symbols that give us a unary
representation of the Natural number.
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First we define the general language
we use to talk about strings.
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Strings Of Symbols.
We take the idea of symbol and sequence of symbols as primitive.
Let be any fixed finite set of symbols. is called an alphabet, or a set of symbols. Examples: = {0,1,2,3,4}= {a,b,c,d, …, z}= all typewriter symbols.
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Strings over the alphabet
A string is a sequence of symbols from . Let s and t be strings. Let st denote the concatenation of s and t, i.e., the string obtained by the string s followed by the string t.
Define + by the following inductive rules: x2 ) x2 +
s,t 2 + ) st 2 +
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Intuitively, + is the set of all finite strings that we can
make using (at least one) letters
from .
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Define be the empty string. I.e.,XY= XY for all strings X and Y. is also called the string of length 0.
Define 0 = { }
Define * = + [ {}
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Intuitively, * is the set of all finite
strings that we can make using letters from including
the empty string.
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The Natural Numbers
= { 0, 1, 2, 3, . . .}
Notice that we include
0 as a Natural number.
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“God Made The Natural Numbers. Everything Else Is The Work Of
Man.”
= { 0, 1, 2, 3, . . .}
Kronecker
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Last Time: Unary Notation
1
2
3
4
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To handle the notation for zero, we introduce a
small variation on unary.
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Peano Unary (PU)
0 01 S02 SS03 SSS04 SSSS05 SSSSS06 SSSSSS0
Giuseppe Peano [1889]
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Each number is a sequence of symbols in {S, 0}+
0 01 S02 SS03 SSS04 SSSS05 SSSSS06 SSSSSS0
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Peano Unary Representation of Natural Number
= { 0, S0, SS0, SSS0, . . .}
0 is a natural number called zero.Set notation: 0 2
If X is a natural number, then SX is a natural number called successor of X.Set notation: X 2 ) SX 2
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Inductive Definition of +
= { 0, S0, SS0, SSS0, . . .}
Inductive definition of addition (+):
X, Y 2 )X “+” 0 = XX “+” SY = S(X”+”Y)
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S0 + S0 = SS0 (i.e., “1+1=2”)
Proof:
S0 + S0 = S(S0 + 0) = S(S0) = SS0
X, Y 2 )X “+” 0 = XX “+” SY = S(X”+”Y)
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Inductive Definition of *
= { 0, S0, SS0, SSS0, . . .}Inductive definition of times (*):
X, Y 2 )X “*” 0 = 0X “*” SY = (X”*”Y) + X
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Inductive Definition of ^
= { 0, S0, SS0, SSS0, . . .}
Inductive definition of raised to the (^):
X, Y 2 )X “^” 0 = 1 [or X0 = 1 ]X “^” SY = (X”^”Y) * X [or XSY = XY * X]
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= { 0, S0, SS0, SSS0, . . .}
Defining < for :
8 x,y 2 “x > y” is TRUE “y < x” is FALSE“x > y” is TRUE “y > x” is FALSE
“x+1 > 0” is TRUE“x+1 > y+1” is TRUE ) “x > y” is TRUE
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= { 0, 1, 2, 3, . . .}
Defining partial minus for :
8 x,y 2 x-0 = x x>y )
(x+1) – (y+1) = x-y
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a = [a DIV b]*b + [a MOD b]
Defining DIV and MOD for :
8 a,b 2 a<b )
a DIV b = 0 a¸b>0 )
a DIV b = 1 + (a-b) DIV b
a MOD b = a – [b*(a DIV b)]
The maximum number of times b goes into a without going over.
The remainder when a is divided by b.
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45 = [45 DIV 10]*10 + [ 45 MOD 10]= 4*10 + 5
Defining DIV and MOD for :
8 a,b 2 a<b )
a DIV b = 0 a¸b>0 )
a DIV b = 1 + (a-b) DIV b
a MOD b = a – [b*(a DIV b)]
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We have defined the Peano
representation of the Naturals, with a notion of +, *, ^,
<, -, DIV, and MOD.
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PU takes size n+1 to represent the
number n.
Higher bases are much more compact.
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1000000 in Peano Unary takes one
million one symbols to write.
1000000 only takes 7 symbols to write in base 10.
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Let’s define base 10 representation of PU numbers.
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Let = {0,
1,2,3,4,5,6,7,8,9} be our symbol
alphabet.
Any string in + will be called a
decimal number.
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Let X be a decimal number. Let’s define
the length of X inductively:
length() = 0
X= aY, a2 , Y2 * ) length(X) = S(length(Y))
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Let X be decimal. Let n (unary) =
length(X).
For each unary i<n, we want to be able to talk about the ith symbol of X.
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The ith symbol of X.Defining rule:
X = PaS where P,S2 * and a2.
i=length(S). ) a is the ith symbol of X.
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For any string X, we can define its length n2 PU. For all i2 PU
s.t. i<n, we can define the ith symbol
ai.
X = an-1 an-2 …a0
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We define a base conversion
function, Base10 to 1
(X), to convert any decimal X to its
unary representation.
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Initial Cases, length(X)=1
Base10 to 1 (0) = 0
Base10 to 1 (1) = S0
Base10 to 1 (2) = SS0
Base10 to 1 (3) = SSS0
……Base10 to 1 (9) = SSSSSSSSS0
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Suppose n=length(X)>S0
For all i<n, let ai be the ith symbol of X.
Hence, X = an-1 an-2 …a0 .
Define Base10 to 1(X) =
i<n Base10 to 1(ai)*10i
where +, *, and ^ are defined over PU
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Example X= 238
Base10 to 1 (238) =
2*100 + 3*10 + 8
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Base10 to 1 (an-1 an-2 … a0) =
an-1*10n-1 + an-2 * 10n-2 +…+ a0 100
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Now we want to go back the other
way.
We want to convert PU to
decimal.
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No Leading Zero:
Let NLZ be the set of decimal
numbers with no leading 0 (leftmost
symbol 0), or the decimal number 0.
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We define Base1 to 10 from PU to NLZ
It will turn out to be the inverse function
to Base10 to 1.
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One digit cases.
Base1 to 10(0) = “0”
Base1 to 10(S0) = “1”
Base1 to 10(SSo) =“2”
….Base1 to 10(SSSSSSSSS0) = “9”
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Suppose X > 9
Let n be the smallest unary number such that 10n > X, 10n-1 · X.
Let d = X DIV 10n-1 [Notice that 1· d · 9]Let Y = X MOD 10n-1
Define Base1 to 10(X) 2 NLZ to be
Base1 to 10 (d) Base1 to 10(Y)
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For each n2 PU define its decimal representation to
beBase1 to 10 (n).
Base1 to 10 goes from PU to NLZ.
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Let’s verify that Base1 to 10 and
Base10 to 1 really are inverse functions.
We need to show X2 NLZ )Base1 to 10 (Base10 to 1 (X)) =
X.
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Clear when X is a single digit.
Base1 to 10 (Base10 to 1 (0) ) = 0
Base1 to 10 (Base10 to 1 (1) ) = 1
Base1 to 10 (Base10 to 1 (2) ) = 2
Base1 to 10 (Base10 to 1 (3) ) = 3
Base1 to 10 (Base10 to 1 (4) ) = 4
Base1 to 10 (Base10 to 1 (5) ) = 5
Base1 to 10 (Base10 to 1 (6) ) = 6
Base1 to 10 (Base10 to 1 (7) ) = 7
Base1 to 10 (Base10 to 1 (8) ) = 8
Base1 to 10 (Base10 to 1 (9) ) = 9
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Let X be a shortest counter-example to the
statement:
X2 NLZ )
Base1 to 10 (Base10 to 1(Z)) =Z
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For all Y shorter than X:
Y2 NLZ )
Base1 to 10 (Base10 to 1(Y)) = Y
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X2 NLZ, n=length(Z) > 1
Suppose X=an-1 an-2 … a0
Base10 to 1(X) = Y =
i<n Base10 to 1(ai)*10i
Base1 to 10 (Y) = ?
n is smallest PU s.t. 10n-1 · Y < 10n
Calculate d = Y DIV 10n-1, Z= Y MOD 10n-1
d= an-1, Z= i<n-1 Base10 to 1(ai)*10i
Output an-1 Base1 to 10 (Z) [Z shorter than X]
= an-1 an-2 … a1 a0
Contradiction of X being a counter-example.
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No counter-example means that
For all Y2 NLZ,Base1 to 10 (Base10 to 1 (Y)) = Y
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So we can move back and forth between
representing a number in PU or in NLZ.
Each string in PU corresponds to one and only one string in NLZ.
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Base X Notation
Let be an alphabet of size X. A base X digit is any element of .
Let S = an-1, an-2, …, a0 be a sequence of base X digits.
Let BaseX to 1(S) = an-1 Xn-1 + … a2X2 + a1X + a0
S is called the base X representation of the number BaseX to 1(S).
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S = an-1, an-2, …, a0
represents the number: an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0
Base 2 [Binary Notation]101 represents 1 (2)2 + 0 (21) + 1 (20)
Base 7015 represents 0 (7)2 + 1 (71) + 5 (70)
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S = an-1, an-2, …, a0
represents the number: an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0
Base 2 [Binary Notation]101 represents 1 (2)2 + 0 (21) + 1 (20) =
Base 7015 represents 0 (7)2 + 1 (71) + 5 (70)=
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Bases In Different Cultures
Sumerian-Babylonian: 10, 60, 360Egyptians: 3, 7, 10, 60Africans: 5, 10French: 10, 20English: 10, 12,20
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Biggest n “digit” number in base X would be:
(X-1)Xn-1 + (X-1)Xn-2 +…+ (X-1)X0
Base 2111 represents 1 (2)2 + 1 (21) + 1 (20)
Base 7666 represents 6 (7)2 + 6 (71) + 6 (70)
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Biggest n “digit” number in base X would be:
(X-1)Xn-1 + (X-1)Xn-2 +…+ (X-1)X0
Base 2111 represents 1 (2)2 + 1 (21) + 1 (20) 111 + 1 = 1000 represents 23
Thus, 111 represents 23 – 1
Base 7666 represents 6 (7)2 + 6 (71) + 6 (70)666 + 1 = 1000 represents 73
Thus, 666 represents 73 - 1
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Biggest n “digit” number in base X would be:
S= (X-1)Xn-1 + (X-1)Xn-2 +…+ (X-1)X0
\Add 1 to get: Xn + 0 Xn-1 + … + 0 X0
Thus, S = Xn - 1
Base 2111 represents 1 (2)2 + 1 (21) + 1 (20) 111 + 1 = 1000 represents 23
Thus, 111 represents 23 – 1
Base 7666 represents 6 (7)2 + 6 (71) + 6 (70)666 + 1 = 1000 represents 73
Thus, 666 represents 73 - 1
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Recall theGEOMETRIC SERIES.
1 + X1 + X11 + X + X22 + X + X 33 + … + X + … + Xn-2n-2 + X + Xn-1n-1 ==
XXn n – 1 – 1 XX - -
11
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(X-1)Xn-1 + (X-1)Xn-2 +…+ (X-1)X0
= Xn - 1
Proof:Factoring out (X-1), we obtain:(X-1) [Xn-1 + Xn-2 + ….. + X + 1] =
(X-1) [(Xn – 1)/(X-1)] =
Xn – 1
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The highest n digit number in base X.
(X-1) ( 1 + X1 + X2 + X3 + … + Xn-1 ) = Xn - 1
Base X. Let S be the sequence of n digits each of which is X-1. S is the largest number of length n that can
be represented in base X.
S = XXn n - 1- 1
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Each of the numbers from 0 to Xn-1 is uniquely
represented by an n-digit number in base X.
We could prove this using our previous
method, but let’s do it another way.
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Our proof introduce an unfamiliar kind of base
representation.
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Plus/Minus Base X
An plus/minus base X digit is any integer –X < a < X
Let S = an-1, an-2, …, a0 be a sequence of plus/minus base X digits. S is said to be the plus/minus base X representation of the number:
an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0
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Does each n-digit number in plus/minus
base X represent a different integer?
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No. Consider plus/minus binary.
5 = 0 1 0 1 = 1 0 -1 -1
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Humm.. So what is it good for?
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Not every number has a unique plus/minus binary representation, but 0 does.
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0 has a unique n-digit plus/minus base X
representation as all 0’s.
Suppose an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0 = S =0, where there is some highest k such that ak 0. Wl.o.g. assume ak > 0.
Because Xk > (X-1)(1 + X1 + X2 + X 3 + … + Xk-1)no sequence of digits from ak-1 to a0 can represent a number as small as –Xk
Hence S 0. Contradiction.
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Each of the numbers from 0 to Xn-1 is uniquely represented by an n-
digit number in base X.
We already know that n–digits will represent something between 0 and Xn – 1.
Suppose two distinct sequences represent the same number:
an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0 =bn-1 Xn-1 + bn-2 Xn-2 + . . . + b0 X0
The difference of the two would be an plus/minus base X representation of 0, but it would have a non-zero digit. Contradiction.
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Each of the numbers from 0 to Xn-1 is uniquely represented by an n-
digit number in base X.
n digits represent up to Xn – 1n-1 digits represents up to Xn-1 - 1
Let k be a number: Xn-1 ≤ k ≤ Xn - 1 So k can be represented with n digits.
For all k: logxk = n – 1
So k uses logxk + 1 digits.
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Fundamental Theorem For Base X:
Each of the numbers from 0 to Xn-1 is uniquely represented by an n-
digit number in base X.
k uses logxk + 1 digits in base X.
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n has length n in unary, but has
length log2n + 1 in
binary.
Unary is exponentially longer
than binary.
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Egyptian Multiplication
The Egyptians used decimal numbers but
multiplied and divided in
binary
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Egyptian Multiplication a times bby repeated doubling
b has some n-bit representation: bn..b0
Starting with a, repeatedly double largest so far to obtain: a, 2a, 4a, …., 2na
Sum together all 2ka where bk = 1
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Egyptian Multiplication 15 times 5by repeated doubling
5 has some 3-bit representation: 101
Starting with 15, repeatedly double largest so far to obtain: 15, 30, 60
Sum together all 2k(15) where bk = 1:
15 + 60 = 75
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Why does that work?
b = b020 + b121 + b222 + … + bn2n
ab = b020a + b121a + b222a + … + bn2na
If bk is 1 then 2ka is in the sum.
Otherwise that term will be 0.
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Wait! How did the Egyptians do the part
where they converted b to binary?
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They used repeated halving to do base conversion. Consider …
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Egyptian Base Conversion
Output stream will print right to left.Input X. Repeat until X=0{
If X is even then Output O; Otherwise {X:=X-1; Output 1}
X:=X/2}
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Egyptian Base Conversion
Output stream will print right to left.Input X. Repeat until X=0{
If X is even then Output O; Otherwise Output 1
X:= X/2}
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Start the algorithm
Repeat until X=0{ If X is even then Output O;
Otherwise Output 1;X:= X/2
}
010101 1
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Start the algorithm
Repeat until X=0{ If X is even then Output O;
Otherwise Output 1;X:= X/2
}
01010 1
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Start the algorithm
Repeat until X=0{ If X is even then Output O;
Otherwise Output 1;X:= X/2
}
01010 01
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Start the algorithm
Repeat until X=0{ If X is even then Output O;
Otherwise Output 1;X:= X/2
}
0101 01
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Start the algorithm
Repeat until X=0{ If X is even then Output O;
Otherwise Output 1;X:= X/2
}
0101 101
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Start the algorithm
Repeat until X=0{ If X is even then Output O;
Otherwise Output 1;X:= X/2
}
010 101
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And Keep Going until 0
Repeat until X=0{ If X is even then Output O;
Otherwise Output 1;X:= X/2
}
0 010101
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Sometimes the Egyptian combined the base conversion by halving and the multiplication by doubling into one algorithm
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Rhind Papyrus (1650 BC)70*13
70140280560
13 * 7063 * 3501 * 910
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Rhind Papyrus (1650 BC)70*13
70140280560
13 * 7063 * 3501 * 910
Binary for 13 is 1101 = 23 + 22 + 20
70*13 = 70*23 + 70*22 + 70*20
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Rhind Papyrus (1650 BC)
173468136
1
2 *
4
8 *
184 48 14
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Rhind Papyrus (1650 BC)
173468136
1
2 *
4
8 *
184 48 14
184 = 17*8 + 17*2 + 14184/17 = 10 with remainder 14
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This method is called “Egyptian Multiplication/Division” or “Russian Peasant Multiplication/Division”.
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Wow. Those Russian peasants were pretty
smart.
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Standard Binary Multiplication= Egyptian Multiplication
X101
* * * * * * * *
* * * * * * * * * * * * * * *
** * * * * * * * * * *
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Egyptian Base 3
We have definedBase 3: Each digit can be 0, 1, or 2Plus/Minus Base 3 uses -2, -1, 0, 1, 2
Here is a new one:Egyptian Base 3 uses -1, 0, 1
Example: 1 -1 -1 = 9 - 3 - 1 = 5
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Unique Representation Theorem for Egyptian Base 3
No integer has 2 distinct, n-digit, Egyptian base-3 representations. We
can represent all integers from -(3n-1)/2 to (3n-1)/2
Proof; If so, their difference would be a non-trivial plus/minus base 3 representation of 0. Contradiction.Highest number = 1111…1 = (3n-1)/2
Lowest number = -1-1-1-1…-1 = -(3n-1)/2
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Unique Representation Theorem for Egyptian Base 3
No integer has 2 distinct, n-digit, Egyptian base-3 representations. We
can represent all integers from -(3n-1)/2 to (3n-1)/2
3n, n-digit, base 3 representations of the numbers from 0 to 3n – 1
Subtract 111111…111 = (3n – 1)/2 from each to get an Egyptian base 3 representation of all the numbers from -(3n-1)/2 to (3n-1)/2.
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How could this be Egyptian? Historically, negative numbers first appear in the writings
of the Hindu mathematician
Brahmagupta (628 AD).
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One weight for each power of 3. Left = “negative”. Right =
“positive”
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References
The Book Of Numbers, by J. Conway and R. Guy
History of Mathematics, Histories of Problems, by The Inter-IREM Commission