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Arithmetic operations in binary

Addition / subtraction

01011

+ 111

?

Method exatly the same as decimal

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Arithmetic operations in binaryAddition

X = xn xi x0

+ Y = yn yi y0

__________________di

di = (xi + yi ) mod r + carry-in

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AdditionFor ( i = 0n ) do

di = (xi + yi + carry-in) mod r

carry-out = (xi + yi + carry-in) div r

End for

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Arithmetic operations in binary Subtraction

Can you write an equivalent method

(algorithm) for subtraction?

Algorithm: a systematic sequence of

steps/operations that describes how tosolve a problem

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Arithmetic operations in binary Subtraction

X = xn xi x0 Y = yn yi y0

___________________________________________________

diFor ( i = 0n ) do

di = ( xi yi bi-1 ) mod rborrow-out = ( xi yi bi-1 ) div r

End for

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Subtraction

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Arithmetic operations in binary Multiplication

For ( i = 0n ) dodi = ( xi v yi + ci-1 ) mod r

ci = ( xi v yi + ci-1 ) div r

End for

Is this correct?

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Negative numbers (4 traditions):

n = number of symbols (digits, bits, etc.)

r = radixradix complement of x is

1)1( ! xrxr nn

e.g. n = 4, r = 107216 --> 9999 - 7216 + 1 = 2784 (10s complement)

n = 4, r = 2

0101 --> 1111 - 0101 + 1 = 1011 (2s complement)

Signed magnitude

Radix complementDiminished radix complement

Excess-b (biased)

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xrn

)1(

e.g. n = 4, r = 10

7216 --> 9999 - 7216 = 2783 (9s complement)

n = 4, r = 20101 --> 1111 - 0101 = 1010 (1s complement)

diminished radix complement is

Note: operation is reversible (ignoring overflow)

i.e. complement of complement returns original pattern

(for both radix and diminished radix forms)

for zero and its complement:

10s complement: 0000 --> 9999 + 1 = 10000 --> 0000

9s complement: 0000 --> 9999 --> 0000

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Arithmetic with complements n = 4r = 2

Excess-8

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Signed number representations[[N]] = rn ( rn N ) = N

( N + [N] ) mod rn = 0

Example: (n = 4)+5 | 0101

-5 | 1011

10000 | 24

N + [N] mod 24 = 0

N = 5 r = 10 N = 32546

[N] = 105 32546 = (67454)10

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Twos complement arithmetic

( an-1, an-2, a0 ) in 2-s complement is

Example: (n = 4)

0101 p 0v23 + 1v22 + 0v2 + 1v1 = 4 + 1 = 51011 p 1v23 + 0v22 + 1v2 + 1v1 = 8 + 2 + 1 = 5

Addition/subtraction (n = 5)

+10 | 01010

+3 | 00011

01101 | 13

!

2

0

1

1 22n

i

i

i

n

naa

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Twos complement arithmetic

Addition/subtraction (n = 5)

+10 | 01010

+7 | 00111Overflow 10001 | 15

+15 | 01010

13 | 10011

Discard 100010 | 2

When can overflowhappen? Only when

both operands have

the same sign and the

sign bit of the result is

different.

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Cyclic representation (n = 4, r = 2)

avoid discontinuity between

0111 and 1000

Add x:

move x positions clockwise

Subtract x:

move x positions counterclockwise

move (16 - x) positions clockwise

(i.e. add radix complement)

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How to detect discontinuity?

5

+ 6

11

0101

+ 0110

1011

no overflow (in 4-bit register)

but, carry into most significant bit is 1 while carry out is 0

-5

+ -6

-11

1011

+ 1010

10101

overflow

but, carry into most significant bit is 0 while carry out is 1

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Same circuitry

signed numbers

add

subtract (use 2s complement of subtrahend)

Intel architecture OF (overflow flag) detects out-of-range resultunsigned numbers

same protocol

but discontinuity is now between 0000 and 1111

detect with overflow for addition

lack of overflow for subtraction

Intel uses CF (carry flag) to detect out-of-range result

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Codes

4-bit codes for digits 0 through 9 (BCD : 8421 weighted)

2421 and Excess-3 are self-complementing

(complement bits to get 9s complement representation)

Biquinary has two ranges 01 and 10

one-bit error produces invalid codeword

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Number representation inside

computer

Integers represented in 2s complement

Single precision 32 bits

Largest positive integer is 231-1 =

2,147,483,647

Smallest negative integer is -231 =

2,147,483,648

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Number representation inside

computer

Floating point

Scientific notation

0.0043271 = 0.43271v10-2

normalized number

The digit after the decimal point is { 0Normalized notation maximizes the use of

significant digits.

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Floating point numbers

Floating point

N = (-1)S v m v rE

S = 0 p positive

S = 1 p negative

m p normalized fraction for radix r = 2

As MSB digit is always 1, no need to explicitly store it

Called hidden bit p gives one extra bit of precision

S E m

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Floating point formats

IEEE format: (-1)Sv(1.m)v2E-127

DEC format: (-1)Sv(0.m)v2E-128

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Floating point formats

Different manufacturers used differentrepresentations

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Mechanical encoder with sequential

sector numbering

At boundaries can get a code differentthan either adjacent sector

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Gray code:

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Gray code algorithm:

input: (binary)

output: (Gray code)0121 bbbb nn .

0121 gggg nn .

!

!

else,1

,0 1iii

bbg

e.g.

n = 3: 6 -> 110 -> 101

n = 4: 10 -> 1010 -> 1111

Alternative algorithm (n bits)

If n = 1 use 0->0 and 1->1

If n > 1

first half is Gray code on (n - 1) bits preceded by 0

second half is reversed Gray code on (n - 1) bits preceded by 1

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Representing non-numeric data

Code: systematic and preferably standardized wayof using a set of symbols for representing

information

Example: BCD (binary coded decimal) for decimal #s

It is a weighted code p the value of a symbol is aweighted sum

Extending number of symbols to include

alphabetic characters

EBCDIC (by IBM): Extended BCD Interchange Code

ASCII: American Standard Code for Information

Interchange

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Codes

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Cyclic codes

A circular shift of any code word produces a

valid code word (maybe the same)

Gray code example of a cyclic code

Code words for consecutive numbers differ by

one bit only

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Ascii, ebcdic codesState codes, e.g.

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Consequences of binary versus one-hot coding:

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n-cubes of codewords:

Hamming distance between x and y is count of positions where

x-bit differs from y-bit

Also equals link count in shortest path from x to y

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Gray code is path that visits each

vertex exactly once

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Error-detecting codes

Problem Solution: parity

concept:

choose code words such thatcorruption generates a non-code word

to detect single-bit error, code words must occupy

non-adjacent cube vertices

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Example: correct one bit errors or detect two-bit errors

Error-correcting codes

minimum distance between code words > 1

Two-bit error from0001011

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Write minimum distance as 2c + d + 1 bits ==>

corrects c-bit errors and

detects (c + d)-bit errors

Example: min distance = 4 = 2(1) + 1 + 1

But also, 4 = 2(0) + 3 + 1

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Why?suppose minimum distance is h

c ch - 2c

y

d = h - 2c -1

pair of closest nodes

maximally distant from left with entering correction zone

2c + d + 1 = 2c + (h - 2c - 1) + 1 = h

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Hamming codes: 12 ! kn

number bit positions 1, 2, 3, ... n from right to left

bit position is a power of 2 => check bit

else => information bite.g. (n = 7)

check bits in positions 1, 2, 4

information bits in positions 3, 5, 6, 7

Create group for each check bitExpress check bit position in binary

Group all information bits whose binary position has a one in same place

e.g. (n = 7)

check information

1 (001) 3 (011), 5 (101), 7 (111)

2 (010) 3 (011), 6 (110), 7 (111)

4 (100) 5 (101), 6 (110), 7 (111)

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Note:Single bit error corrupts one or more parity groups => minimum distance > 1

Two-bit error in locations x, y corrupts at least one parity group =>

minimum distance > 2

Three-bit error (i.e. 1, 4, 5) goes undetected => minimum distance = 3

3 = 2(1) + 0 + 1 = 2(0) + 2 + 1 => can correct 1-bit errors or detect errors

of size 1 or 2.

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Pattern generalizes to longer bit strings: Single bit error corrupts one or more parity groups => minimum distance > 1

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Group 8

Group 4

Group 2

Group 1

Consider two-bit error in positions x and y. To avoid corrupting all groups:

-- avoid group 1: bit 1 (lsb) same for both x and y

-- avoid group 2: bit 2 same for both

-- avoid group 4: bit 3 same for both, etc.

-- forces x = y

So true two-bit error corrupts at least one parity group => min distance > 2

Three-bit error (two msb and lsb) goes undetected => minimum distance = 3

Conclude: process always produces a Hamming code with min distance = 3

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Traditional to permute check bits to far right

Used in memory protection schemes

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Add traditional parity bit (for entire word) to obtain

code with minimum distance = 4

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Number of check bits grows slowly with information bits

For minimum distance = 4: total bits 2

parity bits 1

information bitslg 1

payloadratio 1

kn

p k

i n pn p n n

n n

! !

! !

! !

! !

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Two-dimensional codes (product codes)

min distance is product of

row and column distances

for simple parity on each (below)

min distance is 4

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Scheme for RAID storage

CRC is extension ofHamming codeseach disk is separate row

column is disk block (say 512 characters)

rows have CRC on row disk

columns have even parity

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Checksum codes

mod-256 sum of info bytes becomes checksum byte

mod-255 sum used in IP packets

m-hot codes (m out of n codes)

each valid codes has m ones in a frame of n bits

min distance is 2detect all unidirectional errors

bit flips are all 0 to 1 or 1 to 0

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Serial data transmission

(simple) transmit clock and sync with data (3 lines)

(complex) recover clock and/or sync from data line

Serial line codes:

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NRZ: clock recovery except during long sequences of zeros or ones

NRZ1: nonreturn to zero, invert on one

zero => no change, one => change polarity

RZ: clock recovery except during long sequences of zeroDC balance

Bipolar return to zero (aka alternate mark inversion: send one as +1 or -1)

Manchester: zero => 0 to 1 transition, one => 1 to 0 transition

at interval center

Serial line codes: