COMP250:BinaryNumbers

JérômeWaldispühlSchoolofComputerScience

McGillUniversity(SlidesfromM.Langer)

Decimal(base10)

Digits={0,1,2,3,4,5,7,8,9}

Exampleofnumerals:11,923,5548,etc.

11 = 1 ∗ 10& + 1 ∗ 10(923 = 9 ∗ 10, + 2 ∗ 10& + 3 ∗ 10(

5548 = 5 ∗ 1005 ∗ 10, + 4 ∗ 10& + 8 ∗ 10(

𝑚 =2𝑑 𝑖 ∗ 105�

57(

digit

Binary(base2)

Bits={0,1}

Exampleofnumerals:11,101,1010,etc.

11 = 1 ∗ 2& + 1 ∗ 2(101 = 1 ∗ 2, + 0 ∗ 2& + 1 ∗ 2(

1010 = 1 ∗ 20 + 0 ∗ 2, + 1 ∗ 2& + 0 ∗ 2(

𝑚 =2𝑏 𝑖 ∗ 25�

57(

bit

Relationship

Decimal Binary0 01 12 103 114 1005 1016 1107 1118 1000

Fixedsizerepresentation

Decimal Binary0 000000001 000000012 000000103 000000114 000001005 000001016 000001107 000001118 00001000

Fixednumberofbits(typically8,16,32,64…).

8bitsiscalled“byte”.

Conversion

• Howtoconvertfrombinarytodecimal?

11010 , = 1 ∗ 29 + 1 ∗ 20 + 0 ∗ 2, + 1 ∗ 2& + 0 ∗ 2( = 16 + 8 + 0 + 2 + 0 = 26

Note:Youneedtoknowpowersof2…

• Howtoconvertfromdecimaltobinary?

241 &( =? ? ?

Operationsondecimals

Usethispropertyofanypositiveinteger𝑚 ∶𝑚 = 𝑚%10 + 𝑚 10⁄ ∗ 10

(integer)divisionby10=droppingrightmostdigit

Ex:238/10=23

Multiplicationby10=shiftingleftbyonedigit

Ex:23*10=230

Remainderofintegerdivisionby10=rightmostdigit

Ex:238%10=8

OperationsonbinarySamepropertyholdsforbinary:

𝑚 = 𝑚%2 + 𝑚 2⁄ ∗ 2

Example:𝑚 = 1011 ,𝑚 2⁄ = 0101 ,

𝑚 2⁄ ∗ 2 = 1010 ,𝑚%2 = 0001 ,

Decimal⟶Binary (Algorithm)

Algorithm decimal2binary(m)Input:adecimalmOutput:abinarybi⟵0whilem>0do

b[i]⟵m%2m⟵m/2i⟵i+1

Decimal⟶Binary (Example)i m/2 m%2(b[i])

2410 120 11 60 02 30 03 15 04 7 15 3 16 1 17 0 18 - -

Answer:b[]=…011110001

Whyisthealgorithmworking?

𝑚 = 𝑚 2 ∗ 2 +𝑚%2⁄

…𝑏 3 𝑏 2 𝑏 1 ,

…𝑏 3 𝑏 2 𝑏 1 0 , 𝑏 0 ,

…𝑏 3 𝑏 2 𝑏 1 𝑏[0] ,

Additions

Decimal Binary0+1=1 0+1=11+1=2 1+1=101+2=3 1+10=11

Additions

26+ 15= 41

11010+ 01111= ? ? ? ? ?

Decimal Binary

Additioninbinary

1 1 0 1 0+ 0 1 1 1 1= ? ? ? ?( 1

Additioninbinary

1 1 0 1 0+ 0 1 1 1 1= ? ? ?& 0( 1

Additioninbinary

1 1 0 1 0+ 0 1 1 1 1= ? ?& 0& 0( 1

Additioninbinary

1 1 0 1 0+ 0 1 1 1 1= ?& 1& 0& 0( 1

Additioninbinary

1 1 0 1 0+ 0 0 1 1 1 1= 1 0& 1& 0& 0( 1

Additioninbinary

1 1 0 1 0 = 26+ 0 0 1 1 1 1 = 15= 1 0 1 0 0 1 = 41

2A 29 20 2, 2& 2(1 0 1 0 0 1

= 2A + 20 + 2( = 32 + 8 + 1 = 41

Operationinbinary

Recallgrade-schoolalgorithmforaddition,subtraction,multiplication,anddivision.

Thereisnothingspecialaboutbase10.

Thesealgorithmsworkforbinary(base2),andworkforotherbasestoo!

Representationsize

𝑚 = 2 𝑏 𝑖 ∗ 25BC&

57(

Whatistherelationshipbetween𝑚 andN?

(HowmanybitsNdoweneedtorepresentapositiveintegerm?)

Upperbound

225BC&

57(

= 1 + 2 + 4 +⋯+ 2BC& = 2B − 1

(Thisisaspecialcaseof∑ 𝑥5 = HIC&HC&

BC&57( where𝑥 = 2)

𝑚 = 2 𝑏 𝑖 ∗ 25BC&

57(

≤ 2 1 ∗ 25BC&

57( = 2B − 1 < 2B

log, 𝑚 < 𝑁 (applylogonbothsides)

Lowerbound

Wecanassumethat𝑁 −1istheindex𝑖 oftheleftmostbit𝑏[i]suchthat𝑏[𝑖]=1(weignoreleftmost0’s:00001101).

𝑚 = 2 𝑏 𝑖 ∗ 25BC&

57(

≥ 2BC&

log, 𝑚 ≥ 𝑁 − 1log, 𝑚 + 1 ≥ 𝑁

Howmanybitdoweneed?

log, 𝑚 < 𝑁 ≤ log, 𝑚 + 1Answer: Thelargestintegerlessthanorequaltolog, 𝑚 + 1.

Wewriteitas𝑁 = log, 𝑚 + 1 (a.k.a.”floor”thatmeans“rounddown”)

Examplesm(decimal) m (binary) 𝑁 = log, 𝑚 + 1

0 0 -1 1 12 10 23 11 24 100 35 101 36 110 37 111 48 1000 49 1001 4

Tothinkabout…

• Howarenegativeintegersrepresented?

• Howmanybitsareusedtorepresentint,short,longinacomputer?

• Howarenon-integers(fractionalnumbers)represented?

• Howarecharactersrepresented?