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HD15b.number-theoretic-algorithms
Course: CS 511, Fall 2009School: BU
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NumberSystems NumberTheoryAlgorithms ZephGrunschlag CopyrightZephGrunschlag,20012002. Agenda EuclideanAlgorithmforGCD Decimalnumbers(base10) Binarynumbers(base2) Onescomplement Twoscomplement Generalbasebnumbersystems Addition Multiplication Subtraction1sand2scomplement 2 ArithmeticAlgorithms L11 EuclideanAlgorithm m,n Euclidean Algorithm gcd(m,n) integereuclid(pos.integerm,pos.integern) x=m,y=n...
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NumberSystems
NumberTheoryAlgorithms
ZephGrunschlag
CopyrightZephGrunschlag,20012002.
Agenda
EuclideanAlgorithmforGCD
Decimalnumbers(base10) Binarynumbers(base2)
Onescomplement Twoscomplement
Generalbasebnumbersystems Addition Multiplication Subtraction1sand2scomplement
2
ArithmeticAlgorithms
L11
EuclideanAlgorithm
m,n
Euclidean Algorithm
gcd(m,n)
integereuclid(pos.integerm,pos.integern) x=m,y=n while(y>0) r=xmody x=y y=r returnx
L11 3
gcd(33,77):
EuclideanAlgorithm. Example
r=xmody x 33 y 77
Step 0
L11
4
gcd(33,77):
EuclideanAlgorithm. Example
r=xmody 33mod77 =33 x 33 77 y 77 33
Step 0 1
L11
5
gcd(33,77):
EuclideanAlgorithm. Example
r=xmody 33mod77 =33 77mod33 =11 x 33 77 33 y 77 33 11
Step 0 1 2
L11
6
gcd(33,77):
EuclideanAlgorithm. Example
r=xmody 33mod77 =33 77mod33 =11 33mod11 =0 x 33 77 33 11 y 77 33 11 0
7
Step 0 1 2 3
L11
gcd(244,117):
Step 0
EuclideanAlgorithm. Example
r=xmody x 244 y 117
L11
8
gcd(244,117):
Step 0 1
EuclideanAlgorithm. Example
r=xmody 244mod117=10 x 244 117 y 117 10
L11
9
gcd(244,117):
Step 0 1 2
EuclideanAlgorithm. Example
r=xmody 244mod117=10 117mod10=7 x 244 117 10 y 117 10 7
L11
10
gcd(244,117):
Step 0 1 2 3
EuclideanAlgorithm. Example
r=xmody 244mod117=10 117mod10=7 10mod7=3 x 244 117 10 7 y 117 10 7 3
L11
11
gcd(244,117):
Step 0 1 2 3 4
EuclideanAlgorithm. Example
r=xmody 244mod117=10 117mod10=7 10mod7=3 7mod3=1 x 244 117 10 7 3 y 117 10 7 3 1
L11
12
gcd(244,117): Step 0 1 2 3 4 5
L11
EuclideanAlgorithm. Example
r=xmody 244mod117=10 117mod10=7 10mod7=3 7mod3=1 3mod1=0 x 244 117 10 7 3 1 y 117 10 7 3 1 0
13
Bydefinition244and117arerel.prime.
EuclideanAlgorithmCorrectness
ThereasonthatEuclideanalgorithmworksis gcd(x,y)isnotchangedfromlinetoline.Ifx,y denotethenextvaluesofx,ythen: gcd(x,y)=gcd(y,xmody) =gcd(y,xqy) (theusefulfact) =gcd(y,x)(subtractymultiple) =gcd(x,y)
L11 14
EuclideanAlgorithm RunningTime
EX:Computetheasymptoticrunningtimeofthe Euclideanalgorithmintermsofthenumberof modoperations:
L11
15
EuclideanAlgorithm RunningTime
AssumingmodoperationisO(1): integereuclid(m,n) x=m,y=n while(y>0) r=xmody x=y y=r returnx
L11
O(1)+ ?(O(1)+ O(1) +O(1) (1)) +O(1) =?O(1)
+O
Where?isthenumberofwhileloopiterations.
16
EuclideanAlgorithm RunningTime
Facts:(x=nextvalueofx,etc.) 2. xcanonlybelessthanyatverybeginningof algorithm oncex>y,x=y>y=xmody 4. Whenx>y,twoiterationsofwhileloopguarantee thatnewxis<originalx becausex=y=xmody.Twocases:
I. II.
y>xxmody=xy<x yxxmody<yx
17
L11
EuclideanAlgorithm RunningTime
(1&2)Afterfirstiteration,sizeofxdecreasesby factor>2everytwoiterations. I.e.after2m+1iterations, x<original_x/2m Q:Whenintermsofmdoesthisprocess terminate?
L11
18
EuclideanAlgorithm RunningTime
After2m+1steps,x<original_x/2m A:Whileloopexitswhenyis0,whichisright beforewouldhavegottenx=0.Exiting whileloophappenswhen2m>original_x,so definitelyby: m=log2(original_x) Thereforerunningtimeofalgorithmis: O(2m+1)=O(m)=O(log2(max(a,b)))
L11 19
EuclideanAlgorithm RunningTime
Measuringinputsizeintermsofn=numberofdigitsof max(a,b): n=(log10(max(a,b)))=(log2(max(a,b))) Thereforerunningtimeofalgorithmis: O(log2(max(a,b)))=O(n) (assumednaivelythatmodisanO(1)operation,so estimateonlyholdsforfixedsizeintegerssuchas intsandlongs)
L11 20
NumberSystems
Q:Whatdoesthestringofsymbols 2134 reallymeanasanumberandwhy?
L11
21
NumberSystems
A:2thousands1hundreds3tensand4 =2103+1102+3101+4100 ButontheplanetOgg,theintelligentlifeforms haveonlyonearmwith5fingers.
L11
22
NumberSystems
SoonOgg,numbersarecountedbase5.I.e.on Ogg2134means: 253+152+351+450 Todistinguishbetweenthesesystems,subscripts areused: (2134)10forEarth (2134)5forOgg
L11 23
NumberSystems
DEF:Abasebnumberisastringofsymbols u=akak1ak2a2a1a0 Withtheaiin{0,1,2,3,,b2,b1}. Thestringurepresentsthenumber (u)b=akbk+ak1bk1+...+a1b+a0 NOTE:Whenb>10,runoutofdecimalnumber symbolssoafter7,8,9usecapitalletters, startingfromA,B,C,
L11 24
NumberSystems
EG:base2(binary)101,00010 base8(octal)74,0472 base16(hexadecimal)12F,ABCD Q:Computethebase10versionofthese.
L11
25
NumberSystems
A:base2(binary)101,00010 (101)2=122+021+120=5 (00010)2=0(24+23+22+20)+121=2 base8(octal)74,0472 (74)8=781+480=60 (0472)8=482+781+280=314 base16(hexadecimal)12F,ABCD (12F)16=1162+2161+15160=303 (ABCD)16=10163+11162+12161+13160 L11
26
NumberSystems
Binarymostnaturalsystemforbitstringsand hexadecimalcompactifiesbytestrings(1byte=2 hexadecimals) EGinHTML: <fontcolor="ff00ff">NiceColor</font> Q:Whatcolorwillthisbecome?
L11
27
NumberSystems
A:"ff00ff"representsthergbvalue: Thefirstbyteisforredness,thesecondbyteisfor greenness,andthelastforblueness.TheHTML abovespecifiesthat1516+15=255rednessand bluenessvalues,but016+0=0greenness.Red andbluegivepurple,and255isthetopbrightness sothisisbrightpurple.
L11
28
NumberSystems ReverseConversion
Convertarbitrarydecimalnumbersintovariousbases, (calculatorfunctionstypicallylimitedtobase2,8, 16and10). EG:BackatOgg.Convert646tobase5.Trytodo alloperationsbase5.
L11
29
NumberSystems ReverseConversion
BackatOgg.Convert646tobase5.Trytodoall operationsasanOggian(base5): (646)10=(6)10(10)102+(4)10(10)10+(6)10 Eachquantityeasytoconvertintobase5: (6)10=(11)5since6=5+1 (4)10=(4)5 since4<5 (10)10=(20)5 since10=25+0 SoconvertwholeexpressionanddoOggian arithmetic:
L11 30
NumberSystems ReverseConversion
BackatOgg.Convert646tobase5.Trytodoall operationsasanOggian(base5): (646)10=(11)5(20)52+(4)5(20)5+(11)5 =
L11
31
NumberSystems ReverseConversion
BackatOgg.Convert646tobase5.Trytodoall operationsasanOggian(base5): (646)10=(11)5(20)52+(4)5(20)5+(11)5 =(11)5(400)5+(130)5+(11)5 =
L11
32
NumberSystems ReverseConversion
BackatOgg.Convert646tobase5.Trytodoall operationsasanOggian(base5): (646)10=(11)5(20)52+(4)5(20)5+(11)5 =(11)5(400)5+(130)5+(11)5 =(4400)5+(141)5 =
L11
33
NumberSystems ReverseConversion
BackatOgg.Convert646tobase5.Trytodoall operationsasanOggian(base5): (646)10=(11)5(20)52+(4)5(20)5+(11)5 =(11)5(400)5+(130)5+(11)5 =(4400)5+(141)5 =(10041)5 ThinkinglikeanOggianhurtsbraintoomuch
L11 34
NumberSystems ReverseConversion
Givenanintegernandabasebfindthestringusuchthat(u) b=n. Pseudocode: stringrepresent(pos.integern,pos.integerb) q=n,i=0 while(q>0) ui=qmodb q=q/b i=i+1 returnuiui1ui2u2u1u0
L11 35
NumberSystems ReverseConversion
EG:Convert646toOggian(base5):
i ui=qmodb q=q/b 646
L11
36
NumberSystems ReverseConversion
EG:Convert646toOggian(base5):
i 0 ui=qmodb 646mod5=1 q=q/b 646 646/5=129
L11
37
NumberSystems ReverseConversion
EG:Convert646toOggian(base5):
i 0 1 ui=qmodb 646mod5=1 129mod5=4 q=q/b 646 646/5=129 129/5=25
L11
38
NumberSystems ReverseConversion
EG:Convert646toOggian(base5):
i 0 1 2 ui=qmodb 646mod5=1 129mod5=4 25mod5=0 q=q/b 646 646/5=129 129/5=25 25/5=5
L11
39
NumberSystems ReverseConversion
EG:Convert646toOggian(base5):
i 0 1 2 3 ui=qmodb 646mod5=1 129mod5=4 25mod5=0 5mod5=0 q=q/b 646 646/5=129 129/5=25 25/5=5 5/5=1
L11
40
NumberSystems ReverseConversion
EG:Convert646toOggian(base5):
i 0 1 2 3 4
L11
ui=qmodb
646mod5=1 129mod5=4 25mod5=0 5mod5=0 1mod5=1
q=q/b 646 646/5=129 129/5=25 25/5=5 5/5=1 1/5=0
41
Readinglastcolumninreverse:10041
NumberSystems InClassExercise
Somenumbertheoryfactsarebasedependent.Forexample FirstGradeTeachersRule: Abase10numberisdivisibleby3iffthesumofitsdigitsare. Formally,let n=(ukuk1uk2u2u1u0)10.Then:
n mod 3=
EG:3|12135because3|(1+2+1+3+5=12)
ui
i=0 k
mod 3
L11
42
ArithmeticalAlgorithms
Letswritedownsomefamiliararithmetical algorithms.Conveniently,theyrunthesamein anynumberbase.Insomecases,suchas addition,thereareasymptoticallyfaster approaches,butthesearethesimplest proceduresandtendtobefastestforrelatively small(e.g.<1000bits)numbersizes.
L11
43
ArithmeticalAlgorithms Addition
Numbersareaddedfromleastsignificantdigitto most,whilecarryinganyoverflowresultingfrom addingacolumn:
base10 Carry: x +y
L11
base16
7 + 2
4 9
6 0
3 9 +
A C
4 B
F 0
0 9
44
ArithmeticalAlgorithms Addition
Numbersareaddedfromleastsignificantdigitto most,whilecarryinganyoverflowresultingfrom addingacolumn:
base10 Carry: x +y
L11
base16
1
7 + 2
4 9
6 0
3 9 2 +
A C
4 B
F 0
0 9 9
45
ArithmeticalAlgorithms Addition
Numbersareaddedfromleastsignificantdigitto most,whilecarryinganyoverflowresultingfrom addingacolumn:
base10 Carry: x +y
L11
base16
1
7 + 2
4 9
6 0 7
3 9 2 +
A C
4 B
F 0 F
0 9 9
46
ArithmeticalAlgorithms Addition
Numbersareaddedfromleastsignificantdigitto most,whilecarryinganyoverflowresultingfrom addingacolumn:
base10 Carry: x +y
L11
base16
1
1
7 + 2
4 9 3
6 0 7
3 9 2 +
A C
4 B F
F 0 F
0 9 9
47
ArithmeticalAlgorithms Addition
Numbersareaddedfromleastsignificantdigitto most,whilecarryinganyoverflowresultingfrom addingacolumn:
base10 Carry: x +y
L11
base16
1 1
1
1
7 + 2 0
4 9 3
6 0 7
3 9 2 +
A C 6
4 B F
F 0 F
0 9 9
48
ArithmeticalAlgorithms Addition
Numbersareaddedfromleastsignificantdigitto most,whilecarryinganyoverflowresultingfrom addingacolumn:
base10 Carry: x +y
L11
base16
1 1
1
1
7 + 1 2 0
4 9 3
6 0 7
3 9 2 + 1
A C 6
4 B F
F 0 F
0 9 9
49
ArithmeticalAlgorithms AdditionofPositiveNumbers
stringadd(stringsxkxk1x1x0,ykyk1y1y0,intbase) carry=0,xk+1=yk+1=0 for(i=0tok+1) digitSum=carry+xi+yi zi=digitSummodbase carry=digitSum/base returnzk+1zkzk1z1z0
L11
50
1sComplement 2sComplement
Thebinarynumbersystemmakessomeoperations especiallysimpleandefficientundercertain representations. Twosuchrepresentationsare
1scomplement 2scomplement
Eachmakessubtractionmuchsimpler. Eachhasdisadvantagethatnumberlengthispre determined.
L11 51
1sComplement
Fixkbits.(EG,k=8forbytes) Representnumberswith|x|<2k1 Leftmostbittellsthesign
1negative(butotherbitschangetoo!)
Positivenumbersthesameasstandardbinary
0positive(sopositiveno.sasusual) expansion Negativenumbersgottenbytakingtheboolean complement,hencenomenclature
L11
52
1sComplement Examples
k=8: 00010010represents18 11101101represents18 Notice:whenaddtheserepresentationsasusualget 11111111,i.e.negative00000000or0=0. Guess:addingnumberswithmixedsignworksthe sameasaddingpositivenumbers Tradeoff:0notunique
L11
53
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
0 1
0 1
0 1
1 1
0 0
0 0
1 1
0 1
54
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
0 1
0 1
0 1
1 1
0 0
0 0
1 1
0 1 1
55
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1
0 1
0 1
0 1
1 1
0 0
0 0
1 1 0
0 1 1
56
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1
0 1
0 1
0 1
1 1
0 0
0 0 1
1 1 0
0 1 1
57
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1
0 1
0 1
0 1
1 1
0 0 0
0 0 1
1 1 0
0 1 1
58
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1
0 1
0 1
0 1
1 1 0
0 0 0
0 0 1
1 1 0
0 1 1
59
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1 1
0 1
0 1
0 1 0
1 1 0
0 0 0
0 0 1
1 1 0
0 1 1
60
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1 1 1
0 1
0 1 0
0 1 0
1 1 0
0 0 0
0 0 1
1 1 0
0 1 1
61
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1 1 1
0 1 0
0 1 0
0 1 0
1 1 0
0 0 0
0 0 1
1 1 0
0 1 1
62
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1 1 1
0 1 0
0 1 0
0 1 0
1 1 0
0 0 0
0 0 1
1 1 0
0 1 1
1
63
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1 1 1
0 1 0
0 1 0
0 1 0
1 1 0
0 0 0
0 0 1
1 1 0
1
0 1 1
1
0
64
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1 1 1
0 1 0
0 1 0
0 1 0
1 1 0
0 0 0
0 0 1
1 1 0
1
0 1 1
1
1
0
65
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1 1 1
0 1 0
0 1 0
0 1 0
1 1 0
0 0 0
0 0 1 1
1 1 0
1
0 1 1
1
1
0
66
1sComplement Addition
Additionisthesameasusualbinaryadditionexcept: ifthefinalcarryis1,cyclethecarrytotheleastsignificantdigit: 00010010represents18,11110011represents12 Sum00000110represents6: Carry: x +y presum overflow answer L11
1 1 1 1
0 1 0 0
0 1 0 0
0 1 0 0
1 1 0 0
0 0 0 0
0 0 1 1
1 1 0
1
0 1 1
1
1
0
67
2sComplement
Fixesthenonuniquenessofzeroproblem Addingmixedsignsstilleasy Nocycleoverflow(precomputed) Javasapproach(underthehood) Samefixedlengthk,signconvention,anddefinitionof positivenumbersaswith1scomplement Representnumberswith2(k1)x<2(k1)
EG.Javasbyterangesfrom128to+127
L11
68
2sComplement
Negatives(slightlyharderthan1scomp.): Compute1scomplement Add1 Summarize:x=x+1. 00010010represents18 11101101+1=11101110represents18. Addtogetherwithoutoverflow:00000000 Q:WhataretherangesofJavas32bitintand64bit long?(AllofJavasintegertypesuse2scomplement)
L11
69
2sComplement
A:
2) 32bitints:Largestint=
011111.1=2311=2,147,483,647 Smallestint= 100000.0=231=2,147,483,648 2)64bitlongs:Largestlong= 011111.1=2631=9,223,372,036,854,775,807 Smallestint= 100000.0=263= 9,223,372,036,854,775,808
L11 70
2sComplement Addition
Additionisthesameasusualbinaryadditionno exceptions!! 11101110=(18)10,11110100=(12)10 Sumtogether(11100010)=(30)10:
Carry: x +y
1 1
1 1
1 1
0 1
1 0
1 1
1 0
0 0
L11 71
2sComplement Addition
Additionisthesameasusualbinaryadditionno exceptions!! 11101110=(18)10,11110100=(12)10 Sumtogether(11100010)=(30)10:
Carry: x +y
1 1
1 1
1 1
0 1
1 0
1 1
1 0
0 0 0
L11 72
2sComplement Additi...
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CAS LX 522 Syntax IWeek 9a. A-movement (and a bit more head-movement)NegationnnWeve used negation as a test to see if the verb/auxiliary appears before it or after it as an indication of whether the verb has raised or not. Weve also used adve
BU - LX - 502
Context and meaningCAS LX 5026b. Context and inference 7.4- Nearly everything one reads or hears requiresknowledge of context to interpret. Physical context I don't know why they did it. This can include: Prior conversation And then he ju
BU - LX - 522
CAS LX 522 Syntax IWeek 3a. Categories, features, natural classes, and morphology.Previously, in LX522.n nnSo, here's where we were. We're trying to characterize our knowledge of syntax, using English speaker's knowledge of English as a windo
Pittsburgh - LASA - 98
Cristina Santos University of Toronto LASA 1998 Conference, Chicago The Cultural Politics of the Marginalized Giant: Brazil in Latin American Literary HistoryLiterary history at its height in the nineteenth century tended to contextualize literatur
Pittsburgh - JUL - 2001
2088txt_22to40c5 7/16/01 2:21 PM Page 22Bone, nerve, cartilage. . . all made from scratch from stem cells found where no one thought to look, in muscle. SHOWN HERE : Pitts Johnny Huard regenerated muscle fibers (red) in tissue from an animal that m
Pittsburgh - OCT - 2000
Bora Baysal and Joan Willett-Brozick found a mutation that causes paraganglioma tumors. Thats causing a stir because it looks as if the same defect may encourage the growth of common cancers.12PITTMEDCOVERSTORYWITH HELP FROM ANDEAN MOUNTAIN
BU - PM - 1
BU - PM - 1
Quantum Cryptography with Orthogonal States?In a recent Letter [1], Goldenberg and Vaidman (GV) propose a new quantum cryptographical protocol based on orthogonal states, and claim that it is essentially dierent from all existing quantum cryptosyste
BU - PM - 1
BU - PM - 1
BU - PM - 1
BU - PM - 1
BU - PM - 1
BU - PM - 1
DAMTP/97-124 quant-ph/9711025Unconditionally Secure All-Or-Nothing Disclosure of SecretsAdrian Kent Department of Applied Mathematics and Theoretical Physics, University of Cambridge,quant-ph/9711025 v2 20 Nov 1997Silver Street, Cambridge CB3
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BU - DCC - 2
BU - DCC - 2
BU - DCC - 2
Edward A. Nardell, MD Associate Professor of Medicine Harvard Medical School Tuberculosis Control Officer Massachusetts Department of Public Health President, International Union Against Tuberculosis and Lung Disease, North American RegionComments
BU - LX - 502
CAS LX 502 Semantics3a. A formalism for meaning (contd) 3.2, 3.6Recap!F1 = Rules for generating and interpreting a small fragment of English.!Syntax: Phrase structure rules! !Reviewed on the next slide Idea: All and only sentences generat
BU - LX - 502
CAS LX 502 Semantics2b. A formalism for meaning 2.5, 3.2, 3.6Truth and meaning!The basis of formal semantics: knowing the meaning of a sentence is knowing under what conditions it is true.!Formal semantics, a.k.a. truth conditional semantics
BU - LX - 502
CAS LX 502 Semantics8a. Sense, reference, intension, extension, modality 5.1-2,3-4;7The topic of the class!Weve spent a fair amount of time talking about how we can build up an understanding of the meanings of sentence (or at least the truth co
BU - LX - 502
Your dog ate my homework.CAS LX 5021a. Introduction What does this sentence mean? Is it true? What is the current status of my homework? What was the prior status of my homework? Your dog will have eaten my homework. Your dog must have eat
Pittsburgh - CS - 2740
University of Pittsburgh CS 2740 Knowledge Representation Professor Milos HauskrechtHandout 2 September 10, 2008Problem assignment 1Due: Wednesday, September 17, 2008LispThe goal of this assignment is to practice your lisp programming skills
Pittsburgh - CS - 441
CS 441 Discrete Mathematics for CS Lecture 11Sets and set operationsMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 441 Discrete mathematics for CSM. HauskrechtCourse administrationHomework 3: Due today Homework 4: Due next week
Pittsburgh - CS - 2740
CS 2740 Knowledge Representation Lecture 4Propositional logicMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 2740 Knowledge RepresentationM. HauskrechtAdministration Homework assignment 1 is out Due next week on Wednesday, Septem
Pittsburgh - CS - 441
CS 441 Discrete Mathematics for CS Lecture 8Methods of ProofMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 441 Discrete mathematics for CSM. HauskrechtCourse administration Homework 1 and Homework 2: Due today Homework 3 out to
Pittsburgh - CS - 1571
CS 1571 Introduction to AI Lecture 13Propositional logicMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 1571 Intro to AIM. HauskrechtAnnouncements Homework assignment 5 is out Midterm exam: Wednesday, October 25, 2006 Course web p
Pittsburgh - CS - 3750
Probabilistic Principal Component Analysis and the E-M algorithmThe Minh Luong CS 3750 October 23, 2007Outline Probabilistic Principal Component Analysis Latent variable models Probabilistic PCA Formulation of PCA model Maximum likelihood est
Pittsburgh - CS - 441
CS 441 Discrete Mathematics for CS Lecture 19Summations, CardinalityMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 441 Discrete mathematics for CSM. HauskrechtCourse administration Homework 5 is due today Homework 6 is out Due o
Pittsburgh - CS - 2740
University of Pittsburgh CS 2740 Knowledge Representation Professor Milos HauskrechtHandout 5 October 1, 2008Problem assignment 4Due: Wednesday, October 8, 2008UnicationThe unication process in FOL aims to nd the most general substitution tha
Pittsburgh - CS - 2740
CS 2740 Knowledge representation Lecture 1Knowledge representationMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 2740 Knowledge representationM. HauskrechtCourse administrationInstructor: Milos Hauskrecht 5329 Sennott Square milo
Pittsburgh - CS - 3750
CS 3750 Machine Learning Lecture 11Monte Carlo inferenceMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 3750 Advanced Machine LearningMarkov chain Monte Carlo Likelihood weighting: samples are generated according to Q and every sampl
Pittsburgh - CS - 2710
CS 2710 Foundations of AI Lecture 8Adversarial searchMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 2710 Foundations of AIGame search Game-playing programs developed by AI researchers since the beginning of the modern AI era Progr
Pittsburgh - CS - 1571
CS 1571 Introduction to AI Lecture 12Adversarial searchMilos Hauskrecht milos@cs.pitt.edu 5329 Sennott SquareCS 1571 Intro to AIM. HauskrechtAnnouncements Homework assignment 4 is out Programming and experiments Simulated annealing + Gen
Pittsburgh - MAY - 2005
FEATUREAN EDITOR ON A MISSION BY JESSICA MESMANSO YOU WANT TOCHANGETHE WORLD?HSome people might say Ive reached a point in my life where Im not afraid. But Ive never been afraid, says Catherine DeAngelis.PHOTOGRAPHY |ere comes Dr. De, h
Pittsburgh - EXP - 1593
Pittsburgh - EXP - 1652
Internet WavesDisruptive Technologies> Cool ToolsDigging Digsby and Tweeting Locally> Database ReviewSmart Search Engines Find Best Facts Page 26Page 19Page 41iJune 2008Vol. 25 I Issue 6The Newspaper for Users and Producers of Elect
Pittsburgh - SIS - 793
MARLA MISEKMirror Image InternetThrough the Looking GlassIike the heroine of Lewis Carroll's Alice's Adventures in Wotiderland, who finds herself in a fantastical place where things are not always as they seem, Intemet users generally have no