17 Mathematical Induction

11

Mathematical InductionMathematical Induction

CS/APMA 202CS/APMA 202

Rosen section 3.3Rosen section 3.3

Aaron BloomfieldAaron Bloomfield

22

What is induction?What is induction?

A method of proofA method of proof

It does not generate answers: It does not generate answers: it only can prove themit only can prove them

Three parts:Three parts: Base case(s): show it is true Base case(s): show it is true

for one elementfor one element Inductive hypothesis: assume Inductive hypothesis: assume

it is true for any given elementit is true for any given elementMust be clearly labeled!!!Must be clearly labeled!!!

Show that if it true for the next Show that if it true for the next highest elementhighest element

33

Show that the sum of the first Show that the sum of the first nn odd odd integers is integers is nn22

Example: If Example: If nn = 5, 1+3+5+7+9 = 25 = 5 = 5, 1+3+5+7+9 = 25 = 522

Formally, ShowFormally, Show

Base case: Show that P(1) is trueBase case: Show that P(1) is true

Induction exampleInduction example

n

i

ni1

212

11

11)(21

1

2

i

i

)P( where)P( nnn

)1P(

44

Inductive hypothesis: assume true for Inductive hypothesis: assume true for kk Thus, we assume that P(Thus, we assume that P(kk) is true, or that) is true, or that

Note: we don’t yet know if this is true or not!Note: we don’t yet know if this is true or not!

Inductive step: show true for Inductive step: show true for kk+1+1 We want to show that: We want to show that:

k

i

ki1

212

1

1

2)1(12k

i

ki

Induction example, continuedInduction example, continued

55

12121)1(2 2

1

kkikk

i

121)1(2 22 kkkk

Induction example, continuedInduction example, continued

Recall the inductive hypothesis:Recall the inductive hypothesis:

Proof of inductive step:Proof of inductive step:

k

i

ki1

212

1212 22 kkkk

21

1

)1(12

kik

i

66

What did we showWhat did we show

Base case: P(1)Base case: P(1)If P(If P(kk) was true, then P() was true, then P(kk+1) is true+1) is true

i.e., P(i.e., P(kk) → P() → P(kk+1)+1)

We know it’s true for P(1)We know it’s true for P(1)Because of P(Because of P(kk) → P() → P(kk+1), if it’s true for P(1), then it’s true for P(2)+1), if it’s true for P(1), then it’s true for P(2)Because of P(Because of P(kk) → P() → P(kk+1), if it’s true for P(2), then it’s true for P(3)+1), if it’s true for P(2), then it’s true for P(3)Because of P(Because of P(kk) → P() → P(kk+1), if it’s true for P(3), then it’s true for P(4)+1), if it’s true for P(3), then it’s true for P(4)Because of P(Because of P(kk) → P() → P(kk+1), if it’s true for P(4), then it’s true for P(5)+1), if it’s true for P(4), then it’s true for P(5)And onwards to infinityAnd onwards to infinity

Thus, it is true for all possible values of Thus, it is true for all possible values of nn

In other words, we showed that:In other words, we showed that: )P()1P()P()1P( nnkkk

77

The idea behind inductive proofsThe idea behind inductive proofs

Show the base caseShow the base case

Show the inductive hypothesisShow the inductive hypothesis

Manipulate the inductive step so that you Manipulate the inductive step so that you can substitute in part of the inductive can substitute in part of the inductive hypothesishypothesis

Show the inductive stepShow the inductive step

8

Quick surveyQuick survey

I felt I understood the first example of I felt I understood the first example of induction…induction…

a)a) Very wellVery well

b)b) With some review, I’ll be goodWith some review, I’ll be good

c)c) Not reallyNot really

d)d) Not at allNot at all

9


I felt I could do my own inductive I felt I could do my own inductive proof…proof…





1010

Why speling is not so Why speling is not so important…important…

I cdnuolt blveieetaht I cluod aulaclty uesdnatnrd I cdnuolt blveieetaht I cluod aulaclty uesdnatnrd waht I was rdanieg. The phaonmneal pweor of waht I was rdanieg. The phaonmneal pweor of thehmuan mind. Aoccdrnig to a rscheearch at thehmuan mind. Aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it deosn't mttaer in waht Cmabrigde Uinervtisy, it deosn't mttaer in waht oredr the ltteers in a wrod are, the olny oredr the ltteers in a wrod are, the olny iprmoatnt tihng is taht thefrist and lsat ltteer be iprmoatnt tihng is taht thefrist and lsat ltteer be in the rghit pclae. The rset can be a taotl mses in the rghit pclae. The rset can be a taotl mses andyou can sitll raed it wouthit a porbelm. Tihs andyou can sitll raed it wouthit a porbelm. Tihs is bcuseae the huamn mnid deosnot raed ervey is bcuseae the huamn mnid deosnot raed ervey lteter by istlef, but the wrod as a wlohe. lteter by istlef, but the wrod as a wlohe. Amzanig huh? yaeh and I awlyas thought Amzanig huh? yaeh and I awlyas thought slpeling was ipmorantt.slpeling was ipmorantt.

1111

Second induction exampleSecond induction example

Rosen, section 3.3, question 2:Rosen, section 3.3, question 2: Show the sum of the first Show the sum of the first nn positive even positive even

integers is integers is nn22 + + nn Rephrased: Rephrased:

The three parts:The three parts: Base caseBase case Inductive hypothesisInductive hypothesis Inductive stepInductive step

n

i

nninnn1

22)P( where)P(

1212

Second induction example, Second induction example, continuedcontinued

Base case: Show P(1):Base case: Show P(1):

Inductive hypothesis: Assume Inductive hypothesis: Assume

Inductive step: Show Inductive step: Show

22

11)(2)1P(1

1

2

i

i

k

i

kkik1

22)P(

)1()1(2)1P( 21

1

kkikk

i

1313

2323 22 kkkk

1)1()1(2 22 kkkkk

1)1(2)1(2 2

1

kkikk

i

1)1(2 21

1

kkik

i

Second induction example, Second induction example, continuedcontinued

Recall our inductive Recall our inductive hypothesis:hypothesis:

k

i

kkik1

22)P(

1414

Notes on proofs by inductionNotes on proofs by induction

We manipulate the We manipulate the kk+1 case to make part +1 case to make part of it look like the of it look like the kk case case

We then replace that part with the other We then replace that part with the other side of the side of the kk case case

k

i

kkik1

22)P(

2323 22 kkkk

1)1()1(2 22 kkkkk

1)1(2)1(2 2

1

kkikk

i

1)1(2 21

1

kkik

i

1515

Third induction exampleThird induction example

Rosen, question 7: ShowRosen, question 7: Show

Base case: Base case: nn = 1 = 1

Inductive hypothesis: assume Inductive hypothesis: assume

6

)12)(1(

1

2

nnni

n

i

116

61

6

)12)(11(1

2

1

1

2

i

i

6

)12)(1(

1

2

kkki

k

i

1616

Third induction exampleThird induction example

Inductive step: show Inductive step: show 6

)1)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)32)(2)(1()1(

1

22

kkkik

k

i

6139261392 2323 kkkkkk

)32)(2)(1()12)(1()1(6 2 kkkkkkk

6

)32)(2)(1(

6

)12)(1()1( 2

kkkkkk

k

6

)1)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)12)(1(

1

2

kkki

k

i

1717

Third induction again: what if your Third induction again: what if your inductive hypothesis was wrong?inductive hypothesis was wrong?Show:Show:

Base case: Base case: nn = 1: = 1:

But let’s continue anyway…But let’s continue anyway…Inductive hypothesis: assume Inductive hypothesis: assume

6

)22)(1(

1

2

nnni

n

i

6

71

6

71

6

)22)(11(1

2

1

1

2

i

i

6

)22)(1(

1

2

kkki

k

i

1818

Third induction again: what if your Third induction again: what if your inductive hypothesis was wrong?inductive hypothesis was wrong?Inductive step: show Inductive step: show

6

)2)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)42)(2)(1()1(

1

22

kkkik

k

i

816102614102 2323 kkkkkk

)42)(2)(1()22)(1()1(6 2 kkkkkkk

6

)42)(2)(1(

6

)22)(1()1( 2

kkkkkk

k

6

)2)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)22)(1(

1

2

kkki

k

i

1919

Fourth induction exampleFourth induction example

Rosen, question 14: show that Rosen, question 14: show that nn! < ! < nnnn for for all all nn > 1 > 1

Base case: Base case: nn = 2 = 22! < 22! < 222

2 < 42 < 4

Inductive hypothesis: assume Inductive hypothesis: assume kk! < ! < kkkk

Inductive step: show that (Inductive step: show that (kk+1)! < (+1)! < (kk+1)+1)kk+1+1

)!1( k !)1( kk kkk )1( kkk )1)(1( 1)1( kk

20


I felt I understand induction…I felt I understand induction…





2121

Strong inductionStrong induction

Weak mathematical induction assumes Weak mathematical induction assumes P(P(kk) is true, and uses that (and only that!) ) is true, and uses that (and only that!) to show P(to show P(kk+1) is true+1) is true

Strong mathematical induction assumes Strong mathematical induction assumes P(1), P(2), …, P(P(1), P(2), …, P(kk) are all true, and uses ) are all true, and uses that to show that P(that to show that P(kk+1) is true.+1) is true.

)1P()P(...)3P()2P()1P( kk

2222

Strong induction example 1Strong induction example 1

Show that any number > 1 can be written Show that any number > 1 can be written as the product of primesas the product of primes

Base case: P(2)Base case: P(2) 2 is the product of 2 (remember that 1 is not 2 is the product of 2 (remember that 1 is not

prime!)prime!)

Inductive hypothesis: P(1), P(2), P(3), …, Inductive hypothesis: P(1), P(2), P(3), …, P(P(kk) are all true) are all trueInductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true

2323

Strong induction example 1Strong induction example 1

Inductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true

There are two cases:There are two cases: kk+1 is prime+1 is prime

It can then be written as the product of It can then be written as the product of kk+1+1 kk+1 is composite+1 is composite

It can be written as the product of two composites, It can be written as the product of two composites, a and b, where 2 ≤ a and b, where 2 ≤ aa ≤ ≤ bb < < kk+1+1

By the inductive hypothesis, both P(By the inductive hypothesis, both P(aa) and P() and P(bb) are ) are truetrue

2424

Strong induction vs. non-strong Strong induction vs. non-strong inductioninduction

Rosen, question 31: Determine which Rosen, question 31: Determine which amounts of postage can be written with 5 amounts of postage can be written with 5 and 6 cent stampsand 6 cent stamps

Prove using both versions of inductionProve using both versions of induction

Similar to example 15 (page 250)Similar to example 15 (page 250)

Answer: any postage ≥ 20Answer: any postage ≥ 20

2525

Answer via mathematical inductionAnswer via mathematical induction

Show base case: P(20):Show base case: P(20): 20 = 5 + 5 + 5 + 520 = 5 + 5 + 5 + 5

Inductive hypothesis: Assume P(Inductive hypothesis: Assume P(kk) is true) is true

Inductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true If P(If P(kk) uses a 5 cent stamp, replace that stamp with a ) uses a 5 cent stamp, replace that stamp with a

6 cent stamp6 cent stamp If P(If P(kk) does not use a 5 cent stamp, it must use only 6 ) does not use a 5 cent stamp, it must use only 6

cent stampscent stampsSince Since kk > 18, there must be four 6 cent stamps > 18, there must be four 6 cent stamps

Replace these with five 5 cent stamps to obtain Replace these with five 5 cent stamps to obtain kk+1+1

2626

Answer via strong inductionAnswer via strong induction

Show base cases: P(20), P(21), P(22), P(23), and P(24) Show base cases: P(20), P(21), P(22), P(23), and P(24) 20 = 5 + 5 + 5 + 520 = 5 + 5 + 5 + 5 21 = 5 + 5 + 5 + 621 = 5 + 5 + 5 + 6 22 = 5 + 5 + 6 + 622 = 5 + 5 + 6 + 6 23 = 5 + 6 + 6 + 623 = 5 + 6 + 6 + 6 24 = 6 + 6 + 6 + 624 = 6 + 6 + 6 + 6

Inductive hypothesis: Assume P(20), P(21), …, P(Inductive hypothesis: Assume P(20), P(21), …, P(kk) are ) are all trueall true

Inductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true We will obtain P(We will obtain P(kk+1) by adding a 5 cent stamp to P(+1) by adding a 5 cent stamp to P(kk+1-5)+1-5) Since we know P(Since we know P(kk+1-5) = P(+1-5) = P(kk-4) is true, our proof is complete-4) is true, our proof is complete

2727

Strong induction vs. non-strong Strong induction vs. non-strong induction, take 2induction, take 2

Rosen, section 3.4, example 15: Show Rosen, section 3.4, example 15: Show that every postage amount 12 cents or that every postage amount 12 cents or more can be formed using only 4 and 5 more can be formed using only 4 and 5 cent stampscent stamps

Similar to the previous exampleSimilar to the previous example

2828

Answer via mathematical inductionAnswer via mathematical induction

Show base case: P(12):Show base case: P(12): 12 = 4 + 4 + 412 = 4 + 4 + 4

Inductive hypothesis: Assume P(Inductive hypothesis: Assume P(kk) is true) is trueInductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true If P(If P(kk) uses a 4 cent stamp, replace that stamp with a ) uses a 4 cent stamp, replace that stamp with a

5 cent stamp to obtain 5 cent stamp to obtain PP((kk+1)+1) If P(If P(kk) does not use a 4 cent stamp, it must use only 5 ) does not use a 4 cent stamp, it must use only 5

cent stampscent stampsSince Since kk > 10, there must be at least three 5 cent stamps > 10, there must be at least three 5 cent stampsReplace these with four 4 cent stamps to obtain Replace these with four 4 cent stamps to obtain kk+1+1

Note that only Note that only PP((kk) was assumed to be true) was assumed to be true

2929

Answer via strong inductionAnswer via strong induction

Show base cases: P(12), P(13), P(14), and P(15)Show base cases: P(12), P(13), P(14), and P(15) 12 = 4 + 4 + 412 = 4 + 4 + 4 13 = 4 + 4 + 513 = 4 + 4 + 5 14 = 4 + 5 + 514 = 4 + 5 + 5 15 = 5 + 5 + 515 = 5 + 5 + 5

Inductive hypothesis: Assume P(12), P(13), …, P(Inductive hypothesis: Assume P(12), P(13), …, P(kk) are all true) are all true For For kk ≥≥ 15 15

Inductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true We will obtain P(We will obtain P(kk+1) by adding a 4 cent stamp to P(+1) by adding a 4 cent stamp to P(kk+1-4)+1-4) Since we know P(Since we know P(kk+1-4) = P(+1-4) = P(kk-3) is true, our proof is complete-3) is true, our proof is complete

Note that P(12), P(13), …, P(Note that P(12), P(13), …, P(kk) were all assumed to be true) were all assumed to be true

30


I felt I understand strong vs. weak I felt I understand strong vs. weak induction…induction…





3131

An aside: IOCCCAn aside: IOCCC

The International Obfuscated C Code The International Obfuscated C Code ContestContest– Online at http://www.ioccc.orgOnline at http://www.ioccc.org

C has very terse syntaxC has very terse syntax– So the contest tries to make it terser!So the contest tries to make it terser!

One common method is by modifying One common method is by modifying the whitespacethe whitespace

3232

X XX X

X X X XX X X X

X X X XX X X X

X X X XX X X X

X X X XX X X X

X X X XX X X X

X X X XX X X X

X X X X X XX X X X X X

X XX X X XX XX XX X X XX X

X XXX X XXXXXXXXX X XXX XX XXX X XXXXXXXXX X XXX X

X XXX X XXXX XXXX X XXX XX XXX X XXXX XXXX X XXX X

X XXXX X XX ainma(){ archa XX X XXXX XX XXXX X XX ainma(){ archa XX X XXXX X

X XXXX X oink[9],*igpa, X XXXX XX XXXX X oink[9],*igpa, X XXXX X

X XXXXXX atinla=etcharga(),iocccwa XXXXXX XX XXXXXX atinla=etcharga(),iocccwa XXXXXX X

X XXXX ,apca='A',owla='a',umna=26 XXXX XX XXXX ,apca='A',owla='a',umna=26 XXXX X

X XXX ; orfa(; (atinla+1)&&(!((( XXX XX XXX ; orfa(; (atinla+1)&&(!((( XXX X

X XX atinla-apca)*(apca+umna-atinla) XX XX XX atinla-apca)*(apca+umna-atinla) XX X

X X >=0)+((atinla-owla)*(owla+umna- X XX X >=0)+((atinla-owla)*(owla+umna- X X

X atinla)>=0))); utcharpa(atinla), XX atinla)>=0))); utcharpa(atinla), X

X X atinla=etcharga()); orfa(; atinla+1; X XX X atinla=etcharga()); orfa(; atinla+1; X X

X X ){ orfa( igpa=oink ,iocccwa=( X XX X ){ orfa( igpa=oink ,iocccwa=( X X

X X (atinla- XXX apca)*( XXX apca+umna- X XX X (atinla- XXX apca)*( XXX apca+umna- X X

X atinla)>=0) XXX XXX ; (((( XX atinla)>=0) XXX XXX ; (((( X

X atinla-apca XXXXX XXXXXXX XXXXX )*(apca+ XX atinla-apca XXXXX XXXXXXX XXXXX )*(apca+ X

X umna-atinla XXXXXX )>=0) XXXXXX +((atinla- XX umna-atinla XXXXXX )>=0) XXXXXX +((atinla- X

X owla)*(owla+ XXXX umna- XXXX atinla)>=0)) XX owla)*(owla+ XXXX umna- XXXX atinla)>=0)) X

X &&"-Pig-" XX "Lat-in" XX "COb-fus" XX &&"-Pig-" XX "Lat-in" XX "COb-fus" X

X "ca-tion!!"[ X (((atinla- X apca)*(apca+ XX "ca-tion!!"[ X (((atinla- X apca)*(apca+ X

X umna-atinla) X >=0)?atinla- X apca+owla: XX umna-atinla) X >=0)?atinla- X apca+owla: X

X atinla)-owla X ]-'-')||((igpa== X oink)&&!(*( XX atinla)-owla X ]-'-')||((igpa== X oink)&&!(*( X

X igpa++)='w') X )||! X (*( X igpa X ++)=owla); * XX igpa++)='w') X )||! X (*( X igpa X ++)=owla); * X

X (igpa++)=(( X ( XXX XXX X atinla-apca XX (igpa++)=(( X ( XXX XXX X atinla-apca X

X )*(apca+ X umna XXX - XXX X atinla)>=0) XX )*(apca+ X umna XXX - XXX X atinla)>=0) X

X ?atinla- X apca XXX + XXX owla X :atinla), XX ?atinla- X apca XXX + XXX owla X :atinla), X

X atinla= X X X X etcharga()) XX atinla= X X X X etcharga()) X

X ; orfa( X atinla=iocccwa?(( X (atinla- XX ; orfa( X atinla=iocccwa?(( X (atinla- X

X owla)*(owla+ X umna-atinla)>=0 X )?atinla- XX owla)*(owla+ X umna-atinla)>=0 X )?atinla- X

X owla+apca: X atinla): X atinla; ((( XX owla+apca: X atinla): X atinla; ((( X

X atinla-apca)* X (apca+umna- X atinla)>=0)+( XX atinla-apca)* X (apca+umna- X atinla)>=0)+( X

X (atinla-owla)* X (owla+ X umna-atinla)>= XX (atinla-owla)* X (owla+ X umna-atinla)>= X

X 0)); utcharpa( XX XX atinla),atinla XX 0)); utcharpa( XX XX atinla),atinla X

X =etcharga()); XXXXXXX orfa(*igpa=0, XX =etcharga()); XXXXXXX orfa(*igpa=0, X

X igpa=oink; * igpa; utcharpa( XX igpa=oink; * igpa; utcharpa( X

X *(igpa++))); orfa(; (atinla+1)&&(!((( XX *(igpa++))); orfa(; (atinla+1)&&(!((( X

X atinla-apca )*(apca+ XX atinla-apca )*(apca+ X

X umna- XXXXX XXXXX atinla)>=0 XX umna- XXXXX XXXXX atinla)>=0 X

X )+(( XXXXX atinla- XX )+(( XXXXX atinla- X

XX owla)*( owla+umna- XXXX owla)*( owla+umna- XX

XX atinla)>=0))); utcharpa XXXX atinla)>=0))); utcharpa XX

XX (atinla),atinla= XXXX (atinla),atinla= XX

XX etcharga()); } XXXX etcharga()); } XX

XXXX } XXXXXXXX } XXXX

XXXXXXXXXXXXXXXXXX

#define X#define X#define XX#define XX#define XXX#define XXX#define XXXX#define XXXX#define XXXXX#define XXXXX#define XXXXXX#define XXXXXX#define XXXXXXX#define XXXXXXX#define orfa for#define orfa for#define XXXXXXXXX#define XXXXXXXXX#define archa char#define archa char#define ainma main#define ainma main#define etcharga getchar#define etcharga getchar#define utcharpa putchar#define utcharpa putchar

#include <stdio.h>#define Q r=R[*p++-'0'];while(#define B ;break;casechar*s="Qjou!s\\311^-g\\311^-n\\311^-c\\::^-q-ma%mO1JBHm%BQ-aP1J[O1HB%[Q<nbj\o)*|gps)<<*txjudi)m*|aQdbtf!::::;sfuvso<aQefgbvmu;aQ<m,,a%CQ<csfbla%bQ<aN2!Q\\ndbtf!aP2Q;m>aP2Q<a%!D12J!JGJHJOJQJFJSJJJMHS%HD12D12N3!N4\nJUJT%UQm>aP4HC%T\Qs\\q,,^>m,2<m>aP4HC%SD12N1\nJNQm>s\\..q^aHC%NHb%GN1!D32P3%RN1UP1D12JPQUaP1H\R%PN4\nQ<g\\(aP3Q(^>aP2Q,2<n\\(aP3Q(^>aP4Hb%OD12D12N2!N3\nJVP3Q,,<jg)aP3Q=>n\\\(aP3Q(^*m>g\\(aP3Q(^<fmtf!m,,aHC%QN1!N1\nJ#Qqsjoug)#&e]o#-aP1Q*aHb%#Qqvut)\aP1Q*aHb%FN1\nQm>::::aHC%VP3Q>bupj)hfut)c**aHb%JD12JON1!Qjg)a%LN1UP1D12JIQUa\P1HL%IQ*m>aN2!N2\nP2Q<fmtf!m,,aHC%MN1!N2>P2Q>aN2\nP2Hbdd!b/d";k;char R[4][99];main(c,v)char**v;{char*p,*r,*q;for(q=s;*q;q++)*q>' '&&(*q)--;{FILE*i=fopen(v[1],"r"),*o=fopen(q-3,"w");for(p=s;;p++)switch(*p++){B'M':Q(k=fgetc(i))!=EOF&&k!=*p)*r++=k;if(k==EOF){fputs("}}\n",o);fclose(o);return system(q-6);}*r=0B'P':while(*p!='`')fputc(*p++,o)B'O':Q*r)fputc(*r++,o);p--B'C':k=0;Q k<*p-'0')(*r++=fgetc(i),k++);*r=0 B'I':k= *p;if(**R==k)goto G B'G':k= *p;G:p=s;while(*p!='$'||p[1]!= k)p++;p++B'N':R[*p-'0'][0]++;}}}

a(X){/*/X=-a(X){/*/X=- a(X){/*/X=-a(X){/*/X=-

-1;F;X=--1;F;X=- -1;F;X=--1;F;X=-

-1;F;}/*/-1;F;}/*/ -1;F;}/*/-1;F;}/*/

char*z[]={"char*z[]={","a(X){/*/X=-","-1;F;X=-","-1;F;}/*/","9999999999 :-| ",char*z[]={"char*z[]={","a(X){/*/X=-","-1;F;X=-","-1;F;}/*/","9999999999 :-| ",

"int q,i,j,k,X,O=0,H;S(x)int*x;{X+=X;O+=O;*x+1?*x+2||X++:O++;*x=1;}L(n){for(*","int q,i,j,k,X,O=0,H;S(x)int*x;{X+=X;O+=O;*x+1?*x+2||X++:O++;*x=1;}L(n){for(*",

"z[i=1]=n+97;i<4;i++)M(256),s(i),M(128),s(i),M(64),N;X*=8;O*=8;}s(R){char*r=z","z[i=1]=n+97;i<4;i++)M(256),s(i),M(128),s(i),M(64),N;X*=8;O*=8;}s(R){char*r=z",

"[R];for(q&&Q;*r;)P(*r++);q&&(Q,P(44));}M(m){P(9);i-2||P(X&m?88:O&m?48:32);P(","[R];for(q&&Q;*r;)P(*r++);q&&(Q,P(44));}M(m){P(9);i-2||P(X&m?88:O&m?48:32);P(",

"9);}y(A){for(j=8;j;)~A&w[--j]||(q=0);}e(W,Z){for(i-=i*q;i<9&&q;)y(W|(1<<i++&","9);}y(A){for(j=8;j;)~A&w[--j]||(q=0);}e(W,Z){for(i-=i*q;i<9&&q;)y(W|(1<<i++&",

"~Z));}R(){for(k=J[*J-48]-40;k;)e(w[k--],X|O);}main(u,v)char**v;{a(q=1);b(1);","~Z));}R(){for(k=J[*J-48]-40;k;)e(w[k--],X|O);}main(u,v)char**v;{a(q=1);b(1);",

"c(1);*J=--u?O?*J:*v[1]:53;X|=u<<57-*v[u];y(X);K=40+q;q?e(O,X),q&&(K='|'),e(X","c(1);*J=--u?O?*J:*v[1]:53;X|=u<<57-*v[u];y(X);K=40+q;q?e(O,X),q&&(K='|'),e(X",

",O),R(),O|=1<<--i:J[*J-48+(X=O=0)]--;L(q=0);for(s(i=0);q=i<12;)s(i++),i>4&&N",",O),R(),O|=1<<--i:J[*J-48+(X=O=0)]--;L(q=0);for(s(i=0);q=i<12;)s(i++),i>4&&N",

";s(q=12);P(48);P('}');P(59);N;q=0;L(1);for(i=5;i<13;)s(i++),N;L(2);}",0};";s(q=12);P(48);P('}');P(59);N;q=0;L(1);for(i=5;i<13;)s(i++),N;L(2);}",0};

b(X){/*/X=-b(X){/*/X=- b(X){/*/X=-b(X){/*/X=-

-1;F;X=--1;F;X=- -1;F;X=--1;F;X=-

-1;F;}/*/-1;F;}/*/ -1;F;}/*/-1;F;}/*/

int q,i,j,k,X,O=0,H;S(x)int*x;{X+=X;O+=O;*x+1?*x+2||X++:O++;*x=1;}L(n){for(*int q,i,j,k,X,O=0,H;S(x)int*x;{X+=X;O+=O;*x+1?*x+2||X++:O++;*x=1;}L(n){for(*

z[i=1]=n+97;i<4;i++)M(256),s(i),M(128),s(i),M(64),N;X*=8;O*=8;}s(R){char*r=zz[i=1]=n+97;i<4;i++)M(256),s(i),M(128),s(i),M(64),N;X*=8;O*=8;}s(R){char*r=z

[R];for(q&&Q;*r;)P(*r++);q&&(Q,P(44));}M(m){P(9);i-2||P(X&m?88:O&m?48:32);P([R];for(q&&Q;*r;)P(*r++);q&&(Q,P(44));}M(m){P(9);i-2||P(X&m?88:O&m?48:32);P(

9);}y(A){for(j=8;j;)~A&w[--j]||(q=0);}e(W,Z){for(i-=i*q;i<9&&q;)y(W|(1<<i++&9);}y(A){for(j=8;j;)~A&w[--j]||(q=0);}e(W,Z){for(i-=i*q;i<9&&q;)y(W|(1<<i++&

~Z));}R(){for(k=J[*J-48]-40;k;)e(w[k--],X|O);}main(u,v)char**v;{a(q=1);b(1);~Z));}R(){for(k=J[*J-48]-40;k;)e(w[k--],X|O);}main(u,v)char**v;{a(q=1);b(1);

c(1);*J=--u?O?*J:*v[1]:53;X|=u<<57-*v[u];y(X);K=40+q;q?e(O,X),q&&(K='|'),e(Xc(1);*J=--u?O?*J:*v[1]:53;X|=u<<57-*v[u];y(X);K=40+q;q?e(O,X),q&&(K='|'),e(X

,O),R(),O|=1<<--i:J[*J-48+(X=O=0)]--;L(q=0);for(s(i=0);q=i<12;)s(i++),i>4&&N,O),R(),O|=1<<--i:J[*J-48+(X=O=0)]--;L(q=0);for(s(i=0);q=i<12;)s(i++),i>4&&N

;s(q=12);P(48);P('}');P(59);N;q=0;L(1);for(i=5;i<13;)s(i++),N;L(2);};s(q=12);P(48);P('}');P(59);N;q=0;L(1);for(i=5;i<13;)s(i++),N;L(2);}

c(X){/*/X=-c(X){/*/X=- c(X){/*/X=-c(X){/*/X=-

-1;F;X=--1;F;X=- -1;F;X=--1;F;X=-

-1;F;}/*/-1;F;}/*/ -1;F;}/*/-1;F;}/*/

An aside: IOCCCAn aside: IOCCC

#define _ -F<00||--F-OO--;#define _ -F<00||--F-OO--;int F=00,OO=00;main(){F_OO();printf("%1.3f\n",4.*-F/OO/OO);}F_OO()int F=00,OO=00;main(){F_OO();printf("%1.3f\n",4.*-F/OO/OO);}F_OO(){{ _-_-_-__-_-_-_ _-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-_-_-_-_-__-_-_-_-_-_-_-_-_-_-_-_ _-_-_-_-_-_-_-__-_-_-_-_-_-_-_ _-_-_-__-_-_-_}}

3333

Chess and inductionChess and induction7

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7

Can the knight reach any square in a finite number of moves?

Show that the knight can reach any square (i, j) for which i+j=k where k > 1.

Base case: k = 2

Inductive hypothesis: assume the knight can reach any square (i, j) for which i+j=k where k > 1.

Inductive step: show the knight can reach any square (i, j) for which i+j=k+1 where k > 1.

3434

Chess and inductionChess and induction

Inductive step: show the knight can reach any Inductive step: show the knight can reach any square (square (ii, , jj) for which ) for which ii++jj==kk+1 where +1 where kk > 1. > 1. Note that Note that kk+1 ≥ 3, and one of +1 ≥ 3, and one of ii or or jj is ≥ 2 is ≥ 2 If If ii ≥ 2, the knight could have moved from ( ≥ 2, the knight could have moved from (ii-2, -2, jj+1)+1)

Since Since ii++jj = = kk+1, +1, ii-2 + -2 + jj+1 = +1 = kk, which is assumed true, which is assumed true If If jj ≥ 2, the knight could have moved from ( ≥ 2, the knight could have moved from (ii+1, +1, j-2j-2))

Since Since ii++jj = = kk+1, +1, ii+1 + +1 + jj-2 = -2 = kk, which is assumed true, which is assumed true

3535

We are skipping…We are skipping…

The well-ordering propertyThe well-ordering property

Why mathematical induction is validWhy mathematical induction is valid

These are the last two sub-sections of These are the last two sub-sections of section 3.3section 3.3

3636

Question 40Question 40

Take a pile of Take a pile of nn stones stones Split the pile into two smaller piles of size Split the pile into two smaller piles of size rr

and and ss Repeat until you have Repeat until you have nn piles of 1 stone each piles of 1 stone each

Take the product of Take the product of allall the splits the splits So all the So all the rr’s and ’s and ss’s from ’s from each each splitsplit

Sum up each of these productsSum up each of these products

Prove that this product equals Prove that this product equals 2

)1( nn

3737

10

3

14

7

1 12 2 2 1

23

1 1 1 1 1 1

21

12 2

4 2 1

1 1 1


2

)1( nn

10

2

9*104511112421221

3838


We will show it is true for a pile of We will show it is true for a pile of kk stones, and show it is true for stones, and show it is true for kk+1 stones+1 stones So P(So P(kk) means that it is true for ) means that it is true for kk stones stones

Base case: Base case: nn = 1 = 1 No splits necessary, so the sum of the No splits necessary, so the sum of the

products = 0products = 0 1*(1-1)/2 = 01*(1-1)/2 = 0 Base case provenBase case proven

3939


Inductive hypothesis: assume that P(1), P(2), …, Inductive hypothesis: assume that P(1), P(2), …, P(P(kk) are all true) are all true This is strong induction!This is strong induction!

Inductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true We assume that we split the We assume that we split the kk+1 pile into a pile of +1 pile into a pile of ii

stones and a pile of stones and a pile of kk+1-+1-ii stones stones Thus, we want to show that Thus, we want to show that

((ii)*()*(kk+1-+1-ii) + P() + P(ii) + P() + P(kk+1-+1-ii) = P() = P(kk+1)+1) Since 0 < Since 0 < ii < < k+1k+1, both , both ii and and kk+1-+1-ii are between 1 and are between 1 and kk, inclusive, inclusive

4040


kkkk 22

kkiikikkiiiiki 22222 2222

22

2

2

22222 kkiikikkiiiiki

)1P()1P()P()1(*)( kikiiki

22

)11)(1()1P(

2 kkkkk

2

2

2

)11)(1()1P(

22 iikikkikikik

2)P(

2 iii

Thus, we want to show that Thus, we want to show that ((ii)*()*(kk+1-+1-ii) + P() + P(ii) + P() + P(kk+1-+1-ii) = P() = P(kk+1)+1)

41


I felt I understood the material in this I felt I understood the material in this slide set…slide set…





42


The pace of the lecture for this The pace of the lecture for this slide set was…slide set was…

a)a) FastFast

b)b) About rightAbout right

c)c) A little slowA little slow

d)d) Too slowToo slow

43


How interesting was the material in How interesting was the material in this slide set? Be honest!this slide set? Be honest!

a)a) Wow! That was SOOOOOO cool!Wow! That was SOOOOOO cool!

b)b) Somewhat interestingSomewhat interesting

c)c) Rather bortingRather borting

d)d) ZzzzzzzzzzzZzzzzzzzzzz

17 Mathematical Induction

Documents

Transcript of 17 Mathematical Induction