[PDF] - MA THEMA TICAL INDUCTION Induction and Deduction PDF Document

SLIDE 1 MA THEMA TICAL INDUCTION

Induction

and Deduction

Mathematical

Induction (its strength)

Examples

{ Sum

f

rst n naturual numb ers Prove b y induction that the sum

f

the rst n natural numb ers is n(n + 1)=2. i.e. S (i) : n X i=1 i = n

(n

+ 1) 2 { Sum

f

p

w

ers

f

t w

Prove

b y induction that the sum

f

the rst n p

w

ers

f

t w

(0..n)

is 2 n+1

1.

i.e. S (i) : n X i=0 2 i = 2 n+1

1

{ nth derivative

f

e k x Prove b y induction that the nth derivative

f

e k x is k n e k x . i.e. S (n) : d n dx n e k x = k n e k x 1

SLIDE 2

Arithmetic

and Geometric p rogressions. { Derivations

f

general fo rms { Pro

fs

b y induction { An exp ression fo r the sum

f

the rst n

dd

numb ers and p rove it using Mathematical In- duction. { Pro cedure fo r deriving an exp ression fo r the sum

f

a p

lynomial

series: P n i=0 p(n k ) where p(n k ) is a p

lynomial

in n

f
rder

k : a 1 n k + a 2 n k 1 +

+

a k

General

template fo r Induction p ro

fs:
One

mo re example: Erro r co rrecting co des S(n): C n is any set

f

bit strings

f

length n that is erro r detecting, then C n contains at most 2 n1 strings. { Assume S(n) { Divide C n+1 into t w

sets
f

strings 1x and 0y { x & y

bviously

b e erro r detecting themselves. { By hyp

thesis,

x/y contain no mo re than 2 n1 strings. { Thus maximum total numb er

f

strings is 2

2

n1 = 2 n+11 2

SLIDE 3 Complete (strong) Vs W eak induction

Complete

(o r strong) induction: Uses t w

r

mo re assumptions from the basis up to n.

Examples:

Prove that there exist integers a and b such that 8n 2 Z \ n > : n = 2a + 3b. Prove that there exist no integers x; y ; z such that 8n 2 Z \ n > 2 : x n + y n = z n .

F

allacies and getting ca rried a w a y with induction (All ma rbles a re red, 8n 2 Z : a n = 1)

Structural

induction: Proving p rop erties ab

ut

a structure b y using a basis case and an inductive assumption.

Examples:

In a colony

f

aphids rep ro ducing asexually , let the status

f

an aphid b e rep resented b y the numb er

f

its children. Sho w that the total numb er

f

aphids in the colony is

ne

mo re than the sum

f

the sta- tuses

f

all the aphids. Assume that a pa rticula r implementation

f

Bina ry trees uses the No de with right and left child p

int-

ers. In such a bina ry tree, p rove that the numb er

f

NULL p

inters

is

ne

mo re than the total numb er

f

no des. 3

SLIDE 4 Proving Program Prop erties What a re lo

p

inva riants? A Selection So rt p rogram: (1) for (i = 0; i < n-1; i++) { (2) small = i; (3) for (j = i+1; j < n; j++) (4) if (A[j] < A[small]) (5) small = j; (6) swap(&A[small], &A[i]); (7) } S (k ): If w e reach the test fo r j < n

n

line (3) with j = k then smal l indexes smallest

f

A[i::k

1].

4

SLIDE 5 Basis: S (i + 1): smal l indexes smallest

f

A[i::i]. Induction: Pro

f
f

S (k + 1) assuming S (k ). S (k ): When testing fo r j < n at line (3), smal l indexes smallest

f

A[i::k

1].

No w consider what happ ens in the lo

p

b

dy

when j = k , sp ecically in the \if " statement

f

lines (4) and (5). if A[k ] is less than the smallest

f

A[i + 1::k

1]

then smal l = k . Then smal l indexes smallest

f

A[i + 1::k ]. If A[k ] is not less than the smallest

f

A[i + 1::k

1]

then the value

f

small is unchanged. So small will no w b e the index

f

the smallest

f

A[i + 1::k ]. Thus, in either case, if the lo

p

test is reached subse- quently with the value

f

j incremented from k to k + 1, smal l indexes the smallest

f

A[i + 1::k ] which p roves S (k ) holds fo r all values

f

k

i

+ 1. Question: what is the truth value

f

S (k ) when k > n) 5

SLIDE 6 No w consider the whole SelectionSo rt function. The follo wing assertion is made ab

ut

it. T (m): If w e reach the test i < n

1

at line 1, with i = m, then 1. A[0::m

1]

a re in non-decreasing

rder.

2. All

f

A[m::n

1]

a re greater than

r

equal to any

f

A[0::m

1].

Basis: m = 0. 1. A[0::

1]

a re

bviously

in so rted

rder.

2. All

f

A[0::n

1]
any
f

A[0::

1]

Induction: Pro

f
f

T(m+1) assuming T(m). Consider what happ ens when i = m. W e kno w (b y the IH) that A[0::m

1]

a re in so rted

rder

and no element

f

A[0::m

1]

is greater than any element

f

A[m::n

1].

w e exit the lo

p
f

lines (3){(5), the lo

p

va riable j w

uld

have the value n and so at that p

int

smal l will index the smallest

f

A[m::n

1],

b y the p revious induction. 6

SLIDE 7 No w note t w

things

ab

ut

the situation: 1. By the rst condition

f

the IH, w e kno w that A[0::m

1]

a re already in so rted

rder.

2. By the second condition, w e kno w that smal l is greater than

r

equal to any item in A[0::m

1].

Thus, 1. After sw apping A[m] with A[smal l ], incrementing i to m + 1, p ro ceeding a round the lo

p,

just b efo re the test, w e can assert that A[0::m] is so rted which is condition 1. 2. A[smal l ] is the smallest

f

A[m::n

1]

and that b

th

A[m] and all elements

f

A[0::m

1]

a re less than

r

equal to any element in A[m + 1::n

1]

which is condition 2. Note that when m = n, w e have exited the

uter

lo

p

and so, A[0::n

1]

a re in so rted

rder.

7

SLIDE 8 Pro

f

numb er 2: F acto rial function. (1) int factorial(int n) { (2) int i = 2; (3) int fact = 1; (4) while (i <= n) { (5) fact = fact * i; (6) i++; } (7) return fact; } First { Pro

f
f

termination: Chose E = n

i

+ 1. Second | Pro

f
f

co rrectness: S (k ): If w e reach the test i <= n at line (4), with the value

f

the lo

p

va riable i set to k , then fact will contain the value (k

1)!

Basis: The basis is S (2). i = 2 ! f act = 1 b y assign- ment. Induction: Assume S (k ). Prove S (k + 1). 8

SLIDE 9 Proving p rop erties

f

recursive functions: (1) int fact(int n) (2) if (n <= 1) (3) return 1; (4) else (5) return n * fact(n-1); W e w ant to p rove the follo wing inductive assertion

n

i. S (i) : If the function fact is called with the a rgument i, it returns the value i! Basis: Consider S (1). Line (2) p erfo rms a test which succeeds and so facto rial returns 1 which is indeed 1! Induction: Assume S (k ), i.e. fact(k) returns k !. No w consider what will happ en if fact() is called with a rgument k + 1 with k

1

The test

n

line (2) fails and so the else pa rt is executed. Thus the call

f

facto rial returns the p ro duct

f

(k + 1)

f

act(k ). Ho w ever, b y the IH, fact(k) will return the value k !. Thus the value returned b y the p resent call will b e (k + 1)

k

! which b y the denition

f

the facto rial function is (k + 1)!. QED. 9

SLIDE 10 Recursive Selection so rt: (1) void recSS(int A[], int i, int n) { (2) if (i < n-1) { (3) small=i; (4) for (j = i; j < n; j++) (5) if (A[j] < A[small]) (6) small = j; (7) swap(&A[small], &A[i]); (8) recSS(A, i+1, n); (9) } (10) } S (k ): If the function recSS() is called with i = k , then when it returns it will leave A[k ::n

1]

in non-descending

rder.

Basis: S (n

1).

The IH is fo r S (k ) where

k

< n. W e must sho w that S (k

1)

holds given S (k ). 10

SLIDE 11 Merge So rt: LIST MakeList() { int x; LIST pNewCell; if (scanf(``%d'', &x) == EOF) return NULL; else { pNewCell = (LIST) malloc(sizeof(CELL)); pNewCell->next = MakeList(); pNewCell->element = x; return pNewCell; } } LIST MergeSort(LIST list) { LIST SecondList; if (!list || !list->next) return list; else { SecondList = split(list); return merge(MergeSort(list), MergeSort(SecondList)); } } 11

SLIDE 12 LIST merge(LIST list1, LIST list2) { if (!list1) return list2; if (!list2) return list1; if (list1->element <= list2->element) { list1->next = merge(list1->next, list2); return list1; } else { list2->next = merge(list2->next, list1); return list2; } } LIST split(LIST list) { LIST pSecondCell; if (!list || !list->next) return NULL; else { pSecondCell = list->next; list->next = pSecondCell->next; pSecondCell->next = split(pSecondCell->next); return pSecondCell; } } 12

SLIDE 13 Running Times 1. Log means log 2 in this mo dule. 2. Ab

ut

counting machine instructions... 3. The 90{10 rule: (Eg. Exp eriment using gp rof/gmon under Unix) 4. Why is it p

intless

to count the actual numb er

f

seconds? 5. Benchma rking Vs Analysis 6. Running time is conventionally denoted b y T (n) (no units) A selection so rt p rogram fragment: (2) small = i; (3) for (j = i+1; j < n; j++) { (4) if (A[j] < A[small]) (5) small = j; T

tal

time sp ent in this lo

p

is 4(n

i
1)

+ 1. What is meant b y quadratic and linea r time? 13

SLIDE 14 Big-Oh notation: T (n) is O (f (n)) if there exists some integer n > and constant c > such that 8n > n : T (n)

cf

(n) Example: Find the Big-oh fo r T (n) = (n + 1) 2 (Hint: 3,2

r

1,4) Question: Can T (n) b e considered O (n 2 =1000)? Sho w that: log n is p n etc. 1. Constant facto rs don't matter 2. Lo w

rder

terms don't matter Thumb rule: if lim n!1 h(n) g (n) = then O (g (n) + h(n)) = O (g (n)). The T ransitive La w: if f (n) is O (g (n)) and g (n) is O (h(n)), then f (n) is O (h(n)). Some useful p

ints:

1. if p and q a re p

lynomials

in n and deg r ee(p)

deg

r ee(q ) then p is O (q ) 2. if deg r ee(p) > deg r ee(q ) then p is not O (q ) 3. Every exp

nential

gro ws faster than any p

lynomial

4. No exp

nential

is O (p) where p is a p

lynomial

14

SLIDE 15 Some common exp ressions and their names: BIG-OH INF ORMAL NAME O (1) constant O (log n) loga rithmic O (n) linea r O (n log n) n log n O (n 2 ) quadratic O (n 3 ) cubic O (2 n ) exp

nential

Example: What is the time complexit y

f

the follo wing fragment: (1) scanf(``%d'', &n); (2) for (i = 0; i < n; i++) (3) for (j = 0; j < n; j++) (4) A[i][j] = 0; (5) for (i = 0; i < n; i++) (6) A[i][i] = 1; Discuss p rogram running times fo r: 1. Primitive

p

erations 2. fo r/while lo

ps

(simple and complex) 3. if/switch statements 4. function calls (in lo

ps

(in init/tests/b

dy))

Dra wing a pa rse tree fo r a statement and nding the running time. Use as example the Selection so rt fragment. 15

SLIDE 16 Analysing recursive functions: F

r

the facto rial function, Basis: T (1) = O (1) Induction: T (n) = O (1) + T (n

1)

Let the basis tak e a units

f

time. And the inductive step tak e b + T (n

1)

units

f

time. T (1) = a T (2) = a + b T (3) = a + b + b = a + 2b . . . T (n) = a + (n

1)b

Solving b y rep eated substitution: W e have the sequence

f

equations T (n) = b + T (n

1)

T (n

1)

= b + T (n

2)

. . . T (2) = b + T (1) 16

SLIDE 17 Substituting rep eatedly , w e get T (n) = (n

1)b

+ a Consider T(n) = b + T(n-1) to b e the rst substitution. W e w ant to p rove b y induction

n

the numb er

f

substitutions i (something not stated explicitly in the b

k):

S (i): If 1

i

< n, then T (n) = ib + T (n

i)

so that w e can assert that after (n-1) substitutions, T (n) = (n

1)b

+ T (1). Basis: i = 1, W e kno w T (n) = b + T (n

1)

is true from its denition. Induction: Consider i < n

1.

Then, S (i) : T (n) = ib + T (n

i).

S (i + 1) is the (i + 1)th substitution. So substitute

nce

mo re in S (i), i.e. Substitute fo r T (n

i)

in S (i). T (n) = ib + b + T (n

i
1)

= (i + 1)b + T (n

(i

+ 1)) = S (i + 1) Q.E.D. Analysis

f

Merge so rt: Simple to sho w that merge() tak es O (n) time

n

a list length n 17

SLIDE 18 Consider split() T (0) = O (1) and T (1) = O (1). Let them b e a and b resp. T (n) = O (1) + T (n

2),

i.e. T (n) = c + T (n

2)

Why? T (2) = c + T (0) = a + c T (3) = c + T (1) = b + c T (4) = c + T (2) = a + 2c . . . T (n) = a + cn=2 fo r even n and T (n) = b + c(n

1)=2

fo r

dd

n. Pro

f

b y induction

n

the numb er

f

substitutions: T (n) = c + T (n

2)

= 2c + T (n

4)

. . . S (i): If 1

i

< n=2, then T (n) = ic + T (n

2i)

Substitute

nce

mo re fo r T (n

2i)

to get the (i + 1)th substitution. T (n) = ic + c + T (n

2i
2)

= (i + 1)c + T (n

2(i

+ 1)) = S (i + 1) Q:E :D : 18

SLIDE 19 No w consider the function MergeSo rt() Easy to see that T (n) = 2T (n=2) + O (n), i.e. T (n) = 2T (n=2) + bn Let's b egin the substitutions T (1) = a T (2) = 2T (1) + 2b = 2a + 2b T (4) = 2T (2) + 4b = 4a + 8b T (8) = 2T (4) + 8b = 8a + 24b T (16) = 2T (8) + 16b = 16a + 64b . . . Generalising w e get, T (n) = an + bn log 2 n An induction

n

the numb er

f

substitutions w e need to mak e to T (n) b efo re ending up with T (1). This will

bviously

b e log 2 n since n is halved at each stage. (Ho w many times can w e cut a numb er to half its size b efo re ending up with a numb er <= 1?) 19

SLIDE 20 T

intuit

S (i), consider the follo wing substitutions T (n) = 2T (n=2) + bn = 2(2T (n=4) + bn=2) + bn = 4T (n=4) + 2bn = 4(2T (n=8) + bn=4) + 2bn = 8T (n=8) + 3bn . . . So in general, S (i): If 1

i
log

2 n, then T (n) = 2 i T (n=2 i ) + ibn Pro

f:

Basis: T (n) = 2T (n=2) + bn, true from the denition. Induction: Assume S (i) : T (n) = 2 i T (n=2 i ) + ibn Consider the next substitution into S (i), T (n) = 2 i T (n=2 i ) + ibn = 2 i (2T ((n=2 i )=2) + bn=2 i ) + ibn = 2 i+1 T (n=2 i+1 ) + (i + 1)bn = S (i + 1) Q:E :D : Making log 2 n substitutions to T (n), w e end up with the base case: T (n) = an + bn log 2 n and so T (n) is O (n log n). 20

SLIDE 21 Solving Recurrence Relations Tw

metho

ds: 1. Rep eated substitution 2. Guessing and p roving (Don't freak

ut!!)

Example

f

rst t yp e (Rep eated substitution) T ry T (1) = a; T (n) = T (n

1)

+ g (n) W e can sho w b y rep eated substitution, i times, T (n) = T (n

i)

+ i1 P j =0 g (n

j

) T

get

the basis case (where no mo re subs can b e made, w e must use i = n

1.

This gives: T (n) = T (1) + n2 P j =0 g (n

j

) F

r

facto rial, g (n) = O (1) = b F

r

recursive selection so rt, g (n) = O (n) = bn T ry to derive the complexities

f

the t w

functions.

21

SLIDE 22 Merge so rt: T (n) = 2T (n=2) + g (n) = 4T (n=4) + 2g (n=2) + g (n) = 8T (n=8) + 4g (n=4) + 2g (n=2) + g (n) . . . = 2 i T (n=2 i ) + X j =0 i

12

j g (j =2 j ) The base case when no mo re subs a re to b e made is reached after log 2 n substitutions. So, T (n) = an + log 2 n1 P j =0 2 j g (j =2 j ) F

r

merge so rt, g (n) = bn (recall) So, T (n) = an + log 2 n1 X j =0 bn = an + bn log 2 n is O (n log n) Q:E :D : 22

SLIDE 23 Example

f

second t yp e (b y guessing) T ry to p rove the complexit y

f

merge so rt. Guess that T (n) is b

unded

b y cn log n + d No w p ro ceed along the follo wing lines. Prove the follo wing b y induction. S (n) : F

r

merge so rt, T (n)

f

(n) where f (n) = cn log n+ d where n

1

is a p

w

er

f

2. Basis: S (1) is true if a

d

(Recall that T (1) = a and T (n) = 2T (n=2) + bn. No w assume S (i) fo r 1

i

< n and p rove S (n) as b elo w: W e w ant to p rove T (n)

cn

log n + d fo r some c; d, that is there exists some value

f

c and d fo r which the ab

ve

inequalit y is true; can w e nd them? 23

SLIDE 24 T (n) = 2T (n=2) + bn

2(c

n 2 log n 2 + d) + bn

cn(log

n

log

2) + 2d + bn

cn

log n

cn

+ 2d + bn

cn

log n + d + n(b

c)

+ d Can w e nd a b; c and d such that fo r n

1,

n(b

c)

+ d

0?

If so, the ab

ve

inequalit y will hold. Since n(b

c)

+ d

0;

n(b

c)
d.

Also, since n

1,

this is

nly

true if b

c
d.

Also, since a

d,

assume a = d then w e have b

c
a
r

a + b

c.

Letting c = a + b, and d = a, w e have T (n)

(a

+ b) log n + a Thus T (n) is O (n log n) 24

SLIDE 25 Some common recurrences and their running times: T (n) BIG-OH T (n

1)

+ bn k O (n k +1 ) cT (n

1)

+ bn k ; c > 1 O (c n ) cT (n=d) + bn k ; c > d k O (n log d c ) cT (n=d) + bn k ; c < d k O (n k ) cT (n=d) + bn k ; c = d k O (n k log n) If time p ermits: T ry to guess an p rove an upp er b

und

fo r the follo wing recurrence: Basis: G(1) = 3 Induction: G(n) = (2 ( n 2 ) + 1)G( n 2 ) fo r n > 1. Hints: 1. First expand G(n) b y rep eated substitution to get the dominant term in the exp ression. 2. Don't fo rget the constant term (d) 25

SLIDE 26 Combinato rics Counting assignments { Colours to houses, etc. k n r ul e (Also called selections with replacement) Example: Ho w many bits do y

u

need to uniquely address every b yte in a memo ry bank

f

100 MB? (2 n = 100

1000
1024).

Counting p ermutations { Q (n) Ordered selections without replacement: Q (n; m) = num- b er

f

w a ys w e can select m items from n such that

rder

matters fo r the selected items

nly

. ( Q (n; m) = n(n

1)(n
2)
(n
m

+ 1) = n! (nm)! ) Computing logs

f

la rge facto rials Uno rdered selections (also called combinations)

n

m

=

Q (n;m) Q (m) Example: Numb er

f

p

k

er hands (5 ca rds) 26

SLIDE 27 Imp

rtant

p rop erties

f
n

m

fo

r < m < n;

n

m

=
n
1

m

+
n
1

m

1
n

m

=
n

n

m
P

ascal's triangle:

n

m

=

P [n + 1][m + 1] (mth entry in the nth ro w, numb ering from 0) Computing ratios

f

la rge facto rials Running times

f

functions to compute p ermutations and combinations. (Contrast with running times

f

functions to list all the p ermutations and combinations, Recursive and iterative versions.) Why

n

m

must

yield an integer Shap e

f

the function f (m) =

k

m

27

SLIDE 28 Binomial co-ecients: Consider (x + y ) n fo r n = 1; 2; :::; i (x + y ) n = X m=0 n

n

m

x

m y nm Consider sp ecial cases

f

ab

ve

when y = 0; 1

r

x = y = 1 Multinomials: P ermutations with identical elements. Distribution

f
bjects

to bins Distributing distinguishable

bjects

in k classes 28

SLIDE 29 DIJKSTRA'S SHORTEST-P A TH ALGORITHM This will b e the last lecture from me this semester fo r this course. I will, ho w ever, see y

u

again after the Easter b reak fo r the mid-semester test. The plan

f

this lecture: 1. A ttempt to nd y

ur
wn

sho rtest path algo rithm. 2. Description and discussion

f

Dijkstra's algo rithm 3. Pro

f
f

the algo rithm and why it w

rks.

4. Silent admiration and app reciation

f

the b eaut y , elegance and simplicit y

f

the algo rithm. Example graph: A ! B = 15 B ! C = 12 A ! C = 20 B ! D = 28 D ! E = 24 E ! F = 11 C ! F = 13 29

SLIDE 30 Sta rt

b

y assuming that all

ther

no des a re at distance innit y from the source no de and gradually adjust these distances as w e go

n.

The sho rtest distances a re discovered from the source no de to every

ther

no de, in the

rder
f

nea rness, i.e. nea rest no des a re decided b efo re lo

king

at fa rther no des. Some denitions:

A

no de is called settled if the minimum distance to it from the source is kno wn.

The

sho rtest sp ecial path (SSP) to an unsettled no de, if it exists, is a path that travels

nly

through settled no des except at the last step.

W

e maintain a value dist(u) fo r every no de u such that if the no de gets settled, then dist(u) = length

f

sho rtest path to u. If u is not settled, then dist(u) = length

f

sho rtest sp ecial path to u. 30

SLIDE 31 HO W TO SETTLE A NODE: Adjust the value

f

dist(u) fo r all no des u that remain unsettled and a re aected b y the settled no de. i.e. T

settle

no de v , fo r each a rc v ! u, if dist(v ) + l en(v ! u) < dist(u), then dist(u) := dist(v ) + l en(v ! u) A t the sta rt

f

the algo rithm, no no des a re settled and there exists a sp ecial path

f

length zero from the source no de to itself. A t each pass, settle

ne

unsettled no de with the least value

f

dist(u) as describ ed ab

ve.

The algo rithm terminates when all no des have b een settled. 31

SLIDE 32 IMPORT ANT POINT: A t the b eginning

f

any pass

ver

the no des in the algorithm, the sho rtest

f

all sp ecial paths is THE sho rtest path to the no de it go es to. Note that this is not the case with the

ther

sp ecial paths (non-sho rtest sp ecial paths). That is why this is the no de that is settled next. T

see

why , let the sho rtest

f

all sp ecial paths go to no de v and another sp ecial path, (not the sho rtest

ne)

go to no de u. THERE IS A POSSIBILITY that there might b e a path from v to u, such that going to v and thence to u is sho rter than going to u directly . Ho w ever, THERE IS NO POSSIBILITY that going to u and thence to v is sho rter than going to v directly since going to u is itself longer than going to v . So there is A RISK in settling u b efo re v , but there is NO RISK in settling v b efo re u. Sp end a few minutes

n

this p

int

and understand this completely and y

u

will feel absolutely comfo rtable with the inductive p ro

f.

AL TERNA TEL Y, PONDER: Why should no des b e settled in the

rder
f

nea rness? This is in fact a critical requirement fo r the algo rithm. Consider the scena rio when there exist sp ecial paths

f

length 10 and 20 from the source s to t w

no

des u and v . What gua rantees that the algo rithm will w

rk:

Settling u rst

r

v ? 32

SLIDE 33 PROOF BY INDUCTION: S(n): When there a re k settled no des, a F

r

each settled no de, u; dist(u) gives the minimum distance from s to u and the sho rtest path s ! u consists entirely

f

settled no des. b F

r

each unsettled no de u; dist(u) is the minimum distance

f

any sp ecial path from s ! u

r

1 if no such path exists. First p rove that (a) holds after the (k + 1)th no de, v is settled: Pro

f

is b y contradiction: Assume that dist(u) is not the sho rtest path to v . Then there must have existed some non-sp ecial path that is sho rter than the sp ecial path since b y the Inductive Hyp

thesis

(b), w e kno w that when k no des w ere settled, the sho rtest sp ecial path to v is the sho rtest

f

all sp ecial paths to v . But w e can't have an unsettled w b e

n

a non-sp ecial path to v b ecause (see P

nder

p

int

ab

ve)

w should have b een settled b efo re v . Therefo re (a) holds. Then p rove that (b) also holds after v is settled: Consider an a rbitra ry unsettled no de u after v is settled. Either there is no sp ecial path to u

r

there is

ne.

If the fo rmer, then dist(u) = 1 is indeed the sho rtest

f

sp ecial paths to it b y denition. If the latter, then either v is pa rt

f

the sp ecial path

r

not. If v is pa rt

f

the sp ecial path, then it must b e the p enultimate no de (Why?). In this case, the l en(s ! u) = dist(v ) + l en(v ! u). If v is not pa rt

f

the sp ecial path, then the settling

f

v has no eect

n

the length

f

the path to u. Since the algo rithm sets dist(u) to the smaller

f

the

ld

value

f

dist(u) and dist(v ) + l eng th(v ! u), if settling v changes the sho rtest path to u then it will reset dist(u) acco rdingly . Q.E.D. 33

SLIDE 34 COMMON POINT OF CONFUSION: Do not confuse \Sho rtest

f

all sp ecial paths to A NODE" with \Sho rtest

f

all sp ecial paths." The fo r- mer is used when w e have mo re than

ne

sp ecial path to a given no de (t ypically up

n

a settling a new no de). W e reset the dist() fo r that no de to b e the sho rtest

f

all sp ecial paths to it. The latter is the sho rtest

f

all sp ecial paths to any unsettled no de. This p

ints

to the next no de to b e settled. Final note: Investigate the follo wing w eb site: http://www.MapsOnUs.com T ry planning the sho rtest route from T

m

Bradley Intl Airp

rt,

LA to W all street, NY. What ab

ut

the fastest? Why a re they dierent? 34