15-299 Lecture 11 Feb 17, 1998 Doug Beeferman Carnegie Mellon University Count your blessings... … faster !

15-299 Lecture 11 Feb 17, 1998 Doug Beeferman

Carnegie Mellon University

Count your blessings...

…faster!

Warm-upsWarm-ups

How many different ways areHow many different ways are

there to seat 121 students in a lecture roomthere to seat 121 students in a lecture room

of 214 chairs?of 214 chairs?

Q.Q.

!93

!214)1121214(212213214

!93

!214!121

!93!121

!214!121

121

214

How many different ways areHow many different ways arethere to seat 121 students in a lecture roomthere to seat 121 students in a lecture room

of 214 chairs?of 214 chairs?

There are often multiple ways to count the same setThere are often multiple ways to count the same set

1. Choose which seats are filled, then order the students in them1. Choose which seats are filled, then order the students in them

2. Assign an unfilled seat to each student in a 2. Assign an unfilled seat to each student in a fixed succession, fixed succession, e.g.e.g. alphabetically. alphabetically.

An easier oneAn easier one

How many different subsets ofHow many different subsets of

sleeping students are possible in this classsleeping students are possible in this class

of 121 students?of 121 students?

Q.Q.

How many different subsets ofHow many different subsets ofsleeping students are possible in this classsleeping students are possible in this class

of 121 students?of 121 students?

There are often multiple interpretations of the same countThere are often multiple interpretations of the same count

1212• Subsets of a 121-element setSubsets of a 121-element set

• Binary digit strings of length 121Binary digit strings of length 121

• Outcomes of flipping a penny 121 timesOutcomes of flipping a penny 121 times

• Possible committees drawn from 121 peoplePossible committees drawn from 121 people

kn

k

n xk

nx

0

)1(

Review: Review: The binomial formulaThe binomial formula

)(

0

)( knkn

k

n yxk

nyx

k

k

kn

k

n

xk

n

xk

nx

0

0

)1(

““Closed formClosed form” or” or““Generating formGenerating form” or” or““Generating functionGenerating function””

““Power seriesPower series” (“” (“Taylor seriesTaylor series”) expansion”) expansion

The binomial formulaThe binomial formula, take two, take two

nkk

n

if 0 Since

r

r

r

kr

kkk

nkkkkkkkk r

nr

xxxxkkkk

n

xxxx

321

321

321

321;;;; 321

321

;;;;

)(

Review: Review: The multinomial The multinomial formulaformula

!!!!

!

321 rkkkk

n

Multinomial maniaMultinomial mania

Q.Q.What Is the coefficient of (M•A•G•G•S) in theWhat Is the coefficient of (M•A•G•G•S) in the

expansion of (S+M+A+G)expansion of (S+M+A+G)5 5 ??

The same as the number of arrangementsThe same as the number of arrangementsof “MAGGS”, orof “MAGGS”, or

A.A.

601;2;1;1

5

!2

!5

Explore different possible representations of Explore different possible representations of

the same information or idea,the same information or idea,

and understandand understand

the relationship between them.the relationship between them.

Playing with Playing with the binomial the binomial formulaformula

n

k

n

kn

k

n

k

n

xk

nx

0

0

2

)1(

Can you explain this combinatorially?

Let x=1. We find that

Playing with Playing with the binomial the binomial formulaformula

n

k

n

k

n

0

2

The number ofThe number ofsubsets of an subsets of an nn-element set-element set

The number of The number of kk-element subsets of-element subsets ofan an nn-element set, summed over-element set, summed over

all possible all possible kk. .

Indeed, these mean the same thing!Indeed, these mean the same thing!

Combinatorial proofsCombinatorial proofs

A A combinatorial proofcombinatorial proof demonstrates demonstrates

that each side of an equationthat each side of an equation

corresponds to the size of the same set.corresponds to the size of the same set.

Contrast this to a conventional Contrast this to a conventional algebraic proofalgebraic proof,,

in which symbol manipulation is used toin which symbol manipulation is used to

carry one side to the othercarry one side to the other

MoreMore binomial formulations binomial formulations

1

odd even

0

0

2

)1(0

)1(

nn

k

n

k

kn

k

kn

k

n

k

n

k

n

k

n

xk

nx

Let x= -1. We find that

…or equivalently, that

The odds get evenThe odds get even

n

k

n

k k

n

k

n

odd even

The number ofThe number oflength-length-nn binary strings with binary strings with

an an even number of ones number of ones

The algebra has spoken. But it’s not yetThe algebra has spoken. But it’s not yetindependently clear why these sides count the same thing.independently clear why these sides count the same thing.

Let’s develop a Let’s develop a correspondencecorrespondence from one to the other. from one to the other.

The number ofThe number oflength-length-nn binary strings with binary strings with

an an oddodd number of ones number of ones

More odds and evensMore odds and evensLet Let OOnn be the set of`binary strings of length be the set of`binary strings of length nn with with

an odd number of ones.an odd number of ones.Let Let EEnn be the set of`binary strings of length be the set of`binary strings of length nn with with

an even number of ones.an even number of ones.

We have already presented an algebraicWe have already presented an algebraicproof that proof that OOnn =E=Enn

An elegant combinatorial proof can be had byAn elegant combinatorial proof can be had byputting putting OOn n and and EEn n in in one-to-one correspondence.one-to-one correspondence.

The The correspondence principlecorrespondence principle says that says thatif two sets can be placed in one-to-one if two sets can be placed in one-to-one

correspondence, then they are the same sizecorrespondence, then they are the same size!!

An attempt at a correspondenceAn attempt at a correspondence

Let Let ffnn be the function that takes an be the function that takes an nn-bit bitstring and flips all its bits.-bit bitstring and flips all its bits.

ffnn is clearly a one-to-one and onto function is clearly a one-to-one and onto function for odd for odd nn. . E.g.E.g. in in ff77 we have we have

0010011 0010011 1101100 11011001001101 1001101 0110010 0110010

...but do even ...but do even nn work? In work? In ff66 we have we have

110011 110011 001100 001100101010 101010 010101 010101

Uh oh. Complementing maps evens to evens!Uh oh. Complementing maps evens to evens!

A correspondence that works for all A correspondence that works for all nn

Let Let ffnn be the function that takes an be the function that takes an nn-bit bitstring and flips only -bit bitstring and flips only the first bitthe first bit..

For example,For example,

0010011 0010011 1010011 10100111001101 1001101 0001101 0001101

110011 110011 010011 010011101010 101010 001010 001010

CheckCheck::1. 1. ffnn : O: Onn EEnn??2.2. f fn n is one-to-one? is one-to-one? i.e.i.e. x x y y f fn n (x) (x) f fn n (y)(y)3.3. f fn n is onto? is onto? i.e.i.e. for all y for all y E Enn, , there exists an x there exists an x O Onn such that f such that fn n (x)=y(x)=y

How to countHow to countallocation schemesallocation schemes

Example 1:Example 1:Pirates and gold barsPirates and gold bars

ScenarioScenario: You’re a pirate who has just: You’re a pirate who has just

discovered discovered nn bars of gold (identical and indivisible). bars of gold (identical and indivisible).

Being a generous buc, you decide to split the loot Being a generous buc, you decide to split the loot

between the between the kk distinct shipmates on board. distinct shipmates on board.

How many ways are there to do this? How many ways are there to do this?

Example: n=4, k=3Example: n=4, k=3

… … 15 = allocation schemes!15 = allocation schemes!

Representation: Partition a string of 4 gold bars Representation: Partition a string of 4 gold bars

into 3 substrings by inserting slashes.into 3 substrings by inserting slashes.

13

134

Connecting to a known Connecting to a known representationrepresentation

So the number of allocation schemes So the number of allocation schemes is the is the

same assame as the number of strings of bars and slashes the number of strings of bars and slashes

with with nn bars and bars and k-1k-1 slashes... slashes...……which which is the same asis the same as the number the number

of waysof ways

to choose to choose k-1k-1 positions to make positions to make slashesslashes

from a set of from a set of n+k-1n+k-1 positions, or positions, or

1

1

k

kn

How to count allocation schemesHow to count allocation schemesExample 2:

Solutions to integer equations

Q. How many ways are there to solve:Q. How many ways are there to solve:

0,,

10

321

321

xxx

xxx

662

12

13

1310

A. It’s A. It’s the same as the same as distributing 10 gold bars to 3 pirates!distributing 10 gold bars to 3 pirates!

How to count allocation schemesHow to count allocation schemesExample 3:

Solutions to constrained integer equations

Q. A twist: what if the solutions must be strictly positive?Q. A twist: what if the solutions must be strictly positive?

0,,

10

321

321

xxx

xxx

362

9

13

137

A. First give every “pirate” his required 1 “gold bar”.A. First give every “pirate” his required 1 “gold bar”.Then count the ways to distribute the remaining 10-3=7 “gold bars”:Then count the ways to distribute the remaining 10-3=7 “gold bars”:

How to count pathwaysHow to count pathwaysMeandering in a nameless modern metropolis

Scenario: You’re in a city where all the streets,Scenario: You’re in a city where all the streets,numbered numbered 00 through through xx, run north-south,, run north-south,

and all the avenues, numbered and all the avenues, numbered 00 through through yy, , run east-west. How many [sensible] ways are there run east-west. How many [sensible] ways are there to walk from the corner of 0th St. and 0th avenue to to walk from the corner of 0th St. and 0th avenue to

the opposite corner of the city?the opposite corner of the city?

00

yyxx00

Meandering in a nameless modern metropolis

• All paths require exactly All paths require exactly x+yx+y steps: steps:• xx steps east, steps east, yy steps north steps north• Counting paths is the same as counting Counting paths is the same as counting which of the which of the x+yx+y steps are northward steps are northward steps:steps:

y

yx

yyxx00

00

(i,j)(i,j)

Now, what if we add Now, what if we add the constraint that the constraint that the path must go the path must go through a certain through a certain

intersection, call it intersection, call it (i,j)?(i,j)?

Meandering in a nameless modern metropolis• Given the constraint, we can decompose Given the constraint, we can decompose each valid path into two subpaths:each valid path into two subpaths:

• The subpath from the start to The subpath from the start to (i,j)(i,j)

• The subpath from The subpath from (i,j)(i,j) to to (y,x)(y,x)

yyxx00

00

(i,j)(i,j)

i

ji

•These subpaths These subpaths may be may be independently independently chosen. By the chosen. By the product rule, the product rule, the total path count istotal path count is

)(

)()(

iy

jxiy

)(

)()(

iy

jxiy

i

ji

k

n

k

n

k

n 1

1

1

An important identity for binomial coefficientsAn important identity for binomial coefficients

Combinatorial proof? Consider Combinatorial proof? Consider separating all separating all kk-element subsets of the -element subsets of the

set set {1,2,…,n}{1,2,…,n} into those that include and into those that include and those that exclude those that exclude nnGraphical intuition:Graphical intuition:

Let Let n=x+yn=x+y be the total be the totalsteps needed in the city steps needed in the city walk problem, and let walk problem, and let

k=yk=y be the be thenumber of northward number of northward

steps.steps.There are two cases for There are two cases for

thethevery last step taken.very last step taken.

Toward Toward Pascal’s TrianglePascal’s Triangle

0

1

0

2

0

3

2

2

2

3

1

1

1

2

1

3

2

4

1

4

2

5

3

3

3

4

3

5

1

5

4

4

4

5

5

5

0

0

0

4

0

5

Associate with each intersection the path count from (0,0),Associate with each intersection the path count from (0,0),

k

n

Toward Toward Pascal’s TrianglePascal’s Triangle

1 1 1

1 3

1 2 3

6

4

10

1 4 10

5

1 5

1

1 1 1

Simplifying, we observe startling symmetriesSimplifying, we observe startling symmetries

Pascal’s TrianglePascal’s Triangle

11

1 11 1

1 2 11 2 1

1 3 3 11 3 3 1

1 4 6 4 11 4 6 4 1

1 5 10 10 5 11 5 10 10 5 1

1 6 15 20 15 6 11 6 15 20 15 6 1

Credited toCredited toBlaise Pascal, 1654Blaise Pascal, 1654

““It is extraordinary howIt is extraordinary howfertile in properties thefertile in properties thetriangle is. Everyonetriangle is. Everyonecan try his hand.”can try his hand.”

- Blaise- Blaise

Summing the rows… Summing the rows… gives us powers of 2gives us powers of 2

11

1 + 11 + 1

1 + 2 + 11 + 2 + 1

1 + 3 + 3 + 11 + 3 + 3 + 1

1 + 4 + 6 + 4 + 11 + 4 + 6 + 4 + 1

1 + 5 + 10 + 10 + 5 + 11 + 5 + 10 + 10 + 5 + 1

1 + 6 + 15 + 20 + 15 + 6 + 11 + 6 + 15 + 20 + 15 + 6 + 1

=1=1

=2=2

=4=4

=8=8

=16=16

=32=32

=64=64

n

k

n

k

n

0

2

Summing the diagonals…Summing the diagonals… yields Little Gauss’s formula and more!yields Little Gauss’s formula and more!

11

1 11 1

1 2 11 2 1

1 3 3 11 3 3 1

1 4 6 4 11 4 6 4 1

1 5 10 10 5 11 5 10 10 5 1

1 6 15 20 15 6 11 6 15 20 15 6 1

2

)1(

2

1

1

1

1

nnnii

k

n

k

i

n

ki

n

ki

n

ki

““It is extraordinary howIt is extraordinary howfertile in properties thefertile in properties thetriangle is. Everyonetriangle is. Everyonecan try his hand.”can try his hand.”

- Blaise- Blaise

Try your hand.Try your hand.