combinatorial block designs

8/9/2019 Combinatorial Block Designs

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Combinatorial Block Designs

(Complete/Incomplete Block Designs and Latin Squares)

Shawn Rana

12/04/2014


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Table of Contents

Variables in an Experimental Design.............................................................................................................3

Complete Block Design.................................................................................................................................5

Latin Square..................................................................................................................................................6

Number of Blocks Needed to Test All Varieties Equally................................................................................7

Review and Problem Set 1.9

Incomplete Block Designs(BIBDs)...............................................................................................................11

Necessary Conditions for a BIBD.................................................................................................................13

Sufficient Corollaries for the Existence of a BIBD........................................................................................15

Review and Problem Set 2..18

Complementary Designs.............................................................................................................................20

Symmetric BIBDs.........................................................................................................................................21

Necessary Conditions for a Symmetric BIBD..22

Latin Squares In-Depth................................................................................................................................25

Latin Squares Using Two Variables..............................................................................................................26

Orthogonal Latin Squares...........................................................................................................................29

Conditions for the Existence of Orthogonal Families..................................................................................31

Review and Problem Set 3..33

Finishing Notes35

Sources.36


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Variables in an Experimental Design

Block designs arise from the analysis of combinatorics of various experiments. Many of the scholars,

including R. A. Fisher and F. Yates, first questioned combinatorics through agriculture experiments.

However, through time, the applicability of combinatorics spread through other tests such as drug trials,product tests, and even architectural design.

However, to begin to understand a block design, there must first be an understanding of an experimental

design that compares the effects of differentvarieties or treatments.This can be thought of as a set V.

These varieties or treatments are items that get tested and compared with each other, e.g., different

types of window cleaning solutions, different brands of shoes, different types of fertilizers, or different

doses of pain relieving drugs. Ultimately, the list can go on, but the varietyor treatmentmust be testable

on a certain subject(s).

These specific subjects that get tested on are called experimental units. This can be thought of as a set P,

where P V because the experimental units are made up of the different varieties. Respectable to the

previous examples of possible varietiesor treatments, these can be thought as a type of window

cleaning solution being tested on a specific window in a certain position, a brand of shoe being worn on

a specific foot, a type of fertilizer being tested on a specific plot of land, a dose of drug being tested on a

specific person(the items in bold are the experimental unitsthat get tested on).

When experimental unitsare put together in a group, they become blocks. They are usually put together

due to a similar feature in common. Relating to the examples above: a set of windows are grouped with

the other set of windows to become the windows ofa building, a foot gets grouped together with

another foot (humans have two feet) to become a person, a plot of land gets grouped with the other

plot of lands to become a horizontal row in a field, and a person gets grouped with the other people tobecome a lab group(the items in bold are the blocks made up of the respectable experimental units).

To make this concrete, an example will be used that will better help understand the early stages of

experimental design.

Example: Lets say some technicians wereto test four different operating systems for usability on four

different brand computers in four labs in the library. Design an experimental design that displays each

computer in each room being tested an operating system.

From this information alone, can we determine a broad idea of what the varieties we are testing? How

about the experimental units or blocks?

All of the different varieties of operating systems are the set of varieties, V,{,,,}. (I couldhave easily replaced the operating systems with numbers or letters, but I wanted to signify that anything

can resemble a treatment). The experimental units are made up of each of the specific computers in

each of the labs. Related to the set of varieties, each experimental unit actually consists of a certain

variety. Thus, the experimental units, the set P, is a subset of varieties, the set V because the set P


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consists of only the elements that make up the variety set. The blocks of the experimental design are

also going to be a subset of the varieties, and in this question, the blocks will be made up of the many

computers that make up a lab. One lab is considered one block.

The experimental design will thus look like this:

Experimental Design for Testing Usability of Operating Systems

Lab 1 Lab 2 Lab 3 Lab 4

Computer A

Computer B

Computer C

Computer D

The design above is called an experimental block designwhich is a type of block design. We are testing a

certain operating system on a specific computer in a specific lab and have organized our testing in a

block(design). Now that we have understood an experimental design and the specific variables, a block

design is simply a specific type of design that puts together the experimental units as blocks and assigns

each specific experimental unit a treatment from the set of V. Thus, our blocks; Lab 1, Lab 2, Lab 3, and

Lab 4 are all subsets of our variety set, V, {,, ,}. Respectively, our blocks can also be writtenin set notation where order does matter(in specific cases such as a Latin Square)and repetitions are

possible: Lab 1= {,,,}, Lab 2= {, ,,}, Lab 3= {,,,},

Lab 4= {,,,}.


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Complete Block Design

Also, the experimental design in the last section just so happens to be a complete block design. A

complete block design is a design that has each variety existing in exactly once in each block and an

equal number of times throughout the block itself. Thus, if each block contains all of the set V, and eachelement of Vis used an equal number of times in the design, the experimental design is a complete

block design.

Referring to our block design, all of the operating systems from our set have been used in each lab, and

they have all been used exactly four times throughout all of the labs.

However, for testing purposes, our previous design is a bit flawed. Recall we have four computers, and

they are all different brands. Those four computers are set up around each lab in the library. Looking at

our design, we have Computer C testing operating systemtwice, both in Lab 1 and Lab 2, which omits

Computer C testing operating system. This causes an imbalance of data if we were testing usability,because Computer C will never have tested the usability of . Although each lab gets tested all of thevarieties, not all of the computers get tested all of the varieties.


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Latin Square

In order to get a range of results, each of the different branded computers need to be tested all of the

operating systems and as do all of the labs. The Latin Squarewill display the most varied results because

no row and column of a block design will consist of the same variety more than once. A Latin Square is acomplete block design that has each block containing all of the set Vand each element in Vin a unique

experimental unit. Denoting our computer problem, this means all of the operating systems in our set of

varieties are used up in our labs and each unique computer gets tested all of the varieties. When writing

the blocks in set notation, order does matter in this case.

Fact: A Latin Square Design can always be formed if the number of experimental units making up a

block is equal to the number of total blocks, which is also equal to the number of total varieties.

The above fact can be seen true because if the number of varieties, number of blocks, and number of

experimental units are equal to each other, then each variety will be tested once in each column and

once in each row. This will be better illustrated in the future section of Latin Squares.

Our Latin Square design for testing usability of operating systems is seen below:


Computer A

Computer B

Computer C

Computer D

Seen above, each operating system gets tested once and only once in each block, and each different

brand computer in each lab gets tested a unique operating system. This provides the most accurate

result because there is data being retrieved an equal number of times from each unique computer in

each lab. Also note, there is more than one way to create a Latin square. We could have the blocks of

varieties switched without actually changing the blocks.


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Number of Blocks Needed to Test all Varieties Equally

However, lets say a problem arises and instead of four operating systems, there are five operating

systems to test. How would we design an experiment now, knowing we could only test four of them in

each block?

If each operating system is used r times, we have 5r total operating systems that should be optimally

tested on. We should allot the 5rtotal operating systems in groups of four because there are four unique

computers that make up a lab and five varieties that need to be tested an equal number of times. Thus,

5r must be divisible by four. Let us add another lab from the library that consists of the same four

uniquely branded computers (Computer A, Computer B, Computer C, and Computer D). This would give

us 20 total operating systems to test. So we have 5r total operating systems and this needs to be divided

by four to get the number of labs we need to be able to test 5 operating systems. Thus, because our goal

is to test all of the operating systems an equal number of times to get accurate results, adding Lab 5 will

make sure we get all of the operating systems an equal number of times.

Lets say we had 6 varieties to test instead of 4 or 5. The way we determine the number of extra blocks,

or labs we need is simple. We will have 6r total operating systems to test and this number must be

divisible by four, the number of experimental units making each block up, to determine the total number

of blocks we need for our design. If r= 4 we would have 24 total operating systems and six blocks would

be needed to be able to test all of the varieties. If r= 5 we would have 30 total operating systems to test,

but 30 is not divisible by 4, the number of experimental units making each block up. Because 30 is not

divisible by 4, there is no number of labs that can test 30 operating systems an equal number of times. If

r=6, we would have 36 total operating systems total, and we could arrange those 36 operating systems

into 6 different blocks so that each variety gets tested equally throughout the design. This concept is

important when figuring how many blocks are needed for a block design that offers more varieties than

the number of experimental units making a block.

Example: Design a block design where the technicians were testing 6 different operating systems instead

of four on four different brand computers in the library and write the blocks out in set notation.

Because we know we are going to have 6r operating systems to test, the minimal amount blocks we

need would be 6. Minimally, because realistically, all companies have budget goals that require them to

use only as much as it takes to get the job done, nothing more.

Thus, our set Vnow becomes

{,, ,, , }with the addition of two more operating

systems.


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The block design is shown below:

Lab

1

Lab

2

Lab

3

Lab

4

Lab

5

Lab

6

Computer A

Computer B Computer C

Computer D

Thus, the block notation can be written as such:

Lab 1={,,,}

Lab 2= {, }

Lab 3={,, , }

Lab 4={,,,}

Lab 5={, }

Lab 6={,, , }.The design above meets our goal of testing all six varieties an equal number of times, and without using

too many resources and time; meaning, we could also have completed this design by using more blocks

such as 9 blocks, 12 blocks, 15 blocksetc. The reason the number of blocks should be a multiple of 3

because in 3 blocks we can test all of the varieties an equal number of times (two times exact), and this

cannot be achieved with less than 3 blocks. However, most companies have a budget and time restraint

so spending more money and testing more resources would be unfavorable in a companys point of view.

Moreover, the design above is actually called an incomplete block designbecause not all of V is being

used in each block. According to our problem, this means that we are not using all six of the operating

systems in each lab, we are only using four. Unlike the complete block designwhere each block is all of

V, an incomplete block design uses part of the set. Both designs are geared towards testing each and

every variety equally, not one less and not one more than another. The next topic dives further into the

definition of the incomplete block design.


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Review and Problems

Recall some important definitions and concepts learned from these past sections:

P is the set of experimental units, specific positions of the blocks that will be tested.

V is the set of varieties that get tested.

Blocks are made up of experimental units. Blocks are certain subsets of V.

A Block Design is a design that puts together experimental units from P as blocks and assignseach unit an element from the V set. It is designed so that the experimental units make each

vertical block and each block is a set of varieties.

Set Notation for a block design is displayed through a set or even a matrix (where repetition may

occur and order matters.) For example, Block A consists of: {a, b, c, d}. This can also be written as

a matrix: [a, b, c, d].

A Complete Block Design is a type of block design that has each block testing all of the availablevarieties. Thus, if each block set contains all of the set V, then the block design is complete. Each

variety must be tested.

A Latin Square is a specific type of complete block designbecause each block in the design tests

all of the varieties. However, this design is different from other designs in the way that eachblock contains all of the set V and each v V is in a unique experimental unit. So, each variety is

never tested more than once in a specific position.

An Incomplete Block Design is a block design that has blocks that do not contain all of the set V.If there are more varieties than the number of experimental units making a block, then the block

design is an incomplete block design when constructed.

Problem Set

1.

Design a complete block design that tests comfortability of winter shoes on a dog. The varieties

of shoes are given as, V= {A, B, C, D}. HINT: Knowing a dog is made up of four feet, a block is a

certain dog, and the experimental units making the block are their certain feet position.

2.Design an incomplete block design if there were two more brands of shoes added to the problemabove.


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3.Why would a Sudoku puzzle be considered a type of Latin Square?

4.Turn this block design below into an incomplete block design and all of the varieties (A, B, C, D, E)get tested equally. (The various Bs 1-4 are blocks and Ks 1-4 make up each block.)

B1 B2 B3 B4

K1A E A E

K2B A B A

K3 C B C BK4 D D D D

5.Design a Latin Square with the V set being {1, 2, 3, 4, 5}. Display all the blocks in set notation.

6.How many blocks would I need if I was on a resource constraint, had 7 varieties to test equally,and knew 6 experimental units made each block?


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Incomplete Block Designs (Balanced Incomplete Block

Designs)

Recall from earlier that in certain situations, it is not always possible to test all of the varieties in every

block. This then results as an incomplete block design where each block cannot fit all of the varieties in

all of its experimental units. The specific type of incomplete block design we will be examining is the

balanced incomplete block design. Before I dive into what a balanced incomplete block design is, lets

first state what a balanced block designis.

A balanced block design has the following criterion:

The block design consists of a set of varieties Vwith v 2 elements.

Has an assortment of b > 0 subsets of Vwhich are called blocks.

Each specific block holds knumber of varieties, where k> 0 and k is the same for each block.

Each specific variety in V appears in r blocks where r > 0.

Each variety within a block is paired up with another variety (besides itself) and as such, each

pair of varieties appear simultaneously inblocks. Note thatis uniform through the entire

design and> 0.

Thus, we have five variables that become significant in a block design: b, v, r, k, and. These five

parameters are not all independent; v, k, and determine b and r, and not all combinations of v, k, and

are possible.When v > k, as in when we have more varieties than the number of varieties a block can

hold, the balanced block design becomes a balanced incomplete block design. The block design is alsocalled a BIBD or a (b, v, r, k, )-design. However, this notation does not just apply to incomplete block

designs. It can apply to complete block designs and other designs also if all of the variables can be

determined from the design itself.

An example of a balanced incomplete block design would be the incomplete block design we constructed

with the computer labs and operating systems:

Lab

1

Lab

2

Lab

3

Lab

4

Lab

5

Lab

6

Computer A Computer B

Computer C

Computer D


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This would be considered a (6, 6, 4, 4, 4)-design. We have 6 blocks(b): Lab 1, Lab 2, Lab 3, Lab 4, Lab 5,

and Lab 6. We have 6 varieties (v), {,, ,, , }. Each variety comes up 4 times in all of theblocks(r). Each block can hold up to 4 varieties (k). Notice how k < v; because there are more varieties

than the number of varieties each block can hold, this automatically becomes a BIBD. Also, each pair of

varieties appears 4 times in the entire design (). To clarify, let's pick a variety in Lab 1. The variety

can be paired with three other varieties in its block to form a pair {i, j}. Let's say pairs with to

form the pair That specific pair simultaneously shows up four times across all of the blocksto form, which is 4. Now, this must be true for any and all pairs. One pair cannot show up 3 times, and

another pair show up 4 times. We will prove why this is important in the next section.


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The Necessary (But not Sufficient) Conditions for a BIBD

Theorem 1.1: In a BIBD, bk = vrand (v 1) = r(k 1)

Both conditions above must be met in order for a BIBD to be able to exist. These conditions are

necessary, but not sufficient. For example, there is no such (10, 5, 4, 2, 2)-design that can exist. For

theorem 2.1, bk = vr, which is (10)(2) = (5)(4), which is 20 = 20, so thus, those equal each other. However,

for theorem 2.2, (v 1) =r(k 1), which is (2)((5)-1) (4)((2) 1), which is 8 4.

We will now prove why both of these conditions need to be met. Starting with Theorem 2.1:

Proof of bk = vr:

To prove equal, we must prove both sides to be equivalent. On the left hand side, we havebk. This

represents the product of the number of blocks in a design by the number of varieties a single block can

hold. By multiplying this out, we can achieve all of the elements inside all of the experimental units

making up each block in the design. All of the elements can actually be written out as the elements for

each block:

B1={...}, B2={...}, B3={...},... Bn={...}.

On the right hand side, we have vr.This represents the product of the number of

elements in the V set(so the number of varieties) by the number of different times each

unique variety shows up in the blocks. This is signifying the different reproductions of

each variety that appear through out all of the blocks. Thus, we can write these

elements in blocks similar to bk:

B1={...}, B2={...}, B3={...},... Bn={...}.

This ensures us that we are able to get every variety in all of its different blocks, and thus

giving us all of the elements in the block design in two methods.

Because both the left and the right hand side are able to display all of the elements in

the BIBD, they are both equal.

Q.E.D


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Proof of (v 1) = r(k 1):

To prove equal, we must prove both sides to be equivalent. On the left hand side we

have (v 1). This signifies the number of pairs a particular variety can have. Start with

a particular variety i, which pairs with another variety(not to itself), to form a pair {i, j} ina block. Notice there are (v- 1) choices in place forj with a particular i. Then, these

choices appear like this {i, some j}, and these pairsappear times through out the

design. Thus we get(v 1) number of pairs with a specific variety i.

On the right hand side we have r(k 1). This also signifies the number of pairs a

particular variety can have. Start with a particular variety i. Notice ioccurs rnumber of

times throughout the design. Given i occurring rtimes, i pairs with (k 1)elements in

each block(not including itself). Thus, we get r(k 1) pairs with a specific variety i.

Both the left and right hand side ensures we have an equal number of pairs with a

specific variety i.

Q.E.D

These two conditions are valid in the point that in order to test if a BIBD exists, these two conditions

mustbe satisfied. This says that a (43, 43, 7, 7, 1)-design couldexist; it does not guarantee that the

design doesexist. The further corollaries below will prove more sufficiency regarding the existence of a

BIBD.


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Corollaries that Provide Sufficiency for a BIBD to Exist

Corollary 1.2

The equation (v 1) = r(k 1)also shows us the number of r blocks for any variety i. If we isolaterfor

some i on one side, we can get the number of blocks the particular variety i shows up in:

(v 1) =r(k 1)

r= (v 1)

k 1

Q.E.D

The number ris the same for each particular I and r must be a positive integer for the design to be able

to exist as a balanced incomplete block design. The same is true for b as bmust be an integer when

being solved for.

Corollary 1.3

If an incomplete block design satisfies theorem 1.1 (both conditions must be met for this to be true) then

it should also satisfy:

In relation to b being an integer in relation to v, k, and, this can trivially be done by dividing both sides

by :

We can easily find out how this corollary is derived from the two theorems proved above:

From bk = vr, we get by solving for r.

From (v 1) = r(k 1), we get

by also solving for r.

We can set both of the r's equal to themselves so we get .

When cross-multiplying we get .


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This can be rewritten as from the binomial coefficient with two

from and .

.

By dividing by , we can achieve b,which must be an integer:

Q.E.D

Corollary 1.4

This last corollary further tests the existence of a BIBD:

If a BIBD exists, then and .

Recall from earlier I said that (43, 43, 7, 7, 1)-design couldexist; it does not guarantee that the design

doesexist. This design, when tested, actually passes all of the conditions and corollaries, except for this

one. Thus, a (43, 43, 7, 7, 1)-design does not exist. A BIBD must pass all of these corollaries from the

theorems (and the theorems itself) above to exist.

Example: Now not only are these conditions valuable in testing the existence of a BIBD but are evidently

significant in finding other valuable given certain information. For example, in a track meet of 16sprinters, four sprinters can race in one heat, and there are are 20 heats. What other information is

needed to be able to construct a BIBD?

From this information, we are given:

b= 20 heats(blocks)

v= 16 sprinters(varieties)

k= 4 sprinters competing in one heat(size of block)

Thus, we don't know what ris, which represents the the number of heats any sprinter can be in, and wedon't know what is, which represents the number of times each pair of sprinters can meet at the same

time.


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Thus, we can find these missing parameters using our two theorems using b, v, and k.

Therefore, we are able to construct a (20, 16, 5, 4, 1)-design given our parameters and conditions. In

conclusion, the last corollary finds sufficient existence results for a BIBD, whilst the previous corollaries

were necessary but not sufficient.


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Review and Problems

Recall some important definitions and concepts learned from this section:

A balanced block design has five variables: b, v, r, k,.

b represents the number of blocks, b > 0.

v represents the number of varieties, v 2

r represents the number of blocks each variety shows up in, or the number of replications eachvariety has in the entire block design r > 0.

k represents the max number of varieties each block can hold, k > 0.

represents the number of times each pair of varieties in each block appears at the same time,.> 0.

A balanced incomplete block designhas more varieties than the the number of varieties a blockcan hold, v > k and r > where the number of replications of a variety is greater than the number

of times each pair simultaneously appear.

A pair of varieties is a one variety iin a block paired with k-1 varieties in the block to create a pair

{i, j}.

There are two necessary conditions in order for a BIBD to exist: bk = vr and (v 1) =r(k 1).

Problem Set

1.What does a (7, 7, 3, 3, 1)-design mean?

2.

Does a (15, 9, 5, 3, 2)-design exist?

3.Create a (4, 4, 3, 3, 2)-design and write the blocks in set notation.

4.Aviation engineers want to check engine wear on airplanes. If four engines are used in anairplane, create a block design that tests five varieties of engines (A, B, C, D, E) an equal number


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of times (keeping in remembrance that the engineers want to test as little number of airplanes

as possible due to a budget constraint).

5.Write down all of the blocks in set notation for question 4.

6.

Write down all of the pairs for all of the varieties for question 4.

7.What is the only way to turn a BIBD into a complete block design?

8.Given r=3, k=3, and =1, find the missing parameters


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Complementary Designs

Let me introduce a new problem example to help better display this concept. A generic (7, 7, 3, 3, 1)-

design is shown below. There are seven blocks named B1 through B7; seven varieties: A, B, C, D, E, F, G,

and H; each variety is shown three times, or has three replications; each block can hold up to threevarieties; and no pair of varieties is simultaneously shown more than once.

B1 B2 B3 B4 B5 B6 B7

K1 A A A B B C CK2

B D F D E D EK3 C E G F G G F

The complementof this design is constructed by replacing each block with a block consisting of the

remaining varieties. Thus, this would look like:

B1 B2 B3 B4 B5 B6 B7

K1 D B B A A A AK2

E C C C C B BK3 F F D E D E D

K4 G G E G F F G

From a parameter approach this is a (7, 7, 4, 4, 2)-design. In relation to the original (b,v, r, k,)-design,

the parameters for this complementary design can be labeled as a (b, v, b r, v k, + b 2r)-design.

Thus the complement of a (20, 16, 5, 4, 1)-design is a (20, 16, 15, 12, 11)-design. If a complement of a

BIBD fulfills all of the conditions of a BIBD, then the complement is also a BIBD.


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Symmetric Balanced Incomplete Block Designs

A BIBD is called symmetric if the number of blocks(b) are equal to the number of varieties(v) andif the

number of times a variety shows up throughout the design(r) is equal to the max number of varieties a

block can hold(k). Thus:

A balanced incomplete block design is symmetric if and only if b= vand k = r.

Because b = v and k = r, we can also refer to a symmetric BIBD as a (v, k, )-design because the need to

know what b and r are eliminated if we already know v and k. From the previous section, we had a (7, 7,

3, 3, 1)-design. In this specific BIBD, b = v = 7 and r = k = 3, so in terms of a (v, k, )-design, this is actually

a (7, 3, 1)-design.

However, existence questions do come up with symmetric balanced incomplete block designs. There are

multiple conditions that are necessary but not sufficient in finding if a symmetric BIBD does exist given

certain parameters.


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Necessary Conditions for a Symmetric BIBD

Theorem 1.5: The following determines existence of a (v, k, )-design:

1.

If v is even, then k is the square of an integer.

2.

If v is odd, then the following Diophantine equation has a solution in integersx,y, z, not all of which are 0.

x2= (k- )y

2+ (-1)

(v1)/2z

2

This theorem above is necessary, but not sufficient in finding the existence of a symmetric BIBD. For

example, let us make our parameters for a (v, k, )-design, 16, 6, and 2, respectfully. Following the

theorem, we have v= 16 which is even, and k = 4 which is a square. Thus, this design could exist. To

follow the second subpart of the theorem, let us look at a (111, 11, 1)-design. From the design, vis odd,

and thus when following the Diophantine equation,x2= (k- )y

2+ (-1)

(v1)/2z

2, we get:

x2=10y2z2

This equations solutions arex= 1, y= 1, and z= 3. Thus, in conclusion, this says that a design of these

parameters couldalso exist. However, to provide sufficiency, these next to theorems are valid into

proving the existence of symmetric BIBDs.

Theorem 1.6: For large values of m, and specifically for m = 2k, k 1, there is a (4m 1, 2m1, m1)-

design.

Theorem 1.7: If m 1 is a power of a prime, there is a (m2+ m + 1, m + 1, 1)-design

Both theorems above allow us to construct symmetric BIBDs under certain limits of m. Developing

symmetric BIBDs through the above three theorems ensures us that there is a (v, k,)-design that does

exist through when following the constraints. On an overall idea, these kind of designs are important in

the fields of cryptography in order to break certain codes especially in the breadth of symmetric key

cryptography where block ciphers are used(think lots of plaintext numbers in a box; like the movie, The

Matrix).


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Review and Problems


The complementof a BIBD design is when each column of the design is replaced with varietiesthat have not been used, thus creating a new, unused design.

Parameter-wise: the complement is a (b, v, b r, v k, + b 2r)-design

A balanced incomplete block design is symmetric if and only if b= vand k = r.

The problem of existence also arises with symmetric BIBDs.

Symmetric BIBDs are used in cryptography.

Problem Set

1.What is the complement of this design (if the letters A through G are used as varieties)

B1 B2 B3 B4 B5 B6 B7

K1 D B B A A A AK2

E C C C C B BK3

F F D E D E DK4

G G E G F F G

2.What is the complement of a (7, 7, 3, 3, 1)-design?

3. In your own words, explain how the parameters of a complement design are achieved.

4. Four of the blocks of a (7, 3, 1)-design are {1, 2, 3}, {1, 5, 6}, {2, 5, 7}, and {1, 4, 7}. Find the

remaining blocks.

5. Show that in a (v, k,)-design, any two blocks have exactlyvarieties in common.

6. Show that the following designs do exist or do not exist:

a)

A (32, 16, 8)-design


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b)

A (31, 6, 1)-design

c)

A (57, 12, 3)-design

d)

A (21, 5, 1)-design

e)

A (100, 10, 5)-design

7.Explain why a (43, 43, 7, 7, 1)-design cannot exist.


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Latin Squares In-Depth

This will be an in-depth view on the topic of Latin squares. Recall from our overview in the beginning

section what a Latin square is. A Latin Square is a complete block design that has each block containing

all of the set Vand each element in Vin a unique experimental unit. In simpler terms, each variety that is

tested is in a unique spot in the block design; a certain variety does not appear more than once in the

same row and column.

This can be illustrated in a design below:

A k x k Latin Square

1 2 3 . . . k-1 k

2 3 4 . . . k 1

3 4 5 . . . 1 2

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

k-1 k 1 . . . k-3 k-2

k 1 2 . . . k-2 k-1

The design above displays a Latin square for any variable k. Because the number of varieties, thenumber of blocks, and the number of experimental units making each block are equal to each

other, it is assumed that they are equal to some positive integer k. Thus, the Latin square is

made up of krows and k columns.

Let us analyze this k x k design. Specifically, let us examine the first column and row. All of the

varieties (from 1 through k) are being used in the first column and all of the varieties are also

being used from the first row. However, none of the varieties that are used in the first column

are being used in the same order in any other column and none of the varieties that are used in

the first row are being used in any other row. Thus, each variety is in a unique position in the

Latin square.

Before we get in-depth with the topic of Latin squares, we must first dive into understanding

how block designs can be formed with two variables instead of only one as we have been doing

thus far. Becoming adept at this will help understand families of Latin squares, which will be

explained later.


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Latin Squares Using Two Variables

Thus far, we have been designing block designs with one variable, or specifically one kind of variety:

operating systems, window cleaning solutions, brands of shoes, etc. However, let's say while we were

designing a block design with one type of variety, we also wanted to test another kind of variety in the

same block design.

Example: Some technicians were to test four different operating systems for usability on four different

brands of computers in four labs in the library. However, the technicians also want to test what brand of

hard drive works best with what kind of operating system. Design a block design that best allows to test

each operating system and hard drive in a unique computer and lab so that each operating system

uniquely matches up with a unique hard drive.

What this is essentially asking is to combine two Latin squares, because we must design a block design

that tests each operating system in a unique position in each row and column and also each hard drive in

a unique position in each row and column.

Let us begin this example by designing a Latin square testing four operating systems on four brands of

computers in four labs of the library (refer to the previous Latin square section earlier). The varieties

(operating systems) that will be used are the elements in {1, 2, 3, 4} with each number signifying a

certain operating system.

Latin Square (A) for Testing Usability for Operating Systems

Lab

1

Lab

2

Lab

3

Lab

4Computer A Computer B Computer C Computer D

Before we go further, let me introduce a simple notation to note on the entries in a block design so we

can better understand how a two variable block design comes together. The i, jentry of the block design

above signifies the variety used on computer i in labj. Thus, for example, the variety 2 is used in entry2,1, in entry 1,2, in entry, 4,3, and in entry 3,4. This will be more evident the further we delve into Latin

squares.

Back to the problem, we must also design another Latin square for testing usability for hard drives. Lets

say for the sake of this topic, it is different from the previous Latin square, Latin square(A), because

recall, there are more than one ways to design a Latin square given each variety is in another unique

position not previously used before. Thus, we have:


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Latin Square (B) for Testing Usability for Hard Drives

Lab1

Lab2

Lab3

Lab4

Computer A Computer B Computer C Computer D

This Latin square is completely different from a numerical point of view because the numbers that are

used here were not used in the same positions in Latin square (A). Now, we have two different Latin

squares but we must combine both of these into a single Latin square.

Thus, this can be accomplished by testing each operating system exactly once in combination with each

brand of hard drive. What this entails is that we must build a 4 x 4 array where the row signifies the

computer and the column signifies the lab, and we should place both an operating system and a hard

drive with the corresponding computer and lab.

From the previous entry notation I showed, we will say aij indicates operating system used in entry i, j

and say bij indicates hard drive used in i, j. We want each entry in our new single Latin square design to

look like this:

(aij, bij) which aijand bij are respectfully derived from each Latin square. From the looks of putting both

of these Latin squares together, we will have 4 x 4= 16 different possible ordered pairs (a, b).

The final product of our example is the combination of both Latin squares:

Latin Square (A) for Testing Usability for Operating Systems Latin Square (B) for Testing Usability for Hard Drives

Both of these Latin squares when combined, turn into the one shown on the next page:

Lab1

Lab2

Lab3

Lab4


Lab

1

Lab

2

Lab

3

Lab

4



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Latin Square (C) for Testing the Combined Effects of Operating Systems and Hard Drives on Usability


Computer A

Computer B

Computer C

Computer D

Referring to our problem, this Latin square does not mean the same thing as it would if we would test

both variables one by one. This design signifies the combined effect ofbothvariables on the computers.

Also, the i, j entry is an ordered pair consisting of a computer operating system and hard drive used in

the two Latin squares in computer position i in computer labj.

Now that we understand the concept of two variables in a block design, I will introduce some

terminology that will further our knowledge of Latin squares.


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Orthogonal Latin Squares

As we know, Latin squares are a k x k block design. The orderof a Latin square is the number kwhich

signifies the number of varieties, blocks, and experimental units making up those blocks. Thus for

example, if we examine Latin square A, the order of that Latin square is 4, because it is a 4 x 4 Latin

square. Also, referring to Latin square C, if we have two distinct k x k Latin squares that create k2 ordered

pairs where (aij, bij) are all distinct, then we say the two Latin squares are orthogonal.

For example, the problem we were working on earlier on in the previous section; both of these Latin

squares are orthogonal to each other because when combined to make pairs (aij, bij), they are all

distinct. However, a Latin square cannot be orthogonal alone, it must be combined with another Latin

square for it to be orthogonal with that specific Latin square. An example of two Latin squares that are

notorthogonal are shown below:

Latin Square A Latin Square D

When combined, these two Latin squares form:

Latin Square E

As we can see from Latin square A, an entry is seen as aij where a is the variety used on row i, columnj.

In Latin square D, an entry is seen as bij where b is the variety used on row i, columnj. However,

according to the definition of orthogonality, these Latin squares are not orthogonal because there is

formed pair that appears more than once. The formed pair (2, 4) appears once in row 2, column 1 and

also in row 4, column 3. The same can also be said for pair (4, 3); this pair also appears more than once

in the design, and thus not distinct.

Now what if there were more ways to write a Latin square, than just two? We know two distinct Latin

squares form distinct pairs when combined are orthogonal. However, if we haveA(1),A(2),A(3), ,A(n)

being distinct kx k Latin squares, and every pair of them is orthogonal, then we have a orthogonal

family. An example is provided on the next page:


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Latin Square F Latin Square G Latin Square H

All of these Latin squares are orthogonal to each other and thus create an orthogonal family. However,

you must be thinking: Does there exist an orthogonal family of r Latin squares of order k?

To answer this question, we must prove a theorem in the next section that gives necessary, but not

sufficient conditions for the existence of an orthogonal family of rLatin squares of order k.


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Conditions for the Existence of Orthogonal Families

Theorem 2.1: If there is an orthogonal family of r Latin squares of order k, then r k1.

Proof:

The way we approach this proof is to manipulate entries in orthogonal Latin squares.

Let us suppose that we have an orthogonal family ofAr k x k Latin squares with the order k. Thus,

we have:A(1),A(2),A(3),A(4)A(r).

Recall from our entry notation from earlier, and let us say that aij(x)is the i, j entry for some Latin

squareA(x)inside the orthogonal family ofA(r)Latin squares.

We need to relabel the entries inA(1)

so that the number 1 appears in entry a11(1)and if there was

another numberppreviously in that entry, we need to switch that number with 1. Let us do this

for all ofA(1)

so that 1 is switched withp andpis switched with 1. The reason we are doing this isso we can manually relabel a Latin square to the entries of our choosing. Also, because we have

switched entries around, this Latin square remains orthogonal because all we have done was

switch potential pairs: If there was a pair to be made fromA(1) andA(x), the pair (p, l) would

switch to (1, l) and (1, l) would switch to (p, l).

From this method, we can arrange all of our 1, 1 entries in our Latin squares to be 1 fromA(1)

throughA(r). We can do this for the entire 1, n entries of our Latin squares fromA(1)

throughA(r)

so they all have the same first row: 1 2 3 4 k. In entry notation the first row would look like

this:

a11(1)= a11(2)= a11(3)= a11(4)==a11(r)= 1

a12(1)= a12

(2)= a12(3)= a12

(4)== a12

(r)= 2

a13(1)

= a13(2)

= a13(3)

= a13(4)

== a13(r)

= 3

a14(1)

= a14(2)

= a14(3)

= a14(4)

== a14(r)= 4

.

.

a1k(1)

= a1k(2)

= a1k(3)

= a1k(4)

== a1k(r)

= k

Now that the first row is complete, we can arrange the second row starting with the 2, 1 entry.

Because we already have 1 in the first column, we cannot have 1 again in the 2, 1 entry for itwould violate the definition of a Latin square and would not be orthogonal. By orthogonality,

a21(x)

a21(y)(for some Latin square y), because if they were equal to each other, then we would

have the same variety(v) in both the i, j spot of an entry. When paired up: (a21(x), a21

(y)) = (v, v).

Thus for some v, (a21(x)

, a21(y)

) = (a1v(x)

, a1v(y)

) will infringe upon the definition of orthogonality


32/36

Page 32

because the pair (a21(x)

, a21(y)

) is not distinct. This pair also appears as (a1v(x)

, a1v(y)

). Therefore,

each 2, 1 entry in each Latin square throughA(r) is distinct and different from 1. Because there is

already a 1 in the 1, 1 entry in each Latin square, in order for the block design to confine itself as

a Latin square, 1 cannot appear again in the same column.

Thus, in conclusion, because 1 cannot appear again, there are at most n1 numbers that canappear in the 2, 1 entry, and r k1. This can also be understood by the pigeonhole concept

which states that if there are q objects and there are splaces to put q objects, and q > s, then at

least one spot contains more than one object. We can apply this principle to the second row, first

column entry of each Latin square, and we will also arrive at r k1.

Q.E.D

The condition above is necessary but not sufficient in finding if there exists an orthogonal family of r

Latin squares of order k. In order to find sufficiency, there remains one theorem.

Theorem 2.2: If k> 1 and k=pn, wherepis a prime number and nis a positive integer, then there is a

complete orthogonal family of Latin squares of order k.

Together with both theorems, we may can conclude the existence of orthogonal families. For example,

to see if a family of orthogonal Latin squares of order 3 exists, we know that the number of orthogonal

Latin squares consisting in the family is less than or equal to the order, 3(by the first theorem). We also

know that 3= 31 by the second theorem, so thus, there must exist an orthogonal family of order 3.


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Review and Problems


A Latin Square is a complete block design that has each block containing all of the set Vand eachelement in Vin a unique experimental unit.

A Latin square is a k x k block design where k signifies the order of the Latin square.

The i, j entry signifies i meaning row, andj meaning column.

Two Latin squares can be combined where each pair is an ordered pair (aij, bij). In the orderedpair, aijand bij are respectfully derived from each Latin square.

If we have two distinct k x k Latin squares that create k2 ordered pairs where (aij, bij) are alldistinct, then we say the two Latin squares are orthogonal.

If we haveA(1),A(2),A(3), ,A(n)being distinct kx k Latin squares, and every pair of them isorthogonal, then we have a orthogonal family.

Existence issues arise with orthogonal families of Latin squares.

If there is an orthogonal family of r Latin squares of order k, then r k1.

Problem Set

1.Construct a Latin square of order 10.

2.Construct a second Latin square of order 10 not similar to the first.

3.Does there exist a pair of orthogonal Latin squares of order 12? Why?

4. Supposed that two orthogonal 8 x 8 Latin squares both have 87654321 in the first row. Is it

possible for them to have the same 2, 4 entry?

5.Determine if the pairs on the next page are orthogonal to each other.


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a)

b)

6.Explain whether this is an orthogonal family or not:


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Finishing Notes

Throughout this paper, we taught ourselves the meaning of applying a combinatorial design to

everyday situations. For complete block designs, we can test all of the varieties for a single block,

but for incomplete block designs we cannot. We also learned how significant Latin squares truly

are in real world applications and mathematically speaking, they are truly unique. By going in-

depth with Latin squares, we read upon when there are different Latin squares when combined

to make distinct pairs, they are seen as orthogonal and may have an orthogonal family.

Nonetheless, theorems were proved, and excruciating block designs were created through hours

of simple mouse clicks. However, it is dually noted that I have only scratched the surface of this

topic. It has been a pleasure teaching the topic of block designs to whomever this may interest.

Shawn Rana


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Sources(Peer-Reviewed and Textbook)

1.

Law, Maska, Cheryl E. Praeger, and Sven Reichard. "Flag-Transitive Symmetric 2

Designs."Journal Of Combinatorial Theory - Series A116.5 (2009): 1009-1022.Academic

Search Complete. Web. 4 Dec. 2014.

2.

Kiefer, Jack C. "The Interplay Of Optimality And Combinatorics In Experimental

Design." Canadian Journal Of Statistics9.1 (1981): 1-10.Academic Search Complete. Web. 13

Nov. 2014.

3.

Curtin, Brian, and Ibtisam Daqqa. "The Subconstituent Algebra Of A Latin Square." European

Journal Of Combinatorics30.2 (2009): 447-457.Academic Search Complete. Web. 12 Nov.

2014.

4.

Acha, Chigozie Kelechi. "On Two Methods Of Analysing Balanced Incomplete Block

Designs." Pakistan Journal Of Statistics & Operation Research8.4 (2012): 749-757.Academic

Search Complete. Web. 4 Dec. 2014.

5.

Tesman, Barry. "Combinatorial Design."Applied Combinatorics. By Fred S. Roberts. 2nd ed.

Upper Saddle River: Pearson Prentice Hall, 2005. 489-550. Print.

Websites

1.

www.stat.purdue.edu/~bacraig/notes1/topic13.pdf

2.

www.math.ucdenver.edu/~wcherowi/courses/m6406/bibd1.pdf

3.

www.math.cornell.edu/~web4520/CG10-0.pdf

4.

www.en.wikipedia.org/wiki/Block_design

5.

www.mmorris.public.iastate.edu/stat512/512.6.pdf

6.

http://www.maths.qmul.ac.uk/~rab/DOEbook/doeweb11.pdf

7.

www.math.ucdenver.edu/~wcherowi/courses/m6406/symbibd.pdf

8.

http://mathworld.wolfram.com/BlockDesign.html

9.

http://www.springer.com/cda/content/document/cda_downloaddocument/9780387954875

-c1.pdf?SGWID=0-0-45-101851-p9775353.

10.

http://www.renyi.hu/~dezso/coa/bibdpaper.pdf
http://www.stat.purdue.edu/~bacraig/notes1/topic13.pdfhttp://www.stat.purdue.edu/~bacraig/notes1/topic13.pdfhttp://www.math.ucdenver.edu/~wcherowi/courses/m6406/bibd1.pdfhttp://www.math.ucdenver.edu/~wcherowi/courses/m6406/bibd1.pdfhttp://www.math.ucdenver.edu/~wcherowi/courses/m6406/bibd1.pdfhttp://www.math.ucdenver.edu/~wcherowi/courses/m6406/bibd1.pdfhttp://www.math.cornell.edu/~web4520/CG10-0.pdfhttp://www.math.cornell.edu/~web4520/CG10-0.pdfhttp://www.en.wikipedia.org/wiki/Block_designhttp://www.en.wikipedia.org/wiki/Block_designhttp://www.mmorris.public.iastate.edu/stat512/512.6.pdfhttp://www.mmorris.public.iastate.edu/stat512/512.6.pdfhttp://www.maths.qmul.ac.uk/~rab/DOEbook/doeweb11.pdfhttp://www.maths.qmul.ac.uk/~rab/DOEbook/doeweb11.pdfhttp://www.math.ucdenver.edu/~wcherowi/courses/m6406/symbibd.pdfhttp://www.math.ucdenver.edu/~wcherowi/courses/m6406/symbibd.pdfhttp://mathworld.wolfram.com/BlockDesign.htmlhttp://mathworld.wolfram.com/BlockDesign.htmlhttp://www.springer.com/cda/content/document/cda_downloaddocument/9780387954875-c1.pdf?SGWID=0-0-45-101851-p9775353http://www.springer.com/cda/content/document/cda_downloaddocument/9780387954875-c1.pdf?SGWID=0-0-45-101851-p9775353http://www.springer.com/cda/content/document/cda_downloaddocument/9780387954875-c1.pdf?SGWID=0-0-45-101851-p9775353http://www.springer.com/cda/content/document/cda_downloaddocument/9780387954875-c1.pdf?SGWID=0-0-45-101851-p9775353http://www.springer.com/cda/content/document/cda_downloaddocument/9780387954875-c1.pdf?SGWID=0-0-45-101851-p9775353http://www.renyi.hu/~dezso/coa/bibdpaper.pdfhttp://www.renyi.hu/~dezso/coa/bibdpaper.pdfhttp://www.renyi.hu/~dezso/coa/bibdpaper.pdfhttp://www.springer.com/cda/content/document/cda_downloaddocument/9780387954875-c1.pdf?SGWID=0-0-45-101851-p9775353http://www.springer.com/cda/content/document/cda_downloaddocument/9780387954875-c1.pdf?SGWID=0-0-45-101851-p9775353http://mathworld.wolfram.com/BlockDesign.htmlhttp://www.math.ucdenver.edu/~wcherowi/courses/m6406/symbibd.pdfhttp://www.maths.qmul.ac.uk/~rab/DOEbook/doeweb11.pdfhttp://www.mmorris.public.iastate.edu/stat512/512.6.pdfhttp://www.en.wikipedia.org/wiki/Block_designhttp://www.math.cornell.edu/~web4520/CG10-0.pdfhttp://www.math.ucdenver.edu/~wcherowi/courses/m6406/bibd1.pdfhttp://www.stat.purdue.edu/~bacraig/notes1/topic13.pdf

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