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Algorithm Design

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An Algorithmic Notation – guarded command language

• Provides a framework for recording refinement of specifications into programs

• Consist of :– Sequence– Alternation– Iteration

• Commands are organized into programs by– Assignment statements of the form x := E where x is a program

variable, E an expression– Specifications to be further refined

• In-line or by call to other program

– Skip and abort

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Declarations and commands(1)

• The kind of data available to a program in the guarded command language depend on – the target language,– the given sets of the specification being refined– the abstract data types used in the refinement

• Example of declarations

s : Student given set of specification

where : 0..size type in target language

sarray : array (1..size) of Student array type and given set

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Declarations and commands(2)

• The kind of command available to a program in the guarded command language depend on– Target language– Abstract types used in refinement

• Example of commands

where := 0 assignment statement

sarray(1) := s assignment statement to an

array element

search specification of an operation

to be refined

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Sequence control structure

• A sequence of commands may be presented in:– Horizontal format

x := x + 1; y := y – 1

– Vertical formatx := x + 1;

y := y – 1

• For sequence control structure below

Semicolon acts as separator

A1

A2

A1; A2

In guarded command language

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Alternation control structure(1)

• Each command is prefaced by a boolean expression called a guard; corresponds to boolean expression of target language

• A command is executed only if its guard is true• Bounded by the sign if and fi• The sign □ is the guard command separator• The guard arrow → separates the guard and command• Example

if x ≤ y → z := x □ x ≥ y → z := y fi

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Alternation control structure(2)

• Example when there are several guarded commands

x, y, s : ℤ

if x < y → s := -1

□ x = y → s := 0

□ x > y → s := 1

fi

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Alternation control structure(3)

• In general for if-then

• If-then-else

B

A

T

F if B A ¬B skip fi

B

A1A2

TF

if B A1 ¬B A2 fi

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Iteration control structure (1)

• Each command has a guard• A command will be executed only if the guard is true

• Bounded by the signs do and od

• Example

do z < x z > y → z : z + 1 od

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Iteration control structure (2)

• In general

A

B

A1

B

A2T

F T

Fdo B → A od

A1; do B → A2 od

Initialized do-while

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Method of recording(1)

• At stage of recording an algorithm refinement, record only one step of refinement

• The sign ⊑ to represent ‘is refined by’

A A1

A2

A A1; A2⊑

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Method of recording(2)

A1

A2

B1

A11 A12

A21

A22

B2

A1 if B1⊑ A11 ¬B1 A2 fi

A2 ⊑ A21; do B2 A22 od

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Program states and assignment: Program spaces and program states (1)

• Program spaces– A collection of program variables

• Example: For the following declaration

sarray : array(1..size) of Student;

s : Student;

r : Response;

ectr, tctr : 0..size

• Establish a program space by the following schema

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Program states and assignment: Program spaces and program states (2)

EnrolSpace

sarray : 1..size → Student;

s : Student;

r : Response;

ectr, tctr : 0..size

• This type schema will be referred to in recording the design of a program.

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Program states and assignment: Program spaces and program states (3)

• However EnrolSpace does not establish how input and output of the schema DEnrol to be represented as program variables

• Can introduce such variable, e.g.,

Identify_s_and_rs?, s, s’ : Studentr!, r, r’ : Responses? = sr! = r’

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Program spaces and program states (4)

• To make the schema suitable for algorithm refinement of DEnrol, need to combine it with DEnrol and hide the input and output variables

DEnrolProg ≙ (DEnrol Identify_s_and_r) \ (s?,r!)

DEnrolokProg ≙ (DEnrolok Identify_s_and_r) \ (s?,r!)DNoRoomProg ≙ (DNoRoom Identify_s_and_r) \ (s?,r!)DAlreadyEnrolledProg ≙ (DAlreadyEnrolled Identify_s_and_r) \ (s?,r!)

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Program states and assignment: Using the assignment statement (1)

• Use to refine specification in which just one variable in the program state is to change

• Suppose the program space is defined byPSpace ≙ [x, y, z: 0..maxnat]

the effect of the assignment x := y + z

Assign_1PSpacex’ = y + zy’ = yz’ = z

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Program states and assignment: Using the assignment statement (2)

• In general if the program state is given by schema

GenPSpace

x : T

Others

where x does not occur in Others and E is an expression of type T in the value of x and the components of Others, then effect of x := E is

Assign_2

GenPSpace

Others

x’ = E

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Using the alternation control structure

• Suppose S is a specification on a program space, and A1 and A2 are operations on the same space. Suppose B1 and B2 are boolean expressions defined on the variables in the space, and are computable in the target language, we shall say that

S ⊑if B1 → A1 □ B2 → A2 fi

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Example of refinement by alternation(1)

• Consider the following specification of a program to find the positive difference of two numbers. PSpace has been defined previously

PossDiff

PSpace

z’ = x – y z’ = y – x

x’ = x

y’ = y

The following program as a refinement

if x ≥ y → z := x - y □ y ≥ x → z := y - x fi

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Example of refinement by alternation(2)

• To verify we replace the two assignments by their descriptions

Assign_xmy

PSpace

x ≥ y

z’ = x – y

x’ = x

y’ = y

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Example of refinement by alternation(3)

• To verify we replace the two assignments by their descriptions

Assign_ymx

PSpace

y ≥ x

z’ = y – x

x’ = x

y’ = y

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Example of refinement by alternation(4)

• Now we have

PosDiff ⊑if x ≥ y → Assign_xmy

□ y ≥ x → Assign_ymx

fi

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• Refining a disjunction with an alternation

A1 A2 ⊑ if pre A1 → A1 □ pre A2 → A2 fi

• Refining a disjunction to an if-then-else

A1 A2 ⊑ if pre A1 → A1 □ ¬(pre A1) → A2 fi

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Refinement of the enrol operation(1)

DEnoughRoom ≙DEnrolokProg DAlreadyEnrolledProg

Refine DEnrolProg as

DEnrolProg ⊑if ectr = size →| DNoRoomProg□ ectr < size →| DEnoughRoomfi

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Refinement of the enrol operation(2)

• Refinement of DNoRoomProgDNoRoomProg ⊑r := noroom

• Refinement DEnoughRoom will be dealt later

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Sequence Control Structure: using the structure

• If a sequence of two specification (A1;A2) is offered as a refinement of a third S, we shall record this by writing

S ⊑ A1; A2

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Example of refinement by sequence

• Consider the following program state

XState ≙ [x: 0..maxnat]

and the following specification

UpTwo ≙ [ XState | x’ = x + 2]

with x + 2 ≤ maxnat as its precondition;

refined by the following sequence

x := x + 1; x := x + 1

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Sequence Control Structure

• Replacing commands in a sequence

S ⊑ A1;A2 and A2 ⊑ A3 then we can have S ⊑ A1;A3 • Operations on partitioned states

suppose that PartState ≙ PartA PartB and PartA and PartB have no declarations in common, and OpA is an operation on PartA and OpB is an operation on PartB

the operation OpA OpB on the whole of PartState can be refined by amended operation as follows:

OpAx ≙ [PartState; PartB; OpA]

OpBx ≙ [PartState; PartA; OpB]

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Further refinement of the enrol operation

• We propose a refinement of DEnoughRoom as follows:

DEnoughRoom ⊑where: 0..(size + 1)

DEnoughRoomX

The extension of DEnoughRoom is defined as

DEnoughRoomX

where, where’: 0..(size + 1)

DEnoughRoom

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Further refinement of the enrol operation

• Now we propose that:

DEnoughRoomX ⊑Search;

Decide• Search and Decide are to be specified. Search will set

the variable where to index an occurrence of s in the active part of the array, if there is one, and to ectr + 1 if there isn’t

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Further refinement of the enrol operation

• The following specification is recorded for Search

SearchState

DClass

where: 0..(size + 1)

s: Student

r: Response

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Further refinement of the enrol operation

• The following specification is recorded for Search

Search

SearchState

DClass

(( i: 1..ectr (sarray i) ≠ s where’ = ectr + 1)

((sarray where’) = s where’ 1..ectr ))

s’ = s

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Further refinement of the enrol operation

• For Decide we can supply this definition:

Decide

SearchState

DEnoughRoom

where ≤ ectr (i: 1..ectr (sarray i) = s)

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Further refinement of the enrol operation

• The refinement of Decide by an alternation is

Decide ⊑

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Iteration control structure

• Example of refinement by initialized do-while• E.g., to sum up the elements of an array of integers. The

following axiomatic description for the sum of a sequence of integer

sumseq: seq ℤ → ℤ

(si: seq ℤ

(( si = <> sumseq si = 0)

(si ≠ <> sumseq si = (head si) + sumseq (tail si))))

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Iteration control structure

• Introduce the program space that includes the array to be summed and the variable to hold answer. The variable n is a constant greater than 0, the size of array

SumSpace

intarr: 1..n →ℤs: ℤ

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Iteration control structure

• Now specify the program

Addup

SumSpace

intarr’ = intarr

s’ = sumseq intarr

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Iteration control structure

• Another way: we refine Addup to include counter ctr

Addup ⊑ctr: 0..n

AddupExt• The definition of AddupExt

AddupExt ≙ [Addup; ctr, ctr’ : 0..n]

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Iteration control structure

• The invariant is in the following schema

Addup_invar

SumSpace

ctr, ctr’ : 0..n

s’ = sumseq((1..ctr’) ⊳ intarr’)

ctr’ ≤ n

intarr’ = intarr

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Iteration control structure

• The initialization programAddup_initSumSpacectr, ctr’ : 0..nintarr’ = intarrs’ = 0ctr’ = 0

Sum ⊑ Addup_init; do ctr ≠ n →| Addup_body

od

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Iteration control structure

• The body of while

Addup_body

SumSpace

ctr, ctr’ : 0..n

ctr < n

s = sumseq((1..ctr) ⊳ intarr)

s’ = sumseq((1..ctr’) ⊳ intarr’)

ctr’ ≤ n

intarr’ = intarr

ctr’ = ctr + 1

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Iteration control structure

• The body of while can be simplified as

Addup_body

SumSpace

ctr, ctr’ : 0..n

ctr < n

s’ = s + intarr(ctr’)

intarr’ = intarr

ctr’ = ctr + 1

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Iteration control structure

• Can be implemented by the following program

Addup_body ⊑ctr := ctr + 1;

s := s + intarr(ctr)