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ms

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Chapter 18 Recursive Procedures

One of the most powerful features of Logo is thatyou can divide a project into procedures, each ofwhich is a distinct entity that has its own name. Aprocedure can be called (or used) by any other pro

cedure; it can also call other procedures. Someprocedures call on themselves; these proceduresare recursive. We have already used recursive pro

cedures. For example, p o l y and s p i are both re

cursively defined.

TO POL Y : ST EP : ANGLE

FD : STEP

RT : ANGLE

POLY : STEP : ANG LE This is the recursive call.

END

TO SP I : STEP : ANGLE : INC

FD : STEP

RT : ANGLE

SP I : STEP + : INC : ANGLE : INC

END

p o l y calls p o l y as part of its definition; and s p i calls s p i .

Lets think about this process. Imagine that Logohas an inexhaustible supply of helpers who arecomputer creatures living in the computer. Every

time a procedure is called a helper is called upon tolook up the definition of the procedure. The helperthen begins to carry out the instructions. It doesthis by calling on other helpers. Usually severalhelpers are needed to carry out one procedure.

For example, when p o l y is called its helper calls af d helper. After the f d helper finishes, a r t helperis called. When it finishes, a p o l y helper is called.The second p o l y helper calls a f d helper, a r t helper, and a p o l y helper. Meanwhile, the firstp o l y helper is still around waiting for the second

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p o l y helper to finish. In the process the f d helpers and the r t helpers finish their jobs (wedont know who they call for help). The p o l y helpers never finish; they keep on calling for newp o l y helpers.

When you use p o l y the process continues untilyou type c t r l -g . Not all recursive procedureswork this way. They can be made to stop. In fact,making up appropriate stop rules is a majorpart of writing recursive procedures.

Stopping Recursive ProceduresIn tills section we will discuss recursive proce

dures that stop. We will use as an example a proce

dure that counts backwards from a given number.Assume we have a procedure called c o u n t b a c k and we type

COUNTBACK 55

Logo will respond

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c o u n t b a c k for he lp , b u t th i s t im e g iv e i t i n u m

b e r - l a s in p u t.

COUNTBACK : NUMBER - 1END

Lets try it with a small number like 5.

COUNTBACK 5

Logo responds

5

43

21

0-1

-2

a n d s o o n u n t il y o u t y p e c t r l -g .

The procedure definition is almost complete. It

needs a stop condition. Lets make a stop rule for

c o u n t b a c k . For example, if the input is 0 then

stop.

I F : NUMBER = 0 [S TO P]

Now c o u n t b a c k looks like this:

TO COUNTBACK :NUMBER

IF :NUMBER = 0 [S TO P]

PR : NUMBER

COUNTBACK :NUMBER - 1

END

Try it by typing

COUNTBACK 5

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Logo responds

5

4

321

Everything stops. The last number that is printed

is 1.

Lets now think about this process in more de

tail and try to understand how Logos helpersinteract.

A Model for Recursion: Helpers

There is an inexhaustible supply of helpers readily

available to Logo. Each time a procedure is called,

whether it be p r i n t or m a k e or p o l y or c o u n t -

b a c k , a helper is put on the job. It looks up the

definition of the procedure and starts reading it.

The helper carries out what it can and then calls

other helpers as they are needed. For each

procedure call, a helper is put on the job.

Whenever a helper finishes its job it reports back

to whoever called it and goes away.

Lets revive the image of named things like :s i d e

or:s t a r t as contents of a container. Thus, if a pro

cedure takes an input the helper puts that input

in the container named on the title line. If some

thing is already there, the helper leaves it there,

and puts the new thing on top of it to be available

later. These containers can hold a lot. When the

helper has finished carrying out the procedure it

disposes of the input and then what was under it

in the container is now on top and accessible.

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I

Lets play computer by sketching out what thehelpers do. To do this give c o u n t b a c k a smallnumber as input. So

I COU NTB ACK 2

i Immediately a helper is called on and given the

procedure c o u n t b a c k with an input of 2. Letsdraw a diagram of the interaction.

i C O U N T B A C K 2 COUNTBACK INUMBER

I c o u n t b a c k 2 is called.i The inp ut , n u m b e r , is 2.

| COUNTBACK 2 stops ifNUMBER = 0.

| But NUMBERis 2.

| c o u n t b a c k 2 c o n t i n u e s ,

i 2 p r :n u m b e r

I COUNTBACK INUMBER 1

| c o u n t b a c k l i s ca l led by

COUNTBACK 2.

NUMBER is 1.

COUNTBACK 1 stops ifNUMBER = 0.

But NUMBER is 1.c o u n t b a c k l continues.

1 PR INUMBER

COUNTBACK INUMBER 1

c o u n t b a c k 0 i s ca l led by

COUNTBACK 1.

NUMBERis 0.

COUNTBACK 0 stops ifNUMBER = 0.So it stops.

c o u n t b a c k i now continues, but ithas finished its job and stops.

c o u n t b a c k 2 now continues, but it has? finished its job and stops.

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A helper stops when there is nothing more to do

or when s t o p is encountered.

Lets change the procedure so that the print action occurs in a different place. Make a new proce

dure, CB.

TO CB :NUMBER

I F : NUMBER = 0 [ S TO P ]

CB : N U M B E R - 1

PR : NUMBER

END

Type

CB 5

Logo responds

1

234

5

This is probably a surprise. But remember when

the first CB helper calls the second CB helper, it

waits for the second to do its job, and then the

first helper continues with its own job.

Lets trace through the process giving CB the

number 2 as input.

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CB 2 CB INUMBER

CB 2 is ca l led.

I

I

I

NUMBER is 2.CB 2 stops ifNUMBER = 0.

But n u m b e r is 2 so CB 2 continues.

CB INUMBER 1.

c b l i s c a l le d b y c b 2 .

NUMBER is 1.

CB 1 stops ifNUMBER is 0.

But n u m b e r is 1 so CB l helper continues.CB INUMBER 1

c b 0 is called b y c b l.

NUMBER is 0.

CB 0 stops ifNUMBER = 0.

So it stops.c b l continues.

PR INUMBER

c b l h a s f i n is h e d i t s j o b a n d s t o p s .c b 2 con t inues .

PR INUMBER

c b 2 now has finished its job and stops.

Note that the stop rule has to come before the re

cursion line. Otherwise, the recursion will go onforever and the stop rule will never be reached. Often, the stop rule will be the very first line in theprocedure.

There are other kinds of stop rules we could makeup. For example, in s p i

TO S P I : S T E P : A NG LE : I NC

FD : S TEP

RT : ANGLE

S P I : S T E P + : I NC : A N GL E : I NC

END

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We could decide that s p i should stop if :s t e p is

greater than 200. So, we include the line

I F : S T E P > 2 0 0 [ S T O P ]

After editing s p i it looks like

TO S P I : S T E P : A N GL E : I NC

I F : S T E P > 2 0 0 [ S T O P ]

FD : S T E P

R T : A N G L E

S P I : S T E P + : I NC : A N GL E : I NC

END

Try s p i . Ify o u d o n ' t l ik e t h e s t o p r u le , c h a n g e i t.

Designing a stop rule for p o l y can be a little trick

ier. p o l y completes a figure when the turtle re

turns to its starting state, which may require the

turtle to turn only one complete rotation, that is.

only 360 degrees. But sometimes the turtle turnsa multiple of 360 degrees.

Actually, we only need to know what the turtles

heading was when it started , and then com pare it

to the tu rtle s heading after each u i t t l So before

p o l y is called we have to find the tu rtle 's heading.

HF A

i h e n p o l y LOU UKI o see ii_!t_ ' -

i t 1 e*__.< 0 L Y Lo C j

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------

Now tr y p o l y .

There is a problem here. Each time you call p o l y you have to save the starting heading. It would bebetter to put that action in a procedure. Lets re

name p o l y and call it p o l y i . Then we can make anew p o l y which sets up :s t a r t and then runs

POLYI.

TO P O L Y : S T E P : ANG L E

P 0 L Y 1 : S T E P : AN G L E H E A D I N G

E N D

TO P 0 L Y 1 : S T E P : AN G L E : S T A R T

FD : S T E P

R T : A N G L E

I F H E A D I N G = : S T A RT [ S T O P ]

P 0 L Y 1 : S T E P : AN G L E : S T A R T

END

Now p o l y will do the whole job.

There are many designs you can create with s p i and p o l y . There are many ways of combiningpolygons to make new and dazzling designs.

This introduction to Logo is finished, but we hopeyou will explore other turtle projects on your own.

TheReference Manualdescribes other features ofLogo which you might want to try now.

Have fun.

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Chapter 18 Recursive Procedures

One of the most powerful features of Logo is thatyou can divide a project into procedures, each ofwhich is a distinct entity that has its own name. Aprocedure can be called (or used) by any other pro

cedure; it can also call other procedures. Someprocedures call on themselves; these proceduresare recursive. We have already used recursive pro

cedures. For example, p o l y and s p i are both re

cursively defined.

TO POL Y : ST EP : ANGLE

FD : STEP

RT : ANGLE

POLY : STEP : ANG LE This is the recursive call.

END

TO SP I : STEP : ANGLE : INC

FD : STEP

RT : ANGLE

SP I : STEP + : INC : ANGLE : INC

END

p o l y calls p o l y as part of its definition; and s p i calls s p i .

Lets think about this process. Imagine that Logohas an inexhaustible supply of helpers who arecomputer creatures living in the computer. Every

time a procedure is called a helper is called upon tolook up the definition of the procedure. The helperthen begins to carry out the instructions. It doesthis by calling on other helpers. Usually severalhelpers are needed to carry out one procedure.

For example, when p o l y is called its helper calls af d helper. After the f d helper finishes, a r t helperis called. When it finishes, a p o l y helper is called.The second p o l y helper calls a f d helper, a r t helper, and a p o l y helper. Meanwhile, the firstp o l y helper is still around waiting for the second

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p o l y helper to finish. In the process the f d helpers and the r t helpers finish their jobs (wedont know who they call for help). The p o l y helpers never finish; they keep on calling for newp o l y helpers.

When you use p o l y the process continues untilyou type c t r l -g . Not all recursive procedureswork this way. They can be made to stop. In fact,making up appropriate stop rules is a majorpart of writing recursive procedures.

Stopping Recursive ProceduresIn tills section we will discuss recursive proce

dures that stop. We will use as an example a proce

dure that counts backwards from a given number.Assume we have a procedure called c o u n t b a c k and we type

COUNTBACK 55

Logo will respond

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c o u n t b a c k for he lp , b u t th i s t im e g iv e i t i n u m

b e r - l a s in p u t.

COUNTBACK : NUMBER - 1END

Lets try it with a small number like 5.

COUNTBACK 5

Logo responds

5

43

21

0-1

-2

a n d s o o n u n t il y o u t y p e c t r l -g .

The procedure definition is almost complete. It

needs a stop condition. Lets make a stop rule for

c o u n t b a c k . For example, if the input is 0 then

stop.

I F : NUMBER = 0 [S TO P]

Now c o u n t b a c k looks like this:

TO COUNTBACK :NUMBER

IF :NUMBER = 0 [S TO P]

PR : NUMBER

COUNTBACK :NUMBER - 1

END

Try it by typing

COUNTBACK 5

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Logo responds

5

4

321

Everything stops. The last number that is printed

is 1.

Lets now think about this process in more de

tail and try to understand how Logos helpersinteract.

A Model for Recursion: Helpers

There is an inexhaustible supply of helpers readily

available to Logo. Each time a procedure is called,

whether it be p r i n t or m a k e or p o l y or c o u n t -

b a c k , a helper is put on the job. It looks up the

definition of the procedure and starts reading it.

The helper carries out what it can and then calls

other helpers as they are needed. For each

procedure call, a helper is put on the job.

Whenever a helper finishes its job it reports back

to whoever called it and goes away.

Lets revive the image of named things like :s i d e

or:s t a r t as contents of a container. Thus, if a pro

cedure takes an input the helper puts that input

in the container named on the title line. If some

thing is already there, the helper leaves it there,

and puts the new thing on top of it to be available

later. These containers can hold a lot. When the

helper has finished carrying out the procedure it

disposes of the input and then what was under it

in the container is now on top and accessible.

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I

Lets play computer by sketching out what thehelpers do. To do this give c o u n t b a c k a smallnumber as input. So

I COU NTB ACK 2

i Immediately a helper is called on and given the

procedure c o u n t b a c k with an input of 2. Letsdraw a diagram of the interaction.

i C O U N T B A C K 2 COUNTBACK INUMBER

I c o u n t b a c k 2 is called.i The inp ut , n u m b e r , is 2.

| COUNTBACK 2 stops ifNUMBER = 0.

| But NUMBERis 2.

| c o u n t b a c k 2 c o n t i n u e s ,

i 2 p r :n u m b e r

I COUNTBACK INUMBER 1

| c o u n t b a c k l i s ca l led by

COUNTBACK 2.

NUMBER is 1.

COUNTBACK 1 stops ifNUMBER = 0.

But NUMBER is 1.c o u n t b a c k l continues.

1 PR INUMBER

COUNTBACK INUMBER 1

c o u n t b a c k 0 i s ca l led by

COUNTBACK 1.

NUMBERis 0.

COUNTBACK 0 stops ifNUMBER = 0.So it stops.

c o u n t b a c k i now continues, but ithas finished its job and stops.

c o u n t b a c k 2 now continues, but it has? finished its job and stops.

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A helper stops when there is nothing more to do

or when s t o p is encountered.

Lets change the procedure so that the print action occurs in a different place. Make a new proce

dure, CB.

TO CB :NUMBER

I F : NUMBER = 0 [ S TO P ]

CB : N U M B E R - 1

PR : NUMBER

END

Type

CB 5

Logo responds

1

234

5

This is probably a surprise. But remember when

the first CB helper calls the second CB helper, it

waits for the second to do its job, and then the

first helper continues with its own job.

Lets trace through the process giving CB the

number 2 as input.

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CB 2 CB INUMBER

CB 2 is ca l led.

I

I

I

NUMBER is 2.CB 2 stops ifNUMBER = 0.

But n u m b e r is 2 so CB 2 continues.

CB INUMBER 1.

c b l i s c a l le d b y c b 2 .

NUMBER is 1.

CB 1 stops ifNUMBER is 0.

But n u m b e r is 1 so CB l helper continues.CB INUMBER 1

c b 0 is called b y c b l.

NUMBER is 0.

CB 0 stops ifNUMBER = 0.

So it stops.c b l continues.

PR INUMBER

c b l h a s f i n is h e d i t s j o b a n d s t o p s .c b 2 con t inues .

PR INUMBER

c b 2 now has finished its job and stops.

Note that the stop rule has to come before the re

cursion line. Otherwise, the recursion will go onforever and the stop rule will never be reached. Often, the stop rule will be the very first line in theprocedure.

There are other kinds of stop rules we could makeup. For example, in s p i

TO S P I : S T E P : A NG LE : I NC

FD : S TEP

RT : ANGLE

S P I : S T E P + : I NC : A N GL E : I NC

END

137

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We could decide that s p i should stop if :s t e p is

greater than 200. So, we include the line

I F : S T E P > 2 0 0 [ S T O P ]

After editing s p i it looks like

TO S P I : S T E P : A N GL E : I NC

I F : S T E P > 2 0 0 [ S T O P ]

FD : S T E P

R T : A N G L E

S P I : S T E P + : I NC : A N GL E : I NC

END

Try s p i . Ify o u d o n ' t l ik e t h e s t o p r u le , c h a n g e i t.

Designing a stop rule for p o l y can be a little trick

ier. p o l y completes a figure when the turtle re

turns to its starting state, which may require the

turtle to turn only one complete rotation, that is.

only 360 degrees. But sometimes the turtle turnsa multiple of 360 degrees.

Actually, we only need to know what the turtles

heading was when it started , and then com pare it

to the tu rtle s heading after each u i t t l So before

p o l y is called we have to find the tu rtle 's heading.

HF A

i h e n p o l y LOU UKI o see ii_!t_ ' -

i t 1 e*__.< 0 L Y Lo C j

7/31/2019 Chapter 18 Em

19/19

------

Now tr y p o l y .

There is a problem here. Each time you call p o l y you have to save the starting heading. It would bebetter to put that action in a procedure. Lets re

name p o l y and call it p o l y i . Then we can make anew p o l y which sets up :s t a r t and then runs

POLYI.

TO P O L Y : S T E P : ANG L E

P 0 L Y 1 : S T E P : AN G L E H E A D I N G

E N D

TO P 0 L Y 1 : S T E P : AN G L E : S T A R T

FD : S T E P

R T : A N G L E

I F H E A D I N G = : S T A RT [ S T O P ]

P 0 L Y 1 : S T E P : AN G L E : S T A R T

END

Now p o l y will do the whole job.

There are many designs you can create with s p i and p o l y . There are many ways of combiningpolygons to make new and dazzling designs.

This introduction to Logo is finished, but we hopeyou will explore other turtle projects on your own.

TheReference Manualdescribes other features ofLogo which you might want to try now.

Have fun.