Puissance 4 : méthode joue 1ère version

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En provenance du cours de École Polytechnique Fédérale de Lausanne

Initiation à la programmation (en Java)

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École Polytechnique Fédérale de Lausanne

Initiation à la programmation (en Java)

217 notes

Ce cours initie aux bases de la programmation en utilisant le langage Java : variables, boucles, fonctions, ... Il ne présuppose pas de connaissance préalable. Les aspects plus avancés (programmation orientée objet) sont donnés dans un cours suivant, «Introduction à la programmation orientée objet (en Java)». Il s'appuie sur de nombreux éléments pédagogiques : vidéos sous-titrées, quizz dans et hors vidéos, exercices, devoirs notés automatiquement, notes de cours.

À partir de la leçon

Etude de cas

Nous voulons terminer notre cours avec une étude de cas, la création d'un jeu de « Puissance 4 », nous permettant de revoir tous les concepts abordés au long du cours.

Puissance 4 : introduction5:34

Puissance 4 : premières fonctions17:22

Puissance 4 : méthode joue 1ère version16:43

Puissance 4 : révision de la méthode joue6:36

Puissance 4 : moteur de jeu13:57

Puissance 4 : méthodes estCeGagne et compte15:34

Puissance 4 : finalisation10:05

Rencontrer les enseignants

Jean-Cédric Chappelier
Dr.
School of Computer and Communication Sciences
Jamila Sam
Dr
School of Computer and Communication Sciences

Let's continue coding our Connect Four game

In the previous episodes,

we covered the data structures that we recommend

you use to represent this game;

how to represent the players' colors;

and 2 basic features which are

the initialization of the grid and its display.

Now, let's look at the actual play.

What do we need?

For that, we must be able to ask the user where he wants to play,

as well as verify his play,

validate both the answer

and the data structures.

That is, if the play is legal, make sure that the structures are valid

and put the disk into the grid.

Then, we must ask the other player

to play, and come back,

and alternate between the two players.

And we must decide if the game has ended,

either when one of the two players has won, or when the grid is full.

In this episode

we will focus on the second point,

that is, ask a user for a column

and check if it is valid

and then update the game according to the play that was made.

Now that we know what we want to do

with the method that we will implement,

let's give it a name. We will call the method "joue" (TN: means "play").

Now, let's look at how to program it.

Let's start with its header.

To have an idea of what to put in the header, we should ask ourselves

how the <i>joue</i> would be called in the program.

For example, when we want to play in a certain

column of the grid, for example number 3.

Here, we are in the program,

so the positions will correspond to the positions in Java,

where numbering begins with zero,

and column number 3 would be...here.

Then, we have to specify the color of the player:

is it a red player, or a yellow player?

Here is a typical call that we could make,

which would give the following situation:

in column 3, the player inserts a disk

we would have a red disk which would place itself here.

After that, we would have calls where, typically,

we would have a disk, for example yellow,

that would be placed in column number 2,

that is, the third column,

and then the red player would play again in the same column number 3,

so the fourth column,

which would result in this situation.

There, now it is quite clear to us

how to write the header of our <i>joue</i> method.

So <i>joue</i> will take as first parameter the grid,

which is, as we said, a two-dimensional array of integers,

then it will take a position which will indicate the column,

so another integer here,

and finally we decided to represent the player colors as integers,

so the third argument here will be the color.

That's it for the header of the <i>joue</i> method.

Now, concerning its return value,

we will start with the keyword <i>static</i>

But what is its return type?

To answer this,

note that we did not write "z = joue (...)"

we simply made calls to <i>joue</i> directly

This means that <i>joue</i> returns nothing

and its return type will thus be <i>void</i>.

Now we have written its header,

we can continue with the definition,

with the body of the method.

For that, let's start by asking ourselves

what we really want to do

if, for example, we have a valid play,

where the player chooses column 3. Let's imagine the general case,

where we already had disks in column 3

and the player plays in column 3 again.

What should we do?

The disk should stop at the first empty cell

in that column.

But, first from where? The first from the bottom.

The algorithm will thus search from the

bottom for the first empty spot.

Lines are numbered from zero

up until (size of the grid) - 1:

remember that an array is indexed from zero

to size-1.

Let's start by declaring a variable

which we will call, very creatively, "ligne" (TN: means "line"),

which will start at size of grid minus 1,

and increment it one by one until we find an empty spot.

All we need to do now is write this.

We start by writing a comment

which explains what we want to do:

we iterate over the column from the bottom,

until we find an empty cell.

For that, we will declare a counter variable,

let's call it "ligne" (TN: means "row" or "line").

We will initialize it to the last position of the row array

that is, "grille.length - 1".

So it starts here,

then we will go up as long as we haven't found an empty cell,

as long as the grid, at the position of the row

and column we selected, is not empty.

For this purpose, we had defined <i>VIDE</i>, a value defining an empty cell,

which we can reuse here;

as long as this column is not empty,

we will go up one by one, which we write as " -- ligne ".

There, at this stage we have found an empty cell,

and we can fill it with the color we received.

For that, as usual, we start by writing the comment,

then, in Java, we simply write

" grille[ligne][colonne] = couleur; ", where couleur is the color we received.

This ends our first version of the <i>joue</i> method,

and, as good programmers, we will now test it.

To do that, we write a <i>main</i> method

which will test its workings.

In the <i>main</i> method,

we will declare a grid,

a grid of 6 by 7.

As you saw in a previous episode, we will

initialize the grid. We can start by diplaying it.

Then, we directly test a play

in the third column with the color red,

then we display the grid, which should give us

an output reflecting the fact that we have a red disk here,

since, remember, the index we receive here

corresponds to Java numbering, which begins with zero

So, the output will be like this,

then we continue by placing a yellow disk

in column number 2, that is, the third column

Then we continue to test

by playing another red disk and displaying the grid,

by playing a red disk in the fourth column,

column number 3, which will result in an output

that will look like the grid here.

Now, what happens if we continue to play

like this in column 3, if we add disks,

for example red disks, each time in column 3,

what will our <i>joue</i> method do in this case?

What happens when we play

in a column that is already full?

What will happen is that we initialize

our <i>ligne</i> variable as usual to the last row of our array,

then, while the row, at the corresponding column,

is not empty, i.e here, while

the row is not equal to zero,

this condition will be true, and we will continue doing "-- ligne",

and we will continue going up to position zero.

When <i>ligne</i> equals zero, what happens?

At that moment, this condition will still be true,

position zero is not empty,

and we will do "-- ligne" again.

This will make us go out of the bounds

of our array, outside of our game,

which is at least a conception error,

since we have a loop which takes us out of our array.

So, we will have to correct this,

and stop this <i>while</i> loop,

as soon as we reach the end of the array,

that is, as soon as we reach index 0.

Now we just need to write it.

To do so, we will declare a boolean

which will test whether we have completely filled a column.

We will call it "pleine" (TN: means "full").

In the beginning, the column is not supposed to be full,

so we will declare this variable as <i>false</i>.

Then, in the loop which travels upwards

to find an empty position,

we will change the value of the <i>pleine</i> boolean

and switch it to true,

as soon as we reach position 0.

This is because if we reach position zero in this loop,

it means that position zero was not empty,

so position zero is not empty

only if the rest of the column is also full

So, let's continue writing this. If the line reaches position zero,

then we can say that the column is full,

so "pleine = true".

In this case, we must stop decrementing,

we must not subtract 1 from the row,

so we protect our "-- ligne" instruction

when the row is zero

So we add an <i>else</i>

which protects this decrementation, this "-- ligne"

from the condition, from the fact that the row is empty.

And now we'll correct our loop,

by saying that we continue the loop

as before, that is, we continue to look for an empty cell,

but only while the column is not full.

This symbol, remember, means <i>not</i>,

so while the column is not full,

and we haven't found an empty cell,

we continue as long as the 2 conditions are true.

If it is not full

and we haven't found an empty cell, we go up.

If one of the two conditions, either if it is full,

or we have found an empty cell, are true,

then the loop will stop.

We can end by adding a pair of parentheses here

to make our code more legible.

Finally, since we didn't do that in the beginning, as we should have done,

we will comment what we have just coded.

The idea is to iterate over the column from the bottom

until we find an empty cell, or until we reach the top.

Finally, we must not forget to correct the end which is not valid anymore,

since we cannot guarantee that it is possible to play every time,

since the disk will not be placed if the column is full

So we protect this with the condition:

if the column is not full, so, "if (!pleine)".

now, we can play,

and we can also give an indication:

we can add an instruction to our <i>joue</i> method

so that it can indicate its status to the outside, for example by returning

<i>true</i> to say "I was able to play,

the <i>joue</i> method was able to do what we want".

And conversely, it can return <i>false</i>, to say:

"no, the column was full, the play was not valid,

I was not able to place this disk at the right position".

Finally, as usual, we comment our code to explain,

either to us, later, when we go over the code again

or to anyone else who would read it, what our code does.

Here, if we haven't reached the top, "if (!pleine)",

if we haven't reached the top of the column, if it isn't full,

then, we can play.

Let's not forget, finally, to correct our return type

since we have now decide to return

a boolean to indicate whether the play was valid or not

So we change the return type from <i>void</i> to boolean

There, now we are pretty sure

that we have corrected our first version of the <i>joue</i> method

to have a robust version which accounts for the case where the column is full,

and now we must test this method and for that,

we will write a new <i>main</i> method,

where we will specifically test the case where we always play

in the same column.

As usual, we declare the grid, we initialize it, we display it,

and then here, we will write a loop which will play

10 times in the same column,

in the fourth column here, i.e column number 3.

We write a loop that repeats 10 times,

playing in column 3.

Note that now, since the <i>joue</i> function returns a boolean

indicating whether the play was valide,

we now call <i>joue</i> on the right-hand side of an expression.

So we fetch the return value by writing "valide = joue(...)",

and to see what is going on and to test our program,

in case the play was not valid, we will display

a message telling us that it was not possible to play.

In either case, we display the grid to see what is happening.

What will this give us?

I copied the <i>main</i> here in the bottom, and all of our <i>joue</i> method.

Let's run the <i>main</i>. We start at "i = 0" -- we don't use the <i>i</i>,

it just serves to say that we are executing the loop 10 times.

So here we start at "i = 0", and we always play

in column 3 with red disks.

This will start by placing a disk here,

then another, and so on.

All of these cases aren't especially interesting,

but what will happen when we play the seventh disk?

Once <i>i</i> has a value of 6, here,

we want to play a seventh time

in the same column, which is already full,

and so when <i>i</i> has a value of 6, we will end up here again,

with a value of 3 for the column and the color red,

and here, we will initialize the row to

the size of the grid minus 1.

We will initialize the boolean to <i>false</i>,

so of course, <i>pleine</i> is <i>true</i>, meaning

that the grid, at the position [ligne][colonne], is not empty,

the row is not equal to zero,

so here we jump into the <i>else</i> and we decrement <i>ligne</i>,

which will thus end up here (second row).

Here, both conditions are verified,

so <i>pleine</i> is still <i>true</i>,

and we still have a grid here

which is not empty, and so

we will go up until we reach the position <i>ligne==0</i>.

So when <i>ligne == 0</i>, what will happen?

We always arrive here with a true condition,

and the grid at [0][column] is not empty,

we go in here, but this time the <i>if</i> is true,

the row is zero, and at that moment we change

the value of <i>pleine</i> to <i>true</i>.

We are here, we are at the end of the <i>if</i> bloc,

we come back into our loop,

but this time, the condition, "!pleine",

is false and so the <i>while</i> loop ends,

and we exit the loop.

Then we continue, here, we reach this <i>if</i>,

again, our <i>pleine</i> is not true,

so this condition here is false.

We jump into the <i>else</i> and we return <i>false</i>.

Which means that when we arrive here, when we have a value of 6,

<i>valide</i> will be <i>false</i> and so,

we will display the error message

and this message will be displayed

for <i>i</i> equals 6, 7, 8 and 9, so the error message

indicating it is impossible to play will be displayed 4 times.

That was the test which confirms

that we cannot add disks to a column

that was already full, which is exactly what we wanted.

There, I propose to end this episode here,

and to discuss alternative ways of implementing

the <i>joue</i> method

in the next episodes.

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