Writing Programs Using Top-Down Design and Common Sense

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One way to solve this problem is to declare an array that will have 70 elements to it. That will represent the "track". Since you will need to keep track of both runners' positions, it will have two dimensions, so it will look like this:

int track[2][70];

The runners will be starting at the first square, so initially they will both be at array position 0, right? (The C language begins its arrays at zero.) So initially both arrays will have the runners' positions at the 0th element of the corresponding second dimension of the track. Let's assume that a 1 in the array signifies the position of the runner, and all other elements of the array's second dimension will always be 0.

Let's define a couple of constants first:

#define HARE 0
#define TORTOISE 1

So, initially, the tracks will start with:

track[HARE][0] = 1;
track[TORTOISE][0] = 1;

And the rest of the array will be filled with 0s to make sure it is initialized properly.

Now, if the hare advances nine squares, the array will look like this:

track[HARE][0] = 0;
track[HARE][1] = 0;
track[HARE][2] = 0;
track[HARE][3] = 0;
track[HARE][4] = 0;
track[HARE][5] = 0;
track[HARE][6] = 0;
track[HARE][7] = 0;
track[HARE][8] = 0;
track[HARE][9] = 1;
track[HARE][10] = 0;

It should be easy to see that what you'd need to do to keep track of the position is to zero out the current array position, where the runner is before any advancement or "slips", then count however many squares they advance from there, and put a 1 there.

(I'm assuming, of course, that your instructor intends the 70th square to be to the "right" according to your assignment. And if an animal moves to the "left" or "slips", that means they go to a lower element in our arrays.)

It will also make it easier if we keep track of the current position of both racers. That will just be two int variables, that will index into the particular element on the "track" that each animal is in. Like this:

int hare_position = 0, tortoise_position = 0;

That would be how they both begin. In the example above, assuming the hare was now at the 9th square, (hare_position == 9) would certainly be a true conditional expression assuming everything was working correctly and that the two position markers were kept accurate.

Just to make sure you understand this, if the hare then advances 3 squares, you would only need to do the following:

track[HARE][hare_position] = 0;
hare_position += 3;
track[HARE][hare_position] = 1;

Now that we've got a way to keep track of the race, printing it out shouldn't be much of a problem, either. Each time you print it, you can loop through the entire array, printing the "line" (perhaps the minus character from the keyboard, the "-") when there's a zero in that array element and a "T" or an "H" if there is a 1 in the array's position.

Now we need a way to run the race. Your teacher told you you'll need a loop of some kind. Any kind will do for this particular task. Let's assume it will be a while loop:

int race_is_over = 0;
while (!race_is_over)
{

}

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