Hello World!
I'm going to explain a counting quiz in Alice3.
I have already created the entire program and I'm going to explain it.
In my setup scene I put in 4 fish.
In my program, what I want to do is make some of the fish
invisible by setting their opacity to zero,
and then ask you how many fish you see.
So,  my code
(Go all the way up to the top here.)
First of all I have an integer numRight which I initialized to zero.
and then I've got this whole block,
this whole block of code all the way down,  I'm going to ask the question five times.
and then at the very and I'm going to have the carp tell you how many you got right.
The carp is always going to be visible, because he's going to talk.
We have an integer n is 5, and this for loop
is going to go i is zero; i < N;  i++
That means it's going to execute five times.
i will have the values 0, 1, 2, 3, 4, so that's five.
the counter for the number  visible
isn't initialized to zero, it's initialized to one,
because the carp is always going to be visible.
Next I'm going to have a random int,
have values from 0 to 10 inclusive.
That's actually 11 values because both zero and 10 count.
If that random integer is less than five
I'm going to make the blue tang invisible
by setting its opacity to 0.
and I specified a duration of zero because I want to scene to just change
all at once and you see a different number of fish.
If the random integer is six or up to 10
I'll set the opacity to one
and one to the number of fish that are visible.
So, let's say that number comes out to be 7.
this would be false, so I go to the else,
I set the opacity to 1, that means you'll see the blue tang,
and I add one to the number visible, number visible will now be 2.
I can't copy this statement and paste it because
in this statement I declare the random int variable,
and  down here I have an assignment statement
to give it a new value.
this time I picked from zero to N, where N is 3.
That means you could have a value of 0, 1, or 2.
because this is exclusive so 0, 1, 2 if it's less than one,
in other words, if it's zero,  I make the clown fish invisible.
Otherwise, I make the clown fish visible,
and add one to the number visible.
I have to set the opacity to one each time because
the first time through, they were all visible, but the second time through,
when we asked the second question,
this clown fish might have been made invisible on the previous round.
and then we add one to the number visible.
Then I did copy and paste this statement, and this whole block
and I changed just the fish
from carp, make sure you change it in both places
because you don't want something here that sets the carp to invisible.
and carp2 to be visible.
Then after doing those three fish, and setting them to
visible or not,  I should have the  carp asking this question.
the carp is going to ask the user "How many fish do you see?"
if the answer from the user
is ==, meaning exactly the same as the number visible
I say "Yes, that is right", I add one to the number right.
and otherwise I have the carp say "Uh, I think there are 3."
and notice that there is a space right here after are,
and if we didn't have a space there you would see three or whatever number right there next to that.
And then this whole block is repeated three times
and within that block we set three of the fish
to visible or invisible
and then ask the user how many they see.
The carp is always visible.
At the very end, I tell them how many they got right.
OK. Now you try it.