Scratch

Here's an idea: create two lists, one named "Elements" and one named "Charges."  Populate the "Elements" list with the names of all the ions you want to quiz the user over (so, all relevant element names).  Then, populate the "Charges" list with the charges of each ion, so that the charge indicated by the first member in the "Charges" list corresponds to the element name specified by the first member of the "Elements" list, and so on (this simple data structure is called a parallel array). 

For example:

Elements = { Hydrogen, Lithium, Nitrogen, Oxygen, Fluorine, Sodium, Calcium, ... }
Charges = { 1+, 1+, 3-, 2-, 1-, 1+, 2+, ... }

Now, you want to provide the user with three options to pick from.  Two of the options should be false, but one should be true.  First, we need to select three random elements from the list of elements, but without repetition:

delete (all v) of [Numbers v]
set [i v] to [0]
repeat (length of [Elements v])
change [i v] by (1)
add (i) to [Number v]
end
set [i v] to (pick random (1) to (length of [Numbers v]))
set [WrongElement1 v] to (item (item (i) of [Numbers v]) of [Elements v])
set [WrongCharge1 v] to (item (any v) of [Charges v])
repeat until <not<(WrongCharge1) = (item (item (i) of [Numbers v]) of [Charges v])>>
set [WrongCharge1 v] to (item (any v) of [Charges v])
end
delete (i) of [Numbers v] // Prevents choosing same element twice.  

set [i v] to (pick random (1) to (length of [Numbers v]))
set [WrongElement2 v] to (item (item (i) of [Numbers v]) of [Elements v])
set [WrongCharge2 v] to (item (any v) of [Charges v])
repeat until <not<(WrongCharge2) = (item (item (i) of [Numbers v]) of [Charges v])>>
set [WrongCharge2 v] to (item (any v) of [Charges v])
end
delete (i) of [Numbers v] // Prevents choosing same element twice.  

set [i v] to (item (any v) of [Numbers v])
set [RightElement v] to (item (i) of [Elements v]) // Now we choose the correct one.  
set [RightCharge v] to (item (i) of [Charges v])

Last edited by amcerbu (2013-01-14 21:18:20)

Oh and a question - can scratch take a variable and then search a list to see if the variable is in the list?

KVTaniguchi wrote:

Oh and a question - can scratch take a variable and then search a list to see if the variable is in the list?

ErnieParke is right; there is a shortcut for determining whether a list contains a specific element.  But to find where the element is, you have to iterate through the list, using a loop and an index (the i variable from before). 

delete (all v) of [RandomNumbers v]
repeat (100)
add (pick random (1) to (100)) to [RandomNumbers v] // Fill the list
end
set [i v] to (0)
delete (all v) of [DivisSeven v]
repeat (length of [RandomNumbers v]) // Look for numbers divisible by 7
change [i v] by (1)
if <((item (i) of [RandomNumbers v]) mod (7)) = [0]>
add (i) to [DivisSeven v]
end

Last edited by amcerbu (2013-01-14 23:13:30)

The issue I'm running into with parallel arrays is that every 'slot' in the parallel arrays has to be different from the other slots, otherwise you could end up with a situation where a user chooses an answer that is chemically right:  1: Br 2:1-  but according to how the program is checking it (1:2 is wrong) the user is told he/she is wrong.

ErnieParke wrote:

KVTaniguchi wrote:

Oh and a question - can scratch take a variable and then search a list to see if the variable is in the list?

Yes, yes it can. Just do this:

<[List v] contains (variable)>

I hope that this helps! Also, hello KVTaniguchi and welcome to Scratch! I wish you the best of times here!

HI! Sorry this is not about the topic right now, but I wanted to say that the golden rule is "Thou shalt do unto others as thou would have them do unto you."

mollovespandas wrote:

ErnieParke wrote:

KVTaniguchi wrote:

Oh and a question - can scratch take a variable and then search a list to see if the variable is in the list?

Yes, yes it can. Just do this:

<[List v] contains (variable)>

I hope that this helps! Also, hello KVTaniguchi and welcome to Scratch! I wish you the best of times here!

HI! Sorry this is not about the topic right now, but I wanted to say that the golden rule is "Thou shalt do unto others as thou would have them do unto you."

With all mind, I ask, "Who hast thou quoth?"
Is it Shakespeare, Mark Twain, or even both?
With that out of mind, thank you for that spark,
For it was very kindly of you, I remark.

If you didn't understand the above, basically what I'm saying is:

Who are you quoting? Anyway, thank you for reminding me of that rule.