Subtracts and sums, Liters simulation

iTopGames

Hi guys, i gotta a question about subtracts and sums, imagine have five containers each one with different capacities, so i got some with liquid and i have to spill the liquid inside others and i have to show how many liquid is inside in each one doing subtract and sum, my problem is that i don't know how to put a limit in subtract and sum, i have to make a formula and i am very bad in maths like i always say, if somebody have done something similar please let me know, thanks in advance.

tatiang

Is this a school assignment? As a teacher, I can't help wondering...

I can't quite understand what you're trying to do. Can you explain a little more about what you mean by "limit"? If you can give a sample math problem that shows what you're trying to do (or even just write "I want to add the height of one container's liquid to the height of another but make sure the sum doesn't exceed 50..."), it will make it clearer.

If I had to guess about what you're trying to do, I'd go with the min() and max() functions. Have you worked with those in GameSalad?

iTopGames

Hi @tatiang

Imagine you have 5 containers: the first one with 5 liter capacity, the 2nd one with 4 liters capacity, the 3rd with 3 liters cp, the 4th with 2 L Cp and the last one with 1 L cp.

So the first one is full with 5 liters of water the only 5 liters in the game the other are empty, the game is about pour the water in the others containers:

e.g.: The 5 L cp container pours its water into the 3 L cp container, so the 3 L cp containers is full now and the 5 L cp container has 2 liters of water now.

So the math is case (from 5 to 3) something like this i think, first for each one container i got 2 numbers (index attributes), the content (water) and the capacity, i.e.: 5 L cont and 5 L cap and 3 L cont and 3 L cap.

Where i think that 5 L cont= 5 and 5 L cap=0 (because it's full) and 3 L cont=0 and 3 L cap=3 (Because it's empty)

So here is where i get stuck:

5 L cont= 5 L cont-3 L cap (2 liters)
3 L cont= ?
5 L cap= ?
3 L cap= ?

I think it is a way to do this, but i can't do it all, if you know a way with max and min or if you figure it out how to do it with way let me know please, the solution has to work universally with all the cases e.g.: (from 3 to 1), (from 5 to 2) etc.

tatiang

Okay, that makes sense now. Now I just need to know how the player moves water from one container to another. Does the player choose the "source" container, the "destination" container, and the amount to transfer? Or is it more limited than that (e.g. tap on the source container and then tap on the destination container... only as much water as will fit transfers to the destination container)?

tatiang

Off the top of my head:

5Lcont=5Lcont-3Lcap ... but only if 3L container is empty; otherwise, it's 5Lcont=5Lcont-(3Lcap-3Lcont)
3Lcont=3Lcont+(5Lcont-3Lcap)
5Lcap=5Lcap+(5Lcont-3Lcap)
3Lcap=3Lcap-(5Lcont-3Lcap)

Since you're re-using a difference formula repeatedly, you could do a Change Attribute game.transferAmount to game.5Lcont-game.3Lcap and just use game.transferAmount in place of each (5Lcont-3Lcap) above, but that's assuming these equations all work!

iTopGames

Hi @tatiang

The selection mode is tapping first the "source" container and then "destination" container, only can be selected as "source" the containers which contains water and only can be selected as "destination" the containers who aren't full, the water amount of the water can't be selected, it will only fill the container till its top, so it will subtract the current capacity of the destination to the source and if the content is bigger than the capacity it's going to remains water in the source. It's a imitation of real pouring and filling. Thanks for your time and help my friend.

tatiang

@iTopGames I'm fairly certain those equations will work. Let me know if you run into problems.

iTopGames

@tatiang Ok let me try and i'll let you know what happen, thanks a lot again!

iTopGames

Hi @tatiang

I solved it, i had to add another two index attributes, "Real Capacity" (RCap) and "Beginning Content" (BCont) and Capacity is called now "Current Capacity" (CCap)so it is the equation:

e.g.: From 5 to 3:

BCont = 5LCont

5L Cont = 5L Cont-3L CCap

3L Cont = 3L Cont+(BCont-5L Cont)

3L CCap = 3L RCap-3L Cont

5L CCap= 5L RCap-5L Cont

BCont = 0

Thanks for your help! take care.

tatiang

Glad to hear it!