Exp equation for enemy and player in rpg

Question

I am making an rpg game with specific leveling system I want when a level 1 player fights a level 25 monster to level up to level 20-25 then when he goes and fight the same monster he then levels slowly like level 30 then level 32, 35, 36 on his next 4 fights respectively meaning you must always fight a monster higher than your level else you will level up really slow

At the same time I want a level 1 player to fight a 20000 enemy and then gains like 10000 levels then on his next fight he gains like 50000 then he gains another 50000 meaning that if the player fights a really strong monster he levels up fast until he reaches the level of that monster then he level up real slowly, This system doesn't have to be perfect at all but you get the idea So any help on the equations, I searched really hard and found elementary functions like

Power function:

y = a * x^b + c

Or Linear functions:

y = mx + b

Also exp. functions and log functions but I cant seem to get the desired effect or even get any close to it, Any help or any tip on where I can find my answer is highly appreciated ty

Answer 1

It will be a different solution for each game, for example, some games will only give, you experience if the enemy is on a similar level. You need to start by giving a hard look to your requirements.

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First, we need to start by recognize that the experience gained shall never be negative. Furthermore, I am inferring from your description that when you fight monsters of lower level than you are, it will never result in zero experience, but in increasingly diminishing values instead.

Second, I suspect you want to fix the amount of experience the playing will gain by fighting a monster of the same level.

If we are looking for a function g, that given a difference of levels x (monster_level - player_level) with the description provided, it follows that:

• Lim(g(x))_x->-∞ = 0: When the difference tends to negative infinity, the gained exprience tends to zero
• g(0) = k: When the difference is zero, the gained experience is a constant value k
• dg(x)/dx > 0: The gained experience is a growing function, the greater the difference of levels the greater the gained experience

We cannot represent this with a linear function. To speak is loose terms, a stright line that doesn't cross the origin, and meets the horizontal axis at negative infinity... is an horizontal line. So its slope would be zero. However, we need a slope greater than zero.

It cannot be a quadratic function, because they are not always growing. In fact, you will have a similar problem with higher order functions.

Trigonometric functions are periodic functions, so we can discard those too...

Logarithms are growing on the positive, but decreasing and with an imaginary part on the negative. So, do not base the function on that either...

Exponential functions fit the bill. For example take g(x) = 2^x:

• Lim(2^x)_x->-∞ = 0
• 2^0 = 1
• d/dx {2^x} = log(2) * 2^x > 0

It is very easy to tweak by changing the cosntant 2 to a parameter, but whatever constant we put there, we will get 1 when x = 0. We could consider to add another term, but that would change the limit when x -> -∞. Therefore, what we do is mess with the exponent, in particular:

g(x) = k^(x+1)

With this function, you get g(x) = k when x = 0. The next thing you will want is to tweak how fast it grows, you can do that by adding a factor to x:

g(x) = k^(fx+1)

If I want to make sure that g(a) = b, I do this:

g(x) = k^(fx+1)
g(a) = b
=>
b = k^(fa+1)
=>
log(b) = log(k^(fa+1)
=>
log(b) = (fa+1)*log(k)
=>
log(b)/log(k) = fa+1
=>
log(b)/log(k) - 1 = fa
=>
log(b)/log(k) - 1
------------------ = f
        a

I can rewrite the function like this:

g(x) = k^((log(b)/log(k) - 1)*x/a + 1)
=>
g(x) = k*k^((log(b)/log(k) - 1)*x/a)
=>
g(x) = k*(b/k)^(x/a)

It will have the following properties:

• Lim(g(x))_x->-∞ = 0
• g(0) = k
• g(a) = b
• dg(x)/dx > 0 Where k, a and b. In addition, remember that x is monster_level - player_level.