‹ Back to the Archive

Weaken Defense

2 May 2009 · set down by Guybrush Threepwood · 2 readings

This is a brief article on the effects of weaken defense, an interesting little ability that many creatures have, and many people initially ignore. So, let us look at what it is that weaken defense is explained to do, from there we will expand into what that means, and then into what it is that it actually does.

“Weaken Defence
Creature will lower targets defence attribute for the next two rounds. Effect value is based on attack attribute. Temporary effect, applies next round, lasts 2 rounds.”

Alright, so what does weakening a creatures defense do? Let's look at how damage is calculated from an attack. Damage=Attack-Defense, but can not be reduced past 1. Pretty simple right? This means that if a person attacks for 500 damage and the opponent has 450 defense, the creature will take 500-450=50 damage. On the other hand if a creature attacks for 1 damage (hypothetical, but in fact entirely possible) against a creature with -342 defense the damage dealt would be 1--342=343 damage. So you might be thinking, how is this useful? Let's simply look at an example in which a single creature is attacked by 6 maxed grassan. The first option is for us to set them all to haotic damage. This means that (without vit) the average damage per round will be 600. If we used weaken defense on one instead, what would the result be? Well, the weakener would do 0 damage, but reduce the crit's defense by 100, meaning every other grassan would deal an extra 100 damage. Thus the average damage (after the first round when weaken has not yet come into effect) would be 1000 damage, a good improvement. This is of course ignoring the effect that weaken defense lasts “TWO*” turns. This means that each other grassan's attack is actually increased by 200 damage each turn bringing our average per turn damage (after the second round) to 1500 damage.

We went from 600 damage, to 1500 damage by switching one little creature over to weaken, a good deal yeah? We've over doubled the damage. So the obvious question now is, how do we maximize this function? How many weakeners should we have and how many hitters? Before we go though, we may want to look at what weaken defense ACTUALLY does. In all actuality, weaken defense lasts little sideways 8 turns. That's right, the entire battle. So now on the fifteenth turn your grassan that you've so intelligently switched over to weaken is using weaken for 100 again. The targets defense this turn is? -1400 (remember, the fifteenth hasn't kicked in yet), yeah, that's right, -1400 (assuming it had 0 defense to start with, a poor assumption, but hey in most cases it doesn't compare to -1400). So how much damage do you average this turn? Well, that's 100--1400=1500--> 1500*5=7500 damage. That's a fair amount of damage for one turn. So now with this new assumption let's attempt to maximize our utility in a situation of 6 grassan.

Read on at your own risk. You may find that this contains spoilers. More than likely you won't understand it. If you have a good grasp of basic algebra you can at least use a calculator to do the potion of this that I have used calculus for. Now for the fun stuff

R=rounds d=attack/weaken of each creature h=creatures on attack
W=creatures on weaken Def=Opponent's and creatures defense W=6-h

This is a simple situation when a person attacks with 6 max grassan, it also ignores the effects of Haotic damage. However, since Haotic damage should roughly average out, this should be close.

Damage done during an entire battle to a single creature.

(R*d*h)=damage from attackers before defense is calculated.

(W*d*R*h)=amount of -defense each round, multiplied by the number of times the creature is hit each round. This is the amount of damage a creature would do if it attacked for 0 damage. This is simply the -defense effects of weaken added up. However, this is only the damage for EACH round, not all rounds together. For all rounds together the formula is the summation of this equation:

(W*d*n*h) where n goes from 0 (first round weakens don't apply) To R-1 (last round weakens don't apply). In other words it is some constant (because for each battle W*d*h is constant) times 1+2+3+...R-1. So C(1+2+3...R-1). (1+2+3...R-1) has an equation to solve for this value. It is simply (R+1)*R/2. (This is derived from the fact that you add the last number in the sequence with the first, you get R+1, and then add the second with the second to last, you get R+1. You can do this R/2 times before you get to the middle number(s).)

(W*d*h)*(R+1)*(R/2)=The equation for total damage done in a round from the effects of weaken assuming each attacker does 0 damage themselves.

(Def*R*h/creatures)=The amount of damage the opponents defense prevents each round, though this is the maximum the defense could possibly defend against as defense cannot defend against more than 100% of an attack each turn. This also assumes creatures to have negligible defense compared to their owner. (from what I've noticed, this is true at high levels) Now, if the attacker has greater attack than the defender than the fact that the defense can't block more than 100% is no longer an issue and this is not an estimate, but exact as creatures will not be dealing 1 damage every turn.

I can see a lot of problems people will have with this equation up there. Frankly it's a lot easier to place this at the end than incorporating it into each previous portion and pulling out the commonalities, this part. You can trust me on this one, or you can think about it logically. Yes, weaken defense decreases the creatures defense, but if a creatures starts at 45 defense, and is weakened for 100 they are now at -55. That does not mean that they are not taking 45 less damage per hit. Had it not had that initial 45 defense, it would now have a defense of -100. Thus it takes 45 less damage per hit. Blah blah blah... And so on. This is simply a preemptive attempt to explain what I see as an issue people will have with this. If you still don't believe me, let me know.

(R*d*h)+(((6-h)*d*h))(R+1)*R/2))-(Def*R/creatures)
(6h-h^2)*d)

Now to maximize this function we consider it a function of h to see the number of hitters and weakeners an individual should have. We then attempt to set the derivative to =0.

R*d+(((6-2h)*d))*(R^2+1)/2-(Def*R/creatures)

Lets mess around with this a bit and make it cleaner.

(6*d*(R^2+1)/2-2h*d(R^2+1)/2)+R*d-(def*R/creatures)=0

We will now make the assumption that d (remember, this is attack damage of each creature) is roughly equivalent to the enemies defense. Since most people's attack is greater than defense this seems like the equation would lean more towards the value of R*d-(def*R/creatures) being a positive constant, we'll call it C*R because it is simply some value that increases and decreases with R.

(6*d*(R^2+1)/2-2h*d(R^2+1)/2)+R*C=0

Now we can consider this equation instead, since it is set to 0 as the one below.

(6*(R^2+R)/2-2h*(R^2+R)/2)+C*R/d=0

It's now easy to see that as R gets large, the C*R value on the outside will hardly compare to the remainder of the values, making it nearly negligible. Remember that C is considered to be a small number anyway, and we then divide it again by attack, and that the inner value is being squared whereas it's simply going up linearly. So, as R-->infinity the ideal setup is 3-3. More rounds, the more ideal that you have 3-3. At lower rounds the equation slightly favors hitters over weakeners if C is a positive value, and slightly favors weakeners if C is a negative value. Unless the enemies defense significantly outweighs your attack (or if he's using less creatures, thus having one creature with massive defense) and has little Ve (meaning that the fight will no necessarily last very long) you are better off trying to balance your weaken and attack. Of course you are free to plug in the numbers yourself and see if what I have claimed to be negligible is in fact so. It is also interesting to note that the weaken attack depends only on power, not really on the number of creatures, and the hitters value mostly comes up only depending on the number of attacks. This means that a single creature that could attack for 1 damage multiple times per round would be more useful than a creature attacking once per round with 200 damage. This is because most of the damage is caused by the weaken, especially at the end.

Commentaries — 1
Leave a commentary
Join the discussion

Have a thought or a question? Leave a commentary below.

Login is required to vote.
NA
nadrolski 30 Jul 2010 05:44 Reply
Bravo! Two thumbs up!
0
Kindred chronicles
How to Care For and Raise a Knator
4 September 2009 · set down by Fenrir Greycloth

How to Raise and Care For a Knator By Joykill Before I can even begin about taking care of and raising a Knator, one must first understand a Knator, how they live and what their origins are. Now to start, I believe it is…

Read on ›
Training of a Drachorn
4 September 2009 · set down by Fenrir Greycloth

To Punish a Drachorn: by Malaikat Maut Many individuals believe drachorns to be among the most powerful creatures in the lands of MagicDuel. I have recently had the displeasure of battling such a beast, and can attest th…

Read on ›
Great Tip for Stat Gains
23 March 2009 · set down by Unbelievable Power · 2 commentaries

You need more losses!  That is one of the best tips I can possibly give you to become much stronger. You want as many as you can possibly accumulate. The main reason why this benefits you is due to the bonus XP you will …

Read on ›
Lifesteal
18 March 2009 · set down by Shadow Seeker · 3 commentaries

Lifesteal. Takes a certain percentage of the selected targets and gives it to another random creature on your side. There is multiple or single targets, moving from weak, strong, dying to random. The full range, but none…

Read on ›