# Create a NPC for reset enchant instead the Godly

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• Topic Namn:  A NPC dose customize Reset Enchant
• Describe your suggestion
• What is this? Able to reset enchant with each slot, make more easy to do the enchant instead the godly enchant NPC, also to using the Zeny Coin as well.
• Why do we have to add it? More fair than the godly enchant.
• Who will benefit this? All
• Will this change the system drastic? Don’t  think so.
• Room for improvement?
• Other

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-1

Godly enchanter's only purpose is to remove the randomness from some of the enchants. It is a premium thing that is now not updated and only in the game to be fair to the new players wanting the perfect enchants the same way oldies would have.

Resets lack the main issue that gives Godly its value: randomness. Resets always reset the thing to the initial state. Adding another NPC like that, I doubt would provide anyone with any tangible benefits. Unless you are talking about resetting the enchants that are normally not resettable. If you are, then I am even more vehemently against this.

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the whole point of this suggest is about the godly enchant does not correspond NP2W which todays LRO,

I dont feel any wrong with, the Godly feels more like you feel a man fish, but the customize reset much close to tech a man how to fisher.

we are Increase the success rate for many enchant already but same as we do the customize reset enchant if we able to do that.

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+1

I'm OK with this. It would bring people more motivation to enchant without the need to go by Godly. It does reduce randomness and is imo a good alternative for Godly (if it does not cost too much). Makes this a possible scenario:

• Getting Spell6/6 on a GW (you'll keep these as long as you want Spell enchants)
• Getting Spell6 on the 3rd enchant
• Resetting the 3rd enchant only and try to get Spell7

Better than resetting the 3 enchants and try to get Spell6...

And it still give you chills when you get all the enchants you want rather than just pay to get the max...

• 1

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Player: I want this.. that and this.

Me: $e^{i\pi} + 1 = 0$$e^{i x} = \cos x + i \sin x$$\displaystyle \sum_{n} \frac{1}{n^s} = \prod_{p} {\frac{1}{1 - \frac{1}{p^s}}}$$\displaystyle\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}$${\mathbb{R}} \sim {2^{\mathbb{N}}}$$\displaystyle n! = \int_{0}^{\infty} {x^n e^{-x} \,dx}$

$a^2 + b^2 = c^2$$F(n) = \frac{(\varphi)^n - (-\frac{1}{\varphi})^n}{\sqrt{5}}$$\displaystyle1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots =\frac{\pi^2}{6}$$\displaystyle1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots =\infty$$\displaystyle\pi(x) = \sum_{n = 1}^\infty \frac{\mu(n)}{n} J(\sqrt[n]{x})$$\displaystyle J(x) = Li(x) + \sum_{\rho} Li(x^\rho) - \log 2 + \int_{x}^\infty \frac{dt}{t(t^2 - 1)\log t}$

🤯

• 2

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Ahaha gm lai look alike doraemon. Just enjoy the play. Not all the obstacle should be rid away

• 1

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Come on Lai just solve (- 1) + (- 1) first 🤣

• 1

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