How to reach the max damage of "nunchucks+"? "nunchucks+" or "dagger+"?

This guide will deeply analyzing the usage of "nunchucks+". And it will also comparing the damage with "dagger+".

Introduction

中文版（Chinese ver.）：https://steamcommunity.com/sharedfiles/filedetails/?id=2788696412

Hello everyone! I'm "a void" in Discord. Maybe, thief players who were played the method of dice splitting ( main weapon is "nunchucks+" or "dagger+", other equipment is used to split dice to make the dice usable on your weapon ) have two questions: how to use "nunchucks+"? which one can deal more damage? "nunchucks+" or "dagger+"? i had considered these questions about 1 year ago and i found the answer i needed. now, after seen some players are confused about usage of "nunchucks+" and debate about choosing "nunchucks+" or "dagger+", i felt my thought about these questions could be a conference for Dicey Dungeons players. so, this guide will analyze about how to use "nunchucks+" and if is "dagger+" better than "nunchucks+".

this guide may a little longer. so, first of all, let me give you the conclusion:
1.to reach "nunchucks+"' maximum damage, you need to try to return a dice with each use as many as possible. the best situation is that you can use "nunchucks+" for x-1 times with x dices.
if 1 is more than 3, you need to use 2+1 and 3+1 to maximize your damage with the strategy of 2+1→3+1→2+1→1+1.
if 3 is more than 1 and the amount of 1 and 3 is odd, the strategy is: 2+1→3+1→2+1(left a 2)→3+1→3+2→……（finally, you have 3321）→3+1→3+2→2+1(the order can be exchanged and you'll left a 1).
if 3 is more than 1 and the amount of 1 and 3 is even, the strategy is: 2+1→3+1→2+1(left a 2)→3+1→3+2→……（finally, you have 321）→3+1→2+2(all rolls are used up). just like the gif shown.


too many 1


too many 3


the amount of 1 is equal to the amount of 3

2.if the amount of 3 is a, 2 is b, 1 is c, the max damage formula of "nunchucks+" and "dagger+" are:
dagger+: 4a+3b+2
nunchucks+(c＞a): 6a+3b+c+1/2(-1)^(a+c)-1/2
nunchucks+(a≥c): 4.5a+3b+2.5(c-1)+1/4(-1)^(a+c)+1/4
nunchucks+(if a≥1, (-1)^a=-1, b=0, c=1): 4.5a-0.5 (a is odd)
nunchucks+(if a≥1, (-1)^a=1, b=1, c=0): 4.5a (a is even)
nunchucks+(if all of your rolls are the same one): c+1/2[(-1)^(c)-1] (a=b=0), 2b+(-1)^(b)-1 (a=c=0), 3a+3/2[(-1)^(c)-1] (b=c=0)

when c＞a, if 2a≥c+1 (a+c is odd) or 2a≥c (a+c is even), "nunchucks+"' damage is greater than or equal to "dagger+"'s.
when a≥c, if a+c≥4 (a+c is odd) or a+c≥5 (a+c is even), "nunchucks+"' damage is greater than or equal to "dagger+"'s.
especially, when a≥1, (-1)^a=1(that means a is odd), b=1 and c=0, if a≥5 "nunchucks+"' damage is greater than or equal to "dagger+"'s.
when a≥1, (-1)^a=1(that means a is even), b=1, c=0, if a≥6, "nunchucks+"' damage is greater than or equal to "dagger+"'s.
otherwise, "dagger+"'s damage is greater than "nunchucks+"'.


and now, it's time to analyze these 2 questions deeply.

Ⅰ How to reach the max damage of "nunchucks+"?

In this chapter, I'll introducing "nunchucks+" and analyzing its usage.

Ⅰ- 1 introduction of "nunchucks+"

first of all, let's see "nunchucks" and its upgraded version:


as we can see, the differece between them is that "nunchucks+" can return the dice of difference. that means you can deal extra damage by return dices when you have proper points of dices and the amount of dices is enough.

Ⅰ- 2 the usage of "nunchucks+"

①analysis of creating difference by "nunchucks+"

due to the effect, the usage of "nunchucks+" should be focused on return dice of differece. if your rolls is not only 1 or 2 or 3, "nunchucks+" can be very efficiency.
so, there are 3 ways of creating difference by "nunchucks+": 1(1+2, 2+3) and 2(1+3). and i'll analyzing these 3 ways of creating difference.
firstly, 1+2, the easiest and most commonly used way. it can deal 3 damage and return point 1. when you have lots of 2 and a 1, 2 can be used up with a 1. because the differece is 1. so, if the amount of 2 is n and you have at least a 1, the max damage of "nunchucks+" is at least 3n. so, 1+2 is the most stable way to create difference.

how does 1+2 worked

secondly, 2+3, the controversial way. it can deal 5 damage and return point 1. but the problem is that you have to combined 1 with 2 or 3 to create difference. and this way will consume 2 and 3 at the same time. if you were put this usage into the primary way, you would not return dice with lots of 1 created by 2+3, and your damage would be less than other ways. just like my gif shown to you: but when your dices are abundant and you don't have other ways, 2+3 can deal extra damage. so, if you have other ways, 2+3 is not the primary way.
Last one, 1+3, the most efficient way. it can deal 4 damage and return point 2. due to the difference is 2 and 2+3 is not good at creating difference in most situations, so we will considered about 2+1 primarily. if your dices is abundant, 1+3 can deal 7 damage because 2+1's efficiency is equal to deal 3 damage by only a 2! and "dagger+" can only deal 6 damage by 1+3. so, 1+3 is the most efficient way.



comparing with these 2 gif, you can see the influence about the priority of 3+2 and 3+1. as you can see, 3+2 deal 27 damage while 3+1 deal 33 damage with the same rolls.

in a conclusion, the priority of 3 ways is: 3+1→2+1→3+2. but you can find that you can alwasy left so many 1 with the usage by the priority. or you would get into a loop of "3+1→3+2→3+1→……" with the usage of this priority. why does these situations appeared? the next part will analyze the reason why they appeared.

②the influence of amount

according to the upper analysis, if there are no extreme situations, we can find that we can deal 3 damage with point 2 by "nunchucks+" as same as "dagger+". so, the relationship of amount between 3 and 1 is the key of "nunchucks+"' damage. when 1 is more than 3, the result of creating difference continuously will use up all of 3 and left 1 in your hand. when 3 is more than 1, the result of creating difference continuously with best strategy will left a 1 and lots of 3 in your hand that will lead to a special loop of 3+1→3+2→3+1→……. just like these 2 pictures shown.


too many 1


too many 3

as those pictures shown, the order of usage is different. but rolls is the same and the count of usage is enough. so they deal the same final damage. the max count is the number of usable dices subtract one. if you can't reach the max count, the main reason is that there are so many 1.



these are situations that 1 is more than 3. too many 1 make the best situation unreachable. although the order of uses is a little different, the damage is equal to each other.



these are situations that 1 is more than 3 too. although the order of uses is a little different, the damage is equal to each other.




these are situations that 3 is more than 1. although the order of uses is a little different, the damage is equal to each other.

on the other hand, players who usually use "nunchucks+" may understand it: the odd-even of 1 and 3's sum can influenced "nunchucks+"'s usage. the situation that 1 is more than 3 can be very easy to understand: 1+3 can left so many 1 and the odd-even of rest 1's amount can decided if "nunchucks+" can use up all dices.
and when 3 is more than 1, the special loop will left 2 easiest situations:321 and 3321. their usage are shown by these gif.


how to use 321


how to use 3321

that's why the sum of 1 and 3 could make the "nunchucks+"'s usage different by even and odd. when the sum is even, "nunchucks+" will left 3,2,1 three dices,that means you could use up all dices. when a+c is odd, "nunchucks+" will always left 1 dice, that means you could not use up all dices.


except for all dices is the same point, when rolls of 1,2,3 is more enough, "nunchucks+"' usage will finally be simplified to these 2 situations. as you can see, the odd-even of 1 and 3's sum can influenced that "nunchucks+" can or cannot use up all dices. if the sum is even and 2 special situations isn't appeared, "nunchucks+" can use up all dices and vise verse.
2 special situations: 3 is more, but 2 is zero and 1 is only one; 3 is more, but 1 is zero and 2 is only one. in these situations, you will always left 1 dice by using "nunchucks+" only.


so many 3 and a 1 here


so many 3 and a 2 here

so, the composition of rolls can influence the usage of "nunchucks+". there are 4 conditions you can use up all dices except all rolls are the same:
if 1,2 and 3 are exist, the sum of 1 and 3 is even;
if 1 and 2 are exist, the amount of 1 is even;
if 2 and 3 are exist, the amount of 3 is even and you have two point 2 at least;
if 1 and 3 are exist, the sum of 1 and 3 is even and you have two point 1 at least.
another thing should be noted is that equipments' combination is also influenced if "nunchucks+" use up all rolls. for example, if you have 3, 2, 1, "nunchucks+" and "battle axe+", the max damage is 11, which is higher than 8 by using "nunchucks+" only. as you can see, the proper strategy can make "nunchucks+" deal the max damage.

Ⅰ- 3 conclusion

in a short words, to reach "nunchucks+"'s maximum damage, you need to try to return a dice with each use as many as possible.if the situation is proper and you have x dices to use on "nunchucks+", you need to try your best to use "nunchucks+" for x-1 times. otherwise, when the amount of rolls is a for 3, b for 2 and c for 1, the max times is 3a/2+b+c/2+1/4*(-1)^(c-a)-1/4.

in a conclusion, my "nunchucks+"'s usage is:
1.use all 2 with 1.if you have 2 and 3 but no 1 exist, just use 3+2 to create 1 and using up all 2. then use 3 and 1, there are 2 situations:(1)all of your 3 is use up and 1 is remaining, or (2) 1 has only one and 3 is remaining;
2.if all of your 3 is use up and 1 is remaining, that means 2 and 1 is remained, you just use all 2 with 1 and then use all 1 on it;
3.if 1 has only one and 3 is remaining, you just use all 2 with 1 untill there are only one point 2, and then you can use 3 with 1 and the returned difference untill "nunchucks+" could not use.

Ⅱ The max damage's difference between "nunchucks+" and "dagger+"

to compare with each others, you need to calculate both of their max damage. the calculation of "dagger+"'s damage is very simple to be done, just calculating the sum of different kinds of rolls. but the calculation of "nunchucks+"'s damage is more complex. you have to considered about the situations of all rolls. and different strategies of "nunchucks+"'s usage if also different to others. so, in my opinion, you can dividing and calculating the different stages of "nunchucks+"'s usage.

Ⅱ - 1 thinking

first of all, "nunchucks+" can use up all 2 with 1. you can deal 3b damage in this stage.
then, your strategy is depends on the higher value of a and c. this process will use 3+1. and finally, either 1 or 3 will be used up.
if c＞a（1 is more than 3）, you'll left zero for 3, c-a for 1 , return a dices of point 2 and deal 4a damage in this usage. using up all 2 created by 3+1 will deal 3a damage. finally, trying to use up all 1 will deal c-a damage if possible.
but the odd-even of c-a will determined if you can use up all dices, so you can calculating the damage with the formula 1/2*(-1)^(a+c)-1/2.
so, the final damage is 6a+3b+c+1/2*(-1)^(a+c)-1/2.

if a≥c, you can left c-a+1 for 3 and one for 1 to use up all dices. this process can deal 4(c-1) damage and creating c-1 dices for point 2. then using c-2 dices of 2 and left one 2, you can deal 3(c-2) damage. now, you have a-c+1 for 3 and a special loop of "3+1→3+2→3+1→……" is start. in this process, you can deal 4.5 damage with each uses of "nunchucks+" on average.
if a+c is even, you can left 321 by using a-c dices of point 3 and deal 4.5(a-c)+8 damage. the final damage is 4.5a+3b+2.5c-2.
if a+c is odd, you can left 21 by using up a-c+1 dices of point 3 and deal 4.5(a-c+1)+3 damage. the final damage is 4.5a+3b+2.5c-2.5.
-2 or -2.5 is depends on the odd-even of a+c. you can calculating the damage with formula 1/4*(-1)^(a+c)+1/4 to deal with this judgement.
so, the integrated final damage is 4.5a+3b+2.5(c-1)+1/4*(-1)^(a+c)+1/4.

especially, if a≥1, (-1)^a=-1(that means a is odd), b=0 and c=1, you'll start the special loop of "3+1→3+2→3+1→……". but the actual loop will ended at 3+1, so the formula is not exact and the final damage is 4.5a-0.5.
if a≥1, (-1)^a=1(that means a is even), b=1 and c=0, you'll start the special loop of "3+2→3+1→3+2→……". but the loop will ended at 3+1, so so the formula is not exact and the final damage is 4.5a.

Ⅱ - 2 results

if the amount of 3 is a, 2 is b, c is 1, the max damage equation of these equipments are:

dagger+: 4a+3b+2

nunchucks+(c＞a): 6a+3b+c+1/2(-1)^(a+c)-1/2
nunchucks+(a≥c): 4.5a+3b+2.5(c-1)+1/4(-1)^(a+c)+1/4
nunchucks+(if a≥1, (-1)^a=-1, b=0, c=1): 4.5a-0.5
nunchucks+(if a≥1, (-1)^a=1, b=1, c=0): 4.5a
nunchucks+(if all of your rolls is the same one): c+1/2[(-1)^(c)-1] (a=b=0), 2b+(-1)^(b)-1 (a=c=0), 3a+3/2[(-1)^(c)-1] (b=c=0)


in a conclusion:
when c＞a, if 2a≥c+1 (a+c is odd) or 2a≥c (a+c is even), "nunchucks+"'s damage is greater than or equal to "dagger+"'s.
when a≥c, if a+c≥4 (a+c is odd) or a+c≥5 (a+c is even), "nunchucks+"'s damage is greater than or equal to "dagger+"'s.
especially, when a≥1, (-1)^a=1(that means a is odd), b=1 and c=0, if a≥5 "nunchucks+"' damage is greater than or equal to "dagger+"'s.
when a≥1, (-1)^a=1(that means a is even), b=1, c=0, if a≥6, "nunchucks+"' damage is greater than or equal to "dagger+"'s.
otherwise, "dagger+"'s damage is greater than "nunchucks+"'.

Ⅲ Ending

This guide is whims of mine. Until now, I’d only written a guide of all achievements(Chinese version). The main idea of me is just sharing some of my thoughts and views about playing Dicey Dungeons. When this guide was published, my playing time is near 1600h and I’ve win the dungeon 277 times in ep6 hard mode. I have lots of thoughts about playing dicey dungeons but I’m also busy in my real life that I don’t have so much time to write these things. So, I chose a thought that is easy to understand and easy to write and shared it with others. Hope this guide can help you. If you have any problems or advice about this guide, I’m glad to see you propose and discuss with me. XD
Thank you for your reading!

——The End

[River Exciting]

@jjjonetwo456 it's partially right. because dagger+ can worked better with less rolls or lower values. but we can also have more rolls with higher values.
for example, if your rolls were 321, the max damage of dagger+ is 9 and max damage of nunchucks+ is 8. their damage can be the same with rolls of 33321 of 333321. however, if rolls were 3333321, the max damage of dagger+ is 25 and max damage of nunchucks+ is 26.
tbh, what we need to do is learning to use them with a good strategy. dagger+'s advantage is also its simple and effective strategies. but i would prefer to use nunchucks+ due to its higher damage with our exact strategies. XD

[jjjonetwo456]

simple explination
99% of the time, dagger+ is better then nunchucks+.
for dagger+ the damage will always be #+1
for nunchucks+ its #1+#2+(|#1-#2|)
best case scenario for nunchucks+ is a 3 and a 1 returning a 2
this is 4 damage +2 bonus damage if you put that 2 returned back into the nunchucks
however, a 3 and 1 into dagger+ is 4+2 damage aswell
with nunchucks+, getting a 3 and a 1 is not always the case, you might get 1 and 2, 2 and 3, or doubles
for example, 2 and 1 into nunchucks+ is 3 damage +1 bonus
while 2 and 1 into dagger+ is 3 damage +2 bonus
the only cases where nunchucks+ is better is with fury or if you use the returned dice goes into something else

[River Exciting]

@T-Opp YES! That's what i mean. XD

[T-Opp]

So I'm a little stupid, but if I follow, nunchucks+ does higher damage when the player is willing to engage with maximizing the number of dice returned, in which case the damage more often than not will be 'greater than or equal to'.

Or in layman's: If you're willing to spend a longer time on your turn, even if some turns Nunchucks+ doesn't put out as much damage, over the course of an entire run Nunchucks outdamage Dagger+

Is that so?