## Input on Variation on KRyan’s TWF Elf Barbarian

I really like KRyan’s solution to this character concept: How to optimize a TWF Barbarian Elf

I’m looking to build something similar, but I don’t have all the restrictions that the OP had. For instance, I am planning on using the Arctic Template from Dragon #306 applied to a Wood Elf, giving me +2 Str, +2 Dex, -2 Int, -2 Cha.

With those bonuses, does TWF even make sense anymore? If so, are there changes that would make sense to utilize the STR/DEX synergy?

Getting back into 3.5e after a long time, and I’d forgotten that the limitless options are such a double-edged sword…

## Woocommerce Variation Preset Plugins

I’m not really asking how to develop this. I was wondering if anyone knows of any plugins that could save a preset variation group in Woococmmerce so that you don’t have to go over the redundant process of choosing every single one of them. Thanks.

## Is this variation of the Guidance cantrip balanced? [closed]

As a player who currently can cast Guidance, I find its decision making process a little less than exciting. Essentially, if I’m not concentrating on a spell and there’s no time pressure, cast Guidance. I’d like to make the spell more meaningful. So my thought:

GUIDANCE
Casting Time: 1 action
Range: Touch
Components: V, S
Duration: Concentration, up to 1 minute
Classes: Cleric, Druid

You touch one willing creature. Once before the spell ends, the target can roll one ability check with advantage. The spell then ends. The target cannot benefit from the effects of Guidance until they have completed a short rest.

I think this provides a more meaningful decision point behind the spell, since it can’t be used for every check. It could even provide a meaningful benefit in combat, if you were to use it on, say, your barbarian friend who was planning on grappling someone. I do worry if it is almost too weak, since Working Together wouldn’t stack with it. Would allowing it to be used for attack rolls or saving throws then make it too strong? (Per my understanding, those don’t qualify as ability checks).

## Graph coloring variation

Are there variations of the classic graph coloring problem that the number of neighbors in the same color is limited but not zero (in the original problem – the limit is zero)?

## A variation of the halting problem

Given an infinite set $$S \subseteq \mathbb{N}$$, define the language:

$$L_S = \{ \langle M \rangle : M$$ is a deterministic TM that does not halt on $$\epsilon$$, or, $$T_M \in S\}$$

where $$T_M$$ is the number of steps that $$M$$ takes until it halts with the empty word $$\epsilon$$ as input (or $$\infty$$ if it doesn’t halt).

What are the sets $$S$$ such that $$L_S$$ is decidable?

There are some more trivial cases, if $$S = \{k,k+1,k+2, \dots \}$$ for some $$k \in \mathbb{N}$$ then $$L_S$$ is clearly decidable, as we can simulate $$M$$ on $$\epsilon$$ for $$k-1$$ steps and accept if and only if $$M$$ accepts. though, if we take $$S= \{k,k+2,k+4,\dots \}$$ for some $$k \in \mathbb{N}$$, or even simply taking $$S=\mathbb{N}_{even}$$ or $$S=\mathbb{N}_{odd}$$ this becomes more of a problem, because there is no prevention from it being impossible to have a finite calculation for whether the number of steps until halting will be even in the cases where it halts. Although this seems undecidable I’m not sure how to prove this.

I generally suspect that $$L_S$$ is decidable if and only if $$\mathbb{N} \setminus S$$ is finite

## Dynamic Programming – Thief Variation Probem

I’ve encountered a Dynamic Programming problem which is a variation of the thief one.

Say you are a thief and you are given a number of houses in a row you should rob :

$$House_1,House_2 \dots House_N$$

with each house having the following values : $$(x_i \geq y_i \geq z_i \gt0)$$

You profit X if you rob a house but none of the adjacent houses.

You profit Y if you rob a house and exactly one of the adjacent houses.

You profit Z if you rob a house and both of the adjacent houses.

Cases with houses A-B-C would be :

$$Profit(001)=0+0+C_x$$ $$Profit(101)=A_x+0+C_x$$ $$Profit(110)=A_y+B_y+0$$ $$Profit(111)=A_y+B_z+C_y$$

Where 1 stands for robbing the house and 0 for not robbing the house

Obviously you can’t utilize the Z value for the first and the last house and each set of values is random.

Now the question is : Which houses should you rob to get the maximum profit?

My main issue is that i can’t establish a base case for this problem.

At first i thought of creating a N*M array with M being the maximum amount of houses i can rob from 0-N when every house is not robbed and think like : Rob it – Don’t rob it but came up with nothing.

Any tips or directions would be appreciated.

## Want to add a condition based variation in woocommerce

I want to add a condition based variation in woocommerce or something else similar to below example