Tsuihousareta Meika no Chounan – Baka ni Sareta Hazure Skill de Saikyou e to Nobori Tsumeru - Ch. 6 - Harvesting Plants

[MangaDex]

 

[alu]

Mao Mao would be really envious.
So much poison to test ...

 

[isoycrazy]

Now he just needs to eat lots of horrible, poisonous foods. Luckily women in Mangas have a 50/50 chance of being a Corden Bleugh Chef.

 

[Oklam]

Mao Mao would be really envious.
So much poison to test ...

no point or at least she would need a test subject because it wouldn't work on her


but dang it will be costly let's say: eat 30 shroom then 1 gold coin for testing, do that for every plant and it will add up quickly, unless he strike a deal with the priest or something

 

[WhimsiCat]

Minor correction: It's Orc, not Oak.

but dang it will be costly let's say: eat 30 shroom then 1 gold coin for testing, do that for every plant and it will add up quickly, unless he strike a deal with the priest or something

Depending on how he want to 'sort' this it can certainly take a lot of money.

If he do halfs each time + get lucky with either fully missing or keeping both +1s, it'd take
15
8
4
2

4 more gold coins minimum to identify the 2 exact plants that give str/end.

 

[mahtan]

 

[vanity42]

Shroom shroom.

 

[Baelgun]

Bro in the next chapters

 

[feha]

Figuring out all the plants effects will be costly...

but dang it will be costly let's say: eat 30 shroom then 1 gold coin for testing, do that for every plant and it will add up quickly, unless he strike a deal with the priest or something

That would only work if all plants with an effect provides it in integer quanta, such that the status appraisal shows their distinct improvement as opposed to the accumulation of multiple fractional gains.

Depending on how he want to 'sort' this it can certainly take a lot of money.

If he do halfs each time + get lucky with either fully missing or keeping both +1s, it'd take
15
8
4
2

4 more gold coins minimum to identify the 2 exact plants that give str/end.

Aside from the above possibility of fractional gains, this binary search method would only work if there was only 1 plant that improved stats. Though if you simply recursively spawn additional parallel searches when it split, you it should still work. But worst case you will still end up with nearly one per plant:

1. assuming all plants have an effect, half the plants need to be checked at deepest branch.
2. Then group each tested plant with one that wasn't tested
3. Test half of those groups
4. recursively group each tested group with an untested group and test half of them, like points 2+3 above.
   • until you reach the root node, which when tested allows you to subtract the tested subgroup from the root to know the result for the untested subgroup
   • And you recursively subtract tested node from its parent, to get the value for the untested (sibling) node.
   • Until all untested leaf-nodes has been assigned a value.

You should be able to rather easily calculate the Big-O complexity for this recursive algorithm, and while I am too tired to do it (but generally any binary tree thing ends up with "log()" invovled) myself I would point out that it is easily seen that it holds some pattern akin to 1+\sum_{i=0}^{d-1}(2^i) (where d is depth of tree and d>1, and d is easily calculated from n with a log operation). It should also be possible to express as f(0) = 1; f(d) = 2^(d-1) + f(d-1) where d >= 0.
But still only works if we assume there is no gains below our appraisal-resolution, like mentioned above.

Spoiler: reference image: circular node is tested, square node is derived

edit: so as is standard when stuck awake trying to sleep, my mind decided to calculate it anyway. As was kind of expected ("But worst case you will still end up with nearly one per plant"; the sum&recursive representations implied something like this due to being geometric series with log2(n) iterations/terms), the worst-case did indeed end up as n appraisals. I did kinda expect some slight offset (like +/-1), but no, it is exactly n.
The way it goes, is:

1. We start with the base-case of having 1 plant/node (a tree of a single node)
2. We double/duplicate what we have, creating 2 identical trees; *2 to number of tests
3. We create a new root-node, and attach both trees to it as its 2 branches.
4. We turn the second branch's sub-root node into a square node (untested node); -1 to number of tests
5. We turn the root node into a circle node (tested node); +1 to number of tests, returning to what it was at step 2.
6. Repeat 2-5; doubles the number of plants being tested each iteration, and likewise doubles the number of tests.

Was also easily visible if you for each plant follow it up the tree until it hits a tested-node (half of them already being one themselves) and then sum the number of plants that hit a unique test-node. Not even any need to consider collisions due to the nature of the tree.

 

[Kaiser92]

Figuring out all the plants effects will be costly...

That would only work if all plants with an effect provides it in integer quanta, such that the status appraisal shows their distinct improvement as opposed to the accumulation of multiple fractional gains.

Aside from the above possibility of fractional gains, this binary search method would only work if there was only 1 plant that improved stats. Though if you simply recursively spawn additional parallel searches when it split, you it should still work. But worst case you will still end up with nearly one per plant:

1. assuming all plants have an effect, half the plants need to be checked at deepest branch.
2. Then group each tested plant with one that wasn't tested
3. Test half of those groups
4. recursively group each tested group with an untested group and test half of them, like points 2+3 above.
   • until you reach the root node, which when tested allows you to subtract the tested subgroup from the root to know the result for the untested subgroup
   • And you recursively subtract tested node from its parent, to get the value for the untested (sibling) node.
   • Until all untested leaf-nodes has been assigned a value.

You should be able to rather easily calculate the Big-O complexity for this recursive algorithm, and while I am too tired to do it (but generally any binary tree thing ends up with "log()" invovled) myself I would point out that it is easily seen that it holds some pattern akin to 1+\sum_{i=0}^{d-1}(2^i) (where d is depth of tree and d>1, and d is easily calculated from n with a log operation). It should also be possible to express as f(0) = 1; f(d) = 2^(d-1) + f(d-1) where d >= 0.
But still only works if we assume there is no gains below our appraisal-resolution, like mentioned above.

Spoiler: reference image: circular node is tested, square node is derived

Isekai doping tutorial 👆

 

[lesseffective]

 

[urtoast]

This manga is dope doping!

 

[ciera]

Thanks for the release

 

[WhimsiCat]

snip

Given that he's starting from only knowing that his stat did increase, it's probably best not to assume too many factors until the current test method run into failure case.

 

[Caffeinated_MadMan]

Wait... he can cleanly slice through a goblin and one-shot an ogre with an axe and his stats are only that of a rookie? Wtf can bronze and above do?

I feel the author should have at least made it so the bonuses made him have at least 20 in the strength stats.

 

[PrivatePublic]

Wait... he can cleanly slice through a goblin and one-shot an ogre with an axe and his stats are only that of a rookie? Wtf can bronze and above do?

I feel the author should have at least made it so the bonuses made him have at least 20 in the strength stats.

I mean, he clowned on his brother without stat boosts in the beginning. His brother only won the fight by using his new Holy Sword skill.

Just saying all the training MC did before getting a profession probably had something to do with it.

 

[Tds-PdV]

Now that we know it’s poison nullification, let’s poke the story premise a bit.

A dead leader is a leader that contributes negatively to his subjects.
In fact all opposition in this kind of story setting will gain significant advantage if not outright win when they manage to off the opposition head.
Apparently every many herb in the story has poison in them (along with positive effects).
With so many different poison around, and various risks that the head of a house has….
Rather than a powerful melee class, as the head of a house, things that keep you alive, such as poison resistance or even nullification is 1000% more beneficial because even a decent head that serves for 50+ years is better than a great one (and this case, unlikely to be great) that potentially die in their twenties due to a graze by a poisoned arrow/knife.

So, the MC’s family and the society that upholds that perception are full of idiots (which seems to be the case anyway).
And/Or…. the author just thought that exclusive power-up possibility to my MC due to his capability to nullify poison would be cool and didn’t think much more than that.

IMO, it would have been more coherent if say, the MC’s household got wiped due to an enemy household poisoning them, but he can’t fight back because all he has is poison resistance/nullification and the story begins from there. Oh well, just some whimsical take.

 

[Amplify]

Me reading the description of this manga trying to remember which one it is when a new update drops:

 

[77BLc9JwJGpLVARkamJx]

Would be funny if shitting (or pissing) out the herbs active component that gave him the boost also remove the stat "buff"s.

 

[cryum]

Wait... he can cleanly slice through a goblin and one-shot an ogre with an axe and his stats are only that of a rookie? Wtf can bronze and above do?

I feel the author should have at least made it so the bonuses made him have at least 20 in the strength stats.

His strength is almost double a rookie. Coupled with technique that is well beyond amateur and actual life or death experiences, his pure combat ability is pretty high already.

He just doesn't have the durability to back it up.