Mount & Blade: Warband

Mount & Blade: Warband

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Saikx Dec 17, 2015 @ 1:53pm
How risky is it to have Altimer & Jeremus in one team?
I know they don't like each other, but will it be enough for one of them to leave my team?
I would like it to have a doctor and one guy for the sieges (and such things), but I don't want to spend time with sth. which doesn't work anyway xD
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Showing 1-15 of 21 comments
Rose Dec 17, 2015 @ 1:58pm 
I think you can disable companion disagreements
Tuidjy Dec 17, 2015 @ 2:09pm 
Not at all, if you take the side of the one that is core member of your team, and are willing to let the other go when his time is up. He will be back.

Just a little bit, if you make sure that Klethi and Matheld are NOT on your team, and Leslit and Firentis are.

Quite a bit, if you do not have a plan to handle them.
Death2all546 Dec 17, 2015 @ 2:15pm 
There is no risk, except one of them will eventually leave you because he hates working with the other guy.
|AyyLmao| Dec 17, 2015 @ 2:32pm 
I don't like Artimenis.
lionhart Dec 17, 2015 @ 2:35pm 
If you want to have Artimenner on your team you can dump Jeremus and use Ymira as a medic instead. She actually becomes a better doctor than Jeremus once you level her up a few times.
Preacher Dec 17, 2015 @ 3:22pm 
Actually, I've never had a problem with them. Just keep saying the same thing to both of them.
DecayWolf Dec 17, 2015 @ 4:18pm 
Theoricall speaking it's possible to have all 16 companions, following one comment that I read on internet.

"Each companion likes 1 person, and dislike 2, if you side against him one time, you'll get -1, but if you bring a companion which he likes, he'll get +1... -> +1 + -1 = 0, so basically he'll stay"

He was talking about 11ish part members, but I've taken the liberity following the line of thinking of zero aimed, and it's possible to have the 16, actually there's thousands of possible combinations, however the chain must be perfect and requires all 16 all the times, otherwise the domino effect will happen and you'll end with a shattered party.

I haven't tried it, but theorically speaking based on calculations, it's possible, I've the preadsheet.

http://imgur.com/0wf5OCO

If anybody wants to try it out... But remember, if a single mistake and you won't be abled to have the 16 members. So no mistakes allowed at all. Play attention on each choice and each siding.
Also no turning companions with vassals is allowed with this scheme.
I've started with Lezalith just because I had to start with anyone, it could've started with other character, but then, the chain would be completely different, so it requires to follow every step strictly for this particular chain.
Tuidjy Dec 17, 2015 @ 9:25pm 
The above does not work. Why not? Because you have four opportunities to side with each companion if you bring in both of his enemies, and you end up at -2 if you spread the antagonism.

10 works, because each companion has one enemy and one friend. 16 does not.
DecayWolf Dec 18, 2015 @ 3:56am 
Assuming each companion could have up to -4, instead of -2, then everyone would end with -1...
But then the text that I read about "each companion will have -2 and +1, and at zero they won't leave" would be wrong, since it's -4, not -2.
But then, it doesn't make much sense, since he said "I've 1 friend, and 1 enemy, and it's stable."
Well, -1 + +1 =0
-2 + +1 = -1. So he won't be abled to keep the said guy.

I haven't tried myself, so I couldn't say if it's -2 or -4. That assuming you'll get -2 for each enemy, because the widow pops up two times, thus leading for -4 instead of -2.

However, if you're saying it won't work, because there's 2 enemies, and only 1 friend, it's well distributed, and it will work, because it ends with at 0, a netural stand, cuz each companion will take -1 hit and +1 boon following this chain, making a zero.

So my question is, each enemy counts as -1 or as -2, since there's 2 screens asking which side I stand for each choice.
If each enemy counts as -1, then it will work.
If each enemy counts as -2, it won't work.

...

Let me try demostrate why if one enemy counting as -1, why it will work.
Take lezalith + artimenner for instance, they both have +1 right?
Then I take ymira and side with her, so lezalith will take -1 hit.
+ 1 + -1 = 0, on next choice however, bunduk, I'll side with lezalith and keep the netural stance, bunduk will get -1, which will be countered by his friend, making he also stay at netural stance...

It's a chain and if natural zone theory is correct and if each enemy counts as -1, not -2 then it will work.
Last edited by DecayWolf; Dec 18, 2015 @ 4:24am
EthanT Dec 18, 2015 @ 4:01am 
I have 12 companions, and they all have disagreements. Every time one of them tries to leave I offer them denars and they stay. It's usually like 15 denars. Pocket change.
DecayWolf Dec 18, 2015 @ 4:17am 
I guess you guys did not understand, you cannot simply stack them up, that's not what I said.
You'll have a specific order of the base party of 8 which you must gather first.
You've a specific order of which next companion must be recruited, for instance, in this particular chain, the 9th party member must be ymira. You cannot recruit let's say katrin, it won't work, it follows a chain which must be kept stable.
You have a specific order of siding, at the point it will be evenly distributed.

This chain does not allow any mistake, if your base party is wrong, it won't work.
If you've done any siding different from what I've posted for this specific chain, it won't work either. In the vast majority of cases it would only be possible in new games.
If you haven't recruit the companion in the specific order that I've posted, guess what? it won't work.
It's a very specific and strictly combination, it has thousands of possibilities, but it's always strict in order to have 16 companions, no mistakes were allowed at all.

By the way, this guy made 2 parties of 15 members.
https://forums.taleworlds.com/index.php?topic=323922.0

If he made 2 parties of 15 members, then it means that each enemy counts as -1, instead of -2, which means my spreadsheet will work.But again, no mistakes were allowed, every step must be follow as supreme dogmas.

It's not simply as 'stack up process' guys, it's really not. It has many rules and following sequences.
Last edited by DecayWolf; Dec 18, 2015 @ 4:21am
lionhart Dec 18, 2015 @ 5:03am 
There was a long discussion about this in the wiki:

http://mountandblade.wikia.com/wiki/Talk:Heroes

Someone made a spreadsheet with all of the stable nine and ten member parties. I have used some of the combinations in this spreadsheet (including the ten member party) and they all work:

https://docs.google.com/spreadsheet/ccc?key=0ArGjZ5xv4fgedGVtVDFoenpoMDRmeTVlMUgwMGg1M3c&usp=sharing
DecayWolf Dec 18, 2015 @ 5:29am 
Borcha
Ymira
Rolf
Basheshtur
Matheld
Alayen
Bunduk
Katrin
Nizar
Artimenner

Ymira gets -1 from matheld, +1 from alayen, = 0.
That's pretty much what I've done, with all 16 companions.
The only differece would be in my case scenario I would do something like.

Ymira gets +1 from Alayen, gets -1 from Matheld, and gets -0 from Lezalith, because I'm going side with Ymira against Lezalith, but with Matheld against Ymira, making a zero.
Surely it leads for recruitment order as well, siding chain in order to kept is table...
I call this a chain effect, it must be kept balanced, any mistake and it will break.

___

Anyways, once I start playing VC, I'll use my own chart and see the results from myself.
Because like I said, there's a pattern to follow, with consequential rules to make it happen a 16/16 companion.
It's not simply stack up, it's simply not.
Last edited by DecayWolf; Dec 18, 2015 @ 5:37am
Tuidjy Dec 18, 2015 @ 8:23am 
It is a simple balance of recurring negatives and positives.
- 8 party members can be trouble free
- 10 party members, and you have to pay attention on how you handle them
- 16 party members, and you will be unsuccessfully herding cats
It does not matter what you have read. This is the Internet, one can write things without having a clue about the subject. Test it for yourself, and let us know. In my experience, There is no way to retain a companion while keeping two of his enemies in the party.

You do not have to try for the full 16. You can test it with six companions, and try to keep them all - Matheld, Ymira, Lezalith. NIzar, Arthimener and Alayan. According to your theory, they will all be happy. According to mine, Ymira will bolt.

----

Note that the above six can work, if you take Ymira's side TWICE. When I say that it will not work, I mean, if you treat them in your balanced way.
Last edited by Tuidjy; Dec 18, 2015 @ 9:15am
KiltedLion Dec 18, 2015 @ 9:11am 
I have Borcha,Katrin,Ymira,Matheld,Alayen,Bunduk,Artimenner,Firentis,Jeremis,Deshavi,Marnid and Nazir, but he left....six are disturbed by my party members yet are happy with the other two out of three possible. First run through,still getting to know them....
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Date Posted: Dec 17, 2015 @ 1:53pm
Posts: 21