- A reference guide and manual in Russian for Heroes 3. — General

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Registered: 5/22/2006

Posts: 1,394

#1 5/22/2006 2:35 PM UTC

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Greetings to all fans of the Heroes of Might & Magic III universe!

I would like to invite you to participate in a project dedicated to creating a reference guide for HoMM III.

The topic itself is not new. There are many individual resources dedicated to describing the physics of the Heroes world, but despite all efforts, it is extremely difficult to find an information source that can be unequivocally classified as a reference guide.

This work does not claim to be original and is the result of compiling various official (Player's Manual) and unofficial (forums and websites) sources dedicated to describing the internal world of the game.

The file in Excel and PDF formats can be found at the following link:
http://vk.com/fizmig

At the moment, the work on the project is not yet fully completed, if it is even possible to complete it... :) Therefore, I would be grateful if anyone would like to participate in it and share their opinions, knowledge, and advice. An updated version of the document will be posted at the specified address for all those who wish. A link to the author will definitely be included. :)

Not all the information provided in this reference guide can be considered official, because it contains observations and assessments of players who wished to share their opinions about the features of the game. But, nevertheless, in general, it достаточно realistically reflects the physics of the Heroes world and will undoubtedly be useful to any beginner.

Given the above, any comments, clarifications, and explanations are welcome. Any suggestions regarding a more optimal form of presenting the information provided will also be valuable.

Sincerely,
AmberSoler

Registered: 6/28/2007

Posts: 296

#230 2/20/2008 7:18 AM UTC

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Damn, once again – it's impossible to figure it out without Ambersolero.
Let me rephrase the question like this:
You attack an army of 9 angels with an improved stack. You attack with a hero with the stats 1-1-2-2. Your army can only consist of archers.
Question(s):
What is the minimum number of archers needed to make the angels form three stacks?
How do you calculate the strength of an archer army (which specific numbers should be taken into account) and how do you calculate the strength of an angel army (based on 9 angels or 6 + 3 archangels)?

In reply to Vanechka

Registered: 9/13/2007

Posts: 749

#231 2/20/2008 7:47 AM UTC

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No. Each unit in Heroes 3 has a VALUE parameter!

Can you give an example? How do you calculate the value of each individual monster?

It's not clear about hit points: they are a variable...
In short: how do you understand this?

In short, calculate your value and the value of the neutrals. Divide one by the other. If the result is greater than 2, then they will be in the same stack!

Wouldn't it be easier to say hit points instead of value?:confused:

How do you calculate the strength of an army of crossbowmen (which specific numbers should be taken into account) and how do you calculate the strength of an army of angels (based on 9 angels or 6 + 3 archangels)?

Based on the formula k=A/N=1-1.5 (if you need them to line up in three stacks), the minimum number of crossbowmen should be taken so that A=N, that is, x*10 (hit points of a crossbowman) = 9*200 (hit points of an angel). Hence, x=180 crossbowmen. If archers, then accordingly x*10=9*250, x=225. If with a group stack, then x*10=6*200+3*250, x=195.
These calculations are based on the formula, but I can't check them right now.
Of course, each monster has its own value, because each has a different number of hit points. Dirty, I'm not insisting on my correctness, but I always calculate it this way (although I always estimate it by eye, because it is not so important for online play).

If that were the case, then the Black Dragon would be considered stronger than the Archangel, but that's not the case... And the value of the Archangel is higher.

Can you tell me where this data comes from, and how is the division into stacks calculated according to this value? I've always calculated it one way, and it turns out that it's not the right way? ))

Registered: 6/28/2007

Posts: 296

#232 2/20/2008 8:24 AM UTC

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Here is the very value that was meant.

As I understand it, you, invisible, are using the "health" indicator in your calculations. Okay, when it's explained in Russian with examples, everything becomes clear.
All that remains is to find out what is actually used - value or health.
Because it is critical for me to choose the direction of development on one map: either to create "cheap" units, focusing on the price of one health point, or on the price of this value....

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In reply to Vanechka

Registered: 2/6/2008

Posts: 21

#233 2/20/2008 8:55 AM UTC

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Vanechka;152839

Dude, once again - it's impossible to figure it out without Ambersoler.
Let me rephrase the question like this:
You attack an army of 9 angels with an upgraded stack. You attack with a hero with stats 1-1-2-2. Your army can only consist of crossbowmen.
Question(s):
What is the minimum number of crossbowmen you need to take so that the angels form three stacks?
How to calculate the strength of the crossbowmen army (what specific numbers should be taken into account) and how to calculate the strength of the angel army (based on 9 angels or 6 + 3 archangels)?

1 Most likely, the angels will form a number of stacks equal to the number of stacks of crossbowmen.
2 The strength of the army is most conveniently calculated by simply multiplying the number of units in a stack by the "value", for example, 12 crossbowmen - this is 12*184=2208.
3 The strength of the angel army should definitely be calculated based on 6+3, since the strength of upgraded level 7 creatures increases by one and a half to two times compared to unupgraded ones (taking into account special abilities, such as shooting/resurrection, etc.) or simply by "value" 9*5019=45171; 6*5019+3*8776=56442

Added 3 minutes later

invisible;152850

Example, please! How to calculate the value of each individual monster?

See the table, kindly provided in the previous post

In reply to WiSe-WiT

Registered: 6/28/2007

Posts: 296

#234 2/20/2008 9:05 AM UTC

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WiSe-WiT;152869

1. Most likely, the angels will form stacks equal in number to the number of archer stacks.
2. The strength of an army is most conveniently calculated by simply multiplying the number of units in a stack by the "value" figure. For example, 12 archers is 12*184=2208.
3. The strength of the angel army should definitely be calculated based on 6+3, since the strength of improved level 7 creatures increases by a factor of 1.5 to 2 compared to unimproved ones (taking into account special abilities, such as shooting/resurrection, etc.), or simply by "value": 9*5019=45171; 6*5019+3*8776=56442.

No, the angels will not form stacks equal in number to the number of archer stacks.
The PhysMig wasn't designed for a quick 15-minute calculation, and if there is a formula or dependency provided, it works; you just need to understand what initial data to use in it.

And regarding the strength of the angel army, the trick is this: Invisible counted 195 units across three stacks.
So, the maximum number of archers in three stacks is 195*1.5=293.
Then, with 294 archers, there would be two stacks, right?
It seems so... But no: with two stacks of archers, there are already 4, not 3

...

It turns out to be a rather gloomy and complex calculation, but it is definitely useful...

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In reply to Vanechka

Registered: 9/13/2007

Posts: 749

#235 2/20/2008 9:34 AM UTC

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So, if we take the maximum number of crossbowmen to be three stacks, then 195 x 1.5 = 293.
And if there are 294 crossbowmen, then there will be two stacks, right?

If we calculate precisely, then the formula is not entirely accurate regarding the limits of the ranges. That is, according to the formula k=A/N, if k=1-1.5, there will be three stacks, and if k=1.5-2, there will be two stacks. Therefore, it is unclear how the distribution works when the coefficient k=1.5, that is, to which range it belongs. Perhaps this is why there is a discrepancy. But you can simply test it and clarify the formula: >=, <=, or =. Perhaps the issue is with rounding, that is, 195*1.5=292.5, and you just need to calculate in which direction to round. Just add or subtract one when calculating and test it. If everything matches the formula and the discrepancy is only in +-1, then everything is correct, and you just need to clarify the rounding rules and the equality/inequality signs in the formula.
So, it turns out that it is calculated based on hits, and not on value :), although I still don't understand what the numbers in the value table mean and where they came from? :confused:

In reply to invisible

Registered: 2/6/2008

Posts: 21

#236 2/20/2008 10:18 AM UTC

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invisible;152882

although I still haven't figured out what the numbers in the table of values refer to and where they came from? :confused:

The numbers are taken from a table called "Comparative Characteristics of Units," which is available for download in previous posts. They directly affect the determination of the relative strength of the army. For example, everyone knows that an archangel can easily defeat an archer, but only if there aren't too many of them.
The strength of the army can be calculated by multiplying the number of creatures in a stack by their strength/effectiveness coefficient, which is directly reflected in this table. The table does not claim to be an absolute truth (the report is based on empirical observations and reflects one of the possible, highly probable options, but does not claim to be the truth - author's note).

P.S. If you want, you can try to create your own table, using the Thieves' Guild of the castle, increasing/decreasing the number of different types of troops, and periodically checking the guild. But if the values there are really like in the table, e.g., gold dragon=8613; Azure dragon=78845, then it will take a very long time to do this. :)

Added 5 minutes later

Vanechka;152876

if there is a formula or dependency there, then it works; you just need to understand what to use as input data in it

And regarding the strength of the angel army, the trick is this: invisible counted 195 for three stacks.
Then the maximum number of archers in three stacks is 195 x 1.5 = 293.
So, if there are 294 archers, then there are two stacks, right?
It seems so... But no: with two stacks of archangels, there are already 4, not 3.

I just can't figure out why the number of stacks should correspond to the ratio of the strength of your army to the strength of the enemy army

In reply to WiSe-WiT

Registered: 9/13/2007

Posts: 749

#237 2/20/2008 10:31 AM UTC

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The strength of an army is most conveniently calculated by simply multiplying the number of units in a stack by the "value" figure. For example, 12 archers is 12*184=2208.
3. The strength of an angel army should definitely be calculated based on 6+3, since the strength of improved 7th-level creatures increases by a factor of 1.5 to 2 compared to unimproved ones (taking into account special abilities, such as shooting/resurrection, etc.) or simply by "value" 9*5019=45171; 6*5019+3*8776=56442.

The strength of a unit can be calculated by multiplying the number of creatures in a stack by their strength/effectiveness coefficient, which is directly reflected in this table.

This is all very good:), but I don't see the point of these clever numbers in relation to the question itself:

You are attacking an army of 9 angels with an improved stack. You are attacking with a hero with parameters 1-1-2-2. Your army can only consist of archers.
Question(s):
What is the minimum number of archers needed to make the angels form three stacks?
How do you calculate the strength of an archer army (which specific numbers should be used in the calculation) and how do you calculate the strength of an angel army (based on 9 angels or 6 + 3 archangels)?

I'm not disputing this "value," I just want to know how to use these numbers to calculate the distribution of neutral troops in battle. So far, I'm only reading about clever words and many abstract numbers. :)

P.S. If you want, you can try to create your own table, using the Thieves' Guild of a Castle, increasing/decreasing the number of different types of troops, and periodically checking the guild.

I don't see any point in this yet. :)

I still don't understand why the number of stacks should correspond to the ratio of your army's strength to the enemy army's strength?

Nevertheless, it does correspond, and that is correct. This is to prevent, for example, taking Ciel, 7 butterflies, and going to kill everyone by shooting monsters in one stack with a single arrow.

Registered: 2/6/2008

Posts: 21

#238 2/20/2008 11:44 AM UTC

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invisible;152895

That's all well and good, but I didn't really see the point of these clever numbers in relation to the question itself.

Right now, your question sounds quite specific; earlier, I assumed you'd be satisfied with a hypothetical answer; I misunderstood, sorry.
Regarding your question, it seems to be directed towards AmberSoler. If your numbers don't match up, he's probably using different data with different "values" for the troops than in that table.

Added 31 minutes later
or it's logical to assume that, regarding hits, it's to avoid spamming lightning bolts from the Zealot, but on the other hand, it's ambiguous because, for example, for Solmyr, it's actually a help, not a hindrance, when dividing into stacks in the case of a weak army.

In reply to WiSe-WiT

Registered: 9/13/2007

Posts: 749

#239 2/20/2008 1:48 PM UTC

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Sorry, I was wrong. The strength of the army is indeed calculated not by hits, but by value. The formula is correct; the coefficient k=A/N should be calculated not as the ratio of the total number of hits in the army to the total number of hits of the neutrals, but as the ratio of the total value of the army to the total value of the neutrals. That is, A=value1*n1+value2*n2+...+value7*n7, where value1, value2, ..., value7 are the value of each stack of monsters according to the table, and n1, n2, ..., n7 are the number of monsters in that stack. N is the total value of the attacking neutral. If there is no upgraded stack, it is easy to calculate: multiply the value from the table by the number. But if there is an upgraded stack, then depending on the strength of your army, the number of monsters in the upgraded stack changes, and accordingly, their total value changes (I haven't personally checked this, but I just logically assumed it). Example: We are attacking the same 9 angels with an army consisting of 30 fanatics and 11 champions. The value of the angels is 5019*9=45171 units. The value of our army is 30*750+11*2100=45600. Therefore, k=45600/45171=1.0095. k is greater than 1, but less than 1.5, which means that according to the Physmig table, there will be 3 stacks. If we remove one fanatic, it will be 29*750+11*2100=44850; k=44850/45171=0.9929. k is in the range of 0.67 to 1, and according to Physmig, there should already be 4 stacks. I don't know about offline, but this is very useful sometimes for online play, but it is unrealistic to calculate all this, unless you hire a navigator with a calculator :D

In reply to invisible

Registered: 2/6/2008

Posts: 21

#240 2/20/2008 2:08 PM UTC

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invisible;153007

so, according to the physics table, there will be 3 stacks. If we remove one fanatic, it will be 29*750+11*2100=44850; k=44850/45171=0.9929. k is in the range of 0.67 to 1, and according to the physics, there should already be 4 stacks.

Have you confirmed this in practice, or is it just the algebra of the physics engine? - this is about continuing the discussion about the discrepancy between theoretical and practical results.

In reply to WiSe-WiT

Registered: 9/13/2007

Posts: 749

#241 2/20/2008 2:14 PM UTC

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Have you confirmed this in practice, or is this just theoretical algebra? - This is about continuing the discussion on the discrepancy between theoretical and practical results.

I never write without basis (at least when I'm sober:)), and if I provided specific figures and spent a certain amount of time on calculations, then of course I checked it before writing.