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Messages - flack

#16
TSoG Bugs / Re: Telepath RPG: Servants of God buglist
January 06, 2012, 06:28:03 PM
Not sure if this is a bug. I noticed a couple of locations with straight line paths. These paths however are blocked with those 5 blocks (i.e when you enter the room, there is a path on the other side of those five blocks that you need to cross in order to reach). Now since I mapped out all the possible places I can reach on the first floor in Crypt 4 from the entrance, the paths that are blocked seemed to lead to new places where they can't be reached from any other alternate path. (The reason I think they would be leading to new places is because at location (10n, 4w) the path is blocked in the middle but the top and bottom of the path can be reached via different routes). Are those new places suppose to be reached later in Crypt or are they intended to be like that or they should be dead ends instead? These locations are at (20n, 12w) , (8n, 3w) , (4n, 1w), (9n, 12w), (7n, 12w), (10n, 11w), (20n, 12w), (20n, 14w)
#17
TSoG Wish List / Crypt 4 Ideas
January 01, 2012, 11:36:27 PM
I don't know if it is too late, but if you haven't already completely finished designing crypt 4 can you put a hard maze in it somewhere?

A very hard and long maze that leads to a lot of dead ends and traps with only 1 entrance and some exits each leading to an item or something (e.g. orbs, hidden treasure chests, etc).
1. For example, the maze can be like say on the 5th floor or something and the trap may drop you down to a previous floor and you have to make your way back up.
2. Or traps lead to fights
3. Or just regular traps that hurt you.

If it takes too much time then nvm.
#18
TSoG Wish List / Re: The New Missions Idea thread:
December 19, 2011, 03:54:17 PM
I only suggested it because
A. I like grinding and I think some of the other players do too.
B. I thought it would be easy to make.

I do have better and more sophisticated ideas with new kinds of missions but they might take too long and I think Craig wants to get this game over with. He has promised a lot of deadlines and seemed to have to constantly extend those deadlines.

Personally though I don't mind if it takes a couple of more months to complete this game but other people might not have the patience.
#19
TSoG Bugs / Re: Telepath RPG: Servants of God buglist
December 17, 2011, 07:09:16 PM
Right sorry, didn't see that. My bad.
#20
TSoG Wish List / The New Missions Idea thread:
December 17, 2011, 03:46:15 PM
I have an idea for a new mission:

Have a guy who is looking to collect items off of creatures and asks the protagonist to kill certain amount of creatures for those items and then gives a certain amount of reward to the protagonists.
Each time you do what the guy says he will assign a harder task for the next item.

For example:
Mission 1: Kill 25 Bandits. Reward: 750 Gold.
Mission 2: Kill 50 Bugs. Reward: 2500 Gold.
Mission 3: Kill 200 Wood Creatures. Reward 10000 Gold.
Mission 4: Kill 1000 Shadowlings. Reward 300000 Gold.
Mission 5: Kill 5000 _______. Reward (Rare Item or something).

#21
TSoG Bugs / Re: Telepath RPG: Servants of God buglist
December 17, 2011, 03:27:34 PM
- Then second two bugs Zackirus mentioned.
- The description for learning the move Stipulate is
"As it turns out...BLAH BLAH..., but grows increasingly tiring with" (with what?)
and just ends when the sentence is incomplete.  
- In battle two of Capturing Hellion, on the SECOND turn for the computer Loyalist Avenger 2 moves to the top right corner and continues to stay there.
#22
TSoG Walkthroughs / Re: Leveling Formula
December 11, 2011, 06:03:06 AM
Quote from: Duckling on December 11, 2011, 12:38:26 AM
How did you know?; I'm highly interested in this formula. Personally, I prefer:

C = (n2 - c2 + n - c) × 25 ÷ 2

because it's easy to remember. ... For me.

[spoiler=Xkcd math]But I, too, got your formula, but derived it differently:

[n (n - 1) - c (c - 1)] * 25 / 2
= [[(n - c) (c - 1)] § (n + c (- 1))] * 25 / 2 1
= [(n - c) [(c - 1) § (n + c - 1)]] * 25 / 2 2
= [(n - c) (n + c - 1)] * 25 / 2 3

Note 1: This operation takes the sum of the quotients of the factoring taking place between steps one and two, while the two negative ones are grouped into one bigger negative one. In exchange, the binomial factors are made equivalent.

Note 2: Here, the associativity of the operation is shown, while the summed term swaps itself with the c-1, and closing the metaspace portal created in step one, forever banishing the useless thing into another realm. Meanwhile, the negative one has lost weight.

Note 3: The completed formula. The entire process was mathematically sound; look at the results![/spoiler]

But honestly, I'm wondering how best to describe it such that it's easy to compute quickly and efficiently.
Perhaps it had best be described as...

 25 [c2(-(c - n)2 + (3n - c)) + n2((c - n)2 - (3c + n))]
                                                                                   =   C
                               2 (c - n)2

I think that that's easy enough to remember! (It actually works, if anyone wants to test it out)  ;)
Who could ever want to punch numbers in a calculator when they had a lovely, elegant equation like that to work with?

(But really, the c-n symmetry of the equation astounds me. It's beautiful, and yet baffling at the same time.)

I think you meant 3c - n, because

 25 [c2(-(c - n)2 + (3n - c)) + n2((c - n)2 - (3c - n))]
                                                                                 
                               2 (c - n)2

  25  -c2(c - n)2 + c2(3n - c) + n2(c - n)2 - n2(3c - n)
=   [                                                                        ]  
   2                             (c - n)2

   25             c2(3n-c)             n2(3c-n)
=       [- c2 +             + n2 -                 ]
    2              (c-n)2                (c-n)2

   25                  3nc2-c3 - 3cn2 + n3
=       [n2 - c2 +                              ]
    2                             (c-n)2            

   25                  n3 -c3 - 3cn2 + 3nc2  
=       [n2 - c2 +                              ]
    2                             (c-n)2      

   25                  (n-c)(n2 + nc + c2) - 3cn (n-c)
=       [n2 - c2 +                                             ]
    2                             (c-n)2      

   25                  (n-c)(c2 + nc + n2 - 3cn)
=       [n2 - c2 +                                      ]
    2                             (c-n)2      

   25                  (n-c)(c2 -2nc + n2)
=       [n2 - c2 +                                      ]
    2                             (c-n)2      


   25                  (n-c)(c-n)2
=       [n2 - c2 +               ]
    2                      (c-n)2    

   25                
=       [n2 - c2 + n-c]
    2                        
which is the formula you gave.

Also I would like to point out that
(n2 - c2 + n - c) × 25 ÷ 2 = (n(n+1) - c(c+1)) x 25 ÷ 2 and is not equal to [n (n - 1) - c (c - 1)] * 25 / 2. Unless your definition of n is n+1 and c is c+1.
#23
TSoG Bugs / Re: Telepath RPG: Servants of God buglist
December 09, 2011, 03:39:18 AM
- After I rescue Festus, I talk to him to start the capture Hellion mission. Right after I finished this conversation, I click on Festus again and it seems that I have completed the mission.
- On the world map Helenite Base is there but when I put my mouse cursor over it nothing shows up.
- After I complete the rescue Festus mission, on the world map Helenite Base is gone.
#24
TSoG Wish List / Re: Lizardmen
December 09, 2011, 02:44:12 AM
My question here would be does it make a difference. For instance, lets say there were 10 HELL NO votes, and 11 Yes votes, 0 votes for everything else. What would the outcome be. I know if it was 10 no vs 11 yes then the YES would win. But there are 10 HELL NO votes which seems to be more powerful than just 10 no. So what would the outcome be. If it is the same as before and YES wins, I see no difference between HELL NO and just simply no, vice versa. Anyways I'm going to leave it at that.
#25
TSoG Bugs / Re: Telepath RPG: Servants of God buglist
December 07, 2011, 03:45:51 AM
- Before I even start the rescue Festus mission, fetus is already at the tent and when I talk to him it seemed like I already rescued him.

Rescuing Festus:
 - Battle 2 in the rescue fetus mission when Loyalist Avenger 1 attacks the game freezes. Nobody moves anymore. I can bypass this by killing
   Loyalist Avenger 1 without it attacking me.
 - Battle 3 if I kill anything on the first turn I am defeated. Even pressing done on the first turn without doing anything I am defeated.

- The hidden treasure chest in the northern region of Somnus cannot be opened.
#26
TSoG Wish List / Battle Screens
December 07, 2011, 02:55:15 AM
I find that in all the battles so far in all the telepath games, it is always a single screen battle. Like the maximum number of grids is always limited to one screen. Is it alright to make it into multi-screen battle. Like what I mean by this is that in a SINGLE battle, there are two or three screens instead of 1 and you can navigate across the screen using arrows. (that is the edge of battle screen not just bounded by the edge of the window screen). This allows for harder/longer/more exciting battles. Uh I don't really know how to explain this properly but do you understand what I mean?

Also a small idea for Energy Golem:

Energy Golem 2: Shoots diagonally instead of horizontal/vertical or both.
#27
TSoG Walkthroughs / Re: Leveling Formula
December 06, 2011, 07:20:12 PM
Quote from: Ertxiem on December 06, 2011, 06:49:02 PM
Did you knew that Gauss, when he was aged 10, discovered by himself that to add the numbers between 1 and 60 all he had to do was to multiply 30 by 61? And he explained it's because you have 30 pairs that add 61 (1+60, 2+59, 3+58,...). Pretty amazing for a 10 year old!

After this brief drift only slightly off-topic...

Thanks flack. I forgot to mention that to add the numbers between 1 and n is equivalent to compute n*(n+1)/2.
You can read this expression in 2 ways:
- If n is even, you have n/2 pairs equal to n+1 (1 + n, 2 + n-1, ...);
- If n is odd, you have (n+1)/2 terms equal to n (1 + n-1, 2 + n-2, ..., and the lonely n).

Yes I know. In fact he came up with it in his head without writing anything down in something like 2 minutes to answer his teacher. Did you also know that when Gauss was asked to go to his dieing wife for the last few moments he said "just a minute. Let me finish this problem first."
#28
TSoG Walkthroughs / Re: Leveling Formula
December 06, 2011, 06:28:49 PM
Let s = starting level, t= level you want to train to. Where t>s. We are given that when s = 1, the cost starts at 25.  Hence, the total cost is the following sum:

25s + 25(s+1) + ... +25(t-1) = 25(s + (s+1) + ... + (t-1)) = 25 ([1+ 2 +... + (t-1)] – [1+ 2+ ...(s-1)]) = 25(t(t-1)/2 – s(s-1)/2) = 25(t-s)(t+s-1)/2

This is how you derive the formula.
#29
TSoG Wish List / Re: Lizardmen
December 06, 2011, 02:28:57 PM
What is the difference between the answer "yes/definitely" and "no/hell no" on the poll? They both seem to imply the same result here.
#30
TRPG2 / Re: Maximum Stats? (McCullen's horrible fate)
November 23, 2011, 06:09:52 PM
Yes the cap is 99/99 now. Its been there quite a while actually.