Speed up game play by making all countdowns resolve simultaneously

Pongie
Pongie Posts: 1,412 Chairperson of the Boards
in MPQ Suggestions and Feedback
as the title suggests, is it possible? Reason I ask is no one likes waiting for each countdown to tick down one by one. Especially when there is a board full of them. Having each tick down separately also creates more opportunity for multiple cascades with an unintended side effect of double ticking a countdown if one just happens to drop below one that was resolving. You don't see the strike tiles dealing damage one by one (they all happen at the same time). 

I know now this may require a bit of recode but hopefully some form of stack/queue already exists to add up all the countdown affects and resolve them together in one action, with one cascade. No need for multiple banners. 

Comments

  • Elric_VIII
    Elric_VIII Posts: 103 Tile Toppler
    It might be more effective to have them all resolve quickly, then allow matches to occur, similar to 2* Hawkeye's Speed Shot. Resolving in a certain order can be important if, for example, you have Cho and Switch going off at the same time and they both try to change the same tile, who gets priority?
  • Jaedenkaal
    Jaedenkaal Posts: 3,357 Chairperson of the Boards
    Elric VIII said:
    It might be more effective to have them all resolve quickly, then allow matches to occur, similar to 2* Hawkeye's Speed Shot. Resolving in a certain order can be important if, for example, you have Cho and Switch going off at the same time and they both try to change the same tile, who gets priority?
    He's saying that the game would process all the effects in order, and then just do the graphics all at once, the same way it does everything else. In fact, it probably already DOES do all the calculations behind the scenes, but we still have to play it all out afterwards.

    The one actual change would be postponing board drop until all CDs resolve, which could have some negative impact on existing abilities (Gatling Gun, to name only one)

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