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Home/ Questions/Q 6584391
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T16:31:54+00:00 2026-05-25T16:31:54+00:00

I’m building an app for android which allows objects to be merged together based

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I’m building an app for android which allows objects to be merged together based on certain conditions. At the moment I’m looking to do this using if statements (nested or not) and am just wondering if there is a better way to do this so it is more flexible and easily expanded.

For example, I have 6 objects, 3 of them are red, 2 are blue and 1 is green.

The rules that can be applied to that are:

  • 3 red can be merged
  • 2 blue can be merged
  • 1 red and 1 blue and 1 green can be merged
  • 1 red and 1 blue can be merged
  • 1 red and 1 green can be merged
  • 1 green and 1 blue can be merged

That is a very basic example of it. There will be a lot more than that.

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  1. Editorial Team
    Editorial Team
    2026-05-25T16:31:55+00:00Added an answer on May 25, 2026 at 4:31 pm

    I would probably do something in region of:

    Create a Rule-class, that holds the logic for a given rule (your if-statements). A Rule should have a priority, so that it can be matched up against other rules to determine what rule should be considered first. It also needs a method to which an undetermined number of objects can be passed to, which checks if these objects fulfill the rule.

    Create a RuleEngine, which holds a sorted list (by priority, highest first) of all rules. You then pass the objects you want to match towards the rules to the engine and the engine runs through them, breaking the loop if a rule is fulfilled:

    for( Rule rule : rules ) {
      if( rule.check( objects ) ) {
        //Do what needs to be done
        break;
      }
    }
    

    I don’t know if this makes sense to you – I hope it does.

    Also if we’re talking thousands of rules this solution is not optimal. In that case you would want to determine which rules the object might have a chance to fulfill before checking them (e.g. if you have two red and one green, there’s no need to check the ‘one-of-each’-rule).

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