3
\$\begingroup\$

System

In this system, we'll be rolling only two custom d6s to determine the outcomes. Each d6 can show 3 types of symbols that we are calling A (bad outcome), B (good outcome), and C (null, but can be converted to either) for now. Some faces might have 2 different symbols, like A and C, or also a pair of the same symbols, like 2 Cs.

When you roll the dice, you apply both results. So let's say that I roll an A and a B, what happens is one bad outcome and one good outcome. Two As would give me 2 bad outcomes and so on. If I rolled an A and a C, only a bad outcome would happen, since Cs are worthless unless I have some way to convert it into B (special ability, resource, etc).

Program


function: convertc C:s LIMIT:n {
  loop I over C {
    if I <= LIMIT {
      C: C - 1
      result: 1
    }
  }
  result: 0
}


function: outcome ROLL:s {
  A: 0
  B: 0
  C: 0

  \ FACES \
  loop I over ROLL {  
    if I = 1 { A: A + 1 }
    if I = 2 { B: B + 1 }
    if I = 3 { C: C + 1 }
    if I = 4 { C: C + 1 }
    if I = 5 { C: C + 1 }
    if I = 6 { 
               A: A + 1
               C: C + 1
             }
  }

  \ Conversion of Cs to Bs to represent bonuses \
  if LIMIT > 0 {
    B: B + [convertc C LIMIT]
  }

 \ Sorting outcomes \
  if C = 0 {
    if B = 0 { result: -100 } \ Only As rolled \
    if A = 0 { result:  100 } \ Only Bs rolled \
    if B >= (A * 2) { result: 100 }
    if A >= (B * 2) { result: -100 }
    if B = A { result: 0 }
  }
  if B > 0 {
    if A = 0 { result: 100 }
    if B >= (A * 2) { result: 100 } 
    }
  if A > 0 {
    if B = 0 { result: -100 }
    if A >= (B * 2) { result: -100 }
    }
  if B = A { result: 0 }

  result: 999 \ Default result \

}

output [outcome 2d6]

The program primarily keeps track of three variables: A, B, and C.

To emulate cases where the player would have a number of Cs converted into Bs, the program uses a function where Cs are deducted and added to Bs based on a manually specified LIMIT (ie. the maximum amount of conversions possible).

After that, the program applies sorting rules to narrow down the final result into 3 categories: -100 is mostly bad outcomes, 0 is a balance between good and bad, and 100 is mostly good outcomes. I'm narrowing down to those three just to achieve a solid balance before I can start opening up for more granularity.

Problem

After a long battle I've managed to make the program work somewhat as intended, but I feel like I reached to my limit (both in programming skills and energy). See, the problem is when I'm inputting a LIMIT of conversions over 2, in that case, the results don't change no matter how I setup the amount each C is worth in the dice symbol distribution. If for example I change

if I = 3 { C: C + 1 } to if I = 3 { C: C + 3 }

The output is the same for LIMIT = 3 than it is for LIMIT = 1. It's like it only converts a C only once, regardless if it is worth 2 or more.

Is it a problem on the function that converts the Cs into Bs? Or is it a logic problem? Does anyone knows how to fix this problem?

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

Let's take a look at this function:

function: convertc C:s LIMIT:n {
  loop I over C {
    if I <= LIMIT {
      C: C - 1
      result: 1
    }
  }
  result: 0
}

which you call using

B: B + [convertc C LIMIT]

You have several issues here.

  • When you call convertc, you are sending a number to a parameter of type sequence. This means that inside the scope of convertc, C is a sequence containing one element. Therefore the loop only ever executes once.
  • The C here is not the same as the C in the scope of your outcome function, so any changes to C inside the scope of convertc will not be reflected in the scope of outcome. I'm not sure of your intent here.
  • Functions return immediately when a value is assigned to result, and also you are not accumulating more than one conversion anyways, so only one conversion would ever be done regardless of the value of LIMIT.

You can replace the call to convertc with

CONVERSIONS: [lowest of C and LIMIT]
B: B + CONVERSIONS
C: C - CONVERSIONS
\$\endgroup\$
1
  • \$\begingroup\$ Such a simple but clever solution! I have much to learn, thank you for the insight! \$\endgroup\$
    – Dodecaheye
    Commented Jun 8, 2023 at 0:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .