Having a temporary local counter for equations

Questions
,
1

Hi
In my document, I would like to at some points, have equations with a custom, local numbering

equation (1)

(i) first local equation
(ii) second local equation
(iii) third local equation

equation (2)

that does not interfere with the global numbering of equations in the document. I know how to customize the look of the equation number and how to manually set the counter, but not how to switch back to the original number. Ideally I would be able to assign a fresh counter to equations, or have a block that ignores the document’s equation counter, but it doesn’t seem possible, so I’m contented with this hacky approach that almost works (see screenshot):

#set enum(numbering: "(i)", indent: 1em) // remember the indent you set here, important!
#set text(font:"New Computer Modern")
#set math.equation(numbering: "(1)")

$ a = 1 $<label>

#[
  // #let c = current math equation counter somehow?
  #counter(math.equation).update(0)
  #set math.equation(numbering: "(i)", number-align:left+horizon)
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
  // counter(math.equation).update(c)
]

$ a = 2 $

image

it works as I want it, except that the equation a =2 should be numbered 2. I don’t know how to access the int of the counter (.get().at(0) returns content), or how to create a fresh counter for my block. And ideas on how to achieve this? Cheers!

2

If you wrap everything in a context, you can write

#context [
  #let c = counter(math.equation).get()
  #counter(math.equation).update(0)
  #set math.equation(numbering: "(i)", number-align:left+horizon)
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
  #counter(math.equation).update(c)
]

With a small function definition, you can write

#let eq(body) = context {
  let c = counter(math.equation).get()
  counter(math.equation).update(0)
  body
  counter(math.equation).update(c)
}

$ a = 1 $<label>

#eq[
  #set math.equation(numbering: "(i)", number-align:left+horizon)
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

See Why is the value I receive from context always content?

4 Likes
3

You probably will use this multiple times, in which case this will simplify it even further:

#let eq(body, numbering: "(i)", number-align: left + horizon) = context {
  let eq-counter = counter(math.equation)
  let c = eq-counter.get()
  eq-counter.update(0)
  set math.equation(numbering: numbering, number-align: number-align)
  body
  eq-counter.update(c)
}

$ a = 1 $ <label>

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $
4

Note that the given solutions only work if the eq function isn’t used more than 5 times in the document, as otherwise the layout doesn’t converge. You can read more about why this is the case in Why is State Final not "final"? - #2 by SillyFreak.

The solution I came with is a bit more complex, but works also when used more than 5 times:

#set math.equation(numbering: "(1)")

#let eq(body, numbering: "(i)", number-align: start + horizon) = {
  // Counter for this "eq" function, as each one can contain
  // a different amount of sub-equations.
  let eq-counter = counter("eq")
  eq-counter.step()

  context {
    // Unique counter for sub-equations in this "eq" block.
    let sub-counter = counter("eq/" + str(eq-counter.get().first()))

    // Use the sub-counter to number the sub-equations.
    // We need to add 1 since the counter is only stepped the first time
    // after the first equation (and number) has already been constructed.
    set math.equation(
      numbering: (..) => std.numbering(numbering, sub-counter.get().first() + 1),
      number-align: number-align
    )
      
    // Step the sub-equation counter for each numbered sub-equation.
    show math.equation.where(block: true): it => {
      if it.numbering != none {
        sub-counter.step()
      }
      it
    }

    body

    // Step back main equation counter so that the sub-equations
    // aren't counted towards the main equation number.
    let sub-eqs = sub-counter.final().first()
    counter(math.equation).update(n => n - sub-eqs)
  }
}
2 Likes
5

This is incorrect.

eq function used more than 5 times in the document 
#let eq(body, numbering: "(i)", number-align: left + horizon) = context {
  let eq-counter = counter(math.equation)
  let c = eq-counter.get()
  eq-counter.update(0)
  set math.equation(numbering: numbering, number-align: number-align)
  body
  eq-counter.update(c)
}

$ a = 1 $ <label>

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

That is, unless global set math.equation(numbering: "(1)") is added, which is actually present in the OP. So, good catch.

You can combine .step() and .get() like this:

#let eq(body, numbering: "(i)", number-align: left + horizon) = {
  let eq-counter = counter("eq")
  eq-counter.step()
  context {
    let sub-counter = counter("eq/" + str(eq-counter.get().first()))
    let sub-eq-total = sub-counter.final().first()
    let value = context std.numbering(numbering, ..sub-counter.get())
    set math.equation(
      numbering: (..) => sub-counter.step() + value,
      number-align: number-align,
    )
    body
    counter(math.equation).update(n => n - sub-eq-total)
  }
}
Full example 
#let eq(body, numbering: "(i)", number-align: left + horizon) = {
  let eq-counter = counter("eq")
  eq-counter.step()
  context {
    let sub-counter = counter("eq/" + str(eq-counter.get().first()))
    let sub-eq-total = sub-counter.final().first()
    let value = context std.numbering(numbering, ..sub-counter.get())
    set math.equation(
      numbering: (..) => sub-counter.step() + value,
      number-align: number-align,
    )
    body
    counter(math.equation).update(n => n - sub-eq-total)
  }
}

#set math.equation(numbering: "(1)")

$ a = 1 $ <label>

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $

#eq[
  $ b = 1 $
  $ b = 2 $
  $ b = 3 $
]

$ a = 2 $
1 Like