CoreData sort results by number of matching relationships

Question

I'm not able to figure out how to sort results from CoreData based on the number of matching relationships.

Concrete situation:

I have Recipe(s) that have one or more RecipeIngredient(s). I have a search interface where I can select multiple Ingredient(s) and want to get back Recipe(s) ordered by how many RecipeIngredient(s) match my selection. Basically I want to find the recipes with the highest number of matching ingredients.

CoreData schema:

Here is what I've done until now, (sort descriptor is just a default until I figure out how to handle this):

    func searchRecipes(
        searchRequest: RecipeSearchRequest
    ) -> [CDRecipe] {
        let fetchRequest: NSFetchRequest<CDRecipe> = CDRecipe.fetchRequest()
        var predicates: [NSPredicate] = []
        if searchRequest.searchString != "" {
            predicates.append(NSPredicate(format: "title CONTAINS %@", searchRequest.searchString.lowercased()))
        }
        if searchRequest.savedOnly {
            predicates.append(NSPredicate(format: "saved == %@", "true"))
        }

// HERE
        if !searchRequest.includedIngredientIDs.isEmpty {
            predicates.append(NSPredicate(format: "ANY ingredients.ingredient.identifier IN %@", searchRequest.includedIngredientIDs))
        }
        fetchRequest.predicate = NSCompoundPredicate(type: .and, subpredicates: predicates)
        
        let sortDescriptor = NSSortDescriptor(keyPath: \CDRecipe.title, ascending: true)
        
        
        fetchRequest.sortDescriptors = [sortDescriptor]
        fetchRequest.fetchLimit = searchRequest.limit
        do {
            let recipes = try coreDataStack.managedObjectContext.fetch(fetchRequest)
            return recipes
        } catch {
            print(error.localizedDescription)
            return []
        }
    }

I have also considered pulling all the recipes in my database and doing this sorting in memory. However I have 10K+ recipes and will have more in the future so doing this will would be highly inefficient.