§A. Here is a detailed examination of what can happen when wild cards are used in a sequence. Examples will be of multiplicity 1 for clarity, but are applicable to all multiplicities that are not of mixed modulus.
First of all, any sequence with two adjacent natural cards is valid and unique. For instance, this:
X-X-X-4-7-X-X-X (mod 8)
must mean this:
3-6-1-4-7-2-5-0 (mod 8, inc 3)
Card players desiring simplicity can establish a rule requiring merely that every sequence contain, somewhere, at least two adjacent natural cards. Those players need not read further.
§B. Any sequence can be shifted by adding the same number to every digit. If one sequence is shifted from another, they are equivalent. For instance, all the following are equivalent in modulus 9:
For any shift, there exists an inverse shift to restore the original values.
To reduce a sequence is to shift it so that the first card has the digit 0. Here is the reduction of the sequence above:
§C. A bracket is a sequence that has natural cards at both ends, and nothing but wild cards in the interior. The span of a bracket is a number 1 less than the total number of cards in the sequence, or 1 more than the number of wild cards. Either way, the span is independent of the digits printed on the cards. Conveniently, the span is equal to the number of hyphens when the sequence is written in customary fashion. Examples:
To fill a bracket is to designate natural-card values for all its wild cards. Some brackets have one possible filling, some have many, and some have none.
§D. Important is the mathematical operation known as the greatest common factor (GCF). It takes two positive integers as inputs, and yields the largest integer that evenly divides into both of them. For instance, the GCF of 40 and 55 is 5.
If the GCF of two numbers happens to be 1, the numbers are said to be relatively prime. For example, neither 15 = 3 × 5 nor 14 = 2 × 7 is prime number, but they have no factor in common. Therefore they are relatively prime.
§E. With all this, it becomes possible to determine how many fillings a reduced bracket might have. Calculate the GCF of the span and the modulus; call it g.
Here are examples with modulus 12:
To learn about the fillings of an unreduced bracket, shift it into the reduced state, perform the calculations above, and then perform the inverse shift on any fillings that are generated.
§F. Here is an extended example. With a modulus of 10, consider the following sequence:
The bracket span (5) is not relatively prime to the modulus (10), because of the common factor 5. Hence, unicity should not be expected. Indeed, five fillings exist:
4-5-6-7-8-9 (inc 1)
4-7-0-3-6-9 (inc 3)
4-9-4-9-4-9 (inc 5)
4-1-8-5-2-9 (inc 7)
4-4-3-2-0-9 (inc 9)
Now consider this sequence:
The bracket span (2) is not relatively prime to the modulus (10), because of the common factor 2. Again unicity should not be expected. This time, two fillings exist:
9-2-5 (inc 3)
9-7-5 (inc 8)
Now catenate the two sequences:
and presume the 9 is instead an X:
At this point, ambiguity is resolved, because the bracket between 4 and 5 has a span of 7, which is relatively prime to the modulus 10, and the sole filling must be this:
4-7-0-3-6-9-2-5 (inc 3)
If the 9 had been any other value, this final bracket could not have been filled.