8. Row computation in second order linear dashed lines


We saw that in order to solve the downcast problem it’s necessary to solve the so called downcast characteristic equations, but they assume that you know which row, or more generally which proper dashed subline, the x-th dash belongs to, given x. It may sound strange, but it’s possible to compute a dash row without having an idea of its value. In fact, as we saw in the introductive post From a problem about jogging to dashed lines, row computation and value computation are two well distinct problems. In this post we’ll se how to compute the row for second order linear dashed lines, while later we’ll see how to compute the value, with the downcast technique.

Row computation for second order linear dashed lines

Let’s indicate with T a second order linear dashed line. Let’s assume for simplicity that its rows have indexes 1 and 2. We want to find a condition characterizing the membership of its x-th dash to a specific row. It does not matter if we consider row 1 or 2 because, if the membership condition of row 1 is not satisfied, the dash necessarily belongs to row 2, and vice versa; however, for fixing ideas, we choose row 1.

It can be proved (see Teoria dei tratteggi (Dashed line theory), pagg. 146-150, in Italian) that the membership of the x-th dash of T to row 1 can be deduced from the computation of a particular modulus:

(n_1 x - 1) \mathrm{\ mod\ } (n_1 + n_2) \tag{1}

Let’s see what values this modulus assumes in the case of the dashed line T = (3, 5). In the graphical representation of the dashed line, let’s write the value of (1) in correspondence with the x-th dash. This way we’ll obtain the following picture:

Figure 1: Values of the function (n_1 x – 1) mod (n_1 + n_2) = (3x – 1) mod 8 for the dashed line (3, 5), for several ordinals x

We can note that the smallest vales (from 0 to 4) are placed in the first row, while the biggest ones (from 5 to 7) are placed in the second row. This is not by chance. Generally speaking, it can be proved that the x-th dash belongs to row 1 if and only if modulus (1) assumes a value less than n_2 (in this case n_2 = 5):

\mathrm{t}_T(x) \in T[1] \Leftrightarrow (n_1 x - 1) \mathrm{\ mod\ } (n_1 + n_2) \lt n_2 \tag{2}

This is the reason why the maximum modulus in the first row is 4 = n_2 - 1. Negating this condition, we have the one characterizing the membership to the second row:

\mathrm{t}_T(x) \in T[2] \Leftrightarrow (n_1 x - 1) \mathrm{\ mod\ } (n_1 + n_2) \geq n_2 \tag{3}

It’s worth, to complete the picture, analyzing the question again, from the row 2 perspective. If in (1) we put n_2 x in place of n_1 x - 1, we’ll obtain another significant modulus:

n_2 x \mathrm{\ mod\ } (n_1 + n_2) \tag{4}

Computing this modulus for various xs in the previous dashed line, we’ll obtain what follows:

Figure 2: Values of the function n_2 x mod (n_1 + n_2) = 5x mod 8 for the dashed line (3, 5), for various ordinals x

The depicted situation is the converse of before: now the dashes in the second row (not the first one) have modulus values less than n_1 (not n_2). In general we have that:

\mathrm{t}_T(x) \in T[2] \Leftrightarrow n_2 x \mathrm{\ mod\ } (n_1 + n_2) \lt n_1 \tag{5}

And in consequence

\mathrm{t}_T(x) \in T[1] \Leftrightarrow n_2 x \mathrm{\ mod\ } (n_1 + n_2) \geq n_1 \tag{6}

We have to decide which mudulus to employ, if (1) or (4): in both cases some conditions that characterize the membership of the x-th dash to both rows are obtained. Of course, formulas (2) and (6), both stating the membership to row 1, and formulas (3) and (5), stating the membership to row 2, must be equivalent; the proof, in a slightly general form we’ll see in a moment, is in Teoria dei tratteggi (Dashed line theory), Lemma 6.1 pagg. 147-148, in Italian.

Now let’s try to write the previous formulas in a more general form. Let’s indicate the dashed line row indexes with letters i and j, with the condition we saw in the characteristic equations (in particular in Corollary 1 of Proposition T.4), \{i, j\} = \{1, 2\}.
This way the problem becomes to establish a condition for characterizing the membership of the x-th dash to the generic row i (that can be 1 or 2), indicating with j the other row (respectively, 2 or 1). Thus equations (2) and (5) are generalized by the single equation

\mathrm{t}_T(x) \in T[i] \Leftrightarrow (n_i x - (i \lt j)) \mathrm{\ mod\ } (n_i + n_j) \lt n_j \tag{2 + 5}

Here we have employed the compact notation (i \lt j), thanks to which equation (2) is obtained for i = 1, and (5) for i = 2. We can note that in both cases n_i + n_j = n_1 + n_2 but, since we are using letters, for uniformity of writing we prefer the notation n_i + n_j.

Similarly, equations (3) and (6) can be generalized by the following one:

\mathrm{t}_T(x) \in T[i] \Leftrightarrow (n_j x - (j \lt i)) \mathrm{\ mod\ } (n_i + n_j) \geq n_i \tag{3 + 6}

Starting from this equation in fact we can obtain (3) for i = 2 and (6) for i = 1.

We can state what we have seen so far in the following Theorem:

Computation of the x-th dash row in a second order linear dashed line

Let T be a second order linear dashed line, having \{1, 2\} as indexes set. Let i and j be such that \{i, j\} = \{1, 2\}. Then, for all x \gt 0:

\begin{aligned} \mathrm{t}_T(x) \in T[i] & \Leftrightarrow (n_i x - (i \lt j)) \mathrm{\ mod\ } (n_i + n_j) \lt n_j \\ & \Leftrightarrow (n_j x - (j \lt i)) \mathrm{\ mod\ } (n_i + n_j) \geq n_i \end{aligned} \tag{7}

In particular, the following formulas are valid respectively for rows 1 and 2:

\begin{aligned} \mathrm{t}_T(x) \in T[1] & \Leftrightarrow (n_1 x - 1) \mathrm{\ mod\ } (n_1 + n_2) \lt n_2 \\ & \Leftrightarrow n_2 x \mathrm{\ mod\ } (n_1 + n_2) \geq n_1 \end{aligned} \tag{8}
\begin{aligned} \mathrm{t}_T(x) \in T[2] & \Leftrightarrow n_2 x \mathrm{\ mod\ } (n_1 + n_2) \lt n_1 \\ & \Leftrightarrow (n_1 x - 1) \mathrm{\ mod\ } (n_1 + n_2) \geq n_2 \end{aligned} \tag{9}

Theorem T.2 answers our initial question, that is what row does the x-th dash of a second order linear dashed line T belongs to, yet without the explicit computation of the row. I.e., we have not found a function f(x) that, by means of algebric calculations, returns just the row index of \mathrm{t}_T(x). Instead, the approach is to arbitrarily choose a certain index i, for example i = 1, and ask: “\mathrm{t}_T(x) \in T[i]?”. The answer is given by Theorem T.2. If it’s affermative, it means that indeed the dash belongs to row i; if it’s negative, it means that the dash belongs to the other row, that we indicated with j.

Applying Theorem T.2, let’s establish which row the eighth dash of the dashed line (3, 5) belongs to.

Let’s verify if it belongs to row 1: so let’s apply formula (8) with (n_1, n_2, x) = (3, 5, 8), stopping at the first of the two equivalent forms, obtaining:

\begin{aligned} \mathrm{t}_{(3, 5)}(8) \in T[1] & \Leftrightarrow (3 \cdot 8 - 1) \mathrm{\ mod\ } (3 + 5) \lt 5 \\ & \Leftrightarrow 23 \mathrm{\ mod\ } 8 \lt 5 \\ & \Leftrightarrow 7 \lt 5 \end{aligned}

The condition is false: so the eighth dash does not belong to the first row, hence, since it necessarily belongs to one of the two rows, it belongs to the second one:

Figure 3: The eighth dash of the dashed line (3, 5) belongs to row 2

In the picture we can see that, instead, the seventh dash belongs to the first row. In fact, repeating the previous passages with x = 7, we obtain a true relation:

\begin{aligned} \mathrm{t}_{(3, 5)}(7) \in T[1] & \Leftrightarrow (3 \cdot 7 - 1) \mathrm{\ mod\ } (3 + 5) \lt 5 \\ & \Leftrightarrow 20 \mathrm{\ mod\ } 8 \lt 5 \\ & \Leftrightarrow 4 \lt 5 \end{aligned}

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