# 11. Computation of a dash value in a linear second order dashed line

Prerequirements:

In this post we’ll see some formulas for computing the $x$-th dash value for a second order linear dashed line (we’ll see other formulas in the next posts). The approach we’ll follow is based on some notions which were subject of previous posts:

The whole procedure can be illustrated as follows: Figure 1: Illustration of how to compute the value of a dash in a second order linear dashed line, by the downcast technique: starting from the ordinal x, the row index i is found, hence the ordinal y within the row, lastly the dash value v.

The only element we do not already know, among the ones listed above, is the solution of the downcast characteristic equation. For convenience, we rewrite here the formulation appearing in Corollary 1 of Proposition T.4, in the compact notation:

$$x = y + \left \lfloor \frac{n_i y - (j > i)}{n_j} \right \rfloor \tag{1}$$

The solution of this equation, in the unknown $y$, is given by:

$$y = \left \lceil \frac{n_j x + (j > i)}{n_i + n_j} \right \rceil \tag{2}$$

There are two different proofs of the fact that (2) is the solution of (1). The first (Teoria dei tratteggi (Dashed line theory, in Italian), pages 152-153) is more direct. It shows that (1) implies (2) by applying some integer part properties. The second one (op. cit., pages 153-154) instead is based on the following Lemma:

Lemma for proving the solution of the downcast characteristic equation for linear $\mathrm{t}$, from the second to the first order

Let $T$ be a second order linear dashed line, with indexes $I = \{1, 2\} = \{i, j\}$. Let $t$ be the $x$-th dash of $T$, and $y$ its ordinal within $T[i]$. Then

$$\left \lfloor \frac{n_i y - (j > i)}{n_j} \right \rfloor = \left \lfloor \frac{n_i x - (j > i)}{n_i + n_j} \right \rfloor$$

By this Lemma, the term with the integer part in (1) can be rewritten by putting $x$ in place of $y$ and $n_i + n_j$ in place of $n_j$. So (1) becomes:

$$x = y + \left \lfloor \frac{n_i x - (j > i)}{n_i + n_j} \right \rfloor \tag{1'}$$

Once written in this new form, the downcast characteristic equation is far simpler to be solved, because, contrary to before, the unknown $y$ appears just once: so in order to find the solution it’s enough to bring all terms with $x$ to one side and gather them in a single integer part expression: doing so, in a few passages, equation (2) can be obtained (for details see op. cit., page 154).

As we already noted about the first order $\mathrm{t\_space}$ function, in dashed line theory it’s rather usual to bump into equations with integer parts, like (1), the approximated solution of which can be found simply by removing the integer parts. For example, in the case of (1) we have:

\begin{aligned} x \approx y + \frac{n_i y - (j > i)}{n_j} & \implies \\ x \approx \frac{(n_i + n_j) y - (j > i)}{n_j} & \implies \\ n_j x \approx (n_i + n_j) y - (j > i) & \implies \\ n_j x + (j > i) \approx (n_i + n_j) y & \implies \\ \frac{n_j x + (j > i)}{n_i + n_j} \approx y & \end{aligned}

The result indeed coincides with the correct solution (2), up to the integer part. However it’s a lucky case: generally it’s not known a priori how much the solution found by this technique is similar to the correct solution.

So, in downcast theory terms, we can state the following Theorem:

Solution of the downcast characteristic equation of linear $\mathrm{t}$, from the second to the first order

Let $T$ be a second order linear dashed line, with indexes $I = \{1, 2\} = \{i, j\}$. Then the following function $d_i: \mathbb{N}^{\star} \rightarrow \mathbb{N}^{\star}$ defined by:

$$d_i(x) := \left \lceil \frac{n_j x + (j > i)}{n_i + n_j} \right \rceil \tag{3}$$

is a downcast function of $\mathrm{t}$ from $T$ to $T[i]$.

In particular, for $i = 1$:

$$d_1(x) := \left \lceil \frac{n_2 x + 1}{n_1 + n_2} \right \rceil \tag{4}$$

And for $i = 2$:

$$d_2(x) := \left \lceil \frac{n_1 x}{n_1 + n_2} \right \rceil \tag{5}$$

Let’s verify the previous theorem with reference to the first example of the post The downcast problem. That example was about the downcast of $\mathrm{t}$ from $(3,4)$ to $(3)$, that for convenience we rewrite below:

In this case $(n_1, n_2) = (3, 4)$ and $i = 1$, because the dashed subline we are considering is $(3) = (n_1)$. By placing these values into (4), the formula becomes:

$$d_1(x) = \left \lceil \frac{n_2 x + 1}{n_1 + n_2} \right \rceil = \left \lceil \frac{4 x + 1}{7} \right \rceil$$

Let’s apply the formula for all the first row ordinals that are visible in the figure: 1, 3, 5, 6, 8, 10. So we’ll obtain:

• $d_1(1) = \left \lceil \frac{4 \cdot 1 + 1}{7} \right \rceil = \left \lceil \frac{5}{7} \right \rceil = 1$
• $d_1(3) = \left \lceil \frac{4 \cdot 3 + 1}{7} \right \rceil = \left \lceil \frac{13}{7} \right \rceil = 2$
• $d_1(5) = \left \lceil \frac{4 \cdot 5 + 1}{7} \right \rceil = \left \lceil \frac{21}{7} \right \rceil = 3$
• $d_1(6) = \left \lceil \frac{4 \cdot 6 + 1}{7} \right \rceil = \left \lceil \frac{25}{7} \right \rceil = 4$
• $d_1(8) = \left \lceil \frac{4 \cdot 8 + 1}{7} \right \rceil = \left \lceil \frac{33}{7} \right \rceil = 5$
• $d_1(10) = \left \lceil \frac{4 \cdot 10 + 1}{7} \right \rceil = \left \lceil \frac{41}{7} \right \rceil = 6$

As we expected, the behaviour the function $d_1$ is exactly like the one of the arrow shown in Figure 2.
Instead for $i = 2$, i.e. for the downcast from $(3, 4)$ to $(4)$, formula (5) becomes:

$$d_2(x) = \left \lceil \frac{n_1 x}{n_1 + n_2} \right \rceil = \left \lceil \frac{3 x}{7} \right \rceil$$

Applying this formula to the ordinals of the second row, that are 2, 4, 7 and 9, we again obtain a sequence of consecutive natural numbers (1, 2, 3 and 4), that are the corresponding ordinals in the dashed subline $(4)$. For example

$$d_2(4) = \left \lceil \frac{3 \cdot 4}{7} \right \rceil = \left \lceil \frac{12}{7} \right \rceil = 2$$

in fact the fourth dash of the dashed line $(3,4)$ (the one with value 8) is the second within the dashed subline $(3,4) = (4)$.

Now we have all the tools required for computing the value of the $x$-th dash of a second order linear dashed line $T$, provided that it belongs to row $i$. In fact we can state the following Theorem:

Formula for computing the second order linear $\mathrm{t\_value}$ function

Let $T$ be a second order linear dashed line, with indexes $I = \{1, 2\} = \{i, j\}$. Let $i$ be the row index of the $x$-th dash of $T$, computable by applying Theorem T.2 (Computation of the $x$-th dash row in a second order linear dashed line). Then

$$\mathrm{t\_value}_T(x) = n_i d_i(x) \tag{6}$$

where $d_i(x) = \left \lceil \frac{n_j x + (j > i)}{n_i + n_j} \right \rceil$ is the function defined in Theorem T.7.

Alternatively, in a form in which the downcast technique is more evident:

$$\mathrm{t\_value}_T(x) = \mathrm{t\_value}_{T[i]}(d_i(x)) \tag{6'}$$

The proof is made up of a chain of equalities:

$$\mathrm{t\_value}_T(x) = T(\mathrm{t}_T(x)) = T(\mathrm{t}_{T[i]}(d_i(x))) = T[i](\mathrm{t}_{T[i]}(d_i(x))) = \mathrm{t\_value}_{T[i]}(d_i(x)) = n_i d_i(x)$$

Let’s analyze each step in detail:

• $\mathrm{t\_value}_T(x) = T(\mathrm{t}_T(x))$: it’s the definition of $\mathrm{t\_value}$ (Definition T.8)
• $T(\mathrm{t}_T(x)) = T(\mathrm{t}_{T[i]}(d_i(x)))$: it’s an application of the definition of downcast of $\mathrm{t}$ from $T$ to $T[i]$ (Definition T.10), by which $\mathrm{t}_T(x) = \mathrm{t}_{T[i]}(d_i(x))$
• $T(\mathrm{t}_{T[i]}(d_i(x))) = T[i](\mathrm{t}_{T[i]}(d_i(x)))$: it’s a consequence of the definition of dashed subline: the dashed line function of the dashed subline $T[i]$ is a restriction of the dashed line function of $T$ on the dashes of row $i$, so the two functions coincide on all the dashes of row $i$, and in particular they coincide on its $d_i(x)$-th dash $\mathrm{t}_{T[i]}(d_i(x)))$.
• $T[i](\mathrm{t}_T[i](d_i(x))) = \mathrm{t\_value}_{T[i]}(d_i(x))$: it’s again the definition of $\mathrm{t\_value}$
• $\mathrm{t\_value}_{T[i]}(d_i(x)) = n_i d_i(x)$: it’s the computation of linear first order $\mathrm{t\_value}$, the expression of which is given by Proposition T.1

In the previous example we already computed the number $d(x)$, so the computation of the $\mathrm{t\_value}$ functions turns out to be immediate, by formula (6). However it’s interesting, now, to make the computation of $\mathrm{t\_value}$ starting just from an ordinal, for example starting from $x = 6$. In order to compute it, we can apply formula (8) of Theorem T.2, according to which

$$\mathrm{t}_{(3,4)}(x) \in (3,4) \Leftrightarrow (3x - 1) \mathrm{\ mod\ } (3 + 4) \lt 4$$

For $x = 6$ the right side modulus is $17 \mathrm{\ mod\ } 7 = 3$ that’s lower than 4, so indeed $\mathrm{t}_{(3,4)}(x) \in (3,4) = (3)$. So the value of $i$ to be used in Theorem T.8 is just 1. Hence, by formulas (6) and (4):

$$\mathrm{t\_value}_{(3,4)}(6) = n_1 d_1(6) = 3 d_1(6) = 3 \left \lceil \frac{4 \cdot 6 + 1}{3 + 4} \right \rceil = 3 \left \lceil \frac{25}{7} \right \rceil = 3 \cdot 4 = 12$$

We have thus obtained the value of the sixth dash, shown in Figure 2.

Formula for calculation of linear second order function $\mathrm{t\_value}$ for first row

The formula for calculating $\mathrm{t\_value_T}(x)$ for dashes belonging to the first row of a second order dashed line $T = (n_1, n_2)$, obtained from (6) of Theorem T.8 by replacing $i = 1$, $j = 2$ and $d_i(x)$ with its expression, is the following:

$\mathrm{t\_value_T}(x) = n_1 \biggl \lceil \cfrac{n_2 \cdot x + 1}{n_1 + n_2} \biggr \rceil$