# 3. Periodicity and symmetry in linear dashed lines

Prerequirements:

In this post we’ll talk about two fundamental properties of linear dashed lines: periodicity and symmetry.

## Periodicity and number of dashes in a period

In Table 4 in the post From prime numbers to dashed lines we saw that, in the representation of a linear dashed line, there is a certain outline that repeats itself infinitely many times. This is expressed mathematically saying that a linear dashed line is periodical, in the sense of the following definition:

Periodical dashed line

Let $T$ be a dashed line. If there exists a positive integer $h$ such that, for all $i$, the $i$-th and the $i+h$-th column of $T$ are equal, then:

• $T$ is said periodical
• Indicating with $M$ the smallest possible value of $h$, $M$ is said length of the period
• Any set of $M$ consecutive natural numbers is said period

All the dashed lines we have seen in the post From prime numbers to dashed lines, apart from those of Tables 1, 10 e 11, are periodical. It’s not by chance: these dashed lines are in fact all linear, and it can be proved that a linear dashed line is periodical. We already mentioned this property concerning the dashed line $(2,3,4,5)$ (Table 4), but now we’ll see in detail the underlying mechanism, in the case of the dashed line $(2,3)$. Let’s write in the graphical representation, in place of the dashes, as in Table 13, their ordinal numbers:

Table 14: graphical representation of the linear dashed line $(2, 3)$, where the ordinal numbers of the dashes are written and three periods are highlighted
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1 3 4 6 8 9 11 13 14 16
2 5 7 10 12 15

In this table we can see that the same arrangement of dashes repeats itself every 6 columns, 6 because it’s the least common multiple of the dashed line components, that are 2 and 3. In particular, taking columns 6 by 6 as highlighted in Table 14, the arrangement that repeats itself is the following:

 – – – – –

Also starting from another column, and again taking columns by groups of 6 consecutive ones, we have a repeating outline. For example, if we disregard column 0 and consider the column groups 1-6, 7-12, 13-18, etc., the repeating outline is the following:

 – – – – –

We would have a similar situation starting with any other column, still taking columns by groups of 6 consecutive ones: in any case there will be a repeating arrangement of dashes. The so obtained arrangements are different from each other (as you can see by comparing Tables 15 and 16) but we can note that they contain the same number of dashes, 5 in the example we are considering. The reason for this is that, however we take 6 consecutive integers, three of them are divisible by 2 and two of them are divisible by 3: for example in the set $\{1,2,3,4,5,6\}$ the numbers divisible by 2 are 2, 4 and 6, and the ones divisible by 3 are 3 and 6: this exactly coincides with what’s shown in Table 16. Summarizing, we have that:

• The dashed line $(2,3)$ has an outline repeating every 6 columns, no matter what the starting column is
• Every so obtained outline contains 5 dashes, no matter what the starting column is

Going further, we can observe what follows. Having divided the dashed line into groups of 6 consecutive columns, as in Table 14, we can start from any dash and consider the fifth next dash: this way we get a dash placed in the next group of 6 columns, but in the same position with respect to the repeating arrangement.
For example, with reference to Table 14, starting from the first dash, the dash $(1,1)$ with value 2 placed in the red part, and going ahead by 5 dashes, we’ll get to the dash $(1,4)$ of value 8 in the green part. It’s in the first row too and, with respect to the outline shown in Table 15, it corresponds to the third column too:

 – – – – –

But two dashes placed in two consecutive outline repetitions (like the red one and the green one in Table 14) and that, with respect to it, are placed in the same column (Table 17), certainly have values that differ by the length of the outline itself, 6 in this case. In fact the dash $(1,4)$ has value 8 and the dash $(1,1)$ has value 2 = 8 – 6.
We could also repeat the same argument in the opposite verse: if we start from a dash and move 6 columns (the length of the outline) to the right, we’ll get to a dash placed in the same position with respect to the outline, and 5 dashes after the starting dash, where 5 is the number of dashes contained in the outline.

From the above argument we can conclude that:

• The dashed line $(2,3)$ is periodical according to Definition T.5, because column 0 is equal to column 6, column 1 is equal to column 7, column 2 to column 8, and so on.
• Every group of 6 consecutive columns, starting from any column, is made up of the same outline (arrangement of dashes)
• In particular, every group of 6 consecutive columns is made up of five dashes
• Starting from any dash, the fifth next dash is placed in the same position as the starting one, with respect to the outline

So, referring to Definition T.5, what we called “outline” is a period, and $M = 6$ is the length of the period.

Examining in depth what exposed above and generalizing, the following properties can be proved:

Number of dashes and spaces in a period

In the representation of a periodical dashed line, any two periods have the same number of dashes and the same number of spaces.

We know that the dashed line is periodical, so, by definition, it’s formed by an identically repeated sequence of columns. If we name the $j$-th column as $c_j$, and the period length as $M$, we can represent the dashed line like this (for clarity, rows are omitted):

 $\ldots$ $c_1$ $c_2$ $\ldots$ $c_{M - 2}$ $c_{M - 1}$ $c_M$ $c_1$ $c_2$ $\ldots$ $c_{M - 2}$ $c_{M - 1}$ $c_M$ $\ldots$

We have to prove that, given any two sequences of $M$ conecutive columns of the dashed line, they are formed by the same columns, in the same order or not: if the columns are the same, consequently the two sequences will contain the same number of dashes and spaces. Let’s indicate these two sequences with:

• $P$, starting in a given column $c_p$ and stopping in a given column $c_{p + M - 1}$;
• $Q$, starting in a given column $c_q$ and stopping in a given column $c_{q + M - 1}$.

Let’s suppose $P$ does not follow $Q$, that is $c_p \leq c_q$. The possible cases are:

1. $P$ and $Q$ coincide, that is $c_p = c_q$; so, thay surely contain the same column sequence.
2. The starting columns of $P$ and $Q$ differ by multiples of the period length, that is $c_q = c_{p + wM}$ for a given $w \gt 0$. The dashed line is periodical, so we know that $c_p = c_{p + vM}$ for any $v \gt 0$, so also for $v = w$: then $c_p = c_{p + wM} = c_q$, but it’s also $c_{p+1} = c_{p + wM + 1} = c_{q + 1}$, and so on, then $P$ and $Q$ contain both the same repetead sequence of columns.
3. $P$ and $Q$ share some columns but they don’t coincide, that is $c_p \lt c_q \leq c_{p + M - 1}$. We can represent $P$ and $Q$ like this, highlighting the columns belonging only to $P$ in yellow, those belonging to $Q$ in red, and the shared ones in green:

 $\ldots$ $c_p$ $\ldots$ $c_{q - 1}$ $c_q$ $\ldots$ $c_{p + M - 1}$ $c_{p + M}$ $\ldots$ $c_{q + M - 1}$ $\ldots$

In this way, we have divided the interval into three sections:

• $J$ from $c_p$ to $c_{q - 1}$;
• $K$ from $c_q$ to $c_{p + M - 1}$;
• $L$ from $c_{p + M}$ to $c_{q + M - 1}$.

If we indicate with $T_i$ the number of dashes of a section and with $S_i$ the number of spaces, we have:

• $T_P = T_J + T_K$;
• $T_Q = T_K + T_L$;
• $S_P = S_J + T_K$;
• $S_Q = S_K + T_L$.

But also the sections $J$ and $L$ have, in turn, equal columns, because the dashed line is periodical, so we have $c_p = c_{p + M}$, $c_{p + 1} = c_{p + M + 1}$, and so on, up to $c_{q - 1} = c_{q + M - 1}$. Then, we have also $T_J = T_L$ and $S_J = S_L$, so we get

• $T_P = T_J + T_K$;
• $T_Q = T_K + T_J$;
• $S_P = S_J + S_K$;
• $S_Q = S_K + S_J$.

that is

• $T_P = T_J + T_K = T_Q$;
• $S_P = S_J + S_K = S_Q$.

which means that $P$ and $Q$ have the same number of dashes and spaces.

4. $P$ and $Q$ are distanced, but do not belong to the same period, that is $c_q = c_{p + vM + w}$, for a given $w \gt 0$ and a given $w$ such that $0 \lt w \leq M - 1$. This case can be brought back to the previous one, because the dashed line is periodical: in fact, if we subtract $wM$ from the bounds of $Q$, we get a new interval, which we will name $R$, that starts with $c_{p + w}$, and, due to the periodicity, has the same columns of $Q$. Due to $0 \lt w \leq M - 1$, the starting bound of $R$ is such that $c_p \lt c_{p + w} \lt c_{M - 1}$, then $P$ e $R$ are in the same situation of previous case.

Linear dashed lines are periodical

A linear dashed line $(n_1, n_2, \dots, n_k)$ is periodical, with period length $M = \textrm{MCM}(n_1, n_2, \dots, n_k)$ and number of period dashes $l = \frac{M}{n_1} + \frac{M}{n_2} + \dots + \frac{M}{n_k}$.

Given a single line related to $n_i$, surely the columns $j$ and $j + h_i \cdot n_i$, for each $h_i \gt 0$, are equal, because:

• If the column $j$ is a dash, it’s a multiple of $n_i$, so $k$ exists such that $j = k \cdot n_i$; then we have $j + h_i \cdot n_i = k \cdot n_i + h_i \cdot n_i = (k + h_i) \cdot n_i$. So, also the column $j + h_i \cdot n_i$ is in turn a multiple of $n_i$, then it’s a dash.
• If the column $j$ is not a dash, neither is $j + h_i \cdot n_i$, because, reasoning by absurd, if $j + h_i \cdot n_i$ were multiple of $n_i$, $q$ would exist such that $j + h_i \cdot n_i = q \cdot n_i$, from which we would have $j = q \cdot n_i - h_i \cdot n_i = (q - h_i) \cdot n_i$. So, the column $j$ would be in turn a multiple of $n_i$, that is a dash, which contradicts the hypothesis.

Then, by setting $h_i = 1$, the columns $j$ and $j + h_i$ of the row $n_i$, where $h_i$ is a multiple of $n_i$, are equal; then, if $h$ is at the same time a multiple of $n_1, n_2, \ldots n_k$, the columns $j$ and $j + h$ are in turn equal. So $T$ is surely periodical by definition, and its period is given by the minimum possible $h$, that is by $M = \textrm{LCM}(n_1, ... n_k)$.

Now, we will prove that the number of dashes of a period is $M / n_1 + M / n_2 + ... M / n_k$.
If $c$ is a dash, it’s a multiple of $n_i$, then $v$ exists such that $c = v \cdot n_i$; so $c$ is the $v$-th multiple of $n_i$. The dashed line is periodical, so $c + M$ is a dash, then $w$ exists such that $c + M = (v + w) \cdot n_i$; then $c + M$ is the $v + w$-th multiple of $n_i$. Then, the dash immediately before $c + M$ is the $v + w - 1$-th multiple of $n_i$, then the total dashes in the row of $n_i$ are the last minus the first plus 1, that is $(v + w - 1) - v + 1 = v + w - 1 - v + 1 = w$. From $c + M = (v + w) \cdot n_i$ we have:

$v \cdot n_i + M = (v + w) \cdot n_i$
$v \cdot n_i + M = v \cdot n_i + w \cdot n_i$
$M = w \cdot n_i$
$w = M / n_i$

That is, in the row of $n_i$, if the column $c$ is a dash, there are $M / n_i$ dashes.

Instead, if $c$ weren’t a dash, the reasoning would be analogous, but it would start from the dash immediately after $c$. This dash would be $c + r_i = v \cdot n_i$, for a given $0 < r_i < n_i$; analogously it would be $c + r_i + M = (v + w) \cdot n_i$, and we would get to the same conclusion: the dash immediately before $c + M$ would be the $v + w - 1$-th multiple of $n_i$.

Finally, by summing on all rows, the total number of dashes is $M / n_1 + M / n_2 + \ldots + M / n_k$.

The formula for computing the number of dashes in a period is of particular importance, because it ofter will recur in the formulas of dashed line theory, that we’ll see in the next posts. We are particularly interested in a specific case of this formula, when dashed line components are two by two coprime. In this case their least common multiple coincides with their product: $M = n_1 \cdot \ldots \cdot n_k$. So the following Corollary of Property T.4 is obtained:

Number of dashes in a period of a linear dashed line with two by two coprime components

Let $T = (n_1, \ldots, n_k)$ be a linear dashed line the components of which are two by two coprime. The number of dashes in any period of $T$ is given by the formula

$$\frac{n_1 \cdot \ldots \cdot n_k}{n_1} + \ldots + \frac{n_1 \cdot \ldots \cdot n_k}{n_k} = \sum_{i=1}^k \frac{n_1 \cdot \ldots \cdot n_k}{n_i} \tag{1}$$

In particular, for the orders $k \leq 3$, inverting the order of summation terms, the following formulas are obtained:

• If $k = 1$: $n_1$
• If $k = 2$: $n_1 + n_2$
• If $k = 3$: $n_1 n_2 + n_1 n_3 + n_2 n_3$

We can note that the polynomials generated by (1) are symmetric: in each of them we can exchange in any way two or more variables, always obtaining the same polynomial. For example, in the polynomial $n_1 n_2 + n_1 n_3 + n_2 n_3$, putting $n_1$ in place of $n_2$​, $n_2$​ in place of $n_3$​ and $n_3​$ in place of $n_1​$ we obtain the polynomial $n_3 n_1 + n_3 n_2 + n_1 n_3$ that, taking into account the commutativity of product and sum, is equivalent to the initial polynomial.
But the polynomials generated by (1) are not ordinary ones: they are called elementary symmetric polynomials (v. Definitions and symbols) and they have great importance in algebra. The polynomial obtained above for a certain value of $k$ is denoted with the notation $\sigma_{k-1}(n_1, \ldots, n_k)$.

## Symmetry

Not only linear dashed lines are periodical, but they are also symmetrical, in the sense that, if $M$ is the period length, taken the set of the $M+1$ consecutive columns $\{0,1,\ldots,M-1,M\}$ (that is the period starting with column 0, plus another column), the last column of this set is equal to the first one; the second to last is equal to the second one, and so on.
We can easily view this property by resuming the dashed line $(2,3)$ of Table 14. In this case the period length is $\textrm{LCM}(2, 3) = 2 \cdot 3 = 6$, and we can observe the symmetry in the set of columns $\{0,1,2,3,4,5,6\}$. In fact, column 0 is equal to column 6, column 1 is equal to column 5, column 2 is equal to 4; finally column 3, that obviously is equal to itself, works as center of symmetry of the considered dashed line portion (this center of symmetry exists because the number of considered columns, 7, is odd; otherwise there would still be a symmetry, but without a central column):

Table 18: graphical representation of the dashed line $(2, 3)$, where symmetrical columns of the set $\{0,\ldots,6\}$ are highlighted with the same colour
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

A symmetry of this kind cannot be observed in any set of consecutive columns, for example there is not symmetry in columns $\{1,2,3,4,5,6,7\}$ or $\{0,2,3,4,5,6,7,8\}$. However there is symmetry in other sets of columns, by virtue of periodicity. For example, again with reference to the dashed line $(2,3)$, we know that columns from 0 to 6 are equal, in an orderly fashion, to those ones from 6 to 12, so this portion of the dashed line has the same symmetry:

Tabella 19: graphical representation of the dashed line $(2, 3)$, where symmetrical columns of the set $\{6,\ldots,12\}$ are highlighted with the same colour
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Likewise, we can observe symmetry in any set of columns of the kind $\{k \cdot M, \ldots, (k + 1) \cdot M$, for all $k \geq 0$. But we can do a step further, applying periodicity for extending the portion of considered symmetrical columns. In fact:

• Column 0 is symmetrical to column 6 with respect to the range of columns from 0 to 6, but column 6, by periodicity, is equal to column 12
• Column 1 is symmetrical to column 5 with respect to the range of columns from 0 to 6, but column 5, by periodicity, is equal to column 11
• Column 2 is symmetrical to column 4 with respect to the range of columns from 0 to 6, but column 4, by periodicity, is equal to column 10
• Column 3, by periodicity, is equal to column 9
• Column 8 is symmetrical to column 10 with respect to the range of columns from 6 to 12, but column 10, by periodicity, is equal to column 4
• Column 7 is symmetrical to column 11 with respect to the range of columns from 6 to 12, but column 11, by periodicity, is equal to column 5
• Column 6, by periodicity, is equal to column 0

Taking all these observations as a whole, we can obtain the symmetry of a wider range of columns, from 0 to 12:

Table 20: graphical representation of the dashed line $(2, 3)$, where symmetrical columns of the set $\{0,\ldots,12\}$ are highlighted with the same colour
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

In the same way we can see that any set of columns between two multiples of the period length $M$, is symmetrical. We state the property that linear dashed lines are symmetrical, in this more general form:

Linear dashed lines are symmetrical

Given a linear dashed line having period length $M$, taken any range of consecutive columns between two multiples of $M$ included, they are symmetrical, in the sense that the first of the considered colums is equal to the last one, the second is equal to the last second one, and so on. In other terms, for all $h \geq 0$ and for all $k \gt h$, column $h \cdot M + i$ is equal to column $k \cdot M - i$, for all $0 \leq i \leq (k - h) \cdot M$.

We must prove that:

1. For each $h \geq 0$ and for each $k \gt h$, if by the row of $n_j$ and the column $h \cdot M + i$ there is a dash, then, by the row of $n_j$ and the column $k \cdot M - i$ there is another dash, for each $0 \leq i \leq (k − h) \cdot M$.
2. For each $h \geq 0$ and for each $k \gt h$, if by the row of $n_j$ and the column $h \cdot M + i$ there is not a dash, then, by the row of $n_j$ and the column $k \cdot M - i$ there is not another dash, for each $0 \leq i \leq (k − h) \cdot M$.

Let’s start by proving part 1. By hypothesis, by the row of $n_j$ and the column $h \cdot M + i$ there is a dash, then, for $h = 0$, also in the column $i$ there is a dash, so $i$ is a multiple of $n_j$. Then, $v \gt 0$ exists such that $i = v \cdot n_j$.
By hypothesis, $M$ is (the least common) multiple of all the rows including those of $n_j$, then $w \gt 0$ exists such that $M = w \cdot n_j$. So we have:

$k \cdot M - i = k \cdot w \cdot n_j - v \cdot n_j = n_j \cdot (k \cdot w - v)$

Then, also $k \cdot M - i$ is in turn a multiple of $n_j$, so also in $k \cdot M - i$ there is a dash, which is the thesis.

Let’s proceed with part 2, which instead must be proved by absurd. First step is denying the thesis: then, we will suppose that, by the row of $n_j$ and the column $k \cdot M - i$, there is a dash.

By hypothesis, $k \cdot M - i$ is a dash, so, for $k = 1$, in $M - i$ there is a dash. Due to part 1, by setting $h = 0$ and $k = 1$, we have that, if in the column $i$ there is a dash, then in the column $M - i$ there is in turn another dash.
Then, if in $M - i$ there is a dash, in $M - (M - i) = i$ there is another dash; so, $v \gt 0$ exists such that $i = v \cdot n_j$.
But, by hypothesis, $M$ is (the least common) multiple of all rows, including the one of $n_j$, so $w \gt 0$ exists such that $M = w \cdot n_j$. Then we have:

$h \cdot M + i = h \cdot w \cdot n_j + v \cdot n_j = (h \cdot w + v) \cdot n_j$

Then $h \cdot M + i$ is a multiple of $n_j$, that is, by the row of $n_j$ and the column $h \cdot M + i$ there is a dash, which contradicts the hypothesis; so, the thesis is true