Let $a_{1}={11}^{11}$, $a_{2}={12}^{12}$, $a_{3}={13}^{13}$, and \[a_{n}= \vert a_{n-1}-a_{n-2}\vert+\vert a_{n-2}-a_{n-3}\vert, n \ge 4.\] Determine $a_{{14}^{14}}$.
1. We start by defining the sequence \(a_n\) as given:
   \[
   a_1 = 11^{11}, \quad a_2 = 12^{12}, \quad a_3 = 13^{13}
   \]
   and for \(n \geq 4\),
   \[
   a_n = |a_{n-1} - a_{n-2}| + |a_{n-2} - a_{n-3}|
   \]

2. Define the difference \(\Delta a_n = a_{n+1} - a_n\). Then, we can express the recurrence relation in terms of \(\Delta a_n\):
   \[
   \Delta a_n = |\Delta a_{n-1}| - |\Delta a_{n-3}|
   \]

3. Let \(b_n = \max \{ |\Delta a_n|, |\Delta a_{n+1}|, |\Delta a_{n+2}| \}\). Notice that \(b_n\) is a non-increasing sequence:
   \[
   b_n \geq b_{n+1} \geq b_{n+2} \geq \dots
   \]

4. If there does not exist \(k \in \{1, \ldots, 14^{14} - 2\}\) such that \(a_k = a_{k+1} = 0\), then:
   \[
   b_n \geq b_{n+2} + 1 \quad \forall n \in \{1, \ldots, 14^{14} - 2\}
   \]
   This implies that \(b_{14^{14}} < 0\), which is a contradiction since \(b_n\) represents absolute differences and cannot be negative.

5. Therefore, there must exist some \(k \in \{1, \ldots, 14^{14} - 2\}\) such that \(a_k = a_{k+1} = 0\).

6. Given that the sequence has the form \(0, 0, t, t, t, 0, t, 0, 0, t, \ldots\), and \(t\) divides \(a_n\) for all \(n \in \mathbb{N}\), we conclude that \(t = 1\).

7. We know that \(a_{7l} \equiv 1 \pmod{2}\), so:
   \[
   a_{14^{14}} = 1
   \]

The final answer is \(\boxed{1}\).