Even in the 21st century, there still exists cipher methods that can be executed by pencil and paper and that are concurrently strong enough to resists the computational power of modern PCs quite well. One of those cipher systems is the Double Columnar Transposition Cipher (DCTC), or for the german speaking readers Der Doppelwürfel.
[Double Columnar Transposition Cipher]. The DCTC works as follows: Assume a plaintext $\mathcal{P} = n_1 n_2 ... n_k$. Pick two key words from an (e.g.) english dictionary or even better two short sentences: $\mathcal{K}_1$ and $\mathcal{K}_2$ of length $s_1$ and $s_2$ respectively. To encrypt a message, write the plaintext in a block (or matrix) layout with $s_1$ columns. Now sort the columns regarding the first keyword $\mathcal{K}_1$. E.g. using the plaintext THIS IS IMPORTANT and the first keyword DOUBLE:
$$ \begin{bmatrix}
\textbf{D} & \textbf{O} & \textbf{U} & \textbf{B} & \textbf{L} & \textbf{E} \\
- & - & - & - & - & - \\
\text{T} & \text{H} & \text{I} & \text{S} & \text{I} & \text{S} \\
\text{I} & \text{M} & \text{P} & \text{O} & \text{R} & \text{T}\\
\text{A} & \text{N} & \text{T} & & & \\
\end{bmatrix}
\stackrel{\text{sort}} {\rightarrow}
\begin{bmatrix}
\textbf{B} & \textbf{D} & \textbf{E} & \textbf{L} & \textbf{O} & \textbf{U} \\
- & - & - & - & - & - \\
\text{S} & \text{T} & \text{S} & \text{I} & \text{H} & \text{I} \\
\text{O} & \text{I} & \text{T} & \text{R} & \text{M} & \text{P}\\
& \text{A} & & & \text{N} & \text{T}\\
\end{bmatrix} \stackrel{\text{read column by column}}{\rightarrow}$$
(
The keyword above the text is only displayed for clarity reasons.) After sorting, read the text column by column from the block, thereby skipping the empty places in the last row. In this order write the text again in block-size form, this time with $s_2$ columns. Again, sort the block according to the second keyword $\mathcal{K}_2$, here
CUBE:
$$ \begin{bmatrix}
\textbf{C} & \textbf{U} & \textbf{B} & \textbf{E} \\
- & - & - & - \\
\text{S} & \text{O} & \text{T} & \text{I} \\
\text{A} & \text{S} & \text{T} & \text{I} \\
\text{R} & \text{H} & \text{M} & \text{N} \\
\text{I} & \text{P} & \text{T} &
\end{bmatrix} \stackrel{\text{sort}} {\rightarrow}
\begin{bmatrix}
\textbf{B} & \textbf{C} & \textbf{E} & \textbf{U} \\
- & - & - & - \\
\text{T} & \text{S} & \text{I} & \text{O} \\
\text{T} & \text{A} & \text{I} & \text{S} \\
\text{M} & \text{R} & \text{N} & \text{H} \\
\text{T} & \text{I} & & \text{P}
\end{bmatrix}\stackrel{\text{read column by column}}{\rightarrow}$$
Read the text once again column by column from the block, which is the resulting ciphertex: