Consider generating a sequence from a vocabulary of five elements, $\mathcal{Y} = \{A, B, C, D, E\}$, using a beam size of $k=2$ and a maximum sequence length of $3$. At the first time step, the two tokens with the highest conditional probabilities, such as $A$ and $C$, are chosen. At the second time step, probabilities for all extensions are computed, for example $P(A, y_2 \mid \mathbf{c}) = P(A \mid \mathbf{c})P(y_2 \mid A, \mathbf{c})$ and $P(C, y_2 \mid \mathbf{c}) = P(C \mid \mathbf{c})P(y_2 \mid C, \mathbf{c})$. The algorithm selects the top two combinations overall, such as $A, B$ and $C, E$. This expansion repeats at the third time step by computing $P(A, B, y_3 \mid \mathbf{c})$ and $P(C, E, y_3 \mid \mathbf{c})$, ultimately resulting in the final candidate sequences, like $A, B, D$ and $C, E, D$.