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1{\bf 15.} (This procedure maintains four integers $(A, B, C, D)$ with the
2invariant meaning that \quotation{our remaining job is to output the continued
3fraction for $(Ay + B)/(Cy + D)$, where $y$ is the input yet to come.}) Initially
4set $j \leftarrow k \leftarrow 0$, $(A, B, C, D) \leftarrow (a, b, c, d)$; then
5input $x_j$ and set $(A, B, C, D) \leftarrow (Ax_j + B, A, Cx_j + D, C)$, $j
6\leftarrow j + 1$, one or more times until $C + D$ has the same sign as $C$.
7(When $j > 1$ and the input has not terminated, we know that $1 < y < \infty$;
8and when $C + D$ has the same sign as $C$ we know therefore that $(Ay + B)/(Cy +
9D)$ lies between $(A + B)/(C + D)$ and $A/C$.) Now comes the general step: If no
10integer lies strictly between $(A + B)/(C + D)$ and $A/C$, output $X_k \leftarrow
11\lfloor A/C \rfloor$, and set $(A, B, C, D) \leftarrow (C, D, A - X_ k C, B - X_k
12D)$, $k \leftarrow k + 1$; otherwise input $x_j$ and set $(A, B,C, D) \leftarrow
13(Ax_j + B, A, Cx_j + D,C)$, $j \leftarrow j + 1$. The general step is repeated ad
14infinitum. However, if at any time the \emph{final} $x_j$ is input, the algorithm
15immediately switches gears: It outputs the continued fraction for $(Ax_j +
16B)/(Cx_j + D)$, using Euclid's algorithm, and terminates.
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