The following SDT computes the value of a string of O's and l's interpreted as a positive, binary integer.  B -» Br 0 {B.val = 2 x Bx .val] | Bx 1 {B.val = 2 x Bx .val + 1} j 1 {B.val = 1}  Rewrite...

The following SDT computes the value of a string of O's and l's interpreted as a positive, binary integer.

B -» Br 0 {B.val = 2 x Bx .val] | Bx 1 {B.val = 2 x Bx .val + 1} j 1 {B.val = 1}

Rewrite this SDT so the underlying grammar is not left recursive, and yet the same value of B.val is computed for the entire input string.