This is exactly as the decimal multiply done previously.
Luckily, this is not a jumping jack; shifting bits is slower than sum.
Total number of loop sums (registers) required is the total number of binary places multiplied minus two after the "AND" operation at each intersection are done. So, \(2(n_b-1)\) loop sum registers are required. If the sum registers are staggered with the binary places aligned , then there is no need for shifting bits. \(n_b\) being the length (no. of bits) of the binary numbers being multiplied. The "AND" operation at each intersect is done concurrently, the "sum" operation within a loop is independent of other loops and must be done before the staggered loop sum operation. The summation within the loops are done concurrently followed by the sum across the staggered loops.
Another set of summing circuitry adds the loop sum in a staggered manner. aligned for each binary place. So, multiply can then be achieved with one "AND" and two "SUM" operations.
Good night...z z z
