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Though the multiply instruction became common with the 16-bit generation, [4] at least two 8-bit processors have a multiply instruction: the Motorola 6809, introduced in 1978, [5] and Intel MCS-51 family, developed in 1980, and later the modern Atmel AVR 8-bit microprocessors present in the ATMega, ATTiny and ATXMega microcontrollers.
In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression becomes a ...
5 is halved (2.5) and 6 is doubled (12). The fractional portion is discarded (2.5 becomes 2). The figure in the left column (2) is even, so the figure in the right column (12) is discarded. 2 is halved (1) and 12 is doubled (24). All not-scratched-out values are summed: 3 + 6 + 24 = 33. The method works because multiplication is distributive, so:
This then follows the implementation described above, with modifications in determining the bits of A and S; e.g., the value of m, originally assigned to the first x bits of A, will be now be extended to x+1 bits and assigned to the first x+1 bits of A. Below, the improved technique is demonstrated by multiplying −8 by 2 using 4 bits for the ...
x 1 = x; x 2 = x 2 for i = k - 2 to 0 do if n i = 0 then x 2 = x 1 * x 2; x 1 = x 1 2 else x 1 = x 1 * x 2; x 2 = x 2 2 return x 1. The algorithm performs a fixed sequence of operations (up to log n): a multiplication and squaring takes place for each bit in the exponent, regardless of the bit's specific value. A similar algorithm for ...
Place the result (+3) below the bar. 3x has been divided leaving no remainder, and can therefore be marked as used. The result 3 is then multiplied by the second term in the divisor −3 = −9. Determine the partial remainder by subtracting −4 − (−9) = 5. Mark −4 as used and place the new remainder 5 above it.
The lower bound of multiplications needed is 2mn+2n−m−2 (multiplication of n×m-matrices with m×n-matrices using the substitution method, m⩾n⩾3), which means n=3 case requires at least 19 multiplications and n=4 at least 34. [40] For n=2 optimal 7 multiplications 15 additions are minimal, compared to only 4 additions for 8 multiplications.
Note that the intermediate third multiplication operates on an input domain which is less than two times larger than for the two first multiplications, its output domain is less than four times larger, and base-1000 carries computed from the first two multiplications must be taken into account when computing these two subtractions.