Multiplication CPSC 252 Computer Organization Ellen Walker, Hiram College

Multiplication Multiplication result needs twice as many bits –E.g. 7*7 = 49 (7 is 3 bits, 49 is 6 bits) Multiplication is more complex than addition/subtraction –Develop algorithm, implement using simpler hardware

Multiplication Algorithms Repeated addition –Easy to implement in software –Can only multiply positive numbers directly –Time depends on magnitude of operands Shift-and add –Efficient use of hardware –Time depends only on width of operands –No special case for negative numbers

Shift and Add Algorithm For each bit in multiplier –If bit = 1, add multiplicand to result –Shift multiplicand left by 1 This is the “usual” multiplication algorithm you learned many years ago (but simpler in binary)

Shift and Add example X (included for completeness) ( )

Hardware to Multiply 64-bit register for multiplicand –32 bits for original value –Option to shift left 32 times 64-bit ALU –Add shifted multiplicand to product 32-bit register for multiplier –Shift right after each step (low bits fall off) Control hardware

Multiplication: Implementation Datapath Control

How Long Does it Take? 32 shift/add cycles Each cycle has 3 steps (add, shift left, shift right) Without parallelism, that’s 96 cycles per multiply

Improvements Save hardware by using 1 64-bit register for the product and multiplier –Left part is product (so far) –Right part is unused multiplier –Add a bit to the product and lose a bit from the multiplier each cycle –Shift product right & add instead of shift multiplicand left & add! Shift and add in the same cycle (in parallel) This revision requires only 32 cycles

Final Version What goes here? Multiplier starts in right half of product

Parallel Multiplication Instead of 32 cycles, use 32 adders Each adds 0 or multiplicand Arrange in tree so results cascade properly Result pulls each bit from the appropriate adder

Parallel Multiplication (cont)

x (6 bits) (initial product) Rightmost bit is 1, add multiplicand Shift product right Shift product right Rightmost bit is 1, add multiplicand Shift product right Shift product right Shift product right Shift product right, answer is 105

Negative numbers Convert to positive, multiply, then negate if initial signs were different Or, consider 3 cases: –Both negative: convert to positive and multiply –One negative: make multiplier positive, multiplicand negative –Negative multiplicand: when high bit is 1, shift in 1’s (shift right arithmetic)

MIPS multiplication 2 32-bit registers: Hi and Lo mflo $s0move from lo (to $s0) mfhi $s0move from hi (to $s0) Multiplication Operations mult $s0, $s1{Hi,Lo}= $s0 x $s1 multu $s0, $s1unsigned version