ECE 5367
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Multiplication
Division
ECE 5367
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Multiplication concepts
Multiplicand Multiplier -
Product
1000
1001
1000
0000
0000
1000
1001000ten
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Concepts
• Product bits > either multiplicand or multiplier
bits.
• Product bits = # of multiplicand bits + # of
multiplier bits.
• Multiplication Steps
– Place a copy of multiplicand in proper place if
multiplier digit = 1.
– Place 0 in the proper place if multiplier digit =
0.
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Multiplication hardware – Version1
• Multiplicand is 32 bits.
• Initialize product = 0.
• Need to move
multiplicand left one
digit during each step.
• Here we will move a
total of 32 times.
• Multiplicand – 64 bit to
align with 64 bit product
register.
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Multiplication algorithm
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Multiplication algorithm contd..
Multiply 2ten, 3ten, or 0010two X0011two
Iteratio
n
Step
Multiplie
r
Multiplic
and
Product
0
Initial values
0011
00000010
00000000
1
1a: 1-> Prod = Prod + Mcand
0011
00000010
00000010
2: Shift left Multiplicand
0011
00000100
00000010
3: Shift right multiplier
0001
00000100
00000010
1a: 1-> Prod = Prod + Mcand
0001
00000100
00000110
2: Shift left Multiplicand
0001
00000100
00000110
3: Shift right multiplier
0000
00001000
00000110
1: 0 -> no operation
0000
00001000
00000110
2: Shift left Multiplicand
0000
00000100
00000110
3: Shift right multiplier
0000
00010000
00000110
1: 0 -> no operation
0000
00010000
00000110
2: Shift left Multiplicand
0000
00100000
00000110
3: Shift right multiplier
0000
00100000
00000110
2
3
4
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Second version of algorithm.
• Half of multiplicand bits are wasted. 64-bit ALU
not a good idea.
• Slow algorithm.
• Instead of shifting multiplicand left, shift product
right?
– Multiplicand fixed relative to product.
– Adder need only be 32 bits wide.
– Change halves width of ALU and Mcand.
– 32-bit sum is formed, only left half of 64-bit
product register is changed by the addition.
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Second version - hardware
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Second version of algorithm.
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Second version of algorithm
multiply 0010two X 0011two
Iteration
0
1
2
3
4
Step
Multiplier
Multiplicand
Product
Initial values
0011
0010
00000000
1a:1 => Prod = Prod + Mcand
0011
0010
00100000
2: Shift right product
0011
0010
00010000
3: Shift right multiplier
0001
0010
00010000
1a:1 => Prod = Prod + Mcand
0001
0010
00110000
2: Shift right product
0001
0010
00011000
3: Shift right multiplier
0000
0010
00011000
1: 0=> no operation
0000
0010
00011000
2: Shift right product
0000
0010
00001100
3: Shift right multiplier
0000
0010
00001100
1: 0=> no operation
0000
0010
00001100
2: Shift right product
0000
0010
00000110
3: Shift right multiplier
0000
0010
00000110
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Final version
• Product wasted space = size of the
multiplier.
– As product wasted space disappears, so
does the multiplier : gets shifted out…
• Suggestion: combination of rightmost
half of product with multiplier.
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Final version - hardware
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Final version - algorithm
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Booth recoding
• Observation: many ways to obtain a product.
• Purpose: speed – shifting was faster than addition.
• Also handles signed numbers.
– Replace a string of 1’s in the multiplier with initial
subtract when we first see a 1.
– Later add when we see the bit after the last 1.
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Example
0010two
X0110two
+
0000
shift (0 in multiplier)
- 0010
sub (first 1 in multiplier)
+ 0000
shift (middle of string of 1’s)
+ 0010
add (prior step had last 1)
00001100two
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ECE 5367
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Booth recoding
• Key – classification of numbers into
groups.
End of run
middle of run
01
beginning of run
11
10
current bit
bit to right
Explanation
Example
1
0
Beginning of a run of
1’s
000011110002
1
1
Middle of a run of 1’s
000011110002
0
1
End of a run of 1’s
000011110002
0
0
Middle of a run of 1’s
000011110002
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Booth’s algorithm
• Depending on the current and previous bits of multiplier:
– 00: Middle of a string of 0s so no arithmetic operation.
– 01: End of a string of 1’s, so add Mcand to left half of the
product.
– 10: Beginning of a string of 1’s, so subtract the Mcand from
the left half of the product.
– 11: Middle of a string of 1’s, so no arithmetic operation.
*******when shifting a negative number, always preserve
the sign********
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Booth recoding contd.
Iterati
on
Multiplicand
Booths algorithm
Step
Product
0
0010
Initial values
00000110 0
1
0010
1a:00 -> no operation
00000110 0
0010
2: shift right product
00000011 0
0010
1c:10-> Prod=Prod -Mcand
11100011 0
0010
2: shift right product
11110001 1
0010
1d:11 -> no operation
11110001 1
0010
2: shift right product
11111000 1
0010
1b: 01 -> Prod = Prod + Mcand
00011000 1
0010
2: shift right product
00001100 0
2
3
4
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Multiply 2i via shift
• 5ten X 2ten using a left shift.
• 101two = 5ten
– Shift left 1 bit  1010two = 10ten
– 5ten X 2ten = 10ten
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Division
• Long division
1001ten Quotient
Divisor 1000ten 1001010ten Dividend
-1000
10
101
1010
1000
10ten
Remainder
Dividend = Quotient X Divisor + Remainder
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First version
• 32-bit quotient register.
• Each step moves the divisor to
the right one bit.
– Start with divisor placed in
the left half of the 64-bit
divisor register
• Shift it right 1 bit each
step to align it with the
dividend.
• ***Remainder register is
initialized with the dividend.
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First version
• If remainder is +ve,
– Step 2a generates a
1 in the quotient.
else,
– Step 2b generates a
0 in the quotient and
adds the divisor to
the remainder.
• Final shift aligns the
divisor.
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Divide 710 by 210: 0000 01112 by 00102
Iteration
0
1
2
3
4
5
Step
Quotient
Divisor
Remainder
Initial values
0000
0010 0000
0000 0111
1: Rem = Rem – Div
0000
0010 0000
1110 0111
2b: Rem < 0 => +Div, sll Q, Q0 = 0
0000
0010 0000
0000 0111
3: Shift Div right
0000
0001 0000
0000 0111
1: Rem = Rem – Div
0000
0001 0000
1111 0111
2b: Rem < 0 => +Div, sll Q, Q0 = 0
0000
0001 0000
0000 0111
3: Shift Div right
0000
0000 1000
0000 0111
1: Rem = Rem – Div
0000
0000 1000
1111 1111
2b: Rem < 0 => +Div, sll Q, Q0 = 0
0000
0000 1000
0000 0111
3: Shift Div right
0000
0000 0100
0000 0111
1: Rem = Rem – Div
0000
0000 0100
0000 0011
2a: Rem >= 0 => sll Q, Q0 = 1
0001
0000 0100
0000 0011
3: Shift Div right
0001
0000 0010
0000 0011
1: Rem = Rem – Div
0001
0000 0010
0000 0001
2a: Rem >= 0 => sll Q, Q0 = 1
0011
0000 0010
0000 0001
3: Shift Div right
0011
0000 0001
0000 0001
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Second version
• Only half the divisor
contains useful data at any
one time and so the divisor
can be split in two.
• Shift remainder left and
shift divisor right results in
alignment needed and also
simplifies hardware.
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Final version
• Eliminate quotient
register. Shift quotient
into remainder
instead of shifting in
zeros.
• Algorithm has only 2
steps
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Final version -hardware
• Start by shifting rem. Left.
– This serves to shift both
the remainder and the
quotient.
– Consequence: will shift
rem. one time too many.
• Final correction to shift back
only the rem. half of the
register.
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Divide 000001112 by 00102
Iteration
0
1
2
3
4
Step
Divisor
Remainder
Initial values
0010
00000111
Shift Rem. left by 1
0010
00001110
2: Rem = Rem - Div
0010
11101110
3b: Rem. < 0-> +Div, sll R, R0 = 0
0010
00011100
2: Rem = Rem - Div
0010
11111100
3b: Rem. < 0-> +Div, sll R, R0 = 0
0010
00111000
2: Rem = Rem - Div
0010
00011000
3a: Rem. >= 0-> sll R, R0 = 1
0010
00110001
2: Rem = Rem - Div
0010
00010001
3a: Rem. >= 0-> sll R, R0 = 1
0010
00100011
Shift left half of Rem right by1
0010
00010011