multi-operand addition consider the following addition: sum = a[0]; for (i=1; i

Multi-operand Addition

• Consider the Following Addition:SUM = a[0];for (i=1; i<N; i++) { SUM = SUM + a[i]; }

a[7] a[6] a[5] a[4] a[3] a[2] a[1] a[0]

a[7]+a[6] a[5]+a[4] a[3]+a[2] a[1]+a[0]

a[7]+a[6]+a[5]+a[4] a[3]+a[2]+a[1]+a[0]

a[7]+a[6]+a[5]+a[4]+a[3]+a[2]+a[1]+a[0]

Multi-operand Additiona[7] a[6] a[5] a[4] a[3] a[2] a[1] a[0]

a[7]+a[6] a[5]+a[4] a[3]+a[2] a[1]+a[0]

a[7]+a[6]+a[5]+a[4] a[3]+a[2]+a[1]+a[0]

a[7]+a[6]+a[5]+a[4]+a[3]+a[2]+a[1]+a[0]

• O(lg2N) – Lower Bound – Theoretical Lower Limit• This is “Binary Reduction” Operation• Theoretical Time to Add Two Values

– O(n) – Carry Ripple Operation– O(lg2n) – CLG/CLA tree/Prefix/Carry Skip/Carry Select– O(1) – Avizienis/Takagi Signed Digit Arithmetic

Multiplication

• Multiplication Requires Multi-operand Addition• Dot Product Requires Multi-operand Addition• Defer Carry Assimilation• Represent Intermediate Sums Redundantly

Implementation Serially

Implementation with Pipelining

Parallel Implementation

Parallel Implementation – bit level

Carry Save Adders

• FA Used in This Configuration is Also Known as a 3:2 Compressor

Dot Notation

2:2 Compressor

3:2 Compressor

Example Tree

Example Tree (cont)

Tabular Form Representation

Adder Tree Bus Sizes

Serial Carry Save Adder

Wallace Tree

• Previous Example is 7 Input Wallace Tree

• n-input Wallace Tree Reduces k-bit Inputs to Two

(k + log2n - 1)-bit Outputs

• CSA Reduces Number of Operands by Factor of 1.5

• Smallest Height h(n) For an n-input Tree Can be

Given by a Recurrence Relation

Wallace Tree

• h(n) = 1 + h(2n/3)

• Ignoring Ceiling Operator Write as: h(n) = 1 + h(2n/3)

• Can Get Lower Bound on Tree Height: h(n) log1.5(n/2)

• Equality for n = 2, 3 only

Wallace Tree Height

• Can Also Consider n(h)

– Number of Inputs for a Tree of Height h

• Recurrence is: n(h) = 3n(h-1)/2

• Ignoring Floor Operator Can get Bounds

• Lower Bound: n(h) > 2(3/2)h-1

• Upper Bound: n(h) < 2(3/2)h

• Exact Values for 0 h 20 in Table

Tree Levels

Wallace Versus Dadda Trees• Reduce the Number of Operands at Earliest Opportunity• m Dots Per Column – Apply m/3 Full Adders to Column• Tends to Minimize Overall Delay by Making CPA

CPA as Short as Possible• Delay of Fast CPA is Generally Not Smoothly Increasing

Function of Word Width• EXAMPLE: CLA Has Essentially Same Delay for Widths

of 17-32 Bits• Dadda Tree Reduces Number of Operands to Next Lower

Number Using the Fewest FAs and HAs as Possible• Justification is No Need to Reduce Number of Operands to

Next Lower n(h) in Tree Since A Faster Tree WouldNot Result

Wallace Tree

Dadda Tree

Parallel Counters

• Single-bit Full Adder Referred to as (3:2) Counter (or Compressor)

• Meaning is it “Counts” the Ones in 3 Input Bits

• Can be Generalized to (n : log2(n+1) Counter

• Has n Inputs

• Produces a log2(n+1)-bit Binary Output Representing the

Number of 1’s Among the n Inputs

• Next Example Shows a (10:4) Counter

(10:4) Parallel Counter

Generalized Parallel Counters• Parallel Counter Reduces Number of Dots in a Column

(same Radix Position)

• Output Dots are Placed into Different Positions (one each)

• Can Generalize This Notion

• Generalized Parallel Counter Receives “Dot Patterns” as Input

(not Necessarily in Same Bit Position)

• Converts Them to Other Dot Patterns

(not Necessarily one in Each Column)

• If Output Dot Pattern Has Fewer Dots Than Input, the

Counter is a Compressor and Can be Used for a Tree

Generalized Parallel Counters• Characterized by Number of Dots in Each Input Column

and Output Column

• Book Limits to Class of Counters that Output a Single Dot

in Each Column

• Limitation Allows Output to be Characterized by Single Integer

Representing Number of Columns Spanned by Output

• Input Side is Characterized by Integer Sequence Corresponding

to Number of Inputs in Various Columns

(5,5 : 4) Parallel Counter

• Dot Notation for (5,5 : 4) Counter• (5,5 : 4) Counters to Compress 5 Numbers to 2 Numbers• Can Have Other Forms, eg. ( 4,6 : 4) Counter

• Receives 6 bits of weight 1 and 4 bits of weight 2• Delivers the Weighted Sum in the Form of a 4-bit

Binary Number• This Type Requires Sum of Output Weights to Equal or

Exceed Sum of Input Weights

Generalized Parallel Counters• Powerful Concept – 4-bit Binary Full Adder Can be Viewed as

(2,2,2,3 : 5)-counter• Goal is to Reduce n Numbers to 2 Numbers in Carry-Save Adder• Sometimes Notation of (n : 2)-counter is Used Although it Strictly

Doesn’t Make Sense for n > 3• (n : 2)-counter is Shorthand Notation for a Slice of a Circuit• When Slice is Replicated, n Values are Reduced to 2 Values• Slice i Receives n Input Bits in Position i Plus Transfer (or Carry)

Bits From One or More Positions to Right (i - 1, i - 2, etc.)• Slice i Produces Output Bits in Positions i and i + 1 Plus Transfer

Digits Into Higher Positions (i + 1, i + 2, etc.)

• yj Denotes Number of Transfer bits From Slice i to i + j

(n : 2) Parallel Counters• Must Satisfy This Inequality for Scheme to Work

• 3 Represents Maximum of 2 Output Bits• eg. (7 : 2)-counter can be Built Allowing y1 = 1

- Transfer bit From Position i to i + 1 and y2=2

- Transfer bit into Position i + 2

Adding Multiple Signed Numbers

• Must Sign Extend 2’s Complement Numbers to Final Result Width

• Appears Sign Extension Could Dramatically Increase Complexity

of CSA Tree for Large n

• Trick is to Take Advantage of Fact that all Sign Extension bits are

Identical

• Use a Single Full Adder to do Job of Several Full Adders

• Allows CSA Internal Widths to be Marginally Increased

Hardware Sharing Method

Single Full Adder Used Here With Result Fanned Out

Negative Weight Interpretation

• Recall That 2’s Complement Values May be Interpreted as:2

11

0

| | 2 2n

n in i

i

X x x

• Replace Negative Sign Bit by it’s Complement and Put a -1 in

Sign Column

• Multiple –1’s Can be Combined Each Pair Placed in –1 in Next

Higher Column

• A Solitary –1 in a Column is Replaced by a +1 in That

Column and a –1 in the Next Higher Column

Negative Weight Interpretation

• Complement Three Sign Bits and Place –1’s in Sign Column• Replace Three –1’s by a +1 in Sign Position and Two –1’s in

Next Higher Position• These Two –1’s are Removed and Single –1 is Inserted in

Position k + 1• Latter –1 is in Turn Replaced by a + 1 in Position k + 1 and a – 1 in

Position k + 2• Finally a –1 Moves Out of the Resultant Sum Width and the

Procedure Stops