My question: How is the following table 6.8 calculated for different operands? For example, for 3 operands, how did you calculate:

Number of levels with (3,2) = 1

Number of levels with (4; 2) = 1

Equivalent delay: 1.5

Or how about 9 operands that were calculated:

Number of levels with (3,2) = 4

Number of levels with (4; 2) = 3

Equivalent delay: 4.5

**The textbook also indicates that the equivalent delay of a (4; 2) compressor carry backup tree with 9 operands (4.5) is greater than that of a carry save tree using (3,2) counters. Why this?**

What are the framework conditions, if one is greater than the other?

The following picture shows a (4: 2) compressor from this book.

The truth table for this compressor is:

from where $ a = b = c = 1 $ and $ d = e = f = 0 $, I think these constants are used to generalize the circuit.

An adder tree using (4; 2) compressors has a more regular one

Structure and may have a lower delay than an ordinary CSA tree

of (3,2) counters. Table 6.8 compares the delays of carry-save trees

using either (3,2) counters or (4; 2) compressors. Since the delay of a

(4; 2) Compressor is 1.5 times the number of one (3.2) counter

Stages of (4; 2) compressors in column 3 are multiplied by 1.5 to give

the corresponding delay in column 4. **Note that the corresponding delay of a**

Save tree with (4; 2) compressors (column 4) is not always

smaller than that of a carry save tree with (3,2) counters (column

2). For example, nine partial products (3.2) result in counters

A carry save tree with less delay overall. Various other counters

and compressors can be used in making the addition

Tree for the partial product accumulation; For example, (7.3) counters

Please look at the bold sentence above: