מבנה המחשב – מבוא למחשבים ספרתיים Parallel Prefix Computation.

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Presentation transcript:

מבנה המחשב – מבוא למחשבים ספרתיים Parallel Prefix Computation

The PPC Circuit

Fan-out of a PPC Circuit Upper Bound: There is a constant maximum number of splits per layer. There are O(log n) layers. Therefore, the total number of splits is O(log n).

Fan-out of a PPC circuit (2) Lower Bound: For simplicity, let us renumber the terminals beginning with 1 instead of 0. Observe that the even numbered outputs are split once just before the exit. Also observe that even numbered outputs of each circuit come from an internal circuit.

Fan-out of a PPC circuit (3) Let us look at the n/2 = 2 (log n)-1 output terminal. It will be an Even numbered output terminal of log(n)–2 circuits in the recursive construction.

1 depth Output Terminal Splitting log n -1 depth This is the n/2 th = 2 (log n)-1 output terminal of the full circuit log n - 2 depth 2 depth

Fan-out of a PPC circuit (4) Thus, the right most terminal will split log(n)-2 times and have a fan-out of log n - 1. Since we have shown an example that achieves the Upper Bound, the Lower Bound must be equal to the Upper Bound.

Operator *

PPC for Decimal Addition Reduction to carry bit computation: S[i] = mod( A[i]+B[i]+C[i], 10 ) This is a constant time reduction since the addition of 3 BCD coded decimal digits can be done in constant time. The price is linear since it is a constant price per decimal digit. The problem is reduced to the calculation of the carry bits.

Defining the Alphabet For i=-1: 2·C[0] = 0 or 2 For i≥0: 0 if A[i]+B[i] < 9 (kill carry) 1 if A[i]+B[i] = 9 (propagate carry) 2 if A[i]+B[i] > 9 (generate carry) σ[i] =

The Operator The original * operator is what we need. The meaning of the alphabet is the same as the one used in class: –0 = kill carry –1 = propagate the carry –2 = generate a carry Go through the lecture notes and convince yourselves that it works.

The Circuit The carry bits are calculated using the PPC circuit. We then use a BCD adder for each of the output digits: S[i]=mod(A[i],B[i],C[i],10)

2’s Complement Overflow