Poly SPARC

It is in stark contrast, how the rich differ from the poor. The extravagance is so appalling. I have seen poor schools back home. I have seen their resources and how people study there. It was only recently that I saw a "rich" school. It raised a number of questions in me. What is the need for an XBOX 360 and LCD Tvs in a school? Why do people put money in buying xbox 360s in schools when many schools do not have benches for people to sit on? Who needs to learn the art of bowling and para sailing when other people do not have a foot ball to play with? How is the rich privileged to do what ever they want and the poor need to have do with whatever little they have? Why is it that it is the rich who complain and poor are content with what they have despite the responsibilities they shoulder, burdened upon them by the rich? How can the rich take ownership of the "resources" that is bestowed upon all of us by mother nature? How can mother nature be partial? Can mot...

This is more than stopping Kony. This is a test for humanity.

Project Documentation Introduction Important Terms Needleman-Wunsch Algorithm Parallel Prefix Algorithm Parallel Prefix Methodology VHDL Implementation Data Path Salient Features Model Calculations Download source code Hope this helps you as much as you have helped me. Abishek

MODEL CALCULATIONS The example under consideration is   The Row 0 and Column 0 are assumed to be known. Calculating Row 1 w[j] = max ((T[i-1,j] + f(ai , '-'), (T[i-1,j-1] + f(ai , bj)) w[1] = max((T[0,1] – 1), (T(0,0) + f(a1, b1)) = 0 w[2] = max((T[0,2] – 1), (T(0,1) + f(a1, b2)) = -1 w[3] = max((T[0,3] – 1), (T(0,2) + f(a1, b3)) = -2 w[4] = max((T[0,4] – 1), (T(0,3) + f(a1, b4)) = -3 w[5] = max((T[0,5] – 1), (T(0,4) + f(a1, b5)) = -6 w[6] = max((T[0,6] – 1), (T(0,5) + f(a1, b6)) = -5 w[7] = max((T[0,7] – 1), (T(0,6) + f(a1 , b7)) = -6 w[8] = max((T[0,8] – 1), (T(0,7) + f(a1, b8)) = -7 z[j] = w[j] + y[j] z[1] = w[1] -y[1] = w[1] +1 = 0 z[2] = w[2] -y[2] = w[2] +2 = 0 z[3] = w[3] -y[3] = w[3] +3 = 0 z[4] = w[4] -y[4] = w[4] +4 = 0 z[5] = w[5] -y[5] = w[5] +5 = 0 z[6] = w[6] -y[6] = w[6] +6 = 0 z[7] = w[7] -y[7] = w[7] +7 = 0 z[8] = w[8] -y[8] = w[8] +8 = 0 (Compare with x_reg in the output) T[1,1] = x[1] + g[1] = x[1] - 1 = -1 T[1,2] = x[2] + g[2] = x[2]...

SALIENT FEATURES OF THE PROGRAM Finite state machine cycles through 4 states s0, s1, s2, s3. The states are indicated by a 2 bit counter (counter_2bit) which increments every clock pulse. s0 = 00 s1 = 01 s2 = 10 s3 = 11 One cycle is one pass through the four states of the finite state machine. For a particular row i, consider the scoring matrix RAM (w_ram). Assume the scores for the particular row are already available. Address to the RAM is generated by a 3 bit counter. Every s0, the address to the RAM is incremented. At s1 the read enable of the ram is made high. At s2, the scores for the next row are available in the outputs of the ram. Thus during the calculation of table entries for a particular row, the scores required for the next row are fetched from the memory and are available. While calculating the table entries of a particular row, the table entries of the previous row are already available in the s3 stage of the previous cycle. The parallel prefix module gets the z inp...

The control logic implementing the sequence alignment algorithm is described here.  DATAPATH It is assumed that the scores of alignment are already present in a scoring matrix. The scoring matrix is implemented using a RAM memory. The logic for calculating scores is not implemented. It can be interfaced with the memory so as to write into it. It is designed to give the values of f(ai , bj) for a particular i (row) and all j (columns). In a single read operation it ouputs 8 bytes of data. The scores along with the previous table entries are used to calculate all the w[j] values in parallel. This is done by the w_calculation module. This module receives 8 bytes from the scoring matrices corresponding to f(ai , bj) for the respective row and 8 bytes from the previously calculated row from the table namely T(i-1,j) for all values of j. It calculates the following. w[j] = max ((T[i-1,j] + f(ai , '-'), (T[i-1,j-1] + f(ai , bj)) z[j] = w[j] – y[j] for all 8 values of j. ...

This post describes the VHDL implementation details and abstractions. Kindly look at the post on parallel prefix methodology to understand the context under which these VHDL modules work NODE The basic functional unit of the parallel prefix method is a node. It has two modes of operation namely operation and duplication. The node performs the binary operation on the two inputs given to it when it operates in the operation mode. When in the duplication mode, the output is the first input to the node. It is a combinational circuit. Its VHDL representation is as follows. ------------------------------------------------------------------------------- -- Parallel Prefix : Node .vhd -- Abishek Ramdas -- NYU Poly -- date - 11/26/10 ------------------------------------------------------------------------------- library ieee; use ieee.numeric_std.all; use ieee.std_logic_1164.all; use ieee.std_logic_signed.all; entity node is port ( x_input: in std_lo...