CODEWITHSUNDEEP
verilogxilinxsynthesize
Bharath Kotari
Bharath Kotari

Reputation: 107

How can I improve my code to reduce the synthesis time?

I have written some code in verilog for a median filter using a cumulative histogram method. When I try to synthesize my code in xilinx it's processing up to 1 hour and finally shows an error, "program ran out of memory".

My code is: 

//***** MEDIAN FILTER BY USING CUMULATIVE HISTOGRAM METHOD******//

module medianfilter(median_out,clk,a1,a2,a3,a4,a5,a6,a7,a8,a9);
output median_out;
input [7:0]a1,a2,a3,a4,a5,a6,a7,a8,a9;
integer i,j;
reg[7:0]b[255:0];
reg [7:0]buff[0:8];
input clk;
reg [7:0]median_out;
always@(negedge clk)
begin 
    //**************************************************************************//
    for(i=0;i<256;i=i+1) // initilize the memory bins with zeros
    b[i]=0;
    //*************************************************************************//

    buff[0]=a1;
    buff[1]=a2;
    buff[2]=a3;
    buff[3]=a4;
    buff[4]=a5;
    buff[5]=a6;
    buff[6]=a7;
    buff[7]=a8;
    buff[8]=a9;
    for(i=0;i<9;i=i+1)  // this loop is for cumulative histogram method
    begin
        b[buff[i]]=b[buff[i]]+1;   // incrementing the value in b[i]th  memory address
            for(j=0;j<256;j=j+1)
                if(j>buff[i])
                    b[j]=b[j]+1; // incrementing the bins below b[i]th bin


    end
//**************************************************************************//
    for(i=0;i<256;i=i+1) // loop for finding the median 
    begin
        if(b[i]>4)  ///////// condition for checking median
        begin
            b[i]=1;
            median_out=i;
            i=256; // loop breaks here
        end
    end
//*************************************************************************//
end

endmodule

How can I make the code synthesizable?

Upvotes: 1

Views: 610

Answers (2)

Barry Moss
Barry Moss

Reputation: 529

I agree with everything that @Paebbels had to say here. However, there are some additional considerations. How fast is the data coming in. Do you get a new set of 10 values to sort every clock cycle? If not, you could pipeline the operation and use many fewer adders and fewer register stages, even to the point of using a single adder and storing the results in a block RAM (although this would be much slower). Also, you haven't mentioned which FPGA you are using (although I'm guessing this is a small one). Any FPGA design needs to take into account the available resources on the target device. You could also directly instantiate DSP48 multiplier-accumulators for adders if they are not used elsewhere in your design, once again depending on how many you need and how many are available on the device.

Upvotes: 0

Paebbels
Paebbels

Reputation: 16249

How many adders are generated by your code? I see at least 2,100 8-bit adders which are working in the same cycle.

You should rethink your algorithm: A median filter needs an ordered list of pixel values, so at first you should think about efficient ordering of numbers on FPGAs.

A good approach are sorting networks like:

  • Odd-even-merge sort or
  • Bitonic sort.

Sorting 9 numbers can't be done in one cycle so you need pipelining. (You can do it, but at very low clock speed.)

Our PoC-Library contains pipelined sorting networks,but I have never test these networks with a non-power of two input size!

Upvotes: 3

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