actlib_dataflow_neuro/dataflow_neuro/treegates.act

/*************************************************************************
 *
 *  This file is part of ACT dataflow neuro library
 *
 *  Copyright (c) 2022 University of Groningen - Ole Richter 
 *  Copyright (c) 2022 University of Groningen - Madison Cotteret 
 *  Copyright (c) 2022 University of Groningen - Hugh Greatorex
 *  Copyright (c) 2022 University of Groningen - Michele Mastella
 *  Copyright (c) 2021 Rajit Manohar 
 *
 *  This source describes Open Hardware and is licensed under the CERN-OHL-W v2 or later
 *
 *  You may redistribute and modify this documentation and make products
 *  using it under the terms of the CERN-OHL-W v2 (https:/cern.ch/cern-ohl).
 *  This documentation is distributed WITHOUT ANY EXPRESS OR IMPLIED
 *  WARRANTY, INCLUDING OF MERCHANTABILITY, SATISFACTORY QUALITY
 *  AND FITNESS FOR A PARTICULAR PURPOSE. Please see the CERN-OHL-W v2
 *  for applicable conditions.
 *
 *  Source location: https://git.web.rug.nl/bics/actlib_dataflow_neuro
 *
 *  As per CERN-OHL-W v2 section 4.1, should You produce hardware based on
 *  these sources, You must maintain the Source Location visible in its
 *  documentation.
 *
 **************************************************************************/
import "../../dataflow_neuro/cell_lib_async.act";
import "../../dataflow_neuro/cell_lib_std.act";
import std::channel;
open std::channel;
namespace tmpl {
  namespace dataflow_neuro {

/*
 * Build an OR-gate tree (NOR/NAND/optional INV)
 */


                
export deftype power (bool?! vdd, vss) { }


export template<pint N>
defproc ortree (bool? in[N]; bool! out; power supply)
{
  bool tout;

  { N > 0 : "What?" };

  pint i, end, j;
  i = 0;
  end = N-1;

  pint lenTree2Count, lenTree3Count;
  lenTree2Count = 0;
  lenTree3Count = 0;
    /* Pre"calculate" the number of C cells required, look below if confused */
  *[ i != end ->
     j = 0;
     *[ i < end ->
        j = j + 1;
        [ i+1 >= end ->
          i = end;
          lenTree2Count = lenTree2Count +1;
        [] i+2 >= end ->
          i = end;
          lenTree3Count = lenTree3Count +1;
        [] else ->
          i = i + 2;
          lenTree2Count = lenTree2Count +1;
        ]
      ]
      /*-- update range that has to be combined --*/
      i = end+1;
      end = end+j;
      j = 0;
  ]

  /* array that holds ALL the nodes in the completion tree */
  bool tmp[end+1];
  (k:N:tmp[k] = in[k];)

  /* array to hold the actual C-elments, either A2C or A3C */
  
  [lenTree2Count > 0 ->
    OR2_X1 C2Els[lenTree2Count];
  ]
  
  [lenTree3Count > 0 ->
    OR3_X1 C3Els[lenTree3Count];
  ]
  
  (h:lenTree2Count:C2Els[h].vdd = supply.vdd;)
  (h:lenTree3Count:C3Els[h].vdd = supply.vdd;)

  (h:lenTree2Count:C2Els[h].vss = supply.vss;)
  (h:lenTree3Count:C3Els[h].vss = supply.vss;)

  /* Reset the variables we just stole lol */
  i = 0;
  end = N-1;
  j = 0;
  pint tree2Index = 0;
  pint tree3Index = 0;

  /* Invariant: i <= end */
    
  *[ i != end ->
     /* 
      * Invariant: tmp[i..end] has the current signals that need to be
      * combined together, and "isinv" specifies if they are the inverted
      * sense or not
      */
     j = 0;
     *[ i < end ->
        /*-- there are still signals that need to be combined --*/
        j = j + 1;
        [ i+1 >= end ->
          /*-- last piece: use either a 2 input C-element --*/
          C2Els[tree2Index].a = tmp[i];
          C2Els[tree2Index].b = tmp[i+1];
          C2Els[tree2Index].y = tmp[end+j];
          tree2Index = tree2Index +1;
          i = end;
        [] i+2 >= end ->
          /*-- last piece: use either a 3 input C-element --*/
          C3Els[tree3Index].a = tmp[i];
          C3Els[tree3Index].b = tmp[i+1];
          C3Els[tree3Index].c = tmp[i+2];
          C3Els[tree3Index].y = tmp[end+j];

          tree3Index = tree3Index +1;
          i = end;
        [] else ->
          /*-- more to come; so use a two input C-element --*/
          C2Els[tree2Index].a = tmp[i];
          C2Els[tree2Index].b = tmp[i+1];
          C2Els[tree2Index].y = tmp[end+j];
          tree2Index = tree2Index +1;
          i = i + 2;
        ]
      ]
      /*-- update range that has to be combined --*/
      i = end+1;
      end = end+j;
      j = 0;
  ]
  
    out = tmp[end];
}

export template<pint N>
defproc andtree (bool? in[N]; bool! out; power supply)
{
  bool tout;

  { N > 0 : "What?" };

  pint i, end, j;
  i = 0;
  end = N-1;

  pint lenTree2Count, lenTree3Count;
  lenTree2Count = 0;
  lenTree3Count = 0;
    /* Pre"calculate" the number of C cells required, look below if confused */
  *[ i != end ->
     j = 0;
     *[ i < end ->
        j = j + 1;
        [ i+1 >= end ->
          i = end;
          lenTree2Count = lenTree2Count +1;
        [] i+2 >= end ->
          i = end;
          lenTree3Count = lenTree3Count +1;
        [] else ->
          i = i + 2;
          lenTree2Count = lenTree2Count +1;
        ]
      ]
      /*-- update range that has to be combined --*/
      i = end+1;
      end = end+j;
      j = 0;
  ]

  /* array that holds ALL the nodes in the completion tree */
  bool tmp[end+1];
  (k:N:tmp[k] = in[k];)

  /* array to hold the actual C-elments, either A2C or A3C */
  
    [lenTree2Count > 0 ->
    AND2_X1 C2Els[lenTree2Count];
  ]
  
  [lenTree3Count > 0 ->
     AND3_X1 C3Els[lenTree3Count];
  ]
  
  (h:lenTree2Count:C2Els[h].vdd = supply.vdd;)
  (h:lenTree3Count:C3Els[h].vdd = supply.vdd;)

  (h:lenTree2Count:C2Els[h].vss = supply.vss;)
  (h:lenTree3Count:C3Els[h].vss = supply.vss;)

  /* Reset the variables we just stole lol */
  i = 0;
  end = N-1;
  j = 0;
  pint tree2Index = 0;
  pint tree3Index = 0;

  /* Invariant: i <= end */
    
  *[ i != end ->
     /* 
      * Invariant: tmp[i..end] has the current signals that need to be
      * combined together, and "isinv" specifies if they are the inverted
      * sense or not
      */
     j = 0;
     *[ i < end ->
        /*-- there are still signals that need to be combined --*/
        j = j + 1;
        [ i+1 >= end ->
          /*-- last piece: use either a 2 input C-element --*/
          C2Els[tree2Index].a = tmp[i];
          C2Els[tree2Index].b = tmp[i+1];
          C2Els[tree2Index].y = tmp[end+j];
          tree2Index = tree2Index +1;
          i = end;
        [] i+2 >= end ->
          /*-- last piece: use either a 3 input C-element --*/
          C3Els[tree3Index].a = tmp[i];
          C3Els[tree3Index].b = tmp[i+1];
          C3Els[tree3Index].c = tmp[i+2];
          C3Els[tree3Index].y = tmp[end+j];

          tree3Index = tree3Index +1;
          i = end;
        [] else ->
          /*-- more to come; so use a two input C-element --*/
          C2Els[tree2Index].a = tmp[i];
          C2Els[tree2Index].b = tmp[i+1];
          C2Els[tree2Index].y = tmp[end+j];
          tree2Index = tree2Index +1;
          i = i + 2;
        ]
      ]
      /*-- update range that has to be combined --*/
      i = end+1;
      end = end+j;
      j = 0;
  ]
  
    out = tmp[end];
}
  
/*
 * Build a completion tree using a combination of 2-input and 3-input
 * C-elements 
 */
export template<pint N>
defproc ctree (bool? in[N]; bool! out; power supply)
{
  bool tout;

  { N > 0 : "What?" };

  pint i, end, j;
  i = 0;
  end = N-1;

  pint lenTree2Count, lenTree3Count;
  lenTree2Count = 0;
  lenTree3Count = 0;
  /* Pre"calculate" the number of C cells required, look below if confused */
  *[ i != end ->
     j = 0;
     *[ i < end ->
        j = j + 1;
        [ i+1 >= end ->
          i = end;
          lenTree2Count = lenTree2Count +1;
        [] i+2 >= end ->
          i = end;
          lenTree3Count = lenTree3Count +1;
        [] else ->
          i = i + 2;
          lenTree2Count = lenTree2Count +1;
        ]
      ]
      /*-- update range that has to be combined --*/
      i = end+1;
      end = end+j;
  ]

  /* array that holds ALL the nodes in the completion tree */
  bool tmp[end+1];
  
  // Connecting the first nodes to the input
  (l:N:
    tmp[l] = in[l];
    )

  /* array to hold the actual C-elments, either A2C or A3C */
  [lenTree2Count > 0 ->
    A_2C_B_X1 C2Els[lenTree2Count];
  ]
  
  [lenTree3Count > 0 ->
     A_3C_B_X1 C3Els[lenTree3Count];
  ]
  

  (h:lenTree2Count:C2Els[h].vdd = supply.vdd;)
  (h:lenTree3Count:C3Els[h].vdd = supply.vdd;)

  (h:lenTree2Count:C2Els[h].vss = supply.vss;)
  (h:lenTree3Count:C3Els[h].vss = supply.vss;)

  /* Reset the variables we just stole lol */
  i = 0;
  end = N-1;
  j = 0;
  pint tree2Index = 0;
  pint tree3Index = 0;

  /* Invariant: i <= end */
    
  *[ i != end ->
     /* 
      * Invariant: tmp[i..end] has the current signals that need to be
      * combined together, and "isinv" specifies if they are the inverted
      * sense or not
      */
     j = 0;
     *[ i < end ->
        /*-- there are still signals that need to be combined --*/
        j = j + 1;
        [ i+1 >= end ->
          /*-- last piece: use either a 2 input C-element --*/
          C2Els[tree2Index].c1 = tmp[i];
          C2Els[tree2Index].c2 = tmp[i+1];
          C2Els[tree2Index].y = tmp[end+j];
          tree2Index = tree2Index +1;
          i = end;
        [] i+2 >= end ->
          /*-- last piece: use either a 3 input C-element --*/
          C3Els[tree3Index].c1 = tmp[i];
          C3Els[tree3Index].c2 = tmp[i+1];
          C3Els[tree3Index].c3 = tmp[i+2];
          C3Els[tree3Index].y = tmp[end+j];

          tree3Index = tree3Index +1;
          i = end;
        [] else ->
          /*-- more to come; so use a two input C-element --*/
          C2Els[tree2Index].c1 = tmp[i];
          C2Els[tree2Index].c2 = tmp[i+1];
          C2Els[tree2Index].y = tmp[end+j];
          tree2Index = tree2Index +1;
          i = i + 2;
        ]
      ]
      /*-- update range that has to be combined --*/
      i = end+1;
      end = end+j;
      j = 0;
  ]
  
  out = tmp[end];
  
}

export template<pint N>
defproc vtree (std::data::Mx1of2?<N> in; bool! out; power supply)
{
    // OR layer for making OR between true and false of in (they are then sent to Ctree)
    OR2_X1 OR2_tf[N];
    ctree<N> myctree;
  (l:N:
      OR2_tf[l].a = in.d[l].t;
      OR2_tf[l].b = in.d[l].f;
      OR2_tf[l].y = myctree.in[l];
      OR2_tf[l].vdd = supply.vdd;
      OR2_tf[l].vss = supply.vss;
  )
  myctree.supply = supply;
  out = myctree.out;
}
export template<pint N>
defproc sigbuf (bool? in; bool! out[N]; power supply)
{

  { N >= 0 : "sigbuf: parameter error" };
  { N <= 43 : "sigbuf: parameter error, N too big" };

	  /* -- just use in sized driver here -- */
    [ N <= 4 ->
    BUF_X1 buf1 (.a = in, .y = out[0], .vdd = supply.vdd, .vss = supply.vss);
    [] N >= 5 & N <= 7 -> 
    BUF_X2 buf2 (.a = in, .y = out[0], .vdd = supply.vdd, .vss = supply.vss);
    [] N >= 8 & N <= 10 -> 
    BUF_X3 buf3 (.a = in, .y = out[0], .vdd = supply.vdd, .vss = supply.vss);
   [] N >= 11 & N <= 14 -> 
    BUF_X4 buf4 (.a = in, .y = out[0], .vdd = supply.vdd, .vss = supply.vss);
   [] N >= 15 & N <= 18 -> 
    BUF_X6 buf6 (.a = in, .y = out[0], .vdd = supply.vdd, .vss = supply.vss);
   [] N >= 19 & N <= 29 -> 
    BUF_X8 buf8 (.a = in, .y = out[0], .vdd = supply.vdd, .vss = supply.vss);
   [] N >= 30 & N <= 42 -> 
    BUF_X12 buf12 (.a = in, .y = out[0], .vdd = supply.vdd, .vss = supply.vss);
   ]
   (i:1..N-1:out[i]=out[0];)
}
}}