OpenFPGA/tutorial/02_multimode.ipynb

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    "# Multimode CLB\n",
    "In the I/O section of the architecture, the `<mode>` tag was used to define different modes for the block. One mode was for input, the other mode for output. Modes are also used in CLBs. Open the file `OpenFPGA/tutorial/02_multimode_arch.xml`. It defines two modes for its CLB: either a 6-LUT, or two 5-LUTs with shared inputs.\n",
    "\n",
    "Find the line `<pb_type name=\"clb\" area=\"53894\">` which begins the definition of the CLB. The next few lines define the inputs and outputs: 40 input pins, 20 output pins, and a clock pin. The inputs are marked with `equivalent=\"full\"` which indicates that the inputs have logical equivalence so their locations may be freely swapped. The outputs are `equivalent=\"none\"` so they may not be swapped.\n",
    "\n",
    "The next primitive block is `<pb_type name=\"fle\" num_pb=\"10\">` which is a fracturable logic element that contains the two different modes of this CLB. There are 10 instances of this `fle` in the CLB. Each `fle` has 6 inputs, 2 outputs, and one clock pin.\n",
    "\n",
    "### First Mode\n",
    "`<mode name=\"n2_lut5\">` begins the definition of the first mode. Moving down the architecture file, we see that in this mode, the `fle` contains one `lut5inter` block which contains two `ble5` blocks. The `ble5` blocks have 5 inputs and 1 output. The next group of lines define the LUT and FF in `ble5`.\n",
    "\n",
    "Moving down to the second `<interconnect>` block:\n",
    "```\n",
    "<interconnect>\n",
    "    <direct name=\"direct1\" input=\"lut5inter.in\" output=\"ble5[0:0].in\"/>\n",
    "    <direct name=\"direct2\" input=\"lut5inter.in\" output=\"ble5[1:1].in\"/>\n",
    "    <direct name=\"direct3\" input=\"ble5[1:0].out\" output=\"lut5inter.out\"/>                 \n",
    "    <complete name=\"complete1\" input=\"lut5inter.clk\" output=\"ble5[1:0].clk\"/>                  \n",
    "</interconnect>\n",
    "```\n",
    "we see that the input of `lut5inter` is connected to the inputs of both the first and second `ble5` blocks. Then the two outputs of the two `ble5` blocks go to the two outputs of `lut5inter`. Finally, there is a fully connected crossbar that connects the `lut5inter` clock input to the two `ble5` clock inputs.\n",
    "\n",
    "Finishing out this mode, there is the last interconnect section. Note the line: `<direct name=\"direct1\" input=\"fle.in[4:0]\" output=\"lut5inter.in\"/>`. It shows that out of the 6 `fle` input lines, only the first 5 lines are connected to `lut5inter`. The remaining line is unused.\n",
    "\n",
    "### Second Mode\n",
    "`<mode name=\"n1_lut6\">` begins the definition of the second mode. In this mode, the `fle` block contains one block named `ble6` with 6 inputs and one output. Moving down the architecture, we see that it contains one LUT named `lut6` with 6 inputs and one output, and one flip-flop named `ff`. \n",
    "\n",
    "Moving down to this mode's `<interconnect>` block:\n",
    "```\n",
    "<interconnect>\n",
    "    <direct name=\"direct1\" input=\"fle.in\" output=\"ble6.in\"/>\n",
    "    <direct name=\"direct2\" input=\"ble6.out\" output=\"fle.out[0:0]\"/>\n",
    "    <direct name=\"direct3\" input=\"fle.clk\" output=\"ble6.clk\"/>\n",
    "</interconnect>\n",
    "```\n",
    "we see that the input of `fle` is connected to the input of `ble6`, the output of `ble6` is connected to one output of `fle`, and the clock pins are wired together. The remaining output of `fle` is unused in this mode.\n",
    "\n",
    "The final `<interconnect>` section takes care of the wiring inside the top-level `clb` block. There is a fully connected crossbar that takes as its input the inputs to `clb` as well as the outputs from `flb` in a feedback loop. The crossbar's outputs go to `flb`'s inputs. Then a crossbar connects the clock pins. Now recall that the first mode of `clb` uses both of `flb`'s outputs, but the second mode uses only one outputs. The lines:\n",
    "```\n",
    "    <direct name=\"clbouts1\" input=\"fle[9:0].out[0:0]\" output=\"clb.O[9:0]\"/>\n",
    "    <direct name=\"clbouts2\" input=\"fle[9:0].out[1:1]\" output=\"clb.O[19:10]\"/>\n",
    "```\n",
    "ensure that if a logic block is completely filled by 6-LUTs, then the outputs those 6-LUTs take get evenly distributed across all four sides of the CLB instead of clumped on two sides.\n"
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