kbob

# 12 MHz clock
set_io -nowarn CLK        35

# RS232
set_io -nowarn RX          6
set_io -nowarn TX          9

# LEDs and Button
set_io -nowarn BTN_N      10
set_io -nowarn LEDR_N     11

kbob

`default_nettype none

`define W 24

module top (
        input  CLK,
        output LED1,
        output LED2,
        output LED3,
        output LED4,

kbob

// BEGIN kbob
//    This is identical to https://vulkan-tutorial.com/Drawing_a_triangle/Swap_chain_recreation
//    as of 2018-06-13 except for code between the "BEGIN kbob" and "END kbob" comments.
//    The original code is still there but commented out with "//-".
//
//    Built with MoltenVK 1.1.73 and GLFW 20180519 for MacOS, the original never recreated
//    the swap chain.  It continued to draw at the original 800 by 600 resolution and scaled
//    the result to fit the resized window.
//
//    Also, according to GLFW doc, the glfwGetWindowSize() is not the right function to call.

kbob

#define EXPECT

#include <assert.h>
#include <math.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdio.h>

kbob

#include <assert.h>
#include <stdint.h>
#include <unistd.h>

/* print nine digit positive */
char *p9dp(uint32_t n, char *p)
{
    uint32_t a = n % 10;
    n /= 10;
    if (n) {

kbob

struct TrieNode {
    is_terminal: bool,
    children: [Option<Box<TrieNode>>; 256],
}

impl TrieNode {
    fn new() -> TrieNode {
        let mem : TrieNode = unsafe { std::mem::zeroed() };
        mem
    }

kbob

import itertools

def infinite_combinations(n, iterator):
    seen = []
    for item in iterator:
        for subset in itertools.combinations(seen, n - 1):
            yield subset + (item,)
        seen.append(item)

kbob

#!/usr/bin/python

from itertools import count

def ramanujan_number():
    cubes = (i**3 for i in count(1))
    smaller_cubes = []
    sums = {}
    for n in cubes:
       for m in smaller_cubes:

kbob

#!/usr/bin/python

# https://plus.google.com/u/0/114070220225686847236/posts/d7xrq9ph7WJ

# There are no repeated digits, and no adjacent digits with a
# difference of one, in the number 13524.  How many such arrangements
# of 5 digits are there?  What's the shortest program you can write to
# find out?

import itertools

kbob

#!/usr/bin/python

from itertools import count, cycle, islice, izip, repeat

def fizzbuzz(n):
    class d(int): __add__ = lambda self, other: other
    def i(x): return x
    def f(x): return x + 'Fizz'
    def b(x): return x + 'Buzz'
    def p(x): print x