To fix Mute button keyboard shortcut issue on LXDE

Create a script call “amixertoggle” and save it in /usr/local/bin:

#!/bin/bash

amixer $1 sset Headphone toggle
amixer $1 sset Speaker toggle
amixer $1 sset PCM toggle
amixer $1 sset Master toggle

Edit file $HOME/.config/openbox/lxde-rc.xml and replace block that has “XF86AudioMute” to call our script.  For example:




   
     
              amixertoggle -c 1 toggle
     


(” -c 1″ above is for machine, where the mixer control is actually on card 1.  If doesn’t work, we can try different number)

Reload the modified file lxde-rc.xml by doing:

openbox –reconfigure

That’s it.  Everytime we press “Mute” button on our PC keyboard, the mute/unmute will toggle.

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Raspberry Pi 2

Got this New Raspberry Pi 2 I ordered a few weeks ago.  It’s significantly faster than the first version.  I only wish it had USB 3.0 ports so I can have much bigger and faster storage.

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Cheap ATMEL AVR ISP ICE from QinHeng

I bought this small USB stick from eBay.  It supposedly can do JTAG and many other cool stuff for ATMEL microcontrollers.

Anyway, here is the detail how to use it:

The other end of the stick has dual-line 10-pin connector.  The pins are:

Pin  Purpose
——————————
1    TCK
2    GND
3    TDO
4    VTref
5    TMS
6    nSRT
7    VSupply
8    (nSRT)
9    TDI
10   GND

These pins are actually JTAG pins.  The connections to the Microcontroller is as follow (the pullup resistors are all 4.7k):

The USB id:

$ lsusb

Bus 003 Device 006: ID 1a86:7523 QinHeng Electronics HL-340 USB-Serial adapter

Upon inserting the stick to my PC’s USB (running Ubuntu Linux):

[118845.955500] ch341-uart ttyUSB0: ch341-uart converter now disconnected from ttyUSB0
[118845.955541] ch341 3-2:1.0: device disconnected
[118851.692044] usb 3-2: new full-speed USB device number 6 using ohci-pci
[118851.901261] usb 3-2: New USB device found, idVendor=1a86, idProduct=7523
[118851.901271] usb 3-2: New USB device strings: Mfr=0, Product=2, SerialNumber=0
[118851.901276] usb 3-2: Product: USB2.0-Serial
[118851.904683] ch341 3-2:1.0: ch341-uart converter detected
[118851.940428] usb 3-2: ch341-uart converter now attached to ttyUSB0

To use it as programmer, we can use avrdude:

$ sudo avrdude -c avrisp -P /dev/ttyUSB0 -pm328p

To debug, we can use another opensource, AvaRICE (from ubuntu, just do “apt-get install avarice”).

For example (this is just to show how to use it, as my ICE stick is not connected to any target device yet so it reports it “No configuration available for device ID: ffff”):

$ sudo avarice –jtag /dev/ttyUSB0 -1
AVaRICE version 2.11, Jan 17 2014 02:53:19

Defaulting JTAG bitrate to 250 kHz.

JTAG config starting.
Hardware Version: 0xc3
Software Version: 0x80
Reported JTAG device ID: 0xFFFF
No configuration available for device ID: ffff

The avarice can be set as a gdb server mode, so we can debug the target removely. For example, to make it as gdb-server listening on port 4242.

$ sudo avarice –jtag /dev/ttyUSB0 :4242

From another PC (or the same PC, if host and server is on the same machine), we open gdb, and connect:

$ gdb
GNU gdb (Ubuntu 7.7-0ubuntu3.1) 7.7
Copyright (C) 2014 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type “show copying”
and “show warranty” for details.
This GDB was configured as “i686-linux-gnu”.
Type “show configuration” for configuration details.
For bug reporting instructions, please see:
.
Find the GDB manual and other documentation resources online at:
.
For help, type “help”.
Type “apropos word” to search for commands related to “word”.
(gdb) target remote 127.0.0.1:4242




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Using PicKit2 Programmer to program Atmel’s microcontrollers

Many hobbists are familiar with Arduino kit.  This beast uses various Microcontrollers from Atmel.  For example, Arduino UNO uses AtMega 328p.

Normally, we use the pcb kit to program the chip, as the Arduino (or the clones) use USB to serial chip to translate  USB to serial needed to program Arduino.

For people who has done hacking on PIC microcontrollers from MicroChip and wants to use bare-metal chip, they can program the Atmel chips using the existing cheap PicKit2 programmer.  What they need just the Pickit2 programmer and a software tool called avrdude.

The pin connections:

AVR   - PICKit2 (pin):
----------------------- 
RST   - VPP/MCLR (1)
VDD - VDD Target (2) -- optional if AVR self powered
GND - GND (3)
MISO - PGD (4)
SCLK - PDC (5)
MOSI - AUX (6) 
 
 
Some of command examples are shown below:

To write our firmware to the flash in AtMega 328p:

 
$ avrdude -c pickit2 -p m328p -v -V -U flash:w:"myfirmware.hex":a 
 
To upload bootloader (e.g. to upload ARDUINO bootloader):
 
$ avrdude -c pickit2 -p m328p -v -V -U boot:w:"boot.hex":a 

 

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Left shift by negative number used to test unique characters in a string

What is the outcome of this?

val=-32;
mask = 1<<val;

I tried to execute it on Python, it failed with error message.

But if I do it in C or C++, I get something else.  Turn out, this depends o the compiler and CPU we use.  As I use Pentium4 and GCC to compile this, the following snippet of the generated assembly language is interesting:


.L4:
        movsbl  (%edx), %ecx
        movl    %esi, %edi
        addl    $1, %edx
        subl    $97, %ecx
        sall    %cl, %edi
        testl   %eax, %edi
        jg      .L5
        orl     %edi, %eax

.L2:

What is “sall” above? It is Intel CPU’s mnemonic for “Shift Arithmetic Left for Long Integer”.
In the machine code, -32 is transformed to Second-complement of 32:
32        = 0b00100000
~32      = 0b11011111
~32+1 = 0b11100000 = -32
According Wikipedia,  arithmetic shift left maintains the MSBit, so in this case MSB is always 1. Also, Intel CPUs only shifts the first 5-bits of the shifting bits (in this case, 5-bit LSB of 0b11100000 which is 0), so 1<<-32 is actuall 1<<0 = 1.
What about 1<<-33?
33        = 0b00100001
~33      = 0b11011110
~32+1 = 0b11100001 = -33
The first five bits is 1, so 1<<-33 = 1<<1 = 2
What about 1<<-31?
31        = 0b00011111
~31      = 0b11100000
~31+1 = 0b11100001 = -31
The first five bits is 1, so 1<<-31 = 1<<1 = 0x00000010
What about 1<<-7?
7        = 0b00000111
~7      = 0b11111000
~7+1 = 0b11111001 = -7
The first five bits is 1, so 1<<-7 = 1<<25 = 0x02000000
Now, how do we use this to check a uniqueness of characters in a string? 
Alphabet ‘A’ to ‘Z’ has 26 characters. Alphabet ‘a’ to ‘z’ has also 26 characters.
The difference between the character i in a string is: str[i] – ‘a’.  Now this is the interesting part: if we declare a 32-bit variable (which is enough to cover ‘A’ to ‘Z’) where each bit corresponds to the occurence of each character in alphabet, we can determine if the character has been used before and tell the string has unique characters or not (e.g, bit 1 to tell if character ‘A’ has been used or not, etc.)
‘A’ – ‘a’ = -32
‘Z’ – ‘a’ = -7
‘a’ – ‘a’ = 0
‘Z’ – ‘a’ = 25
if we shift 1 to -32, it is the same as 1<<0.  If we shift 1 to -33 it is the same as 1<<1, so on.
1<<-32 = 0x01
1<<-31 = 0x02
1<<-7 = 1<<25 = 0x2000000
1<<0 = 0x01
1<<1 = 0x02
1<<25 = 0x2000000
so:
 1<<('A'-'a') == 1<<('a'-'a') = 0x01
 1<<('B'-'a') == 1<<('b'-'a') = 0x02
1<<('Z'-'a') == 1<<('z'-'a') = 0x2000000
So the code is:
for (i=0; i<strlen(s); i++)
{
    val = s[i] – ‘a’;
    if (flags & (1 << val))
        return 0; /* the string is not unique */
    flags |= 1 << val;
}
return 1;
Notice this algorithm might not work on different machine with different compiler (as shifting by negative number is undefined in standard C).

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The Use of c++decl

cdecl and c++decl are two nice tools to check syntax or to generate declaration of C and C++ respectively.

for example, using c++decl (the text in bold is the one we type):

$ cdecl

Type `help’ or `?’ for help
c++decl>


c++decl> declare foo as pointer to member of class X function (arg1, arg2) returning reference to class Y
class Y &(X::*foo)(arg1, arg2)
c++decl>

c++decl> declare foo as reference to class X
class X &foo
c++decl>
c++decl> cast foo into reference to class X
(class X &)foo
c++decl>
We can also use them to check a declaration syntax, for example:
c++decl> explain int (*a)[5]
declare a as pointer to array 5 of int
c++decl>
c++decl> explain long **foo[10]
declare foo as array 10 of pointer to pointer to long
c++decl>

c++decl> explain int*&(foo);
declare foo as reference to pointer to int
c++decl>

The function can also be used to translate an english declaration to C (or C++) declaration.  For example:
$ c++decl declare “i as array 5 of pointer to function(int,int) returning reference to class X”
class X &(*i[5])(int , int )
$
$ c++decl declare “i as array 5 of pointer to function(reference to class X,constant reference to int) returning reference to class X”
class X &(*i[5])(class X &, int & const )
$
$ c++decl declare “const_ptr as const pointer to const int”
const int * const const_ptr
$ c++decl declare “const_ptr as const pointer to const reference to class X”
class X & const * const const_ptr
$

c++decl> declare myfunc as const pointer to function (const pointer to int, reference to class X) returning void
void (* const myfunc)(int * const , class X &)
c++decl>

c++decl> declare myfunctpr as array of pointer to function (void) returning int
int (*myfunctpr[])(void )
c++decl>
declare myptr as pointer to array 10 of struct MyStruct
struct MyStruct (*myptr)[10]
c++decl>
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Odd Numbers with Parity

#!/usr/bin/python

table=[]
for n in range(0,256):
p=False
while n:
p = not p
n &= n-1
table.append(p)

parity_odd = []
print "Odd numbers with Parity:"
for i in range(0,256):

if (i%2) and (table[i]): parity_odd.append(i) print i,


Result:

Odd numbers with Parity:
1 7 11 13 19 21 25 31 35 37 41 47 49 55 59 61 67 69 73 79 81 87 91 93 97 103 107 109 115 117 121 127 131 133 137 143 145 151 155 157 161 167 171 173 179 181 185 191 193 199 203 205 211 213 217 223 227 229 233 239 241 247 251 253


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Fastest Fibonacci Sequencer

This algorithm is based on Fast-doubling  as explained in Fast Fibonacci algorithms.

Given F(k) and F(k+1), we can calculate these equations:

F(2k) = F(k)[2F(k+1)−F(k)]
F(2k+1) = F(k+1)2+F(k)2.

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#include 
#include
#include

#ifdef DEBUG
#define DBGMSG(x) printf x
#else
#define DBGMSG(x)
#endif

typedef unsigned long long num_type;

void _fib(int n, num_type *fn, num_type *fn_1);

num_type fibonacci(uint n, num_type *s)
{
int i;
num_type fn=0, fn_1=1;

if (!s)
return 0;

if (n < 2)
{
return 1;
}
for(i=1; i<n; i++)
{
_fib(i, &fn, &fn_1);
DBGMSG(("i=%u, n=%u, fn=%llu, fn+1=%llu\n", i, n, fn, fn_1));
s[i] = fn;
}
return s[n];
}

/*
* This is base on matrix exponentiation algorithm:
* where:
* f(2k) = f(k)*{ 2*f(k+1) - f(k)} ---> c
* f(2k+1) = f(k)^2 + f(k+1)^2 ---> s
*
* if we plugin k = n/2, we get f(n) and f(n+1)
*/
void _fib(int n, num_type *fn, num_type *fn_1)
{
num_type c,d;
num_type a, b;

DBGMSG( ("%s: Initially n=%u, fn=%llu, fn_1=%llu\n", __FUNCTION__, n, *fn, *fn_1) );
if (n==0)
{
*fn = 0; // fn
*fn_1 = 1; // fn+1
return;
}
else {
a = *fn;
b = *fn_1;
}
_fib(n>>1, &a, &b);
DBGMSG (("%s: local fn=%llu, fn_1=%llu\n", __FUNCTION__, a, b));
c = a * (2*b - a);
d = b*b + a*a;
DBGMSG( ("%s: c=%llu, d=%llu\n", __FUNCTION__, c, d) );
if (n % 2 == 0)
{
// n is even
*fn = c;
*fn_1 = d;
}
else {
*fn = d;
*fn_1 = c+d;
}
}

int main(int argc, char *argv[])
{
uint n;
uint f;
int i;
num_type *result;
char *sbuf;
int status = 0;

if (argc > 1)
n = atoi(argv[1]);
else {
printf("Enter a number: ");
sbuf = (char *)malloc(50);
status = (getline(&sbuf, &n, stdin) > 0);
if (status == 0) {
n = atoi(sbuf);
free(sbuf);
}
else {
printf("Error in input\n");
return (1);
}
}
if (n <= 0)
{
printf("What have yo done???\n");
return(0);
}
if (status == 0)
{
result = (num_type *)malloc(n*sizeof(unsigned long long));
f = fibonacci(n, result);
printf("Fibonacci sequence: ");
for(i=0; i<n; i++)
printf("%llu ", result[i]);
printf("\n");
if (result)
free(result);
}
}
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Fastest bit counting

The best bit counting algorithm as far as I know is the one invented by folks at Stanford University, which is always O(1).

int bitcount(int n)
{
    int cnt = 0;

    n = n – ((n >> 1) & 0x55555555);
    n = (n & 0x33333333) + ((n >> 2) & 0x33333333);

    return (((n + (n >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}

For example, after I compile it to x86 Assembly:
    .cfi_startproc
    movl    4(%esp), %eax
    movl    %eax, %edx
    sarl    %edx
    andl    $1431655765, %edx
    subl    %edx, %eax
    movl    %eax, %edx
    sarl    $2, %eax
    andl    $858993459, %edx
    andl    $858993459, %eax
    addl    %edx, %eax
    movl    %eax, %edx
    sarl    $4, %edx
    addl    %edx, %eax
    andl    $252645135, %eax
    imull   $16843009, %eax, %eax
    sarl    $24, %eax
    ret
    .cfi_endproc
But, how does actually the algorithm work?
Ok, let try a simple one.  Assume n is a 4-bit number, and the bits can be represented as a set such that n= {a,b,c,d}., where a,b,c.. can only have either 0 or 1.  The decimal value of the n is: 8*a + 4*b + 2*c + d.  Total number of bit one is: a + b + c +d.  
For example, for n=0b1101, a=1, b=1, c=0, d=1 (which in decimal is 8*1 + 4*1 + 2*0 + 1 = 13), and total number of bit one is a+b+c+d = 1 + 1 + 0 + 1 = 3.  So far so good?
Now, we know that to count the 1-bits is as simple as: a+b+c+d.  But, wait…. n itself is not a+b+c+d, but 8*a + 4*b + 2*c + d.  Ok, let’s conquer this step-by-step.  
If we shift n one bit to the right the LSbit is gone and other numbers just divided by two, so n becomes 4*a + 2*b + c, right? Now substract this to the original n.
  
n      = 8*a + 4*b + 2*c + d
n>>1   = 0 +   4*a + 2*b + c
—————————– –
       = 0   + 4*a + 2*b + c + d

That’s a good direction!  Now if  (n>>1)  is written in the 4-bit nibble it is actually 0 + 4a + 2b + c.  If I just want to subtract 4a and c, we have to mask out 2*b.  so we mask it with binary 0101 (0x5), so we get only 4a + c:
n              = 8*a + 4*b + 2*c + d
(n>>1)&0x5     = 4*a +   0 + c
———————————- –
n -(n>>1)&0x05 = 4*a + 4*b + c + d

Now store this result back to n, so from now on n is 4*a + 4*b + c + d
To get c + d only, we mask n with 0b11 or n & 0x03
if we shift n above once, we get 0 + 2a + 2b + 0, but if shift it again it becomes a + b!
To make sure we only get the lowest two bits (a + b), we mask it to 0x03 or:

n & 0x03       = c + d
(n >> 2)&0x03) = a + b
———————— +

Nice! now we have this expression: a + b + c +d .

so, for 4-bit bit counting, we can use the expression:

n = n – (n>>1) & 0x05
bits = (n & 0x3) + (n>>2) & 0x3
Proof: as example above, n=13 (0b1101).  so a=1,b=1, c=0, d=1
n = 0b1101 – (0b0110) & 0b0101 = 0b1101 – 0b100 = 13 – 4 = 9 = 0b1001
then the next step:
bits =(0b1001 & 0b0011)  +  (0b1001>>2) & 0b0011, or bits = 0b0001 + (0b0010) & 0b0011
 or bits = 1 + 2 = 3 !!
For 32-bit, it is based on the same idea, except we have to do more.
Say n  has set of coefficients {a[0], a[1], …., a[31]} to represent the number, so n = a[0]*2^31 + a[1]*2^30 + …. + a[15]*2^0
The mask should be 32-bit, so instead of 0x5, we use 0x55555555 = 0b0101010101…0101
n                   = 2^31*a[0] + … + 2^1*a[30] + 2^0*a[31]
(n>>1)&0b0101..0101 = 2^30*a[0] + … + 2^2*a[28] + 2^0*a[30]
————————————————————– –
n -(n>>1)&0x055555555 = 2^31*a[0] – 2^28*a[0] + (2^30*a[1] – 2^26*a[1]) + … + 4*a[29] – 2*a[30] + a[31]


Let’s review binary arithmetics.

23 – 22 = 8 – 4 = 4

216 – 215 = 65536 – 32768 = 32768
or: 2(a+1) – 2a = 2a

2(a+2) – 2a = 2a * 22 – 2a = 2a (4 – 1) = 3*2a

So the result is:

a[0] * (2^31 – 2^28) + a[1] * (2^30 – 2^26) + ….. + a[30] (4 -2) + a[31]
= 3*2^28*a[0] + .3*2^26*a[1].. + 2*a[30] + a[31]

n – n(>>1) & 0x055555555 = 3*2^28*a[0] + .3*2^26*a[1].. + 2*a[30] + a[31]

stored this as a new n.

n>>2 = 2^24*a[0] + 2^22*a[1] + …2*a[28] + a[29]

The rest is actually manipulation to count this a[0] + a[1] + …  + a[31]


A variant, but this one is invented by folks at MIT:

int bitcount(unsigned int n)
{
    int cnt = 0;
    register unsigned int tmp;
                     
    tmp = n – ((n >> 1) & 033333333333) – ((n >> 2) &    011111111111);

    return ((tmp + (tmp >>3)) & 0030707070707) % 63;
}

It uses the similar method. but with different approach (notice the number 01…, 0333… and 00307… are in octal.  We could use Hex version but then it is harder to remember)


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Audio on Laptop HP dv7-6c80us not working

Problem:

HP-PAVILION-DV7:~$ aplay -l
aplay: device_list:268: no soundcards found…
HP-PAVILION-DV7:~$ 

HP-PAVILION-DV7:~$ lsmod | grep snd
snd_hda_codec_hdmi     41154  0 
snd_hda_codec_idt      50341  1 
snd_hda_intel          52267  0 
snd_hda_codec         188738  3 snd_hda_codec_hdmi,snd_hda_codec_idt,snd_hda_intel
snd_hwdep              13602  1 snd_hda_codec
snd_pcm               102033  3 snd_hda_codec_hdmi,snd_hda_codec,snd_hda_intel
snd_page_alloc         18710  2 snd_pcm,snd_hda_intel
snd_seq_midi           13324  0 
snd_seq_midi_event     14899  1 snd_seq_midi
snd_rawmidi            30095  1 snd_seq_midi
snd_seq                61560  2 snd_seq_midi_event,snd_seq_midi
snd_seq_device         14497  3 snd_seq,snd_rawmidi,snd_seq_midi
snd_timer              29433  2 snd_pcm,snd_seq
snd                    69141  11 snd_hwdep,snd_timer,snd_hda_codec_hdmi,snd_hda_codec_idt,snd_pcm,snd_seq,snd_rawmidi,snd_hda_codec,snd_hda_intel,snd_seq_device,snd_seq_midi
soundcore              12680  1 snd

HP-PAVILION-DV7:~$ cat /proc/asound/pcm 
00-00: 92HD81B1X5 Analog : 92HD81B1X5 Analog : playback 1 : capture 1

HP-PAVILION-DV7:~$  lspci -v


00:1b.0 Audio device: Intel Corporation 6 Series/C200 Series Chipset Family High Definition Audio Controller (rev 05)
Subsystem: Hewlett-Packard Company Device 165b
Flags: bus master, fast devsel, latency 0, IRQ 54
Memory at c6500000 (64-bit, non-prefetchable) [size=16K]
Capabilities: [50] Power Management version 2
Capabilities: [60] MSI: Enable+ Count=1/1 Maskable- 64bit+
Capabilities: [70] Express Root Complex Integrated Endpoint, MSI 00
Capabilities: [100] Virtual Channel
Capabilities: [130] Root Complex Link
Kernel driver in use: snd_hda_intel

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