Related
In a bash script that I'm writing for Linux/Solaris I need to decode more than a hundred thousand base64-encoded text strings, and, because I don't wanna massively fork a non-portable base64 binary from awk, I wrote a function that does the decoding.
Here's the code of my base64_decode function:
function base64_decode(str, out,i,n,v) {
out = ""
if ( ! ("A" in _BASE64_DECODE_c2i) )
for (i = 1; i <= 64; i++)
_BASE64_DECODE_c2i[substr("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/",i,1)] = i-1
i = 0
n = length(str)
while (i <= n) {
v = _BASE64_DECODE_c2i[substr(str,++i,1)] * 262144 + \
_BASE64_DECODE_c2i[substr(str,++i,1)] * 4096 + \
_BASE64_DECODE_c2i[substr(str,++i,1)] * 64 + \
_BASE64_DECODE_c2i[substr(str,++i,1)]
out = out sprintf("%c%c%c", int(v/65536), int(v/256), v)
}
return out
}
Which works fine:
printf '%s\n' SmFuZQ== amRvZQ== |
LANG=C command -p awk '
{ print base64_decode($0) }
function base64_decode(...) {...}
'
Jane
jdoe
SIMPLIFIED REAL-LIFE EXAMPLE THAT DOESN'T WORK AS EXPECTED
I want to get the givenName of the users that are members of GroupCode = 025496 from the output of ldapsearch -LLL -o ldif-wrap=no ... '(|(uid=*)(GroupCode=*))' uid givenName sn GroupCode memberUid:
dn: uid=jsmith,ou=users,dc=example,dc=com
givenName: John
sn: SMITH
uid: jsmith
dn: uid=jdoe,ou=users,dc=example,dc=com
uid: jdoe
givenName:: SmFuZQ==
sn:: RE9F
dn: cn=group1,ou=groups,dc=example,dc=com
GroupCode: 025496
memberUid:: amRvZQ==
memberUid: jsmith
Here would be an awk for doing so:
LANG=C command -p awk -F '\n' -v RS='' -v GroupCode=025496 '
{
delete attrs
for (i = 2; i <= NF; i++) {
match($i,/::? /)
key = substr($i,1,RSTART-1)
val = substr($i,RSTART+RLENGTH)
if (RLENGTH == 3)
val = base64_decode(val)
attrs[key] = ((key in attrs) ? attrs[key] SUBSEP val : val)
}
if ( /\nuid:/ )
givenName[ attrs["uid"] ] = attrs["givenName"]
else
memberUid[ attrs["GroupCode"] ] = attrs["memberUid"]
}
END {
n = split(memberUid[GroupCode],uid,SUBSEP)
for ( i = 1; i <= n; i++ )
print givenName[ uid[i] ]
}
function base64_decode(...) { ... }
'
On BSD and Solaris the result is:
Jane
John
While on Linux it is:
John
I don't know where the issue might be; is there something wrong with the base64_decode function and/or the code that uses it?
Your function generates NUL bytes when its argument (encoded string) ends with padding characters (=s). Below is a corrected version of your while loop:
while (i < n) {
v = _BASE64_DECODE_c2i[substr(str,1+i,1)] * 262144 + \
_BASE64_DECODE_c2i[substr(str,2+i,1)] * 4096 + \
_BASE64_DECODE_c2i[substr(str,3+i,1)] * 64 + \
_BASE64_DECODE_c2i[substr(str,4+i,1)]
i += 4
if (v%256 != 0)
out = out sprintf("%c%c%c", int(v/65536), int(v/256), v)
else if (int(v/256)%256 != 0)
out = out sprintf("%c%c", int(v/65536), int(v/256))
else
out = out sprintf("%c", int(v/65536))
}
Note that if the decoded bytes contains an embedded NUL then this approach may not work properly.
Problem is within base64_decode function that outputs some junk characters on gnu-awk.
You can use this awk code that uses system provided base64 utility as an alternative:
{
delete attrs
for (i = 2; i <= NF; i++) {
match($i,/::? /)
key = substr($i,1,RSTART-1)
val = substr($i,RSTART+RLENGTH)
if (RLENGTH == 3) {
cmd = "echo " val " | base64 -di"
cmd | getline val # should also check exit code here
}
attrs[key] = ((key in attrs) ? attrs[key] SUBSEP val : val)
}
if ( /\nuid:/ )
givenName[ attrs["uid"] ] = attrs["givenName"]
else
memberUid[ attrs["GroupCode"] ] = attrs["memberUid"]
}
END {
n = split(memberUid[GroupCode],uid,SUBSEP)
for ( i = 1; i <= n; i++ )
print givenName[ uid[i] ]
}
I have tested this on gnu and BSD awk versions and I am getting expected output in all the cases.
If you cannot use external base64 utility then I suggest you take a look here for awk version of base64 decode.
This answer is for reference
Here's a working base64_decode function (thanks #MNejatAydin for pointing out the issue(s) in the original one):
function base64_decode(str, out,bits,n,i,c1,c2,c3,c4) {
out = ""
# One-time initialization during the first execution
if ( ! ("A" in _BASE64) )
for (i = 1; i <= 64; i++)
# The "_BASE64" array associates a character to its base64 index
_BASE64[substr("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/",i,1)] = i-1
# Decoding the input string
n = length(str)
i = 0
while ( i < n ) {
c1 = substr(str, ++i, 1)
c2 = substr(str, ++i, 1)
c3 = substr(str, ++i, 1)
c4 = substr(str, ++i, 1)
bits = _BASE64[c1] * 262144 + _BASE64[c2] * 4096 + _BASE64[c3] * 64 + _BASE64[c4]
if ( c4 != "=" )
out = out sprintf("%c%c%c", bits/65536, bits/256, bits)
else if ( c3 != "=" )
out = out sprintf("%c%c", bits/65536, bits/256)
else
out = out sprintf("%c", bits/65536)
}
return out
}
WARNING: the function requires LANG=C
It also doesn't check that the input is a valid base64 string; for that you can add a simple condition like:
match( str, "^([a-zA-Z/-9+]{4})*([a-zA-Z/-9+]{2}[a-zA-Z/-9+=]{2})?$" )
Interestingly, the code is 2x faster than base64decode.awk, but it's only 3x faster than forking the base64 binary from inside awk.
notes:
In a base64 encoded string, 4 bytes represent 3 bytes of data; the input have to be processed by groups of 4 characters.
Multiplying and dividing an integer by a power of two is equivalent to do bitwise left and right shifts operations.
262144 is 2^18, so N * 262144 is equivalent to N << 18
4096 is 2^12, so N * 4096 is equivalent to N << 12
64 id 2^6, so N * 4096 is equivalent to N << 6
65536 is 2^16, so N / 65536 (integer division) is equivalent to N >> 16
256 is 2^8, so N / 256 (integer division) is equivalent to N >> 8
What happens in printf "%c", N:
N is first converted to an integer (if need be) and then, WITH LANG=C, the 8 least significant bits are taken in for the %c formatting.
How the possible padding of one or two trailing = characters at the end of the encoded string is handled:
If the 4th char isn't = (i.e. there's no padding) then the result should be 3 bytes of data.
If the 4th char is = and the 3rd char isn't = then there's 2 bytes of of data to decode.
If the fourth char is = and the third char is = then there's only one byte of data.
Is there any way to convert a signed integer into an array of bytes in NXC? I can't use explicit type casting or pointers either, due to language limitations.
I've tried:
for(unsigned long i = 1; i <= 2; i++)
{
MM_mem[id.idx] = ((val & (0xFF << ((2 - i) * 8)))) >> ((2 - i) * 8));
id.idx++;
}
But it fails.
EDIT: This works... It just wasn't downloading. I've wasted about an hour trying to figure it out. >_>
EDIT: In NXC, >> is a arithmetic shift. int is a signed 16-bit integer type. A byte is the same thing as unsigned char.
NXC is 'Not eXactly C', a relative of C, but distinctly different from C.
How about
unsigned char b[4];
b[0] = (x & 0xFF000000) >> 24;
b[1] = (x & 0x00FF0000) >> 16;
b[2] = (x & 0x0000FF00) >> 8;
b[3] = x & 0xFF;
The best way to do this in NXC with the opcodes available in the underlying VM is to use FlattenVar to convert any type into a string (aka byte array with a null added at the end). It results in a single VM opcode operation where any of the above options which use shifts and logical ANDs and array operations will require dozens of lines of assembly language.
task main()
{
int x = Random(); // 16 bit random number - could be negative
string data;
data = FlattenVar(x); // convert type to byte array with trailing null
NumOut(0, LCD_LINE1, x);
for (int i=0; i < ArrayLen(data)-1; i++)
{
#ifdef __ENHANCED_FIRMWARE
TextOut(0, LCD_LINE2-8*i, FormatNum("0x%2.2x", data[i]));
#else
NumOut(0, LCD_LINE2-8*i, data[i]);
#endif
}
Wait(SEC_4);
}
The best way to get help with LEGO MINDSTORMS and the NXT and Not eXactly C is via the mindboards forums at http://forums.mindboards.net/
Question originally tagged c; this answer may not be applicable to Not eXactly C.
What is the problem with this:
int value;
char bytes[sizeof(int)];
bytes[0] = (value >> 0) & 0xFF;
bytes[1] = (value >> 8) & 0xFF;
bytes[2] = (value >> 16) & 0xFF;
bytes[3] = (value >> 24) & 0xFF;
You can regard it as an unrolled loop. The shift by zero could be omitted; the optimizer would certainly do so. Even though the result of right-shifting a negative value is not defined, there is no problem because this code only accesses the bits where the behaviour is defined.
This code gives the bytes in a little-endian order - the least-significant byte is in bytes[0]. Clearly, big-endian order is achieved by:
int value;
char bytes[sizeof(int)];
bytes[3] = (value >> 0) & 0xFF;
bytes[2] = (value >> 8) & 0xFF;
bytes[1] = (value >> 16) & 0xFF;
bytes[0] = (value >> 24) & 0xFF;
I'm using the following keys to calculate the correct handshake response string:
Key1: 18x 6]8vM;54 *(5: { U1]8 z [ 8
Key2: 1_ tx7X d < nw 334J702) 7]o}` 0
Key3: 54:6d:5b:4b:20:54:32:75
I've calculated Key1 and Key2's values:
Key1: 0947fa63 (hex)
Key2: 0a5510d3
However I'm not sure on what to do next, from what I can gather, you concatenate them and MD5 it, but that doesn't seem to work out i.e.
MD5 hashing: 0947fa630a5510d3546d5b4b20543275
Help!
This is the python code for creating the response hash:
from hashlib import md5
import struct
....
hashed = md5(struct.pack('>II8s', num1, num2, key3)).digest()
In the example num1 and num2 are the numeric values of key1 and key2. key3 is the actual textual string (raw bytes) received.
The struct.pack() call is using big endian mode (for the numeric values) and packing them 4 bytes for each number followed by the 8 byte key3 string (bytes).
See the Documentation for the python struct module.
The C version would look more like this:
/* Pack it big-endian */
buf[0] = (num1 & 0xff000000) >> 24;
buf[1] = (num1 & 0xff0000) >> 16;
buf[2] = (num1 & 0xff00) >> 8;
buf[3] = num1 & 0xff;
buf[4] = (num2 & 0xff000000) >> 24;
buf[5] = (num2 & 0xff0000) >> 16;
buf[6] = (num2 & 0xff00) >> 8;
buf[7] = num2 & 0xff;
strncpy(buf+8, headers->key3, 8);
buf[16] = '\0';
md5_buffer(buf, 16, target);
target[16] = '\0';
md5_buffer is in glibc.
For further reference you can look at working implementations (where the above code came from) of websockify (disclaimer: I wrote websockify).
Here's my version:
https://github.com/boothead/stargate/blob/master/stargate/handshake.py#L104
If you use stargate then all of that nasty stuff is done for you :-)
I need to find whether a number is divisible by 3 without using %, / or *. The hint given was to use atoi() function. Any idea how to do it?
The current answers all focus on decimal digits, when applying the "add all digits and see if that divides by 3". That trick actually works in hex as well; e.g. 0x12 can be divided by 3 because 0x1 + 0x2 = 0x3. And "converting" to hex is a lot easier than converting to decimal.
Pseudo-code:
int reduce(int i) {
if (i > 0x10)
return reduce((i >> 4) + (i & 0x0F)); // Reduces 0x102 to 0x12 to 0x3.
else
return i; // Done.
}
bool isDiv3(int i) {
i = reduce(i);
return i==0 || i==3 || i==6 || i==9 || i==0xC || i == 0xF;
}
[edit]
Inspired by R, a faster version (O log log N):
int reduce(unsigned i) {
if (i >= 6)
return reduce((i >> 2) + (i & 0x03));
else
return i; // Done.
}
bool isDiv3(unsigned i) {
// Do a few big shifts first before recursing.
i = (i >> 16) + (i & 0xFFFF);
i = (i >> 8) + (i & 0xFF);
i = (i >> 4) + (i & 0xF);
// Because of additive overflow, it's possible that i > 0x10 here. No big deal.
i = reduce(i);
return i==0 || i==3;
}
Subtract 3 until you either
a) hit 0 - number was divisible by 3
b) get a number less than 0 - number wasn't divisible
-- edited version to fix noted problems
while n > 0:
n -= 3
while n < 0:
n += 3
return n == 0
Split the number into digits. Add the digits together. Repeat until you have only one digit left. If that digit is 3, 6, or 9, the number is divisible by 3. (And don't forget to handle 0 as a special case).
While the technique of converting to a string and then adding the decimal digits together is elegant, it either requires division or is inefficient in the conversion-to-a-string step. Is there a way to apply the idea directly to a binary number, without first converting to a string of decimal digits?
It turns out, there is:
Given a binary number, the sum of its odd bits minus the sum of its even bits is divisible by 3 iff the original number was divisible by 3.
As an example: take the number 3726, which is divisible by 3. In binary, this is 111010001110. So we take the odd digits, starting from the right and moving left, which are [1, 1, 0, 1, 1, 1]; the sum of these is 5. The even bits are [0, 1, 0, 0, 0, 1]; the sum of these is 2. 5 - 2 = 3, from which we can conclude that the original number is divisible by 3.
A number divisible by 3, iirc has a characteristic that the sum of its digit is divisible by 3. For example,
12 -> 1 + 2 = 3
144 -> 1 + 4 + 4 = 9
The interview question essentially asks you to come up with (or have already known) the divisibility rule shorthand with 3 as the divisor.
One of the divisibility rule for 3 is as follows:
Take any number and add together each digit in the number. Then take that sum and determine if it is divisible by 3 (repeating the same procedure as necessary). If the final number is divisible by 3, then the original number is divisible by 3.
Example:
16,499,205,854,376
=> 1+6+4+9+9+2+0+5+8+5+4+3+7+6 sums to 69
=> 6 + 9 = 15 => 1 + 5 = 6, which is clearly divisible by 3.
See also
Wikipedia/Divisibility rule - has many rules for many divisors
Given a number x.
Convert x to a string. Parse the string character by character. Convert each parsed character to a number (using atoi()) and add up all these numbers into a new number y.
Repeat the process until your final resultant number is one digit long. If that one digit is either 3,6 or 9, the origional number x is divisible by 3.
My solution in Java only works for 32-bit unsigned ints.
static boolean isDivisibleBy3(int n) {
int x = n;
x = (x >>> 16) + (x & 0xffff); // max 0x0001fffe
x = (x >>> 8) + (x & 0x00ff); // max 0x02fd
x = (x >>> 4) + (x & 0x000f); // max 0x003d (for 0x02ef)
x = (x >>> 4) + (x & 0x000f); // max 0x0011 (for 0x002f)
return ((011111111111 >> x) & 1) != 0;
}
It first reduces the number down to a number less than 32. The last step checks for divisibility by shifting the mask the appropriate number of times to the right.
You didn't tag this C, but since you mentioned atoi, I'm going to give a C solution:
int isdiv3(int x)
{
div_t d = div(x, 3);
return !d.rem;
}
bool isDiv3(unsigned int n)
{
unsigned int n_div_3 =
n * (unsigned int) 0xaaaaaaab;
return (n_div_3 < 0x55555556);//<=>n_div_3 <= 0x55555555
/*
because 3 * 0xaaaaaaab == 0x200000001 and
(uint32_t) 0x200000001 == 1
*/
}
bool isDiv5(unsigned int n)
{
unsigned int n_div_5 =
i * (unsigned int) 0xcccccccd;
return (n_div_5 < 0x33333334);//<=>n_div_5 <= 0x33333333
/*
because 5 * 0xcccccccd == 0x4 0000 0001 and
(uint32_t) 0x400000001 == 1
*/
}
Following the same rule, to obtain the result of divisibility test by 'n', we can :
multiply the number by 0x1 0000 0000 - (1/n)*0xFFFFFFFF
compare to (1/n) * 0xFFFFFFFF
The counterpart is that for some values, the test won't be able to return a correct result for all the 32bit numbers you want to test, for example, with divisibility by 7 :
we got 0x100000000- (1/n)*0xFFFFFFFF = 0xDB6DB6DC
and 7 * 0xDB6DB6DC = 0x6 0000 0004,
We will only test one quarter of the values, but we can certainly avoid that with substractions.
Other examples :
11 * 0xE8BA2E8C = A0000 0004, one quarter of the values
17 * 0xF0F0F0F1 = 10 0000 0000 1
comparing to 0xF0F0F0F
Every values !
Etc., we can even test every numbers by combining natural numbers between them.
A number is divisible by 3 if all the digits in the number when added gives a result 3, 6 or 9. For example 3693 is divisible by 3 as 3+6+9+3 = 21 and 2+1=3 and 3 is divisible by 3.
inline bool divisible3(uint32_t x) //inline is not a must, because latest compilers always optimize it as inline.
{
//1431655765 = (2^32 - 1) / 3
//2863311531 = (2^32) - 1431655765
return x * 2863311531u <= 1431655765u;
}
On some compilers this is even faster then regular way: x % 3. Read more here.
well a number is divisible by 3 if all the sum of digits of the number are divisible by 3. so you could get each digit as a substring of the input number and then add them up. you then would repeat this process until there is only a single digit result.
if this is 3, 6 or 9 the number is divisable by 3.
Here is a pseudo-algol i came up with .
Let us follow binary progress of multiples of 3
000 011
000 110
001 001
001 100
001 111
010 010
010 101
011 000
011 011
011 110
100 001
100 100
100 111
101 010
101 101
just have a remark that, for a binary multiple of 3 x=abcdef in following couples abc=(000,011),(001,100),(010,101) cde doest change , hence, my proposed algorithm:
divisible(x):
y = x&7
z = x>>3
if number_of_bits(z)<4
if z=000 or 011 or 110 , return (y==000 or 011 or 110) end
if z=001 or 100 or 111 , return (y==001 or 100 or 111) end
if z=010 or 101 , return (y==010 or 101) end
end
if divisible(z) , return (y==000 or 011 or 110) end
if divisible(z-1) , return (y==001 or 100 or 111) end
if divisible(z-2) , return (y==010 or 101) end
end
C# Solution for checking if a number is divisible by 3
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
int num = 33;
bool flag = false;
while (true)
{
num = num - 7;
if (num == 0)
{
flag = true;
break;
}
else if (num < 0)
{
break;
}
else
{
flag = false;
}
}
if (flag)
Console.WriteLine("Divisible by 3");
else
Console.WriteLine("Not Divisible by 3");
Console.ReadLine();
}
}
}
Here is your optimized solution that every one should know.................
Source: http://www.geeksforgeeks.org/archives/511
#include<stdio.h>
int isMultiple(int n)
{
int o_count = 0;
int e_count = 0;
if(n < 0)
n = -n;
if(n == 0)
return 1;
if(n == 1)
return 0;
while(n)
{
if(n & 1)
o_count++;
n = n>>1;
if(n & 1)
e_count++;
n = n>>1;
}
return isMultiple(abs(o_count - e_count));
}
int main()
{
int num = 23;
if (isMultiple(num))
printf("multiple of 3");
else
printf(" not multiple of 3");
return 0;
}
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
I made the ultimate laugh generator using these rules. Can you implement it in your favorite language in a clever manner?
Rules:
On every iteration, the following transformations occur.
H -> AH
A -> HA
AA -> HA
HH -> AH
AAH -> HA
HAA -> AH
n = 0 | H
n = 1 | AH
n = 2 | HAAH
n = 3 | AHAH
n = 4 | HAAHHAAH
n = 5 | AHAHHA
n = 6 | HAAHHAAHHA
n = 7 | AHAHHAAHHA
n = 8 | HAAHHAAHHAAHHA
n = 9 | AHAHHAAHAHHA
n = ...
Lex/Flex
69 characters. In the text here, I changed tabs to 8 spaces so it would look right, but all those consecutive spaces should be tabs, and the tabs are important, so it comes out to 69 characters.
#include <stdio.h>
%%
HAA|HH|H printf("AH");
AAH|AA|A printf("HA");
For what it's worth, the generated lex.yy.c is 42736 characters, but I don't think that really counts. I can (and soon will) write a pure-C version that will be much shorter and do the same thing, but I feel that should probably be a separate entry.
EDIT:
Here's a more legit Lex/Flex entry (302 characters):
char*c,*t;
#define s(a) t=c?realloc(c,strlen(c)+3):calloc(3,1);if(t)c=t,strcat(c,#a);
%%
free(c);c=NULL;
HAA|HH|H s(AH)
AAH|AA|A s(HA)
%%
int main(void){c=calloc(2,1);if(!c)return 1;*c='H';for(int n=0;n<10;n++)printf("n = %d | %s\n",n,c),yy_scan_string(c),yylex();return 0;}int yywrap(){return 1;}
This does multiple iterations (unlike the last one, which only did one iteration, and had to be manually seeded each time, but produced the correct results) and has the advantage of being extremely horrific-looking code. I use a function macro, the stringizing operator, and two global variables. If you want an even messier version that doesn't even check for malloc() failure, it looks like this (282 characters):
char*c,*t;
#define s(a) t=c?realloc(c,strlen(c)+3):calloc(3,1);c=t;strcat(c,#a);
%%
free(c);c=NULL;
HAA|HH|H s(AH)
AAH|AA|A s(HA)
%%
int main(void){c=calloc(2,1);*c='H';for(int n=0;n<10;n++)printf("n = %d | %s\n",n,c),yy_scan_string(c),yylex();return 0;}int yywrap(){return 1;}
An even worse version could be concocted where c is an array on the stack, and we just give it a MAX_BUFFER_SIZE of some sort, but I feel that's taking this too far.
...Just kidding. 207 characters if we take the "99 characters will always be enough" mindset:
char c[99]="H";
%%
c[0]=0;
HAA|HH|H strcat(c, "AH");
AAH|AA|A strcat(c, "HA");
%%
int main(void){for(int n=0;n<10;n++)printf("n = %d | %s\n",n,c),yy_scan_string(c),yylex();return 0;}int yywrap(){return 1;}
My preference is for the one that works best (i.e. the first one that can iterate until memory runs out and checks its errors), but this is code golf.
To compile the first one, type:
flex golf.l
gcc -ll lex.yy.c
(If you have lex instead of flex, just change flex to lex. They should be compatible.)
To compile the others, type:
flex golf.l
gcc -std=c99 lex.yy.c
Or else GCC will whine about ‘for’ loop initial declaration used outside C99 mode and other crap.
Pure C answer coming up.
MATLAB (v7.8.0):
73 characters (not including formatting characters used to make it look readable)
This script ("haha.m") assumes you have already defined the variable n:
s = 'H';
for i = 1:n,
s = regexprep(s,'(H)(H|AA)?|(A)(AH)?','${[137-$1 $1]}');
end
...and here's the one-line version:
s='H';for i=1:n,s = regexprep(s,'(H)(H|AA)?|(A)(AH)?','${[137-$1 $1]}');end
Test:
>> for n=0:10, haha; disp([num2str(n) ': ' s]); end
0: H
1: AH
2: HAAH
3: AHAH
4: HAAHHAAH
5: AHAHHA
6: HAAHHAAHHA
7: AHAHHAAHHA
8: HAAHHAAHHAAHHA
9: AHAHHAAHAHHA
10: HAAHHAAHHAHAAHHA
A simple translation to Haskell:
grammar = iterate step
where
step ('H':'A':'A':xs) = 'A':'H':step xs
step ('A':'A':'H':xs) = 'H':'A':step xs
step ('A':'A':xs) = 'H':'A':step xs
step ('H':'H':xs) = 'A':'H':step xs
step ('H':xs) = 'A':'H':step xs
step ('A':xs) = 'H':'A':step xs
step [] = []
And a shorter version (122 chars, optimized down to three derivation rules + base case):
grammar=iterate s where{i 'H'='A';i 'A'='H';s(n:'A':m:x)|n/=m=m:n:s x;s(n:m:x)|n==m=(i n):n:s x;s(n:x)=(i n):n:s x;s[]=[]}
And a translation to C++ (182 chars, only does one iteration, invoke with initial state on the command line):
#include<cstdio>
#define o putchar
int main(int,char**v){char*p=v[1];while(*p){p[1]==65&&~*p&p[2]?o(p[2]),o(*p),p+=3:*p==p[1]?o(137-*p++),o(*p++),p:(o(137-*p),o(*p++),p);}return 0;}
Javascript:
120 stripping whitespace and I'm leaving it alone now!
function f(n,s){s='H';while(n--){s=s.replace(/HAA|AAH|HH?|AA?/g,function(a){return a.match(/^H/)?'AH':'HA'});};return s}
Expanded:
function f(n,s)
{
s = 'H';
while (n--)
{
s = s.replace(/HAA|AAH|HH?|AA?/g, function(a) { return a.match(/^H/) ? 'AH' : 'HA' } );
};
return s
}
that replacer is expensive!
Here's a C# example, coming in at 321 bytes if I reduce whitespace to one space between each item.
Edit: In response to #Johannes Rössel comment, I removed generics from the solution to eek out a few more bytes.
Edit: Another change, got rid of all temporary variables.
public static String E(String i)
{
return new Regex("HAA|AAH|HH|AA|A|H").Replace(i,
m => (String)new Hashtable {
{ "H", "AH" },
{ "A", "HA" },
{ "AA", "HA" },
{ "HH", "AH" },
{ "AAH", "HA" },
{ "HAA", "AH" }
}[m.Value]);
}
The rewritten solution with less whitespace, that still compiles, is 158 characters:
return new Regex("HAA|AAH|HH|AA|A|H").Replace(i,m =>(String)new Hashtable{{"H","AH"},{"A","HA"},{"AA","HA"},{"HH","AH"},{"AAH","HA"},{"HAA","AH"}}[m.Value]);
For a complete source code solution for Visual Studio 2008, a subversion repository with the necessary code, including unit tests, is available below.
Repository is here, username and password are both 'guest', without the quotes.
Ruby
This code golf is not very well specified -- I assumed that function returning n-th iteration string is best way to solve it. It has 80 characters.
def f n
a='h'
n.times{a.gsub!(/(h(h|aa)?)|(a(ah?)?)/){$1.nil?? "ha":"ah"}}
a
end
Code printing out n first strings (71 characters):
a='h';n.times{puts a.gsub!(/(h(h|aa)?)|(a(ah?)?)/){$1.nil?? "ha":"ah"}}
Erlang
241 bytes and ready to run:
> erl -noshell -s g i -s init stop
AHAHHAAHAHHA
-module(g).
-export([i/0]).
c("HAA"++T)->"AH"++c(T);
c("AAH"++T)->"HA"++c(T);
c("HH"++T)->"AH"++c(T);
c("AA"++T)->"HA"++c(T);
c("A"++T)->"HA"++c(T);
c("H"++T)->"AH"++c(T);
c([])->[].
i(0,L)->L;
i(N,L)->i(N-1,c(L)).
i()->io:format(i(9,"H"))
Could probably be improved.
Perl 168 characters.
(not counting unnecessary newlines)
perl -E'
($s,%m)=qw[H H AH A HA AA HA HH AH AAH HA HAA AH];
sub p{say qq[n = $_[0] | $_[1]]};p(0,$s);
for(1..9){$s=~s/(H(AA|H)?|A(AH?)?)/$m{$1}/g;p($_,$s)}
say q[n = ...]'
De-obfuscated:
use strict;
use warnings;
use 5.010;
my $str = 'H';
my %map = (
H => 'AH',
A => 'HA',
AA => 'HA',
HH => 'AH',
AAH => 'HA',
HAA => 'AH'
);
sub prn{
my( $n, $str ) = #_;
say "n = $n | $str"
}
prn( 0, $str );
for my $i ( 1..9 ){
$str =~ s(
(
H(?:AA|H)? # HAA | HH | H
|
A(?:AH?)? # AAH | AA | A
)
){
$map{$1}
}xge;
prn( $i, $str );
}
say 'n = ...';
Perl 150 characters.
(not counting unnecessary newlines)
perl -E'
$s="H";
sub p{say qq[n = $_[0] | $_[1]]};p(0,$s);
for(1..9){$s=~s/(?|(H)(?:AA|H)?|(A)(?:AH?)?)/("H"eq$1?"A":"H").$1/eg;p($_,$s)}
say q[n = ...]'
De-obfuscated
#! /usr/bin/env perl
use strict;
use warnings;
use 5.010;
my $str = 'H';
sub prn{
my( $n, $str ) = #_;
say "n = $n | $str"
}
prn( 0, $str );
for my $i ( 1..9 ){
$str =~ s{(?|
(H)(?:AA|H)? # HAA | HH | H
|
(A)(?:AH?)? # AAH | AA | A
)}{
( 'H' eq $1 ?'A' :'H' ).$1
}egx;
prn( $i, $str );
}
say 'n = ...';
Python (150 bytes)
import re
N = 10
s = "H"
for n in range(N):
print "n = %d |"% n, s
s = re.sub("(HAA|HH|H)|AAH|AA|A", lambda m: m.group(1) and "AH" or "HA",s)
Output
n = 0 | H
n = 1 | AH
n = 2 | HAAH
n = 3 | AHAH
n = 4 | HAAHHAAH
n = 5 | AHAHHA
n = 6 | HAAHHAAHHA
n = 7 | AHAHHAAHHA
n = 8 | HAAHHAAHHAAHHA
n = 9 | AHAHHAAHAHHA
Here is a very simple C++ version:
#include <iostream>
#include <sstream>
using namespace std;
#define LINES 10
#define put(t) s << t; cout << t
#define r1(o,a,c0) \
if(c[0]==c0) {put(o); s.unget(); s.unget(); a; continue;}
#define r2(o,a,c0,c1) \
if(c[0]==c0 && c[1]==c1) {put(o); s.unget(); a; continue;}
#define r3(o,a,c0,c1,c2) \
if(c[0]==c0 && c[1]==c1 && c[2]==c2) {put(o); a; continue;}
int main() {
char c[3];
stringstream s;
put("H\n\n");
for(int i=2;i<LINES*2;) {
s.read(c,3);
r3("AH",,'H','A','A');
r3("HA",,'A','A','H');
r2("AH",,'H','H');
r2("HA",,'A','A');
r1("HA",,'A');
r1("AH",,'H');
r1("\n",i++,'\n');
}
}
It's not exactly code-golf (it could be made a lot shorter), but it works. Change LINES to however many lines you want printed (note: it will not work for 0). It will print output like this:
H
AH
HAAH
AHAH
HAAHHAAH
AHAHHA
HAAHHAAHHA
AHAHHAAHHA
HAAHHAAHHAAHHA
AHAHHAAHAHHA
ANSI C99
Coming in at a brutal 306 characters:
#include <stdio.h>
#include <string.h>
char s[99]="H",t[99]={0};int main(){for(int n=0;n<10;n++){int i=0,j=strlen(s);printf("n = %u | %s\n",n,s);strcpy(t,s);s[0]=0;for(;i<j;){if(t[i++]=='H'){t[i]=='H'?i++:t[i+1]=='A'?i+=2:1;strcat(s,"AH");}else{t[i]=='A'?i+=1+(t[i+1]=='H'):1;strcat(s,"HA");}}}return 0;}
There are too many nested ifs and conditional operators for me to effectively reduce this with macros. Believe me, I tried. Readable version:
#include <stdio.h>
#include <string.h>
char s[99] = "H", t[99] = {0};
int main()
{
for(int n = 0; n < 10; n++)
{
int i = 0, j = strlen(s);
printf("n = %u | %s\n", n, s);
strcpy(t, s);
s[0] = 0;
/*
* This was originally just a while() loop.
* I tried to make it shorter by making it a for() loop.
* I failed.
* I kept the for() loop because it looked uglier than a while() loop.
* This is code golf.
*/
for(;i<j;)
{
if(t[i++] == 'H' )
{
// t[i] == 'H' ? i++ : t[i+1] == 'A' ? i+=2 : 1;
// Oh, ternary ?:, how do I love thee?
if(t[i] == 'H')
i++;
else if(t[i+1] == 'A')
i+= 2;
strcat(s, "AH");
}
else
{
// t[i] == 'A' ? i += 1 + (t[i + 1] == 'H') : 1;
if(t[i] == 'A')
if(t[++i] == 'H')
i++;
strcat(s, "HA");
}
}
}
return 0;
}
I may be able to make a shorter version with strncmp() in the future, but who knows? We'll see what happens.
In python:
def l(s):
H=['HAA','HH','H','AAH','AA','A']
L=['AH']*3+['HA']*3
for i in [3,2,1]:
if s[:i] in H: return L[H.index(s[:i])]+l(s[i:])
return s
def a(n,s='H'):
return s*(n<1)or a(n-1,l(s))
for i in xrange(0,10):
print '%d: %s'%(i,a(i))
First attempt: 198 char of code, I'm sure it can get smaller :D
REBOL, 150 characters. Unfortunately REBOL is not a language conducive to code golf, but 150 characters ain't too shabby, as Adam Sandler says.
This assumes the loop variable m has already been defined.
s: "H" r: "" z:[some[["HAA"|"HH"|"H"](append r "AH")|["AAH"|"AA"|"A"](append r "HA")]to end]repeat n m[clear r parse s z print["n =" n "|" s: copy r]]
And here it is with better layout:
s: "H"
r: ""
z: [
some [
[ "HAA" | "HH" | "H" ] (append r "AH")
| [ "AAH" | "AA" | "A" ] (append r "HA")
]
to end
]
repeat n m [
clear r
parse s z
print ["n =" n "|" s: copy r]
]
F#: 184 chars
Seems to map pretty cleanly to F#:
type grammar = H | A
let rec laugh = function
| 0,l -> l
| n,l ->
let rec loop = function
|H::A::A::x|H::H::x|H::x->A::H::loop x
|A::A::H::x|A::A::x|A::x->H::A::loop x
|x->x
laugh(n-1,loop l)
Here's a run in fsi:
> [for a in 0 .. 9 -> a, laugh(a, [H])] |> Seq.iter (fun (a, b) -> printfn "n = %i: %A" a b);;
n = 0: [H]
n = 1: [A; H]
n = 2: [H; A; A; H]
n = 3: [A; H; A; H]
n = 4: [H; A; A; H; H; A; A; H]
n = 5: [A; H; A; H; H; A]
n = 6: [H; A; A; H; H; A; A; H; H; A]
n = 7: [A; H; A; H; H; A; A; H; H; A]
n = 8: [H; A; A; H; H; A; A; H; H; A; A; H; H; A]
n = 9: [A; H; A; H; H; A; A; H; A; H; H; A]