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Joined: Jan 2006
Posts: 15
Pikka bird
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OP
Pikka bird
Joined: Jan 2006
Posts: 15 |
Please, I have need of some script that generate succession of 30 in number 0 and 1. And in this succession it musn't have pieces that are repeated 3 times in turn. It musn't have pieces exemplarily like this: 000, 111, 010101, 100110011001 and ...and so on and so forth 
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Joined: Aug 2005
Posts: 525
Fjord artisan
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Fjord artisan
Joined: Aug 2005
Posts: 525 |
Now just explain that in english.
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Joined: Oct 2005
Posts: 1,741
Hoopy frood
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Hoopy frood
Joined: Oct 2005
Posts: 1,741 |
So you want to take 1073741824 unique binary numbers, remove any that have the same sequence 3 times in a row? Is that right? Or maybe you want binary numbers with 30 digits that don't have the same sequence 3 times in a row?
Please give an example of what you want.
-genius_at_work
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Joined: Jan 2006
Posts: 15
Pikka bird
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OP
Pikka bird
Joined: Jan 2006
Posts: 15 |
No, I just want to generate a row of 0 and 1. Not other figures. The combinations like 000, 010101, 100110011001 are forbidden. There is element "01" 3 times in 010101 and this is forbidden.And i don't know how to make it in mirc scripting  In Pascal the code is like this: program genstr;
type strarr = array[1..30] of integer;
var i,j,k : integer;
br, start: integer;
B : strarr;
function min (r: integer; s: integer):integer;
begin;
if r >= s then min := s
else min := r;
end;
function noseq (var A: strarr; usedlen:integer;el:integer ): boolean;
var i, j, k, t, q, br, l, flag: integer;
helparr : array [1..10] of integer;
begin
A[usedlen] := el;
q:= usedlen div 3;
for j:= 1 to q do
for i:= 1 to j do
begin
br:= 0;
for l := 1 to j do helparr [l] := 2;
flag := 0;
for k := i to usedlen do
begin
if (k mod q) = 0 then t:= q
else t := k mod q;
if k mod q = 1 then
begin
if flag = q then
begin
br := br+1;
if br= 2 then noseq := false
else br := 0;
end
else flag := 0;
end;
if A[k]= helparr[t] then flag := flag + 1;
helparr [t] := A[k];
end;
if br < 2 then noseq := true;
end;
end;
procedure newel (var A:strarr;var i:integer;start: integer; el:integer);
begin
if noseq (A, i, el) = true then
begin
A[i] := el;
i:= i+1;
if i < start then newel (A, i , start, (random(256)mod 2));
end
else
begin
if noseq(A, i, ((el + 1) mod 2)) = true then
begin
A[i]:= (el + 1) mod 2;
i := i+ 1;
if i < start then newel (A, i , start, (random(256) mod 2));
end
else
begin
i:=i - 1;
newel(A, i , start, (A[i] mod 2));
end
end;
end;
begin
br := 0;
i := 1;
write ('String generation is on');writeln('');
repeat
k:=( random (256) mod 2);
start := i;
newel(B, i, start, k);
br:= br + 1;
until (i > 29) or (br > 1000 );
if br > 1000 then write (' Can not generate str')
else begin
for j := 1 to 30 do write (B[j]);
end;
end.
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Joined: Jan 2006
Posts: 15
Pikka bird
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OP
Pikka bird
Joined: Jan 2006
Posts: 15 |
Or maybe you want binary numbers with 30 digits that don't have the same sequence 3 times in a row?
-genius_at_work This is right  but the row of digits must be only with 0 and 1. Some other like 2, 3, 4, 5, 6, 7, 8, 9 are unneeded.
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Joined: Jan 2006
Posts: 15
Pikka bird
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OP
Pikka bird
Joined: Jan 2006
Posts: 15 |
or some like this in C: #include <stdlib.h>
#include <stdio.h>
#include <string.h>
typedef unsigned short USHORT;
#define BIN_LEN 30
#define FALSE 0
#define TRUE !FALSE
int SearchThreeRepeats( USHORT i, char *BinStr )
{
USHORT p = 1;
char Buf[(BIN_LEN / 3) + 1],
OldBuf[(BIN_LEN / 3) + 1] = "";
USHORT j;
for ( j = 0; j < BIN_LEN; j += i ) {
strncpy( Buf, BinStr + j, i );
Buf[i] = 0;
if ( strcmp( OldBuf, Buf ) == NULL )
p++;
else
p = 1;
strcpy( OldBuf, Buf );
if ( p == 3 )
return( TRUE );
}
return( FALSE );
}
int CheckRepeatTriad( char *BinStr )
{
USHORT i;
for ( i = 1; i <= BIN_LEN / 3; i++ ) {
if ( SearchThreeRepeats( i, BinStr ) )
return( TRUE );
}
return( FALSE );
}
int main(int argc, char* argv[])
{
USHORT i;
char BinStr[BIN_LEN + 1] = "";
char c[2];
randomize();
do {
while ( TRUE ) {
for ( i = 0; i < BIN_LEN; i++ ) {
itoa( rand() % 2, c, 2 );
BinStr[i] = c[0];
}
BinStr[BIN_LEN] = 0;
if ( ! CheckRepeatTriad( BinStr ) )
break;
}
printf( "Binary String = %s\nNew String (Y/N)?\n", BinStr );
} while ( getch() != 'Y' );
return 0;
} (sorry for posting other codes :tongue:)
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Joined: Oct 2005
Posts: 1,741
Hoopy frood
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Hoopy frood
Joined: Oct 2005
Posts: 1,741 |
This should be simple to create. If someone hasn't already done it when I get off work, I will.
-genius_at_work
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Joined: Sep 2003
Posts: 4,230
Hoopy frood
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Hoopy frood
Joined: Sep 2003
Posts: 4,230 |
alias make.bin {
var %bin = $+($rand(0,1),$rand(0,1))
var %size = 30
var %i = 3
while (%i <= %size) {
var %new = $rand(0,1)
if ($invalid.binary($+(%bin,%new))) {
var %failsize = $v1
var %new = $iif(%new,0,1)
if ($invalid.binary($+(%bin,%new))) {
if ($v1 > %failsize) { var %failsize = $v1 }
dec %i %failsize
var %bin = $left($+(%bin,%new),$calc(-1 * %failsize))
var %new
}
}
var %bin = $+(%bin,%new)
inc %i
}
return %bin
}
alias -l invalid.binary {
var %max = $int($calc($len($1) / 3))
var %c = 1
while (%c <= %max) {
if (($right($1,%c) == $left($right($1,$calc(%c * 2)),%c)) && ($right($1,%c) == $left($right($1,$calc(%c * 3)),%c))) { return %c }
inc %c
}
}usage //echo $make.bin * code was tested to work * but i dont garrentee it works as specified, i couldnt be shagged writting a checking script, but since the invalid.binary routine should check it at 30 chars anyway I would assume its valid results. Someone might like to validate it if they can be shagged.
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Joined: Nov 2004
Posts: 842
Hoopy frood
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Hoopy frood
Joined: Nov 2004
Posts: 842 |
Someone might like to validate it if they can be shagged.
[00:23:03] 100110110010011001101011001101
[00:23:07] 100101001100101001001101100110
[00:23:11] 001001101001010011001101101011
[00:23:13] 110110101100110110011011010010
[00:23:15] 101101100101100101101100100110
[00:23:18] 011001101011001100101101100110
[00:23:19] 001011011001001010010110101101
[00:23:21] 001001101100100110010010100101
[00:23:41] 110110011011010110011010110011
[00:23:42] 001010010011010110010100100110
[00:23:42] 011001010011011010110011001011
[00:23:42] 011001100101001101011010010110
[00:23:43] 101101001100101001100100101100
[00:23:43] 001010010100110010100110010010
[00:23:44] 010100110110100100110110010110
[00:23:44] 110010100110110011011010110010
[00:23:45] 110110100110110010110110011011
[00:23:45] 001001100100110010110011001010
[00:23:46] 001100110110010011001001010010
Works flawlessly for me ...
What do you do at the end of the world? Are you busy? Will you save us?
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Joined: Oct 2005
Posts: 1,741
Hoopy frood
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Hoopy frood
Joined: Oct 2005
Posts: 1,741 |
Looks like my work here is done.. :P
-genius_at_work
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Joined: Jan 2006
Posts: 15
Pikka bird
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OP
Pikka bird
Joined: Jan 2006
Posts: 15 |
Hehe Thank you very! It's very good work.
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Joined: Jan 2006
Posts: 15
Pikka bird
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OP
Pikka bird
Joined: Jan 2006
Posts: 15 |
and by the way... in my test sometimes there aren't 30 digits. alias makebin {
var %zzz = $make.bin
if $len( %zzz ) == 30 {
echo 12 $len( %zzz ) -> %zzz
}
else $makebin
}
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Joined: Sep 2003
Posts: 4,230
Hoopy frood
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Hoopy frood
Joined: Sep 2003
Posts: 4,230 |
Your test makes no sence for a start so is inconclusive as a result. Your echoing in color 12 if it matches 30 characters. If not your acttually running the generation of a new binary number and then attempting to execute it as a command. Try this code. alias makebin {
if (!$0) { .timer.makebin -m 100000 1 makebin - }
else {
var %zzz = $make.bin
echo -s $calc(100000 - $timer($ctimer).reps) : %zzz : $len(%zzz)
if ($len(%zzz) != 30) { .timer.makebin off }
}
}This should not lag or freeze up mirc and should attempt 100,000 generations, stopping if it locates one thats not 30 in length. * PS: i ran this and it did the whole 100,000 with out error (well ok truth its still doing em but i cant be bothered waiting)
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