
June 3rd, 2007, 06:05 PM
#16
Re: The haystack is growing...
But... For me, the really interesting question regarding this sequence (and this somehow brings it close to the MUpuzzle in Gödel, Escher Bach...) is: Will this sequence always produce growing strings  or is there a chance that they get shorter along the way?
What I mean is: It's pretty obvious that a row in this sequence can be shorter than its predecessor  for example, the successor of "1111111" will be "71", or the successor of "222333" will be "3233"... But: Will this ever happen with the sequence starting with the seed "1"? And if not: Why not?

June 4th, 2007, 04:30 AM
#17
Re: The haystack is growing...
Originally Posted by gstercken
Nice sequence... The next one is "1113213211", right?
Good cracking !
Regards,
Ramkrishna Pawar

June 4th, 2007, 07:22 AM
#18
Re: The haystack is growing...
Originally Posted by gstercken
Nice sequence... The next one is "1113213211", right?
Well, I knew you know the answer. Let's see if anybody can find the next one.
Will this ever happen with the sequence starting with the seed "1"? And if not: Why not?
Allow me to ask a simpler question: will a string ever encounter a 4?

June 4th, 2007, 07:39 AM
#19
Re: The haystack is growing...
31131211131221
I do not think so (given the starting seed). Without giving away the algorythm, just one key word "Consecutive"
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June 4th, 2007, 07:54 AM
#20
Re: The haystack is growing...
Yes, you got it right.
I checked until the string with length 234241786. No 4 in it.

June 4th, 2007, 07:57 AM
#21
Re: The haystack is growing...
Originally Posted by cilu
Yes, you got it right.
I checked until the string with length 234241786. No 4 in it.
But perhaps with a length of 234241787???
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June 4th, 2007, 08:47 AM
#22
Re: The haystack is growing...
13211311123113112211
I dont think it'll ever have a 4 .... It's just not in..
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June 4th, 2007, 08:54 AM
#23
Re: The haystack is growing...
Another interesting question is can an element in the sequence be collapsed to that it is an compacted representation of the same value?
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June 4th, 2007, 09:04 AM
#24
Re: The haystack is growing...
Originally Posted by TheCPUWizard
Another interesting question is can an element in the sequence be collapsed to that it is an compacted representation of the same value?
I dont know..
Considering the sequence, wont it break when you try to compact any part of it ???
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June 4th, 2007, 10:12 AM
#25
Re: The haystack is growing...
Originally Posted by TheCPUWizard
Another interesting question is can an element in the sequence be collapsed to that it is an compacted representation of the same value?
Not sure I'm following... Can you give an example?

June 4th, 2007, 05:45 PM
#26
Re: The haystack is growing...

June 4th, 2007, 07:17 PM
#27
Re: The haystack is growing...
Originally Posted by gstercken
Interesting... Now I wonder what a proof for that would look like?
<thinking>
// Same for the steadily growing strings, btw... Did your empirical approach support that too, Marius?
Just solve the following for "t":
Code:
D(t+1) = (sigma(K=1,LOG(D(t)*10)LOG(D(t)*10)%1,((D(t)D(t)%10^(LOG(D(t))
LOG(D(t))%1)+sigma(S=1,LOG(D(t))LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)
LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*
10%10^R)%10^(R+1)))/10)(sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*
10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)
%10^(S1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)
*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)(
sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10
^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^(S1))%10^S+.5)*2*(D(t)
D(t)%10^(S1))%10^S))(D(t)D(t)%10^(LOG(D(t))LOG(D(t))%1)+sigma(S=1,
LOG(D(t))LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)
*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)
(sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%
10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^(S1))%10^S+1)%(((
sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%
10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)(sigma(R=1,LOG(D(t)*
10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*
10D(t)*10%10^R)%10^(R+1)))/10)%10^(S1))%10^S+.5)*2*(D(t)D(t)%10^(S
1))%10^S))%10^(K1))%10^K/10^(K1)*100^(2*sigma(N=1,K,(((D(t)D(t)%
10^(LOG(D(t))LOG(D(t))%1)+sigma(S=1,LOG(D(t))LOG(D(t))%1,(((sigma(R=1,
LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10
(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)(sigma(R=1,LOG(D(t)*10)LOG(D(t)*
10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%
10^(R+1)))/10)%10^(S1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%
11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%
10^(R+1)))/10)(sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)
*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^(S1))
%10^S+.5)*2*(D(t)D(t)%10^(S1))%10^S))(D(t)D(t)%10^(LOG(D(t))LOG(D(t))
%1)+sigma(S=1,LOG(D(t))LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)
%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^
(R+1)))/10)(sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%
10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^(S1))%10^
S+1)%(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^
(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)(sigma(R=1,
LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/
10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^(S1))%10^S+.5)*2*(D(t)D(t)%
10^(S1))%10^S))%10^(N1))%10^N+1)%(((D(t)D(t)%10^(LOG(D(t))LOG(D(t))%
1)+sigma(S=1,LOG(D(t))LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*
10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^
R)%10^(R+1)))/10)(sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10
D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^
(S1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10
D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)
(sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%
10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^(S1))%10^S+.5)*2*
(D(t)D(t)%10^(S1))%10^S))(D(t)D(t)%10^(LOG(D(t))LOG(D(t))%1)+
sigma(S=1,LOG(D(t))LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10
)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^
R)%10^(R+1)))/10)(sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*
10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/
10)%10^(S1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,
ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%
10^(R+1)))/10)(sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*
10D(t)*10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))
/10)%10^(S1))%10^S+.5)*2*(D(t)D(t)%10^(S1))%10^S))%10^(N1))%10^
N+.5))))/100)+(sigma(K=1,LOG(D(t)*10)LOG(D(t)*10)%1,100^(1+sigma(N=
1,K1,2*((((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*
10%10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)(
sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%
10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^(N1))%10^N/10^(N
1)+1)%(((sigma(R=1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%
10^(R+1))%10^(R+2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)(sigma(R=
1,LOG(D(t)*10)LOG(D(t)*10)%11,ABS((D(t)*10D(t)*10%10^(R+1))%10^(R+
2)/10(D(t)*10D(t)*10%10^R)%10^(R+1)))/10)%10^(N1))%10^N/10^(N1)+
.5)))))/10)
(Yes, that is really the algorythm!!!)
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June 4th, 2007, 10:01 PM
#28
Re: The haystack is growing...
I made this C++ code to display the first n elements of the series.
Code:
#include <string>
#include <vector>
#include <fstream>
#define LIMIT 20
using namespace std;
string parse(string str1);
string eval(vector<int> vec);
int main(int argc, char* argv[])
{
int i=1;
string str1="1";
ofstream logData;
logData.open("CG.txt",ios::out);
for(i=1;i<LIMIT;i++) {
printf("%s\n",str1.c_str());
logData << str1.c_str() << "\n";
str1 = parse(str1);
}
logData.close();
return 0;
}
string parse(string str1) {
int len = str1.length();
int j;
vector<int> numList;
for(int i=0;i<len;i++) {
j = str1[i]  '0';
numList.push_back(j);
}
return eval(numList);
}
string eval(vector<int> vec) {
int len = vec.size();
int cnt1=1;
vector<int> vec_new;
for(int i=1;i<=len;i++) {
if(vec[i1]==vec[i]) {
cnt1++;
}
if(vec[i1]!=vec[i]) {
vec_new.push_back(cnt1);
vec_new.push_back(vec[i1]);
cnt1=1;
}
}
string str;
for(i=0;i<vec_new.size();i++) {
str+=vec_new[i]+'0';
}
return str;
}
enjoy
/** The only stupid question is the one you never ask. */

June 4th, 2007, 10:25 PM
#29
Re: The haystack is growing...
Sweet
Although I would NOT have put the call to eval inside of parse. This is because parse now does not return a "parsed" string, it returns an "evaluated" one (which is at odds with the function name.....
[picky, picky, picky, picky, picky, picky, picky]
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June 4th, 2007, 10:33 PM
#30
Re: The haystack is growing...
/** The only stupid question is the one you never ask. */
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