Discussion:
The Birth Of AI, and Computer Consciousness.
(too old to reply)
Corey White
2016-03-17 14:34:39 UTC
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I've been writing programs that solve themselves. It works on the same principle used in other machine learning systems. A program can try to find its way through a maze and fail dozens of times, but eventually it solves the problem. Well, instead of learning this way, computers can solve problems together and solve problems even faster than could be done with any one machine. It is called cluster computing.

If it takes a computer 1,000 tries to solve a maze successfully, all you need to do is run 1,000 instances of the program on 1,000 different computers. That's cluster computing. It also makes writing software easier. Instead of designing algorithms, you just set a test condition, and tell the computers to randomly guess the answer and check if it fits the answer. Just like when someone wins the lottery, if a million computers are working together to find a solution it ceases to become guess work.

I'm working on a program now to map numbers to words with brute force. I'm trying to prove that numbers are basically just variables, and any problem can have more than one answer. I'm sure its true, but the education system doesn't want people to know this.
Mike Duffy
2016-03-17 15:06:26 UTC
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Post by Corey White
I'm working on a program now to map numbers to words with brute force.
I've been toying with the idea for years to do something similar, I think,
if I understand what you are saying. But my idea was just to convert text
to equivalent IPA (International Phonetic Alphabet) encoding in order to
run time-domain fourier analysis on poetry to see if the entropy content
changes when prose is lyricized.
Post by Corey White
I'm trying to prove that numbers are basically just variables,
and any problem can have more than one answer.
These two statements seem unrelated, but true. Although I would say that
variables are basically just numbers.

Are you the same Corey I used to kibbutz with years ago? My memory modules
are failing and I can't remember if he was White or not. If you are,
welcome back. It's been moribund around here. I've practically given up on
running into anyone who has magickal beliefs anywhere similar to my own.
--
http://mduffy.x10host.com/index.htm
Corey White
2016-03-19 22:21:11 UTC
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Yes, it is me. I've been working on this idea some more, and hope you might be able to help.

I'm trying to brute force one possible answer to a problem. The problem is to assign numbers to the first ten words that we normally use to talk about the numbers one through ten. To do this each letter is given one unique number. The numerical values for all the letters in a word are added together. So o+n+e=1, and t+w+o=2. That's all there is too it!

Here is a chart I was able to create with the following program. It solves the problem for the numbers 0-9, but unfortunately I am still waiting to see if the computer can crank out the data for "ten". Any comments or help would be appreciated!!!
tnx.

zero 25 40 -50 -15
one -15 -24 40
two 56 -39 -15
three 56 -83 -50 40 40
four 46 -15 23 -50
five 46 17 -98 40
six 49 17 -60
seven 49 40 -98 40 -24
eight 40 17 -22 -83 56
nine -24 17 -24 40
ten ? ? ?
(Numbers next to words relate to the letters in the words)

//C++ Program Follows
#include <iostream>
#include <cstdlib>
#include <ctime>

using namespace std;

int main()
{

srand((int) time(0));
int total = 0;
int breakpoint = 0;
int range = 0;
while(breakpoint!=-1){
cout<<"Brakepoint: ";
cin>>breakpoint;
cout<<"Range: ";
cin>>range;

int a = 999001;
int b = 999002;
int c = 999003;
int d = 999004;
int e = 999005;
int f = 999006;
int g = 999007;
int h = 999008;
int i = 999009;
int j = 999010;
int k = 999011;
int l = 999012;
int m = 999013;
int n = 999014;
int o = 999015;
int p = 999016;
int q = 999017;
int r = 999018;
int s = 999019;
int t = 999020;
int u = 999021;
int v = 999026;
int w = 999027;
int x = 100028;
int y = 100029;
int z = 100030;

while(1){

total=-1;
int cnt=0;

while(total != 0){
cnt++;
if(cnt>500){
goto esc;
}
z = (rand() % range)-(range/2);
e = (rand() % range)-(range/2);
r = (rand() % range)-(range/2);
o = (rand() % range)-(range/2);
total = z+e+r+o;
}
if(breakpoint==0){
cout<<"Done @ 0: "<<total<<endl;
break;
}
cnt=0;

while(total != 1 || n==z || z==o || o==r || r==e){
cnt++;
if(cnt>500){
goto esc;
}
n = (rand() % range)-(range/2);
total = o+n+e;

}
if(breakpoint==1){
cout<<"Done @ 1: "<<total<<endl;
break;
}
cnt=0;

while(total != 2 || t==w || w==o || o==n || n==e || e==z || z==r){
cnt++;
if(cnt>500){
goto esc;
}
t = (rand() % range)-(range/2);
w = (rand() % range)-(range/2);
total=t+w+o;

}


cnt=0;

if(breakpoint==2){
cout<<"Done @ 2: "<<total<<endl;
break;
}

while(total !=3 || t==h || h==r || r==e || e==w || w==o || o==n){
cnt++;
if(cnt>500){
goto esc;
}
h = (rand() % range)-(range/2);

total=t+h+r+e+e;
}
if(breakpoint==3){
cout<<"Done @ 3: "<<total<<endl;
break;
}
cnt=0;

while(total !=4 || f==o || o==u || u==r || r==n || n==e || e==t || t==w || w==h){

cnt++;
if(cnt>500){
goto esc;
}

f = (rand() % range)-(range/2);
u = (rand() % range)-(range/2);
total=f+o+u+r;

}

cnt=0;
if(breakpoint==4){
cout<<"Done @ 4: "<<total<<endl;
break;
}

while(total !=5 || f==i || i==v || v==e || e==o || o==n || n==t || t==w || w==h || h==r || r==f || f==u){
cnt++;
if(cnt>500){
goto esc;
}

i = (rand() % range)-(range/2);
v = (rand() % range)-(range/2);
total=f+i+v+e;


}
cnt=0;
if(breakpoint==5){
cout<<"Done @ 5: "<<total<<endl;
break;
}
while(total !=6 || s==x || x==f || f==i || i==v || v==e || e==o || o==n || n==t || t==w || w==h || h==r || r==f || f==u){
cnt++;
if(cnt>500){
goto esc;
}
s = (rand() % range)-(range/2);
x = (rand() % range)-(range/2);
total=s+i+x;


}
cnt=0;
if(breakpoint==6){
cout<<"Done @ 6: "<<total<<endl;
break;
}
total=s+e+v+e+n;
if(total!=7){
goto esc;
}
if(breakpoint==7){
cout<<"Done @ 7: "<<total<<endl;
break;
}
while(total !=8 || v==g || g==s || s==x || x==f || f==i || i==v || v==e || e==o || o==n || n==t || t==w || w==h || h==r || r==f || f==u){
cnt++;
if(cnt>500){
goto esc;
}
g = (rand() % range)-(range/2);
total=e+i+g+h+t;
cnt++;
}
cnt=0;
if(breakpoint==8){
cout<<"Done @ 8 :"<<total<<endl;
break;
}
total=n+i+n+e;
if(total!=9){
goto esc;
}
if(breakpoint==9){
cout<<"Done @ 9: "<<total<<endl;
break;
}
total=t+e+n;
if(total!=10){
goto esc;
}
if(breakpoint==10){
cout<<"Done @ 10: "<<total<<endl;
break;
}

esc:
int foo=1;
}
cout<<"ZERO: "<<z<<" "<<e<<" "<<r<<" "<<o<<" "<<z+e+r+o<<endl;
cout<<"ONE: "<<o<<" "<<n<<" "<<e<<" "<<o+n+e<<endl;
cout<<"TWO: "<<t<<" "<<w<<" "<<o<<" "<<t+w+o<<endl;
cout<<"THREE: "<<t<<" "<<h<<" "<<r<<" "<<e<<" "<<e<<" "<<t+h+r+e+e<<endl;
cout<<"FOUR: "<<f<<" "<<o<<" "<<u<<" "<<r<<" "<<f+o+u+r<<endl;
cout<<"FIVE: "<<f<<" "<<i<<" "<<v<<" "<<e<<" "<<f+i+v+e<<endl;
cout<<"SIX: "<<s<<" "<<i<<" "<<x<<" "<<s+i+x<<endl;
cout<<"SEVEN: "<<s<<" "<<e<<" "<<v<<" "<<e<<" "<<n<<" "<<s+e+v+e+n<<endl;
cout<<"EIGHT: "<<e<<" "<<i<<" "<<g<<" "<<h<<" "<<t<<" "<<e+i+g+h+t<<endl;
cout<<"NINE: "<<n<<" "<<i<<" "<<n<<" "<<e<<" "<<n+i+n+e<<endl;
cout<<"TEN: "<<t<<" "<<e<<" "<<n<<" "<<t+e+n<<endl;

}
return 0;
}
Mike Duffy
2016-03-19 23:55:36 UTC
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Post by Corey White
Yes, it is me. I've been working on this idea some more,
and hope you might be able to help.
I'm trying to brute force one possible answer to a problem.
The problem is to assign numbers to the first ten words that
we normally use to talk about the numbers one through ten.
To do this each letter is given one unique number.
The numerical values for all the letters in a word are added together.
So o+n+e=1, and t+w+o=2. That's all there is to it!
Here is a chart I was able to create with the following program.
It solves the problem for the numbers 0-9, but unfortunately
I am still waiting to see if the computer can crank out the
data for "ten". Any comments or help would be appreciated!!!
Good grief!! You're using a 'monte carlo' algorithm to solve a system of
linear equations!

https://en.wikipedia.org/wiki/Monte_Carlo_method
https://en.wikipedia.org/wiki/Linear_algebra


Without performing an exhaustive analysis, It seems at first glance that:

1) Using a Monte Carlo method to guess at the values gains you nothing over
using a nested series of expansions. Also, it prevents you from suspending
and restarting the simulation. But then again, it does not cost anything in
terms of the average time to find a solution.

2) There are exactly twenty-six letters in the English alphabet. But all of
the numbers up to 'twenty five' have less than eleven letters. Thus you are
not guaranteed to be able to come up with a solution for 'ten'. (Because
you started at 'zero', not 'one'). This restriction is due to the 'degrees
of freedom' within the linear system.

https://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29

Normally, a linear will be guranteed a solution if there are more degrees
of freedom than the number of linear equations defining the system. In this
case, there are nominally 26 degrees of freedom, with only 11 linear
equations.

But, I assume that you want a *diophantine* solution. The only easy way to
do this is to run a simulation such as you have done. But as I pointed out,
you have no a priori indication of whether or not it is a solvable problem.

https://en.wikipedia.org/wiki/Undecidable_problem


I had a similar problem when I wrote the Javascript program for my 'phone
poem generator:

http://mduffy.x10host.com/poetry.htm#quickIDX16

(New users will need to pass the poetry 'exam' first.)

I had no way to know if the algorithm would handle all 10-digit numbers, so
I had to write a program that actually calculated the poem factors and
re-evaluated the phone number. It took a few days to run the simulation to
10 digits (Phone number = 999-999-9999). I have no idea at what point the
algorithm fails.


In your case, I find my wondering where you got 999001 to 999027 and 100028
to 100030. Does it not make more sense to start with -13 to +13? Or perhaps
+/- a half of the fifth power of 26 (Because of a max of 5 chars in numbers
0 to 10.)
--
http://mduffy.x10host.com/index.htm
Corey White
2016-03-20 03:02:39 UTC
Permalink
Raw Message
Post by Mike Duffy
Post by Corey White
Yes, it is me. I've been working on this idea some more,
and hope you might be able to help.
I'm trying to brute force one possible answer to a problem.
The problem is to assign numbers to the first ten words that
we normally use to talk about the numbers one through ten.
To do this each letter is given one unique number.
The numerical values for all the letters in a word are added together.
So o+n+e=1, and t+w+o=2. That's all there is to it!
Here is a chart I was able to create with the following program.
It solves the problem for the numbers 0-9, but unfortunately
I am still waiting to see if the computer can crank out the
data for "ten". Any comments or help would be appreciated!!!
Good grief!! You're using a 'monte carlo' algorithm to solve a system of
linear equations!
https://en.wikipedia.org/wiki/Monte_Carlo_method
https://en.wikipedia.org/wiki/Linear_algebra
1) Using a Monte Carlo method to guess at the values gains you nothing over
using a nested series of expansions. Also, it prevents you from suspending
and restarting the simulation. But then again, it does not cost anything in
terms of the average time to find a solution.
2) There are exactly twenty-six letters in the English alphabet. But all of
the numbers up to 'twenty five' have less than eleven letters. Thus you are
not guaranteed to be able to come up with a solution for 'ten'. (Because
you started at 'zero', not 'one'). This restriction is due to the 'degrees
of freedom' within the linear system.
https://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29
Normally, a linear will be guranteed a solution if there are more degrees
of freedom than the number of linear equations defining the system. In this
case, there are nominally 26 degrees of freedom, with only 11 linear
equations.
But, I assume that you want a *diophantine* solution. The only easy way to
do this is to run a simulation such as you have done. But as I pointed out,
you have no a priori indication of whether or not it is a solvable problem.
https://en.wikipedia.org/wiki/Undecidable_problem
I had a similar problem when I wrote the Javascript program for my 'phone
http://mduffy.x10host.com/poetry.htm#quickIDX16
(New users will need to pass the poetry 'exam' first.)
I had no way to know if the algorithm would handle all 10-digit numbers, so
I had to write a program that actually calculated the poem factors and
re-evaluated the phone number. It took a few days to run the simulation to
10 digits (Phone number = 999-999-9999). I have no idea at what point the
algorithm fails.
In your case, I find my wondering where you got 999001 to 999027 and 100028
to 100030. Does it not make more sense to start with -13 to +13? Or perhaps
+/- a half of the fifth power of 26 (Because of a max of 5 chars in numbers
0 to 10.)
--
http://mduffy.x10host.com/index.htm
Here is a chart with the solution. I suspect this could be extended out as far as you want. You just need a supercomputer built out of raspberry pi's. (or leave your computer running when you leave the house)

zero 43 57 -56 -44
one -44 -12 57
two -35 81 -44
three -35 -20 -56 57 57
four 62 -44 42 -56
five 62 -24 -90 57
six -5 -24 35
seven -5 57 -90 57 -12
eight 57 -24 30 -20 -35
nine -12 -24 -12 57
ten -35 57 -12
(Numbers next to words are assigned to corresponding letters in the words)

This is the same program I wrote that solved the problem: http://ideone.com/3Gbjjr
Mike Duffy
2016-03-20 16:33:44 UTC
Permalink
Raw Message
Post by Corey White
Post by Mike Duffy
Post by Corey White
Yes, it is me. I've been working on this idea some more,
and hope you might be able to help.
I'm trying to brute force one possible answer to a problem.
The problem is to assign numbers to the first ten words that
we normally use to talk about the numbers one through ten.
To do this each letter is given one unique number.
The numerical values for all the letters in a word are added together.
So o+n+e=1, and t+w+o=2. That's all there is to it!
Here is a chart I was able to create with the following program.
It solves the problem for the numbers 0-9, but unfortunately
I am still waiting to see if the computer can crank out the
data for "ten". Any comments or help would be appreciated!!!
Good grief!! You're using a 'monte carlo' algorithm to solve a system of
linear equations!
https://en.wikipedia.org/wiki/Monte_Carlo_method
https://en.wikipedia.org/wiki/Linear_algebra
1) Using a Monte Carlo method to guess at the values gains you nothing over
using a nested series of expansions. Also, it prevents you from suspending
and restarting the simulation. But then again, it does not cost anything in
terms of the average time to find a solution.
2) There are exactly twenty-six letters in the English alphabet. But all of
the numbers up to 'twenty five' have less than eleven letters. Thus you are
not guaranteed to be able to come up with a solution for 'ten'. (Because
you started at 'zero', not 'one'). This restriction is due to the 'degrees
of freedom' within the linear system.
https://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29
Normally, a linear will be guranteed a solution if there are more degrees
of freedom than the number of linear equations defining the system. In this
case, there are nominally 26 degrees of freedom, with only 11 linear
equations.
But, I assume that you want a *diophantine* solution. The only easy way to
do this is to run a simulation such as you have done. But as I pointed out,
you have no a priori indication of whether or not it is a solvable problem.
https://en.wikipedia.org/wiki/Undecidable_problem
I had a similar problem when I wrote the Javascript program for my 'phone
http://mduffy.x10host.com/poetry.htm#quickIDX16
(New users will need to pass the poetry 'exam' first.)
I had no way to know if the algorithm would handle all 10-digit numbers, so
I had to write a program that actually calculated the poem factors and
re-evaluated the phone number. It took a few days to run the simulation to
10 digits (Phone number = 999-999-9999). I have no idea at what point the
algorithm fails.
In your case, I find my wondering where you got 999001 to 999027 and 100028
to 100030. Does it not make more sense to start with -13 to +13? Or perhaps
+/- a half of the fifth power of 26 (Because of a max of 5 chars in numbers
0 to 10.)
--
http://mduffy.x10host.com/index.htm
Here is a chart with the solution.
I suspect this could be extended out as far as you want.
zero 43 57 -56 -44
one -44 -12 57
two -35 81 -44
three -35 -20 -56 57 57
four 62 -44 42 -56
five 62 -24 -90 57
six -5 -24 35
seven -5 57 -90 57 -12
eight 57 -24 30 -20 -35
nine -12 -24 -12 57
ten -35 57 -12
I do not agree. Eventually, you will have more equations than variables.
'eleven' is easy, because it adds one new variable (L) to an existing
solution. It's trivial to solve for L = -35:

eleven 57 -35 57 -90 57 -35

'twelve' is easy as well because you add one new variable (W) with one new
equation. It's trivial to solve for W = 58:

twelve -35 58 57 -35 -90 57.

But now, you have a big problem with thirteen. You need to add a new
equation to your existing system without adding any new variables, because
'thirteen' is spelled using the a subset of the existing variables:

thirteen -35 -20 -24 -56 -35 57 57 -12 = -68, NOT 13.

So now, you need to search for a completely new set of diaphantine
solutions. I am guessing that some might exist if you use values higher in
magnitude than 90, because you already have 16 variables in play
(zerontwhfuivsxgl) with only 14 equations.

'fourteen' will be problematic though, because 'four'=4, thus 'teen'=10.
But if 'teen'=10 and 'ten'=10, this forces 'e' to zero. You've reached the
end of the road for this puzzle, unless you can find a solution for
'thirteen' with E=0.

(And if you can then find a solution for 'fourteen' as well, you already
have 'sixteen', 'seventeen', 'eighteen', and 'nineteen'.)
--
http://mduffy.x10host.com/index.htm
Corey White
2016-03-20 16:45:45 UTC
Permalink
Raw Message
Post by Mike Duffy
Post by Corey White
Post by Mike Duffy
Post by Corey White
Yes, it is me. I've been working on this idea some more,
and hope you might be able to help.
I'm trying to brute force one possible answer to a problem.
The problem is to assign numbers to the first ten words that
we normally use to talk about the numbers one through ten.
To do this each letter is given one unique number.
The numerical values for all the letters in a word are added together.
So o+n+e=1, and t+w+o=2. That's all there is to it!
Here is a chart I was able to create with the following program.
It solves the problem for the numbers 0-9, but unfortunately
I am still waiting to see if the computer can crank out the
data for "ten". Any comments or help would be appreciated!!!
Good grief!! You're using a 'monte carlo' algorithm to solve a system of
linear equations!
https://en.wikipedia.org/wiki/Monte_Carlo_method
https://en.wikipedia.org/wiki/Linear_algebra
1) Using a Monte Carlo method to guess at the values gains you nothing over
using a nested series of expansions. Also, it prevents you from suspending
and restarting the simulation. But then again, it does not cost anything in
terms of the average time to find a solution.
2) There are exactly twenty-six letters in the English alphabet. But all of
the numbers up to 'twenty five' have less than eleven letters. Thus you are
not guaranteed to be able to come up with a solution for 'ten'. (Because
you started at 'zero', not 'one'). This restriction is due to the 'degrees
of freedom' within the linear system.
https://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29
Normally, a linear will be guranteed a solution if there are more degrees
of freedom than the number of linear equations defining the system. In this
case, there are nominally 26 degrees of freedom, with only 11 linear
equations.
But, I assume that you want a *diophantine* solution. The only easy way to
do this is to run a simulation such as you have done. But as I pointed out,
you have no a priori indication of whether or not it is a solvable problem.
https://en.wikipedia.org/wiki/Undecidable_problem
I had a similar problem when I wrote the Javascript program for my 'phone
http://mduffy.x10host.com/poetry.htm#quickIDX16
(New users will need to pass the poetry 'exam' first.)
I had no way to know if the algorithm would handle all 10-digit numbers, so
I had to write a program that actually calculated the poem factors and
re-evaluated the phone number. It took a few days to run the simulation to
10 digits (Phone number = 999-999-9999). I have no idea at what point the
algorithm fails.
In your case, I find my wondering where you got 999001 to 999027 and 100028
to 100030. Does it not make more sense to start with -13 to +13? Or perhaps
+/- a half of the fifth power of 26 (Because of a max of 5 chars in numbers
0 to 10.)
--
http://mduffy.x10host.com/index.htm
Here is a chart with the solution.
I suspect this could be extended out as far as you want.
zero 43 57 -56 -44
one -44 -12 57
two -35 81 -44
three -35 -20 -56 57 57
four 62 -44 42 -56
five 62 -24 -90 57
six -5 -24 35
seven -5 57 -90 57 -12
eight 57 -24 30 -20 -35
nine -12 -24 -12 57
ten -35 57 -12
I do not agree. Eventually, you will have more equations than variables.
'eleven' is easy, because it adds one new variable (L) to an existing
eleven 57 -35 57 -90 57 -35
'twelve' is easy as well because you add one new variable (W) with one new
twelve -35 58 57 -35 -90 57.
But now, you have a big problem with thirteen. You need to add a new
equation to your existing system without adding any new variables, because
thirteen -35 -20 -24 -56 -35 57 57 -12 = -68, NOT 13.
So now, you need to search for a completely new set of diaphantine
solutions. I am guessing that some might exist if you use values higher in
magnitude than 90, because you already have 16 variables in play
(zerontwhfuivsxgl) with only 14 equations.
'fourteen' will be problematic though, because 'four'=4, thus 'teen'=10.
But if 'teen'=10 and 'ten'=10, this forces 'e' to zero. You've reached the
end of the road for this puzzle, unless you can find a solution for
'thirteen' with E=0.
(And if you can then find a solution for 'fourteen' as well, you already
have 'sixteen', 'seventeen', 'eighteen', and 'nineteen'.)
--
http://mduffy.x10host.com/index.htm
That's why I just need to use a super computer cluster of raspberry pi's
The One
2016-03-20 19:59:21 UTC
Permalink
Raw Message
Post by Mike Duffy
Post by Corey White
Post by Mike Duffy
Post by Corey White
Yes, it is me. I've been working on this idea some more,
and hope you might be able to help.
I'm trying to brute force one possible answer to a problem.
The problem is to assign numbers to the first ten words that
we normally use to talk about the numbers one through ten.
To do this each letter is given one unique number.
The numerical values for all the letters in a word are added together.
So o+n+e=1, and t+w+o=2. That's all there is to it!
Here is a chart I was able to create with the following program.
It solves the problem for the numbers 0-9, but unfortunately
I am still waiting to see if the computer can crank out the
data for "ten". Any comments or help would be appreciated!!!
Good grief!! You're using a 'monte carlo' algorithm to solve a system of
linear equations!
https://en.wikipedia.org/wiki/Monte_Carlo_method
https://en.wikipedia.org/wiki/Linear_algebra
1) Using a Monte Carlo method to guess at the values gains you nothing over
using a nested series of expansions. Also, it prevents you from suspending
and restarting the simulation. But then again, it does not cost anything in
terms of the average time to find a solution.
2) There are exactly twenty-six letters in the English alphabet. But all of
the numbers up to 'twenty five' have less than eleven letters. Thus you are
not guaranteed to be able to come up with a solution for 'ten'. (Because
you started at 'zero', not 'one'). This restriction is due to the 'degrees
of freedom' within the linear system.
https://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29
Normally, a linear will be guranteed a solution if there are more degrees
of freedom than the number of linear equations defining the system. In this
case, there are nominally 26 degrees of freedom, with only 11 linear
equations.
But, I assume that you want a *diophantine* solution. The only easy way to
do this is to run a simulation such as you have done. But as I pointed out,
you have no a priori indication of whether or not it is a solvable problem.
https://en.wikipedia.org/wiki/Undecidable_problem
I had a similar problem when I wrote the Javascript program for my 'phone
http://mduffy.x10host.com/poetry.htm#quickIDX16
(New users will need to pass the poetry 'exam' first.)
I had no way to know if the algorithm would handle all 10-digit numbers, so
I had to write a program that actually calculated the poem factors and
re-evaluated the phone number. It took a few days to run the simulation to
10 digits (Phone number = 999-999-9999). I have no idea at what point the
algorithm fails.
In your case, I find my wondering where you got 999001 to 999027 and 100028
to 100030. Does it not make more sense to start with -13 to +13? Or perhaps
+/- a half of the fifth power of 26 (Because of a max of 5 chars in numbers
0 to 10.)
--
http://mduffy.x10host.com/index.htm
Here is a chart with the solution.
I suspect this could be extended out as far as you want.
zero 43 57 -56 -44
one -44 -12 57
two -35 81 -44
three -35 -20 -56 57 57
four 62 -44 42 -56
five 62 -24 -90 57
six -5 -24 35
seven -5 57 -90 57 -12
eight 57 -24 30 -20 -35
nine -12 -24 -12 57
ten -35 57 -12
I do not agree. Eventually, you will have more equations than variables.
'eleven' is easy, because it adds one new variable (L) to an existing
eleven 57 -35 57 -90 57 -35
'twelve' is easy as well because you add one new variable (W) with one new
Hunt for hidden gold, the hardy boys, 58 line 1
Post by Mike Duffy
twelve -35 58 57 -35 -90 57.
But now, you have a big problem with thirteen. You need to add a new
equation to your existing system without adding any new variables, because
thirteen -35 -20 -24 -56 -35 57 57 -12 = -68, NOT 13.
So now, you need to search for a completely new set of diaphantine
solutions. I am guessing that some might exist if you use values higher in
magnitude than 90, because you already have 16 variables in play
(zerontwhfuivsxgl) with only 14 equations.
'fourteen' will be problematic though, because 'four'=4, thus 'teen'=10.
But if 'teen'=10 and 'ten'=10, this forces 'e' to zero. You've reached the
end of the road for this puzzle, unless you can find a solution for
'thirteen' with E=0.
(And if you can then find a solution for 'fourteen' as well, you already
have 'sixteen', 'seventeen', 'eighteen', and 'nineteen'.)
--
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Corey White
2016-03-19 22:35:31 UTC
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The source code can be more easily viewed here:
http://ideone.com/3Gbjjr
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