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**seerj**- Replies: 0

I have

(( n - 1)/(n+1))^n² by using root criteria for series convergence I arrived to this:

((n-1)/(n+1))^n I don't remember how to trasform it in

(1-2/n)^n -> e^-2

Does it correct do this:

(n-1)/(n+1) -> [I don't remember this!! ] (n-1-1)/n and a then (n-2)/n (1-2/n) ??

Thx 2 all

this is your code. I tried with it but after 6 hours it has crashed

#include <iostream>

#include <fstream>

#include <vector>

#include <math.h>

#include <windows.h>

using namespace std;

void testVals(vector<int> base, vector<bool> use, vector<int> test);

bool compare(int x, int y);

int main()

{

int start = GetTickCount();

for (int size = 2; size <= 32; size++)

{

cout << "Testing list size " << size << "..." << endl;

vector<int> base;

vector<bool> used;

vector<int> test;

for (int gen = 1; gen <= size; gen++)

{

base.push_back(gen);

used.push_back(false);

}

testVals(base, used, test);

}

int end = GetTickCount();

cout << "Time: " << (end - start) / 1000.0 << endl;

return 0;

}

void testVals(vector<int> base, vector<bool> used, vector<int> test) {

if (test.size() == base.size())

{

bool result = compare(test[test.size()-1], test[0]);

if (result)

{

ofstream o("asdf.txt", ios::app);

o << "Size: " << test.size() << endl;

for (int x = 0; x < test.size(); x++)

{

o << test[x] << " ";

}

o << endl;

}

return;

}

for (int x = 0; x < used.size(); x++)

{

if (used[x] == false)

{

if (x != 0)

{

if (!compare(test[test.size()-1], base[x])) {

continue;

}

}

used[x] = true;

test.push_back(base[x]);

testVals(base, used, test);

test.erase(test.end()-1);

used[x] = false;

}

}

}

bool compare(int x, int y)

{

double temp = sqrt((double)x + y);

const double amount = 0.00001;

if (temp + amount > (int) temp && temp - amount < (int)temp)

{

return true;

}

return false;

}

Yes they are right.

So u can create a matrix associated to the system. Where in the left side there are the values of the variables and on the right side there are the known values.

| 3 2 -2 | 1| |3 2 -2 | 1 |

| 2 1 4 | 0| by using gauss method u'll have a matrix like this |0 -1/3 16/3 | -2/3|

| 5 3 -5 | 2| |0 0 -7 | 1 |

so u have to create a new system with the new values

| 3x +2y -2z=1

| -1/3y +16/3z=-2/3

| -7z=1

and ull have that z= -1/7 y=-2/7 x=3/7

Tell me if u need help

PS: I used the Gauss-Jordan algorithm -->http://mathworld.wolfram.com/Gauss-JordanElimination.html

If Ididn't wrong the values are:

x=3/7

y=-2/7

z=-1/7

Tell me if they are right please.

Ricky wrote:

O my.. i let my program running all night long but it crashed at size 43!!!!!! :\\\

Crashed? What language did you program it in?

I should have size 62 by the next time I post.

I used visual c++ 6.

The contest is ended. However i say thank u!

Now it's hard to explain why 32 is the minimum number..but u can find an explanation from

http://www.geocities.com/dharwadker/hamilton/main.html with the Hamilton Algorithm..

Look at here:

http://www.mat.uniroma2.it/~tauraso/Natale/congettura.txt

is made by my teacher.

O my.. i let my program running all night long but it crashed at size 43!!!!!! :\\\

this is the winner of the contest:

size=61

1,3,6,10,15,21,4,5,11,14,50,31,18,7,2,23,58,42,22,59,41,8,28,53,47,34, 30,51,49,32,17,19,45,36,13,12,37,27,54,46,35,29,52,48,33,16,20,44,56, 25,39,61,60,40,9,55,26,38,43,57,24

krassi_holmz wrote:

I've developed another algoritm. It generates all square chains and checks if they're circular.

Size 36:

{1, 3, 6, 19, 30, 34, 2, 23, 26, 10, 15, 21, 4, 32, 17, 8, 28, 36, 13, 12, 24, 25, 11, 5, 20, 29, 7, 18, 31, 33, 16, 9, 27, 22, 14, 35}

~5min.

It seems that for all n > 32 there exist a circular sequence. Strange.

Very good Krassi, thanks.

Anyone know why 32 is the minimum number?

I'm running a program made with the Ricky's algorithm but till now [it has been passed 2 hours] it reached n=41..

however THANK U ALL FRIENDS!!

NICE WORK GUY!!

Can u find a list greater than 59?

Ahah hi!. Nice thing, I'm your new friend .

my friend is winning the contest ..He's the only person who shown why 32 is the minimum number.

But there is another boy who found a 59 lenght sequence!!!!!

1,3,6,10,15,21,4,5,11,14,50,31,18,7,2,23,58,42,39,25,56,8,17,32,49,51, 13,36,28,53,47,34,30,19,45,55,26,38,43,57,24,12,37,44,20,29,52,48,33,16, 9,40,41,59,22,27,54,46,35

Krassi did u find an algorithm to create any circular sequences?

krassi_holmz wrote:

But is really 32 the minimum?

Yes My teacher [who is a math researcher] write : MINIMUM N is 32

Now he's asking why..

And he would see other sequences longer than tat boy's one

n<31 He was near the solution...I dunno how. As soon as my uni starts i ask to my friend to show me it

SOMEONE SUCCESFUL in this problem.

The solution is LENGHT = 32 . 32 is the minimum number!

His circular sequence is:

1,8,28,21,4,32,17,19,30,6,3,13,12,24,25,11,5,31,18,7,29, 20,16,9,27,22,14,2,23,26,10,15

lenght=32

His second list:

1,3,6,19,30,34,2,7,18,31,33,16,9,40,24,25,39,10,15,21, 28,36,13,23,26,38,11,5,20,29,35,14,22,27,37,12,4,32,17,8

lenght=40

But the challenge isn't finished

krassi_holmz wrote:

For my new algoritm I should learn more about the diophante equations.

Sorry if I'm late, but u have been fantastic!!!! Thanks for ur work!

Thank u again Krassi and JohnFranklin

krassi_holmz wrote:

I wrote my program in Mathematica language and it has some extra properties:

Oh Mathematica! I thought something like C/C++. Coz I dunno how to check if the sum of 2 consecutive numbers is perfect square??!

When the sum of 2 numbers is a perfect square?

if (a+b) = ?!?!

seerj wrote:

krassi_holmz wrote:arctan(x) also has two horizontal asymptotes.

But I think that these functions cannot be adapted for this graph, because it is not symmetric.Yes I quote him! You can use arctan(x) by creating your own function

For example try to plot a function like this:2*atan(x-3)+3

Or wait also tanh(x) is good. I think is perfect for your case

Try to plot it

tanh(x-3/5)+1

krassi_holmz wrote:

arctan(x) also has two horizontal asymptotes.

But I think that these functions cannot be adapted for this graph, because it is not symmetric.

Yes I quote him! You can use arctan(x) by creating your own function

For example try to plot a function like this:

2*atan(x-3)+3

MARGIN+PICNUM*(MARGIN+WIDTH)+MARGIN+WIDTH=PICW/2

Let m MARGIN p PICNUM x WIDTH q PICW

Then

m+p*(m+x)+m+x=q/2

px+x=q/2 -m -pm -m

x ( p+1 )=q/2-2m-pm

x= (q-4m-2pm)/2*(p+1)

x= (q-2m*(2+p)) / (2*( p+1 ))

WIDTH=( PICW-2MARGIN*(2+PICNUM))/(2*( PICNUM+1))

Ehy seems to be cool.

If u want can u send me the algorithm? [if u want for sure!]

on leopardus@inwind.it

However thank u...

We need to understand which is this minimum n....

krassi_holmz wrote:

hristo, where are you from?

I have a friend called Hristo and he's hungarian. I think that he comes from Hungary. Is it right?

krassi_holmz wrote:

So the other solutions are not solutions.

That's because I calculated them by hand.

Now I'll do some progarmming...

You're great! Thank u in advance. So Did u understand the problem now? It wasn't so difficult but is so difficult to understand which is the minimum n ^^

Thank u all for the work!

I try to ask but i think that the tips was n<50

So all the circular sequences with n<50 like krassi is doing . [only that 11+15 isn't a perfect square..so the sequence isn't correct ;| ]

krassi_holmz wrote:

1,3,6,10,15,11,5,4,12,13,23,2,7,9,16,20,29,27,22,14,35

length=21

Seems to be good coz the number are all less than 50, but

Let ps=perfect square

1+3 ps

3+6 ps

6+10 ps

10+15 ps

15+11= 26 isn't a perfect square!

11+5 ps

5+4 ps

4+12 ps

12+13 ps

13+23 ps

23+2 ps

2+7 ps

Almost good. But 15+11 isn't right btw

katy2 wrote:

thanks for the answers, GOD...but i still do not understand this quote :"whose product is a minimum" of #2

what is a minimum?and how do you know whether the product is a min or max???

U can decide is a x1 is a minimum also by studying the graph of the function.

For example in a parabola with vertex in (-5,0) -5 is a min! and max doesn't exist.

please, not the smallest solution. Krassi if u made an algorithm we need at least to find the longest circular sequence! [NOT THE SMALLEST] the smallest thing is n, not the solution. If my geek teacher said that a tip is n<50 there is any reason. What I proposed is a particular problem, and the math scientists found this minimum n. There is a conjecture for which does exist a particular n and further it does always exist a circular sequence with this minimum n, but they can't explain why.

But they found this minimum n.

Even if we can't find it, we can try to find the longest circular sequence at least!

I think now it's ok

In faith