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Hi! I’m the “MathWizHoax” and I like to leave my account logged in on public terminals!!!!

Getting from 1 point to another in less than 3 moves!!

When trying to find if you can get from one vertex to another in 3 or less moves you can find it out with matrix algetbra. Here is a 10×10 matrix that can get from anypoint to anypoint in 3 or less moves.

A+A^2+A^3

A= 0 1 0 0 1 0 0 1 0 0
1 0 0 1 0 0 1 0 0 0
0 0 0 1 0 0 1 0 0 1
0 1 1 0 0 0 0 1 0 0
1 0 0 0 0 1 0 0 1 0
0 0 0 0 1 0 0 1 1 0
0 1 1 0 0 0 0 0 0 1
1 0 0 1 0 1 0 0 0 0
0 0 0 0 1 1 0 0 0 1
0 0 1 0 0 0 1 0 1 0
A^2

3 0 0 2 0 2 1 0 1 0
0 3 2 0 1 0 0 2 0 1
0 2 3 0 0 0 1 1 1 1
2 0 0 3 0 1 2 0 0 1
0 1 0 0 3 1 0 2 1 1
2 0 0 1 1 3 0 0 1 1
1 0 1 2 0 0 3 0 1 1
0 2 1 0 2 0 0 3 1 0
1 0 1 0 1 1 1 1 3 0
0 1 1 1 1 1 1 0 0 3
A^3
0 6 3 0 6 1 0 7 2 2
6 0 1 7 0 3 6 0 2 2
3 1 2 6 1 2 6 0 1 5
0 7 6 0 3 0 1 6 2 2
6 0 1 3 2 6 2 1 5 1
1 3 2 0 6 2 1 6 5 1
0 6 6 1 2 1 2 3 1 5
7 0 0 6 1 6 3 0 2 2
2 2 1 2 5 5 1 2 2 5
2 2 5 2 1 1 5 2 5 2
A^2+A^3=
3 6 3 2 6 3 1 7 3 2
6 3 3 7 1 3 6 2 2 3
3 3 5 6 1 2 7 1 2 6
2 7 6 3 3 1 3 6 2 3
6 1 1 3 5 7 2 3 6 2
3 3 2 1 7 5 1 6 6 2
1 6 7 3 2 1 5 3 2 6
7 2 1 6 3 6 3 3 3 2
3 2 2 2 6 6 2 3 5 5
2 3 6 3 2 2 6 2 5 5

The matrix A represents the line graph that each vertex has 3 lines coming off it. A tells you how many ways of walk 1 from each vertex to another

Finding the Square root of 2

Here is some code I wrote to find a close appoximation of the square root of 2.
u1=2;
u2=1;
l1=1;
l2=1;
ub=u1/u2;
lb=l1/l2;
n=3;
d=2;
k=2;
for i=1:k
if (n^2/d^2)>(u1/u2)
u1=n;
u2=d;
n=2*n-1;
d=2*d;

elseif (n^2/d^2)<(l1/l2)
l1=n;
l2=d;
n=2*n;
d=2*d+1;
else
l1=n;
l2=d;
n=2*n;
d=2*d+1;
end
end

Some Quick notes on 4 peg ToH

Here are some quick results that I have found. This is a spread sheet with the number of moves that it took for 4 peg and 3 peg. On the second sheet you can see some unproved formulas.
http://spreadsheets.google.com/ccc?key=pRxUDwKj7i9opcHqK-Dz7cw

Tower of Hanoi: Grey Code

When using Grey Code to determine the minimum number of moves in the Tower if Hanoi, it is easy to see that you can only use each subset once.

So for 4 disks 3 pegs the gery code would be;

Start {0 0 0 0};

Move the first disk {0 0 0 1}

Move the second disk{0 0 1 1};

Move the first disk onto the second disk  {0 0 1 0};

Move the third disk to the empty peg now{0 1 1 0};

Move the first disk onto the forth disk  {0 1 1 1};

Move the second disk onto the third disk  {0 1 0 1};

Move the first disk on top of the second disk{0 1 0 0}:

Now move the forth disk{1 1 0 0};

Move the first disk onto the forth disk{1 1 0 1};

Move the second disk to the empty peg{1 1 1 1} ;

Move the first disk onto the second disk  {1 1 1 0};

Move the third disk   onto the forth disk  {1 0 1 0};

Now move the first onto the empty peg{1 0 1 1};

 Move the second onto the third{1 0 0 1};

 And  finaly move the first onto the second disk{1 0 0 0}

As you can easily see in order for this to be the minimum number of moves you have to only use each subset once.

Tower of Hanoi

I have been doing this puzzle fore a couple of hours now and have found some methods that can really help complete the puzzle in the minimum amount of moves. This only works with 3 pegs. I will be building a 4 peg setup over the weekend and i will see how that works. We will say the amount of disks you have is and the size of your pile is Here are some methods;

if = 2x-1 then start at the far right peg

if = 2x then start at the middle peg

if = 2x-1 then start on the peg you want to move the pile to

if = 2x then start away from the peg you want to move the pile to

Breaking Down Exponents??

I have this idea that i was working on in high school about breaking down exponents to calculate them faster. I have found a very weak linear connection to exponents. 

 I will post a little more on this when I remember how I did it.

Free Cell

Good Morning my fellow Mathletes.

 This is the first post of many post to come. I have a few problems to deal with. After watching a person play over 40,000 games of Free cell, it makes me wonder how many different possibilities there could be. I think you can you could figure this out with a simple Combination, that’s not my concern. While some games take as long as an hour to figure out you are not going to beat that said game, then on the other hand, some are finished in as little as 66 seconds. It is mind boggling to see someone finish a game so quickly. So are some games easier than others, or can you actually look at a game before you start and know exactly what you are going to do?

So how many of the games cannot be finished out of the amount of different combinations of cards. I am unsure if there is even a way to tell if it is unbeatable.

Let me know what you think.

 

AP