rooks and squares

pascal roca

  1. How many different ways to arrange rooks on chess board knowing the condition that none of rook should lie on a square which is controlled by an other one  ?
  2. The first solution which comes up immediately is to put all the rooks on the principal diagonal
  3. then if you choose the first column you have eight possible choices, for the second column 7 choices and so on, the number of arrangements is finally 8! respect to the condition above

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Chess Rook Paths and squares

pascal roca How many paths can a Rook travels from upper Left Corner to lower right corner : Well you can find a solution at  http://mathacadabra.com/Items2013/GeneratingFunctionAdventuresIV.aspx it needs a little bit some explanations first about the case 8 * 8  : rookPath

  1.  First the rook will start at a8,  then to reach a8, there is only one path : staying at the same place
  2. the chess rook can move horizontally or vertically from 1 up to 8 (including is initial position)
  3. the table sum up the number of paths could be used by the  rook in order to reach the case, for example if  the rook want to reach e5 ( 838) , the rook can come only vertically from the case above e4 ( e5, e6, e7, e8) or horizontally from the left of e4 ( a4, b4, c4, d4 ) therefore to get the total number of paths to reach e4 is :  8+28+94+289+289+94+28 = 838 paths  for a square 5 * 5, we can calculate all the paths recursively
  4. Actually there are 470010 paths for a square 8*8
  5. In the Link provided above to the website , they are trying to calculate the generating function for the diagonal which is the number of paths for a square N * N.pascal roca
  6. let’s demonstrate after reading the post :(  http://mathacadabra.com/Items2013/GeneratingFunctionAdventuresIV.aspx ) that the 2 variables generating function is  (1 – s – t  + st) / ( 1 – 2 *s  -2*t  + 3 *st )  for a square N*N
  7. pascal roca

Serie of 1

S = 1 + 11 + 111 + 1111 + …………….+ 1(N times 1 )

How to calculate  the sum of N terms  ?

you can start by looking at 9 * S = 9 + 99 + 999 + 9999 + …….

Which could be rewriten as : 9 * S = (10 -1) + (100 – 1 ) + (1000 -1) + …….. +

9S +1 =  1 + 10 + 10 ^2 + 10  ^ 3 + …. + 10 ^ N +  ( – N )

9S+1 + N  =  1 + 10 + 10 ^2 + 10  ^ 3 + …. + 10 ^ N = ( 1 – 10 ^ N+1) / (1 -10)

John von Neumann and the fly

The story was that John von Neumann was asked about the following problem :

  • 2 bikes are headed toward each other at the same speed 60 km/h, they are separated each other by 2 kilometers. A Fly leaves the front of the first bike and travels at 90 km/h (a very fast one) towards the second one then back to the first bike and so on
  • how far does the fly travel ?

i am going to explain the methodical way used by  John von Neumann :

Bikes are traveling at the speed 1 km/minute and the fly 1.5 km/minute

let look at the fist trip when the fly collide with the second bike before going back to the first bike:

we call d(Fly), the distance corresponding to the travel of the fly, and d(Bike 2) the one corresponding to the Bike 2 :

we have  d(Fly) + d(Bike 2) = d(Total) = 2 

at time t, they collide : t *  ( V(Fly) + V(Bike 2) )d(Total)

then td(Total) / ( V(Fly) + V(Bike 2) )  = d(Total) / ( 1 + 1.5 ) =  2/5 * d(Total)

then it comes  d(Fly) = (1.5 * d(Total)) * 2/5  = 6/5  (1)

and d(Bike 2) =  d(Total)) * 2/5 = 4/5 = d(Bike 1)  (2)

Now for the second trip the distance to go back to the first bike leaving the front of second bike is  d(Fly) –  d(Bike 1)   = 2/5

but we don’t have to redo the calculation again, we have to infer 2/5 in d(Total) for equation (1) and (2) which is 1/5 of total distance ( 2 km ),  then  d(Fly) = 6/ 5*5  and so on  :

then we can extract an infinite serie for the fly travel distance :

d(Fly n steps) = 6/5 ( 1 + 1/5 + 1/(5*5 ) + 1/ (5*5*5) + … + 1/(5*5*5 *   n times))

d(Fly) = d(Fly n steps) when n → ∞

d(Fly) = 6/5 * 1/(1 – 1/5) = 3/2 km

 

Basic functional equation

function

find at least on real function which is defined by  : f( k * x) = f (x) for all x Real and k real

  • Obviously all constants satisfies the equation above , but you can say more if f is continue at x=0 , then f is a constant function
  • a non trivial example of a solution of equation above could be : x → sin( 2π * ln(x) )           with k = e, you can notice that this function is not continue at x = 0

Chess Board and rectangles

chessBorad

How many rectangles you can see ?

  • you have on the upper side of the chess board 8 indivisible squares ,  7 points belongs to the edge and to the frontiers separating the squares , plus 2 points at the corners .  then if you want define a side of a rectangle you need to choose 2 points out of the 9 points belonging to the upper side. the order doesn’t matter , we will use combination : http://en.wikipedia.org/wiki/Combination :  finally we have 9! / ( 2! * 7!) combinations
  • you can apply the same reasoning on the left side of the Chess Board
  • at the end you have 9! / ( 2! * 7!) * 9! / ( 2! * 7!) rectangles and squares (which are technically rectangle) but if you want only and only rectangle you can subtract the  number of squares with the formula defined in the previous post about Chess Board and squares : https://pascalroca.com/2015/03/01/chess-board-and-squares/