help me with the program

[sahu]

N-Queens :
You are given a N x N chessboard & you are required to place N queens
on it so that they can't attack each-other. You just need to find out
the valid positions of the queens and the total number of such valid
combinations. A sample I/p and o/p is given below (where N will be
input by the user).

Sample I/p :
Enter the no. of queens: 4
Sample o/p:
1,3,0,2
2,0,3,1
Total no. of possible combinations = 2

Explanation: In the aforesaid example 4 is the value of N input by the
user(4 x 4 chessboard & 4 queens). The o/p prints the position no.s of
the four queens on the chessboard on screen (position no.s can have
value from 0 to N-1, refer the figure below).

0 1 2 3
- Q - -
- - - Q
Q - - -
- - Q -

The second line shows next such valid combination. Finally it prints
the total no. of such valid combinations on screen.

[Barry Schwarz]

On 4 Sep 2005 11:02:20 -0700, "sahu" <sw********@gmail.com> wrote:

N-Queens :

snip

Your third posting in five minutes with absolutely no lines of code
whatsoever. You must have confused this with the
do.my.homework.for.me group.
<<Remove the del for email>>

[Wolfgang Riedel]

sahu wrote:

N-Queens :
You are given a N x N chessboard & you are required to place N queens
on it so that they can't attack each-other. You just need to find out
the valid positions of the queens and the total number of such valid
combinations.

<snip>

start with the easy case n == 2

[Michael Wojcik]

In article <43**************@retarus.de>, Wolfgang Riedel <wo*************@retarus.de> writes:

sahu wrote:

N-Queens :

start with the easy case n == 2

The *really* easy case is n == 0, and the solution for that one also
works for any multiple of 0.

--
Michael Wojcik mi************@microfocus.com

[Wolfgang Riedel]

Michael Wojcik wrote:

In article <43**************@retarus.de>, Wolfgang Riedel <wo*************@retarus.de> writes:

sahu wrote:

N-Queens :

start with the easy case n == 2

The *really* easy case is n == 0, and the solution for that one also
works for any multiple of 0.

if n == 0 and n == 2 have the same number of solutions, it should be save to
assume
this holds for all (at least even) values of n.
Of course N-Janus is another story...

[Lawrence Kirby]

On Tue, 06 Sep 2005 10:09:50 +0200, Wolfgang Riedel wrote:

Michael Wojcik wrote:

In article <43**************@retarus.de>, Wolfgang Riedel <wo*************@retarus.de> writes:

> sahu wrote:
> >
> > N-Queens :
>
> start with the easy case n == 2

The *really* easy case is n == 0, and the solution for that one also
works for any multiple of 0.

if n == 0 and n == 2 have the same number of solutions, it should be save to
assume
this holds for all (at least even) values of n.

Not safe and in fact wrong.

Lawrence

[Eric Sosman]

Wolfgang Riedel wrote:

Michael Wojcik wrote:

In article <43**************@retarus.de>, Wolfgang Riedel <wo*************@retarus.de> writes:

sahu wrote:

N-Queens :

start with the easy case n == 2

The *really* easy case is n == 0, and the solution for that one also
works for any multiple of 0.

if n == 0 and n == 2 have the same number of solutions, it should be save to
assume
this holds for all (at least even) values of n.
Of course N-Janus is another story...

It seems to me that the case n==0 has one solution,
n==1 also has one, and n==2 has none. Fitting these three
observations with a quadratic gives this general formula
for the number of solutions as a function of n:

S(n) = 1 + n/2 - n*n/2

This useful formula is a great time-saver. For example,
we need not write a computer program to discover that the
number of solutions on the standard n==8 chessboard is
minus twenty-seven.

--
Eric Sosman
es*****@acm-dot-org.invalid

[Wolfgang Riedel]

Eric Sosman wrote:

Wolfgang Riedel wrote:

Michael Wojcik wrote:

In article <43**************@retarus.de>, Wolfgang Riedel <wo*************@retarus.de> writes:

sahu wrote:

> N-Queens :

start with the easy case n == 2

The *really* easy case is n == 0, and the solution for that one also
works for any multiple of 0.

if n == 0 and n == 2 have the same number of solutions, it should be save to
assume
this holds for all (at least even) values of n.
Of course N-Janus is another story...

It seems to me that the case n==0 has one solution,
n==1 also has one, and n==2 has none. Fitting these three
observations with a quadratic gives this general formula
for the number of solutions as a function of n:

S(n) = 1 + n/2 - n*n/2

This useful formula is a great time-saver. For example,
we need not write a computer program to discover that the
number of solutions on the standard n==8 chessboard is
minus twenty-seven.

while I still think, you can place nothing on no chessboard, even no no queen -
for n == 3, there are as well 0 solutions.
So the formula should be:

S(n) = (n**4 - 13n**3 + 46n*2 -48n) / -14

and we see one of those 'off by one' features
(they just slip by and are so easy to miss,
not that I want to introduce TDD into this serious debate):

-27 is true for n == 9
for n == 8, there is simply no solution

Wolfgang