Tricky for loop

Posted on 2008-03-20 15:01 By Rogerb (

42)

Categories:

How do I ...? with Maple

Hi Guys, i have a forloop problem.

I have a nxs matrix, where s=sum((k),k=1..n-1), so when n=3, s=3, when n=4,s=6 and so on.

I need to code each column number lexicographically for cartesian pair (i,j) where i<j, so when n=3, the columns would be

(1,2),(1,3),(2,3)

and when n=4 we have

(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)

I am able to create this lexicographical order independently but where i am stuck, and wondering whether it can be done is to associate the cartesian pair, with the true column number, so with the previous example

1 2 3 4 5 6

(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)

Is it possible?

not tricky, and no loops!

Comment on 2008-03-20 16:00 from Doug Meade ( Maple Rating 4

722)

Here is how I would approach this problem. First, s is the sum of the first n-1 integers:

s := n -> n*(n-1)/2;
                                     1          
                                n -> - n (n - 1)
                                     2

The list of ordered pairs (i,j) with i<j, i=1..n, and j=1..s can be constructed with

P := n -> [seq( seq( [r,c], c=r+1..s(n) ), r=1..n )];

P(3);
                          [[1, 2], [1, 3], [2, 3]]

P(4);
  [[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 3], [2, 4], [2, 5], [2, 6], 
    [3, 4], [3, 5], [3, 6], [4, 5], [4, 6]]

To find the index in this list, use the member command with the optional third argument.

member( 1, [0,1,2], 'p' );
                                     true
p;
                                      2

These can all be put together in a single procedure, as follows:

LookUp := proc(e,S)
  local p;
  if not member(e,S,'p') then p:=0 end if;
  return p
end proc:

LookUp( [4,5], 5 );
                                      25
LookUp( [5,5], 5 );
                                      0

You might want to have a different behavior when the element is not in the set. I chose to have the LookUp procedure return 0.

Does this do what you have requested?

Doug

---------------------------------------------------------------------
Douglas B. Meade  <><
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu/~meade/

Possible solution

Comment on 2008-03-20 16:07 from Scott03 ( Maple Rating 4

468)

A quick bit of code to get the sequence you had in your example would be the following:

[seq(seq([i, j], i = 1 .. j-1), j = 2 .. 4)];

If you want to place this inside of a Maple table structure, you could call

Test := table([seq(seq([i, j], i = 1 .. j-1), j = 2 .. 4)]);

which would allow you to call

Test[2]

[1,3]

Another option is to us an array which would be something like the following:

Test2 := Array([seq(seq([i, j], i = 1 .. j-1), j = 2 .. 4)]);

Test2[2]

[1,3]

Scott

Using a table

Comment on 2008-03-21 04:03 from John Fredsted ( Maple Rating 4

566)

The association may also be obtained as follows: Define the procedure

assoc := (n::posint) -> table([seq(seq([i,j],j = i+1..n),i = 1..n)]):

Then, for instance,

t := assoc(4):
t[1],t[2],t[3],t[4],t[5],t[6];
         [1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]

Hi these loops are all

Comment on 2008-03-21 06:16 from Rogerb (

42)

Hi these loops are all greats thanks!

Tricky for loop

not tricky, and no loops!

Possible solution

Using a table

Hi these loops are all

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