MaplePrimes - answers and comments on Question, Permute_locate

105 Solutions

Robert Israel — Wed, 24 Feb 2010 02:02:01 Z

Actually I get 105 solutions.

> with(Optimization): 
  cons:= {seq(add(A[i][j]*x[i],i=1..nops(A))=1,j=1..8)}:
  avoidcons:= {}:
  sols:= NULL;
  do
    try
      sol:= Maximize(0, cons union avoidcons, assume=binary)[2];
    catch:
      break
    end try;
    sols:= sols,select(t -> subs(sol,x[t])=1,{$1..28});
    avoidcons:= avoidcons union
       {add(subs(sol, x[i])*(1-2*x[i])+x[i], i=1..28)>=1};
  end do:  
  sort([sols],(a,b) -> add((a[i]-b[i])*30^(4-i),i=1..4)<0);

[{1, 14, 23, 28}, {1, 14, 24, 27}, {1, 14, 25, 26}, {1, 15, 20, 28}, {1, 15, 21, 27}, {1, 15, 22, 26}, 
{1, 16, 19, 28}, {1, 16, 21, 25}, {1, 16, 22, 24}, {1, 17, 19, 27}, {1, 17, 20, 25}, {1, 17, 22, 23}, 
{1, 18, 19, 26}, {1, 18, 20, 24}, {1, 18, 21, 23}, {2, 9, 23, 28}, {2, 9, 24, 27}, {2, 9, 25, 26}, 
{2, 10, 20, 28}, {2, 10, 21, 27}, {2, 10, 22, 26}, {2, 11, 19, 28}, {2, 11, 21, 25}, {2, 11, 22, 24}, 
{2, 12, 19, 27}, {2, 12, 20, 25}, {2, 12, 22, 23}, {2, 13, 19, 26}, {2, 13, 20, 24}, {2, 13, 21, 23}, 
{3, 8, 23, 28}, {3, 8, 24, 27}, {3, 8, 25, 26}, {3, 10, 16, 28}, {3, 10, 17, 27}, {3, 10, 18, 26}, 
{3, 11, 15, 28}, {3, 11, 17, 25}, {3, 11, 18, 24}, {3, 12, 15, 27}, {3, 12, 16, 25}, {3, 12, 18, 23}, 
{3, 13, 15, 26}, {3, 13, 16, 24}, {3, 13, 17, 23}, {4, 8, 20, 28}, {4, 8, 21, 27}, {4, 8, 22, 26}, 
{4, 9, 16, 28}, {4, 9, 17, 27}, {4, 9, 18, 26}, {4, 11, 14, 28}, {4, 11, 17, 22}, {4, 11, 18, 21}, 
{4, 12, 14, 27}, {4, 12, 16, 22}, {4, 12, 18, 20}, {4, 13, 14, 26}, {4, 13, 16, 21}, {4, 13, 17, 20}, 
{5, 8, 19, 28}, {5, 8, 21, 25}, {5, 8, 22, 24}, {5, 9, 15, 28}, {5, 9, 17, 25}, {5, 9, 18, 24}, 
{5, 10, 14, 28}, {5, 10, 17, 22}, {5, 10, 18, 21}, {5, 12, 14, 25}, {5, 12, 15, 22}, {5, 12, 18, 19}, 
{5, 13, 14, 24}, {5, 13, 15, 21}, {5, 13, 17, 19}, {6, 8, 19, 27}, {6, 8, 20, 25}, {6, 8, 22, 23}, 
{6, 9, 15, 27}, {6, 9, 16, 25}, {6, 9, 18, 23}, {6, 10, 14, 27}, {6, 10, 16, 22}, {6, 10, 18, 20}, 
{6, 11, 14, 25}, {6, 11, 15, 22}, {6, 11, 18, 19}, {6, 13, 14, 23}, {6, 13, 15, 20}, {6, 13, 16, 19}, 
{7, 8, 19, 26}, {7, 8, 20, 24}, {7, 8, 21, 23}, {7, 9, 15, 26}, {7, 9, 16, 24}, {7, 9, 17, 23}, 
{7, 10, 14, 26}, {7, 10, 16, 21}, {7, 10, 17, 20}, {7, 11, 14, 24}, {7, 11, 15, 21}, {7, 11, 17, 19}, 
{7, 12, 14, 23}, {7, 12, 15, 20}, {7, 12, 16, 19}]

partitioning

Joe Riel — Wed, 24 Feb 2010 03:56:43 Z

The problem is equivalent to partitioning a set of 2*m objects into m-partitions of 2 objects. That can be easily done recursively:

part2 := proc(L :: list)
local n,k,p;
    n := nops(L);
    if n = 2 then
        return [[L]];
    end if;
    return [seq(seq([ [L[1], L[k]], p[] ]
                     , p in procname(subsop(1=NULL,k=NULL,L)))
                , k=2..n )]
end proc:

 part2([seq(1..4)]);
            [[[1, 2], [3, 4]], [[1, 3], [2, 4]], [[1, 4], [2, 3]]]

The conversion of the partitioned integers to the desired form is straightforward.

Permute_locate

brian bovril — Thu, 25 Feb 2010 14:39:26 Z

Thanks Robert and Joe.

I don't really undersand the methodology fully Robert, but it works. Brilliant!

rsolve

brian bovril — Sat, 27 Feb 2010 23:22:46 Z

ok, how would i modify the above recursion formula to calculate the number of partitions into triples from

permute([1, 1, 1, 1, 0, 0, 0, 0 ,0 ,0 ,0 ,0], 12);

sols

brian bovril — Sun, 28 Feb 2010 20:42:43 Z

cool bananas

rsolve

Joe Riel — Thu, 25 Feb 2010 20:33:17 Z

Note that you can use a recurrence to solve for the number of partitions. That is, given a list of 2*m elements, the number of partitions into pairs can be expressed as the recurrence

req := f(k) = (2*k-1)*f(k-1):

Use ?rsolve to solve this


f := rsolve({req, f(0)=1}, f(m));
                                  m
                                 2  GAMMA(m + 1/2)
                            f := -----------------
                                         1/2
                                       Pi


eval(f, m=4);
                                      105