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how do you check whether the attempted computation of topological sorting is indeed one


By : B. Brewer
Date : September 17 2020, 02:00 AM
hop of those help? You need to test for every node n in the ordered list that each node from an incoming edge in appears before n in the ordered list:
code :
isTopologicallySorted(list, graph)
    for each n in list
        for each in in graph[edges incoming at n]
            if list.indexof(in.source) > list.indexof(n) then return false
    return true


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Topological sorting in sql


By : Stirling
Date : March 29 2020, 07:55 AM
hop of those help? I am resolving dependency between some objects in a table. I have to do something with objects in order their dependency. For example, the first object doesn't depend on any object. The second and third ones depends on first one and so on. I have to use topological sorting. Could someone show the sample of implementation so sorting in t-sql. I have a table: , It seams, it works:
code :
declare @step_no int

declare @dependency table 
(
  DependencyId  int
  ,ObjectId     int 
  ,ObjectName   varchar(100)
  ,DependsOnObjectId int 
  ,[rank]       int         NULL
  ,degree       int         NULL
);

insert into @dependency values (5, 5, 'Obj 5', 2, NULL, NULL)
insert into @dependency values (6, 6, 'Obj 6', 7, NULL, NULL)
insert into @dependency values (2, 2, 'Obj 2', 1, NULL, NULL)
insert into @dependency values (3, 3, 'Obj 3', 1, NULL, NULL)
insert into @dependency values (1, 1, 'Obj 1', 1, NULL, NULL)
insert into @dependency values (4, 4, 'Obj 4', 2, NULL, NULL)
insert into @dependency values (7, 7, 'Obj 7', 2, NULL, NULL)


update @dependency set rank = 0
-- computing the degree of the nodes
update  d set d.degree = 
    (
        select count(*) from @dependency t
        where t.DependsOnObjectId = d.ObjectId 
        and t.ObjectId <> t.DependsOnObjectId
    )
from @dependency d


set @step_no = 1
while 1 = 1
begin
    update @dependency set rank = @step_no where degree = 0

    if (@@rowcount = 0) break
    update @dependency set degree = NULL where rank = @step_no

    update d set degree = (
        select count(*) from @dependency t
        where t.DependsOnObjectId = d.ObjectId and t.ObjectId != t.DependsOnObjectId
        and t.ObjectId in (select tt.ObjectId from @dependency tt where tt.rank = 0))
    from @dependency d
    where d.degree is not null

    set @step_no = @step_no + 1
end

select * from @dependency order by rank

What is the difference between sorting and topological-sorting?


By : Praveen Kumar
Date : March 29 2020, 07:55 AM
I wish this helpful for you At an abstract level they are connected: As Saeed and Stefan say, it's the difference between a total order and a partial order. That is a fantastically concise description, but sometimes not helpful when you're learning.
A total order means that, in the absence of repeats, when you sort something, you're going to get one unique proper answer. If you sort 3, 6, 2 in ascending order, you had better get one answer: 2, 3, 6.

topological sorting using dfs


By : Alex
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I have done a few changes and now your program finishes successfully on ideone The most significant change is that you did not initialize matrix and v(even without this change the program still finished successfully but the output was only 0-s). I did not see the error you are talking about. The reason for getting only 0-s when you did not initialize v is obvious - all the values where non-zero and so all nodes where considered not visited.
EDIT: I also changed line 27 where you seemed to have forgotten " = true;"

Topological sorting in PHP


By : Kerem Cevren
Date : March 29 2020, 07:55 AM
Hope this helps You get duplicate lines because there are duplicate edges. I'm no graph theory thug but I'm pretty sure this is not legal :
code :
0 => 
array (
  0 => 'nominal',
  1 => 'subtotal',
),
2 => 
array (
  0 => 'nominal',
  1 => 'subtotal',
),
...
if ($id==$e[0] && !in_array($e[1], $nodes[$id]['out']))
{
  $nodes[$id]['out'][]=$e[1];
}
if ($id==$e[1] && !in_array($e[0], $nodes[$id]['in'])) // Not needed but cleaner
{
  $nodes[$id]['in'][]=$e[0];
}

Topological sorting asm x86


By : patrickbromley
Date : March 29 2020, 07:55 AM
hop of those help? Some bugs I caught by reading the source (I don't claim any completeness of the list, there may be more):
code :
LEA DI,size
MOV DX,[DI]
CMP [BX+SI],1
MUL size
CALL visitar
MOV CL,1
PUSH CX
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