I need to traverse all vertices that are connected by edges where the property 'dependence' is 'true'
This is what I have so far:
SELECT
FROM (TRAVERSE *
FROM (SELECT outE() FROM 9:5)
WHILE (#class = 'E' AND dependence = 'yes') OR #class = 'V')
WHERE #class = 'V'
Although im not sure if is the best way to do it, this seems to work fine following only the paths where edges have 'dependence' = 'yes'.
Now, There could be more than one path generated, and I need to get the last vertex from each path/branch.
traverserdVertex(-1) should return the last one, but im guessing that is from the whole traversal so is no good. (and it looks like there's a bug because it retrieves more than one)
The outer SELECT returns the whole bag of vertices so I'm thinking that maybe finding the ones that doesn't have an outgoing edge with dependence='yes' might solve it, although I'm not sure how to do it nicely.
SOLUTION:
SELECT
FROM (TRAVERSE *
FROM (SELECT outE() FROM 9:5)
WHILE (#class = 'E' AND dependence = 'yes') OR #class = 'V')
WHERE #class = 'V' AND NOT (outE() contains (dependence='yes'))
This effectively returns the last vertex from each branch. I'm open to any other option, I'm wondering if it could be improved.
I tried with an example by building the following graph
The javascript function "myFunction" has three parameters which are ridVertex, property and value
var g=orient.getGraph();
var previous=[];
var currently=[];
var node=[];
var b=g.command("sql","select from v where #rid =" + ridVertex);
if(b.length>0){
previous.push(b[0]);
node.push(b[0]);
do{
for(i=0;i<previous.length;i++){
var edges=g.command("sql","select expand(outE()) from V where #rid = "+ previous[i].getId());
var myVertex=[];
for(j=0;j<edges.length;j++){
var edge=edges[j];
var dependence=edge.getProperty(property);
if(dependence==value){
var vIn=edge.getProperty("in");
myVertex.push(vIn);
}
}
if(myVertex.length>0){
setPaths();
}
}
change();
}while(previous.length>0);
}
return node;
function setPaths(){
for (m = 0; m < node.length; m++) {
var lastId=node[m].getId().toString();
var idOut=previous[i].getId().toString();
if (lastId==idOut) {
for(r=0;r<myVertex.length;r++){
var vertex=myVertex[r];
node.push(vertex);
currently.push(vertex);
}
node.splice(m,1);
break;
}
}
}
function change(){
previous=[];
for (indice=0;indice<currently.length;indice++)
previous.push(currently[indice]);
currently=[];
}
Using the following command
select expand(result) from (select myFunction("#9:0","dependence","yes") as result)
the paths are A -> D and A -> B -> C -> G and then will be returned the verteces D and G
The following is a slight simplification of #sebastian's solution, using Allesandro's graph (with dependentOn.value being 0 or 1):
select from
(TRAVERSE * FROM (select from Circle where name="A")
while (#CLASS="dependentOn" and value=1) OR #CLASS="Circle")
where #CLASS='Circle' AND NOT (outE().value contains 1)
----+-----+------+----+--------------
# |#RID |#CLASS|name|in_dependentOn
----+-----+------+----+--------------
0 |#11:6|Circle|G |[#12:4]
1 |#11:3|Circle|D |[#12:1]
Related
I have the following Table:
It represents cases on which a certain Team is working on over the Time until the case is closed.
And there is also a Date Table over column Date.
I would like to cumulative count the open cases until the selected date.
So I used this measure:
CountOpen =
VAR CurrentDate = MAX('Date'[Date])
VAR Closed =
CALCULATE(
DISTINCTCOUNT(Tabelle1[case]),
ALL('Date'),'Date'[Date]<=CurrentDate,Tabelle1[Status_Open]="0")
VAR OpenAll =
CALCULATE(
DISTINCTCOUNT(Tabelle1[case]),
ALL('Date'),'Date'[Date]<=CurrentDate,Tabelle1[Status_Open]="1")
RETURN OpenAll-Closed
And it works for the overall view. But for the view within the Dimension CurrentTeam it's not correct:
It should be:
a = 0
b = 1
c = 0
So... this is actually quite tricky, you have to pick the latest status per case up to the selected date. In my solution I create a table, with a column R which ranks the cases per date, then in the result I filter for those depending on which team you have selected.
Measure is below:
VAR CurrentDate = MAX('Date'[Date])
VAR CurrentTeam = SELECTEDVALUE(Tabelle1[CurrentTeam])
VAR tbl =
SUMMARIZE(
FILTER(ALL('Tabelle1'), 'Tabelle1'[Date] <= CurrentDate),
Tabelle1[case],
Tabelle1[CurrentTeam],
Tabelle1[Status_Open],
Tabelle1[Date],
"R",
VAR c = MAX(Tabelle1[case])
VAR d = LASTDATE(Tabelle1[Date])
RETURN
CALCULATE(DISTINCTCOUNT(Tabelle1[Date]),
ALLSELECTED(Tabelle1),
Tabelle1[case] = c,
Tabelle1[Date] >= d)
)
RETURN SUMX(
FILTER(tbl,
[R] = 1 &&
(ISBLANK(CurrentTeam) || [CurrentTeam] = CurrentTeam) &&
[Status_Open])
, 1) + 0 //+0 is here to show 0 where it would be blank
I need to find the shortest set of paths to connect each element of Set A with at least one element of Set B. Repetitions in A OR B are allowed (but not both), and no element can be left unconnected. Something like this:
I'm representing the elements as integers, so the "cost" of a connection is just the absolute value of the difference. I also have a cost for crossing paths, so if Set A = [60, 64] and Set B = [63, 67], then (60 -> 67) incurs an additional cost. There can be any number of elements in either set.
I've calculated the table of transitions and costs (distances and crossings), but I can't find the algorithm to find the lowest-cost solution. I keep ending up with either too many connections (i.e., repetitions in both A and B) or greedy solutions that omit elements (e.g., when A and B are non-overlapping). I haven't been able to find examples of precisely this kind of problem online, so I hoped someone here might be able to help, or at least point me in the right direction. I'm not a graph theorist (obviously!), and I'm writing in Swift, so code examples in Swift (or pseudocode) would be much appreciated.
UPDATE: The solution offered by #Daniel is almost working, but it does occasionally add unnecessary duplicates. I think this may be something to do with the sorting of the priorityQueue -- the duplicates always involve identical elements with identical costs. My first thought was to add some kind of "positional encoding" (yes, Transformer-speak) to the costs, so that the costs are offset by their positions (though of course, this doesn't guarantee unique costs). I thought I'd post my Swift version here, in case anyone has any ideas:
public static func voiceLeading(from chA: [Int], to chB: [Int]) -> Set<[Int]> {
var result: Set<[Int]> = Set()
let im = intervalMatrix(chA, chB: chB)
if im.count == 0 { return [[0]] }
let vc = voiceCrossingCostsMatrix(chA, chB: chB, cost: 4)
// NOTE: cm contains the weights
let cm = VectorUtils.absoluteAddMatrix(im, toMatrix: vc)
var A_links: [Int:Int] = [:]
var B_links: [Int:Int] = [:]
var priorityQueue: [Entry] = []
for (i, a) in chA.enumerated() {
for (j, b) in chB.enumerated() {
priorityQueue.append(Entry(a: a, b: b, cost: cm[i][j]))
if A_links[a] != nil {
A_links[a]! += 1
} else {
A_links[a] = 1
}
if B_links[b] != nil {
B_links[b]! += 1
} else {
B_links[b] = 1
}
}
}
priorityQueue.sort { $0.cost > $1.cost }
while priorityQueue.count > 0 {
let entry = priorityQueue[0]
if A_links[entry.a]! > 1 && B_links[entry.b]! > 1 {
A_links[entry.a]! -= 1
B_links[entry.b]! -= 1
} else {
result.insert([entry.a, (entry.b - entry.a)])
}
priorityQueue.remove(at: 0)
}
return result
}
Of course, since the duplicates have identical scores, it shouldn't be a problem to just remove the extras, but it feels a bit hackish...
UPDATE 2: Slightly less hackish (but still a bit!); since the requirement is that my result should have equal cardinality to max(|A|, |B|), I can actually just stop adding entries to my result when I've reached the target cardinality. Seems okay...
UPDATE 3: Resurrecting this old question, I've recently had some problems arise from the fact that the above algorithm doesn't fulfill my requirement |S| == max(|A|, |B|) (where S is the set of pairings). If anyone knows of a simple way of ensuring this it would be much appreciated. (I'll obviously be poking away at possible changes.)
This is an easy task:
Add all edges of the graph in a priority_queue, where the biggest priority is the edge with the biggest weight.
Look each edge e = (u, v, w) in the priority_queue, where u is in A, v is in B and w is the weight.
If removing e from the graph doesn't leave u or v isolated, remove it.
Otherwise, e is part of the answer.
This should be enough for your case:
#include <bits/stdc++.h>
using namespace std;
struct edge {
int u, v, w;
edge(){}
edge(int up, int vp, int wp){u = up; v = vp; w = wp;}
void print(){ cout<<"("<<u<<", "<<v<<")"<<endl; }
bool operator<(const edge& rhs) const {return w < rhs.w;}
};
vector<edge> E; //edge set
priority_queue<edge> pq;
vector<edge> ans;
int grade[5] = {3, 3, 2, 2, 2};
int main(){
E.push_back(edge(0, 2, 1)); E.push_back(edge(0, 3, 1)); E.push_back(edge(0, 4, 4));
E.push_back(edge(1, 2, 5)); E.push_back(edge(1, 3, 2)); E.push_back(edge(1, 4, 0));
for(int i = 0; i < E.size(); i++) pq.push(E[i]);
while(!pq.empty()){
edge e = pq.top();
if(grade[e.u] > 1 && grade[e.v] > 1){
grade[e.u]--; grade[e.v]--;
}
else ans.push_back(e);
pq.pop();
}
for(int i = 0; i < ans.size(); i++) ans[i].print();
return 0;
}
Complexity: O(E lg(E)).
I think this problem is "minimum weighted bipartite matching" (although searching for " maximum weighted bipartite matching" would also be relevant, it's just the opposite)
I'm trying to use class behavior in lua where there is a ship that has two other classes pos and vector
But I cannot get it working like I assumed I would be able to
Point = {x=0, y=0}
function Point:new(p)
p = p or {}
return p
end
Ship =
{
pos = {Point:new{x=0,y=0}},
vector = {Point:new{x=0,y=0}} -- I thought this would be sufficient for access being available for vector
}
-- create new ship class
function Ship:new(pos)
p = p or {}
p.pos = pos
p.vector = Point:new{x=0,y=0} -- I need to do this or accessing vector will crash (The problem)
return p
end
-- Create new ship...
plrShip = Ship:new{}
plrShip.pos.x = 300
plrShip.pos.y = 300
If anyone knows how to make the above code cleaner/better I would be thankful
You can use metatables to set default fields. (I've made some assumptions about what you're trying to do. If this doesn't work for you, please add some clarification to your question.)
local Point = {x=0, y=0}
Point.__index = Point
function Point:new(p)
p = p or {}
setmetatable(p, self)
return p
end
-- create new ship class
local Ship = {}
Ship.__index = Ship
function Ship:new(pos)
setmetatable(pos, Point)
local p = {pos = pos, vector = Point:new()}
setmetatable(p, self)
return p
end
-- Create new ship...
local plrShip = Ship:new{}
plrShip.pos.x = 300
plrShip.pos.y = 300
I found solution to do this, it still not perfect but only I got working was this modified code:
Ship =
{
pos = Point:new{x=0,y=0},
vector = Point:new{x=0,y=0}
}
function Ship:new()
p = p or {}
p.pos = self.pos
p.vector = self.vector
return p
end
plrShip = Ship:new()
plrShip.pos.x = 300
plrShip.pos.y = 300
I figured out how to compare dates in Google Sheets but when I try to enter more dates for some reason all the cells that were green and red become all red. Also how can I make two cells red if only one cell has a date?
Example: In cell D18 the Due date is 4-18-2014 and in cell E18 the cell is blank. I want to make both cells red so I would know that I should find out why is that cell red.
This is the code I have so far:
function onEdit() {
var ss = SpreadsheetApp.getActiveSpreadsheet();
var s = ss.getSheetByName('Copy of Project Sheet 1');
var values1Rule1 = s.getRange('E2:E1000').getValues();
var values2Rule1 = s.getRange('D2:D1000').getValues();
var range3Rule1 = s.getRange('D2:E2');
var color1 = 'Red';
var color2 = 'Green';
for (var row in values1Rule1) {
for (var col in values1Rule1[row]) {
if (values1Rule1[row][col] > values2Rule1[row][col]) s.getRange(s.getRange('D2').offset(row, col, 1, 2).getA1Notation()).setBackgroundColor(color1);
else if (values1Rule1[row][col] < values2Rule1[row][col]) s.getRange(s.getRange('D2').offset(row, col, 1, 2).getA1Notation()).setBackgroundColor(color2);
else s.getRange(s.getRange('D2').offset(row, col, 1, 2).getA1Notation()).setBackgroundColor('white'); }}
};
All you need to do is add this condition as an OR clause in your red condition, e.g.
if (values1Rule1[row][col] > values2Rule1[row][col] || values1Rule1[row][col] === '')
But there's lots of "minor" problems with your code. First of all, you're doing way too many API calls unnecessarily. This is a big performance issue. For example, when you offset, you already have the new range, there's no need to getA1Notation then get the range again, you could do:
s.getRange('D2').offset(row, col, 1, 2).setBackgroundColor(color1);
But that's still two calls, getting D2, then offseting. You could get the desired range at once:
s.getRange(row+1, 4, 1, 2).setBackgroundColor(color1);
I'd go even further and build a matrix of colors and set it all at once after the loop:
s.getRange('D2:E1000').setBackgroundColors(colors);
But even better, inside an onEdit you should only work on what has just being edited, instead of triggering a full recalculation of your colors because the user edited something on another column or another sheet entirely.
I think your code should be something like this:
function onEdit(e) {
var ss = e.source;
var s = ss.getActiveSheet();
if( s.getName() !== 'Copy of Project Sheet 1' ) return; //only interested in one sheet
var r = s.getActiveRange();
var c = r.getColumn();
if( c !== 4 && c !== 5 ) return; //only interested in changes on columns D or E
r = r.offset(0, c === 4 ? 0 : -1, 1, 2);
var v = r.getValues()[0];
r.setBackgroundColor(v[1] === '' || v[1] > v[0] ? 'red' : v[1] < v[0] ? 'green' : 'white');
}
--edit
You can not run this function manually directly, because it needs a parameter that is passed only when it runs automatically. But you can emulate it with a test function, like this:
function testEdit() { onEdit({source:SpreadsheetApp.getActive()}); }
I need to merge two tables, with the contents of the second overwriting contents in the first if a given item is in both. I looked but the standard libraries don't seem to offer this. Where can I get such a function?
for k,v in pairs(second_table) do first_table[k] = v end
Here's what i came up with based on Doug Currie's answer:
function tableMerge(t1, t2)
for k,v in pairs(t2) do
if type(v) == "table" then
if type(t1[k] or false) == "table" then
tableMerge(t1[k] or {}, t2[k] or {})
else
t1[k] = v
end
else
t1[k] = v
end
end
return t1
end
Wouldn't this work properly?
function merge(t1, t2)
for k, v in pairs(t2) do
if (type(v) == "table") and (type(t1[k] or false) == "table") then
merge(t1[k], t2[k])
else
t1[k] = v
end
end
return t1
end
For numeric-index table merging:
for k,v in pairs(secondTable) do table.insert(firstTable, v) end
Doug Currie's answer is the simplest for most cases. If you need more robust merging of tables, consider using the merge() method from the Penlight library.
require 'pl'
pretty.dump(tablex.merge({a=1,b=2}, {c=3,d=4}, true))
-- {
-- a = 1,
-- d = 4,
-- c = 3,
-- b = 2
-- }
Here's iterative version for deep merge because I don't like potential stack overflows of recursive.
local merge_task = {}
function merge_to_left_o(orig, new)
merge_task[orig] = new
local left = orig
while left ~= nil do
local right = merge_task[left]
for new_key, new_val in pairs(right) do
local old_val = left[new_key]
if old_val == nil then
left[new_key] = new_val
else
local old_type = type(old_val)
local new_type = type(new_val)
if (old_type == "table" and new_type == "table") then
merge_task[old_val] = new_val
else
left[new_key] = new_val
end
end
end
merge_task[left] = nil
left = next(merge_task)
end
end
I preferred James version for its simplicity and use it in my utils.lua - i did add a check for table type for error handling.
function merge(a, b)
if type(a) == 'table' and type(b) == 'table' then
for k,v in pairs(b) do if type(v)=='table' and type(a[k] or false)=='table' then merge(a[k],v) else a[k]=v end end
end
return a
end
Thanks for this nice function which should be part of the table class so you could call a:merge(b) but doing table.merge = function(a, b) ... did not work for me. Could even be compressed to a one liner for the real nerds :)
Like Doug Currie said, you can use his function, but there is a problem with his method. If first_table has things in it's k index, the function will over write it.
I'm assuming you're trying to merge these tables, not overwrite index's and value's. So this would be my method, it's very similar but is used for merging tables.
for _, v in pairs(second_table) do table.insert(first_table, v) end
The only problem with this solution is that the index is set as numbers, not as strings. This will work with tables with numbers as the index, and for tables with strings as their index, use Doug Currie's method.
Doug Currie's method:
for k,v in pairs(second_table) do first_table[k] = v end
Extending this great answer, https://stackoverflow.com/a/1283399/1570165, I would like to go with a (pure) functional approach like this one below:
-- example values
local t1 = { a = 0, b = 2 }
local t2 = { a = 1, c = 3 }
-- merge function that takes functional approach
local merge = function(a, b)
local c = {}
for k,v in pairs(a) do c[k] = v end
for k,v in pairs(b) do c[k] = v end
return c
end
-- t1 and t2 value still same after merge
print(merge(t1, t2)) -- { a = 1, b = 2, c = 3 }
print(t2) -- { a = 1, c = 3 }
print(t1) -- { a = 0, b = 2 }
for k,v in pairs(t2) do t1[k] = v end
key for string solution