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• catching a deadlock in a simple odd-even sending

  - by user562264

  I'm trying to solve a simple problem with MPI, my implementation is MPICH2 and my code is in fortran. I have used the blocking send and receive, the idea is so simple but when I run it it crashes!!! I have absolutely no idea what is wrong? can anyone make quote on this issue please? there is a piece of the code: integer,parameter::IM=100,JM=100 REAL,ALLOCATABLE ::T(:,:),TF(:,:) CALL MPI_COMM_RANK(MPI_COMM_WORLD,RNK,IERR) CALL MPI_COMM_SIZE(MPI_COMM_WORLD,SIZ,IERR) prv = rnk-1 nxt = rnk+1 LIM = INT(IM/SIZ) IF (rnk==0) THEN ALLOCATE(TF(IM,JM)) prv = MPI_PROC_NULL ELSEIF(rnk==siz-1) THEN NXT = MPI_PROC_NULL LIM = LIM+MOD(IM,SIZ) END IF IF (MOD(RNK,2)==0) THEN CALL MPI_SEND(T(2,:),JM+2,MPI_REAL,PRV,10,MPI_COMM_WORLD,IERR) CALL MPI_RECV(T(1,:),JM+2,MPI_REAL,PRV,20,MPI_COMM_WORLD,STAT,IERR) ELSE CALL MPI_RECV(T(LIM+2,:),JM+2,MPI_REAL,NXT,10,MPI_COMM_WORLD,STAT,IERR) CALL MPI_SEND(T(LIM+1,:),JM+2,MPI_REAL,NXT,20,MPI_COMM_WORLD,IERR) END IF as I understood even processes are not receiving anything while the odd ones finish sending successfully, in some cases when I added some print to observe what is going on I saw that the variable NXT is changing during the sending procedure!!! for example all the odd process was sending message to process 0 not their next one!

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• How can i get rid of 'ORA-01489: result of string concatenation is too long' in this query?

  - by core_pro

  this query gets the dominating sets in a network. so for example given a network A<----->B B<----->C B<----->D C<----->E D<----->C D<----->E F<----->E it returns B,E B,F A,E but it doesn't work for large data because i'm using string methods in my result. i have been trying to remove the string methods and return a view or something but to no avail With t as (select 'A' as per1, 'B' as per2 from dual union all select 'B','C' from dual union all select 'B','D' from dual union all select 'C','B' from dual union all select 'C','E' from dual union all select 'D','C' from dual union all select 'D','E' from dual union all select 'E','C' from dual union all select 'E','D' from dual union all select 'F','E' from dual) ,t2 as (select distinct least(per1, per2) as per1, greatest(per1, per2) as per2 from t union select distinct greatest(per1, per2) as per1, least(per1, per2) as per1 from t) ,t3 as (select per1, per2, row_number() over (partition by per1 order by per2) as rn from t2) ,people as (select per, row_number() over (order by per) rn from (select distinct per1 as per from t union select distinct per2 from t) ) ,comb as (select sys_connect_by_path(per,',')||',' as p from people connect by rn > prior rn ) ,find as (select p, per2, count(*) over (partition by p) as cnt from ( select distinct comb.p, t3.per2 from comb, t3 where instr(comb.p, ','||t3.per1||',') > 0 or instr(comb.p, ','||t3.per2||',') > 0 ) ) ,rnk as (select p, rank() over (order by length(p)) as rnk from find where cnt = (select count(*) from people) order by rnk ) select distinct trim(',' from p) as p from rnk where rnk.rnk = 1`

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• SQL SERVER – Introduction to PERCENT_RANK() – Analytic Functions Introduced in SQL Server 2012

  - by pinaldave

  SQL Server 2012 introduces new analytical functions PERCENT_RANK(). This function returns relative standing of a value within a query result set or partition. It will be very difficult to explain this in words so I’d like to attempt to explain its function through a brief example. Instead of creating a new table, I will be using the AdventureWorks sample database as most developers use that for experiment purposes. Now let’s have fun following query: USE AdventureWorks GO SELECT SalesOrderID, OrderQty, RANK() OVER(ORDER BY SalesOrderID) Rnk, PERCENT_RANK() OVER(ORDER BY SalesOrderID) AS PctDist FROM Sales.SalesOrderDetail WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY PctDist DESC GO The above query will give us the following result: Now let us understand the resultset. You will notice that I have also included the RANK() function along with this query. The reason to include RANK() function was as this query is infect uses RANK function and find the relative standing of the query. The formula to find PERCENT_RANK() is as following: PERCENT_RANK() = (RANK() – 1) / (Total Rows – 1) If you want to read more about this function read here. Now let us attempt the same example with PARTITION BY clause USE AdventureWorks GO SELECT SalesOrderID, OrderQty, ProductID, RANK() OVER(PARTITION BY SalesOrderID ORDER BY ProductID ) Rnk, PERCENT_RANK() OVER(PARTITION BY SalesOrderID ORDER BY ProductID ) AS PctDist FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY PctDist DESC GO Now you will notice that the same logic is followed in follow result set. I have now quick question to you – how many of you know the logic/formula of PERCENT_RANK() before this blog post? Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Function, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

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• CSS - image overflow tricks

  - by rnk

  I'm stucking with some design techniques where I want to make image item box like this Where the actual image can be upto maximum height of 300px and width of 225px. The width and height of the item box including the image and the text is 190px x 190px I'm using this image http://farm8.staticflickr.com/7122/7424355198_72620895bd_m.jpg I tried using overflow: hidden for the image item to hide it's height below to show the image text. But I'm getting only like this http://jsfiddle.net/Dkh4q/ Could anyone tell the mistake I've done? Thanks! UPDATE For more information about the expected result, if you can login to zerply, then please check this http://zerply.com/christievdc/portfolio for example.

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