I'm looking for the name (and for the code : in PL/SQL or PG/SQL) of the algorithm which is finding all couple (2-uplets) of a set.
Example :
A - B - C
Result :
1 : A - B
2 : A - C
3 : B - C
I know that the powerset algorithm do this part of the job, but I'm looking for an optimised couple finder algorithm.
Link for the powerset pg/sql algorithm : https://www.postgresql.org/message-id/20060924054759.GA71934%40winnie.fuhr.org
Have you considered something like
Select A.x, B.x
From YourTable as A, YourTable as B
Where A.key <> B.key
You mention SQL so this might be preferable. Note that the number of rows in the cross product is about the same as the number of pairs, so it isn't terribly inefficient.
I've build the solution :
CREATE OR REPLACE FUNCTION twouplets(a anyarray)
RETURNS SETOF anyarray AS
$BODY$
DECLARE
retval a%TYPE;
size integer := array_upper(a, 1);
i integer;
j integer;
BEGIN
i := 0;
j := 1;
FOR i IN 1 .. size LOOP
FOR j IN 1 .. size-i LOOP
retval := '{}';
retval := array_append(retval, a[i]);
retval := array_append(retval, a[i+j]);
RETURN NEXT retval;
END LOOP;
END LOOP;
RETURN;
END;
$BODY$
LANGUAGE plpgsql IMMUTABLE STRICT
COST 100
ROWS 1000;
ALTER FUNCTION twouplets(anyarray)
OWNER TO postgres;
Related
I'm learning algorithms and I'm trying to make an algorithm that extracts numbers lets say n in [1..100] from a string. Hopefully I get an easier algorithm.
I tried the following :
procedure ReadQuery(var t : tab); // t is an array of Integer.
var
x,v,e : Integer;
inputs : String;
begin
//readln(inputs);
inputs:='1 2 3';
j:= 1;
// make sure that there is one space between two integers
repeat
x:= pos(' ', inputs); // position of the space
delete(inputs, x, 1)
until (x = 0);
x:= pos(' ', inputs); // position of the space
while x <> 0 do
begin
x:= pos(' ', inputs); //(1) '1_2_3' (2) '2_3'
val(copy(inputs, 1, x-1), v, e); // v = value | e = error pos
t[j]:=v;
delete(inputs, 1, x); //(1) '2_3' (2) '3'
j:=j+1; //(1) j = 2 (2) j = 3
//writeln(v);
end;
//j:=j+1; // <--- The mistake were simply here.
val(inputs, v, e);
t[j]:=v;
//writeln(v);
end;
I get this result ( resolved ) :
1
2
0
3
expected :
1
2
3
PS : I'm not very advanced, so excuse me for reducing you to basics.
Thanks for everyone who is trying to share knowledge.
Your code is rather inefficient and it also doesn't work for strings containing numbers in general.
A standard and performant approach would be like this:
type
TIntArr = array of Integer;
function GetNumbers(const S: string): TIntArr;
const
AllocStep = 1024;
Digits = ['0'..'9'];
var
i: Integer;
InNumber: Boolean;
NumStartPos: Integer;
NumCount: Integer;
procedure Add(Value: Integer);
begin
if NumCount = Length(Result) then
SetLength(Result, Length(Result) + AllocStep);
Result[NumCount] := Value;
Inc(NumCount);
end;
begin
InNumber := False;
NumCount := 0;
for i := 1 to S.Length do
if not InNumber then
begin
if S[i] in Digits then
begin
NumStartPos := i;
InNumber := True;
end;
end
else
begin
if not (S[i] in Digits) then
begin
Add(StrToInt(Copy(S, NumStartPos, i - NumStartPos)));
InNumber := False;
end;
end;
if InNumber then
Add(StrToInt(Copy(S, NumStartPos)));
SetLength(Result, NumCount);
end;
This code is intentionally written in a somewhat old-fashioned Pascal way. If you are using a modern version of Delphi, you wouldn't write it like this. (Instead, you'd use a TList<Integer> and make a few other adjustments.)
Try with the following inputs:
521 cats, 432 dogs, and 1487 rabbits
1 2 3 4 5000 star 6000
alpha1beta2gamma3delta
a1024b2048cdef32
a1b2c3
32h50s
5020
012 123!
horses
(empty string)
Make sure you fully understand the algorithm! Run it on paper a few times, line by line.
I need to write a PL/SQL program to print out the first 10 numbers of Lazy Caterer's sequence. I'm having trouble. I don't understand how to do this
Lazy caterer's sequence has the formula F(1)=2, F(n)=F(n-1)+n.
E.g., F(2) = F(1) + 2 = 2+2=4; F(3) = F(2)+3 = 4+3=7
Please use a loop. The numbers you print out should look like
2
4
7
...
56
DECLARE
n NUMBER := 1;
BEGIN
FOR i IN 1..10 LOOP
n := n + i;
DBMS_OUTPUT.PUT_LINE(n);
END LOOP;
END;
/
I cheated and looked up Lazy Caterer's algorithm on Wikipedia. This is an implementation of the simplistic version of the formula:
create or replace function lc_algo
(n in number)
return number as
begin
return (power(n,2) + n + 2) / 2;
end;
/
I admit it doesn't use a loop, but it's hard to see why anybody would need to. However, if your assignment insists on loops (because performance is not your bag) here you go:
create or replace function lc_algo
(n in number)
return number
as
rv simple_integer := 1;
begin
for i in 1..n loop
rv := rv + i;
end loop;
return rv;
end;
/
To get the first ten numbers (with whatever version):
select lc_algo(level) -- assuming you don't want to start with f(0)
from dual
connect by level <= 10
/
How can I transpose matrix in ADA?. I´ve tried:
procedure transpose(A: in out matrix) is
B : matrix(1..A'Last(2),1..A'Last(1));
begin
for i in A'Range(1) loop
for j in A'Range(2) loop
B(j,i):= A(i,j);
end loop;
end loop;
A := B;
end transpose;
but it doesn´t work when A isn´t a square Matrix.
Any help would be appreciated.
As a procedure this can never work for non-square matrices because the output is a different constrained type from the input. However you can return B from a suitable function.
function transpose(A: in matrix) return matrix is
B : matrix(A'Range(2),A'Range(1));
begin
for i in A'Range(1) loop
for j in A'Range(2) loop
B(j,i):= A(i,j);
end loop;
end loop;
return B;
end transpose;
The easiest way to create a matrix of the right constrained type for the result is a declare block:
declare
Transposed : Matrix := Transpose(A);
begin
-- operations on the transposed matrix
end;
You can transpose a matrix by creating a record like:
type Matrix_Type is record
Data : array (1..MAX_SIZE, 1..MAX_SIZE) of Float;
Last_Row : Positive range 1 .. MAX_SIZE;
Last_Column : Positive range 1 .. MAX_SIZE;
end record;
This record can support matrices of any m by n size up to n,m <= MAX_SIZE
Your procedure becomes:
procedure transpose(A: in out Matrix_Type) is
B : Matrix_Type;
begin
for i in 1..A.Last_Row loop
for j in 1..A.Last_Column loop
B(j,i):= A(i,j);
end loop;
end loop;
B.Last_Column := A.Last_Row;
B.Last_Row := A.Last_Column;
A := B;
end transpose;
The easiest way I know of is to use the Fortran convention on an intermediate array-type. (This is because Fortran convention is column-major while in Ada, it is row-major.)
Function Transpose(M : Matrix) return Matrix is
subtype Constrained is Matrix(M'Range(2), M'Range(1));
Type Xposed is new Matrix with Convention => Fortran;
Temp : Xposed := Xposed(M);
Result : Constrained with Import, Address => Temp'Address; --'
begin
Return Result;
end Transpose;
The following pl/sql program generates an error on execution on line the sum :=temp*sum; encountered symbol ; when expecting ( . Please explain my mistake.
declare
n number;
temp number;
sum number := 1;
begin
n := &n;
temp := n;
while temp>0 loop
sum := temp*sum;
temp := temp-1;
end loop;
dbms_output.put_line('Factorial of '||n||' is '||sum);
end;
/
Maybe not the answer to your question, but there is no need for PL/SQL here:
select round(exp(sum(ln(level))))
from dual
connect by level <= 5;
where 5 is your number (5!).
Additionally, if you like to operate faster in PL/SQL use pls_integer instead of number.
UPDATE
So according to comments I felt free to test:
create or replace package test_ is
function by_query(num number) return number deterministic;
function by_plsql(num number) return number deterministic;
end test_;
/
create or replace package body test_ is
function by_query(num number) return number deterministic
is
res number;
begin
select round(exp(sum(ln(level))))
into res
from dual
connect by level <= num;
return res;
end;
function by_plsql(num number) return number deterministic
is
n number := 0;
begin
for i in 1..num loop
n := n + ln(i);
end loop;
return round(exp(n));
end;
end test_;
So there are two functions with different content. Test query:
declare
dummy number;
begin
for i in 1..10000 loop
dummy := test_.by_query(5);
end loop;
end;
0.094 sec.
declare
dummy number;
begin
for i in 1..10000 loop
dummy := test_.by_plsql(5);
end loop;
end;
0.094 sec.
You'll say I am cheater and using deterministic keyword but here it is obvious and is needed by logic. If I remove it, the same scripts are working 1.7 sec vs 1.3 sec, so procedure is only a bit faster, there is no even double-win in performance. The totally opposite effect you will get if you use the function in a query so it is a fair trade.
Sum is reserved word in sql. Change variable name like
declare
n number;
temp number;
sum_ number := 1;
begin
n := &n;
temp := n;
while temp>0 loop
sum_ := temp*sum_;
temp := temp-1;
end loop;
dbms_output.put_line('Factorial of '||n||' is '||sum_);
end;
/
declare
n number;
i number;
sum_of_log_10s number;
exponent number;
base number;
begin
n := &n;
i := 1;
sum_of_log_10s := 0;
while i <= n loop
-- do stuff
sum_of_log_10s := sum_of_log_10s + log(10,i);
i := i + 1;
end loop;
dbms_output.put_line('sum of logs = '||sum_of_log_10s);
exponent := floor(sum_of_log_10s);
base := power(10,sum_of_log_10s - exponent);
dbms_output.put_line(n||'! = '||base||' x 10^'||exponent);
end;
I came up with this code that I like even better than #smnbbrv's answer. It's a great way to check the speed of a machine. I've been using a variation of this since my Atari 800
ALTER SESSION FORCE PARALLEL DDL PARALLEL 16;
ALTER SESSION FORCE PARALLEL DML PARALLEL 16;
ALTER SESSION FORCE PARALLEL QUERY PARALLEL 16;
with t as (
select /*+materialize*/
rownum i
from dual connect by rownum < 100000 -- put number to calculate n! here
)
,t1 as (
select /*+parallel(t,16)*/ /*+materialize*/
sum(log(10,i)) logsum
from t
)
select
trunc(power(10,(mod(logsum,1))),3) ||' x 10^'||trim(to_char(floor(logsum),'999,999,999,999')) factorial
-- logsum
from t1
;
-- returns 2.824 x 10^456,568
Here is the simple code for finding factorial of number at run time...
declare
-- it gives the final answer after computation
fac number :=1;
-- given number n
-- taking input from user
n number := &1;
-- start block
begin
-- start while loop
while n > 0 loop
-- multiple with n and decrease n's value
fac:=n*fac;
--dbms_output.put(n||'*');
n:=n-1;
end loop;
-- end loop
-- print result of fac
dbms_output.put_line(fac);
-- end the begin block
end;
I have a project where the program must accept 10 words and display the words in descending order (alphabetical order from Z-A)
using bubble sorting.
Here's what I know so far:
Program sample;
uses crt;
TYPE
no._list=ARRAY(1...10)OF REAL;
CONST
no.:no._list=(20.00,50.50.35.70....);
VAR
x:INTEGER;
small:REAL;
BEGIN clrscr:
small:=no.(1);
FOR x:=2 TO 10 DO
IF small>number(x);
writeln('Smallest value in the array of no.s is',small:7:2);
END
I really don't know how to do this though and could use some help.
Here's a video by Alister Christie on Bubble sort describing the principle :
http://codegearguru.com/index.php?option=com_content&task=view&id=64&Itemid=1
The algorithm in Pascal can be found # http://delphi.wikia.com/wiki/Bubble_sort
function BubbleSort( list: TStringList ): TStringList;
var
i, j: Integer;
temp: string;
begin
// bubble sort
for i := 0 to list.Count - 1 do begin
for j := 0 to ( list.Count - 1 ) - i do begin
// Condition to handle i=0 & j = 9. j+1 tries to access x[10] which
// is not there in zero based array
if ( j + 1 = list.Count ) then
continue;
if ( list.Strings[j] > list.Strings[j+1] ) then begin
temp := list.Strings[j];
list.Strings[j] := list.Strings[j+1];
list.Strings[j+1] := temp;
end; // endif
end; // endwhile
end; // endwhile
Result := list;
end;