# SQL Server triangular upper matrix

MTRIU

Updated: 31 January 2012

Use the scalar function MTRIU to return the upper triangular part of the string representation of a matrix.
MTRIU expects a string representation of the matrix, with columns separated by commas and rows separated by semi-colons.
Syntax
SELECT [wctMath].[wct].[MTRIU](
<@Matrix, nvarchar(max),>)
Arguments
@Matrix
a string representation of a matrix.
Return Types
[nvarchar](max)
Remarks
·         The string representations of @Matrix must only contain numbers, commas (to separate the columns), and semi-colons to separate the rows.
·         Consecutive commas will generate an error.
·         Consecutive semi-colons will generate an error.
·         Non-numeric data between commas will generate an error
·         Non-number data between semi-colons will generate an error
·         To convert non-normalized data to a string format, use the Matrix2String or the Matrix2String_q function.
·         To convert normalized data to a string format, use the NMatrix2String or the NMatrix2String_q function.
Examples
Let’s assume that we had the following matrix, A, and we want to return the upper triangular part.
A = [-79,-45,9,9,-91,-5;68,46,9,81,-61,35;83,-25,80,-67,-22,-38;77,40,-24,69,73,-20;-17,-72,-9,-72,-6,-34;64,-47,48,-54,18,11;-4,-36,7,-56,-34,-3;-41,90,78,-43,38,64;-60,-85,-31,-83,-96,-36;-40,31,-93,-62,64,10]
We could enter the following SQL to perform the calculation.
DECLARE @A as varchar(max)

SET @A = '-79,-45,9,9,-91,-5;68,46,9,81,-61,35;83,-25,80,-67,-22,-38;77,40,-24,69,73,-20;-17,-72,-9,-72,-6,-34;64,-47,48,-54,18,11;-4,-36,7,-56,-34,-3;-41,90,78,-43,38,64;-60,-85,-31,-83,-96,-36;-40,31,-93,-62,64,10'

SELECT wct.MTRIU(@A) as [U]
This produces the following result.
U
-----------------------------------------------------------------------------
-79,-45,9,9,-91,-5;0,46,9,81,-61,35;0,0,80,-67,-22,-38;0,0,0,69,73,-20;0,0,0,0,-6,-34;0,0,0,0,0,11;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0;0,0,0,0,0,0

Of course, this is a little hard to read. Since the result is a string, we can reformat the solution to make it easier to read. Simply by changing the SELECT statement:
SELECT l.StringSegment as [U]
FROM wctString.wct.SPLIT(';',(SELECT wct.MTRIU(@A))) l
This produces the following result:
U
-----------------------------------------
-79,-45,9,9,-91,-5
0,46,9,81,-61,35
0,0,80,-67,-22,-38
0,0,0,69,73,-20
0,0,0,0,-6,-34
0,0,0,0,0,11
0,0,0,0,0,0
0,0,0,0,0,0
0,0,0,0,0,0
0,0,0,0,0,0

Which is a little bit easier to follow

However, we can use the table-valued function MATRIX, to format the result in third-normal form where it is even easier to see the output.

SELECT *
FROM wct.MATRIX((SELECT wct.MTRIU(@A))) l
This produces the following result.
RowNum      ColNum              ItemValue
----------- ----------- ----------------------
0           0                    -79
0           1                    -45
0           2                      9
0           3                      9
0           4                    -91
0           5                     -5
1           0                      0
1           1                     46
1           2                      9
1           3                     81
1           4                    -61
1           5                     35
2           0                      0
2           1                      0
2           2                     80
2           3                    -67
2           4                    -22
2           5                    -38
3           0                      0
3           1                      0
3           2                      0
3           3                     69
3           4                     73
3           5                    -20
4           0                      0
4           1                      0
4           2                      0
4           3                      0
4           4                     -6
4           5                    -34
5           0                      0
5           1                      0
5           2                      0
5           3                      0
5           4                      0
5           5                     11
6           0                      0
6           1                      0
6           2                      0
6           3                      0
6           4                      0
6           5                      0
7           0                      0
7           1                      0
7           2                      0
7           3                      0
7           4                      0
7           5                      0
8           0                      0
8           1                      0
8           2                      0
8           3                      0
8           4                      0
8           5                      0
9           0                      0
9           1                      0
9           2                      0
9           3                      0
9           4                      0
9           5                      0

And, if we wanted to see the result in a row/column presentation, we could use the following SQL.

SELECT [0],[1],[2],[3],[4],[5]
FROM (
SELECT *
FROM wct.MATRIX((SELECT wct.MTRIU(@A)))
) M PIVOT(
MAX(ItemValue)
FOR colnum IN([0],[1],[2],[3],[4],[5])
) AS pvt
ORDER BY rownum
This produces the following result.

In this example, we insert the matrix values into a table, #m, which is in ‘spreadsheet’ format, and we use the MATRIX2SRTING function to convert the table values into a string format to be used by the MTRIU function.
SELECT *
INTO #m
FROM (
SELECT -79,-45, 9, 9,-91, -5 UNION ALL
SELECT 68, 46, 9, 81,-61, 35 UNION ALL
SELECT 83,-25, 80,-67,-22,-38 UNION ALL
SELECT 77, 40,-24, 69, 73,-20 UNION ALL
SELECT -17,-72, -9,-72, -6,-34 UNION ALL
SELECT 64,-47, 48,-54, 18, 11 UNION ALL
SELECT -4,-36, 7,-56,-34, -3 UNION ALL
SELECT -41, 90, 78,-43, 38, 64 UNION ALL
SELECT -60,-85,-31,-83,-96,-36 UNION ALL
SELECT -40, 31,-93,-62, 64, 10
) n(x0,x1,x2,x3,x4,x5)

SELECT [0],[1],[2],[3],[4],[5]
FROM (
SELECT *
FROM wct.MATRIX((SELECT wct.MTRIU(wct.MATRIX2STRING('#m','*','',NULL))))
) M PIVOT(
MAX(ItemValue)
FOR colnum IN([0],[1],[2],[3],[4],[5])
) AS pvt
ORDER BY rownum

This produces the following result.