Generating a list of random numbers, summing to a fixed amount using SQL - sql

Question

This question is about generating N random numbers whose sum is M(constant) using SQL.
For example we have a number M=9.754. We need 10 random numbers whose sum is 9.754.
It should be done in SQL Server environment.
Can anybody help me?

Answer 1

While the #Squirrel answer is interesting but numbers here is more random
here is the code:
DECLARE #s INT=1,
#k FLOAT=0,
#final FLOAT=9.917,
#sum FLOAT =0,
#min FLOAT=1,
#max FLOAT=9.917
BEGIN
WHILE (#sum <> #final)
BEGIN
WHILE (#s <= 10)
BEGIN
SET #k =
(
SELECT ROUND(RAND(CHECKSUM(NEWID())) * (#max - #min) + #min,3)
);
PRINT (CONCAT('random: ',#k));
IF(#sum+#k <=#final)
SET #sum+=#k;
SET #max=#final-#sum;
PRINT (CONCAT('computed sum: ',#k));
IF(#max>1) SET #min=1 ELSE SET #min=0;
IF(#sum=#final)
BREAK;
SET #s = #s + 1;
SET #k = #k + 0;
END;
PRINT (CONCAT('final', #final))
PRINT (CONCAT('sum', #sum))
IF(#sum<>#final)--force stop if after 10 try the sum not match with final
BEGIN
PRINT(CONCAT('final random number:',#final-#sum))
SET #sum=#final;
END;
SET #s=0;
IF(#sum=#final)
BEGIN
PRINT('****************************DONE****************************')
BREAK;
END
END;
PRINT ('end');
END;

Answer 2

Interesting requirement.
Query below uses tally table / number table to generate 10 random numbers after that find the ratio. Final query check for case where sum of the numbers is not equal to #m and make the adjustment on the biggest number.
declare #m decimal(10,3) = 9.754,
#n int = 10;
with
-- using recursive CTE to generate a number table
numbers as
(
select n = 1
union all
select n = n + 1
from numbers
where n < #n
),
-- generate random positive numbers using newid()
-- Note : 100 is chosen arbitrary
num as
(
select n = abs(checksum(newid())) % 100
from numbers
),
-- calculate the ratio
ratio as
(
select r,
rn = row_number() over (order by r desc),
sum_r = sum(r) over()
from
(
select r = convert(decimal(10,3), n * 1.0 / sum(n) over() * #m)
from num
) r
)
-- sum(r) may not equal to #m due to rounding
-- find the difference and adjust it to the biggest r
select r, rn, sum_r,
adj_r = r + case when rn = 1 then #m - sum_r else 0 end,
sum_adj_r = sum(r + case when rn = 1 then #m - sum_r else 0 end) over()
from ratio
dbfiddle demo