I’ve stumbled upon this very interesting question on Stack Overflow, recently. Its title is:

In this article, we’ll compare the user’s imperative approach to the extremely elegant (Oracle) SQL approach. We’ll be making use of any combination of these awesome SQL features:

- Window functions
- FIRST and LAST functions, using Oracle-specific syntax
`LATERAL JOIN`

(or`APPLY`

- Recursive SQL

### The problem

The user alhashmiya who had asked this question, was looking for a solution to the problem of finding the “closest” sum of elements in a subset of numbers A to a set of “expected” sums B. More concretely, alhasmiya had the following two tables:

ID ASSIGN_AMT -------------- 1 25150 2 19800 3 27511

And…

ID WORK_AMT ------------ 1 7120 2 8150 3 8255 4 9051 5 1220 6 12515 7 13555 8 5221 9 812 10 6562

The `ASSIGN_AMT`

value is the “expected” sum. What alhashmiya was looking for is the sum of a subset of `WORK_AMT`

values A, such that this sum is as close as possible to any of the “expected” sums. There are two ways to understand this problem:

- The possible “closest” sums are restricted to be the sums obtained in a strictly defined order, e.g. ordered by
`ID`

. An application of this understanding is to find out the exact moment when a well-defined, ordered running total (e.g. bank account balance) has exceeded a certain threshold - The possible “closest” sums are unrestricted. Any unordered subset qualifies to calculate such a sum. An application of this understanding is to find a combination of discrete values to reach a target value as closely as possible, e.g. to optimise a trade.

The second understanding is called the “subset sum problem”, for which there are only exponential algorithms if you’re looking for an exact solution. **It is important to notice that this algorithm will NOT scale well at all, regardless of the solution technique!**

But let’s look at the simpler understanding first:

## 1. Calculating a sum of an ordered subset of values

By strictly ordered we mean (in the sense of the question) that we want to order all `WORK_AMT`

values, e.g. by `ID`

, and allow only for sums that can appear in this particular order. I.e.

ID WORK_AMT SUBSET_SUM ------------------------ 1 7120 7120 (= WORK_AMT) 2 8150 15270 (= previous SUBSET_SUM + WORK_AMT) 3 8255 23525 (= previous SUBSET_SUM + WORK_AMT) 4 9051 32576 (= ...) 5 1220 33796 6 12515 46311 7 13555 59866 8 5221 65087 9 812 65899 10 6562 72461

The above `SUBSET_SUM`

value is the sum of all `WORK_AMT`

values with `ID <= "current" ID`

. We’ve seen this before on this blog, this is called a running total, and it is best calculated using window functions as follows:

WITH VALS (ID, WORK_AMT) AS ( SELECT 1 , 7120 FROM DUAL UNION ALL SELECT 2 , 8150 FROM DUAL UNION ALL SELECT 3 , 8255 FROM DUAL UNION ALL SELECT 4 , 9051 FROM DUAL UNION ALL SELECT 5 , 1220 FROM DUAL UNION ALL SELECT 6 , 12515 FROM DUAL UNION ALL SELECT 7 , 13555 FROM DUAL UNION ALL SELECT 8 , 5221 FROM DUAL UNION ALL SELECT 9 , 812 FROM DUAL UNION ALL SELECT 10, 6562 FROM DUAL ) SELECT ID, WORK_AMT, SUM (WORK_AMT) OVER (ORDER BY ID) AS SUBSET_SUM FROM VALS ORDER BY ID

The above window function calculates the sum of all `WORK_AMT`

values that are in the “window” of values where the `ID`

is smaller or equal to the current `ID`

.

### Finding the “closest” of these sums with quantified comparison predicates

Now, the task at hand is to find for each value `ASSIGN_AMT`

in `25150`

, `19800`

, and `27511`

the closest value of `SUBSET_SUM`

. In a way, what we are trying to do is we want to minimise the expression `ABS(SUBSET_SUM - ASSIGN_AMT)`

.

However, the `MIN()`

aggregate function won’t help us here, because that will simply return the minimum value of this difference. We want the value of `SUBSET_SUM`

that produces this difference in the first place.

One solution would be to use a **quantified comparison predicate**, a rarely used and not well-known comparison operator that works in all SQL databases:

-- The common table expressions are the same as -- in the previous examples WITH ASSIGN(ID, ASSIGN_AMT) AS ( SELECT 1, 25150 FROM DUAL UNION ALL SELECT 2, 19800 FROM DUAL UNION ALL SELECT 3, 27511 FROM DUAL ), VALS (ID, WORK_AMT) AS ( SELECT 1 , 7120 FROM DUAL UNION ALL SELECT 2 , 8150 FROM DUAL UNION ALL SELECT 3 , 8255 FROM DUAL UNION ALL SELECT 4 , 9051 FROM DUAL UNION ALL SELECT 5 , 1220 FROM DUAL UNION ALL SELECT 6 , 12515 FROM DUAL UNION ALL SELECT 7 , 13555 FROM DUAL UNION ALL SELECT 8 , 5221 FROM DUAL UNION ALL SELECT 9 , 812 FROM DUAL UNION ALL SELECT 10, 6562 FROM DUAL ), SUMS (ID, WORK_AMT, SUBSET_SUM) AS ( SELECT VALS.*, SUM (WORK_AMT) OVER (ORDER BY ID) FROM VALS ) -- This is now the interesting part, where we -- calculate the closest sum SELECT ASSIGN.ID, ASSIGN.ASSIGN_AMT, SUBSET_SUM FROM ASSIGN JOIN SUMS ON ABS (ASSIGN_AMT - SUBSET_SUM) <= ALL ( SELECT ABS (ASSIGN_AMT - SUBSET_SUM) FROM SUMS )

The **quantified comparison predicate** reads intuitively. We’re looking for *that specific SUBSET_SUM* whose difference to the “expected”

`ASSIGN_AMT`

is smaller or equal to all the other possible differences.The above query yields:

ID ASSIGN_AMT SUBSET_SUM -------------------------- 1 25150 23525 2 19800 23525 3 27511 23525

In this case, it’s always the same.

You may have noticed that the solution is not entirely correct in the event where `ASSIGN_AMT`

is allowed to be zero (let’s ignore the possibility of negative values) – in case of which we’ll produce duplicate values in the `JOIN`

. This can be achieved when replacing:

UNION ALL SELECT 4 , 0 FROM DUAL

Now, the result is:

ID ASSIGN_AMT SUBSET_SUM -------------------------- 1 25150 23525 2 19800 23525 2 19800 23525 3 27511 23525

One solution is to remove those duplicates using `DISTINCT`

(which is an anti-pattern. See #6 in this list). A better solution is to make the `JOIN`

predicate unambiguous by comparing `ID`

values as well, i.e.:

SELECT ASSIGN.ID, ASSIGN.ASSIGN_AMT, SUBSET_SUM FROM ASSIGN JOIN SUMS ON (ABS (ASSIGN_AMT - SUBSET_SUM), SUMS.ID) <= ALL ( SELECT ABS (ASSIGN_AMT - SUBSET_SUM), ID FROM SUMS )

The above unfortunately doesn’t work in Oracle (yet), which will report an error:

ORA-01796: this operator cannot be used with lists

Oracle supports comparing tuples / row value expressions only with equal comparators, not with less than / greater than comparators – which is a shame. The same query runs smoothlessly in PostgreSQL.

### Finding the “closest” of these sums with Oracle’s FIRST function

Oracle has a very interesting function to keep the first (or last) values in a set of aggregate values of a group given any particular ordering, and calculating an aggregate function only on those values within the group. The following SQL statement will illustrate this:

SELECT ASSIGN.ID, ASSIGN.ASSIGN_AMT, MIN (SUBSET_SUM) KEEP ( DENSE_RANK FIRST ORDER BY ABS (ASSIGN_AMT - SUBSET_SUM) ) AS CLOSEST_SUM FROM ASSIGN CROSS JOIN SUMS GROUP BY ASSIGN.ID, ASSIGN.ASSIGN_AMT

Essentially, we’re grouping all values from the `SUMS`

table for each `ASSIGN_AMT`

. For each of those groups, we’ll look for the `"FIRST"`

row that appears when ordering rows within the group by our criteria `ABS(ASSIGN_AMT - SUBSET_SUM)`

. We `"KEEP"`

only those rows in the group, and retain from those rows the minimum `SUBSET_SUM`

.

This query will again yield:

ID ASSIGN_AMT SUBSET_SUM -------------------------- 1 25150 23525 2 19800 23525 3 27511 23525

This is an extremely nice functionality that can come in handy every once in a while.

Remember that we’ve seen a similar feature recently on this blog, when we were looking for `FIRST_VALUE`

(or `LAST_VALUE`

) within the `PARTITION`

of a window. In standard SQL, a similar thing can be achieved using window functions as such:

SELECT DISTINCT ASSIGN.ID, ASSIGN.ASSIGN_AMT, FIRST_VALUE (SUBSET_SUM) OVER ( PARTITION BY ASSIGN.ID ORDER BY ABS (ASSIGN_AMT - SUBSET_SUM) ) AS CLOSEST_SUM FROM ASSIGN CROSS JOIN SUMS

Unfortunately, these solutions all produce duplicates, which we have to remove either via `GROUP BY`

(`KEEP`

solution), or via `DISTINCT`

(`FIRST_VALUE`

solution).

### Finding the “closest” of these sums with LATERAL JOIN

A cleaner solution that doesn’t rely on the removal of duplicates is using Oracle 12c’s new `FETCH FIRST`

clause along with `CROSS JOIN LATERAL`

(or `CROSS APPLY`

, which is the same)

SELECT ASSIGN.ID, ASSIGN.ASSIGN_AMT, CLOSEST_SUM FROM ASSIGN CROSS JOIN LATERAL ( SELECT SUBSET_SUM AS CLOSEST_SUM FROM SUMS ORDER BY ABS (ASSIGN.ASSIGN_AMT - SUBSET_SUM) FETCH FIRST 1 ROW ONLY ) SUMS

What does this mean? We’re essentially joining for each value in `ASSIGN`

only the `FIRST 1 ROW`

in `SUMS`

, ordered by the usual criteria. We need `LATERAL`

(or `APPLY`

), because this allows us to reference columns from the *left side* of the `JOIN`

expression also in the *right side*, which wouldn’t be possible otherwise.

The same functionality is supported in SQL Server (only using `CROSS APPLY`

), or in PostgreSQL (only using `CROSS JOIN LATERAL`

).

Lateral can be very useful whenever the right hand side of a `JOIN`

depends on the left hand side. Unlike ordinary joins, this means that the `JOIN`

order will be set in stone from left to right, and the optimiser has a reduced set of join algorithm options. This is useful in examples like these (with `ORDER BY`

and `FETCH FIRST`

), or when joining unnested table-valued functions, which we’ll cover in a future blog post.

## 2. Calculating a sum of any subset of values

So far, we’ve worked on a simplified version of the problem. This is probably not what alhashmiya meant on their Stack Overflow question. They probably wanted to solve the *Subset sum problem*, finding the “closest” sum of *any* subset of `WORK_AMT`

values.

We’ll use recursive SQL to calculate all the possible sums. First off, let’s remember how recursive SQL works:

In recursive SQL, we need a `UNION ALL`

query in a common table expression (`WITH`

clause in Oracle, or `WITH RECURSIVE`

clause in PostgreSQL). The first subquery of `UNION ALL`

generates the “seed row(s)” of the recursion, and the second subqeury of `UNION ALL`

generates the recursion based on a `SELECT`

that selects from the table being declared, recursively.

So, a naive solution to this subset sum problem can be seen here:

-- Repetition of the previous data WITH ASSIGN (ID, ASSIGN_AMT) AS ( SELECT 1, 25150 FROM DUAL UNION ALL SELECT 2, 19800 FROM DUAL UNION ALL SELECT 3, 27511 FROM DUAL ), WORK (ID, WORK_AMT) AS ( SELECT 1 , 7120 FROM DUAL UNION ALL SELECT 2 , 8150 FROM DUAL UNION ALL SELECT 3 , 8255 FROM DUAL UNION ALL SELECT 4 , 9051 FROM DUAL UNION ALL SELECT 5 , 1220 FROM DUAL UNION ALL SELECT 6 , 12515 FROM DUAL UNION ALL SELECT 7 , 13555 FROM DUAL UNION ALL SELECT 8 , 5221 FROM DUAL UNION ALL SELECT 9 , 812 FROM DUAL UNION ALL SELECT 10, 6562 FROM DUAL ), -- A new way of calculating all possible sums by -- Recursively adding up all the sums SUMS (SUBSET_SUM, MAX_ID) AS ( SELECT WORK_AMT, ID FROM WORK UNION ALL SELECT WORK_AMT + SUBSET_SUM, GREATEST (MAX_ID, WORK.ID) FROM SUMS JOIN WORK ON SUMS.MAX_ID < WORK.ID ) -- The same technique to match the "closest" sum -- As before SELECT ASSIGN.ID, ASSIGN.ASSIGN_AMT, MIN (SUBSET_SUM) KEEP ( DENSE_RANK FIRST ORDER BY ABS (ASSIGN_AMT - SUBSET_SUM) ) AS CLOSEST_SUM FROM SUMS CROSS JOIN ASSIGN GROUP BY ASSIGN.ID, ASSIGN.ASSIGN_AMT

The recursion is simple. In the first subquery of the recursion (the “seed row”), we select each individual row in `WORK`

:

SELECT WORK_AMT, ID FROM WORK

In the second subquery of the recursion (the “recusion rows”), we join the value of the previous recursion step (`SUMS`

) with all the remaining values (`WORK`

), i.e. all the values that have a higher `ID`

:

SELECT WORK_AMT + SUBSET_SUM, GREATEST (MAX_ID, WORK.ID) FROM SUMS JOIN WORK ON SUMS.MAX_ID < WORK.ID

In this combination, we calculate the intermediate sum (which is also a running total, by the way) and we calculate the highest summed-up ID thus far, to reduce the number of combinations. The latter, we can do because summing is commutative.

The main difference in this solution compared to previous approaches is the fact that we’re now generating a lot (a huge lot) more different values in the `SUMS`

table.

After a still acceptable 0.112s for 10 different `WORK_AMT`

values, the database calculated:

ID ASSIGN_AMT CLOSEST_SUM --------------------------- 1 25150 25133 2 19800 19768 3 27511 27488

But beware, as soon as you start adding values to the `VALS`

table, this algorithm starts exploding in time and space complexity. Running the same query with the following, only slightly bigger `WORK`

table already requires 16.3 seconds to yield a result:

WORK(ID, WORK_AMT) AS ( SELECT 1 , 7120 FROM DUAL UNION ALL SELECT 2 , 8150 FROM DUAL UNION ALL SELECT 3 , 8255 FROM DUAL UNION ALL SELECT 4 , 9051 FROM DUAL UNION ALL SELECT 5 , 1220 FROM DUAL UNION ALL SELECT 6 , 12515 FROM DUAL UNION ALL SELECT 7 , 13555 FROM DUAL UNION ALL SELECT 8 , 5221 FROM DUAL UNION ALL SELECT 9 , 812 FROM DUAL UNION ALL SELECT 10, 6562 FROM DUAL UNION ALL SELECT 11, 1234 FROM DUAL UNION ALL SELECT 12, 61 FROM DUAL UNION ALL SELECT 13, 616 FROM DUAL UNION ALL SELECT 14, 2456 FROM DUAL UNION ALL SELECT 15, 5161 FROM DUAL UNION ALL SELECT 16, 414 FROM DUAL UNION ALL SELECT 17, 516 FROM DUAL UNION ALL SELECT 18, 617 FROM DUAL UNION ALL SELECT 19, 146 FROM DUAL ),

And the result would be:

ID ASSIGN_AMT CLOSEST_SUM --------------------------- 1 25150 25150 2 19800 19800 3 27511 27511

Want proof about the actual sum? That’s easy as well with recursive SQL:

-- Repetition of the previous data WITH ASSIGN (ID, ASSIGN_AMT) AS ( SELECT 1, 25150 FROM DUAL UNION ALL SELECT 2, 19800 FROM DUAL UNION ALL SELECT 3, 27511 FROM DUAL ), WORK (ID, WORK_AMT) AS ( SELECT 1 , 7120 FROM DUAL UNION ALL SELECT 2 , 8150 FROM DUAL UNION ALL SELECT 3 , 8255 FROM DUAL UNION ALL SELECT 4 , 9051 FROM DUAL UNION ALL SELECT 5 , 1220 FROM DUAL UNION ALL SELECT 6 , 12515 FROM DUAL UNION ALL SELECT 7 , 13555 FROM DUAL UNION ALL SELECT 8 , 5221 FROM DUAL UNION ALL SELECT 9 , 812 FROM DUAL UNION ALL SELECT 10, 6562 FROM DUAL ), -- A new way of calculating all possible sums by -- Recursively adding up all the sums SUMS (SUBSET_SUM, MAX_ID, CALC) AS ( SELECT WORK_AMT, ID, TO_CHAR(WORK_AMT) FROM WORK UNION ALL SELECT WORK_AMT + SUBSET_SUM, GREATEST (MAX_ID, WORK.ID), CALC || '+' || WORK_AMT FROM SUMS JOIN WORK ON SUMS.MAX_ID < WORK.ID ) -- The same technique to match the "closest" sum -- As before SELECT ASSIGN.ID, ASSIGN.ASSIGN_AMT, MIN (SUBSET_SUM) KEEP ( DENSE_RANK FIRST ORDER BY ABS (ASSIGN_AMT - SUBSET_SUM) ) AS CLOSEST_SUM, MIN (CALC) KEEP ( DENSE_RANK FIRST ORDER BY ABS (ASSIGN_AMT - SUBSET_SUM) ) AS CALCULATION FROM SUMS CROSS JOIN ASSIGN GROUP BY ASSIGN.ID, ASSIGN.ASSIGN_AMT

The above now yields:

ID ASSIGN_AMT CLOSEST_SUM CALCULATION ------------------------------------------------------------ 1 25150 25133 7120 + 8150 + 9051 + 812 2 19800 19768 1220 + 12515 + 5221 + 812 3 27511 27488 8150 + 8255 + 9051 + 1220 + 812

## Conclusion

SQL is powerful. Extremely powerful. In this article, we’ve used various techniques to calculate the subset sum problem, or a simplification thereof. We’ve shown how to solve this problem in Oracle or PostgreSQL using a combination of these awesome SQL features:

- Window functions
`KEEP FIRST`

(in Oracle only)`LATERAL JOIN`

(or`APPLY`

- Recursive SQL

Did you like this article? Would you like to learn more about advanced SQL? Contact us to enquire about our advanced SQL training sessions, where we help you understand the simplicity of set-oriented thinking and calculations with SQL.

We’d like to point out that all of these solutions can be written in Java using jOOQ in a type safe manner as well.

Finally, a lot of this grounds is covered in more detail in any of the following articles. Enjoy learning, and enjoy SQL!

- Don’t Miss out on Awesome SQL Power with FIRST_VALUE(), LAST_VALUE(), LEAD(), and LAG()
- You Probably don’t Use SQL INTERSECT or EXCEPT Often Enough
- Use this Neat Window Function Trick to Calculate Time Differences in a Time Series
- Do You Really Understand SQL’s GROUP BY and HAVING clauses?
- Probably the Coolest SQL Feature: Window Functions

My advice for all people reading this article will be following: While provided solution is very elegant, there are however many ways to skin a cat. Using pl/sql or java to solve you problem will be a valid solution.

Thanks for the outstanding tutorial and explanation. I needed to do exactly this and so far, I’ve learned a lot. Especially helpful was the calculation column that showed how the SUM was obtained. I have a further requirement however. What if instead of the amount of the assignment, I wanted a list of the work_id’s used to calculate the sum? I’ve figured out how to add-in the

work_ids || ‘+’ || work_id

part, but the MIN / KEEP / DENSE_RANK part is throwing me off. When I try to add that, it just gives me a much larger resultset (each assign_id having many many rows).

Thoughts? I need the ID because i’ll need to update the “work” records to tie them together to (to persist the aggregation).

Thanks!!!

Hey, great to hear that this extremely rare problem / solution found its way your way, and that you could figure out the issue you were having 🙂

nevermind, figured it out. i had the pk column added into the group by at the very bottom. realized i didn’t need it cause we had an aggregate function applied to it in the select list. all the extra rows are gone now. thanks once again!

hey, I’m back! this time with thoughts on performance.

my understanding (with approach #2) is that the number of possible permutations is 2^n where n represents the number of records you are summing (in this case, it would be 10 for the 10 assignments).

i ran approach #2 across our dataset. in simple examples, it runs fast/very fast. in the example where n was 16, it ran in 1 min and 49 seconds. (not too bad but definitely the upper limit on what is acceptable because we would have to do this multiple times per day). for that effort, it had to try 65,536 possibilities. Unfortunately, in one of our use cases, i’m seeing where n would be 68. That’s 295,147,905,179,352,825,856 possibilities. I let the query run for 30 mins before killing the session. long ops seemed to indicate it was going to take over 3 hours. No way we can accept that for one occurrence.

in our situation, we are looking for “perfect matches”. we are dealing with transactions and bank deposits, so the sums will always be equal if the data exists (or it wont match perfectly if we haven’t yet received transactions that need to be counted). is there a way to modify our approach so that if we find a perfect match, we’ll drop out and not calculate the remaining possibilities? i’m assuming i’d be able to do that if i was looping through results in pl/sql instead of letting the recursive query do the work, but i’d rather not take that approach since figuring out a way to ensure all combos are tried seems difficult (1 query seems much cleaner).

additionally, it might be helpful to know how the execution of this query occurs. In our use-case, we actually ave a SUM()/group by in one of our WITH tables. We actually need to do a preliminary sum that we then sum again together to equal the total amount. the tables that contain this data are going to be arbitrarily large. i have them indexed which makes things quick for now, but i can’t tell if those “early” calculations are being done tons of times (one time for each recursive call). if it is constantly recalculating them, it might be easier for me to do the calc once and store it in a table and have our WITH query reference that table instead.

lastly, oracle’s explain plan seems to have trouble predicting the cost of the total query. it seems to understand the basics of what’s going on in the WITH queries (indexes to use/etc), but can’t predict how much strain the recursion part will introduce. i was hoping to be able to fine tune this, but unfortunately, it looks like your warnings about exponential growth and scalability were correct.

i guess the main question is, knowing that we have perfect matches and not closest sum, is there someway we can cheat around the performance limits? any direction (links/etc) would be helpful. obviously, i realize i asked a lot of questions and i don’t expect you to do our work for me!

thanks!

Hey there. I’m available for hire, if you like 🙂

Please, enquire with sales@datageekery.com for details.