# Optimization by puzzle

Given a query routine that takes a name and may return several, write a routine that takes a single name and returns a set of names for which each of the following is true:

1. For each name in the set, query has been called exactly once.
2. All the results from the calls to query are included in the set
3. the parameter to the routine is not included in the set

You may assume the following:

1. Calls to query are idempotent1.
2. There is a finite number of values for names.
3. Names are less-than-comparable value-types (i.e. you can store them in an std::set) and are not expensive to copy
4. query results never contain their argument2

This is almost exactly the problem I had to solve recently: a tool was taking several minutes to perform a routine task that, in my opinion, should take milliseconds. Several other issues were involved as well, but this one has the bonus of being fun.

I should make this an interview question.

The way this ends up working is as follows:

1. We create three sets: one for the results, one for the things we’ve checked and one for the things that remain to_check.
2. We insert the value we got as a parameter in the to_check set.
3. As long as there are things left to check:
1. run query for each value in to_check
2. insert the results from the query in the results set
3. After iterating over each of the values, insert the values from to_check into the checked set,
4. clear the to_check set
5. fill to_check with the set difference between the results and the checked sets

Or, in C++:

 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 template < typename T, typename F > set< T > foo(T t, F query) { set< T > results; set< T > checked; set< T > to_check; to_check.insert(t);   do { for (typename set< T >::const_iterator check(to_check.begin()); check != to_check.end(); ++check) { typename F::result_type query_results(query(*check)); results.insert(query_results.begin(), query_results.end()); } checked.insert(to_check.begin(), to_check.end()); to_check.clear(); set_difference(results.begin(), results.end(), checked.begin(), checked.end(), inserter(to_check, to_check.end())); } while (!to_check.empty()); return results; }

Insertion into a set is $O(\lg{n})$ so lines 43 and 45 are both $O(n\lg{n})$. Line 46 should be $O(c)$ but is probably $O(n)$. Line 47 is $O(n)$ so the whole things boils down to $O(n\lg{n})$ complexity.

In order to play with the code a bit, I put it on GitHub as a Gist, with a test case (Query fails if you call it more than once with the same value):

1. so you really do need to call them only once []
2. i.e. for the case at hand, we’re querying a directed acyclic graph, so our first argument will never be seen in any of the query results, although any given value may appear more than once in query results []
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## About rlc

Software Analyst in embedded systems and C++, C and VHDL developer, I specialize in security, communications protocols and time synchronization, and am interested in concurrency, generic meta-programming and functional programming and their practical applications. I take a pragmatic approach to project management, focusing on the management of risk and scope. I have over two decades of experience as a software professional and a background in science.
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