Many SQL implementations don’t have loops, making some kinds of analysis very difficult. Postgres, SQL Server, and several others have the next best thing — recursive CTEs!



We’ll use them to solve the <a href="https://en.wikipedia.org/wiki/Travelling_salesman_problem" target="_blank" rel="noreferrer noopener" aria-label=" (opens in a new tab)">Traveling Salesman Problem</a> and find the shortest round-trip route through several US cities.



<figure class="wp-block-image fancybox"><img decoding="async" src="https://cdn.sisense.com/wp-content/uploads/image-01-13.png" alt="Route map" class="wp-image-78389"/></figure>



<h2 class="wp-block-heading">With Recursive</h2>



Normal CTEs are great at helping to organize large queries. They are a simple way to make temporary tables you can access later in your query.



<a href="https://www.postgresql.org/docs/8.4/queries-with.html" target="_blank" rel="noreferrer noopener" aria-label=" (opens in a new tab)">Recursive CTEs</a> are more powerful &#8211; they reference themselves and allow you to explore hierarchical data. While that may sound complicated, the underlying concept is very similar to a for loop in other programming languages. 



These CTEs have two parts — an&nbsp;anchor&nbsp;member and a&nbsp;recursive&nbsp;member. The&nbsp;anchor&nbsp;member selects the starting rows for the recursive steps. 



The&nbsp;recursive&nbsp;member generates more rows for the CTE by first joining against the&nbsp;anchor&nbsp;rows, and then joining against rows created in previous recursions. The&nbsp;recursive&nbsp;member comes after a&nbsp;union all&nbsp;in the CTE definition.



Here’s a simple recursive CTE that generates the numbers 1 to 10. The&nbsp;anchor&nbsp;member selects the value 1, and the&nbsp;recursive&nbsp;member adds to it up to the number 10:



<pre class="wp-block-code"><code>with recursive incrementer(prev_val) as (
 select 1 -- anchor member
 union all
 select -- recursive member
 incrementer.prev_val + 1
 from incrementer
 where prev_val &lt; 10 -- termination condition
)
select * from incrementer</code></pre>



The first time the recursive CTE runs it generates a single row&nbsp;1&nbsp;using the&nbsp;anchor&nbsp;member. In the second execution, the&nbsp;recursive&nbsp;member joins against the&nbsp;1&nbsp;and outputs a second row,&nbsp;2. In the third execution the&nbsp;recursive&nbsp;step joins against both rows&nbsp;1&nbsp;and&nbsp;2&nbsp;and adds the rows&nbsp;2&nbsp;(a duplicate) and&nbsp;3.



Recursive CTEs also only return distinct rows. Even though our CTE above creates many rows with the same value, only a distinct set of rows will be returned.



Notice how the CTE specifies its output as the named value&nbsp;prev_val. This lets us refer to the output of the previous recursive step.



And at the very end there is a termination condition to halt the recursion once the sum gets to 10. Without this condition, the CTE would enter an infinite loop!



Under the hood, the database is building up a table named after this recursive CTE using unions:



<figure class="wp-block-image fancybox"><img decoding="async" src="https://cdn.sisense.com/wp-content/uploads/image-02-13.png" alt="Recursive union" class="wp-image-78383"/></figure>



 Recursive CTEs can also have many parameters. Here’s one that takes the sum, double, and square of starting values of 1, 2 and 3: 



<pre class="wp-block-code"><code>with recursive cruncher(inc, double, square) as (
 select 1, 2.0, 3.0 -- anchor member
 union all
 select -- recursive member
 cruncher.inc + 1,
 cruncher.double * 2,
 cruncher.square ^ 2
 from cruncher
 where inc &lt; 10
)
select * from cruncher</code></pre>



With recursive CTEs, we can solve the Traveling Salesman Problem.



<h2 class="wp-block-heading">Finding the Shortest Path</h2>



There are many algorithms for finding the shortest round-trip path through several cities. We’ll use the simplest: brute force. Our recursive CTE will enumerate all possible routes and their total distances. We’ll then sort to find the shortest.



First, a list of cities with our customers, along with their latitudes and longitudes:



<pre class="wp-block-code"><code>create table places as (
 select
 'Seattle' as name, 47.6097 as lat, 122.3331 as lon
 union all select 'San Francisco', 37.7833, 122.4167
 union all select 'Austin', 30.2500, 97.7500
 union all select 'New York', 40.7127, 74.0059
 union all select 'Boston', 42.3601, 71.0589
 union all select 'Chicago', 41.8369, 87.6847
 union all select 'Los Angeles', 34.0500, 118.2500
 union all select 'Denver', 39.7392, 104.9903
)</code></pre>



And we’ll need a distance function to compute how far two lat/lons are from each other (thanks to <a href="https://stackoverflow.com/questions/10034636/postgresql-latitude-longitude-query" target="_blank" rel="noreferrer noopener" aria-label=" (opens in a new tab)">strkol on stackoverflow.com</a>):\



<pre class="wp-block-code"><code>create or replace function lat_lon_distance(
 lat1 float, lon1 float, lat2 float, lon2 float
) returns float as $$
declare
 x float = 69.1 * (lat2 - lat1);
 y float = 69.1 * (lon2 - lon1) * cos(lat1 / 57.3);
begin
 return sqrt(x * x + y * y);
end
$$ language plpgsql</code></pre>



 Our CTE will use San Francisco as its anchor city, and then recurse from there to every other city: 



<pre class="wp-block-code"><code>with recursive travel(places_chain, last_lat, last_lon,
 total_distance, num_places) as (
 select -- anchor member
 name, lat, lon, 0::float, 1
 from places
 where name = 'San Francisco'
 union all
 select -- recursive member
 -- add to the current places_chain
 travel.places_chain || ' -> ' || places.name,
 places.lat,
 places.lon,
 -- add to the current total_distance
 travel.total_distance + 
 lat_lon_distance(last_lat, last_lon, places.lat, places.lon),
 travel.num_places + 1
 from
 places, travel
 where
 position(places.name in travel.places_chain) = 0
)</code></pre>



The parameters in the CTE are:



<ul><li>places_chain: The list of places visited so far, which will be different for each instance of the recursion</li><li>last_lat and last_lon: The latitude and longitude of the last place in the&nbsp;places_chain</li><li>total_distance: The distance traveled going from one place to the next in the&nbsp;places_chain</li><li>num_places: The number of places in&nbsp;places_chain&nbsp;— we’ll use this to tell which routes are complete because they visited all cities</li></ul>



In the&nbsp;recursive&nbsp;member, the&nbsp;where&nbsp;clause ensures that we never repeat a place. If we’ve already visited Denver,&nbsp;position(&#8230;)&nbsp;will return a number greater than 0, invalidating this instance of the recursion.



We can see all possible routes by selecting all 8-city chains:



<pre class="wp-block-code"><code>select * from travel where num_places = 8</code></pre>



We need to add in the distance from the last city back to San Francisco to complete the round-trip. We could hard code San Francisco’s lat/lon, but a join is more elegant. Once that’s done we sort by distance and show the smallest:



<pre class="wp-block-code"><code>select
 travel.places_chain || ' -> ' || places.name,
 total_distance + lat_lon_distance(
 travel.last_lat, travel.last_lon,
 places.lat, places.lon) as final_dist
from travel, places
where
 travel.num_places = 8
 and places.name = 'San Francisco'
order by 2 -- ascending!
limit 1</code></pre>



 Even though this query is significantly more complicated than the&nbsp;incrementer&nbsp;query earlier, the database is doing the same things behind the scenes. The top branch is the creating the CTE’s rows, the bottom branch is the final join and sort: 



<figure class="wp-block-image fancybox"><img decoding="async" src="https://cdn.sisense.com/wp-content/uploads/image-03-12.png" alt="Complicated recursive union" class="wp-image-78395"/></figure>



Run this query and you’ll see the shortest route takes 6671 miles and visits the cities in this order:



<pre class="wp-block-preformatted"> San Francisco -&gt; Seattle -&gt; Denver -&gt;
Chicago -&gt; Boston -&gt; New York -&gt; Austin -&gt;
Los Angeles -&gt; San Francisco </pre>



Thanks to recursive CTEs, we can solve the Traveling Salesman Problem in SQL!

Solving the Traveling Salesman Problem with Postgres Recursive CTEs

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