Recursion
This concept guide helps you have a better understanding of the logic behind recursion and covers cases that allow you to successfully write recursively defined relations in Rel. It also sheds light on common challenges you may encounter when using recursion and provides examples on how you can look out for and resolve them.
Introduction
A program or a query is called recursive when a relation is defined in terms of itself. For many computational problems, recursion helps to reduce complexity and to express solutions in a cleaner and more understandable way. A common side effect is that recursive solutions are often shorter, more compact, and easier to maintain.
Rel supports recursive computations, which is also a key ingredient for making the language Turing complete. This means Rel can be used to perform any computation that any other programming language supports.
With the capability to express queries recursively, Rel is able to perform a variety of computations that at first glance you would not expect from a database language:

Iterative algorithms can be expressed as recursive Rel programs; for example, Fibonacci or Prim’s Minimal spanning tree.

Relations requiring a variable recursion depth can be expressed more compactly and efficiently; for example, transitive closure, PageRank, and moving averages.
To define recursive computations, you need the following ingredients:
 A base case.
 A recursive rule.
The base case is the starting point of the recursive computation. The recursive rule is the heart of the recursive computation, which references itself.
Termination of Recursive Computations
Computationally, you need a third ingredient in order to have a successful recursive algorithm: termination conditions, which ensure that the recursive computation doesn’t run forever.
Generally, a recursive computation in Rel terminates when a fixpoint is reached, meaning that the elements in a relation don’t change anymore from one iteration to the next. See the Behind the Scenes section for more details.
More formally, this means the solution to the recursive computation is a relation (or a set of relations) R
, which is a fixpoint of the following definition:
def R(x...) = base_case(x...) or recursive_rule[R](x...)
A recursive computation defining an infinite relation, or infinite function, such as the Fibonacci sequence, currently requires a domain restriction for the computation to terminate. See the examples below.
For certain recursive computations, no explicit domain restriction is needed because the definitions naturally reach a fixpoint. Graph reachability (see below) is one such problem because the output domain, which is all possible pairs of graph nodes, is finite, and the recursion is monotonically increasing.
Recursion is said to be monotonically increasing if at each iteration you can only add new elements and never remove any. The query optimizer will automatically determine whether or not a recursively defined relation is monotonic. If the recursion is found to be monotonic, the system uses a more efficient seminaive algorithm to find a fixpoint. A more detailed discussion on whether or not a recursion is monotonic can be found in section Recursive Rules That Eliminate Elements.
Introductory Examples
Recursive Computation on Graphs
One of the most fundamental graph queries asks if there is a path between two nodes. Such questions appear frequently in the real world. In terms of international travel, this question could be:
Can I fly from Providence to Honolulu?
To answer that question in Rel, you first need a few airports:
// model
def airport = {
"Providence (PVD)"; "New York (JFK)"; "Boston (BOS)";
"San Francisco (SFO)"; "Los Angeles (LAX)"; "Honolulu (HNL)";
}
You also need a flight network:
// model
def directional_flight = {
("Providence (PVD)", "Boston (BOS)"); ("Boston (BOS)", "San Francisco (SFO)");
("San Francisco (SFO)", "Honolulu (HNL)"); ("New York (JFK)", "Los Angeles (LAX)");
}
def flight(a, b) = directional_flight(a, b) or directional_flight(b, a)
// integrity constraint
ic flight_between_airports(a, b) = flight(a, b) implies airport(a) and airport(b)
The network introduces directional flights, directional_flight
, and bidirectional flights, flight
.
The integrity constraint checks that flights can only occur between airports.
Now, you can calculate all pairs of airports that are directly or indirectly connected to each other:
// model
def connected(a, b) = flight(a, b) // the base case
def connected(a, b) = exists(c : connected(a, c) and flight(c, b))
This is the recursive part of the problem.
The original question now reads:
// read query
def output =
if connected("Providence (PVD)", "Honolulu (HNL)") then
"yes"
else
"no"
end
// This integrity constraint will fail if we don't compute that
// PVD and HNL are connected:
ic {equal(output, "yes")}
Suppose you want to ask another question and check whether you can fly from Boston (BOS)
to Los Angeles (LAX)
:
// read query
def output =
if connected("Boston (BOS)", "Los Angeles (LAX)") then
"yes"
else
"no"
end
// This integrity constraint will fail if we compute that
// BOS and LAX are connected:
ic {equal(output, "no")}
The answer is no, because in this toy world LAX and JFK are connected to each other but isolated from all the other airports.
For the mathematical enthusiasts, the problem of computing all pairs of connected nodes in a graph is called the transitive closure problem.
In this example, the edges of the graph are defined in the flight
relation.
Single Recursion
One of the simplest mathematical recursion formulas is the definition of the factorial (opens in a new tab) $n! = n \cdot (n1) \cdot \ldots \cdot 1$ for $n\in \mathbb{N}_+$, which fulfills the recursive rule: $n! = n \cdot (n1)!$. In Rel, this reads:
// read query
def F[1] = 1 // the base case
def F[n] = n * F[n  1], n <= 10
def output = F
There are several interesting aspects worth mentioning about this recursively defined relation:
Relation F
has arity 2, where the first entry is the iteration step n
, and the second entry is the value of n!
.
The definition also contains all ingredients for a successful recursion:
 Base case:
def F[1] = 1
is the starting point and translates to1! = 1
.  Recursive rule:
def F[n] = n * F[n  1]
translates to $n! = n\cdot(n1)!$.  Termination condition:
n<=10
makes the output domain finite, forces the fixpoint to be reached when $n$ hits 10, and consequently terminates the recursive evaluation. In the future, once support for demand transformations is implemented, this explicit statement of the termination condition will not be required anymore.
Multiple Recursion
Fibonacci Sequence
Multiple recursion, where the recursive rule contains multiple selfreferences, can also be easily expressed in Rel.
The most famous example is the Fibonacci sequence (opens in a new tab) $F(n) = F(n1) + F(n2)$ with $F(0)=0$ and $F(1)=1$. In Rel, this relation reads:
// read query
def F[0] = 0 // base cases
def F[1] = 1 // base cases
def F[n] = F[n1] + F[n2], n<=10
def output = F
As you can see, all ingredients for a successful recursive computation are present.
Connected Nodes in a Graph
In the previous international travel example, you could have defined the connected
relation using multiple recursions:
// query
def connected(a, b) = flight(a, b) // the base case
def connected(a, b) = connected[a].connected(b) // recursive rule
Here, the recursive rule also contains two selfreferences and reads:
Two airports a, b
are connected if there exists a third airport that is connected to both airports.
Negation in Recursion
Recursive dependencies can be made as complex as needed and can go well beyond just asking for a single element of a previous iteration, such as F[n  1]
.
The Recamán sequence (opens in a new tab) demonstrates a more complex recursive computation. The sequence is defined as:
The recursive case requires you to check that the potential next value isn’t already in the sequence, and if it is, to assign a different value.
The Recamán sequence can be visualized in a very neat way by connecting subsequent values with an arch:
(CC license) (opens in a new tab)
One way to define the Recamán sequence in Rel is as follows:
// read query
@inline
def reject(n,x) = x < 0 or exists(m: recaman(m, x) and m < n)
def recaman[0] = 0
def recaman[n](x) = recaman[n  1]  n = x and not reject(n, x) and 0 < n < 20
def recaman[n](x) = recaman[n  1] + n = x and reject(n, x  2 * n) and 0 < n < 20
def output = recaman
// This IC verifies the correctness of the results by comparing with handcomputed results.
ic recaman_sequence = equal(
recaman[_],
{0;1;3;6;2;7;13;20;12;21;11;22;10;23;9;24;8;25;43;62;}
)
Only the first 20 terms were calculated. Several logical elements were used here to express the recursive Recamán rule:
 Negation.
 Existential quantification.
 Disjunction.
Additionally, you used the @inline
functionality to factor out the condition evaluation in a separate relation, reject
, helping to keep the main recursive rule compact and readable.
Finally, adding an integrity constraint (IC) checks that the set of computed values is correct. Note that this IC does not check that the values are in the correct order, but it still gives additional confidence in the results.
Advanced Examples
Rel supports complex recursive rules that may involve multiple relations that depend on each other and may be recursive themselves. This is known as mutual recursion.
Before diving into them, look at relations that are not monotonically increasing in cardinality as the iteration advances. This is possible because elements can not only be added but also deleted during an iterative step.
Recursive Rules That Eliminate Elements
For all the examples seen so far, the number of elements — also known as cardinality — in the recursively defined relation is larger at the end of the computation. However, Rel can also delete, not just add, elements during an iteration, so it is also possible to finish with fewer elements in the relation than at the start.
The following Rel program has two recursive rules for D
.
The first adds the elements 1 through 10, if D
is empty.
The second keeps only elements that are smaller than or equal to 3, or different from the maximum element in D
.
The effect is that the
largest element is iteratively removed until only the elements 1; 2; 3
remain:
// read query
def D(n) = range[1, 10, 1](n) and empty(D)
def D(n) = D(n) and (n <= 3 or n != max[D])
def output = D
Notice that the program contains only the two recursive rules and does not explicitly state a base case.
This means that the base case is, by default, an empty relation D
.
The first rule adds the numbers 1 through 10 if and only if D
is empty.
The key difference between the two recursive rules is: The first rule probes the nonexistence of elements whereas the second rule depends on an aggregated view of existing elements. Either of these aspects, negation or aggregation, is enough to achieve a nonmonotonic recursion. In the absence of both features, recursion is always monotonic.
Asymptotic Convergence: PageRank
Another class of recursive formulas that can easily be implemented in Rel is infinite series that converge asymptotically. That is in contrast to the example in the Recursive Computation on Graphs section, where you reached convergence after a finite number of steps.
In practice, you can define a convergence criterion, where you stop the iteration if the value(s) between consecutive iterations is closer than a small, userdefined $ϵ$:
$\ a_{n+1}  a_n \^2 < ϵ \quad \text{for}\ n>N_\text{converged} \enspace,$
where $N_\text{converged}$ is the number of iterations needed to achieve the desired accuracy.
Many machine learning algorithms fall into this category, where you declare the model as trained once the model parameters and/or the accuracy of the model change only marginally between iterations.
Take the PageRank (opens in a new tab) algorithm with damping as an example. It is defined as:
$PR_a = \frac{1d}{N} + d \sum_{b \in \mathcal{S}_a} \frac{PR_b}{L_b} \enspace ,$
$PR_a$ is the page rank of website $a$, $L_b$ is the number of outbound links from site $b$, $\mathcal{S}_a$ is the collection of websites that have a link to website $a$, $N$ is the total number of websites, and $d$ is a damping parameter, which is usually set around 0.85.
First, define an undirected weighted graph, inspired by Wikipedia (opens in a new tab):
// model
module graphA
def edge = {
("N1", "B"); ("N1", "E");
("N2", "B"); ("N2", "E");
("N3", "B"); ("N3", "E");
("N4", "E"); ("N5", "E");
("E", "F"); ("F", "E");
("E", "B"); ("E", "D");
("D", "A"); ("D", "B");
("B", "C"); ("C", "B");
("F", "B"); }
def node = x : edge(x, _) or edge(_, x)
end
Here is a visualization of the graph:
Here, the size of a node is proportional to its PageRank.
The PageRank algorithm assumes that sink nodes, which have no outgoing edges, are connected to all other nodes in the graph. You can derive a new graph from the original as follows:
// model
module graphB
def node = graphA:node
def edge = graphA:edge
def sink(n in node) = empty(graphA:edge[n])
def edge(a in node, b in node) = sink(a) // add new edges
end
In Rel, the PageRank algorithm may read:
// model
// parameters
def eps = 10.0^(10)
def damping = 0.85
def node = graphB:node
def edge = graphB:edge
def node_count = count[node]
def outdegree[x] = count[edge[x]]
def MAX_ITER = 100
// a version of sum that gives 0.0 for the empty set,
// needed if there are edges with no incoming nodes:
@inline
def sum_default[R] = sum[R] <++ 0.0
// pagerank(iteration, site, rank)
def pagerank[0, a in node] = 1.0 / node_count //base case: equal rank
def pagerank[n, a in node] = // recursive rule
(1damping)/node_count + damping * sum_default[
pagerank[n1, b]/outdegree[b] for b where edge(b, a)
],
pagerank(n1, a, _) // grounding `n` and `a`
and not converged(n1) // termination condition
and n<=MAX_ITER // safeguard
// track convergence:
def converged(n) =
forall(a in node : (pagerank[n, a]  pagerank[n1, a])^2 < eps)
and range(1, MAX_ITER, 1, n)
The relation converged
stores the iteration numbers for which the convergence criterion is fulfilled.
The not converged(n1)
condition in the main recursive definition for pagerank
ensures that the iteration stops once convergence is reached.
A safeguard condition is also added that stops the recursive computation if 100 iterations have been reached.
Notice that the relations pagerank
and converged
depend on each other and must be solved together.
You can use the sum_default
relation to ensure that the sum over an empty set returns 0.0 instead of false
,
which is the default behavior of sum[]
.
This is needed if there are no sink nodes in the original graph, and in case of a node with no incoming edges.
Now you can see the results. After 62 iterations, you have reached the desired accuracy level with the following results:
// read query
def max_iteration = max[n : pagerank(n, _, _)]
def did_converge = "yes", converged = max_iteration
def ranks = pagerank[max_iteration]
def sumranks = sum[pagerank[max_iteration]]
max_iteration
did_converge
ranks
sumranks
Since PageRank corresponds to a probability distribution over nodes, the ranks should sum to approximately 1.0.
Prim’s Minimum Spanning Tree
Prim’s algorithm is a greedy algorithm (opens in a new tab) to calculate a graph’s minimum spanning tree (MST) (opens in a new tab):
 Initialize a tree with a single node, chosen arbitrarily from the graph.
 Grow the tree by one edge: Of the edges that connect the tree to nodes not yet in the tree, find a minimumweight edge, and add it to the tree.
 Repeat step 2, until all nodes are in the tree.
The input to the algorithm is an edgeweighted undirected graph, which is assumed to be connected; that is, it has no disjoint components. This one is inspired by www.geeksforgeeks.org (opens in a new tab):
// model
def graph:edge_weight = {
(0, 1, 4); (1, 2, 9); (2, 3, 7); (3, 4, 9);
(4, 5, 10);(5, 6, 2); (6, 7, 1); (7, 8, 7);
(7, 0, 8); (1, 7, 11); (5, 2, 4);(5, 3, 14);
(2, 8, 2); (6, 8, 6);
}
def graph:node(x) = graph:edge_weight(x,_,_) or graph:edge_weight(_,x,_)
// make graph undirected by adding reverse edges
// (makes the code a little simpler):
def graph:edge_weight(x, y, w) = graph:edge_weight(y, x, w)
The relation graph:edge_weight
contains the edge information.
If graph:edge_weight(x,y,w)
holds, there is an edge of weight w
from x
to y
.
The relation graph:node
is the set of nodes.
It is possible to implement Prim’s algorithm with one recursive definition. To demonstrate the interplay of several recursive definitions, the code below uses three: Two of them track the edges and nodes in the MST, and one tracks the nodes not yet in the MST:
// read query
def mst:node = 0 // initialize MST with an arbitrary node
// collect all nodes that have been already reached by the MST
def mst:node(x) = mst:edge(x, _) or mst:edge(_, x)
// main recursive rule
def mst:edge = just_one[x, y :
mst:node(x) and
unvisited(y) and
graph:edge_weight[x, y] = lightest_edge_weight[unvisited, mst:node]
]
@inline
def just_one[R] = (top[1,R])[1]
// don't remove edges that have been added to the MST:
def mst:edge = mst:edge
// list of nodes not in the MST yet:
def unvisited(x) = graph:node(x) and not mst:node(x)
// smallest weight of an edge between visited and unvisited nodes:
@inline
def lightest_edge_weight[S, T] = min[v : v=graph:edge_weight[S, T]]
// check that all nodes are visited:
ic { empty(unvisited) }
// the number of edges in a spanning tree is the number of nodes minus one:
ic { count[mst:edge] = count[graph:node]  1 }
def output:edge = mst:edge
def output:weight = sum[mst:edge <: graph:edge_weight]
The main recursive part of this algorithm involves the three relations mst:node
, mst:edge
, and unvisited
.
At each iteration, mst:edge
includes all the edges already added (mst:edge(x, y) = mst:edge(x, y)
),
plus one new edge, which is the lightest edge connecting a visited node, mst:node(x)
, to an unvisited node, unvisited(y)
.
The just_one
higherorder inline relation makes sure only one edge is added at a time.
In general, there could be two such edges, with the same minimum weight, leading to the same unvisited node;
adding both edges would create a cycle, which must be avoided for the result to be a tree.
The recursive dependencies are quite nontrivial here because the three relations refer to each other in a cyclical way:
unvisited
depends on mst:node
, which depends on mst:edge
, which in turn depends on mst:node
and unvisited
again.
The only direct recursive dependency is def mst:edge = mst:edge
.
This rule ensures that edges, once inserted, are not removed in a future iteration.
This safeguard is needed because the edge tuple (x,y)
in the other mst:edge
rule
requires that y
is unvisited; that is, not in mst:node
.
Without this safeguard, new edges would be removed from mst[:edge]
in future iterations, when y
is in visited
.
Now, the condition that y
should not be in mst:node
might be fulfilled again,
entering a cycle that never ends.
See the Common Pitfalls section for more details on neverending recursive computations.
This safeguard prevents all that.
The relation unvisited
is an example of a relation that decreases in size as the recursive evaluation proceeds.
See the Recursive Rules That Eliminate Elements section.
Initially, unvisited
contains all the nodes in the graph,
but once the recursive evaluation is done, unvisited
is actually empty,
which you can also double check with an integrity constraint, ic {not unvisited(_)}
.
The edge weights of the MST are not stored in mst
. Instead, you can look them up in graph[:edge_weight]
.
You can visualize the minimal spanning tree you just calculated:
Logging Recursive Progress
Normally, you can only access the relation after the recursive evaluation has finished, i.e., converged. It would be great to also be able to see the intermediate progress to better understand and debug the logic you write.
Take the example above where elements are successively removed until only three elements are left.
See the Recursive Rules That Eliminate Elements section.
If you remove the condition, n<=3
, you create a recursive algorithm that never converges.
See the Common Pitfalls section.
To see if this is actually what is happening, you can create a logging relation Log
that records the size of D
after each iteration.
To avoid the endless loop, you can add a condition that forces the recursive evaluation to terminate after 10 iterations have been reached:
// model
// number of iterations
def Niter = 10
// endless loop
def D(n) = n=range[1, 3, 1] and not D(_) and continue
def D(n) = D(n) and n!=max[D] and continue
// logging
def Log = 0,0 : not Log(_, _) // dummy initialization
def Log(n, d_size) =
n = max[first[Log]]+1 and
d_size = sum[count[D]; 0] and
continue
def Log(x, y) = Log(x, y) and x > 0 // don't remove entries
// termination condition
def continue = count[Log] < Niter
Note that the relation was modified from the original above and 1;2;3
was inserted instead of 1;2;...;10
.
You can check whether Log
really contains the size/cardinality of relation D
after each iteration:
// read query
Log
This confirms that D
starts with three elements and decreases to zero, i.e., empty, at which point the first recursive rule holds again, and D
is again initialized with 1;2;3
.
This cycle repeats forever, except that in this case you terminate the iteration when the relation continue
turns false, which is the case when the logging relation Log
contains at least Niter=10
entries.
Common Pitfalls
Probably the single most common problem users will face when using recursion is nonconvergent computation.
There are two main types of nonconvergence:

Unbounded domain: Recursive computations where there is no fixpoint or a fixpoint that takes infinitely many iterations to reach. PageRank is an example where the fixpoint lies at infinity. The Fibonacci Sequence and calculating the lengths of all paths in a cyclical graph are examples of computations with no fixpoint.
This requires restricting the domain to make sure that the recursive computation terminates after a finite number of iterations.
This will be addressed in the future with the demand transformation feature. See the Running to Infinity section.

Oscillations: No fixpoints exist even though the output domain is finite. This can occur if the recursive computation gets stuck in a cycle. This situation is much harder to spot and can be quite subtle as the discussion in the MST example above demonstrated. In the future, the query evaluator will be able to detect this behavior and report a failure when you repeat a solution you have seen before. See the Stuck in a Cycle section.
Running to Infinity
Factorial
How can you end up with an infinite loop? One obvious way is when you forget to restrict the output domain in a recursive rule that explores a larger and larger domain space as the recursive computation progresses. An example is the factorial $n!$. Imagine that you write the factorial but don’t state at which $n$ to stop:
// query
def F[0] = 1
def F[n] = n * F[n1]
Since Rel computes fixpoints in a bottomup fashion, deriving new facts from existing ones, the iteration will go on forever, as $n$ grows larger and larger. In the future, this issue will be solved with ondemand evaluation, which only performs computations needed when a specific set of values are requested.
Lengths of All Paths in a Cyclical Graph
In graph problems, infinite recursion can occur even though the graph itself is finite. For cyclical graphs, for instance, calculating the lengths of all paths will not terminate and will run forever because the cyclical nature of the graph ensures that you can always find a path of arbitrary length by passing through the loop, which exists in the graph, multiple times.
To demonstrate that point, you can construct a minimal graph with three nodes
and one loop between nodes 1
and 2
:
// model
def cyclical_graph:node = {1; 2; 3}
def cyclical_graph:edge = {(1, 2); (2, 1); (2, 3)}
// path_length(start_node, end_node, distance)
def path_length = cyclical_graph[:edge], 1 // base case
def path_length(a, c, len) =
path_length(a, b, len1) and
path_length(b, c, len2) and
len = len1 + len2 and
len <= 4 // termination to avoid infinite iteration
from b, len1, len2
To avoid an endless loop, you can insert the condition len <= 4
.
Inspecting the relation path_length
shows that between each node pair you have multiple path lengths increasing in steps of two,
which is the length of the loop in the cyclical graph:
// read query
path_length
Stuck in a Cycle
There is another way that you can get stuck in an infinite recursive computation. Elements can be added and removed during the iteration. Hence, it is possible to end up in a cycle that never ends.
Take the example in the Recursive Rules That Eliminate Elements section where you start with the elements 1 to 10 and iteratively remove the largest element, but only if it is larger than 3. Suppose you take this condition away:
// query
def D = n : n=range[1, 10, 1] and not D(_)
def D(n) = D(n) and n != max[D]
You will end up in an endless cycle.
Why?
Because once you remove the last element from D
, you are again fulfilling the initial condition used to populate the relation in the first place.
Now, you are back at the beginning, and the cycle repeats itself forever.
This recursive computation has no fixpoint and will never terminate.
As mentioned above, this problem will be solved in the future and Rel will be able to recognize when a solution has already been visited, but the recursive computation has not converged.
Behind the Scenes: Recursion and Fixpoints
This section goes into more detail in how recursive rules are evaluated in Rel. You don’t need to understand this to successfully write recursively defined relations. It might be, however, very helpful for writing complex recursive algorithms and understanding why Rel returns what it does.
At a high level, Rel solves the recursive computation for $R$ by starting at step 0 with the base case(s), $R_0$, which may be the empty set (in case no explicit base case is defined). The relation $R_0$ represents the starting point of the recursive computation. If $f$ stands for a function mapping relation $S$ to relation $T$, then you can express the converged relation $R$ in the following way:
$R = f(R_{N1}) = (f \circ f)(R_{N2}) = \underbrace{(f \circ \ldots \circ f)}_{N}(R_{0})$,
where after $N$ iterative applications of $f$ you have converged on a final result by reaching the fixpoint, $f(R) = R$, and the iteration stops as a repetitive application of $f$ doesn’t change the content of $R$.
Rel solves the recursive computations in a bottomup fashion where you can start with the base case(s) and then iteratively apply the recursive rule until you reach a fixpoint. Hence, it is currently difficult to implement a topdown algorithm. Many naive implementations of divideandconquer algorithms (opens in a new tab), such as mergesort (opens in a new tab), FFT (opens in a new tab), and Tower of Hanoi (opens in a new tab), are solved in a topdown fashion.
Besides the common pitfalls discussed above, you also need to be aware of scenarios where you might have multiple fixpoints.
Multiple Fixpoints
A function $f$ can have multiple fixpoints, that is, more than one value of $x$ where $f(x)=x \enspace.$ This is easy to see with mathematical functions, such as polynomials. Take, for instance, the function $f(x) = 2(x1)^2+2 \enspace.$ It has the fixpoints $x=0$ and $x=\frac{3}{2}$, which are the two points where the function intersects the $y=x$ line. Whether iterative methods can find these values may depend on the initial conditions chosen.
This applies to relations in general, and it arises in Rel, particularly when using negation.
Negations and Cycles
Consider the following example, featuring recursion and negation:
// read query
def man = "Dilbert"
def single(x) = man(x) and not husband(x)
def husband(x) = man(x) and not single(x)
def output:single = single
def output:husband = husband
There are two different least fixpoints:
 Dilbert is single and not a husband:
single(Dilbert) and not husband(Dilbert)
 Dilbert is a husband and not single:
husband(Dilbert) and not single(Dilbert)
They are incomparable: Neither is smaller than the other. Rel computes one of them, but it can be argued that they are both equally valid.
Note that single
depends on husband
, which in turn depends on single
again, so you have a
dependency cycle, with intervening negations.
Models with a cyclical definition that involves negation are called nonstratifiable and should be avoided when possible because they can lead easily to multiple fixpoints as it is the case here. Furthermore, the cycles can lead to nontermination where Rel will throw an exception.
In general, negations and aggregations should be used with caution in recursive definitions, since they can result in nonmonotonic recursion and unintended behaviors.
SymmetryInduced Fixpoints
Symmetry is a fundamental concept in mathematics and physics
and is relevant to properties like the law of conservation of energy (opens in a new tab).
Symmetries in the problem definition are a common cause for multiple fixpoints.
The multiple fixpoints in the Dilbert example above come from
the symmetry between the definitions of husband
and single
.
In more complex situations, the underlying symmetry might not be so easy to spot. For example:
// query
def circle = 1, 0
def circle = y, x from x, y where circle(x, y)
The computation reaches the fixpoint after four iterations. This is because the recursion returns to $(1,0)$ after visiting the points $(0,1), (1,0)$, and $(0,1)$.
The reason for this cyclical behavior is that you have encoded a rotational symmetry in your problem. With each iteration, you perform a $90^\circ$ rotation starting at $(1,0)$.
Hence, once you visit all the points of the rotational group $C_4$, modulo the starting point,
you will have reached the fixpoint, and no new elements will be added to circle
.
This toy example shows how the underlying symmetry in the algorithm forces you to stay within a restricted domain, and that all elements in your relation are elements of that domain. Only symmetrybreaking operations will be able to escape this domain.
Symmetries are often beneficial, because this restriction can be exploited and can lead to significant speedups in calculations without loss of generality.
Summary
In summary, a recursive definition is one that references the relation being defined in the body of the definition. To successfully write a recursive algorithm in Rel, you need a base case, a recursive rule, and the termination conditions that set the fixpoint to terminate the algorithm.
In Rel, you can express recursive computations for simple mathematical formulas, for computations involving several relations including multiple selfreferences, and for computations involving negation and aggregation statements. Understanding how recursion works helps to write logic that does not lead to programs that never terminate. It is also useful to know the algorithm for how recursive logic is evaluated in Rel, particularly when writing complex recursive algorithms.
See Also
For further readings on the Rel features used in this concept guide, see:
 The Integrity Constraints concept guide.
 The Modules concept guide.