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Relations · step 13 of 13

Symmetric closure vs transitive closure

Reflexive and symmetric closures are one-step operations: add the diagonal, or add R⁻¹, and you are finished — the added pairs never create a new obligation. Transitive closure is different. Each shortcut you add can create fresh chains needing further shortcuts, so the process must iterate to a fixed point. That difference is why closure algorithms exist for transitivity and not for the other two.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

  1. Symmetric closure — one unionR ∪ R⁻¹. Adding (b,a) never creates a demand for anything else, so no iteration is needed.
  2. Transitive closure — iterate to a fixed pointAdd all forced shortcuts, then repeat on the enlarged relation until a pass adds nothing.
  3. Watch the interactionThe symmetric closure of a transitive relation need not be transitive, so "symmetric transitive closure" must be computed by closing repeatedly under both.
  4. Bound the workOn n elements the transitive closure stabilises within n − 1 rounds, or in one Warshall sweep.

Worked example

For R = {(1,2),(2,3)} on {1,2,3}, compute the symmetric closure and then the transitive closure of that result.

  1. Symmetric closure: add reverses, giving {(1,2),(2,1),(2,3),(3,2)}.
  2. Now transitive closure: (1,2)+(2,1) forces (1,1); (2,1)+(1,2) forces (2,2); (2,3)+(3,2) forces (2,2); (3,2)+(2,3) forces (3,3).
  3. Also (1,2)+(2,3) forces (1,3) and (3,2)+(2,1) forces (3,1).
  4. Another pass adds nothing new.

Answer. All nine pairs — the closure is the total relation on {1,2,3}, showing how quickly transitive closure expands after symmetrising.

Where marks get dropped

These are the specific errors that cost credit on symmetric closure vs transitive closure questions — QED's rubric penalises each of them separately.

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Symmetric closure vs transitive closure — frequently asked questions

What is the equivalence closure?

The smallest equivalence relation containing R: take R ∪ R⁻¹ ∪ Δ and then the transitive closure. This is exactly how union-find algorithms build connected components.

Why can’t transitive closure be done in one pass?

Because adding (a,c) may combine with an existing (c,d) to force (a,d), which was not a chain in the original relation. Warshall’s ordering of intermediates is what tames this into a single sweep.

Does the order of closures matter?

Yes. Closing transitively then symmetrically may leave a non-transitive relation, which is why the equivalence closure ends with the transitive step.

The rest of Relations

Properties of relations, composition, and representations. Each subtopic below has its own method, worked example and mark-losing traps.

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