Games automata play

Value iteration for parity games

December 25, 2017 | by Nathanaël Fijalkow

This post presents a generic value iteration algorithm for parity games parametrised by universal trees. As special cases this extends the small progress measure of Jurdziński and the succinct progress measure of Jurdziński and Lazić. The complexity of the algorithm is the size of the universal tree, motivating constructing small universal trees.

Universal trees

We fix two parameters: $n$, which will be the number of leaves, and $h$, which will be the height. All trees we consider have height $h$, and more precisely all the leaves have depth exactly $h$.

We look at totally ordered trees with unbounded degree, meaning that each node may have arbitrarily many children and the children of a node are totally ordered.

We say that a tree embeds into another if the first one can be obtained by removing nodes from the second. Note that since all trees we consider have height $h$, the root of the first tree is mapped to the root of the second tree. We say that a tree $T$ is $(n,h)$-universal if all trees with at most $n$ leaves each of depth exactly $h$ embed into $T$.

An example of a $(n,h$)-universal tree is the complete tree of height $h$ with each node of degree $n$. It has $n^h$ leaves.

The naive $(5,2)$-universal tree has 25 branches.
A tree with $5$ leaves and height $2$, and one possible embedding into the universal tree above.

Tree labeling as winning witnesses

Consider a parity game with $n$ vertices and using priorities in $[1,d]$ (without loss of generality $d$ is even). Let $h = d/2$.

We consider labelings of trees by vertices. The labels are on the leaves of the tree, and a leaf may be labeled with several vertices. Hence a labeling is a function $\mu : V \to L(T) \cup \{\bot\}$, where $L(T)$ is the set of leaves of $T$.

It induces a set of orders on vertices: for $p \in [1,d]$ and $v,v’$ two vertices, we have $v \ge_p v’$ if the ancestor at level $p/2$ of $v$ is to the left of the ancestor at the same level of $v’$ (levels are counted from the bottom up). The element $\bot$ is the smallest element for all orders $\ge_p$. The strict version is $>_p$. A picture is worth a thousand words:

$u >_1 v >_1 w$ and $u =_3 v >_3 w$.

Note that $\ge_{2p}$ and $\ge_{2p+1}$ coincide.

Theorem: For all $(n,d/2)$-universal tree $T$, there exists a labeling $\mu : V \to L(T) \cup \{\bot\}$ such that:

  • $\mu(v) \neq \bot$ if, and only if, Eve wins from $v$,
  • if $v \in V_E$ has priority $p$, then there exists $(v,v’) \in E$ such that $v \ge_p v’$, strict if $p$ odd,
  • if $v \in V_A$ has priority $p$, then for all $(v,v’) \in E$ we have $v \ge_p v’$, strict if $p$ odd.

Proof: This is a rephrasing of Jurdziński’s original proof for the small progress measure algorithm. The existence is proved by analysing Zielonka’s algorithm, which is explained in this later post.

Indeed, the algorithm yields a labeling $\mu : V \to [1,n]^{d/2}$ satisfying the properties above. We see $[1,n]^{d/2}$ as a leaf in the naive universal tree above, given by its path from the root.

Removing all leaves that are not labeled by some vertex, we obtain a tree with at most $n$ leaves and height $d/2$. By universality it embeds into $T$, implying that it also satisfies the properties above.

The fact that such a tree witnesses winning relies on the observation that the last two properties rule out odd cycles in the underlying positional strategy.

Remark: Mikołaj Bojańczyk observed that in this theorem, it is not necessary to talk about linear preorders, i.e. trees. They could be replaced by general preorders. I do not know yet how this will help, but it sounds like a smart observation.

Generic value iteration algorithm

We fix a $(n,h)$-universal tree $T$. The value iteration algorithm constructs a labeling $\mu : V \to L(T) \cup \{\bot\}$ as in the theorem above. The first labeling assigns all vertices to the maximal (i.e., leftmost) leaf in $T$. At each step, the value iteration algorithm picks a vertex $v$, and checks whether the local conditions defined in conditions 2. and 3. are satisfied by $\mu$ for $v$. If they are not, the value of $\mu(v)$ is decreased by the minimal possible amount to satisfy them, giving rise to $\lift_v(\mu)$. If this is not possible, $\mu(v)$ is assigned $\bot$ (losing).

The correctness follows from the two observations: let $\mu_*$ be a labeling as in the theorem above, for the order $\ge_d$:

  • the initial labeling is pointwise larger than $\mu_*$,
  • if $\mu$ is pointwise larger than $\mu_*$, then so is $\lift_v(\mu)$.


The algorithm given above lifts a vertex $v$ at most $|T|$ many times, which is the number of leaves of $T$. Hence the total number of lifts is $O(n |T|)$, hence so is (roughly) the complexity of the algorithm, provided lifts can be performed in reasonable time.

Theorem: (Jurdziński) There exists a $(n,h)$-universal tree with $n^h$ leaves.

Theorem: (Jurdziński and Lazić) There exists a $(n,h)$-universal tree with $n^{\log(h)}$ leaves.

Question: What is the size of the smallest $(n,h)$-universal tree? If it is polynomial in $n$ and $h$ and such a tree is constructible in polynomial time, then this gives rise to a polynomial time algorithm for parity games.

Below we show the smallest $(5,2)$-universal tree. It has 11 leaves, which is less than the naive one (25 branches).

The smallest $(5,2)$-universal tree has 11 leaves.

Upper bound: The construction we present here is a streamlined version of Jurdziński’s and Lazić’s. It is very marginally smaller.

We construct an $(n,h)$-universal tree with $f(n,h)$ leaves by induction, where

The inductive construction.

To construct the $(n,h)$-universal tree $T$, let:

  • $T_\text{middle}$ a $(n,h-1)$-universal tree;
  • $T_\text{left}$ a $(\lfloor n/2 \rfloor,h)$-universal tree;
  • $T_\text{right}$ a $(n - 1 - \lfloor n/2 \rfloor,h)$-universal tree.

Now construct $T$ as in the drawing above.

We argue that it is $(n,h)$-universal. Consider a $(n,h)$-tree $t$. The question is where to cut in the middle. Let $v_1,\ldots,v_m$ be the children of the root of $t$, and let $n(v_i)$ be the number of leaves below $v_i$. Choose the unique $v_p$ such that


To embed $t$ into $T$, map $v_p$ to the root of $T_\text{middle}$, and then proceed by induction.

Lower bound: We can show that any $(n,h)$-universal tree has at least $g(n,h)$ leaves by induction, where

This will be posted here someday…