Correctness

In the previous section we constructed the threshold gameG_bofG, which we obtained

by augmenting the parity game with weights G with the memory structure M =

(M, init, upd). Each vertex(v, o, r)ofG_bconsists of a vertex v fromG, a counter value o that counts the number of times the cost of some request has exceeded the bound b, and a request function r : D → {⊥} ∪I that stores information about the open requests and the costs they have incurred so far.

Moreover, we claimed that the threshold problem reduces to solving the constructed parity game. In this section, we show that this is indeed the case, i.e., this section is

dedicated to showing the following theorem. Recall that, in : Section 3.1, we de- : Page 31

fined Costv∗(_σ) = sup

ρCost(ρ), where ρ ranges over all plays starting in v

∗ _{and con-}

sistent with σ.

Theorem 4.8. Let v∗ be a vertex inG. Player 0 has a strategy σ inG with Costv∗(_σ) ≤ b if

and only if she winsG_bfrom(v∗, init(v∗)).

We split the proof into several lemmas. To this end, recall that we defined G_b = (A0, Parity(Ω0)). LetA0 = (V0, V₀0, V₁0, E0), i.e., in particular, let V0 =V×M and V_i0 =

Vi×M for i ∈ {0, 1}. Moreover, for the remainder of this section, fix some vertex v∗ of G.

The direction from right to left is relatively straightforward: Since G_b is a parity game, if Player 0 wins G_b from (v∗, init(v∗)), then she does so with a positional strat-

egy σ0, due to: Proposition 2.18. This strategy assigns to each vertex(v, o, r)of Player 0 : Sec. 2.3, Page 20

in G_b a unique successor(v0, o0, r0), where the values of o0 and r0 are deterministic up- dates of o and r via the update function upd. Hence, σ0 can be interpreted as picking only a successor vertex v0 of v with respect to the current memory state(o, r). Thus, the choices of σ0 can be mimicked inG.

Lemma 4.9. If Player 0 wins G_b from (v∗, init(v∗)), then she has a strategy σ in G with

Proof. Let σ0: V₀0 →V0 be a positional winning strategy for Player 0 from(v∗, init(v∗))

inG_b. SinceG_bis a classical parity game, such a strategy exists due to Proposition 2.18. We define the finite-state strategy σ for Player 0 inG as the unique strategy induced by the memory structureMand the next-move function nxt defined as nxt(v, m) =v0, if σ0(v, m) = (v0, m0)for some m0. It remains to show Costv∗(_σ) ≤b.

Let ρ=v0v1v2· · · be a play inG that begins in v∗ and that is consistent with σ. Let

ρ0 =ext(ρ) = (v0, o0, r0)(v1, o1, r1)(v2, o2, r2) · · · be the unique extended play inG_b.

A simple induction shows that ρ0 is consistent with σ0. As ρ0 moreover starts in(v∗, init(v∗)) by definition, it is winning for Player 0, i.e., it satisfies the parity con- dition. This in particular implies that the overflow counter along ρ0 never saturates, i.e., that we have oj < n for all j ∈ N, due to Remark 4.6. Hence, the plays ρ and ρ0 coincide on their color sequences. Since ρ0 is winning for Player 0 inG_b, it satisfies the parity condition, which in turn implies that ρ satisfies the parity condition. It remains to show that almost all requests in ρ are answered with cost at most b.

As argued above, the overflow counter along ρ0 eventually stabilizes at some value less than n. Moreover, since ρ satisfies the parity condition, after some finite prefix, all requests are answered. Hence, there exists a position j such that oj0 =o_j <n and such

that Cor(ρ, j0) <∞ for every j0 > j. Assume towards a contradiction that there exists some j0 ≥j with b <Cor(ρ, j0) <∞. Then, the play ρ0 contains some overflow position at some point after j0 due to Remark 4.7.1. This, however, contradicts the choice of j. Hence, we obtain that almost all requests in ρ are answered with cost at most b.

We now turn our attention to the other direction of the statement, i.e., we aim to show that, if Player 0 has a strategy σ in G with Costv∗(_σ) ≤ b, then she wins G_b

from(v∗, init(v∗)). We show this claim via contraposition: Assume that Player 0 does not win G_b from (v∗, init(v∗)). Since G_b is a parity game, it is determined due to : Proposition 2.18. Hence, Player 1 winsG_b from(v∗, init(v∗)), say with the positional

: Sec. 2.3, Page 20

strategy τ0. We obtain that such a positional strategy for him exists due to Proposi- tion 2.18. We construct a strategy τ for Player 1 inGthat enforces Cost(ρ) >b for each play ρ starting in v∗ and consistent with τ. Since this implies Costv∗(_σ) > b for each

strategy σ of Player 0, this suffices to show the desired statement.

Recall that the overflow counter along each play starting in (v∗, init(v∗)) is mono- tonically increasing and bounded from above by the number n of vertices inG. Hence, the value of the overflow counter either stabilizes at some value less than n, or it even- tually saturates at value n. In the former case, τ0 has to ensure that the resulting play violates the parity condition. Hence, it suffices to mimic the moves made by τ0 in G

ad infinitum in this case, which results in a play with infinitely many unanswered requests inG. In the latter case, however, mimicking τ0 does not yield a strategy inG

with the desired property, as τ0 does not necessarily prescribe “meaningful” moves inG_bonce the overflow counter saturates. In order to leverage τ0 even after saturation of the overflow counter, we intervene whenever the overflow counter is incremented, by resetting it to the smallest possible value from which τ0 is still winning.

Formally, we define the set R that contains all vertices (v, o, r) that are visited by Def.R

some play that starts in (v∗, init(v∗)) and that is consistent with τ0. Recall that we defined rv as the function denoting the requests opened by visiting the vertex v in

: Section 4.1.2. Given a vertex v, we then define ov = min({n} ∪ {o| (v, o, rv) ∈ R}). : Page 88 In particular, we have ov∗ =0, since(v∗, init(v∗)) = (v∗, 0, r_v) ∈ R.

Lemma 4.10. The strategy τ0is winning for Player 1 from(v, ov, rv)inGbfor all v∈V.

Proof. If ov = n, then all plays starting in (v, ov, rv) violate the parity condition by construction ofA0_{. Thus, for the remainder of this proof, assume o}

v <n. Let(v, ov, rv)ρ be a play starting in(v, ov, rv)that is consistent with τ0. Moreover, let π be a play prefix starting in(v∗, init(v∗)), consistent with τ0, and ending in(v, ov, rv). Such a play prefix exists due to definition of ov and due to the assumption of ov<n.

Since τ0 is positional, the play πρ starts in (v∗, init(v∗)) and is consistent with τ0. Thus, πρ violates the parity condition, which implies that ρ violates the parity condi- tion due to prefix-independence.

We now define a new memory structure M0 implementing the strategy τ. Recall

that we defined the memory structure M = (M, init, upd) during the construction

of the threshold game G_b in : Section 4.1.2. Using the components of that memory : Page 88

structure, we define M0 = M= [n+1] ×R, init0 =init, and upd0((o, r),(v, v0)) =

(

(o, r0) if upd((o, r),(v, v0)) = (o, r0), and

(ov0, r0) if upd((o, r),(v, v0)) = (o+1, r0).

and combine these elements into the memory structureM0 _{= (M}0_{, init}0_{, upd}0₎

.

In the second case of the definition of upd0, we have r0 = rv0 by definition of upd.

Finally, we define the next-move function nxt0 via nxt0(v, m) =v0, if τ0(v, m) = (v0, m0)

for some m0 ∈ M and let τ be the finite-state strategy implemented byM0_{and nxt. We}

claim that for each play ρ starting in v∗ that is consistent with τ we have Cost(ρ) >b. To show this claim, let ρ=v0v1v2· · · be some play inG that starts in v∗ and that is consistent with τ. Moreover, let

ρ0 =extM0(_ρ) = (v₀, o₀, r₀)(v₁, o₁, r₁)(v₂, o₂, r₂) · · ·

be the extended play of ρ with respect toM0. We say that j is a reset position if j=0 Def.reset position

or if

upd((oj−1, rj−1),(vj−1, vj)) = (oj−1+1, rj) ,

i.e., if the second case in the definition of upd0 is applied.

The play ρ0 is not necessarily a play in G_b, since G_b is defined with respect to M

instead ofM0, but every infix of ρ0 that starts at a reset position and does not contain another one, is a play infix in G_b that is consistent with τ0. At every reset position, instead of incrementing the overflow counter, we set it to ov.

Our first aim is to show oj < n for all j by analyzing the structure of ρ0. Intuitively, this then implies that the strategy τ always uses “meaningful” moves of τ0 for its choice of move and thus allows us to subsequently argue that τ is indeed winning for Player 1.

To this end, we first show that, even though ρ0 is, in general, not a play in G_b, every vertex (vj, oj, rj)of ρ0 is in R. This implies that, intuitively, the positional strategy τ0 prescribes a “useful” move from every(vj, oj, rj)in V10.

Lemma 4.11. For each j∈N we have o_j <n.

Proof. We first show that for each j ∈ _{N we have} (vj, oj, rj) ∈ R by induction over j. For j=0, this is clear, as we obtain(v0, o0, r0) = (v∗, init(v∗)) ∈ Rdue to the definition ofR.

For j > 0, the induction hypothesis yields v0_j−1 = (vj−1, oj−1, rj−1) ∈ R. Thus, by definition of R, there exists a play prefix π starting in (v∗, init(v∗)) = (v0, o0, r0), consistent with τ0 and ending in v0_j−1. If v0_j−1 ∈ V₀0, then π· (vj, o, rj) is consistent with τ0 for some o∈ {0, . . . , n}, which implies(vj, o, rj) ∈ R. If o =oj, then(vj, oj, rj) ∈

R clearly holds true. If o 6=oj, however, then oj = ovj and rj = rvj, i.e., (vj, oj, rj) ∈ R by definition of ovj. Otherwise, i.e., if v

0

j−1 ∈ V10, then we have τ(π) = (vj, o, rj) for some o∈ {0, . . . , n}. Similar reasoning as in the previous case yields(vj, oj, rj) ∈ Rin this case as well.

In a second step, we show that we do not “skip” values of the overflow counter usingM0, i.e., we show oj+1≤oj+1 for all j∈N. To this end, let j∈ N and, towards a contradiction, assume oj+1 >oj+1. Since oj+16=oj, the second case in the definition of upd0 is applied, which implies

• oj+1 =ovj+1, as well as

• upd((oj, rj),(vj, vj+1)) = (oj+1, rvj+1).

We show(vj+1, oj+1, rvj+1) ∈ R, which implies oj+1 = ovj+1 ≤ oj+1, i.e., the desired contradiction, due to definition of ovj+1.

Recall that we showed(vj, oj, rj) ∈ Rabove. If(vj, oj, rj)is a vertex of Player 0, then we directly obtain(vj+1, oj+1, rvj+1) ∈ Rdue to upd((oj, rj),(vj, vj+1)) = (oj+1, rvj+1), which implies that there is an edge leading from(vj, oj, rj)to(vj+1, oj+1, rvj+1)inA

0_.

If, however, the vertex (v_j, oj, rj) belongs to Player 1, then we have τ(vj, oj, rj) =

(vj+1, oj+1, rj+1). By definition of τ, this yields τ0(vj, oj, rj) = (vj+1, o, r)for some o ∈N and some request function r. By definition of the arenaA0 _{= A × Mwe have(o, r) =}

upd((o_j, rj),(vj, vj+1)) = (oj+1, rvj+1). Thus, we obtain(vj+1, oj+1, rvj+1) ∈ R, which yields the desired contradiction as argued above. This concludes the proof of oj+1 ≤ oj+1 for all j∈N.

Finally, we prove the claim of this lemma by showing the following property by induction over j:

If oj = k, then for every k0 ≤ k there is a reset position jk0 ≤ j with

oj_k0 =k0.

Let us first argue that this indeed implies oj <n for all j∈N. Towards a contradiction,

value k in the range 0 ≤k≤n for the overflow counter. Thus, two such positions j0, j00 share the same vertex vj0 =v_j00, but have differing values of the overflow counter o_j06=

oj00. By construction of M0, however, we obtain o_j0 = o_v

j0 = ovj00 = oj00, i.e., the desired contradiction. It remains to show that the property above indeed holds true.

For the induction start, we have o0 = 0 and pick j0 = 0, which is a reset position. Now, let j > 0 and let oj = k. If k ≤ oj−1, then the induction hypothesis yields the necessary positions. Hence, assume we have k> oj−1, which implies k=oj−1+1 due to oj+1≤ oj+1, which we showed above. Then, j is a reset position. Hence, we define jk =j and obtain the remaining jk0for k0 <k via the induction hypothesis.

It remains to show that we indeed have Cost(ρ) > b. As argued above, this then directly implies that for each strategy σ of Player 0 we have Costv∗(_σ) >b, concluding

the proof of the direction from left to right of Theorem 4.8.

Proof of Theorem 4.8. The direction from right to left is encapsulated in Lemma 4.9. For the direction from left to right, recall that we defined a strategy τ for Player 1, that we picked ρ beginning in v∗ and consistent with τ arbitrarily and that we de- fined ρ0 = extM0(ρ) = (v0, o0, r0)(v1, o1, r1)(v2, o2, r2) · · ·. First assume that the over- flow counter of ρ0 eventually stabilizes, i.e., there exists some j∈ N such that o_j0 = o_j

for all j0 > j. Then, there exists a suffix of ρ0 that is consistent with τ0, which there- fore violates the parity condition. Hence, it suffices to note that the color sequences induced by ρ0 and by ρ coincide in this case due to Lemma 4.11 and due to the con- struction ofG_b, in which vertices of the form(v, o, r)inherit the coloring of the vertex v for o < n. Thus, ρ violates the parity condition and, in turn, also the parity condition with weights with respect to any bound.

Now assume that the overflow counter of ρ0 does not stabilize. Then, there are

infinitely many reset positions in ρ0. We obtain that there exist a request that incurs cost greater than b between any two adjacent such positions as a direct consequence

of : Remark 4.7.2. Hence, we obtain Cost(ρ) > b, which concludes this direction of : Sec. 4.1, Page 91

the proof.

Having thus reduced the threshold problem for parity games with weights to that of solving classical parity games, we discuss the implications of this reduction on the complexity of the former problem in the following section.