Constrained School Choice



One easily shows that that Oτ (P, 1) NW(P) ∩ IR(P). Hence, by Lemma A.7 there
exist a set of students
CI = {i1, . . . , ip} and a set of schools CS = {s1, . . . , sp} such that
for each
ir CI, sr Pir sr+1 = τ (Q)(ir). This proves (a). Obviously,

As(Q) = τ(Q)(s) for any Q QI.                          (8)

Hence, by Lemma A.7, for each school s CS, As(Q) = τ(Q)(s) = qs. Hence, (b)
follows. Since a student is part of exactly 1 cycle, (c) follows.                           

For any student ir CI define Qr := (sr, Q-ir). By Step 1, τ (Q)(ir) = Qir = sr+1

(modulo p). Since Qirr = sr Pir sr+1 and Q Eτ (P, 1), τ(Qr)(ir) = ir.

Step 2 Let r {1, . . . , p}. For each s S\sr+1, As(Qr) = As(Q), and Asr+1(Qr) =
Asr+1(Q) - 1.

In view of (8) and ir τ (Q)(Sr+1), we are done if we prove that for each s S\sr+1,
τ(Qr)(s) = τ (Q)(s), and τ (Qr)(sr+1) = τ (Q)(sr+1)\ir.

For each school s = sr, sr+1, only the qs (see Step 1(b)) students in τ (Q)(s) list school
s in Qr. Note also that only the qsr+1 - 1 (see Step 1(b)) students in τ (Q)(sr+1)\ir list
school
sr+1 in Qr. Since Qr QI (1) and τ(Qr) NW (Qr), τ (Qr)(s) = τ (Q)(s) for each
school
s = sr. Finally, note that only the qsr + 1 students in τ (Q)(sr) ir list school sr
in Qr. Since Qrr = sr, τ(Qr)(ir) = ir, and τ(Qr) NW(Qr), τ(Qr)(sr) = τ(Q)(sr).    

Step 3 For any r {1, . . . ,p}, Asr(Qr) = Asr(Q).

We are done if we show the following claim.

Claim: For each integer l, we have

[ sr V(Q, l) if and only if sr V(Qr, l) ],                                     1

[ if sr is in a cycle C of G(Q, l) then C is also a cycle of G(Qr, l)], and (9)
[
if sr is in a cycle C of G(Qr, l) then C is also a cycle of G(Q, l)].       

Proof of Claim: We distinguish among six cases.

Case 1: l < min{σ(Q, ir), σ(Qr, ir)}. Then, (9) follows from Lemma B.3(a).

Case 2: σ(Q,ir) ≤ l < σ(Qr,ir). Since Qirr = sr and τ(Qr)(ir) = ir, sr P(Qr,l,ir).

Since σ(Qr, sr) = σ(Qr, ir) - 1, sr V(Qr, l).  By Lemma B.7, sr V(Q, l) and

F (Q, l, sr) = F (Qr, l, sr) (as directed paths). So, (9) holds.

Case 3: σ(Q, ir) < l = σ(Qr, ir). Since σ(Qr, ir) = σ(Qr, sr) + 1, F(Qr, l - 1, sr) is the
last cycle of
sr under TT C(Qr). By Case 2, this is a cycle of sr under TT C(Q). In fact,

41



More intriguing information

1. Pupils’ attitudes towards art teaching in primary school: an evaluation tool
2. Getting the practical teaching element right: A guide for literacy, numeracy and ESOL teacher educators
3. Pricing American-style Derivatives under the Heston Model Dynamics: A Fast Fourier Transformation in the Geske–Johnson Scheme
4. The name is absent
5. Menarchial Age of Secondary School Girls in Urban and Rural Areas of Rivers State, Nigeria
6. FISCAL CONSOLIDATION AND DECENTRALISATION: A TALE OF TWO TIERS
7. Behaviour-based Knowledge Systems: An Epigenetic Path from Behaviour to Knowledge
8. Assessing Economic Complexity with Input-Output Based Measures
9. AN EMPIRICAL INVESTIGATION OF THE PRODUCTION EFFECTS OF ADOPTING GM SEED TECHNOLOGY: THE CASE OF FARMERS IN ARGENTINA
10. Tariff Escalation and Invasive Species Risk
11. Dual Inflation Under the Currency Board: The Challenges of Bulgarian EU Accession
12. The name is absent
13. The name is absent
14. Implementation of Rule Based Algorithm for Sandhi-Vicheda Of Compound Hindi Words
15. Structural Conservation Practices in U.S. Corn Production: Evidence on Environmental Stewardship by Program Participants and Non-Participants
16. The name is absent
17. ‘Goodwill is not enough’
18. Implementation of a 3GPP LTE Turbo Decoder Accelerator on GPU
19. Demand Potential for Goat Meat in Southern States: Empirical Evidence from a Multi-State Goat Meat Consumer Survey
20. The name is absent