Revisiting the coupon collector’s problem to unveil users’ online sessions in networked systems

Abstract

Accuratecomprehension of users’ behavior is paramount for understanding the dynamics of several systems, such as e-commerce platforms, social networks, and mobile computing. To this end, several strategies have been proposed to obtain data sets based on the capture of usage information, which can then serve for user analytics. A popular strategy consists of taking periodic snapshots of online users, a practical instance of the coupon collector’s problem tailored to users monitoring in networked systems. Due to system-specific limitations, however, users may fail to appear in some snapshots, although online. To bridge this gap, we present a methodology to correct ill-collected snapshots and build more accurate data sets. In summary, we formally model user snapshotting as an instance of the coupon collector’s problem, estimate the probability that some users are missing in a given snapshot following a Bernoulli process, and correct those snapshots should the probability exceed a given threshold.

Appendix

1.1 A Formal Proof of Theorems 1 and 2

Theorem 1

Let c be the budget required for capturing snapshots from G, and let L(v_i,b_k) be the acquaintance list returned by b_k ∈ B upon request from v_i ∈ V^′. Considering that crawler v_i makes only one acquaintance list request per snapshot, we then have $s_{t} = \bigcup _{i=1}^{c} L(v_{i}, b_{i}) = V_{t} \iff L_{max} \to \infty$, c ≥|B|, and each bootstrap entity is queried at least once.

Proof

A ⇒ B. Suppose that each user v_i ∈ V is regarded as currently online by a single bootstrap entity b_k ∈ B only. We thus have from Definition 1 that $\bigcup _{k=1}^{|B|} e(b_{k}) = V_{t}$ and $\bigcap _{k=1}^{|B|} e(b_{k}) = \emptyset$. From Definition 3, we have $L(v_{i}, b_{k}) \subseteq e({b_{k}})$. Now consider the restrictions that a crawler v_i ∈ V^′ makes only one acquaintance list request per snapshot. In this case, obtaining a full snapshot of the system (s_t = V_t) requires contacting each of the bootstrap entities in the system, and that L(v_i,b_k) = e(b_k). The condition for L(v_i,b_k) = e(b_k) being true regardless of the size of a bootstrap acquaintance list is that $L_{max} \to \infty$, whereas the condition for $\bigcup _{i=1}^{c} L(v_{i}, b_{i}) = V_{t}$ is that there is at least one crawler for each bootstrap entity (i.e., c ≥|B|), and that each bootstrap entity be queried by a crawler at least once.

B ⇒ A. Suppose that $L_{max} \to \infty$. In this case, L(v_i,b_k) = e(b_k). Suppose also that c = |B| and that each bootstrap entity is queried at least once when building a snapshot set. In this case, we have $\bigcup _{i=1}^{c} L(v_{i}, b_{i}) = V_{t}$, which also respects the restriction that each crawler makes only one acquaintance list request per snapshot. □

Theorem 2

Let c be the budget required for capturing snapshots from G, and let L(vi′,v_j) be the acquaintance list returned by v_j ∈ V_t upon request from a instrumented client vi′∈ V^′. Considering that client vi′ makes one acquaintance list request per snapshot, we then have $s_{t} = \bigcup _{i=1}^{c} L(v'_{i}, v_{i}) = V_{t} \iff \bigcup _{j=1}^{|V_{t}|} e({v_{j}}) = V_{t}$, $L_{max} \to \infty$, c ≥|V_t ∖ V^′|, and each user v_i is queried at least once.

Proof

A ⇒ B. Suppose that each ordinary user v_i ∈ V is regarded as online by at least one user v_j ∈ V (i.e., E_t = V_t × V_t). This situation may lead to several possible configurations in the acquaintance graph; it ranges from one in which the graph forms a single cycle, to another one in which just a single user v_i is aware of all other users v_k in the system, and some other user v_j is aware of v_i as being online. We thus have from Definition 2 that $\bigcup _{j=1}^{|V_{t}|} e(v_{j}) = V_{t}$ and $\bigcap _{j=1}^{|V_{t}|} e(v_{j}) = \emptyset$. From Definition 4, we have $L(v_{i}, v_{j}) \subseteq e({v_{j}})$. Now consider the restrictions that a crawler v_i ∈ V^′ makes only one acquaintance list request per snapshot. In this case, obtaining a full snapshot of the system (s_t = V_t) requires contacting each of the users in the system, and that L(vi′,v_j) = e(v_j). The condition for L(vi′,v_j) = e(v_j) being true regardless of the size of a user’s acquaintance list is that $L_{max} \to \infty$. The condition for $\bigcup _{i=1}^{c} L(v'_{i}, v_{i}) = V_{t}$ is that (i) there is at least one crawler for each ordinary user (c ≥|V_t ∖ V^′|), (ii) that each ordinary user be queried by a crawler at least once, and (iii) that every user be regarded as currently online by at least another user in the system (which results in $\bigcup _{j=1}^{|V_{t}|} e({v_{j}}) = V_{t}$).

B ⇒ A. Suppose that $L_{max} \to \infty$. In this case, L(vi′,v_i) = e(v_i). Suppose also that c ≥|V_t ∖ V^′|, that each ordinary user is queried at least once when building a snapshot set, and that every user be regarded as currently online by at least another user in the system (i.e., $\bigcup _{j=1}^{|V_{t}|} e({v_{j}}) = V_{t}$). In this case, querying all ordinary users using the budget we have and making the union of the peer lists obtained, we obtain a full snapshot of the system ($\bigcup _{i=1}^{c} L(v'_{i}, v_{i}) = V_{t}$). Observe that since we have a crawler for each ordinary user, the restriction that each crawler makes only one acquaintance list request per snapshot can thus be respected. □