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Influence Maximization and Outbreak Detection (#12)

* add influence_maximization and figures, edit .gitignore to ignore jekyll cache

* fix repeated the

* change the file name with dash, add outbreak detection

*  finished outbreak detection

* proof reading done

* update with suggestions

* Add sidenote for other evaluation methods besides sketches, make Hill Climbing Algorithm more precise
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layout: post
title: Contents
---
<span class="newthought">These notes</span> form a concise introductory course on machine learning with large-scale graphs. They mirror the topics topics covered by Stanford [CS224W](https://cs224w.stanford.edu), and are written by the CS 224W TAs.
<span class="newthought">These notes</span> form a concise introductory course on machine learning with large-scale graphs. They mirror the topics topics covered by Stanford [CS224W](https://cs224w.stanford.edu), and are written by the CS 224W TAs.
{% include marginnote.html id='mn-construction' note='The notes are still under construction! They will be written up as lectures continue to progress. If you find any typos, please let us know, or submit a pull request with your fixes to our [GitHub repository](https://github.com/snap-stanford/cs224w-notes).'%}
You too may help make these notes better by submitting your improvements to us via [GitHub](https://github.com/snap-stanford/cs224w-notes). Note that submitting substantial improvements will result in *bonus points* being added to your overall grade!
Starting with the Fall 2019 offering of CS 224W, the course covers three broad topic areas for understanding and effectively learning representations from large-scale networks: preliminaries, network methods, and machine learning with networks. Subtopics within each area correspond to individual lecture topics.
Starting with the Fall 2019 offering of CS 224W, the course covers three broad topic areas for understanding and effectively learning representations from large-scale networks: preliminaries, network methods, and machine learning with networks. Subtopics within each area correspond to individual lecture topics.
## Preliminaries
@ -19,8 +19,8 @@ Starting with the Fall 2019 offering of CS 224W, the course covers three broad t
1. [Structural Roles in Networks](): RolX, Granovetter, the Louvain algorithm
2. [Spectral Clustering](network-methods/spectral-clustering): Graph partitions and cuts, the Laplacian, and motif clustering
3. [Influence Maximization](): Influential sets, submodularity, hill climbing
4. [Outbreak Detection](): CELF, lazy hill climbing
3. [Influence Maximization](network-methods/influence-maximization): Influential sets, submodularity, hill climbing
4. [Outbreak Detection](network-methods/outbreak-detection): CELF, lazy hill climbing
5. [Link Analysis](): PageRank and SimRank
6. [Network Effects and Cascading Behavior](network-methods/network-effects-and-cascading-behavior): Decision-based diffusion, probabilistic contagion, SEIZ
7. [Network Robustness](): Power laws, preferential attachment

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---
layout: post
title: Influence Maximization
---
## Motivation
Identification of influential nodes in a network has important practical uses. A good example is "viral marketing", a strategy that uses existing social networks to spread and promote a product. A well-engineered viral marking compaign will identify the most influential customers, convince them to adopt and endorse the product, and then spread the product in the social network like a virus.
The key question is how to find the most influential set of nodes? To answer this question, we will first look at two classical cascade models:
- Linear Threshold Model
- Independent Cascade Model
Then, we will develop a method to find the most influential node set in the Independent Cascade Model.
## Linear Threshold Model
In the Linear Threshold Model, we have the following setup:
- A node $$v$$ has a random threshold $$\theta_{v} \sim U[0,1]$$
- A node $$v$$ influenced by each neighbor $$w$$ according to a weight $$b_{v,w}$$, such that
$$
\sum_{w\text{ neighbor of v }} b_{v,w}\leq 1
$$
- A node $$v$$ becomes active when at least $$\theta_{v}$$ fraction of its neighbors are active. That is
$$
\sum_{w\text{ active neighbor of v }} b_{v,w}\geq\theta_{v}
$$
The following figure demonstrates the process:
![linear_threshold_model_demo](../assets/img/influence_maximization_linear_threshold_model_demo.png?style=centerme)
*(A) node V is activated and influences W and U by 0.5 and 0.2, respectively; (B) W becomes activated and influences X and U by 0.5 and 0.3, respectively; (C) U becomes activated and influences X and Y by 0.1 and 0.2, respectively; (D) X becomes activated and influences Y by 0.2; no more nodes can be activated; process stops.*
## Independent Cascade Model
In this model, we model the influences (activation) of nodes based on probabilities in a directed graph:
- Given a directed finite graph $$G=(V, E)$$
- Given a node set $$S$$ starts with a new behavior (e.g. adopted new product and we say they are active)
- Each edge $$(v, w)$$ has a probability $$p_{vw}$$
- If node $$v$$ becomes active, it gets one chance to make $$w$$ active with probability $$p_{vw}$$
- Activation spread through the network
Note:
- Each edge fires only once
- If $$u$$ and $$v$$ are both active and link to $$w$$, it does not matter which tries to activate $$w$$ first
## Influential Maximization (of the Independent Cascade Model)
### Definitions
- **Most influential Set of size $$k$$** ($$k$$ is a user-defined parameter) is a set $$S$$ containing $$k$$ nodes that if activated, produces the largest expected{% include sidenote.html id='note-most-influential-set' note='Why "expected cascade size"? Due to the stochastic nature of the Independent Cascade Model, node activation is a random process, and therefore, $$f(S)$$ is a random variable. In practice, we would like to compute many random simulations and then obtain the expected value $$f(S)=\frac{1}{\mid I\mid}\sum_{i\in I}f_{i}(S)$$, where $$I$$ is a set of simulations.' %} cascade size $$f(S)$$.
- **Influence set $$X_{u}$$ of node $$u$$** is the set of nodes that will be eventually activated by node $$u$$. An example is shown below.
![influence_set](../assets/img/influence_maximization_influence_set.png?style=centerme)
*Red-colored nodes a and b are active. The two green areas enclose the nodes activated by a and b respectively, i.e. $$X_{a}$$ and $$X_{b}$$.*
Note:
- It is clear that $$f(S)$$ is the size of the union of $$X_{u}$$: $$f(S)=\mid\cup_{u\in S}X_{u}\mid$$.
- Set $$S$$ is more influential, if $$f(S)$$ is larger
### Problem Setup
The influential maximization problem is then an optimization problem:
$$
\max_{S \text{ of size }k}f(S)
$$
This problem is NP-hard [[Kempe et al. 2003]](https://www.cs.cornell.edu/home/kleinber/kdd03-inf.pdf). However, there is a greedy approximation algorithm--**Hill Climbing** that gives a solution $$S$$ with the following approximation guarantee:
$$
f(S)\geq(1-\frac{1}{e})f(OPT)
$$
where $$OPT$$ is the globally optimal solution.
### Hill Climbing
**Algorithm:** at each step $$i$$, activate and pick the node $$u$$ that has the largest marginal gain $$\max_{u}f(S_{i-1}\cup\{u\})$$:
- Start with $$S_{0}=\{\}$$
- For $$i=1...k$$
- Activate node $$u\in V\setminus S_{i-1}$$ that $$\max_{u}f(S_{i-1}\cup\{u\})$$
- Let $$S_{i}=S_{i-1}\cup\{u\}$$
**Claim:** Hill Climbing produces a solution that has the approximation guarantee $$f(S)\geq(1-\frac{1}{e})f(OPT)$$.
### Proof of the Approximation Guarantee of Hill Climbing
**Definition of Monotone:** if $$f(\emptyset)=0$$ and $$f(S)\leq f(T)$$ for all $$S\subseteq T$$, then $$f(\cdot)$$ is monotone.
**Definition of Submodular:** if $$f(S\cup \{u\})-f(S)\geq f(T\cup\{u\})-f(T)$$ for any node $$u$$ and any $$S\subseteq T$$, then $$f(\cdot)$$ is submodular.
**Theorem [Nemhauser et al. 1978]:**{% include sidenote.html id='note-nemhauser-theorem' note='also see this [handout](http://web.stanford.edu/class/cs224w/handouts/CS224W_Influence_Maximization_Handout.pdf)' %} if $$f(\cdot)$$ is **monotone** and **submodular**, then the $$S$$ obtained by greedily adding $$k$$ elements that maximize marginal gains satisfies
$$
f(S)\geq(1-\frac{1}{e})f(OPT)
$$
Given this theorem, we need to prove that the largest expected cascade size function $$f(\cdot)$$ is monotone and submodular.
**It is clear that the function $$f(\cdot)$$ is monotone based on the definition of $$f(\cdot)$${% include sidenote.html id='note-monotone' note='If no nodes are active, then the influence is 0. That is $$f(\emptyset)=0$$. Because activating more nodes will never hurt the influence, $$f(U)\leq f(V)$$ if $$U\subseteq V$$.' %}, and we only need to prove $$f(\cdot)$$ is submodular.**
**Fact 1 of Submodular Functions:** $$f(S)=\mid \cup_{k\in S}X_{k}\mid$$ is submodular, where $$X_{k}$$ is a set. Intuitively, the more sets you already have, the less new "area", a newly added set $$X_{k}$$ will provide.
**Fact 2 of Submodular Functions:** if $$f_{i}(\cdot)$$ are submodular and $$c_{i}\geq0$$, then $$F(\cdot)=\sum_{i}c_{i} f_{i}(\cdot)$$ is also submodular. That is a non-negative linear combination of submodular functions is a submodular function.
**Proof that $$f(\cdot)$$ is Submodular**: we run many simulations on graph G (see sidenote 1). For the simulated world $$i$$, the node $$v$$ has an activation set $$X^{i}_{v}$$, then $$f_{i}(S)=\mid\cup_{v\in S}X^{i}_{v}\mid$$ is the size of the cascades of $$S$$ for world $$i$$. Based on Fact 1, $$f_{i}(S)$$ is submodular. The expected influence set size $$f(S)=\frac{1}{\mid I\mid}\sum_{i\in I}f_{i}(S)$$ is also submodular, due to Fact 2. QED.
**Evaluation of $$f(S)$$ and Approximation Guarantee of Hill Climbing In Practice:** how to evaluate $$f(S)$$ is still an open question. The estimation achieved by simulating a number of possible worlds is a good enough evaluation [[Kempe et al. 2003]](https://www.cs.cornell.edu/home/kleinber/kdd03-inf.pdf):
- Estimate $$f(S)$$ by repeatedly simulating $$\Omega(n^{\frac{1}{\epsilon}})$$ possible worlds, where $$n$$ is the number of nodes and $$\epsilon$$ is a small positive real number
- It achieves $$(1\pm \epsilon)$$-approximation to $$f(S)$$
- Hill Climbing is now a $$(1-\frac{1}{e}-\epsilon)$$-approximation
### Speed-up Hill Climbing by Sketch-Based Algorithms
**Time complexity of Hill Climbing**
To find the node $$u$$ that $$\max_{u}f(S_{i-1}\cup\{u\})$$ (see the algorithm above):
- we need to evaluate the $$X_{u}$$ (the influence set) of each of the remaining nodes which has the size of $$O(n)$$ ($$n$$ is the number of nodes in $$G$$)
- for each evaluation, it takes $$O(m)$$ time to flip coins for all the edges involved ($$m$$ is the number of edges in $$G$$)
- we also need $$R$$ simulations to estimate the influence set ($$R$$ is the number of simulations/possible worlds)
We will do this $$k$$ (number of nodes to be selected) times. Therefore, the time complexity of Hill Climbing is $$O(k\cdot n \cdot m \cdot R)$$, which is slow. We can use **sketches** [[Cohen et al. 2014]](https://www.microsoft.com/en-us/research/wp-content/uploads/2014/08/skim_TR.pdf) to speed up the evaluation of $$X_{u}$$ by reducing the evaluation time from $$O(m)$$ to $$O(1)$${% include sidenote.html id='note-evaluate-influence' note='Besides sketches, there are other proposed approaches for efficiently evaluating the influence function: approximation by hypergraphs [[Borgs et al. 2012]](https://arxiv.org/pdf/1212.0884.pdf), approximating Riemann sum [[Lucier et al. 2015]](https://people.seas.harvard.edu/~yaron/papers/localApproxInf.pdf), sparsification of influence networks [[Mathioudakis et al. 2011]](https://chato.cl/papers/mathioudakis_bonchi_castillo_gionis_ukkonen_2011_sparsification_influence_networks.pdf), and heuristics, such as degree discount [[Chen et al. 2009]](https://www.microsoft.com/en-us/research/wp-content/uploads/2016/02/weic-kdd09_influence.pdf).'%}.
**Single Reachability Sketches**
- Take a possible world $$G^{i}$$ (i.e. one simulation of the graph $$G$$ using the Independent Cascade Model)
- Give each node a uniform random number $$\in [0,1]$$
- Compute the **rank** of each node $$v$$, which is the **minimum** number among the nodes that $$v$$ can reach in this world.
*Intuition: if $$v$$ can reach a large number of nodes, then its rank is likely to be small. Hence, the rank of node $$v$$ can be used to estimate the influence of node $$v$$ in $$G^{i}$$.*
However, influence estimation based on Single Reachability Sketches (i.e. single simulation of $$G$$ ) is inaccurate. To make a more accurate estimate, we need to build sketches based on many simulations{% include sidenote.html id='note-sketches' note='This is similar to take an average of $$f_{i}(S)$$ in sidenote 1, but in this case, it is achieved by using Combined Reachability Sketches.' %}, which leads to the Combined Reachability Sketches.
**Combined Reachability Sketches**
In Combined Reachability Sketches, we simulate several possible worlds and keep the smallest $$c$$ values among the nodes that $$u$$ can reach in all the possible worlds.
- Construct Combined Reachability Sketches:
- Generate a number of possible worlds
- For node $$u$$, assign uniformly distributed random numbers $$r^{i}_{v}\in[0,1]$$ to all $$(v, i)$$ pairs, where $$v$$ is the node in $$u$$'s reachable nodes set in the world $$i$$.
- Take the $$c$$ smallest $$r^{i}_{v}$$ as the Combined Reachability Sketches
- Run Greedy for Influence Maximization:
- Whenever the greedy algorithm asks for the node with the largest influence, pick node $$u$$ that has the smallest value in its sketch.
- After $$u$$ is chosen, find its influence set $$X^{i}_{u}$$, mark the $$(v, i)$$ as infected and remove their $$r^{i}_{v}$$ from the sketches of other nodes.
Note: using Combined Reachability Sketches does not provide an approximation guarantee on the true expected influence but an approximation guarantee with respect to the possible worlds considered.

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---
layout: post
title: Outbreak Detection in Networks
---
## Introduction
The general goal of outbreak detection in networks is that given a dynamic process spreading over a network, we want to select a set of nodes to detect the process efficiently. Outbreak detection in networks has many applications in real life. For example, where should we place sensors to quickly detect contaminations in a water distribution network? Which person should we follow on Twitter to avoid missing important stories?
The following figure shows the different effects of placing sensors at two different locations in a network:
![sensor_placement](../assets/img/outbreak_detection_sensor_placement.png?style=centerme)
*(A) The given network. (B) An outbreak $$i$$ starts and spreads as shown. (C) Placing a sensor at the blue position saves more people and also detects earlier than placing a sensor at the green position, though costs more.*
## Problem Setup
The outbreak detection problem is defined as below:
- Given: a graph $$G(V,E)$$ and data on how outbreaks spread over this $$G$$ (for each outbreak $$i$$, we knew the time $$T(u,i)$$ when the outbreak $$i$$ contaminates node $$u$$).
- Goal: select a subset of nodes $$S$$ that maximize the expected reward:
$$
\max_{S\subseteq U}f(S)=\sum_{i}p(i)\cdot f_{i}(S)
$$
$$
\text{subject to cost }c(S)\leq B
$$
where
- $$p(i)$$: probability of outbreak $$i$$ occurring
- $$f_{i}(S)$$: rewarding for detecting outbreak $$i$$ using "sensors" $$S$${% include sidenote.html id='note-outbreak-detection-problem-setup' note='It is obvious that $$p(i)\cdot f_{i}(S)$$ is the expected reward for detecting the outbreak $$i$$' %}
- $$B$$: total budget of placing "sensors"
**The Reward** can be one of the following three:
- Minimize the time to detection
- Maximize the number of detected propagations
- Minimize the number of infected people
**The Cost** is context-dependent. Examples are:
- Reading big blogs is more time consuming
- Placing a sensor in a remote location is more expensive
## Outbreak Detection Formalization
### Objective Function for Sensor Placements
Define the **penalty $$\pi_{i}(t)$$** for detecting outbreak $$i$$ at time $$t$$, which can be one of the following:{% include sidenote.html id='note-outbreak-detection-penalty-note' note='Notice: in all the three cases detecting sooner does not hurt! Formally, this means, for all three cases, $$\pi_{i}(t)$$ is monotonically nondecreasing in $$t$$.'%}
- **Time to Detection (DT)**
- How long does it take to detect an outbreak?
- Penalty for detecting at time $$t$$: $$\pi_{i}(t)=t$$
- **Detection Likelihood (DL)**
- How many outbreaks do we detect?
- Penalty for detecting at time $$t$$: $$\pi_{i}(0)=0$$, $$\pi_{i}(\infty)=1$${% include sidenote.html id='note-penalty-dl' note='this is a binary outcome: $$\pi_{i}(0)=0$$ means we detect the outbreak and we pay 0 penalty, while $$\pi_{i}(\infty)=1$$ means we fail to detect the outbreak and we pay 1 penalty. That is we do not incur any penalty if we detect the outbreak in finite time, otherwise we incur penalty 1.'%}
- **Population Affected (PA)**
- How many people/nodes get infected during an outbreak?
- Penalty for detecting at time $$t$$: $$\pi_{i}(t)=$$ number of infected nodes in the outbreak $$i$$ by time $$t$$
The objective **reward function $$f_{i}(S)$$ of a sensor placement $$S$$** is defined as penalty reduction:
$$
f_{i}(S)=\pi_{i}(\infty)-\pi_{i}(T(S,i))
$$
where $$T(S,i)$$ is the time when the set of "sensors" $$S$$ detects the outbreak $$i$$.
### Claim 1: $$f(S)=\sum_{i}p(i)\cdot f_{i}(S)$$ is monotone{% include sidenote.html id='note-monotone' note='For the definition of monotone, see [Influence Maximization](influence-maximization)' %}
Firstly, we do not reduce the penalty, if we do not place any sensors. Therefore, $$f_{i}(\emptyset)=0$$ and $$f(\emptyset)=\sum_{i}p(i)\cdot f_{i}(\emptyset)=0$$.
Secondly, for all $$A\subseteq B\subseteq V$$ ($$V$$ is all the nodes in $$G$$), $$T(A,i)\geq T(B,i)$$, and
$$
\begin{align*}
f_{i}(A)-f_{i}(B)&=\pi_{i}(\infty)-\pi_{i}(T(A,i))-[\pi_{i}(\infty)-\pi_{i}(T(B,i))]\\
&=\pi_{i}(T(B,i))-\pi_{i}(T(A,i))
\end{align*}
$$
Because $$\pi_{i}(t)$$ is monotonically nondecreasing in $$t$$ (see sidenote 2), $$f_{i}(A)-f_{i}(B)<0$$. Therefore, $$f_{i}(S)$$ is nondecreasing. It is obvious that $$f(S)=\sum_{i}p(i)\cdot f_{i}(S)$$ is also nondecreasing, since $$p(i)\geq 0$$.
Hence, $$f(S)=\sum_{i}p(i)\cdot f_{i}(S)$$ is monotone.
### Claim 2: $$f(S)=\sum_{i}p(i)\cdot f_{i}(S)$$ is submodular{% include sidenote.html id='note-submodular' note='For the definition of submodular, see [Influence Maximization](influence-maximization)' %}
This is to proof for all $$A\subseteq B\subseteq V$$ $$x\in V \setminus B$$:
$$
f(A\cup \{x\})-f(A)\geq f(B\cup\{x\})-f(B)
$$
There are three cases when sensor $$x$$ detects the outbreak $$i$$:
1. $$T(B,i)\leq T(A, i)<T(x,i)$$ ($$x$$ detects late): nobody benefits. That is $$f_{i}(A\cup\{x\})=f_{i}(A)$$ and $$f_{i}(B\cup\{x\})=f_{i}(B)$$. Therefore, $$f(A\cup \{x\})-f(A)=0= f(B\cup\{x\})-f(B)$$
2. $$T(B, i)\leq T(x, i)<T(A,i)$$ ($$x$$ detects after $$B$$ but before $$A$$): $$x$$ only helps to improve the solution of $$A$$ but not $$B$$. Therefore, $$f(A\cup \{x\})-f(A)\geq 0 = f(B\cup\{x\})-f(B)$$
3. $$T(x, i)<T(B,i)\leq T(A,i)$$ ($$x$$ detects early): $$f(A\cup \{x\})-f(A)=[\pi_{i}(\infty)-\pi_{i}(T(x,t))]-f_{i}(A)$$$$ \geq [\pi_{i}(\infty)-\pi_{i}(T(x,t))]-f_{i}(B) = f(B\cup\{x\})-f(B)$${% include sidenote.html id='note-submodularity-proof1' note='Inequality is due to the nondecreasingness of $$f_{i}(\cdot)$$, i.e. $$f_{i}(A)\leq f_{i}(B)$$ (see Claim 1).'%}
Therefore, $$f_{i}(S)$$ is submodular. Because $$p(i)\geq 0$$, $$f(S)=\sum_{i}p(i)\cdot f_{i}(S)$$ is also submodular.{% include sidenote.html id='note-submodularity-proof1' note='Fact: a non-negative linear combination of submodular functions is a submodular function.'%}
We know that the Hill Climbing algorithm works for optimizing problems with nondecreasing submodular objectives. However, it does not work well in this problem:
- Hill Climbing only works for the cases that each sensor costs the same. For this problem, each sensor has cost $$c(s)$$.
- Hill Climbing is also slow: at each iteration, we need to re-evaluate marginal gains of all nodes. The run time is $$O(\mid V\mid\cdot k)$$ for placing $$k$$ sensors.
Hence, we need a new fast algorithm that can handle cost constraints.
## CELF: Algorithm for Optimziating Submodular Functions Under Cost Constraints
### Bad Algorithm 1: Hill Climbing that ignores the cost
**Algorithm**
- Ignore sensor cost $$c(s)$$
- Repeatedly select sensor with highest marginal gain
- Do this until the budget is exhausted
**This can fail arbitrarily bad!** Example:
- Given $$n$$ sensors and a budget $$B$$
- $$s_{1}$$: reward $$r$$, cost $$B$$
- $$s_{2}$$,..., $$s_{n}$$: reward $$r-\epsilon$$, cost $$\epsilon$$ ($$\epsilon$$ is an arbitrary positive small number)
- Hill Climbing always prefers $$s_{1}$$ to other cheaper sensors, resulting in an arbitrarily bad solution with reward $$r$$ instead of the optimal solution with reward $$\frac{B(r-\epsilon)}{\epsilon}$$, when $$\epsilon \rightarrow 0$$.
### Bad Algorithm 2: optimization using benefit-cost ratio
**Algorithm**
- Greedily pick the sensor $$s_{i}$$ that maximizes the benefit to cost ratio until the budget runs out, i.e. always pick
$$
s_{i}=\arg\max_{s\in(V\setminus A_{i-1})}\frac{f(A_{i-1}\cup\{s\})-f(A_{i-1})}{c(s)}
$$
**This can fail arbitrarily bad!** Example:
- Given 2 sensors and a budget $$B$$
- $$s_{1}$$: reward $$2\epsilon$$, cost $$\epsilon$$
- $$s_{2}$$: reward $$B$$, cost $$B$$
- Then the benefit ratios for the first selection are: 2 and 1, respectively
- This algorithm will pick $$s_{1}$$ and then cannot afford $$s_{2}$$, resulting in an arbitrarily bad solution with reward $$2\epsilon$$ instead of the optimal solution $$B$$, when $$\epsilon \rightarrow 0$$.
### Solution: CELF (Cost-Effective Lazy Forward-selection)
**CELF** is a two-pass greedy algorithm [[Leskovec et al. 2007]](https://www.cs.cmu.edu/~jure/pubs/detect-kdd07.pdf):
- Get solution $$S'$$ using unit-cost greedy (Bad Algorithm 1)
- Get solution $$S''$$ using benefit-cost greedy (Bad Algorithm 2)
- Final solution $$S=\arg\max[f(S'), f(S'')]$$
**Approximation Guarantee**
- CELF achieves $$\frac{1}{2}(1-\frac{1}{e})$$ factor approximation.
CELF also uses a lazy evaluation of $$f(S)$$ (see below) to speedup Hill Climbing.
## Lazy Hill Climbing: Speedup Hill Climbing
### Intuition
- In Hill Climbing, in round $$i+1$$, we have picked $$S_{i}=\{S_{1},...,S_{i}\}$$ sensors. Now, pick $$s_{i+1}=\arg\max_{u}f(S_{i}\cup \{u\})-f(S_{i})$$
- By submodularity $$f(S_{i}\cup\{u\})-f(S_{i})\geq f(S_{j}\cup\{u\})-f(S_{j})$$ for $$i<j$$.
- Let $$\delta_{i}(u)=f(S_{i}\cup\{u\})-f(S_{i})$$ and $$\delta_{j}(u)=f(S_{j}\cup\{u\})-f(S_{j})$$ be the marginal gains. Then, we can use $$\delta_{i}$$ as upper bound on $$\delta_{j}$$ for ($$j>i$$)
### Lazy Hill Climbing Algorithm:
- Keep an ordered list of marginal benefits $$\delta_{i-1}$$ from previous iteration
- Re-evaluate $$\delta_{i}$$ only for the top nodes
- Reorder and prune from the top nodes
The following figure show the process.
![lazy_evaluation](../assets/img/outbreak_detection_lazy_evaluation.png?style=centerme)
*(A) Evaluate and pick the node with the largest marginal gain $$\delta$$. (B) reorder the marginal gain for each sensor in decreasing order. (C) Re-evaluate the $$\delta$$s in order and pick the possible best one by using previous $$\delta$$s as upper bounds. (D) Reorder and repeat.*
Note: the worst case of Lazy Hill Climbing has the same time complexity as normal Hill Climbing. However, it is on average much faster in practice.
## Data-Dependent Bound on the Solution Quality
### Introduction
- Value of the bound depends on the input data
- On "easy data", Hill Climbing may do better than the $$(1-\frac{1}{e})$$ bound for submodular functions
### Data-Dependent Bound
Suppose $$S$$ is some solution to $$f(S)$$ subjected to $$\mid S \mid\leq k$$, and $$f(S)$$ is monotone and submodular.
- Let $$OPT={t_{i},...,t_{k}}$$ be the optimal solution
- For each $$u$$ let $$\delta(u)=f(S\cup\{u\})-f(S)$$
- Order $$\delta(u)$$ so that $$\delta(1)\geq \delta(2)\geq...$$
- Then, the **data-dependent bound** is $$f(OPT)\leq f(S)+\sum^{k}_{i=1}\delta(i)$$
Proof:{% include sidenote.html id='note-data-dependent-bound-proof' note='For the first inequality, see [the lemma in 3.4.1 of this handout](http://web.stanford.edu/class/cs224w/handouts/CS224W_Influence_Maximization_Handout.pdf). For the last inequality in the proof: instead of taking $$t_{i}\in OPT$$ of benefit $$\delta(t_{i})$$, we take the best possible element $$\delta(i)$$, because we do not know $$t_{i}$$.'%}
$$
\begin{align*}
f(OPT)&\leq f(OPT\cup S)\\
&=f(S)+f(OPT\cup S)-f(S)\\
&\leq f(S)+\sum^{k}_{i=1}[f(S\cup\{t_{i}\})-f(S)]\\
&=f(S)+\sum^{k}_{i=1}\delta(t_{i})\\
&\leq f(S)+\sum^{k}_{i=1}\delta(i)
\end{align*}
$$
Note:
- This bound hold for the solution $$S$$ (subjected to $$\mid S \mid\leq k$$) of any algorithm having the objective function $$f(S)$$ monotone and submodular.
- The bound is data-dependent, and for some inputs it can be very "loose" (worse than $$(1-\frac{1}{e})$$)

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