A reduction is a way of transforming/converting one problem into another problem.

Suppose you have a problem A which you do not know how to solve. However, you know an algorithm to solve problem B. If you can "transform" an instance α of problem A into an instance β of problem B, you can use the known algorithm for B to solve the "transformed" instance β, and obtain the solution for α from the solution β, by "reversing the transformation". We then say that A reduces to B.

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Through reductions, we have another tool to solve problems.

Furthermore, with an efficient, polynomial-time "transformation", reductions provide the ability to compare the "hardness" of two problems A and B.

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Given two decision problems A and B, a polynomial-time reduction from A to B,
denoted A ≤p B, is a transformation from instance α of A to instance β of B such that:

1. α is a YES-instance for A if and only if β is a YES-instance for B.
(A YES-instance is an instance of the decision problem whose answer is YES/True).
2. The transformation takes polynomial time in the size of α.

With a polynomial-time reduction, we will be able to claim:
If B is "easily solvable", then so is A.
If A is "hard", then so is B.

However, if B is "hard", it does not imply that A is "hard".
Likewise, if A is "easy", it does not mean that B is "easy".

Pro-tip 2: We designed this visualization and this e-Lecture mode to look good on 1366x768 resolution or larger (typical modern laptop resolution in 2021). We recommend using Google Chrome to access VisuAlgo. Go to full screen mode (F11) to enjoy this setup. However, you can use zoom-in (Ctrl +) or zoom-out (Ctrl -) to calibrate this.

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In this (static) visualization page (no animation), we want to show the (well-known) network of reductions from different NP-complete decision problems to another. In theory, since all these problems are NP-complete, they can all be reduced to one another, but the transformation can become very convoluted for some problems.

This network of reductions shown in this page is a combination from "Introductions to Algorithms (CLRS, 4th edition)", Karp's 21 NP-complete problems, "Computers and Intractability (Garey and Johnson, 79), etc.

The content of this visualization will be added over time as more and more NP-complete problems are discussed in NUS algorithm classes.

Pro-tip 3: Other than using the typical media UI at the bottom of the page, you can also control the animation playback using keyboard shortcuts (in Exploration Mode): Spacebar to play/pause/replay the animation, / to step the animation backwards/forwards, respectively, and -/+ to decrease/increase the animation speed, respectively.

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A vertex contains the (abbreviated) name of an NP-complete decision problem. Hover over it to see its full name. You can also click on the vertex to open the relevant slide that formally describes the problem.

A directed edge between vertices opens the slide which contains the proof of reduction from one problem to the other, in the direction of the arrow, as found in many computational complexity resources (books/Internet/etc). The reduction proof is generally if and only if (iff). Green arrow indicates that we have digitize the reduction proof. Black arrow indicates that we have not done so. We hope to add as much content to this page gradually over time.

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The CIRCUIT-SATISFIABILITY problem (C-SAT) is the decision problem of determining whether a given Boolean circuit - essentially a Directed Acyclic Graph - (using AND ∧, OR ∨, NOT ¬ gates) has an assignment of its n binary inputs that makes the output True (1). In other words, it asks whether the n inputs to a given Boolean circuit can be consistently set to True (1) or False (0) such that the circuit outputs True (1). If that is the case, the circuit is called satisfiable. Otherwise, the circuit is called unsatisfiable.

YES-instance (Sample certificate: False, True, True, False)

Important: L ≤p C-SAT for every language L ∈ NP.

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Definitions: Literal is a Boolean variable or its negation, e.g., xi, i.
Clause: A disjunction (OR ∨) of literals, e.g, Cj = x1 ∨ x̄2 ∨ x3.

The CONJUNCTIVE-NORMAL-FORM-SATISFIABILITY (CNF-SAT), or sometimes just called as SATISFIABILITY (SAT) problem, is the decision problem of determining whether a given formula Φ that is conjunction (AND ∧) of clauses, e.g., Φ = C1 ∧ C2 ∧ C3 ∧ C4, has a satisfying truth assignment.

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3-CNF-SAT(isfiability), usually just called as 3-SAT problem, is a CNF-SAT where each clause contains exactly 3 literals corresponding to different variables.

Example: Φ = (x̄1 ∨ x2 ∨ x3) ∧ (x1 ∨ x̄2 ∨ x3) ∧ (x̄1 ∨ x2 ∨ x4) is a YES-instance.
(Sample certificate: x1 = x2 = x4 = True, x3 = False).

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A clique is a set of vertices C of graph G = (V, E) such that ∃ an edge between every pair of distinct vertices in the set (u, v ∈ C, ∃ (u, v)). Set C is a complete subgraph of G. Then the size of the clique is the number of vertices in the set C.

The CLIQUE decision problem asks the following question on a graph G = (V, E):
Does there exist a complete subgraph in G of at least size k?

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Given a graph G = (V, E), a subset C ⊆ V is said to be a vertex cover if for each edge (u, v) ∈ E, either u ∈ C and/or v ∈ C.

Given a graph G = (V, E), the VERTEX-COVER (VC) problem asks:
Does there exist a vertex cover of size ≤ k?

Note that we have built a specialized mvc visualization page that discusses this problem and various strategies to attack this problem in a much detailed manner.

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Given an undirected unweighted graph G = (V, E),
the HAMILTONIAN-CYCLE (HC, sometimes abbreviated as HAM-CYCLE) problem asks:
Does there exist a (simple) cycle passing through all vertices exactly once?

Left/YES-instance (Sample certificate: 0-1-3-4-2-0) ------- ------- ------- Right/NO-instance

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Given an undirected complete graph with non-negative weight on edges and b ∈ R+,
the TRAVELING-SALESPERSON(/previously MAN)-PROBLEM (TSP) problem asks:
Does there exist a tour of cost at most b?

YES-instance for b = 108 but NO-instance for b = 107 as OPT = 108

Note that we have built a specialized tsp visualization page that discusses this problem and various strategies to attack this problem in a much detailed manner.

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Given a multiset S of N (usually non-negative) Integers: {A1, A2, ..., AN},
Is there exist a subset of S which sums to W?

Example: N = 5, S = {5, 1, 5, 1, 4}, and W = 7,
then it is a YES-instance with certificate indices {0, 1, 3} or values {5, 1, 1} that sums to 7.

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Given n items described by non-negative Integer pairs (w1, v1), (w2, v2), ..., (wn, vn), capacity W and threshold V, is there a subset of item with total weight at most W and total value at least V?

YES-instance: n = 5: {(12, 4), (1, 2), (4, 10), (1, 1), (2, 2)}, W = 15, and V = 15, certificate: take everything except item (12, 4) with total weight 8 and total value 15.

NO-instance: use the same instance as above but V = 16.

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Given a graph G = (V, E), a subset X ⊆ V is said to be an independent set if for each pair of vertices u, v ∈ X, then (u, v) ∉ E.

Given a graph G = (V, E), the INDEPENDENT-SET (IS) problem asks:
Does there exist an independent set of size ≥ k?

Note that we have built a specialized mvc visualization page that discusses this problem and various strategies to attack this problem in a much detailed manner (remember that a set of vertices is IS if and only if its complement is a VC).

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Given a set T of non-negative Integers, can we partition T into two sets of equal sum?

YES-instance: T = [18, 2, 8, 5, 7, 24], certificate: S1 = [18, 5, 7, 2] and S2 = [8, 24]
NO-instance: T = [1, 2]

PS: Obviously if the sum of T is odd, we cannot partition T into two sets with equal sum (immediately a NO instance). Without loss of generality, PARTITION problem is usually asked on instances when the sum of T is even.

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Dominating-Set 决策问题是，给定一个整数 k，是否存在一个在 G 中的支配集，其大小最多为 k

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Given a set of sets S = {S1, S2, ..., Sn}, a set H is said to be a hitting set of S
if for all Si, H ∩ Si ≠ ∅ (i.e., H has a non-empty intersection with all Si).

Given the set of sets S and a non-negative integer k, the HITTING-SET (HS) problem asks:
Is there exists a hitting set of S of size ≤ k?.

Example: S = {S1, S2, S3, S4}, S1 = {2, 3}, S2 = {6, 7}, S3 = {1, 4, 5, 7}, S4 = {3, 7, 8}, k = 2, then it is a YES-instance with certificate H = {3, 7}.

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C-SAT ≤p CNF-SAT reduction is shown in CLRS Chapter 34.

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CNF-SAT ≤p 3-SAT reduction is shown in CLRS Chapter 34.

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The key idea (major hint) is that 3-CNF-SAT has k clauses (disjunction of 3 literals) but IS is on a graph G = (V, E). How about creating k triangles (with 3 vertices = 3 literals each) and do something involving a variable and its negation.

Can you work out the details of the poly-time reduction and its proof yourself?
Try first before clicking next.

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For each clause, create 3 vertices in G, one for each literal.
We connect these 3 literals of a clause into a triangle sub-graph.
We also connect a literal to each of its negation (in any other clauses).
The IS criteria will ensure that we will only pick exactly one variable per triangle and we will not pick a variable in one clause and its negation in any other clause.

Take note that this reduction runs in poly-time.

Theorem: YES-instance for 3-SAT if and only YES-instance for IS.

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The key idea (major hint) is to that VC has a graph G with V vertices and E edges but HC has n sets of Integers. How about converting the edges into sets of size two?

Can you work out the details of the poly-time reduction and its proof yourself?
Try first before clicking next.

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Take an instance (G = (V, E), k) of VC.
Then for each edge e = (u, v), e ∈ E, we create a set Se = {u, v} of HC.
Thus the transformed instance of HC is: ({Se | e ∈ E}, k).

Take note that this reduction runs in poly-time, in O(E).

It is easy to see that a YES-instance for VC if and only YES-instance for HC.

Proof omitted: The vertices in C of VC correspond to elements in H of HS and vice versa.

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The key idea (major hint) is to build a weighted complete graph G' such that the HC in G turns out to be the least cost tour in G'.

Can you work out the details of the poly-time reduction and its proof yourself?
Try first before clicking next.

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Let G = (V, E) be an instance α of HC. We build an instance β of TSP as follows: create a complete graph G' on the same V vertices, but for each pair u, v ∈ V: if u, v ∈ E, then w(u, v) = 1, else w(u, v) = 2 (or anything greater than 1).

Take note that this reduction runs in poly-time, to be precise O(n2) as at most C(n, 2) edges are added from G to G'.

Theorem: G has a Hamiltonian cycle if and only if G' has a TSP tour of cost at most n.

Proof: Split into two parts:
• (if) G' has a TSP tour of cost at most n (YES-instance of TSP) → G has a Hamiltonian cycle (YES-instance of HC)
• (only if) G has a Hamiltonian cycle (YES-instance of HC) → G' has a TSP tour of cost at most n (YES-instance of TSP)

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The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

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The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

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The key idea (major hint) is to that PARTITION has n Integers only but KNAPSACK has n pair of Integers. How about duplicating the Integers? Also PARTITION has a specific target value (half of total of the n Integers) that can also be used as appropriate parameter for KNAPSACK.

Can you work out the details of the poly-time reduction and its proof yourself?
Try first before clicking next.

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Given a PARTITION instance α with T = [w1, w2, ..., wn] with total sum S = ∑i=[1..n] wi, construct a KNAPSACK instance β: {(w1, w1), (w2,w2), ..., (wn,wn)}
with capacity W = S/2 and threshold V = S/2.

Take note that this reduction runs in poly-time, to be precise O(n · log (n · wmax)) as it just copies n weights to n (weight, weight-as-value) pairs. If we assume n · wmax fits in standard 32/64-bit signed Integers, then log (n · wmax)) is just 32/64 respectively and this reduction runs in \$O(n).

Theorem: YES-instance for PARTITION if and only YES-instance for KNAPSACK.

Proof: Split into two parts:
• YES-instance for PARTITION → YES-instance for KNAPSACK.
• YES-instance for KNAPSACK → YES-instance for PARTITION.

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YES-instance for PARTITION → YES-instance for KNAPSACK.

Proof:
• We simply use one subset, e.g., subset S1 (S2 is also OK) in PARTITION for KNAPSACK.
• Subset S1 has total weight of S/2 and total value of S/2 (actually, S2 too).
• It is also a YES-instance for KNAPSACK.

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YES-instance for KNAPSACK → YES-instance for PARTITION.

Proof:
• YES-instance for KNAPSACK means that there is a subset Z with weight no more than S/2 with value at least S/2.
• Since weight equals value in the transformed instances from α to β, the only way that can happen is if the weight and value of subset Z are both exactly equal to S/2.
• So, the same subset Z (and T\\Z) can be used for as YES-instance for PARTITION.

You have reached the last slide. Return to 'Exploration Mode' to start exploring!

Note that if you notice any bug in this visualization or if you want to request for a new visualization feature, do not hesitate to drop an email to the project leader: Dr Steven Halim via his email address: stevenhalim at gmail dot com.

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#### 关于

VisuAlgo最初由副教授Steven Halim于2011年构思，旨在通过提供自学、互动式学习平台，帮助学生更深入地理解数据结构和算法。

VisuAlgo涵盖了Steven Halim博士与Felix Halim博士、Suhendry Effendy博士合著的书《竞技编程》中讨论的许多高级算法。即使过去十年，VisuAlgo仍然是可视化和动画化这些复杂算法的独家平台。

VisuAlgo仍然在不断发展中，正在开发更复杂的可视化。目前，该平台拥有24个可视化模块。

VisuAlgo配备了内置的问题生成器和答案验证器，其“在线测验系统”使学生能够测试他们对基本数据结构和算法的理解。问题根据特定规则随机生成，并且学生提交答案后会自动得到评分。随着越来越多的计算机科学教师在全球范围内采用这种在线测验系统，它可以有效地消除许多大学标准计算机科学考试中手工基本数据结构和算法问题。通过给通过在线测验的学生分配一个小但非零的权重，计算机科学教师可以显著提高学生对这些基本概念的掌握程度，因为他们可以在参加在线测验之前立即验证几乎无限数量的练习题。每个VisuAlgo可视化模块现在都包含自己的在线测验组件。

VisuAlgo已经被翻译成三种主要语言：英语、中文和印尼语。此外，我们还用各种语言撰写了关于VisuAlgo的公开笔记，包括印尼语、韩语、越南语和泰语：

id, kr, vn, th.

#### 团队

Associate Professor Steven Halim, School of Computing (SoC), National University of Singapore (NUS)
Dr Felix Halim, Senior Software Engineer, Google (Mountain View)

CDTL TEG 1: Jul 2011-Apr 2012: Koh Zi Chun, Victor Loh Bo Huai

Jul 2012-Dec 2013: Phan Thi Quynh Trang, Peter Phandi, Albert Millardo Tjindradinata, Nguyen Hoang Duy
Jun 2013-Apr 2014 Rose Marie Tan Zhao Yun, Ivan Reinaldo

CDTL TEG 2: May 2014-Jul 2014: Jonathan Irvin Gunawan, Nathan Azaria, Ian Leow Tze Wei, Nguyen Viet Dung, Nguyen Khac Tung, Steven Kester Yuwono, Cao Shengze, Mohan Jishnu

Jun 2014-Apr 2015: Erin Teo Yi Ling, Wang Zi
Jun 2016-Dec 2017: Truong Ngoc Khanh, John Kevin Tjahjadi, Gabriella Michelle, Muhammad Rais Fathin Mudzakir
Aug 2021-Apr 2023: Liu Guangyuan, Manas Vegi, Sha Long, Vuong Hoang Long, Ting Xiao, Lim Dewen Aloysius

Optiver: Aug 2023-Oct 2023: Bui Hong Duc, Oleh Naver, Tay Ngan Lin

Aug 2023-Apr 2024: Xiong Jingya, Radian Krisno, Ng Wee Han

List of translators who have contributed ≥ 100 translations can be found at statistics page.

NUS教学与学习发展中心（CDTL）授予拨款以启动这个项目。在2023/24学年，Optiver的慷慨捐赠将被用来进一步开发 VisuAlgo。

#### 使用条款

VisuAlgo并不是一个完成的项目。Steven Halim副教授仍在积极改进VisuAlgo。如果您在使用VisuAlgo时发现任何可视化页面/在线测验工具中的错误，或者您想要请求新功能，请联系Steven Halim副教授。他的联系方式是将他的名字连接起来，然后加上gmail dot com。