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.


Remarks: By default, we show e-Lecture Mode for first time (or non logged-in) visitor.
If you are an NUS student and a repeat visitor, please login.

🕑

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.


Pro-tip 1: Since you are not logged-in, you may be a first time visitor (or not an NUS student) who are not aware of the following keyboard shortcuts to navigate this e-Lecture mode: [PageDown]/[PageUp] to go to the next/previous slide, respectively, (and if the drop-down box is highlighted, you can also use [→ or ↓/← or ↑] to do the same),and [Esc] to toggle between this e-Lecture mode and exploration mode.

🕑

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.

🕑

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.

🕑

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.

🕑

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.

🕑

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.

🕑

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).

🕑

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?

🕑

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.

🕑

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

🕑

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.

🕑

Given a multiset S of N (usually non-negative) Integers: {A1, A2, ..., AN},
the SUBSET-SUM problem asks:
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.

🕑

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.

🕑

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).

🕑

This page is a stub to explain 3-D-MATCHING (3DM) problem.

🕑

This page is a stub to explain PARTITION-INTO-TRIANGLES (PIT) problem.

🕑

This page is a stub to explain FEEDBACK-EDGE-SET (FES) problem.

🕑

This page is a stub to explain SET-COVER (SC) problem.

🕑

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.

🕑

给定一个图 G = (V, E),支配集 D 是顶点的一个子集,对于所有不在 D 中的其他顶点 u,存在某个在 D 中的顶点 v,使得 (u, v) 是 G 中的一条边(即,u 与 v 相邻)。


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

🕑

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}.

🕑

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

🕑

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

🕑

This page is a stub to explain 3-SAT ≤p CLIQUE reduction.

🕑

This page is a stub to explain 3-SAT ≤p SUBSET-SUM reduction.

🕑

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.

🕑

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.


Proof: To be added.

🕑

To be added

🕑

This page is a stub to explain CLIQUE ≤p VC reduction.

🕑

This page is a stub to explain VC ≤p HC reduction.

🕑

To be added

🕑

To be added

🕑

To be added

🕑

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.

🕑

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.

🕑

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.

🕑

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)

🕑

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.

🕑

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.

🕑

To be added

🕑

To be added

🕑

To be added

🕑

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.

🕑

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.

🕑

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.

🕑

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.


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.

🕑
关于 团队 使用条款
隐私政策

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。

虽然VisuAlgo主要面向新加坡国立大学(NUS)的学生,包括各种数据结构和算法课程(例如CS1010/等价课程,CS2040/等价课程(包括IT5003),CS3230,CS3233和CS4234),但它也是全球好奇心的宝贵资源,促进在线学习。

最初,VisuAlgo并不适用于智能手机等小触摸屏,因为复杂的算法可视化需要大量的像素空间和点击拖动交互。为了获得最佳用户体验,建议使用最低分辨率为1366x768的屏幕。然而,自2022年4月以来,VisuAlgo的移动(精简)版本已经推出,使得在智能手机屏幕上使用VisuAlgo的部分功能成为可能。

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

项目领导和顾问(2011年7月至今)
Associate Professor Steven Halim, School of Computing (SoC), National University of Singapore (NUS)
Dr Felix Halim, Senior Software Engineer, Google (Mountain View)

本科生研究人员 1
CDTL TEG 1: Jul 2011-Apr 2012: Koh Zi Chun, Victor Loh Bo Huai

最后一年项目/ UROP学生 1
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

本科生研究人员 2
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

最后一年项目/ UROP学生 2
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

本科生研究人员 3
Optiver: Aug 2023-Oct 2023: Bui Hong Duc, Oleh Naver, Tay Ngan Lin

最后一年项目/ UROP学生 3
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.

致谢
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慷慨地向全球计算机科学界提供免费服务。如果您喜欢VisuAlgo,我们恳请您向其他计算机科学学生和教师宣传它的存在。您可以通过社交媒体平台(如Facebook、YouTube、Instagram、TikTok、Twitter等)、课程网页、博客评论、电子邮件等方式分享VisuAlgo。

数据结构与算法(DSA)的学生和教师可以直接在课堂上使用本网站。如果您从本网站截取屏幕截图或视频,可以在其他地方使用,但请引用本网站的URL(https://visualgo.net)和/或下面的出版物列表作为参考。但请不要下载VisuAlgo的客户端文件并将其托管在您的网站上,因为这构成了抄袭行为。目前,我们不允许他人分叉此项目或创建VisuAlgo的变体。个人使用离线副本的客户端VisuAlgo是可以接受的。

请注意,VisuAlgo的在线测验组件具有重要的服务器端元素,保存服务器端脚本和数据库并不容易。目前,普通公众只能通过“培训模式”访问在线测验系统。“测试模式”提供了一个更受控制的环境,用于在新加坡国立大学的真实考试中使用随机生成的问题和自动验证。


出版物列表

这项工作曾在2012年国际大学生程序设计竞赛(波兰,华沙)的CLI研讨会上和2012年国际信息学奥林匹克竞赛(意大利,锡尔米奥内-蒙蒂基亚里)的IOI会议上展示过。您可以点击此链接阅读我们2012年关于该系统的论文(当时还没有称为VisuAlgo),以及此链接阅读2015年的简短更新(将VisuAlgo与之前的项目关联起来)。


错误报告或新功能请求

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。

隐私政策

版本 1.2 (更新于2023年8月18日星期五)。

自2023年8月18日(星期五)起,我们不再使用 Google Analytics。因此,我们现在使用的所有 cookies 仅用于此网站的运营。即使是首次访问的用户,烦人的 cookie 同意弹窗现在也已关闭。

自2023年6月7日(星期五)起,由于 Optiver 的慷慨捐赠,全世界的任何人都可以自行创建一个 VisuAlgo 账户,以存储一些自定义设置(例如,布局模式,默认语言,播放速度等)。

此外,对于 NUS 学生,通过使用 VisuAlgo 账户(一个 NUS 官方电子邮件地址,课堂名册中的学生姓名,以及在服务器端加密的密码 - 不存储其他个人数据),您同意您的课程讲师跟踪您的电子讲义阅读和在线测验培训进度,这是顺利进行课程所必需的。您的 VisuAlgo 账户也将用于参加 NUS 官方的 VisuAlgo 在线测验,因此,将您的账户凭据传递给他人代您进行在线测验构成学术违规。课程结束后,您的用户账户将被清除,除非您选择保留您的账户(OPT-IN)。访问完整的 VisuAlgo 数据库(包含加密密码)的权限仅限于 Halim 教授本人。

对于全球其他已经给 Steven 写过信的 CS 讲师,需要一个 VisuAlgo 账户(您的(非 NUS)电子邮件地址,您可以使用任何显示名称,以及加密密码)来区分您的在线凭据与世界其他地方。您的账户将具有 CS 讲师特定的功能,即能够查看隐藏的幻灯片,这些幻灯片包含了在隐藏幻灯片之前的幻灯片中提出的问题的(有趣的)答案。您还可以访问 VisuAlgo 在线测验的 Hard 设置。您可以自由地使用这些材料来增强您的数据结构和算法课程。请注意,未来可能会有其他 CS 讲师特定的功能。

对于任何拥有 VisuAlgo 账户的人,如果您希望不再与 VisuAlgo 工具有关联,您可以自行删除您的账户。