Download disjoint sets - Electrical and Computer Engineering

ECE 250 Algorithms and Data Structures
Disjoint sets
Douglas Wilhelm Harder, M.Math. LEL
Department of Electrical and Computer Engineering
University of Waterloo
Waterloo, Ontario, Canada
ece.uwaterloo.ca
[email protected]
© 2006-2013 by Douglas Wilhelm Harder. Some rights reserved.
Disjoint sets
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Outline
In this topic, we will cover disjoint sets, including:
– A review of equivalence relations
– The definition of a Disjoint Set
– An efficient data structure
• A general tree
– An optimization which results in
• Worst case O(ln(n)) height
• Average case O(a(n))) height
• Best case Q(1) height
– A few examples and applications
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Definitions
Recall the properties of an equivalence relation:
– a ~ a for all a
– a ~ b if and only if b ~ a
– If a ~ b and b ~ c, it follows that a ~ c
An equivalence relation partitions a set into distinct equivalence
classes
Each equivalence class may be represented by a single object: the
representative object
– Another descriptive term for the sets in such a partition is disjoint sets
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Implicitly Defined Relations
For example, big-Q defines an equivalence class of functions which
grow at the same rate
– We choose a single function to represent the class
e.g., we represent all functions with quadratic growth with n2
Another example: partition the numbers from 1 to 20 according to
the relation a ~ b if a and b share the same factors of 2:
{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}, {2, 6, 10, 14, 18}, {4, 12, 20}, {8}, {16}
These equivalence relations are implicitly defined by the relation
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Explicitly Defined Disjoint Sets
Alternatively, a partition or collection of disjoint sets may be used to
explicitly define an equivalence relation:
– a ~ b if and only if a and b are in the same partition
For example, the 10 numerals
1, 2, 3, 4, 5, 6, 7, 8, 9, 0
can be partitioned into the three sets
{1, 2, 3, 5, 7}, {4, 6, 9, 0} , {8}
Therefore, 1 ~ 2, 2 ~ 3, etc.
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Explicitly Defined Disjoint Sets
Consider simulating a device and tracking the connected
components in a circuit
This forms an equivalence relation:
a ~ b if a and b are
connected
http://www.morphet.org.uk/ferro/s6mono.html
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Operations on Disjoint Sets
There are two operations we would like to perform on disjoint sets:
– Determine if two elements are in the same disjoint set, and
– Take the union of two disjoint sets creating a single set
We will determine if two objects are in the same disjoint set by
defining a function which finds the representative object of one of
the disjoint sets
– If the representative objects are the same, the objects are in the same
disjoint set
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Implementation
Given two elements a and b, they are in the same set if
find( a ) == find( b )
What find returns is irrelevant so long as:
– If a and b are in the same set, find( a ) == find( b )
– If a and b are not in the same set, find( a ) != find( b )
We will have find return an integer
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Implementation
Here is a poor implementation:
– Have two arrays and the second array stores the representative object
– Finding the representative object is Q(1)
– However, taking the union of two sets is Q(n)
• It would be necessary to check each array entry
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Implementation
As an alternate implementation, let each disjoint set be represented
by a general tree
– The root of the tree is the representative object
To take the union of two such sets, we will simply attach one tree to
the root of the other
Find and union are now both O(h)
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Implementation
Normally, a node points to its children:
We are only interested in the root; therefore, our interest is in storing
the parent
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Implementation
For simplicity, we will assume we are creating disjoint sets the n
integers
0, 1, 2, ..., n – 1
We will define an array
parent = new size_t[n];
for ( int i = 0; i < n; ++i ) {
parent[i] = i;
}
If parent[i] == i, then i is a root node
– Initially, each integer is in its own set
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Implementation
We will define the function
size_t Disjoint_set::find( size_t i ) const {
while( parent[i] != i ) {
i = parent[i];
Tfind(n) = O(h)
}
return i;
}
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Implementation
Initially, you will note that
find( i ) != find( j )
for i != j, and therefore, we begin with each integer being in its
own set
We must next look at the union operation
– how to join two disjoint sets into a single set
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Implementation
This function is also easy to define:
void set_union( size_t i, size_t j ) {
i = find( i );
j = find( j );
Tset_union(n) = 2Tfind(n) + Q(1)
= O(h)
if ( i != j ) {
// slightly sub-optimal...
parent[j] = i;
}
}
The keyword union is reserved in C++
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Example
Consider the following disjoint set on the ten decimal digits:
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Example
If we take the union of the sets containing 1 and 3
set_union(1, 3);
we perform a find on both entries and update the second
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Example
Now, find(1) and find(3) will both return the integer 1
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Example
Next, take the union of the sets containing 3 and 5,
set_union(3, 5);
we perform a find on both entries and update the second
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Example
Now, if we take the union of the sets containing 5 and 7
set_union(5, 7);
we update the value stored in find(7) with the value find(5):
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Example
Taking the union of the sets containing 6 and 8
set_union(6, 8);
we update the value stored in find(8) with the value find(6):
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Example
Taking the union of the sets containing 8 and 9
set_union(8, 9);
we update the value stored in find(8) with the value find(9):
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Example
Taking the union of the sets containing 4 and 8
set_union(4, 8);
we update the value stored in find(8) with the value find(4):
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Example
Finally, if we take the union of the sets containing 5 and 6
set_union(5, 6);
we update the entry of find(6) with the value of find(5):
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Optimizations
To optimize both find and set_union, we must minimize the height
of the tree
– Therefore, point the root of the shorter tree to the root of the taller tree
– The height of the taller will increase if and only if the trees are equal in
height
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Worst-Case Scenario
Let us consider creating the worst-case disjoint set
As we are always attaching the tree with less height to the root of
the tree with greater height, the worst case must occur when both
trees are equal in height
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Worst-Case Scenario
Thus, building on this, we take the union of two sets with one
element
– We will keep track of the number of nodes at each depth
1
1
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Worst-Case Scenario
Next, we take the union of two sets, that is, we join two worst-case
sets of height 1:
1
2
1
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Worst-Case Scenario
And continue, taking the union of two worst-case trees of height 2:
1
3
3
1
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Worst-Case Scenario
Taking the union of two worst-case trees of height 3:
1
4
6
4
1
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Worst-Case Scenario
And of four:
1
5
10
10
5
1
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Worst-Case Scenario
And finally, take the union of two worst-case trees of height 5:
– These are binomial trees
1
6
15
20
15
6
1
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Worst-Case Scenario
From the construction, it should be clear that this would define
Pascal’s triangle
– The binomial coefficients
1
m  0 or m  n

n 
    n  1   n  1 
0mn
 m  
 

m
m

1
 


n!

m ! n  m  !
1
1
1
1
1
1
1
5
4
3
2
1
6
15
10
6
3
1
20
10
4
1
15
5
1
6
1
1
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Worst-Case Scenario
Thus, suppose we have a worst-case tree of height h
– We need the number of nodes and the average depth of a node
– Using Maple
> sum( binomial( h, k ), k = 0..h );
> sum( k*binomial( h, k ), k = 0..h );
we get:
h
   2h  n
k
k 0  
h

h
k    h2h1
k
k 0  
h

– Therefore, the average depth is
h2 h1 h lg( n)
 
h
2
2
2
– The height and average depth of the worst case are O(ln(n))
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Best-Case Scenario
In the best case, all elements point to the same entry with a resulting
height of Q(1):
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Average-Case Scenario
What is the average case?
Could it be any better than O(ln(n))?
– is there something better?
To answer this, I created a program which, given the integers from 0
to 2n – 1 continued to randomly choose numbers until all entries
were in a single large set
– For each n, I did this multiple times and found the mean (average)
height
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Average-Case Scenario
The resulting graph shows the average height of a randomly
generated disjoint set data structure with 2n elements
215 =32768
This suggests that the average height of such a tree is o(ln(n))
– See reference [1] for a detailed analysis
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Average-Case Scenario
The actual asymptotic behaviour is O(a(n)) where a(n) is the inverse
of the function A(n, n) where A(m, n) is the Ackermann function:
n 1
if m  0


A( m, n)  
A( m  1,1)
if m  0 and n  0
A( m  1, A( m, n  1)) if m  0 and n  0

The first values are:
A(0, 0) = 1, A(1, 1) =3, A(2, 2) = 7, A(3, 3) = 61
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Average-Case Scenario
However, A(4, 4) = 2A(3, 4) – 3 where A(3,4) is the
19729-decimal-digit number
A(3, 4) = 2003529930406846464979
072351
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872692299860827905317169187657886090894181705799340489021844155979109267686279659758395248392673488363474565168701616624064242424122896111801061568234253939218005248345472377921991122859591419187749179382334001007812832650671028178139602912091472010094787875255126337288422235386949006792766451163475810119387531965724212147603828477477457170457861041738574791130190858387789015233434301300528279703858035981518292960030568261209195094373732545417105638388704752895056396102984364136093564163258940813798151169333861979733982167076100460798009601602482309694304380695662012321365014054958625061528258803302
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065587754782025245478065730134227747061674424196895261316427410469547462148375628829977180418678508454696561915090869587425118443583730659095146098045124740941137389992782249298336779601101538709612974970556630163730720275073475992294379239382442742118615823616131788639255309511718842129850830723825972914414225157940388301135908333165185823496722125962181250705811375949552502274727467436988713192667076929919908446716122873885845758462272657333075373557282395161696417519867501268174542932373829414382481437713986190671665757294580780482055951188168718807521297183263644215533678775127476694079011705750
981957508456356521738954417987507452385445520013357203333237989507439390531291821225525983379090946363020218535384885482506289771561696386071238277172562131346054940177041358173193176337013633225281912754719144345092071184883836681817426334294961187009150304916533946476371776643912079834749462739782217150209067019030246976215127852195614207080646163137323651785397629209202550028896201297014137964003805573494926907353514596120867479654773369295877362863566014376796403843079686413856344780132826128458918489852804804884418082163942397401436290348166545811445436646003249061876303950235640204453074821024
136689519664422133920075747912868380517515063466256939193774028351207566626082989049187728783385217852279204577184696585527879044756219266399200840930207567392536373562839082981757790215320210640961737328359849406665214119818381088451545977289516457213189779790749194101314836854463961690460703010759681893374121757598816512700076126278916951040631585763753478742007022205107089125761236165802680681585849985263146587808661680073326467683020639169720306489440562819540619068524200305346315662189132730906968735318164109451428803660599522024824888671155442910472192913424834643870536850864874909917881267056
566538719104972182004237149274016446094345984539253670613221061653308566202118896823400575267548610147699368873820958455221157192347968688816085363161586288015039594941852948922707441082820716930338781808493620401825522227101098565344481720747075601924591559943107294957819787859057894005254012286751714251118435643718405356302418122547326609330271039796809106493927272268303541046763259135527968383770501985523462122285841055711992173171796980433931770775075562705604783177984444763756025463703336924711422081551997369137197516324130274871219986340454824852457011855334267526471597831073124566342980522145
549415625272402891533335434934121786203700726031527987077187249123449447714790952073476138542548531155277330103034247683586549609372232400715451812973269208105842409055772564580368146223449318970813889714329983134761779967971245378231070373915147387869211918756670031932128189680332269659445928621060743882741691946516226763254066507088107103039417886056489376981673415902592519461182364294565266937220315550470021359884629275801252771542201662995486313032491231102962792372389976641680349714122652793190763632613681414551637665655983978848938173308266877990196288693229659737995193162118721545528739417024
366988559388879331674453336311954151840408828381519342123412282003095031334105070476015998798547252919066522247931971544033179483683737322082188577334162385644138070054191353024594391350255453188645479625226025176292837433046510236105758351455073944333961021622967546141578112719700173861149427950141125328062125477581051297208846526315809480663368767014731073354071771087661593585681409821296773075919738297344144525668877085532457088895832099382343210271822411476373279135756861542125284965790333509315277692550584564401055219264450531207375628774499816364633283581614033017581396735942732769044892036188
038675495575180689005853292720149392350052584514670698262854825788326739873522045722823929020714482221988558710289699193587307427781515975762076402395124386020203259659625021257834995771008562638611823381331850901468657706401067627861758377277289589274603940393033727187385053691295712671506689668849388088514294360996201296675907922508227531381284985152690293170026313632894209579757795932763553116206675348865131732387243874806351331451264488996758982881292548007642518658649024111112730135719718138160258317850693224400799865663537154408845486639318170839573578079905973083909488180406093595919090747396
090441015051632174968141210076571917748376735575100073361692238653742907945780320004233745280756615304292901449578062963413838355178359976470885134900485697369796523869584599459559209070905895689145114141268450546211794502661175016692826025095077077821195043261738322356243760177679936279609936897519139496503335850715541843645685261667424368892037103749532842592713161053783498074073915863381796765842525803673720646935124865223848134166380806150570482905989069645193644001859712042572300731641000991698752426037736217776343062161674488493081092990100951797454156425120482208671458684925513244426677712786
372821133153622430109182439124338021404624222334915355951689081628848798998827363044537243217428021575577796702166631704796972817248339284101564227450727177926939992974030807277039501358154514249404902653610582540937311465310494338248437971860693721444460082679800247122948940576185389220342560830269705287662137737359439422411470707407290272546130735854174569141944648762435768239706570318416846754073346634629367398362000404140071405427763248013274220268539369886978760700959004868465062677136307097982100655728510130660101078063374334477307347865388174268123074376606664331277535646657860371519292276844
045827328324380821284121877613204246046490080105473142674926082692215563740548624171703102791999694264562095561981645454766204502241144940474934983220680719135276798674781345820385957041346617793722853494003163159954409368408957253343870298671782977037333280680176463950209002394193149911500910527682111951099906316615031158558283558260717941005252858361136996130344279017381178741206128818206202326384986151565645123004779296756361834576810504334176954306753804111392855379252924134733948105053202570872818630729115891133594201476187266429156403637192760230628384065042544174233546454998705531872688792642
410214736369862546374715974435494344389973005174252511087735788639094681209667342815258591992485764048805507132981429935991146323991911395992675257635900744657281019180584180734222773472139772321823177171691640010882611254909336118678057572239101818616854910850088527227437421208652485237245624869766224538481929867112945294551549703058591930719849710541418163696897613112674402700964866754593456705993699546450055892162804797636568613331656390739570327203438917541526750091501119885687270884819553167693168127289214303137681801644547736751835349785792427646335416243360112596025210950161226411034608346564
823559793427405686884922445874549377675212032470380303549115754483129527589193989368087632768543876955769488142284431199859570072752139317683783177033913042306095899913731468456901042209516196707050642025673387344611565527617599272715187766001023894476053978951694570880272873622512107622409181006670088347473760515628553394356584375627124124445765166306408593950794755092046393224520253546363444479175566172596218719927918657549085785295001284022903506151493731010700944615101161371242376142672254173205595920278212932572594714641722497732131638184532655527960427054187149623658525245864893325414506264233
7885651464670604298564781968461593663288954299780722542264790400616019751975007460545150060291806638271497016110987951336633771378434416194053121445291855180136575558667615019373029691932076120009255065081583275508499340768797252369987023567931026804136745718956641431852679054717169962990363015545645090044802789055701968328313630718997699153166679208958768572290600915472919636381673596673959975710326015571920237348580521128117458610065152598883843114511894880552129145775699146577530041384717124577965048175856395072895337539755822087777506072339445587895905719156733
Thus, A(4, 4) – 3, in binary, is 1 followed by this many zeros....
http://xkcd.com/207/
Disjoint sets
40
Average-Case Scenario
Therefore, we (as engineers) can, in clear conscience, state that the
average run-time is Q(1) as there are no physical circumstances
where the average depth could by anything more than 4
Disjoint sets
41
Optimizations
Another optimization is that, whenever find is called, update the
object to point to the root
size_t Disjoint_set::find( size_t n ) {
if ( parent[n] == n ) {
return n;
} else {
parent[n] = find( parent[n] );
return parent[n];
}
}
The next call to find(n) is Q(1); the cost is O(h) memory
Disjoint sets
42
Lotto 6/49 problem
How do you randomly select 6 numbers from 1-49 without ever
selecting the same number twice?
int array[49] = {1, 2, 3, 4, ..., 49};
for ( int i = 0;
size_t index
std::cout <<
array[index]
}
i < 6; ++i ) {
= mrand48() % (49 - i);
array[index];
= array[49 - i - 1];
Disjoint sets
43
Lotto 6/49 problem
Let’s play Lotto 2/20: Pick a random number from 0 to 19, say 7:
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Move the object in the last entry to its
1
2
3
4
5
6
7
20
9
10 11 12 13 14 15 16 17 18 19 20
Now pick a random number from 0 to 18, and so on
Disjoint sets
44
Lotto 6/49 problem
This is an easy way to randomly permute a list:
– The randomly chosen entry is moved to the last position
int array[N];
for ( size_t i = 0; i < N; ++i ) {
array[i] = i;
}
for ( size_t rng = N; rng >= 2; --rng ) {
// pick an entry from 0 to rng - 1
size_t index = mrand48() % rng;
// swap it and the last entry--it's okay if index == rng - 1
std::swap( array[index], array[rng - 1] );
}
Disjoint sets
45
Implementation Summary
We now have two exceptionally fast operations: both find and union
will run in Q(1) time on average, and O(ln(n)) in the worst-case
scenario
Disjoint sets
46
Application: Image Processing
One common application is in image processing
Suppose you are attempting to recognize similar features within an
image
Within a photograph, the same object may be separated by an
obstruction; e.g., a road may be split by
– a telephone pole in an image
– an overpass on an aerial photograph
Disjoint sets
47
Application: Image Processing
Consider the following image of the author climbing up the Niagara
Escarpment at Rattlesnake Point
Suppose we have a program
which recognizes skin tones
Disjoint sets
48
Application: Image Processing
A first algorithm may make an initial pass and recognize five
different regions which are recognized as exposed skin
– the left arm and hand are separated by a watch
Each region would be represented by a
separate disjoint set
Disjoint sets
49
Application: Image Processing
Next, a second algorithm may take sets which are close in proximity
and attempt to determine if they are from the same person
In this case, the algorithm takes the union of:
– the red and yellow regions, and
– the dark and light blue regions
Disjoint sets
50
Application: Image Processing
Finally, a third algorithm may take more distant sets and, depending
on skin tone and other properties, may determine that they come
from the same individual
In this example, the third pass may, if
successful, take the union of the red, blue,
and green regions
Disjoint sets
51
Application: Maze Generation
Another fun application is in the generation of mazes
Impress your (non-engineering) friends
– They’ll never guess how easy this is...
Disjoint sets
52
Application: Maze Generation
Here we have a maze which spans
a 500 × 500 grid of
squares where:
– There is one unique solution
– Each point can be reached by
one unique path from the start
Ref: Lance Hampton http://littlebadwolf.com/mazes/
Disjoint sets
53
Application: Maze Generation
Zooming in on the maze, you will note that it is rather complex
and seemingly
random
Ref: Lance Hampton http://littlebadwolf.com/mazes/
Disjoint sets
54
Application: Maze Generation
Finding the solution is a problem for a different lecture
– Backtracking algorithms
We will look at creating the
maze using disjoint sets
Ref: Lance Hampton http://littlebadwolf.com/mazes/
Disjoint sets
55
Application: Maze Generation
What we will do is the following:
– Start with the entire grid subdivided into squares
– Represent each square as a separate disjoint set
– Repeat the following algorithm:
• Randomly choose a wall
• If that wall connects two disjoint set of cells, then remove the wall and union
the two sets
– To ensure that you do not randomly remove the same wall twice, we
can have an array of unchecked walls
Disjoint sets
56
Application: Maze Generation
Let us begin with an entrance, an exit, and a disjoint set of 20
squares and 31 interior walls
Disjoint sets
57
Application: Maze Generation
First, we select 6 which joins cells B and G
– Both have height 0
Disjoint sets
58
Application: Maze Generation
Next we select wall 18 which joins regions J and O
Disjoint sets
59
Application: Maze Generation
Next we select wall 9 which joins the disjoint sets E and J
– The disjoint set containing E has height 0, and therefore it is attached to J
Disjoint sets
60
Application: Maze Generation
Next we select wall 11 which joins the sets identified by B and H
– H has height 0 and therefore we attach it to B
Disjoint sets
61
Application: Maze Generation
Next we select wall 20 which joins disjoint sets L and M
– Both are height 0
Disjoint sets
62
Application: Maze Generation
Next we select wall 17 which joins disjoint sets I and N
– Both are height 0
Disjoint sets
63
Application: Maze Generation
Next we select wall 7 which joins the disjoint set C and the disjoint
set identified by B
– C has height 0 and thus we attach it to B
Disjoint sets
64
Application: Maze Generation
Next we select wall 19 which joins the disjoint set K to the disjoint
sent identified by L
– Because K has height 0, we attach it to L
Disjoint sets
65
Application: Maze Generation
Next we select wall 23 and join the disjoint set Q with the set
identified by L
– Again, Q has height 0 so we attach it to L
Disjoint sets
66
Application: Maze Generation
Next we select wall 12 which joints the disjoint sets identified by B
and I
– They both have the same height, but B has more nodes, so we add I to
the node B
Disjoint sets
67
Application: Maze Generation
Selecting wall 15 joints the sets identified by B and L
– The tree B has height 2 while L has height 1 and therefore we attach L
to B
Disjoint sets
68
Application: Maze Generation
Next we select wall 5 which joins disjoint sets A and F
– Both are height 0
Disjoint sets
69
Application: Maze Generation
Selecting wall 30 also joins two disjoint sets R and S
Disjoint sets
70
Application: Maze Generation
Selecting wall 4 joints the disjoint set D and the disjoint set identified
by J
– D has height 0, J has height 1, and thus we add D to J
Disjoint sets
71
Application: Maze Generation
Next we select wall 10 which joins the sets identified by A and B
– A has height 1 while B has height 2, so we attach A to B
Disjoint sets
72
Application: Maze Generation
Selecting wall 31, we union the sets identified by R and T
– T has height 0 so we attach it to I
Disjoint sets
73
Application: Maze Generation
Selecting wall 27 joins the disjoint sets identified by J and R
– They both have height 1, but J has more elements, so we add R to J
Disjoint sets
74
Application: Maze Generation
Selecting wall 8 joins sets identified by B and J
– They both have height 2 so we note that J has fewer nodes than B, so
we add J to B
Disjoint sets
75
Application: Maze Generation
Finally we select wall 23 which joins the disjoint set P and the
disjoint set identified by B
– P has height 0, so we attach it to B
Disjoint sets
76
Application: Maze Generation
Thus we have a (rather trivial) maze where:
– there is one unique solution, and
– you can reach any square by a unique path from the starting point
Disjoint sets
77
Application: Maze Generation
You may also note that the average depth is 1.6 whereas the
average depth of the worst-case disjoint tree is 2:
Disjoint sets
78
Application: Maze Generation
For fun, the following C code generates mazes of arbitrary length:
char M[2],A,Z,E=40,J[40],T[40];main(C){for(*J=A=scanf("%d",&C);
-E;
J[
E]
=T
[E
]= E)
printf("._"); for(;(A-=Z=!Z) || (printf("\n|"
)
,
A
=
39
,C
-)
;
Z
||
printf
(M
))M[Z]=Z[A-(E
=A[J-Z])&&!C
&
A
==
T[
A]
|6<<27<rand()||!C&!Z?J[T[E]=T[A]]=E,J[T[A]=A-Z]=A,"_.":" |"];}
It does not use disjoint sets...
Disjoint sets
79
Application: Maze Generation
$ gcc maze.c
maze.c: In function ‘main’:
maze.c:1: warning: incompatible implicit declaration of built-in function ‘scanf’
maze.c:3: warning: incompatible implicit declaration of built-in function ‘printf’
$ ./a.out
30
._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
|_._. ._| |_. ._._. . . | . ._| |_. ._._._|_._. | |_. |_. | |_. | | | ._|_. | |
|_._._. |_._._| . ._|_| . | ._._._._._._._| ._._._| | . |_. | ._|_._._. | | | |
| ._| | . . . . |_._._|_| |_| | | |_._. | | . . . ._|_| | |_. ._._._| ._._. ._|
|_._._._|_|_|_|_| . . . . | | ._. ._._|_. ._|_| |_| ._. . . ._._. |_. . ._|_. |
| ._| . |_. . . |_| |_| |_|_._. |_._._._| ._._|_. |_._|_|_| |_. |_._. |_| | | |
| |_._|_. ._|_| . |_| | | | |_. ._. ._. ._._. |_. . . ._. |_._|_. |_. | ._. | |
|_._. |_._. |_. |_._._| | | . . . | ._| ._|_. | |_| |_._|_| | ._._| |_._| |_| |
|_._. . ._._._| ._. . | ._._| | | |_| ._|_. | ._|_. ._._._._|_._._._._. ._._._|
| ._. |_._._._|_. |_|_|_. ._|_| |_._| . . ._|_._._| ._| ._|_._._._. . ._._. ._|
|_._|_| ._._._| |_. |_. | |_._._| . | |_|_._._|_._. | ._. . ._._. | |_| | |_| |
| | | |_. . ._. ._| ._|_._| | ._. |_. |_._. . . . . | | ._| ._. | ._._._. ._| |
| . ._| ._|_. | ._| | ._. ._. . |_|_._._. | | | | | ._|_._| |_. | |_. |_._|_. |
| |_._._| . |_|_| | |_. |_. |_|_. . ._| |_| |_|_| |_._| ._| ._| |_| | . ._._._|
| | | . ._|_._._._._| | ._|_._._|_|_._._. |_. | |_. . ._| | . |_| ._._| |_. ._|
|_._. |_._. | . |_. ._| ._. . . ._._._. | | ._|_. |_| | | |_|_|_. |_. | ._._._|
|_. . . . |_._|_. . |_._| | |_| . |_._. | |_|_._. | ._._| | ._._._. | |_._| | |
| |_|_| |_._._. |_| ._. ._._| ._| ._._| | | . . . ._._._._|_| | | ._| ._._| | |
| ._. . . ._|_._._|_. | | |_._|_._._|_._|_| |_|_|_. | | . | | ._. | |_. ._|_. |
|_|_. |_|_. . |_. |_._|_._._. ._._. . . ._|_._._| | ._._| ._|_. | | | |_. |_. |
|_. . ._| ._| ._._._. . | | | |_._. |_|_| |_. . . |_| | |_| . ._|_._. . | ._._|
|_. |_. | ._| . | | | |_|_. ._._._|_._| . . |_| | . | ._| ._|_| |_._. |_|_._. |
|_._._|_| ._|_| ._._| |_. | . ._._._|_._|_| |_._|_| | ._|_._._. ._. |_| | |_. |
|_._._. |_| |_._._._| . | | |_|_._. | . . ._| | ._| . ._._|_._. | |_. . ._| ._|
|_. ._._|_. | |_. ._._|_. |_. |_. ._. | |_|_. ._| |_|_. | |_. ._| ._. | ._._| |
|_._._. | . | | |_| |_. |_. |_|_. ._|_|_._. ._| ._. | | | | | ._|_._| |_._._| |
| . | ._._|_._. ._._|_. ._._| ._| . . |_. | | ._. | ._._|_. |_._. ._| | | |_. |
| | ._| ._._. . ._| | ._._._. | |_| | | |_._. |_._| ._. ._._._._._._|_._. |_. |
| | ._| ._. | | ._|_._|_. . . | | |_| ._. |_._._. |_._|_. ._|_. . ._| . | ._._|
| | | ._._|_|_|_._. ._._|_|_|_. | | ._. |_._| | |_|_. | |_._. ._|_._._|_| ._| |
|_|_|_._._|_._._._._._._._._._._._._._|_|_._._._._._._._._._._|_._._._._._._._|
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Application: Maze Generation
The ECE 250 web site has a Q(mn) algorithm for generating an
m × n maze:
http://ece.uwaterloo.ca/~dwharder/aads/Algorithms/Maze_generation
The actual maze generation code is quite short:
Disjoint_sets rooms( m*n );
int number_of_walls = 2*m*n - m - n;
bool is_wall[number_of_walls];
Permutation untested_walls( number_of_walls );
for ( int i = 0; i < number_of_walls; ++i ) {
is_wall[i] = true;
}
while ( rooms.disjoint_sets() > 1 ) {
int wall = untested_walls.next();
int room[2];
find_adjacent_rooms( room, wall, n );
if ( rooms.find( room[0] ) != rooms.find( room[1] ) ) {
is_wall[wall] = false;
rooms.set_union( room[0], room[1] );
}
}
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Summary
In this topic, we have covered disjoint sets
– Equivalence relations and the definition of a disjoint set
– An efficient data structure
• A general tree
– An optimization which results in
• Worst case O(ln(n)) height
• Average and best cases Q(1) height
– A few examples and applications
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References
[1]
[2]
Cormen, Leiserson, and Rivest, Introduction to Algorithms, McGraw Hill,
1990, Ch.22, pp.440-461.
Weiss, Data Structures and Algorithm Analysis in C++, 3rd Ed., Addison
Wesley, Ch.8, pp.315-337.
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83
References
Wikipedia, http://en.wikipedia.org/wiki/Disjoint_set_(data_structure)
[1]
[2]
Cormen, Leiserson, and Rivest, Introduction to Algorithms, McGraw Hill, 1990, Ch.22, pp.440461.
Weiss, Data Structures and Algorithm Analysis in C++, 3rd Ed., Addison Wesley, Ch.8, pp.315337.
These slides are provided for the ECE 250 Algorithms and Data Structures course. The
material in it reflects Douglas W. Harder’s best judgment in light of the information available to
him at the time of preparation. Any reliance on these course slides by any party for any other
purpose are the responsibility of such parties. Douglas W. Harder accepts no responsibility for
damages, if any, suffered by any party as a result of decisions made or actions based on these
course slides for any other purpose than that for which it was intended.