© 2007 Seth James Nielson DFS and A*. © 2007 Seth James Nielson It is a dark time for searching Admiral…

© 2007 Seth James Nielson

DFS and A*


It is a dark time for searching…

Admiral PiettLord Vader, if the Millennium Falcon went into light-speed, it'll be on the other side of the galaxy by now.

Darth VaderAlert all commands. Calculate every possible destination along their last known trajectory.

Admiral PiettYes, my Lord. We'll find them. Darth VaderDon't fail me again, Admiral.


What’s the Admiral to do?

The fleet is in the Anoat system

It has neighboring systemsThose systems have

neighboring systems (etc.) How can we help the Admiral

not receive a Sith sponsored telekinetic larynx massage?


Anoat


Does this look familiar?


for each vertex v in V color[v] = white d[v] = INFINITY p[v] = NULL color[s] = gray d[s] = 0 Queue.clear() Queue.put(s) while (!Queue.empty()) v = Queue.get() for each u adjacent to v if (color[u] == white) color[u] = gray d[u] = d[v] + 1 p[u] = v Queue.put(u) color[v] = black


From a Tree View

Anoat

1 42 3

1196 8 10 125 7

1 2 3 4 5 6 7 8 9 10 11 12

Ano

at


Admiral Piett’s Problem

Observe the mem requirement

bd, whereb = branching factord = depth

Example: b = 4d = 2, mem = 16d = 10, mem = 1,048,576


Switch in a StackUse a stack instead of a

queueDescendant before siblingUse recursion for “built-in”

stackThis is called “Depth-First”


DFS(G) Code

1. for each vertex u in V2. color[u] = white3. p[u] = null4. for each vertex u in V5. if (color[u] == white)6. DFS-VISIT(u)


DFS-VISIT(u) Code

1. color[u] = gray2. for each v adjacent to u3. if (color[v] == white)4. p[v] = u5. DFS-VISIT(v)6. color[u] = black


Code NotesNot shortest pathDoesn’t compute distanceCreate “forest”, cover all

nodesOptions (see CLRS 541)

Node discovery time - d[v]Node completion time - f[v]


Parenthesis TheoremIn a dfs tree, one of the

following holds:[d[u],f[u]] & [d[v],f[v]] are

disjoint <--> neither u or v descends from the other

[d[u],f[u]] within [d[v],f[v]] and u descends from v

Same as above, switch u,v


Parenthesis Theorem explained

If v descends from u thend[u]<d[v]<f[v]<f[u]

Rewrite d[u] as (uf[u] as u)Then creates well formed

expression (balanced parens)e.g., (u (v (x x) v) u)


Two More Theorems

1. White Path Theorem2. DFS on undirected graphs

See p545-546 in text


An Interesting DFS Use

DFS can be used to generate perfect mazes

Perfect maze has 1, and only 1, path between any 2 points


Can DFS help Piett?Let’s see how DFS works

on the Piett Problem…


Anoat

1 2 3

96 8 105 7

`

1235678910A

noat


DFS and MemoryAdmiral Piett is relieved to

see that DFS will not exceed the memory resources of his Star Destroyer.

But he can’t shake the feeling there might be something better…..


Introducing A*Also known as “best-first”Requires

Node evaluation heuristicData struct that orders

nodes using true cost from start + estimated cost to end

True cost so far required for a guaranteed shortest path


Admissible Heuristicd_estimate <= d_actualSatisfy triangle inequality

Very rarely a problemNot covered hereSee online sources such as

Wikipedia for more info


A-Maze-Ing ExampleSearch for exit from startAssume closer nodes betterVarious heuristics (G=Goal)

Manhattan f(x,y) = |xG-x|+|yG-y|Diagonal f(x,y) = max(|xG-x|,|yG-

y|)Straight-line

f(x,y) = SQRT( (xG-x)2 + (yG-y)2 )


Priority QueueThe data structure A*

needsLike a queue, but

dequeues the best (max or min on some function) element


Visual Example

0,0 - 0+5=51,0 - 1+4=51,1 - 2+3=51,2 - 3+2=5

2,2 - 4+1=5

2,1 - 5+2=7

3,1 - 6+1=7

0,2 - 4+3=7


Piett and A*Piett can’t use the same

heruisticsHowever, he does have other

info (intel, stats, trace data) to rate the various nodes

Can use these, if they produce an admissible heuristic

Do you see why A* is called a informed search?


HeapGood for Priority QueuesAlmost complete binary

treesProperty:

Node’s value > value of all descendants (for max heaps)

Min heap swaps < for >


Max Heap Example


Heap Implementation

Single arrayOperations (1-indexed):

parent(i) = floor(i/2)left(i) = 2*iright(i) = 2*i + 1

14 10 8 716 9 3


NotesSubtree of heap = heapNodes stored in array in

layersHighest nodes first, etc


Maintaining Heap Property

MAX-HEAPIFYGiven heap A and node i

i does not satisfy heap propBUT, left and right subtrees do

Traverse down switching nodes until property is satisfied

Running time is O(log(n))


MAX-HEAPIFY(A, i)1. l = left(i)2. r = right(i)3. if (l<=A.size && A[l] > A[i])4. largest = l5. else6. largest = i7. if (r<=A.size && A[r]>A[largest])8. largest = r9. if (largest != i)10. exchange(A,i,largest)11. MAX-HEAPIFY(A,largest)


Building Max-Heap Given arbitrary array,

notice leaves satisfy heap property

BUILD-MAX_HEAP(A)1. A.size = A.length2. For (i=floor(A.length/2); i>0;

i--)3. MAX-HEAPIFY(A,i)


Getting the Max Easy: HEAP-MAXIMUM(A)

1. return A[1]


DequeueingMust maintain heap propIdea:

Exchange A[1] with last node

Remove (new) last nodeMAX-HEAPIFY(A,1)


HEAP-EXTRACT-MAX(A)

1. if (A.size < 1)2. error “heap underflow”3. max = A[1]4. A[1] = A[A.size]5. A.size = A.size-16. MAX-HEAPIFY(A,1)7. return max


Increasing Node Value

Sometimes, need to increase “priority” (like in OS)

After increasing value, need to “bubble” it up in the tree until heap property is satisfied


HEAP-INCREASE-KEY(A,i,key)

If (key < A[i]) error “new key smaller”A[i] = keyWhile (i>1 && A[parent(i)] <

A[i]) exchange(A, i, parent(i)) i = parent(i)


Heap InsertionAdd element at end of ASet value to -INFINITYIncrease value to correct

value


MAX-HEAP-INSERT(A,key)

1. heap-size[A] = heap-size[A]+12. A[heap-size[A]] = -INFINITY3. HEAP-INCREASE-KEY(A,heap-

size[A],key)


Summary - DFSLower memory

requirementsUses a stack (usually

recursion) instead of queue


Summary - A*Requires context and infoUses a priority queueCalled an informed searchTo get shortest path, add

true cost so far to estimate to goal


Summary - HeapsAlmost complete binary

treesSatisfy heap propertyCan be used as priority

queue


Summary - Admiral Piett

Never found the FalconBut learned a lot about

search(Was spared because

Vader had reached his “bag limit”)


Epilogue

LANDO: Luke, we’re ready for take offLUKE: Good luck, LandoLANDO: (into comlink) When we find Jabba the Hut and that bounty hunter, we'll contact youReturn of the Search!coming soon to a classroom near you

Remember, the Search will be with you. Always!

Documents

© 2007 Seth James Nielson DFS and A*. © 2007 Seth James Nielson It is a dark time for searching Admiral…