Non-MST
Ed
ge
Discount
path
:
Devise
a
0
(IvI)-time
als
for
consider
a
directed
graph
the
following
prob
lem
:
G
=
(V
,
E)
w/
pos
.
weights
((i
,
i)
Input
:
Graph
G
=
(V
,
E)
w/
distinct
on
its
edges
.
Given
a
path
=
vo-
edges
weights
Wi;
for
edge
V
,
...
->
UK
,
the
discounted
cost
(i
,
i)
=
E
.
of
the
path
is
the
sum
of
Solution
:
Find
some
edge
lengths
of
all
but
the
longest
edge
.
(i
,
i)
EE
S
.
E
.
Li
,
j)
is
Not
part
Formally
,
Of
any
MST
.
k
-
1
1)
Give
a
succinct
and
precise
cost(p)
=
((Vi
,
vi
+
1)
-
maxc(Vi
,
vit)
description
of
the
9l9
.:
(It
is
the
sum
of
cost
of
all
edges
The
idea
is
to
find
the
minus
maximum
costed
ge(
heaviest
edge
in
some
cycle
Devise
an
alg
,
to
compute
the
smallest
of
the
graph
.
Run
DFS
on
discounted
cost
path
from
a
vertex
the
graph
until
we
s
to
vertext
Give
a
succinct
and
encounter
a
back
edge
U-V
,
Precise
descr
.
of
your
alg
.
Proof
of
discovering
a
cycle
.
We
will
correctness
and
run-time
analysis
find
the
heaviest
edge
are
not
necessary
.
The
runtime
of
Cycle
and
return
it
.
This
Your
alg
.
Should
be
o((IVI
+
1E1)
10g
(v))
.
the
heaviest
edge
,
Check
Minimizing
the
discounted
cost
of
all
edges
along
the
path
a
path
is
the
same
as
minimizing
~
to u
in
the
tree
.
This
the
cost
of
a
path
wherein
one
edge
can
be
&
one
effiently
by
can
be
skipped
at
zero
cost
.
Since
popping
the
call
stack
we
are
looking
for
p
2X
act
1y
for
Explore
to
retrace
the
One
"skipped
edge"
,
wh
create
Path
up
the
tree
.
a
layered
graph
G'as
follows
:
For
2)
Briefly
justify
the
each
vertex
v
Ev
,
there
would
be
~
un-time
of
the
alg.
two
Copies
Vo
,
Vi
.
For
each
U-V
,
add
Each
edge
encountered
edges
No
-
Vo
,
No e
VI
,
and
U
,
-V
.
during
DFS
in
an
undir
.
Weights
of
the
edges
are
graph
is
either
a
tree
edge
((u0
+
Vol
=
c (U
,
evi)
=
c(ueV)
and
or
a
back-edge
.
Since
there
C
(Uo
,
V
,
)
=
0
.
Essentially
G'consists
are
only
Iv1-1
free
edges
,
of
2
separate
copies
of
G
,
w/
DFS
will
encounter
a
zero-weight
"skip"
edges
from
one
back-edge
in
O
(1v1)
time
.
Copy
to
other
.
We
construct
the
graph
G'and
use
Dij
.
Alg
.
to
find
Np-completeness
reductions
the
shortest
path
from
So
to
+i.
You
may
assume
that
the
following
problems
are
NPC
:
set
ordering
Rudrata
(Hamiltonian)
path
,
Input
:
A
family
of
subsets
S
,
....
Sn
Rudrata
(Hamiltonian)
Cycle
,
of
E1
,
...,
m3
.
Vertex
cover
,
Ind
.
Set
,
Solution
:
An
ordering
of
the
sets
3
-
Coloring
,
3-SAT
and
Int
So
that
consequetive
sets
in
the
programming
.
For
each
of
orderings
have
intersection
exactly
the
following
problems
,
fill
One
.
Formally
,
an
ordering
of
the
sets
in
the
details
of
the
proof
that
is
given
by
a
permutation
It
:
(n]
=
[n]
,
they
are
in
NPC
.
So
that
for
each
it
E1
,
...,
n-
13
,
1)
Tricl ique
:
Given
a
graph
IsnicunSicil
=
1
.
Proof
,
it
is
clear
that
G
=
(V
,
E)
,
a
subsets
<
Vof
the
set-ordering
problem
is
in
NP
.
Now
Vertices
is
called
a
clique
,
if
we
will
use
a
poly
,
time
reduction
to
every
Pair
of
vertices
ins
are
show
that
the
problem
is
indeed
NPC
.
conne
Che
&
by
an
edge
,
i
.
e
.,
for
Given
an
instance
!
of
problem
every
U
,
v
ES
we
have
(u
,
vI
E
.
We
will
construct
an
instance
T
Here
is
the
Tricl ique
problem
.
of
the
problem
as
follows
.
Input
:
A
graph
G
=
(V
,
El
Given
an
instance
G
=
(V
,
E)
of
solution
:
A
partition
of
the
Rudrata
Path
,
define
an
instance
vertices
into
3
disjoint
sets
of
set
ordering
as
follows
:
Let
S
,
USzUSy
=
VS
.
t
.,
each
of
the
the
universe
be
the
set
of
edges
Sets
S
1,
S2
,
So
is
a
clique
,
Proof
,
E
and
we
will
have
ach
Si
be
a
it
is
clear
that
the
Tricl ique
subset
of
this
universe
.
For
each
problem
is
in
NP
.
Now
we
will
use
a
vertex
itv
,
define
Si
=
Eedges
incident
polynomial
time
reduction
to
show
at
vertex
j3
.
that
the
problem
is
indeed
NPC
.
This
proof
that
this
is
a
valid
Given
a
instance
of
of
the
reduction
as
follows
:
Problem
we
will
SUPP
Ore
there
is
a
Rudrata
path
construct
an
instance
I
of
in
the
original
graph
,
then
it
gives
the
problem
as
follows
.
an
ordering
it
:
[n]
-
[n]
of
the
Given
an
instance
G
=
(V
,
E)
vertices
of
the
graph
.
The
corr
of
3-coloring
,
define
an
orderings
of
sets
satisfies
the
instance
G'
=
IV'sE')
of
Cond
.,
Sin
ce
Salis
1 Stcitic
contains
Tricl ique
as
follows
:
Set
Precisely
the
edge
from
Vertex
TT(i)
v
=
V
.
For
e
ach
Pair
(u
,
v)
to
Vertex
It
(i
+
1)
.
Conversely
,
suppose
define
,
la
,
use
Es
(4
,
v)
there
is
an
ordering
It
:
[n]e
<
n]
of
↓
E
.
(In
other
words
,
edge
sets
.
Since
Sicil
Sici
+
i)
=
1
then
it
C
(u
,
v)
exists
in
E'if
and
only
means
that
the
set
of
edges
incident
not
exist
in
E
.
at
Vertices
iT
(i)
and
T
(i
+
1)
,
have
if
it
does
f
that
this
is
a
valid
exactly
one
common
edge
.
This
implies
reduction
as
follows
:
that
IT(i)
--
Tli+1)
is
an
edge
of
Given
a
3-coloring
G
=
(V,E)
,
be
vertices
let
Si
,
S2
,
Sy
the
set
colore
&
red
,
b
lue
,
and
green
very
-
respecti
w/in
e
ach
color
class
,
there
are
no
edges
in
E
...
the
Sets
Si
,
Sa
,
5
are
Cliques
in
the
graph
G'
=
(V'
,
E')
.
conversly
,
S
upp
OSe
S
,
Sc
,
S3
are
cliques
in
G'
=
(V'
,
E'l
Then
consider
the
coloring
that
assigns
vertices
in
these
sets
reen
red
,
blue
,
and
9
respectively
Since
each
Si
is
a
clique
in
G
,
there
are
no
edges
w/in
any
S
:
Hence
this
is
a
valid
3
coloring
of
graph
G
.
-
-o
0-0
I
-
&
-
b
↳
↓
↳
-
-
~
-
do
O
El
-
Trian gul ar
Removal
Reduction
a
following
2
problems
:
In
the
triangular
removal
problem
,
the
input
is
a
graph
G
=
(V
,
E)
.
The
goal
is
1)
Matching
to
remove
the
smallest
#
of
edges
from
Input
:
Graph
G
=
(V
,
E)
and
a
#
so
as
to
delete
all
triangles
in
the
positive
integer
k
g
raph
.
Formally
,
here
is
the
def
.
of
Sol
:
A
set
of
k
edges
e
,
...,
ek
,
triangle
freeness
.
Definition
:
A
graph
no
pair
of
which
share
a
vertex
.
is
triangle
-
free
if
there
are
NO
3
2)
Indpendent
Set
Vertices i
,
j
,
keV
S
.
t
.
(i
,
j)
,
(Jsk)
,
Input
:
Graph
G'
=
(V'
,
E')
and
a
and
(k
,
i)
are
all
edges
.
Here
is
the
Positive
integer
K
for mal
def
.
of
triangle
removal
Problem
:
Sol
:
A
set
of
K
vertices
SCV'
,
no
Input
:
An
unweighted
undir
.
graph
G
=
(v
,
E)
pair
of
which
are
connected
by
an
Sol
:
Find
the
smallest
set
of
edges
E'such
edge
.
that
deleting
E'makes
the
graph
become
The
problem
Independent
Set
triangle-free
·
We
will
now
deve
an
approx
.
is
believed
to
not
reduce
in
&
lg
,
for
the
problem
.
First
write
a
linear
poly
.
time
to
Bipartite
programming
relaxation
for
the
Matching
because
Independent
trial-removal
problem
.
Set
is
a
NPC
problem
While
What
are
the
variables
of
the
linear
program
Bipartite
matching
is
a
(LP)
?
(write
out
succinctly
and
precisely
,
lem
in
p
.
Since
the
In
do
Prob
the
intended
meaning
of
the
variables
in
Set
is
a
harder
problem
than
your
LP)
Xe
for
each
edge
ef
E
.
The
intended
Bipartite
Matching
,
Provid
a
meaning
is
that
Xe
is
the
amount
of
Bipartite
matching
w
ould
edges
e
that
is
deleted
.
Show
that
p
=
Np
,
which
What
is
the
obj
.
func
.?
is
currently
not
believed
.
Minimize
e
on
the
other
hand
,
Matching
reduces
in
P
·
lynomial
time
What
are
the
constraints
of
the
LP
?
to
Ind
.
Set
.
Reduction
:
Given
&
Xe11
for
all
e
E
an
instan
ce
d
of
the
prob
nem
Xa
+
Xb
+
Xc])
for
all
triangles
a
,
b
,
C
.
Matching
we
will
construct
Describe
a
rounding
scheme
to
obtain
an
instance
↑
Or
a
50
,
13
sol
·
from
the
opt
.
Sol
.
to
the
above
If
Xe
y
,
round
it
to
1
.
0
.
W
.
round
to
0
.
E'
is
the
set
of
edges
e
W/Xe
=
1
.
i
Briefly
justify
why
the
rounding
Scheme
Prod
.
a
Sol
.
to
the
Trian gula r
Reduction
:
Removal
problem
.
For
any
triangle
(a
,
b
,
c)
in
G
,
since
Xa
+
Xb
+
X
,
=
1
,
at
least
one
edges
&
EEa
,
b
,
13
has
Xe
?
"
,
so
we
delete
at
least
one
of
edges
in
the
triangle
·
So
there
can't
be
a
triangle
in
E
'
Wh
at
is
the
approx
.
ratio
of
your
alg
.?
3
.
b/c
Xe
is
scaled
up
by
a
factor
of
at
most
3
during
rounding
.
Edit
Distance
a
...
n
Y
[1
,
...,
my
,
recall
that
the
edit
dist
.
is
the
smallest
#
of
keystrokes
needed
to
edit
the
S
tring
X
into
String y
.
Here
each
insertion
and
deletion
of
a
character
takes
one
key
stroke
.
We
devised
a
dynamic
programming
alg
.
for
the
problem
wherein
the
subproblems
are
e
#
D[i
,
j]
=
minimum
#
of
keystrokes
nee
de
&
to
edit
the
Prefix
[1
,
...,
i]
into
the
Prefix
11
,
....
j]
.
Suppose
We
had
a
special
keeboard
in
which
·
each
Insertion
takes
z
keystrokes
·
each
insertion
takes
3
keystrokes
·
each
substitution
takes
a
keystrokes
1)
write
down
the
modified
recurrence
relation
ED[i
,
i]
E
ED[i
-
1
,
j]
+
3
-
intersection
ED[i
,
j]
=
min
ED[i
,
j
-
13
+
2
ED
Siri
,
j-1]
+
4
.
1x[i]
*
y[i]]
write
down
the
modified
base
case
for
ED(i
,
j]
ED[i
,
0]
=
3i
vi
ED[O
,
j]
=
2
;
j
Minimum
Median
Te
-
the
followin
S
problem
919.
for
the
graph
.
Hence
TT
(1)
-
-
T(z)
--
...
(n)
is
Input
:
A
graph
G
=
(V
,
E)
and
a
Rudrata
Path
in
the
original
graph
.
weights
on
the
edges
[Welec
E3
.
Solution
:
A
spanning
tree
T
that
All
pairs
shortest
paths
:
minimizes
the
median
cost
(T)
,
In
the
lecture
,
we
have
seen
the
Where
median
cost
(1)
is
defined
Floyd-Warshall
a19
,
for
computing
as
median
cost
(T)
=
median
of
the
distances
blu
every
pair
of
vertices
weights
of
the
weights
of
the
in
an
input
graph
.
The
Floyd
Warshall
edge
5
in
the
tree
T
.
The
runtime
of
alg
.
is
a
dynamic
programming
alg
.
Your
alg
.
must
be
polynomial
in
the
~
/
runtime
O(n3)
.
In
this
question
,
#
of
vertices
n
=
IVI
.
we
will
devise
a
simplier
dynamic
Give
a
succinct
and
precise
descr
.
programming
based
a
19.
for
all
pairs
of
your
a
eal's
or
prim's
shortest
paths
,
wh
slightly
worse
prove
the
correctness
o
f
your
a1g
.
runtime
Girlhahihugraphe
The
C
ut
property
stills
hold
.
If
(i
,
j)
.
Consider
the
following
sub-problem
:
+
lighest
edge
in
a
cut
is
not
ne
D
[i
,
j
,
t]
=
length
of
shortest
path
from
in
a
MMT
,
replace
it
/
that
and
i
to
;
w/
at
most
2t
hops
.
(Here
"hop"
You
get
a
better
MMT
::
any
MST
refers
to
the
#
of
edges
on
the
path)
.
is
a
MMT
.
Write
the
recurrence
relation
for
D[i
,
j
,
t]
D[i
,
j
,
+]
=
MinED(i
,
k
,
t-1
+
D(k
,
i
-
>
write
psuedocode
for
the
alg.
to
compute
all-pair
distances
,
including
the
initialization
.
For
i
,
jEV
,
D(ii]
=
SWijifiRE
For
t
=
1
+
o
Di
,
j
,
+]onForiinForinForKin
S
return
D(i
,
j(logn]]
KEV
for
all
i
,
j
.
What
is
the
runtime
of
your
alg
.?
O(n3
logn)
where
n
=
IVl
.
-
the
problem
End.
set
as
follows
.
For
each
&
of
$
,
create
a
edge
an
in
6'
of
T
.
For
each
Pair
of
edges
(e
,
22)
that
Share
vertices
,
add
an
edge
e'
=e
blu
the
corr
.
Vertices
VI'
,
↓
in
G'of
Y
.
NP-completeness
Reduction
:
1)
Hitting
Set
:
Input
:
Positive
integers
n
,
m
,
k
and
a
family
of
sub
El
,...,
m3
Sol
:
A
subset
H
Sets
Si
.
Sno
e
size
k
that
overlaps
every
one
o
the
Sets S
s
null
Q
u
Proof
it
is
clear
that
Hitting
Set
Problem
is
in
NP
.
Now
we
will
use
a
polynomial
time
reduction
to
show
that
the
problem
is
indeed
NPC
.
Give
an
instance
6
of
the
problem
Vertex
cover
we
will
construct
an
instance
↑
of
the
Problem
Hitting
set
as
follows
.
We
want
to
reduce
from
VC
to
Its
so
we
construct
an
instance
of
As
to
solve
a
given
instance
of
VC
.
Let
G
=
(V
,
El
be
the
input
graph
to
Vertex
cover
.
T
~
or
each
edge
(u
,
v)
EE
,
add
a
Subset
Si
=
Eu
,
v3
to
the
Hitting
set
instance
.
Finding
a
vertex
cover
of
s
ize
K
is
the
same
as
finding
a
hitting
Set
HCV
of
size
on
inputs
S
...,
SiEl
Crepe
integers
in
the
Hitting
Set
instances)
.
Proof
:
A
Vertex
cover
SCV
has
the
property
that
for
all
(u
,
v)
E
E
at
least
one
u
,
v
is
in
S
:
S
intersects
w/
each
of
Si
and
is
a
hitting
set
.
Now
suppose
we
have
a
hitting
Set
H
.
For
each
(u
,
v]
E
E
,
Eu
,
v3
10
::
I
covers
all
edges
and
is
a
vertex
cover
.
Max
Flow
:
There
are
connected
,
undir
.
graphs
G
S
.
t
.
dec
.
the
capacity
Of
any
single
edge
in
6
doesn't
Chan
aX
-
>
ow
the
value
of
the
min
cut
There
are
connected
,
undir
.
graphs
G
S
.
t
.
in2
.
the
capacit
a
of
any
singled
a
of
the
max
Chan
-
ow
.
True
There
are
dir
.
graphs
G
S
.
t
.
des
the
capacity
of
any
single
edge
in
a
does
not
change
the
Val
.
of
max
flow
.
True
CT
↓
rueFas
can't
have
exact
distinct
optimal o
a
Bakery
((P/dual)
:
John
owns
a
bakery
which
prod
cakes
,
cookies
,
and
scones
.
Jon
makes
$10
for
each
cake
,
$1
for
each
cookie
,
and
$2
for
each
scone
sold
.
Ing
of
200
units
of
flour
,
100
units
of
sugar
,
and
so
eggs
.
LP
:
max
10
a
+
b
+
2C
dual
:
min
zood
+
100e
+
50f
one
cake
need
a
a
+
b
+
3C
1200
5d
+
2e
+
5f
-
10
5
·
e
cookie
needs
o
2a
+
2b
+
c
1100
d
+
2e
+
f
-
I
&
2)
I
le
f
·
one
S
59
+
b
+
2150
3d
+
e
+
2
-2
3f
,
I Scone
need
s
a
b
,
]
O
d
,
e
,
f
>
0
