ACM today… ! (2009). DP for NP!. DP for APSP!. Max Flow. Knapsack. FloydWarshall. Harbin, China. 2010 finals. Jotto!. A wordguessing game similar to mastermind…. Sophs. Jrs. Srs. Profs. Chalk 1. Chalk 0. Chalk 1. Chalk 1. Quine 1. Quine 1. Quine 2. Quine 2. ?. ?. ?. ?.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
DP for NP!
DP for APSP!
Max Flow
Knapsack
FloydWarshall
Harbin, China
2010 finals
A wordguessing game similar to mastermind…
Sophs
Jrs
Srs
Profs
Chalk 1
Chalk 0
Chalk 1
Chalk 1
Quine 1
Quine 1
Quine 2
Quine 2
?
?
?
?
next guesses?
v1 = 100
Maximizing Candy!
w1 = 2
i = 1
Suppose you can consume 13 candywt units of candy.
v2 = 120
w2 = 3
i = 2
What choice will maximize your candyvalue experience?
v3 = 230
w3 = 5
i = 3
(1) if you can take fractional parts of candy packages…?
v4 = 560
w4 = 7
(2) if you can take any number of whole candy packages…?
i = 4
v5 = 675
(3) if you can take 0 or 1 of each whole candy package…?
w5 = 9
i = 5
v1 = 100
Maximizing Candy!
w1 = 2
i = 1
vpw1 = 50
Suppose you can consume 13 candywt units of candy.
v2 = 120
w2 = 3
i = 2
vpw2 = 40
What choice will maximize your candyvalue experience?
v3 = 230
w3 = 5
GO WITH GREED!
i = 3
vpw3 = 46
(1) if you can take fractional parts of candy packages…?
v4 = 560
w4 = 7
(2) if you can take any number of whole candy packages…?
i = 4
vpw4 = 80
v5 = 675
(3) if you can take 0 or 1 of each whole candy package…?
w5 = 9
i = 5
vpw5 = 75
v1 = 100
Knapsack problem
w1 = 2
"unbounded"
i = 1
vpw1 = 50
Suppose you can consume 13 candywt units of candy.
v2 = 120
w2 = 3
(2) if you can take any number of whole candy packages…?
i = 2
vpw2 = 40
v3 = 230
w3 = 5
i = 3
vpw3 = 46
IDEA:
Consider all possible weights (integers) from 0 up to the weight you can carry 
For each one choose the best from all N items
v4 = 560
w4 = 7
i = 4
vpw4 = 80
v5 = 675
w5 = 9
i = 5
vpw5 = 75
v1 = 100
Knapsack problem
w1 = 2
"unbounded"
i = 1
vpw1 = 50
Suppose you can consume 13 candywt units of candy.
v2 = 120
w2 = 3
(2) if you can take any number of whole candy packages…?
i = 2
vpw2 = 40
v3 = 230
TOTAL WEIGHT
w3 = 5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
i = 3
vpw3 = 46
0
0
v4 = 560
100
120
200
230
300
560
560
675
680
775
800
875
w4 = 7
i = 4
vpw4 = 80
max total value
v5 = 675
w5 = 9
max
V(wwi) + vi
V(w) =
i
i = 5
vpw5 = 75
v1 = 100
Knapsack problem
w1 = 2
"01 problem"
i = 1
vpw1 = 50
Suppose you can consume 13 candywt units of candy.
v2 = 120
w2 = 3
(3) if you can take 0 or 1 of each whole candy package…?
i = 2
vpw2 = 40
v3 = 230
w3 = 5
i = 3
vpw3 = 46
IDEA:
v4 = 560
Do the same thing as before,
but consider sublists of items that grow oneatatime
w4 = 7
i = 4
vpw4 = 80
v5 = 675
w5 = 9
i = 5
vpw5 = 75
v1 = 100
Knapsack problem
w1 = 2
"01 problem"
i = 1
vpw1 = 50
Suppose you can consume 13 candywt units of candy.
(3) if you can take 0 or 1 of each whole candy package…?
v2 = 120
V(i1,w)
w2 = 3
V(i,w) =
max
i = 2
vpw2 = 40
V(i,wwi) + vi
i
v3 = 230
TOTAL WEIGHT
w3 = 5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
i = 3
vpw3 = 46
0
01
v4 = 560
w4 = 7
012
i = 4
vpw4 = 80
0123
01234
v5 = 675
012345
w5 = 9
i = 5
Total Items
vpw5 = 75
max total value
exact number of cows we need, indexed from 1 .. 100
Input
but the sumofsquares of their indices must be < 30
3 30
Input
the number of different subsets of cows satisfying that constraint.
4
total
14
1 2 3
1 2 4
2 3 4
1 3 4
21
29
26
exact number of cows we need, indexed from 1 .. 100
Input
but the sumofsquares of their indices must be < 30
3 30
"use it or lose it"
Input
analysis
the number of different subsets of cows satisfying that constraint.
4
total
14
1 2 3
1 2 4
2 3 4
1 3 4
21
You're either going to USE the first cow  or you're not. Sum up both cases!
29
26
exact number of cows we need, indexed from 1 .. 100
Input
but the sumofsquares of their indices must be < 30
3 30
# of len3 subsets, totaling under 30, starting from 1
N(3,30,1) =
N(2,29,2) + N(3,30,2)
Input
# of len2 subsets, totaling under 29, starting from 2
# of len3 subsets, totaling under 30, starting from 2
the number of different subsets of cows satisfying that constraint.
4
"use it or lose it"
total
14
1 2 3
1 2 4
2 3 4
1 3 4
21
29
26
but it's much too slow!
Pythonfunction decorators
plain
allpairsshortestpaths
Directed graph as adjacency matrix:
dst
Directed graph:
"to"
1
2
3
4
100
inf
100
0
14
1
2
14
1
14
inf
0
50
2
50
src
"from"
inf
inf
0
14
3
10
14
inf
10
inf
0
4
4
3
14
0 intermediate nodes
dst
dst
"to"
"to"
1
2
3
4
1
2
3
4
inf
100
inf
100
0
14
0
14
1
1
14
14
inf
0
50
inf
0
50
2
2
src
src
"from"
"from"
inf
inf
inf
0
14
inf
0
14
3
3
24
inf
10
inf
0
10
inf
0
4
4
1 intermediate node(s)!
0 intermediate nodes
1
dst
dst
"to"
"to"
1
2
3
4
1
2
3
4
inf
64
inf
100
0
14
0
14
1
1
14
14
inf
0
50
inf
0
50
2
2
src
src
"from"
"from"
inf
inf
inf
0
14
inf
0
14
3
3
24
24
10
inf
0
10
inf
0
4
4
1 intermediate node(s)
2 intermediate node(s)!
1
1
2
Allpairs shortest paths:
dst
dst
"to"
"to"
1
2
3
4
1
2
3
4
28
42
28
42
0
14
0
14
1
1
14
14
38
0
28
inf
0
28
2
2
src
src
"from"
"from"
24
inf
38
0
14
inf
0
14
3
3
24
24
10
38
0
10
38
0
4
4
3 intermediate node(s)
4 intermediate node(s) ~ done!
1
2
3
1
2
3
4
dst
T[src][dst][k]
"to"
1
2
3
4
minimum distance from src to dst using intermediate nodes 1..k
28
42
0
14
1
T[src][dst][k1]
inf
14
inf
0
28
2
min
=
src
T[src][k][k1] +
T[k][dst][k1]
14 + 10
"from"
inf
inf
0
14
3
24
10
38
0
4
24
4 intermediate node(s)
1
2
3
4
Floyd Warshall…
not a huge amount of extra work to keep an extra table (pred) of the first intermediate vertex on the path from i to j
These two are available if you haven't submitted them successfully before…
hurdles
Input
number of edges
number of "tasks"
5 6 3
1 2 12
3 2 8
1 3 5
2 5 3
3 4 4
2 4 8
3 4
1 2
5 1
from node 1 to node 2 the cost ("height") is 12
12
2
1
Output
3
8
5
What is the minimum cost along
any one of the edges required
to get from start to end ?
5
3
8
4
4
8
1
4
not possible!
dining
Input
total # of foods
total # of drinks
4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3
Likes
1 2
2 3
1 3
1 3
3 1
1 2
1 2
3
0
# of foods cow[i] likes
foods
drinks
1
# of drinks cow[i] likes
2
Output
3
What is a cowsatisfying assignment here?
3
# of cows that can receive both a food and a drink they like…
foods
drinks
each can be used only once
FordFulkerson algorithm
12
B
D
20
16
sink
F
A
10
4
9
7
source
4
13
C
E
14
capacity
The problem how much traffic (flow) can get from the source to the sink ?
16
13





10
12



4


14



9


20



7

4






Max Flow
The problem how much traffic can get from the source to the sink ?
12
B
D
20
16
sink
F
A
10
4
9
7
source
4
13
“Capacity Graph”
C
E
14
TO
C
capacity
A
B
C
D
E
F
A
B
C
FROM
D
E
F
16
13





10
12



4


14



9


20



7

4






Find a path in C via BFS
The problem how much traffic can get from the source to the sink ?
Need to subtract this flow from C !
12
B
D
20
16
sink
F
A
10
4
9
7
source
4
13
C
E
14
TO
C
capacity
A
B
C
D
E
F
A
B
C
FROM
D
E
F
12
0



12

0
12



0


0


12
0


12



0

0



12


Create F
Create a FLOW GRAPH with the minimum weight from that path
12
B
D
12
20
16
sink
12
12
F
A
10
4
9
7
source
4
13
“Flow Graph”
C
E
14
TO
F
capacity
A
B
C
D
E
F
A
B
flow in one direction is negative flow in the other direction
C
FROM
D
E
F
4
13



12

10
0



4


14


12
9


8



7

4



12


R = C  F
Use the RESIDUAL GRAPH with the rest of the capacity after that flow
0
B
D
12
8
4
sink
12
12
F
A
10
4
9
7
source
4
13
“Residual Graph”
C
E
14
TO
R
capacity
A
B
C
D
E
F
A
B
reverse edges allow the "undoing" of previous flow!
C
FROM
D
E
F
Continue running this on the residual capacity until BFS fails…
12/12
B
D
19/20
11/16
sink
F
A
0/10
1/4
0/9
7/7
source
4/4
12/13
C
E
11/14
max flow:23
Continue until there is no longer a path from A to F !
1. Set F to all 0
The max flow algorithm
2. Compute R = C  F
3. BFS in R from source to sink.
3a. If no path, you’re done.
3b. If a path, add the available capacity to F; goto (2).
FloydFulkerson
12
B
D
3
20
16
sink
2
4
1
0
F
A
10
4
2
9
1
7
2
source
4
4
13
C
5
E
14
capacity
flow
residual
“Residual Graph”
3
B
D
12
18
9
Don't need to keep R around explicity: Keep only the current flow (F) and the original capacity (C).
4
2
A
1
9
2
5
6
F
7
2
5
C
E
4
11
9
A little bit of name contention…
edmonds_karp
This is the algorithm.
defmax_flow(C, source, sink):
n = len(C) # C is the capacity matrix
F = [[0] * n for i in range(n)] # F is the flow matrix
# residual capacity from u to v is C[u][v]  F[u][v]
whileTrue:
path = BFS(C, F, source, sink)
if not path: break # no path  we're done!
# find the path's flow, that is, the "bottleneck"
edges = [C[u][v]F[u][v] for u,v in path]
path_flow = min( edges )
print"Augmenting by", path_flow
for u,v in path: # traverse path to update flow
F[u][v] += path_flow # forward edge up
F[v][u] = path_flow # backward edge down
return sum([F[source][i] for i inrange(n)]) # out from source
A brief BFS algorithm using the Capacity matrix
defBFS(C, F, source, sink):
queue = [source] # the BFS queue
paths = {source: []} # stores 1 path per graph node
while queue:
u = queue.pop(0) # next node to explore (expand)
for v inrange(len(C)): # for each possible next node
# path from u to v? and not yet at v?
if C[u][v]  F[u][v] > 0 and v notin paths:
paths[v] = paths[u] + [(u,v)]
if v == sink:
return paths[v]
queue.append(v) # go from v in the future
returnNone
And the code needed to run it…
if __name__ == "__main__":
# make a capacity graph
# node A B C D E F
C = [ [ 00, 16, 13, 00, 00, 00 ], # A
[ 00, 00, 10, 12, 00, 00 ], # B
[ 00, 04, 00, 00, 14, 00 ], # C
[ 00, 00, 9, 00, 00, 20 ], # D
[ 00, 00, 00, 7, 00, 4 ], # E
[ 00, 00, 00, 00, 00, 00 ] ] # F
print"C is", C
source = 0 # A
sink = 5 # F
max_flow_value = max_flow( C, source, sink )
print"max_flow_value is", max_flow_value
Linked at the ACM website by the slides…
A wordguessing game similar to mastermind…
Sophs
Jrs
Srs
Profs
Chalk 1
Chalk 0
Chalk 1
Chalk 1
Quine 1
Quine 1
Quine 2
Quine 2
?
?
?
?
next guesses?
v1 = 100
Knapsack problem
w1 = 2
"01 problem"
i = 1
vpw1 = 50
Suppose you can consume 13 candywt units of candy.
(3) if you can take 0 or 1 of each whole candy package…?
v2 = 120
w2 = 3
V(i,w) =
i = 2
vpw2 = 40
v3 = 230
TOTAL WEIGHT
w3 = 5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
i = 3
vpw3 = 46
0
01
v4 = 560
w4 = 7
012
i = 4
vpw4 = 80
0123
01234
v5 = 675
012345
w5 = 9
i = 5
Total Items
vpw5 = 75
max total value