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. ?. ?. ?. ?.
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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?
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
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
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
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
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
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!
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 !
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?
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