Strongly Polynomial Mincost flow

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Strongly Polynomial Min-cost flow

• Evà Tardos algorithm
• William Fahle CS6381

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Original problem statement

• min ?ij ci,jfi,j / minimize sum of costsflows
• ?j(fi,j - fj,i) qi ? i?N / each vertex has a
  flow
• li,j ? fi,j ? ui,j / bounded as usual

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Dual of problem out of kilter
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Complementary slackness

• fi,j gt li,j ? xi,j 0 xi,j gt 0 ? fi,j li,j
• fi,j lt li,j ? yi,j 0 yi,j gt 0 ? fi,j ui,j
• Combining, we have
• ?i - ? j lt ci,j ? xi,j gt 0 ? fi,j li,j
• ?i - ? j gt ci,j ? yi,j gt 0 ? fi,j ui,j
• lij lt fij lt uij ? xi,j yi,j 0 ? ?i - ? j
  ci,j

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Modified problem statement

• min ?ij ci,jfi,j / minimize sum of costsflows
• ?j(fi,j - fj,i) 0 ? i?N / flow conservation
• li,j ? fi,j ? ui,j / bounded as usual
• flows are 0 rather than qi, without loss

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Modified circulation problem

• Circulation polytope P(l, u)
• Bounded polyhedron around feasible solutions
• Find a feasible circulation
• No or many feasible circulations
• Strongly polynomial algorithm known
• Minimum cost circulation highest cost C
• Find minimum cost circulation
• Out-of-kilter is O(A3C)
• Edmonds-Karp scaling makes it log C instead.
• Our challenge is to be independent of C

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Tight arc definition

• An arc (i,j) is tight if lij fij uij
• - There may be no tight arcs to start with
• - Goal is to increase number of tight arcs by
  one each iteration will be proven
• - The set of tight arcs is denoted T(l, u)

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Equivalent cost functions

• c and c' are (l,u)-equivalent if there is a
  potential function ? such that for each non-tight
  edge (i,j), we have
• c'(i,j)c(i,j)?i - ?j

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Lemma 2.1

• Let c and c' be (l,u)-equivalent cost functions
• A feasible circulation f is c-minimal if and only
  if it is c'-minimal
• We rely on the fact that f is a circulation, and
  that the peeling process can be used.

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Proof

• The middle term is 0 because it is a circulation,
  and the second term can be peeled off to simple
  cycle flows. These sum to zero as well, being
  cycle circulations.
• Thus, c'f cf is a constant independent of f.

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Special (l,u)-equivalent cost fn.

• So, it doesnt matter what (l,u)-eqivalent cost
  function is used to solve the minimization.
• One with handy properties lies in the subspace
• Qf?REf is a circulation and f(e)0 for tight
  e
• We say such a cost function fits P(l, u).

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Almost complementary slackness

• For some ?gt0, we have

?i-?j ci,j
2?
li,j
fi,j
ui,j
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Main idea contributed by Tardos

• Let ?gt0. Let (f,?) be functions such that
• ci,j - ?i ?j gt ? ?fi,j li,j
• ci,j - ?i ?j lt -? ?fi,j ui,j
• That is, outside the small box implies on the
  upper or lower bounds lines.
• Such functions satisfy almost complementary
  slackness.

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Lemma 2.2

• Let f be a feasible circulation, c' a cost
  function, ? a potential, and ?gt0 a real number
  such that
• If c'(i,j) ?(i)- ?(j) ? ? then f(i,j) l(i,j),
• If c'(i,j) ?(i)- ?(j) ? -? then f(i,j) u(i,j).
• Then for any c'-minimal circulation f'
• c'(i,j) ?(i)- ?(j) ? N? implies f(i,j)
  f'(i,j).

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Proof

• We can assume f'?f, otherwise reverse everything
• Suppose c'i0,j0 ?i0 - ?j0 ? N?
• If the inner block is lt -?, f' ? fu ?f', so
  instead let c'i0,j0 ?i0 - ?j0 ? N? hence
  fi,jli,j.
• Thus f'-f is nonnegative, and ? a cycle C
  strictly positive. Decrease every edge by the
  smallest value on that cycle.

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Proof (cont.)

• For every edge of C, fi,j lt f'i,j ? ui,j
• so therefore, f'i,j ?i - ?j ? -?.
• The cost of this new circulation - ? is
• c'f?c'f' - ??f'i,j(i,j)?C c'f' -
  ??f'i,jpi-pj(i,j)?C
• ?c'f' - ?(f'i0,j0pi0-pj0-(C-1) ?)
• ?c'f' - ?(N ? - (N-1) ?), because CltN and
  ?gt0.
• c'f' - ? ? lt c'f'
• But, f' was supposed to be optimal.
  Contradiction.

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The algortihm

• Overview Check if the cost function is (l,u)
  equivalent to the 0 vector. Stop if so.
  Otherwise, find the cost function which fits
  P(l,u) and is (l,u) equivalent to c. Apply
  out-of-kilter with this new cost function to find
  new tight edge(s) and set l,u to f for these.
• Then repeat until equivalent to the 0-vector.
• At the end, all solutions are feasible.

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The algorithm (continued)

• Check for (l,u)-equivalence to the 0 vector.
• Each iterative step, find c projection of c onto
  the subspace with c0 on tight edges.
• Let E be the set of all edges not tight.
• c I-Et(EEt)-1Ec
• c' c?kN?A for kmax element of c.
• Finally, take the ceiling of each element of c'.

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Algorithm (cont)

• 2b. Use the out-of-kilter to find the potentials
  and flows.
• 3. Set the new bounds accordingly
• If c'i,j pi pj lt N, leave l,u alone
• otherwise, set them to fi,j

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Example

• f,c,l,u

Node-arc incidence matrix E
2,2,2,2
a
b
2,2,2,3
-3,0,3,4
1,1,1,4
1,1,1,4
d
c
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Example (cont)

• We use the LaGrange method from before
• Subtract result from identity matrix
• Aij1/4 for all i,j
• Scale by 4?5
• Result is c'0,9,9,9,9

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• f,c,l,u

2,2,2,2
a
b
2,2,2,3
-3,0,3,4
1,1,1,4
1,1,1,4
d
c
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New potentials based on costs

• f,c,l,u

2,0,2,2
a
b
9
9
2,9,2,3
-3,9,3,4
1,9,1,4
1,9,1,4
d
c
0
0
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Select new l,u

• If c'i,j-?i?j lt N, l'i,jli,j and u'i,j
  ui,j
• Else l'i,j u'i,j fi,j.
• The Else arcs are newly created tight arcs.
• Theorem there is at least one Else arc.

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Proof

• We show that there is at least one more tight arc
  at each step. That is, there is at least one
  (i,j) that is not tight with the old bounds, such
  that c'i,j-?i?j gt N is true.
• Let c c'i,j-?i?j if (i,j) wasnt tight before
• 0 otherwise.
• We show max(i,j)?Ac i,j ? N

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Proof (cont)

• Recall that c' fits P(l,u), and that it is a
  circulation with c'i,j0 if (i,j) was already
  tight.
• Claim c' perpendicular to (c- c').
• Proof in notes product sums to 0 orthogonal.
• So they form a right triangle, and the
  square-root of A from earlier falls out leaving
  maxci,j ? N.

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New bounds based on potentials
2,0,2,2

• d-c has exceeded bounds!
• 9 - ?d ?c 9 gt 4
• d-c sets lu1.
• It is now tight.

a
b
9
9
2,9,2,3
-3,9,3,4
1,9,1,4
1,9,1,1
d
c
0
0
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Final result

• Iterative step lather, rinse, repeat

2,2,2,2
a
b

• All arcs are tight
• All solutions are given as new l,u
• In this case, only one solution is optimal

2,2,2,2
-3,0,3,3
1,1,1,1
1,1,1,1
d
c