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| == Flow and Cut== | | == Problem 1== |
| | Prove the following identity: |
| | *<math>\sum_{k=1}^n k{n\choose k}= n2^{n-1}</math>. |
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| |
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| === Flows ===
| | (Hint: Use double counting.) |
| An instance of the maximum flow problem consists of:
| |
| * a directed graph <math>G(V,E)</math>;
| |
| * two distinguished vertices <math>s</math> (the '''source''') and <math>t</math> (the '''sink'''), where the in-degree of <math>s</math> and the out-degree of <math>t</math> are both 0;
| |
| * the '''capacity function''' <math>c:E\rightarrow\mathbb{R}^+</math> which associates each directed edge <math>(u,v)\in E</math> a nonnegative real number <math>c_{uv}</math> called the '''capacity''' of the edge.
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| The quadruple <math>(G,c,s,t)</math> is called a '''flow network'''.
| | == Problem 2 == |
| | (Erdős-Spencer 1974) |
|
| |
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| A function <math>f:E\rightarrow\mathbb{R}^+</math> is called a '''flow''' (to be specific an '''<math>s</math>-<math>t</math> flow''') in the network <math>G(V,E)</math> if it satisfies:
| | Let <math>n</math> coins of weights 0 and 1 be given. We are also given a scale with which we may weigh any subset of the coins. Our goal is to determine the weights of coins (that is, to known which coins are 0 and which are 1) with the minimal number of weighings. |
| * '''Capacity constraint:''' <math>f_{uv}\le c_{uv}</math> for all <math>(u,v)\in E</math>.
| |
| * '''Conservation constraint:''' <math>\sum_{u:(u,v)\in E}f_{uv}=\sum_{w:(v,w)\in E}f_{vw}</math> for all <math>v\in V\setminus\{s,t\}</math>.
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| The '''value''' of the flow <math>f</math> is <math>\sum_{v:(s,v)\in E}f_{sv}</math>.
| | This problem can be formalized as follows: A collection <math>S_1,S_1,\ldots,S_m\subseteq [n]</math> is called '''determining''' if an arbitrary subset <math>T\subseteq[n]</math> can be uniquely determined by the cardinalities <math>|S_i\cap T|, 1\le i\le m</math>. |
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| Given a flow network, the maximum flow problem asks to find the flow of the maximum value.
| | * Prove that if there is a determining collection <math>S_1,S_1,\ldots,S_m\subseteq [n]</math>, then there is a way to determine the weights of <math>n</math> coins with <math>m</math> weighings. |
| | * Use pigeonhole principle to show that if a collection <math>S_1,S_1,\ldots,S_m\subseteq [n]</math> is determining, then it must hold that <math>m\ge \frac{n}{\log_2(n+1)}</math>. |
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| |
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| The maximum flow problem can be described as the following linear program.
| | (This gives a lower bound for the number of weighings required to determine the weights of <math>n</math> coins.) |
| :<math>
| |
| \begin{align}
| |
| \text{maximize} \quad& \sum_{v:(s,v)\in E}f_{sv}\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align} f_{uv}&\le c_{uv} &\quad& \forall (u,v)\in E\\
| |
| \sum_{u:(u,v)\in E}f_{uv}-\sum_{w:(v,w)\in E}f_{vw} &=0 &\quad& \forall v\in V\setminus\{s,t\}\\
| |
| f_{uv}&\ge 0 &\quad& \forall (u,v)\in E
| |
| \end{align}
| |
| \end{align}
| |
| </math> | |
|
| |
|
| === Cuts ===
| |
| {{Theorem|Definition|
| |
| :Let <math>(G(V,E),c,s,t)</math> be a flow network. Let <math>S\subset V</math>. We call <math>(S,\bar{S})</math> an '''<math>s</math>-<math>t</math> cut''' if <math>s\in S</math> and <math>t\not\in S</math>.
| |
| :The '''value''' of the cut (also called the '''capacity''' of the cut) is defined as <math>\sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv}</math>.
| |
| }}
| |
|
| |
|
| A fundamental fact in flow theory is that cuts always upper bound flows.
| | == Problem 3 == |
|
| |
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| {{Theorem|Lemma|
| | A set of vertices <math>D\subseteq V</math> of graph <math>G(V,E)</math> is a [http://en.wikipedia.org/wiki/Dominating_set ''dominating set''] if for every <math>v\in V</math>, it holds that <math>v\in D</math> or <math>v</math> is adjacent to a vertex in <math>D</math>. The problem of computing minimum dominating set is NP-hard. |
| :Let <math>(G(V,E),c,s,t)</math> be a flow network. Let <math>f</math> be an arbitrary flow in <math>G</math>, and let <math>(S,\bar{S})</math> be an arbitrary <math>s</math>-<math>t</math> cut. Then
| |
| ::<math>\sum_{v:(s,v)}f_{sv}\le \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv}</math>,
| |
| :that is, the value of any flow is no greater than the value of any cut.
| |
| }}
| |
| {{Proof|By the definition of <math>s</math>-<math>t</math> cut, <math>s\in S</math> and <math>t\not\in S</math>.
| |
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| Due to the conservation of flow,
| | * Prove that for every <math>d</math>-regular graph with <math>n</math> vertices, there exists a dominating set with size at most <math>\frac{n(1+\ln(d+1))}{d+1}</math>. |
| :<math>\sum_{u\in S}\left(\sum_{v:(u,v)\in E}f_{uv}-\sum_{v:(v,u)\in E}f_{vu}\right)=\sum_{v:(s,v)\in E}f_{sv}+\sum_{u\in S\setminus\{s\}}\left(\sum_{v:(u,v)\in E}f_{uv}-\sum_{v:(v,u)\in E}f_{vu}\right)=\sum_{v:(s,v)\in E}f_{sv}\,.</math>
| |
| On the other hand, summing flow over edges,
| |
| :<math>\sum_{v\in S}\left(\sum_{u:(u,v)\in E}f_{uv}-\sum_{u:(v,u)\in E}f_{vu}\right)=\sum_{u\in S,v\in S\atop (u,v)\in E}\left(f_{uv}-f_{uv}\right)+\sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu}=\sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu}\,.</math>
| |
| Therefore,
| |
| :<math>\sum_{v:(s,v)\in E}f_{sv}=\sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu}\le\sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}\le \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv}\,,</math>
| |
| }}
| |
|
| |
|
| === Augmenting paths ===
| | * Try to obtain an upper bound for the size of dominating set using Lovász Local Lemma. Is it better or worse than previous one? |
| {{Theorem|Definition (Augmenting path)|
| |
| :Let <math>f</math> be a flow in <math>G</math>. An '''augmenting path to <math>u_k</math>''' is a sequence of distinct vertices <math>P=(u_0,u_1,\cdots, u_k)</math>, such that
| |
| :* <math>u_0=s\,</math>;
| |
| :and each pair of consecutive vertices <math>u_{i}u_{i+1}\,</math> in <math>P</math> corresponds to either a '''forward edge''' <math>(u_{i},u_{i+1})\in E</math> or a '''reverse edge''' <math>(u_{i+1},u_{i})\in E</math>, and
| |
| :* <math>f(u_i,u_{i+1})<c(u_i,u_{i+1})\,</math> when <math>u_{i}u_{i+1}\,</math> corresponds to a forward edge <math>(u_{i},u_{i+1})\in E</math>, and
| |
| :* <math>f(u_{i+1},u_i)>0\,</math> when <math>u_{i}u_{i+1}\,</math> corresponds to a reverse edge <math>(u_{i+1},u_{i})\in E</math>.
| |
| :If <math>u_k=t\,</math>, we simply call <math>P</math> an '''augmenting path'''.
| |
| }}
| |
|
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|
| Let <math>f</math> be a flow in <math>G</math>. Suppose there is an augmenting path <math>P=u_0u_1\cdots u_k</math>, where <math>u_0=s</math> and <math>u_k=t</math>. Let <math>\epsilon>0</math> be a positive constant satisfying | | == Problem 4 == |
| *<math>\epsilon \le c(u_{i},u_{i+1})-f(u_i,u_{i+1})</math> for all forward edges <math>(u_{i},u_{i+1})\in E</math> in <math>P</math>;
| | Let <math>H(W,F)</math> be a graph and <math>n>|W|</math> be an integer. It is known that for some graph <math>G(V,E)</math> such that <math>|V|=n</math>, <math>|E|=m</math>, <math>G</math> does not contain <math>H</math> as a subgraph. Prove that for <math>k>\frac{n^2\ln n}{m}</math>, there is an edge <math>k</math>-coloring for <math>K_n</math> that <math>K_n</math> contains no monochromatic <math>H</math>. |
| *<math>\epsilon \le f(u_{i+1},u_i)</math> for all reverse edges <math>(u_{i+1},u_i)\in E</math> in <math>P</math>.
| |
| Due to the definition of augmenting path, we can always find such a positive <math>\epsilon</math>.
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|
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| Increase <math>f(u_i,u_{i+1})</math> by <math>\epsilon</math> for all forward edges <math>(u_{i},u_{i+1})\in E</math> in <math>P</math> and decrease <math>f(u_{i+1},u_i)</math> by <math>\epsilon</math> for all reverse edges <math>(u_{i+1},u_i)\in E</math> in <math>P</math>. Denote the modified flow by <math>f'</math>. It can be verified that <math>f'</math> satisfies the capacity constraint and conservation constraint thus is still a valid flow. On the other hand, the value of the new flow <math>f'</math>
| | Remark: Let <math>E=\binom{V}{2}</math> be the edge set of <math>K_n</math>. "An edge <math>k</math>-coloring for <math>K_n</math>" is a mapping <math>f:E\to[k]</math>. |
| :<math>\sum_{v:(s,v)\in E}f_{sv}'=\epsilon+\sum_{v:(s,v)\in E}f_{sv}>\sum_{v:(s,v)\in E}f_{sv}</math>.
| |
| Therefore, the value of the flow can be "augmented" by adjusting the flow on the augmenting path. This immediately implies that if a flow is maximum, then there is no augmenting path. Surprisingly, the converse is also true, thus maximum flows are "characterized" by augmenting paths.
| |
|
| |
|
| {{Theorem|Lemma|
| | == Problem 5 == |
| :A flow <math>f</math> is maximum if and only if there are no augmenting paths.
| |
| }}
| |
| {{Proof|We have already proved the "only if" direction above. Now we prove the "if" direction.
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| Let <math>S=\{u\in V\mid \exists\text{an augmenting path to }u\}</math>. Clearly <math>s\in S</math>, and since there is no augmenting path <math>t\not\in S</math>. Therefore, <math>(S,\bar{S})</math> defines an <math>s</math>-<math>t</math> cut. | | Let <math>G(V,E)</math> be a cycle of length <math>k\cdot n</math> and let <math>V=V_1\cup V_2\cup\dots V_n</math> be a partition of its <math>k\cdot n</math> vertices into <math>n</math> pairwise disjoint subsets, each of cardinality <math>k</math>. |
| | For <math>k\ge 11</math>, show that there must be an independent set of <math>G</math> containing precisely one vertex from each <math>V_i</math>. |
|
| |
|
| We claim that
| | Hint: you can use Lovász Local Lemma. |
| :<math>\sum_{v:(s,v)}f_{sv}= \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv}</math>, | |
| that is, the value of flow <math>f</math> approach the value of the cut <math>(S,\bar{S})</math> defined above. By the above lemma, this will imply that the current flow <math>f</math> is maximum.
| |
| | |
| To prove this claim, we first observe that
| |
| :<math>\sum_{v:(s,v)}f_{sv}= \sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu}</math>.
| |
| This identity is implied by the flow conservation constraint, and holds for any <math>s</math>-<math>t</math> cut <math>(S,\bar{S})</math>.
| |
| | |
| We then claim that
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| *<math>f_{uv}=c_{uv}</math> for all <math>u\in S,v\not\in S, (u,v)\in E</math>; and
| |
| *<math>f_{vu}=0</math> for all <math>u\in S,v\not\in S, (v,u)\in E</math>.
| |
| If otherwise, then the augmenting path to <math>u</math> apending <math>uv</math> becomes a new augmenting path to <math>v</math>, which contradicts that <math>S</math> includes all vertices to which there exist augmenting paths.
| |
| | |
| Therefore,
| |
| :<math>\sum_{v:(s,v)}f_{sv}= \sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu} = \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv}</math>.
| |
| As discussed above, this proves the theorem.
| |
| }}
| |
| | |
| == Max-Flow Min-Cut ==
| |
| | |
| === The max-flow min-cut theorem ===
| |
| {{Theorem|Max-Flow Min-Cut Theorem|
| |
| :In a flow network, the maximum value of any <math>s</math>-<math>t</math> flow equals the minimum value of any <math>s</math>-<math>t</math> cut.
| |
| }}
| |
| | |
| {{Proof|
| |
| Let <math>f</math> be a flow with maximum value, so there is no augmenting path.
| |
| | |
| Again, let
| |
| :<math>S=\{u\in V\mid \exists\text{an augmenting path to }u\}</math>.
| |
| As proved above, <math>(S,\bar{S})</math> forms an <math>s</math>-<math>t</math> cut, and
| |
| :<math>\sum_{v:(s,v)}f_{sv}= \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv}</math>,
| |
| that is, the value of flow <math>f</math> equals the value of cut <math>(S,\bar{S})</math>.
| |
| | |
| Since we know that all <math>s</math>-<math>t</math> flows are not greater than any <math>s</math>-<math>t</math> cut, the value of flow <math>f</math> equals the minimum value of any <math>s</math>-<math>t</math> cut.
| |
| }}
| |
| | |
| === Flow Integrality Theorem ===
| |
| {{Theorem|Flow Integrality Theorem|
| |
| :Let <math>(G,c,s,t)</math> be a flow network with integral capacity <math>c</math>. There exists an integral flow which is maximum.
| |
| }}
| |
| {{Proof|
| |
| Let <math>f</math> be an integral flow of maximum value. If there is an augmenting path, since both <math>c</math> and <math>f</math> are integral, a new flow can be constructed of value 1+the value of <math>f</math>, contradicting that <math>f</math> is maximum over all integral flows. Therefore, there is no augmenting path, which means that <math>f</math> is maximum over all flows, integral or not.
| |
| }}
| |
| | |
| === Applications: Menger's theorem ===
| |
| Given an undirected graph <math>G(V,E)</math> and two distinct vertices <math>s,t\in V</math>, a set of edges <math>C\subseteq E</math> is called an '''<math>s</math>-<math>t</math> cut''', if deleting edges in <math>C</math> disconnects <math>s</math> and <math>t</math>.
| |
| | |
| A simple path from <math>s</math> to <math>t</math> is called an '''<math>s</math>-<math>t</math> path'''. Two paths are said to be '''edge-disjoint''' if they do not share any edge.
| |
| | |
| {{Theorem|Theorem (Menger 1927)|
| |
| :Let <math>G(V,E)</math> be an arbitrary undirected graph and <math>s,t\in V</math> be two distinct vertices. The minimum size of any <math>s</math>-<math>t</math> cut equals the maximum number of edge-disjoint <math>s</math>-<math>t</math> paths.
| |
| }}
| |
| {{proof|
| |
| Construct a directed graph <math>G'(V,E')</math> from <math>G(V,E)</math> as follows: replace every undirected edge <math>uv\in E</math> that <math>s,t\not\in\{u,v\}</math> by two directed edges <math>(u,v)</math> and <math>(v,u)</math>; replace every undirected edge <math>sv\in E</math> by <math>(s,v)</math>, and very undirected edge <math>vt\in E</math> by <math>(v,t)</math>. Then assign every directed edge with capacity 1.
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| It is easy to verify that in the flow network constructed as above, the followings hold:
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| *An integral <math>s</math>-<math>t</math> flow corresponds to a set of <math>s</math>-<math>t</math> paths in the undirected graph <math>G</math>, where the value of the flow is the number of <math>s</math>-<math>t</math> paths.
| |
| *An <math>s</math>-<math>t</math> cut in the flow network corresponds to an <math>s</math>-<math>t</math> cut in the undirected graph <math>G</math> with the same value.
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| The Menger's theorem follows as a direct consequence of the max-flow min-cut theorem.
| |
| }}
| |
| | |
| === Applications: König-Egerváry theorem ===
| |
| Let <math>G(V,E)</math> be a graph. An edge set <math>M\subseteq E</math> is called a '''matching''' if no edge in <math>M</math> shares any vertex. A vertex set <math>C\subseteq V</math> is called a '''vertex cover''' if for any edge <math>uv\in E</math>, either <math>u\in C</math> or <math>v\in C</math>.
| |
| | |
| {{Theorem|Theorem (König 1936)|
| |
| :In any bipartite graph <math>G(V_1,V_2,E)</math>, the size of a ''maximum'' matching equals the size of a ''minimum'' vertex cover.
| |
| }}
| |
| We now show how a reduction of bipartite matchings to flows.
| |
| | |
| Construct a flow network <math>(G'(V,E'),c,s,t)</math> as follows:
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| * <math>V=V_1\cup V_2\cup\{s,t\}</math> where <math>s</math> and <math>t</math> are two new vertices.
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| * For ever <math>uv\in E</math>, add <math>(u,v)</math> to <math>E'</math>; for every <math>u\in V_1</math>, add <math>(s,u)</math> to <math>E'</math>; and for every <math>v\in V_2</math>, add <math>(v,t)</math> to <math>E'</math>.
| |
| * Let <math>c_{su}=1</math> for every <math>u\in V_1</math> and <math>c_{vt}=1</math> for every <math>v\in V_2</math>. Let <math>c_{uv}=\infty</math> for every bipartite edges <math>(u,v)</math>.
| |
| | |
| {{Theorem|Lemma|
| |
| :The size of a maximum matching in <math>G</math> is equal to the value of a maximum <math>s</math>-<math>t</math> flow in <math>G'</math>.
| |
| }}
| |
| {{proof|
| |
| Given an integral <math>s</math>-<math>t</math> flow <math>f</math> in <math>G'</math>, let <math>M=\{uv\in E\mid f_{uv}=1\}</math>. Then <math>M</math> must be a matching since for every <math>u\in V_1</math>. To see this, observe that there is at most one <math>v\in V_2</math> that <math>f_{uv}=1</math>, because of that <math>f_{su}\le c_{su}=1</math> and conservation of flows. The same holds for vertices in <math>V_2</math> by the same argument. Therefore, each flow corresponds to a matching.
| |
| | |
| Given a matching <math>M</math> in bipartite graph <math>G</math>, define an integral flow <math>f</math> as such: for <math>uv\in E</math>, <math>f_{uv}=1</math> if <math>uv\in M</math> and <math>f_{uv}=0</math> if otherwise; for <math>u\in V_1</math>, <math>f_{su}=1</math> if <math>uv\in M</math> for some <math>v</math> and <math>f_{su}=0</math> if otherwise; for <math>v\in V_2</math>, <math>f_{vt}=1</math> if <math>uv\in M</math> for some <math>u</math> and <math>f_{vt}=0</math> if otherwise.
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| It is easy to check that <math>f</math> is valid <math>s</math>-<math>t</math> flow in <math>G'</math>. Therefore, there is an one-one correspondence between flows in <math>G'</math> and matchings in <math>G</math>. The lemma follows naturally.
| |
| }}
| |
| | |
| We then establish a correspondence between <math>s</math>-<math>t</math> cuts in <math>G'</math> and vertex covers in <math>G</math>.
| |
| | |
| Suppose <math>(S,\bar{S})</math> is an <math>s</math>-<math>t</math> cut in <math>G'</math>.
| |
| {{Theorem|Lemma|
| |
| :The size of a minimum vertex cover in <math>G</math> is equal to the value of a minimum <math>s</math>-<math>t</math> cut in <math>G'</math>.
| |
| }}
| |
| {{proof|
| |
| Let <math>(S,\bar{S})</math> be an <math>s</math>-<math>t</math> cut of minimum capacity in <math>G'</math>. Then <math>\sum_{u\in S, v\not\in S\atop (u,v)\in E'}c_{uv}</math> must be finite since <math>S=\{s\}</math> gives us an <math>s</math>-<math>t</math> cut whose capacity is <math>|V_1|</math> which is finite. Therefore, no edge <math>uv\in E</math> has <math>u\in V_1\cap S</math> and <math>v\in V_2\setminus S</math>, i.e., for all <math>uv\in E</math>, either <math>u\in V_1\setminus S</math> or <math>v\in V_2\cap S</math>. Therefore, <math>(V_1\setminus S)\cup(V_2\cap S)</math> is a vertex cover in <math>G</math>, whose size is
| |
| :<math>|(V_1\setminus S)\cup(V_2\cap S)|=|V_1\setminus S|+|V_2\cap S|=\sum_{u\in V_1\setminus S}c_{su}+\sum_{v\in V_2\cap S}c_{ut}=\sum_{u\in S,v\not\in S\atop (u,v)\in E'}c_{uv}</math>.
| |
| The last term is the capacity of the minimum <math>s</math>-<math>t</math> cut <math>(S,\bar{S})</math>.
| |
| }}
| |
| | |
| The König-Egerváry theorem then holds as a consequence of the max-flow min-cut theorem.
| |
| | |
| == Duality ==
| |
| | |
| Consider the following LP:
| |
| :<math>
| |
| \begin{align}
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| \text{minimize} && 7x_1+x_2+5x_3\\
| |
| \text{subject to} &&
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| x_1-x_2+3x_3 &\ge 10\\
| |
| &&
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| 5x_1-2x_2-x_3 &\ge 6\\
| |
| && x_1,x_2,x_3 &\ge 0
| |
| \end{align}
| |
| </math>
| |
| | |
| Let <math>OPT</math> be the value of the optimal solution. We want to estimate the upper and lower bound of <math>OPT</math>.
| |
| | |
| Since <math>OPT</math> is the minimum over the feasible set, every feasible solution forms an upper bound for <math>OPT</math>. For example <math>\boldsymbol{x}=(2,1,3)</math> is a feasible solution, thus <math>OPT\le 7\cdot 2+1+5\cdot 3=30</math>.
| |
| | |
| For the lower bound, the optimal solution must satisfy the two constraints:
| |
| :<math>
| |
| \begin{align}
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| x_1-x_2+3x_3 &\ge 10,\\
| |
| 5x_1-2x_2-x_3 &\ge 6.\\
| |
| \end{align}
| |
| </math>
| |
| Since the <math>x_i</math>'s are restricted to be nonnegative, term-by-term comparison of coefficients shows that
| |
| :<math>7x_1+x_2+5x_3\ge(x_1-x_2+3x_3)+(5x_1-2x_2-x_3)\ge 16.</math>
| |
| The idea behind this lower bound process is that we are finding suitable nonnegative multipliers (in the above case the multipliers are all 1s) for the constraints so that when we take their sum, the coefficient of each <math>x_i</math> in the sum is dominated by the coefficient in the objective function. It is important to ensure that the multipliers are nonnegative, so they do not reverse the direction of the constraint inequality.
| |
| | |
| To find the best lower bound, we need to choose the multipliers in such a way that the sum is as large as possible. Interestingly, the problem of finding the best lower bound can be formulated as another LP:
| |
| | |
| ::<math>
| |
| \begin{align}
| |
| \text{maximize} && 10y_1+6y_2\\
| |
| \text{subject to} &&
| |
| y_1+5y_2 &\le 7\\
| |
| &&
| |
| -y_1+2y_2 &\le 1\\
| |
| &&3y_1-y_2 &\le 5\\
| |
| && y_1,y_2&\ge 0
| |
| \end{align}
| |
| </math>
| |
| | |
| Here <math>y_1</math> and <math>y_2</math> were chosen to be nonnegative multipliers for the first and the second constraint, respectively. We call the first LP the '''primal program''' and the second LP the '''dual program'''. By definition, every feasible solution to the dual program gives a lower bound for the primal program.
| |
| | |
| === LP duality ===
| |
| Given an LP in canonical form, called the '''primal''' LP:
| |
| :<math>
| |
| \begin{align}
| |
| \text{minimize} && \boldsymbol{c}^T\boldsymbol{x}\\
| |
| \text{subject to} &&
| |
| A\boldsymbol{x} &\ge\boldsymbol{b}\\
| |
| && \boldsymbol{x} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| | |
| the '''dual''' LP is defined as follows:
| |
| :<math>
| |
| \begin{align}
| |
| \text{maximum} && \boldsymbol{b}^T\boldsymbol{y}\\
| |
| \text{subject to} &&
| |
| A^T\boldsymbol{y} &\le\boldsymbol{c}\\
| |
| && \boldsymbol{y} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| | |
| We then give some examples.
| |
| | |
| ;Surviving problem (diet problem)
| |
| Let us consider the surviving problem. Suppose we have <math>n</math> types of natural food, each containing up to <math>m</math> types of vitamins. The <math>j</math>th food has <math>a_{ij}</math> amount of vitamin <math>i</math>, and the price of the <math>j</math>th food is <math>c_j</math>. We need to consume <math>b_i</math> amount of vitamin <math>i</math> for each <math>1\le i\le m</math> to keep a good health. We want to minimize the total costs of food while keeping healthy. The problem can be formalized as the following LP:
| |
| :<math>
| |
| \begin{align}
| |
| \text{minimize} \quad& c_1x_1+c_2x_2+\cdots+c_nx_n\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align} a_{i1}x_{1}+a_{i2}x_{2}+\cdots+a_{in}x_{n} &\le b_{i} &\quad& \forall 1\le i\le m\\
| |
| x_{j}&\ge 0 &\quad& \forall 1\le j\le n
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| The dual LP is
| |
| :<math>
| |
| \begin{align}
| |
| \text{maximize} \quad& b_1y_1+b_2y_2+\cdots+b_ny_m\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align} a_{1j}y_{1}+a_{2j}y_{2}+\cdots+a_{mj}y_{m} &\le c_{j} &\quad& \forall 1\le j\le n\\
| |
| y_{i}&\ge 0 &\quad& \forall 1\le i\le m
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| The problem can be interpreted as follows: A food company produces <math>m</math> types of vitamin pills. The company wants to design a pricing system such that
| |
| * The vitamin <math>i</math> has a nonnegative price <math>y_i</math>.
| |
| * The price system should be competitive to any natural food. A costumer cannot replace the vitamins by any natural food and get a cheaper price, that is, <math>\sum_{i=1}^my_ja_{ij}\le c_j</math> for any <math>1\le j\le n</math>.
| |
| * The company wants to find the maximal profit, assuming that the customer only buy exactly the necessary amount of vitamins (<math>b_i</math> for vitamin <math>i</math>).
| |
| | |
| ;Maximum flow problem
| |
| In the last lecture, we defined the maximum flow problem, whose LP is
| |
| :<math>
| |
| \begin{align}
| |
| \text{maximize} \quad& \sum_{v:(s,v)\in E}f_{sv}\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align} f_{uv}&\le c_{uv} &\quad& \forall (u,v)\in E\\
| |
| \sum_{u:(u,v)\in E}f_{uv}-\sum_{w:(v,w)\in E}f_{vw} &=0 &\quad& \forall v\in V\setminus\{s,t\}\\
| |
| f_{uv}&\ge 0 &\quad& \forall (u,v)\in E
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| where directed graph <math>G(V,E)</math> is the flow network, <math>s\in V</math> is the source, <math>t\in V</math> is the sink, and <math>c_{uv}</math> is the capacity of directed edge <math>(u,v)\in E</math>.
| |
| | |
| We add a new edge from <math>t</math> to <math>s</math> to <math>E</math>, and let the capacity be <math>c_{ts}=\infty</math>. Let <math>E'</math> be the new edge set. The LP for the max-flow problem can be rewritten as:
| |
| :<math>
| |
| \begin{align}
| |
| \text{maximize} \quad& f_{ts}\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align} f_{uv}&\le c_{uv} &\quad& \forall (u,v)\in E\\
| |
| \sum_{u:(u,v)\in E'}f_{uv}-\sum_{w:(v,w)\in E'}f_{vw} &\le0 &\quad& \forall v\in V\\
| |
| f_{uv}&\ge 0 &\quad& \forall (u,v)\in E'
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| The second set of inequalities seem weaker than the original conservation constraint of flows, however, if this inequality holds at every node, then in fact it must be satisfied with equality at every node, thereby implying the flow conservation.
| |
| | |
| To obtain the dual program we introduce variables <math>d_{uv}</math> and <math>p_v</math> corresponding to the two types of inequalities in the primal. The dual LP is:
| |
| :<math>
| |
| \begin{align}
| |
| \text{minimize} \quad& \sum_{(u,v)\in E}c_{uv}d_{uv}\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align} d_{uv}-p_u+p_v &\ge 0 &\quad& \forall (u,v)\in E\\
| |
| p_s-p_t &\ge1 \\
| |
| d_{uv} &\ge 0 &\quad& \forall (u,v)\in E\\
| |
| p_v&\ge 0 &\quad& \forall v\in V
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| It is more helpful to consider its integer version:
| |
| :<math>
| |
| \begin{align}
| |
| \text{minimize} \quad& \sum_{(u,v)\in E}c_{uv}d_{uv}\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align} d_{uv}-p_u+p_v &\ge 0 &\quad& \forall (u,v)\in E\\
| |
| p_s-p_t &\ge1 \\
| |
| d_{uv} &\in\{0,1\} &\quad& \forall (u,v)\in E\\
| |
| p_v&\in\{0,1\} &\quad& \forall v\in V
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| In the last lecture, we know that the LP for max-flow is totally unimordular, so is this dual LP, therefore the optimal solutions to the integer program are the optimal solutions to the LP.
| |
| | |
| The variables <math>p_v</math> defines a bipartition of vertex set <math>V</math>. Let <math>S=\{v\in V\mid p_v=1\}</math>. The complement <math>\bar{S}=\{v\in V\mid p_v=1\}</math>.
| |
| | |
| For 0/1-valued variables, the only way to satisfy <math>p_s-p_t\ge1</math> is to have <math>p_s=1</math> and <math>p_t=0</math>. Therefore, <math>(S,\bar{S})</math> is an <math>s</math>-<math>t</math> cut.
| |
| | |
| In an optimal solution, <math>d_{uv}=1</math> if and only if <math>u\in S,v\in\bar{S}</math> and <math>(u,v)\in E</math>. Therefore, the objective function of an optimal solution <math>\sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv}</math> is the capacity of the minimum <math>s</math>-<math>t</math> cut <math>(S,\bar{S})</math>.
| |
| | |
| === Duality theorems ===
| |
| Let the primal LP be:
| |
| :<math>
| |
| \begin{align}
| |
| \text{minimize} && \boldsymbol{c}^T\boldsymbol{x}\\
| |
| \text{subject to} &&
| |
| A\boldsymbol{x} &\ge\boldsymbol{b}\\
| |
| && \boldsymbol{x} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| | |
| Its dual LP is:
| |
| :<math>
| |
| \begin{align}
| |
| \text{maximum} && \boldsymbol{b}^T\boldsymbol{y}\\
| |
| \text{subject to} &&
| |
| A^T\boldsymbol{y} &\le\boldsymbol{c}\\
| |
| && \boldsymbol{y} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| | |
| {{Theorem|Theorem|
| |
| : The dual of a dual is the primal.
| |
| }}
| |
| {{Proof|
| |
| The dual program can be written as the following minimization in canonical form:
| |
| :<math>
| |
| \begin{align}
| |
| \min && -\boldsymbol{b}^T\boldsymbol{y}\\
| |
| \text{s.t.} &&
| |
| -A^T\boldsymbol{y} &\ge-\boldsymbol{c}\\
| |
| && \boldsymbol{y} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| Its dual is:
| |
| :<math>
| |
| \begin{align}
| |
| \max && -\boldsymbol{c}^T\boldsymbol{x}\\
| |
| \text{s.t.} &&
| |
| -A\boldsymbol{x} &\le-\boldsymbol{b}\\
| |
| && \boldsymbol{x} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| which is equivalent to the primal:
| |
| :<math>
| |
| \begin{align}
| |
| \min && \boldsymbol{c}^T\boldsymbol{x}\\
| |
| \text{s.t.} &&
| |
| A\boldsymbol{x} &\ge\boldsymbol{b}\\
| |
| && \boldsymbol{x} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| }}
| |
| | |
| We have shown that feasible solutions of a dual program can be used to lower bound the optimum of the primal program. This is formalized by the following important theorem.
| |
| | |
| {{Theorem|Theorem (Weak duality theorem)|
| |
| :If there exists an optimal solution to the primal LP:
| |
| ::<math>
| |
| \begin{align}
| |
| \min && \boldsymbol{c}^T\boldsymbol{x}\\
| |
| \text{s.t.} &&
| |
| A\boldsymbol{x} &\ge\boldsymbol{b}\\
| |
| && \boldsymbol{x} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| :then,
| |
| ::<math>
| |
| \begin{align}
| |
| \begin{align}
| |
| \min && \boldsymbol{c}^T\boldsymbol{x}\\
| |
| \text{s.t.} &&
| |
| A\boldsymbol{x} &\ge\boldsymbol{b}\\
| |
| && \boldsymbol{x} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| &\begin{align}
| |
| \ge\\
| |
| \\
| |
| \\
| |
| \end{align}&\quad
| |
| \begin{align}
| |
| \max && \boldsymbol{b}^T\boldsymbol{y}\\
| |
| \text{s.t.} &&
| |
| A^T\boldsymbol{y} &\le\boldsymbol{c}\\
| |
| && \boldsymbol{y} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| }}
| |
| {{proof|
| |
| Let <math>\boldsymbol{x}</math> be an arbitrary feasible solution to the primal LP, and <math>\boldsymbol{y}</math> be an arbitrary feasible solution to the dual LP.
| |
| | |
| We estimate <math>\boldsymbol{y}^TA\boldsymbol{x}</math> in two ways. Recall that <math>A\boldsymbol{x} \ge\boldsymbol{b}</math> and <math>A^T\boldsymbol{y} \le\boldsymbol{c}</math>, thus
| |
| :<math>\boldsymbol{y}^T\boldsymbol{b}\le\boldsymbol{y}^TA\boldsymbol{x}\le\boldsymbol{c}^T\boldsymbol{x}</math>.
| |
| | |
| Since this holds for any feasible solutions, it must also hold for the optimal solutions.
| |
| }}
| |
| | |
| A harmonically beautiful result is that the optimums of the primal LP and its dual are equal. This is called the strong duality theorem of linear programming.
| |
| | |
| {{Theorem|Theorem (Strong duality theorem)|
| |
| :If there exists an optimal solution to the primal LP:
| |
| ::<math>
| |
| \begin{align}
| |
| \min && \boldsymbol{c}^T\boldsymbol{x}\\
| |
| \text{s.t.} &&
| |
| A\boldsymbol{x} &\ge\boldsymbol{b}\\
| |
| && \boldsymbol{x} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| </math>
| |
| :then,
| |
| ::<math>
| |
| \begin{align}
| |
| \begin{align}
| |
| \min && \boldsymbol{c}^T\boldsymbol{x}\\
| |
| \text{s.t.} &&
| |
| A\boldsymbol{x} &\ge\boldsymbol{b}\\
| |
| && \boldsymbol{x} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| &\begin{align}
| |
| =\\
| |
| \\
| |
| \\
| |
| \end{align}&\quad
| |
| \begin{align}
| |
| \max && \boldsymbol{b}^T\boldsymbol{y}\\
| |
| \text{s.t.} &&
| |
| A^T\boldsymbol{y} &\le\boldsymbol{c}\\
| |
| && \boldsymbol{y} &\ge \boldsymbol{0}
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| }}
| |
| | |
| | |
| == Unimodularity ==
| |
| | |
| === Integer Programming===
| |
| Consider the '''maximum integral flow''' problem: given as input a flow network <math>(G(V,E),c,s,t)</math> where for every <math>uv\in E</math> the capacity <math>c_{uv}</math> is integer. We want to find the integral flow <math>f:E\rightarrow\mathbb{Z}</math> with maximum value.
| |
| | |
| The mathematical programming for the problem is:
| |
| :<math>
| |
| \begin{align}
| |
| \text{maximize} \quad& \sum_{v:(s,v)\in E}f_{sv}\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \\
| |
| \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align} f_{uv}&\le c_{uv} &\quad& \forall (u,v)\in E\\
| |
| \sum_{u:(u,v)\in E}f_{uv}-\sum_{w:(v,w)\in E}f_{vw} &=0 &\quad& \forall v\in V\setminus\{s,t\}\\
| |
| f_{uv}&\in\mathbb{N} &\quad& \forall (u,v)\in E
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| where <math>\mathbb{N}</math> is the set of all nonnegative integers. Compared to the LP for the max-flow problem, we just replace the last line <math> f_{uv}\ge 0</math> with <math> f_{uv}\in\mathbb{N}</math>. The resulting optimization is called an '''integer programming (IP)''', or more specific '''integer linear programming (ILP)'''.
| |
| | |
| Due to the Flow Integrality Theorem, when capacities are integers, there must be an integral flow whose value is maximum among all flows (integral or not). This means the above IP can be efficiently solved by solving its LP-relaxation. This is usually impossible for general IPs.
| |
| | |
| Generally, an IP of canonical form is written as
| |
| :<math>
| |
| \begin{align}
| |
| \text{maximize} \quad& \boldsymbol{c}^T\boldsymbol{x}\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align}
| |
| A\boldsymbol{x} &\ge\boldsymbol{b} \\
| |
| \boldsymbol{x}&\ge \boldsymbol{0}\\
| |
| \boldsymbol{x}&\in\mathbb{Z}^n
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| | |
| Consider the '''3SAT''' problem. Each instance is a '''3CNF(conjunctive normal form)''': <math>\bigwedge_{i=1}^m(\ell_{i_1}\vee\ell_{i_2}\vee\ell_{i_3})</math>, where each <math>(\ell_{i_1}\vee\ell_{i_2}\vee\ell_{i_3})</math> is a '''clause''' and each <math>\ell_{i_r}\in\{x_{j},\neg x_{j}\mid 1\le j\le n\}</math>, called a '''literal''', is either a boolean variable or a negation of a boolean variable. We want to determine whether there exists an truth assignment of the <math>n</math> boolean variables <math>x_1,\ldots,x_n</math> such that the input formula is satisfied (i.e., is true).
| |
| | |
| The following IP solves 3SAT:
| |
| :<math>
| |
| \begin{align}
| |
| \text{maximize} \quad& \sum_{i=1}^mz_i\\
| |
| \begin{align}
| |
| \text{subject to} \\
| |
| \\
| |
| \\
| |
| \\
| |
| \end{align}
| |
| \quad &
| |
| \begin{align}
| |
| z_i &\le y_{i_1}+y_{i_2}+y_{i_3} &\quad& \forall 1\le i\le m\\
| |
| y_{i_r}&\le x_{j} &\quad& \text{if }\ell_{i_r}=x_{j} \\
| |
| y_{i_r}&\le 1-x_{j} &\quad& \text{if }\ell_{i_r}=\neg x_{j} \\
| |
| z_i, x_j&\in\{0,1\} &\quad& \forall 1\le i\le m, 1\le j\le n
| |
| \end{align}
| |
| \end{align}
| |
| </math>
| |
| Since 3SAT is NP-hard (actually it is the first problem known to be NP-hard), generally IP is NP-hard.
| |
| | |
| === Integrality of polytopes ===
| |
| A point in an <math>n</math>-dimensional space is integral if it belongs to <math>\mathbb{Z}^n</math>, i.e., if all its coordinates are integers.
| |
| | |
| A polyhedron is said to be '''integral''' if all its vertices are integral.
| |
| | |
| An easy observation is that an integer programming has the same optimal solutions as its LP-relaxation when the polyhedron defined by the LP-relaxation is integral.
| |
| | |
| {{Theorem|Theorem (Hoffman 1974)|
| |
| : If a polyhedron <math>P</math> is integral then for all integer vectors <math>\boldsymbol{c}</math> there is an optimal solution to <math>\max\{\boldsymbol{c}^T\boldsymbol{x}\mid \boldsymbol{x}\in P\}</math> which is integral.
| |
| }}
| |
| {{Proof|
| |
| There always exists an optimal solution which is a vertex in <math>P</math>. For integral <math>P</math>, all vertices are integral.
| |
| }}
| |
| | |
| === Unimodularity and total unimodularity ===
| |
| {{Theorem|Definition (Unimodularity)|
| |
| :An <math>n\times n</math> integer matrix <math>A</math> is called '''unimodular''' if <math>\det(A)=\pm1</math>.
| |
| :An <math>m\times n</math> integer matrix <math>A</math> is called '''total unimodular''' if every square submatrix <math>B</math> of <math>A</math> has <math>\det(B)\in\{1,-1,0\}</math>, that is, every square, nonsingular submatrix of <math>A</math> is unimodular.
| |
| }}
| |
| | |
| A totally unimodular matrix defines a integral convex polyhedron.
| |
| {{Theorem|Theorem|
| |
| :Let <math>A</math> be an <math>m\times n</math> integer matrix.
| |
| :If <math>A</math> is totally unimodualr, then for any integer vector <math>\boldsymbol{b}\in\mathbb{Z}^n</math> the polyhedron <math>\{\boldsymbol{x}\in\mathbb{R}^n\mid A\boldsymbol{x}=\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}\}</math> is integral.
| |
| }}
| |
| {{Proof|
| |
| Let <math>B</math> be a basis of <math>A</math>, and let <math>\boldsymbol{b}'</math> be the corresponding coordinates in <math>\boldsymbol{b}</math>. A basic solution is formed by <math>B^{-1}\boldsymbol{b}'</math> and zeros. Since <math>A</math> is totally unimodular and <math>B</math> is a basis thus nonsingular, <math>\det(B)\in\{1,-1,0\}</math>. By [http://en.wikipedia.org/wiki/Cramer's_rule Cramer's rule], <math>B^{-1}</math> has integer entries, thus <math>B^{-1}\boldsymbol{b}'</math> is integral. Therefore, any basic solution of <math>A\boldsymbol{x}=\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}</math> is integral, which means the polyhedron <math>\{\boldsymbol{x}\in\mathbb{R}^n\mid A\boldsymbol{x}=\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}\}</math> is integral.
| |
| }}
| |
| | |
| Our next result is the famous Hoffman-Kruskal theorem.
| |
| {{Theorem|Theorem (Hoffman-Kruskal 1956)|
| |
| :Let <math>A</math> be an <math>m\times n</math> integer matrix.
| |
| :If <math>A</math> is totally unimodualr, then for any integer vector <math>\boldsymbol{b}\in\mathbb{Z}^n</math> the polyhedron <math>\{\boldsymbol{x}\in\mathbb{R}^n\mid A\boldsymbol{x}\ge\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}\}</math> is integral.
| |
| }}
| |
| {{Proof|
| |
| Let <math>A'=\begin{bmatrix}A & -I\end{bmatrix}</math>. We claim that <math>A'</math> is also totally unimodular. Any square submatrix <math>B</math> of <math>A</math> can be written in the following form after permutation:
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| :<math>B=\begin{bmatrix}
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| C & 0\\
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| D & I
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| \end{bmatrix}</math>
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| where <math>C</math> is a square submatrix of <math>A</math> and <math>I</math> is identity matrix. Therefore,
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| :<math>\det(B)=\det(C)\in\{1,-1,0\}</math>,
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| thus <math>A'</math> is totally unimodular.
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| Add slack variables to transform the constraints to the standard form <math>A'\boldsymbol{z}=\boldsymbol{b},\boldsymbol{z}\ge\boldsymbol{0}</math>. The polyhedron <math>\{\boldsymbol{x}\mid A\boldsymbol{x}\ge\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}\}</math> is integral if the polyhedron <math>\{\boldsymbol{z}\mid A'\boldsymbol{z}=\boldsymbol{b}, \boldsymbol{z}\ge \boldsymbol{0}\}</math> is integral, which is implied by the total unimodularity of <math>A'\,</math>.
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| }}
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