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| Author | SHA1 | Date | |
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| 02a75eec17 | |||
| 2ba95ac69b | |||
| d37c54f387 | |||
| b95f42e73e | |||
| affe315163 |
@@ -17,9 +17,8 @@ jobs:
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- uses: http://localhost:3000/sxlxc/gitea-release-action@v1
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- uses: http://localhost:3000/sxlxc/gitea-release-action@v1
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with:
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with:
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body: ''
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body: ''
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prerelease: true
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name: PDF
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name: PDF
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token: ${{ secrets.RELEASE_TOKEN }}
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token: ${{ secrets.RELEASE_TOKEN }}
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tag_name: latest
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tag_name: latest
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files: |-
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files: |-
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./*.pdf
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./*.pdf
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60
notes.tex
60
notes.tex
@@ -9,7 +9,7 @@
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\maketitle
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\maketitle
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\section{Introduction}
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\section{Introduction}
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Geelen and Kapadia design randomized polynomial time algorithms for computing girth and cogirth of perturbed graphic matroids \cite{geelen_computing_2018}. They leave an open problem that if there are deterministic polynomial time algorithms. We solve the cogirth part using base packing techniques.
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Geelen and Kapadia design randomized polynomial time algorithms for computing girth and cogirth of perturbed graphic matroids \cite{geelen_computing_2018}. They leave an open problem for the existence of deterministic polynomial time algorithms. We solve the cogirth part using base packing techniques.
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% We want to use basepacking on perturbed graphic matroid. Basepacking works for deletion closed matroid classes with constant cogirth-packing gap.
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% We want to use basepacking on perturbed graphic matroid. Basepacking works for deletion closed matroid classes with constant cogirth-packing gap.
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A binary matroid is a low rank perturbed graphic matroid (PGM) if it has a binary representation $A+P$, where $A$ is the incidence matrix of a graph and $P$ is a binary matrix with rank at most a constant $r$.
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A binary matroid is a low rank perturbed graphic matroid (PGM) if it has a binary representation $A+P$, where $A$ is the incidence matrix of a graph and $P$ is a binary matrix with rank at most a constant $r$.
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@@ -35,10 +35,10 @@ A=
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\end{array},
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\end{array},
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\]
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\]
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where $T$ indexes $t$ new columns and $\set{s}$ indexes a new row.
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where $T$ indexes $t$ new columns and $\set{s}$ indexes a new row.
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The matroid $M(A)$ is the linear matroid on the matrix $A\in \F_2^{(V(G)+S)\times (E(G)+T)}$.
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The matroid $M(A)$ is the binary matroid on the matrix $A\in \F_2^{(V(G)+S)\times (E(G)+T)}$.
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\subsection{previous works}
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\subsection{previous works}
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The cogirth problem on $(1,t)$-signed-grafts can be considered as variation of graph min-cut under congruency constraints.
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The cogirth problem on $(1,t)$-signed-grafts can be considered as a variation of graph min-cut under congruency constraints.
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\begin{problem}[$t$-dimensional even cut, \cite{geelen_computing_2018}]
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\begin{problem}[$t$-dimensional even cut, \cite{geelen_computing_2018}]
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\label{prob:tdimevencut}
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\label{prob:tdimevencut}
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Let $G=(V,E)$ be a graph and let $\ell:V\to \F_2^{t}$ be a $t$-dimensional coloring on vertices. Given a edge set $C\subset E$ and a coloring $D\in \F_2^{1\times t}$, find a non-empty vertex set $X\subset V$ that minimizes the smaller value of the following two:
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Let $G=(V,E)$ be a graph and let $\ell:V\to \F_2^{t}$ be a $t$-dimensional coloring on vertices. Given a edge set $C\subset E$ and a coloring $D\in \F_2^{1\times t}$, find a non-empty vertex set $X\subset V$ that minimizes the smaller value of the following two:
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@@ -49,11 +49,11 @@ Let $G=(V,E)$ be a graph and let $\ell:V\to \F_2^{t}$ be a $t$-dimensional color
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\end{problem}
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\end{problem}
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Consider a special case of \autoref{prob:tdimevencut} that $\ell=D=0$ for all vectices and $C=E$. Then the $\delta(X)$ achieving the minimum value of case 2 is exactly the max-cut of $G$.
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Consider a special case of \autoref{prob:tdimevencut} that $\ell=D=0$ for all vectices and $C=E$. Then the $\delta(X)$ achieving the minimum value of case 2 is exactly the max-cut of $G$.
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This observation suggests that one cannot deal with the two cases separately.
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This observation suggests that one cannot solve the two cases separately in polynomial time.
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Another interesting special case is that $C=\emptyset$ and $D=0$. The problem becomes graph min-cut with congruency constrants,
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Another interesting special case is that $C=\emptyset$ and $D=0$. The problem becomes graph min-cut with congruency constrants,
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which is a special case of submodular function minimization under congruency constraints (SFMC) studied by Nägele \etal \cite{nagele_submodular_2019}.
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which is a special case of submodular function minimization under congruency constraints (SFMC) studied by Nägele \etal \cite{nagele_submodular_2019}.
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They show that SFMC with constant number of constraints can be solved in polynomial time if the modular is prime. However, the objective in case 2 of \autoref{prob:tdimevencut} is not submodular so their method does not generalize.
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They show that SFMC with constant number of constraints can be solved deterministically in polynomial time if the modular is prime. However, the objective in case 2 of \autoref{prob:tdimevencut} is not submodular so their method does not generalize.
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\subsection{proof outline}
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\subsection{proof outline}
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We prove that if $M(A)$ is a binary matroid on $(1,t)$-signed-graft and $t$ is a constant then $M(A)/T$ has constant cogirth-packing gap,
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We prove that if $M(A)$ is a binary matroid on $(1,t)$-signed-graft and $t$ is a constant then $M(A)/T$ has constant cogirth-packing gap,
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@@ -64,21 +64,36 @@ We first prove that for any binary matroid with constant gap, adding another row
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Contracting a set $T$ from a matroid $M$ can be done by contracting elements in $T$ one by one, $M/e_1/\dots/e_t$. We fix an arbitrary order for elements in $T$ and let $T_i$ be the set of first $i$ elements. We prove that if $(M/T_i)|E(G)$ has constant gap then so does $(M/T_{i+1})|E(G)$.
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Contracting a set $T$ from a matroid $M$ can be done by contracting elements in $T$ one by one, $M/e_1/\dots/e_t$. We fix an arbitrary order for elements in $T$ and let $T_i$ be the set of first $i$ elements. We prove that if $(M/T_i)|E(G)$ has constant gap then so does $(M/T_{i+1})|E(G)$.
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\section{Proof of constant gap}
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\section{Proof of constant gap}
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\paragraph{notations and useful lemmas}
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Let $M=(E,\mathcal I)$ be a matroid. We write $\lambda(M)$ for the cogirth of $M$ and write $\sigma(M)$ for the strength of $M$.
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It is known that the strength of $M$ equals the maximum fractional base packing number and $\sigma(M)=\min_{F\subset E} \frac{|E-F|}{r(E)-r(F)}$.
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In the proof we frequently use the following result on binary matroids.
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The \emph{cocircuit space} of a binary matroid $M=(E,\mathcal I)$ is the subspace of $\F_2^{E}$ that are generated by the incidence vectors of cocircuits of $M$.
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\begin{lemma}[\cite{Oxley_2011}, Chapter 9]
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Let $M=(E,\mathcal I)$ be a binary matroid and let $A\in \F_2^{r(M)\times |E|}$ be its binary representation.
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Then the row space of $A$ equals the cocircuit space of $M$.
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Any non-zero vector in the cocircuit space represents a dependent set of $M$.
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\end{lemma}
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We first show that adding one additional row keeps the gap constant.
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We first show that adding one additional row keeps the gap constant.
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\begin{lemma}
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\begin{lemma}
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Let $M$ be a binary matroid with binary representation $B\in \F_2^{n\times m}$. If $M$ has constant gap, then $M\left(\begin{bmatrix}B\\ \sigma\end{bmatrix}\right)$ has constant gap for any row vector $\sigma\in \F_2^{m}$.
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Let $M$ be a binary matroid with binary representation $B\in \F_2^{n\times m}$. If $M$ has constant gap, then $M\left(\begin{bmatrix}B\\ \sigma\end{bmatrix}\right)$ has constant gap for any row vector $\sigma\in \F_2^{m}$.
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\end{lemma}
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\end{lemma}
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\begin{proof}
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\begin{proof}
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Let $M'$ be $M\left(\begin{bmatrix}B\\ \sigma\end{bmatrix}\right)$ and let the ground set be $E$.
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Let $M'$ be $M\left(\begin{bmatrix}B\\ \sigma\end{bmatrix}\right)$ and let the ground set be $E$.
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$\lambda(M')\leq \lambda(M)$ (row space).
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We have $\lambda(M')\leq \lambda(M)$ since the row space becomes larger.
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Let $F^*\in \argmin_{F\subset E} \frac{|E-F|}{r'(E)-r'(F)}$.
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Let $F^*\in \argmin_{F\subset E} \frac{|E-F|}{r'(E)-r'(F)}$. Let $r$ be the rank of $M$ and let $r'$ be the rank of $M'$.
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We can assume that $r(E)-r(F^*)\geq 1$ since otherwise we have $r'(E)-r'(F^*)\leq 1$ which implies the gap of $M'$ is 1.
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\begin{equation*}
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\begin{equation*}
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\begin{aligned}
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\begin{aligned}
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\sigma(M') &= \frac{|E-F^*|}{r'(E)-r'(F^*)}\\
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\sigma(M') &= \frac{|E-F^*|}{r'(E)-r'(F^*)}
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&\geq \frac{|E-F^*|}{r(E)+1-r(F^*)}\\
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\geq \frac{|E-F^*|}{r(E)+1-r(F^*)}\\
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&\geq \frac{|E-F^*|}{2(r(E)-r(F^*)}\\
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&\geq \frac{|E-F^*|}{2(r(E)-r(F^*))}
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&\geq \frac{\sigma(M)}{2}
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\geq \frac{\sigma(M)}{2}
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\end{aligned}
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\end{aligned}
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\end{equation*}
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\end{equation*}
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Thus we have $\frac{\lambda(M')}{\sigma(M')}\leq \frac{2\lambda(M)}{\sigma(M)}$.
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Thus we have $\frac{\lambda(M')}{\sigma(M')}\leq \frac{2\lambda(M)}{\sigma(M)}$.
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@@ -93,14 +108,16 @@ If $M$ and deletion minors of $M$ have constant gap, then $M(A')/\tau$ has const
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\paragraph{Graphic case}
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\paragraph{Graphic case}
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Before proving this lemma consider an easier case where $M$ is graphic. The new element $\tau$ identifies a vertex set $T$.
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Before proving this lemma consider an easier case where $M$ is graphic. The new element $\tau$ identifies a vertex set $T$.
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A cut $\delta(X)$ is even if $|X\cap T|$ is even. The minimum cocircuit of $M(A')/\tau$ is the minimum even cut in $(G,T)$.
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Note that for a cut $\delta(X)$ one can sum up the rows labelled by $v\in X$ in the incidence matrix and get an indicator vector of $\delta(X)$.
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A cut $\delta(X)$ is even if $|X\cap T|$ is even.
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Then the minimum cocircuit of $M(A')/\tau$ is the minimum even cut in $(G,T)$.
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Let $G=(V,E)$ be a graph and let $T\subset V$ be a vertex set with even size.
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An edge set $F\subset E$ is a \emph{$T$-join} if $X$ induces a subgraph in which the set of vertices with odd degree is exactly $T$.
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An cut $\delta(X)$ is a \emph{$T$-cut} if $|X\cap T|$ odd. Note that any $T$-cut hits every $T$-join. For proofs, we refer to Section~12.4 in \cite{Korte_Vygen_2018}.
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For a spanning tree $B$, let $C(B,\tau)$ be the fundamental circuit in $B\cup \set{\tau}$. Then $C(B,\tau)\cap E$ is a $T$-join, since it induces a subgraph in which the set of vertices with odd degree is exactly $T$.
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For a spanning tree $B$, let $C(B,\tau)$ be the fundamental circuit in $B\cup \set{\tau}$. Then $C(B,\tau)\cap E$ is a $T$-join, since it induces a subgraph in which the set of vertices with odd degree is exactly $T$.
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% Now we try to characterize the set of bases of $M(A')/\tau$.
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% A graph contains a $T$-join if and only if every component contains an even number of vertices in $T$.
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One can see that the set of bases is $\set{B-e}$ for all spanning tree $B$ and all $e\in C(B,\tau)\cap B$.
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One can see that the set of bases is $\set{B-e}$ for all spanning tree $B$ and all $e\in C(B,\tau)\cap B$.
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Note that the minimum base hitting set of $M(A')/t$ is the minimum even cut of $(G,T)$.
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% Note that the minimum base hitting set of $M(A')/t$ is the minimum even cut of $(G,T)$.
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To show that gap we prove the followings,
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To show that gap we prove the followings,
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@@ -115,14 +132,14 @@ The second inequality holds since the integrality gap for graph $k$-cut is 2.
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For the last inequality, we claim that $x=\chi_J + \sigma(M(A')/\tau)|_{E\setminus J}$ is a feasible solution to 3-cut LP. First it is easy to see that $x(e)\leq 1$ for any edge $e$.
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For the last inequality, we claim that $x=\chi_J + \sigma(M(A')/\tau)|_{E\setminus J}$ is a feasible solution to 3-cut LP. First it is easy to see that $x(e)\leq 1$ for any edge $e$.
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Consider the size of $J\cap B$ for a spanning tree $B$. If this is at least 2, then $x$ satisfies the 3-cut constraint for $B$; Otherwise, $B\setminus J$ should be a base of $M'/\tau$ since $T$-cut intersects every $T$-join, and we have $\sigma(M'/T)|_{B\setminus J}\geq 1$ as a constraint in the base hitting set LP of $M(A')/\tau$.
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Consider the size of $J\cap B$ for a spanning tree $B$. If this is at least 2, then $x$ satisfies the 3-cut constraint for $B$; Otherwise, $B\setminus J$ should be a base of $M'/\tau$ since $T$-cut intersects every $T$-join, and we have $\sigma(M'/T)|_{B\setminus J}\geq 1$ as a constraint in the base hitting set LP of $M(A')/\tau$.
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To finish the proof consider the connections in the minimum cut of $G$, the minimum $T$-cut of $G$ and the minimum even cut of $G$. Note that if $\delta(X)$ is a $T$-cut then $|X\cap T|$ is odd. Then the minimum cut is either an $T$-cut or an even cut.
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To finish the proof consider the connections in the minimum cut of $G$, the minimum $T$-cut of $G$ and the minimum even cut of $G$. Recall that the minimum $T$-cut is the minimum cut $\delta(X)$ with $|X\cap T|$ odd. Then the minimum cut is either a $T$-cut or an even cut.
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\begin{enumerate}
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\begin{enumerate}
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\item If $J$ is the mincut. $\lambda(M'/t)\leq 2(\lambda(G)+\sigma(M'/t))\leq 2(2\sigma(G)+\sigma(M'/t)) \leq 6\sigma(M'/t)$
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\item If $J$ is the mincut. $\lambda(M'/t)\leq 2(\lambda(G)+\sigma(M'/t))\leq 2(2\sigma(G)+\sigma(M'/t)) \leq 6\sigma(M'/t)$
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\item If $\lambda(M'/t)$ is the mincut. $\lambda(M'/t)=\lambda(G)\leq 2\sigma(G)\leq 2\sigma(M'/t)$
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\item If $\lambda(M'/t)$ is the mincut. $\lambda(M'/t)=\lambda(G)\leq 2\sigma(G)\leq 2\sigma(M'/t)$
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\end{enumerate}
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\end{enumerate}
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For either case, we have constant gap.
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For either case, we have constant gap.
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Now we generalize this idea to binary matroids. Before the proof we need some useful lemmas.
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Now we generalize this idea to binary matroids.
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Note that we need constant gap for all deletion minors of $M$ since we are going to use the gap of 2-cocycle LP of $M$.
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Note that we need constant gap for all deletion minors of $M$ since we are going to use the gap of 2-cocycle LP of $M$.
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\begin{lemma}[$(1,d)$-good implies $(k,kd)$-good]
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\begin{lemma}[$(1,d)$-good implies $(k,kd)$-good]
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@@ -132,7 +149,7 @@ Note that we need constant gap for all deletion minors of $M$ since we are going
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Instead of $T$-cut we find the minimum hitting set of $C(B,\tau)$ for all base $B$. An useful lemma is the following. We write $M+f$ for an one element extension on the binary representation of $M$.
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Instead of $T$-cut we find the minimum hitting set of $C(B,\tau)$ for all base $B$. An useful lemma is the following. We write $M+f$ for an one element extension on the binary representation of $M$.
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\begin{lemma}\label{TcutTjoin}
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\begin{lemma}\label{TcutTjoin}
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Let $M$ be a binary matroid and let $M'$ be $(M+f)/f$ for a new element $f$.
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Let $M$ be a binary matroid and let $M'$ be $(M+f)/f$ for a new element $f$.
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The cogirth of $M$ is either the minimum hitting set of $\{C(B,f)\setminus f|\forall B\}$ or the cogirth of $M'$.
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The cogirth of $M$ is either the minimum hitting set of $\{C(B,f)\setminus f|\forall B\in\mathcal B(M)\}$ or the cogirth of $M'$.
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\end{lemma}
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\end{lemma}
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\begin{proof}
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\begin{proof}
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@@ -143,7 +160,7 @@ $f$-extension can be seen as adding a new column $v$ to the binary matrix $A$. N
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Any cocycle has an indicator vector $yA$ and thus we can use a row vector $y$ to represent any cocycle.
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Any cocycle has an indicator vector $yA$ and thus we can use a row vector $y$ to represent any cocycle.
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We say a cocycle is even if $yv=0$ and odd if $yv=1$.
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We say a cocycle is even if $yv=0$ and odd if $yv=1$.
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First consider the cogirth of $M'$. It follows from definitions that the minimum hitting set of $\{B-e |\forall B ,\forall e\in B\cap C(B,f)\}$ is exactly the minimum cocircuit of $M'$, which is the minimum set in the cocircuit space with $yv=0$. So the cogirth of $M'$ is an even cocycle.
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First consider the cogirth of $M'$. It follows from definitions that the minimum hitting set of $\{B-e |\forall B\in\mathcal{B}(M),\forall e\in B\cap C(B,f)\}$ is exactly the minimum cocircuit of $M'$, which is the minimum set in the cocircuit space with $yv=0$. So the cogirth of $M'$ is the minimum even cocycle.
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On the other hand, consider a set $X$ which fails to hit all $\{C(B,f)\setminus f|\forall B\}$. Then there must be a set $F\subseteq E\setminus X$ that span the new element $f$. On matrix, this is equivalent to the existence of a indicator vector $\chi_F$ such that $A\chi_F=v$. Now we apply the following lemma which is easy to prove.
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On the other hand, consider a set $X$ which fails to hit all $\{C(B,f)\setminus f|\forall B\}$. Then there must be a set $F\subseteq E\setminus X$ that span the new element $f$. On matrix, this is equivalent to the existence of a indicator vector $\chi_F$ such that $A\chi_F=v$. Now we apply the following lemma which is easy to prove.
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% Using Farkas' lemma on $\mathbb F_2$, this implies that there dose not exists $y$ that $\supp(yA)\subset X\land yv=1$. \note{chatGPT says one can use Farkas lemma on finite field like this, need to verify.}
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% Using Farkas' lemma on $\mathbb F_2$, this implies that there dose not exists $y$ that $\supp(yA)\subset X\land yv=1$. \note{chatGPT says one can use Farkas lemma on finite field like this, need to verify.}
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@@ -151,7 +168,7 @@ On the other hand, consider a set $X$ which fails to hit all $\{C(B,f)\setminus
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Let $A$ be a $n\times m$ binary matrix and $b$ be a $n$-dimensional binary vector. Either there is a $x\in \F_2^m$ such that $Ax=b$, or there is $y\in \F_2^n$ such that $y^TA=0$ and $y^Tb=1$.
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Let $A$ be a $n\times m$ binary matrix and $b$ be a $n$-dimensional binary vector. Either there is a $x\in \F_2^m$ such that $Ax=b$, or there is $y\in \F_2^n$ such that $y^TA=0$ and $y^Tb=1$.
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\end{lemma}
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\end{lemma}
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Then $X$ fails to be a hitting set is equivalent to the fact that there does not exist $y$ satisfying $\supp(yA)\subset X\land yv=1$.
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Then $X$ fails to be a hitting set is equivalent to the fact that there does not exist $y$ satisfying $\supp(yA)\subset X\land yv=1$.
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So $X$ hits all $\{C(B,f)\setminus f|\forall B\}$ iff there is such a $y$. Minimizing $X$ pushes it to $\supp(yA)$ which is an odd cocycle. Hence, the minimum hitting set for $\{C(B,f)\setminus f|\forall B\}$ is an odd cocycle.
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So $X$ hits all $\{C(B,f)\setminus f|\forall B\}$ iff there is such a $y$. Minimizing $X$ pushes it to $\supp(yA)$ which is an odd cocycle. Hence, the minimum hitting set for $\{C(B,f)\setminus f|\forall B\}$ is the minimum odd cocycle.
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The theorem then follows directly from the fact that any cocycle is either odd or even.
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The theorem then follows directly from the fact that any cocycle is either odd or even.
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\end{proof}
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\end{proof}
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@@ -162,7 +179,6 @@ We follow the same framework as the graphic case. Let $c$ be the 2-hitting set g
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\begin{equation}\label{eq1}
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\begin{equation}\label{eq1}
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\lambda(M'/\tau)\leq \lambda_2(M)\leq c\sigma_2(M)\leq c(|J|+\sigma(M'/T)|_{E\setminus J})
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\lambda(M'/\tau)\leq \lambda_2(M)\leq c\sigma_2(M)\leq c(|J|+\sigma(M'/T)|_{E\setminus J})
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\end{equation}
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\end{equation}
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where $J$ is the minimum hitting set of $\{C(B,f)\setminus f|\forall B\}$ and $\lambda_2(M)$ ($\sigma_2(M)$) is the integral (fractional) 2-hitting set of bases of $M$.
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where $J$ is the minimum hitting set of $\{C(B,f)\setminus f|\forall B\}$ and $\lambda_2(M)$ ($\sigma_2(M)$) is the integral (fractional) 2-hitting set of bases of $M$.
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We then apply \autoref{TcutTjoin}.
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We then apply \autoref{TcutTjoin}.
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2
ref.bib
2
ref.bib
@@ -30,3 +30,5 @@
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year = {2019},
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year = {2019},
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pages = {1351--1386},
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pages = {1351--1386},
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}
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}
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@book{Oxley_2011, title={Matroid Theory}, ISBN={9780198566946}, url={https://academic.oup.com/book/34846}, DOI={10.1093/acprof:oso/9780198566946.001.0001}, publisher={Oxford University Press}, author={Oxley, James}, year={2011}, month=feb }
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@book{Korte_Vygen_2018, address={Berlin, Heidelberg}, series={Algorithms and Combinatorics}, title={Combinatorial Optimization: Theory and Algorithms}, volume={21}, rights={https://www.springer.com/tdm}, ISBN={9783662560389}, url={https://link.springer.com/10.1007/978-3-662-56039-6}, DOI={10.1007/978-3-662-56039-6}, publisher={Springer Berlin Heidelberg}, author={Korte, Bernhard and Vygen, Jens}, year={2018}, collection={Algorithms and Combinatorics}, language={en} }
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Reference in New Issue
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