Discrete Math Quiz: Bipartite Graphs And Matchings
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Bipartite Graphs And MatchingsQuestion 1 of 15

In a bipartite graph G=(XY,E)G = (X \cup Y, E), let SXS \subseteq X with S=4|S| = 4. If N(S)=3|N(S)| = 3, where N(S)N(S) denotes the neighborhood of SS in YY, and GG has a matching that saturates all vertices in XSX \setminus S, what is the maximum possible size of a matching in GG?

X1|X| - 1
X4|X| - 4
X3|X| - 3
min(X,Y)1\min(|X|, |Y|) - 1
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Discrete Math Quiz

Discrete Math Quiz: Bipartite Graphs And Matchings

Practice Bipartite Graphs And Matchings in Discrete Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Bipartite Graphs And Matchings, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

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Question 1

In a bipartite graph G=(XY,E)G = (X \cup Y, E), let SXS \subseteq X with S=4|S| = 4. If N(S)=3|N(S)| = 3, where N(S)N(S) denotes the neighborhood of SS in YY, and GG has a matching that saturates all vertices in XSX \setminus S, what is the maximum possible size of a matching in GG?

  1. X1|X| - 1 (correct answer)
  2. X4|X| - 4
  3. X3|X| - 3
  4. min(X,Y)1\min(|X|, |Y|) - 1
Explanation: When analyzing matching problems in bipartite graphs, you need to understand how Hall's Marriage Theorem and its implications work. This question tests your ability to find the maximum matching when there's a bottleneck in the graph structure. The key insight is that we have a subset SXS \subseteq X with S=4|S| = 4 vertices, but their neighborhood N(S)N(S) has only 3 vertices. This creates a fundamental limitation: at most 3 vertices from SS can be matched, since there are only 3 possible partners for them in YY. This means at least 1 vertex from SS must remain unmatched in any matching. Since GG has a matching that saturates all vertices in XSX \setminus S (meaning all vertices in XX except those in SS are matched), we can match X4|X| - 4 vertices from XSX \setminus S plus at most 3 vertices from SS. This gives us a maximum matching of size (X4)+3=X1(|X| - 4) + 3 = |X| - 1, making choice A correct. Choice B (X4|X| - 4) incorrectly assumes no vertices from SS can be matched. Choice C (X3|X| - 3) mistakenly thinks all 4 vertices in SS can be matched despite the neighborhood constraint. Choice D (min(X,Y)1\min(|X|, |Y|) - 1) applies a general upper bound without considering the specific bottleneck created by SS and N(S)N(S). Remember: In bipartite matching problems, always check for Hall's condition violations. When N(S)<S|N(S)| < |S| for some subset SS, you've found the bottleneck that determines the maximum matching size.

Question 2

A company wants to assign 8 employees to 6 projects. Each employee is qualified for exactly 3 projects, and each project requires exactly 4 qualified employees to choose from. If this situation is modeled as a bipartite graph, what can be concluded about the existence of an assignment where each project gets exactly one employee?

  1. Such an assignment is impossible because there are more employees than projects
  2. Such an assignment is guaranteed to exist by the given degree conditions
  3. Such an assignment may exist depending on which specific qualifications each employee has (correct answer)
  4. Such an assignment requires additional constraints beyond those given in the problem
Explanation: This asks for a matching that saturates all 6 projects. By the handshaking lemma, the degree conditions (8×3 = 24 total edges = 6×4) are consistent. However, Hall's condition for the project side requires checking that every subset of projects has sufficiently many qualified employees. The given constraints don't guarantee this - it depends on the specific qualification pattern.

Question 3

Let GG be a bipartite graph with parts AA and BB where A=B=6|A| = |B| = 6. If GG has exactly 20 edges and every vertex has degree at least 2, which of the following statements about perfect matchings in GG is necessarily true?

  1. GG must contain at least two edge-disjoint perfect matchings
  2. GG cannot contain a perfect matching due to insufficient connectivity
  3. GG may or may not contain a perfect matching depending on edge distribution (correct answer)
  4. GG must contain exactly one perfect matching due to the degree constraints
Explanation: With 20 edges and minimum degree 2, we have average degree 20/6 ≈ 3.33 per vertex. While this suggests good connectivity, Hall's condition depends on the specific edge distribution. We can construct examples where Hall's condition fails (violating perfect matching existence) and others where it holds, both satisfying the given constraints.

Question 4

Let M1M_1 and M2M_2 be two different maximum matchings in a bipartite graph GG. If M1=M2=k|M_1| = |M_2| = k and M1M2=k2|M_1 \cap M_2| = k-2, how many connected components does the symmetric difference M1M2M_1 \triangle M_2 induce in GG?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: Since |M₁ ∩ M₂| = k-2, there are 2 edges in M₁ but not M₂, and 2 edges in M₂ but not M₁. The symmetric difference M₁ ⊕ M₂ contains exactly 4 edges. In a bipartite graph, the symmetric difference of two matchings forms a collection of alternating paths and cycles. With 4 edges total, we must have exactly 2 connected components (either 2 paths of length 2, or 1 path of length 3 and 1 isolated edge, etc.).

Question 5

In a bipartite graph G=(AB,E)G = (A \cup B, E) where A=n|A| = n and B=n1|B| = n-1, every vertex in AA has degree 2 and every vertex in BB has degree greater than 2. If GG has a matching of size n1n-1, which statement about the structure of GG must be true?

  1. Every vertex in BB has degree exactly 3 to satisfy edge count requirements
  2. There exists a vertex in AA whose removal creates a graph with no perfect matching
  3. The matching of size n1n-1 must be unique due to the degree constraints
  4. Hall's condition holds with equality N(S)=S|N(S)| = |S| for at least one proper subset SAS \subseteq A (correct answer)
Explanation: Since |A| = n and |B| = n-1, a maximum matching has size n-1, leaving exactly one vertex in A unmatched. For this maximum to be achieved, Hall's condition |N(S)| ≥ |S| must hold for all subsets S ⊆ A. Given the tight constraint (only n-1 vertices in B to match n vertices in A), there must exist at least one subset S where |N(S)| = |S|, creating a bottleneck that forces exactly one vertex in A to remain unmatched.

Question 6

A graph GG has vertex set V={a,b,c,d,e,f}V = \{a, b, c, d, e, f\} and edge set E={(a,d),(a,e),(b,d),(b,f),(c,e),(c,f)}E = \{(a,d), (a,e), (b,d), (b,f), (c,e), (c,f)\}. If GG is bipartite with bipartition (X,Y)(X, Y), what is the size of the maximum matching in GG?

  1. 2
  2. 3 (correct answer)
  3. 4
  4. The graph cannot be bipartite with the given edge set
Explanation: First, we verify the graph is bipartite by finding a valid bipartition. Setting X={a,b,c}X = \{a, b, c\} and Y={d,e,f}Y = \{d, e, f\} works since all edges connect vertices from different sets. To find the maximum matching, we can use the fact that in a bipartite graph, we want to match as many vertices as possible. We can match: (a,d)(a,d), (b,f)(b,f), and (c,e)(c,e), giving us 3 edges. This is maximum because we have 3 vertices in each partition, and we've successfully matched all of them. Choice A (2) represents finding only a partial matching. Choice C (4) is impossible since we only have 6 vertices total. Choice D is incorrect because the bipartition shown is valid.

Question 7

In a bipartite graph G=(XY,E)G = (X \cup Y, E), suppose there exists a matching MM that covers all vertices in XX but leaves exactly 2 vertices in YY unmatched. If Y=X+2|Y| = |X| + 2, and we add one new vertex xx' to XX along with edges connecting xx' to the 2 unmatched vertices in YY, what can we conclude about perfect matchings in the resulting graph GG'?

  1. GG' definitely has a perfect matching extending the original matching MM (correct answer)
  2. GG' has no perfect matching because the new vertex creates an imbalance
  3. GG' may or may not have a perfect matching depending on the original graph structure
  4. GG' has exactly two perfect matchings using different edges from xx'
Explanation: In the original graph, MM covers all of XX and all but 2 vertices of YY. After adding xx' to XX and connecting it to the 2 unmatched vertices in YY, we have X{x}=X+1|X \cup \{x'\}| = |X| + 1 and Y=X+2|Y| = |X| + 2, so the new graph has equal-sized partitions. We can extend MM to a perfect matching by adding one edge from xx' to either of the 2 previously unmatched vertices in YY. Choice B is wrong because the sizes are now balanced. Choice C is incorrect because we can always extend the existing matching MM. Choice D is wrong because while xx' could connect to either unmatched vertex, this doesn't guarantee exactly two perfect matchings total.

Question 8

A bipartite graph GG with bipartition (A,B)(A, B) has the property that for every subset SAS \subseteq A with S3|S| \leq 3, we have N(S)S|N(S)| \geq |S|. However, there exists a subset TAT \subseteq A with T=4|T| = 4 such that N(T)=3|N(T)| = 3. If A=6|A| = 6 and B=6|B| = 6, what is the maximum possible size of a matching in GG?

  1. 3
  2. 4
  3. 5 (correct answer)
  4. 6
Explanation: By Hall's theorem, since there exists TAT \subseteq A with T=4|T| = 4 and N(T)=3<4|N(T)| = 3 < 4, there is no matching that saturates all of AA. However, we can still find a large matching. The bottleneck is the 4-subset TT that can reach only 3 vertices in BB. At most 3 vertices from TT can be matched. The remaining 64=26 - 4 = 2 vertices in ATA \setminus T could potentially be matched to vertices in BN(T)B \setminus N(T). Since BN(T)=63=32|B \setminus N(T)| = 6 - 3 = 3 \geq 2, we can match these 2 vertices as well, giving a total matching size of 3+2=53 + 2 = 5. Choice A severely underestimates. Choice B (4) doesn't account for matching vertices outside TT. Choice D (6) ignores the Hall condition violation.

Question 9

Consider a bipartite graph G=(XY,E)G = (X \cup Y, E) where X=5|X| = 5 and Y=4|Y| = 4. Every vertex in XX has degree 3, and every vertex in YY has degree at least 2. Which statement about maximum matchings in GG is necessarily true?

  1. The maximum matching has exactly 4 edges and saturates all vertices in YY (correct answer)
  2. The maximum matching has exactly 3 edges due to degree constraints
  3. The maximum matching could have 5 edges if the degree distribution allows it
  4. The maximum matching size depends on the specific edge configuration
Explanation: Since Y=4|Y| = 4 and we're looking at a bipartite graph, the maximum possible matching size is min(X,Y)=min(5,4)=4\min(|X|, |Y|) = \min(5, 4) = 4. The degree conditions ensure sufficient connectivity: total degree from XX is 5×3=155 \times 3 = 15, and since every vertex in YY has degree at least 2, there are enough edges for a complete matching of YY. By König's theorem and Hall's theorem, these degree conditions guarantee that all vertices in YY can be matched. Choice B underestimates the matching size. Choice C is impossible since Y=4|Y| = 4 limits us to at most 4 edges. Choice D is incorrect because the degree constraints are sufficient to guarantee the maximum.

Question 10

A bipartite graph G=(LR,E)G = (L \cup R, E) has the property that every vertex in LL has degree 2 and every vertex in RR has degree 3. If L=9|L| = 9, what is R|R|, and what is the size of the maximum matching in GG?

  1. R=6|R| = 6 and maximum matching size is 6 (correct answer)
  2. R=6|R| = 6 and maximum matching size is 9
  3. R=9|R| = 9 and maximum matching size is 6
  4. R=12|R| = 12 and maximum matching size is 9
Explanation: Using the handshaking lemma for bipartite graphs: the total degree from LL equals the total degree from RR. Total degree from LL is 9×2=189 \times 2 = 18. If every vertex in RR has degree 3, then R×3=18|R| \times 3 = 18, so R=6|R| = 6. For the maximum matching size: since L=9|L| = 9 and R=6|R| = 6, the maximum possible matching size is min(9,6)=6\min(9,6) = 6. The degree conditions ensure this maximum is achievable. Choice B has the correct R|R| but wrong matching size (can't exceed min(L,R)\min(|L|,|R|)). Choice C has wrong R|R|. Choice D has both wrong R|R| and impossible matching size.

Question 11

Let GG be a bipartite graph with bipartition (X,Y)(X, Y) where X=7|X| = 7 and Y=5|Y| = 5. If GG has a matching of size 4, and we know that removing any single vertex from YY reduces the maximum matching size to at most 3, what can we conclude about the structure of GG?

  1. Every vertex in YY has degree exactly 1 in the matching
  2. There exists exactly one vertex in YY that is critical for the matching
  3. At least two vertices in YY are essential for achieving matching size 4 (correct answer)
  4. The graph has a unique maximum matching of size 4
Explanation: Since the maximum matching has size 4 and removing any single vertex from YY reduces this to at most 3, this means that no single vertex in YY is solely responsible for the fourth matched edge. If only one vertex in YY were essential, removing just that vertex would drop the matching size to 3, but removing others would maintain size 4. Since removing ANY vertex from YY causes this drop, multiple vertices in YY must be critical for achieving the full matching size of 4. Choice A confuses matching edges with vertex degrees. Choice B contradicts the given condition that removing ANY vertex causes the reduction. Choice D makes an unjustified claim about uniqueness.

Question 12

Consider two bipartite graphs G1=(X1Y1,E1)G_1 = (X_1 \cup Y_1, E_1) and G2=(X2Y2,E2)G_2 = (X_2 \cup Y_2, E_2) where X1=X2=4|X_1| = |X_2| = 4, Y1=Y2=4|Y_1| = |Y_2| = 4, and both have perfect matchings. If we form a new bipartite graph HH by taking the disjoint union and adding exactly one edge between Y1Y_1 and X2X_2, what can we conclude about the maximum matching size in HH?

  1. The maximum matching size is at most 7 because the new edge disrupts the bipartite structure
  2. The maximum matching size could be 9 if the new edge creates additional matching possibilities
  3. The maximum matching size is exactly 9 due to the additional connectivity
  4. The maximum matching size is 8, same as the sum of the original maximum matchings (correct answer)
Explanation: When analyzing matching problems in graph theory, you need to understand how structural changes affect maximum matching sizes. A matching is a set of edges with no shared vertices, and in bipartite graphs, the maximum matching size is constrained by the smaller partition. Let's trace through what happens when we form graph HH. Initially, G1G_1 and G2G_2 each have perfect matchings of size 4 (matching all vertices in both partitions). In their disjoint union, you could achieve a total matching of 8 edges by using both perfect matchings simultaneously, since the graphs share no vertices. Now we add one edge between Y1Y_1 and X2X_2. This creates a single connected component, but critically, it doesn't increase the maximum matching size. Here's why: HH now has 8 vertices that could potentially be matched from the "left" side (X1Y1X_1 \cup Y_1) and 8 vertices on the "right" side (Y2X2Y_2 \cup X_2). However, any maximum matching can still use at most 8 edges total, achieving perfect matching of all 16 vertices. Option A is wrong because adding one edge between partitions doesn't disrupt the bipartite structure - HH remains bipartite with partitions (X1Y1)(X_1 \cup Y_1) and (X2Y2)(X_2 \cup Y_2). Option B is incorrect because 9 edges would require 18 vertices, but we only have 16. Option C fails for the same reason - we cannot exceed 8 edges with 16 vertices. Study tip: In matching problems, always count vertices in each partition first. The maximum matching size cannot exceed half the total vertices in any bipartite graph.

Question 13

In a bipartite graph G=(AB,E)G = (A \cup B, E) with A=6|A| = 6 and B=6|B| = 6, suppose that for some subset SAS \subseteq A with S=4|S| = 4, the neighborhood N(S)N(S) has exactly 3 vertices. What can be concluded about perfect matchings in GG?

  1. GG has a perfect matching because A=B|A| = |B|
  2. GG has no perfect matching due to Hall's condition violation (correct answer)
  3. GG may or may not have a perfect matching depending on other subsets
  4. GG has at least one maximum matching of size 5
Explanation: Hall's Marriage Theorem states that a bipartite graph G=(AB,E)G = (A \cup B, E) has a matching that saturates AA if and only if for every subset SAS \subseteq A, we have N(S)S|N(S)| \geq |S|. Here, we have a subset SAS \subseteq A with S=4|S| = 4 but N(S)=3|N(S)| = 3, violating Hall's condition since 3<43 < 4. This means there is no matching that saturates AA, and therefore no perfect matching exists. Choice A ignores the Hall condition violation. Choice C is incorrect because finding one violating subset is sufficient to conclude no perfect matching exists. Choice D is wrong because the maximum matching size could be less than 5 due to this bottleneck.

Question 14

A bipartite graph has vertex sets U={u1,u2,u3,u4}U = \{u_1, u_2, u_3, u_4\} and V={v1,v2,v3,v4}V = \{v_1, v_2, v_3, v_4\}. If the graph has a perfect matching M1={(u1,v2),(u2,v4),(u3,v1),(u4,v3)}M_1 = \{(u_1,v_2), (u_2,v_4), (u_3,v_1), (u_4,v_3)\} and another perfect matching M2={(u1,v3),(u2,v1),(u3,v4),(u4,v2)}M_2 = \{(u_1,v_3), (u_2,v_1), (u_3,v_4), (u_4,v_2)\}, what is the minimum number of edges the graph must contain?

  1. 4
  2. 6
  3. 8 (correct answer)
  4. At least 10 to ensure both perfect matchings exist
Explanation: The graph must contain all edges from both perfect matchings M1M_1 and M2M_2. Since M1={(u1,v2),(u2,v4),(u3,v1),(u4,v3)}M_1 = \{(u_1,v_2), (u_2,v_4), (u_3,v_1), (u_4,v_3)\} and M2={(u1,v3),(u2,v1),(u3,v4),(u4,v2)}M_2 = \{(u_1,v_3), (u_2,v_1), (u_3,v_4), (u_4,v_2)\}, and these two sets are disjoint (no edge appears in both matchings), the graph must have at least M1+M2=4+4=8|M_1| + |M_2| = 4 + 4 = 8 edges. Choice A represents only one perfect matching. Choice B incorrectly assumes some edges are shared between the matchings. Choice D overestimates by assuming additional edges are needed beyond those in the two given perfect matchings.

Question 15

Consider a complete bipartite graph K3,4K_{3,4} with vertex sets A={a1,a2,a3}A = \{a_1, a_2, a_3\} and B={b1,b2,b3,b4}B = \{b_1, b_2, b_3, b_4\}. How many different maximum matchings does this graph have?

  1. 12
  2. 24 (correct answer)
  3. 36
  4. The number depends on the specific labeling of vertices
Explanation: In K3,4K_{3,4}, the maximum matching size is min(3,4)=3\min(3,4) = 3, so we need to match all 3 vertices from set AA to 3 of the 4 vertices in set BB. First, choose which 3 vertices from BB to use: (43)=4\binom{4}{3} = 4 ways. Then, for each choice, arrange the 3 vertices from AA to match with the chosen 3 vertices from BB: 3!=63! = 6 ways. Total: 4×6=244 \times 6 = 24 maximum matchings. Choice A (12) might result from incorrectly calculating (43)×3=12\binom{4}{3} \times 3 = 12. Choice C (36) could come from incorrect overcounting. Choice D is wrong because the structure of K3,4K_{3,4} determines the count regardless of labeling.