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IAL 2021 June D1 Q7

A Level / Edexcel / D1

IAL 2021 June Paper · Question 7

题目

Problem

Figure 5 shows a weighted graph that contains 12 arcs and 8 vertices.

It is given that

  • no two arcs have the same weight
  • xx and yy are positive integers
  • arc CD is not in the minimum spanning tree for the graph

(a) Explain why y<x+7y < x + 7

(2)

It is also given that when Prim’s algorithm, starting at A, is applied to the weighted graph, AB is the first arc selected.

(b) Show that y>2xy > 2x and write down and simplify two further constraints on the values of xx and yy.

(3)

(c) Represent these four constraints on Diagram 1 in the answer book.

(4)

(d) Using Diagram 1 only, write down the possible pairs of values that xx and yy can take in the form (x,y)(x, y).

(2)

The minimum spanning tree for the weighted graph in Figure 5 has total weight 73.

Six of the seven arcs in the minimum spanning tree are AB, AD, BC, CE, EF and GH.

(e) Determine the value of xx and the value of yy. You must make your method and working clear.

(4)
题目中文翻译

图 5 显示了一个加权图,包含 12 条弧和 8 个顶点。

已知

  • 没有两条弧具有相同的权重
  • xxyy 是正整数
  • 弧 CD 不在图的最小生成树中

(a) 解释为什么 y<x+7y < x + 7

还已知当从 A 开始的 Prim 算法应用于加权图时,AB 是选择的第一条弧。

(b) 证明 y>2xy > 2x 并写出并简化关于 xxyy 值的两个进一步约束。

(c) 在答案本的图 1 上表示这四个约束。

(d) 仅使用图 1,写出 xxyy 可以取的可能值对,形式为 (x,y)(x, y)

图 5 中加权图的最小生成树总权重为 73。

最小生成树中七条弧中的六条是 AB、AD、BC、CE、EF 和 GH。

(e) 确定 xx 的值和 yy 的值。必须清楚说明方法和计算过程。

解答

(a)

解法一

思路

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D 只与 A、C 相连。既然 CD 不在最小生成树中,为了连接 D,AD 必须在树中;又因为各边权互不相同,所以 CD 的权必须大于 AD 的权。

答题过程

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The only arcs incident to D are AD and CD. Since CD is not in the minimum spanning tree, AD must be in the tree. Therefore

2y+x>3y7.2y+x>3y-7.

Hence

y<x+7.\boxed{y<x+7}.

(b)

解法一

思路

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从 A 开始运行 Prim 算法时,第一条边只能从与 A 相连的 AB、AC、AD、AE 中选取。AB 首先被选且所有边权互不相同,所以 AB 的权分别严格小于另外三条边的权。

答题过程

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Since AB is selected first,

4x+1<2y+1,4x+1<8x3,4x+1<3y7.\begin{align*} 4x+1<&\,2y+1,\\ 4x+1<&\,8x-3,\\ 4x+1<&\,3y-7. \end{align*}

Therefore

y>2x,\boxed{y>2x},

and the two further constraints are

x>1and3y>4x+8.\boxed{x>1} \qquad\text{and}\qquad \boxed{3y>4x+8}.

(c)

解法一

思路

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画出四条边界直线 y=x+7y=x+7y=2xy=2xx=1x=13y=4x+83y=4x+8。由于四个不等式都是严格不等式,边界不属于可行域;取同时满足 y<x+7y<x+7y>2xy>2xx>1x>13y>4x+83y>4x+8 的交集。

答题过程

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Draw the boundary lines

y=x+7,y=2x,x=1,3y=4x+8.y=x+7, \qquad y=2x, \qquad x=1, \qquad 3y=4x+8.

The feasible region is the intersection satisfying

y<x+7,y>2x,x>1,3y>4x+8.y<x+7, \qquad y>2x, \qquad x>1, \qquad 3y>4x+8.

(d)

解法一

思路

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读出可行域内部的所有整点。边界上的点不能计入,因为原来的四个不等式都是严格不等式。

答题过程

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The possible integer pairs are

(2,6),(2,7),(2,8),(3,7),(3,8),(3,9),(4,9),(4,10),(5,11).\boxed{ \begin{gathered} (2,6),(2,7),(2,8),\\ (3,7),(3,8),(3,9),\\ (4,9),(4,10),\\ (5,11). \end{gathered} }

(e)

解法一

思路

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已知的六条边还没有把 H 所在部分与其余顶点连起来。比较可用的 EH、EG、FG,EH 的权最小,所以第七条边是 EH。把七条边的权相加并令总权为 73,再从上一问的九组整数解中筛选即可。

答题过程

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In the feasible region,

(2y+4)(y+1)=y+3>0.(2y+4)-(y+1)=y+3>0.

Also, since xx and yy are integers, x>1x>1 and y<x+7y<x+7 give x2x\geq2 and yx+6y\leq x+6. Hence

(5x+1)(y+1)=5xy4x6>0.(5x+1)-(y+1)=5x-y\geq4x-6>0.

Thus EH is lighter than both EG and FG, so the remaining arc in the minimum spanning tree is EH.

Therefore the total weight is

(4x+1)+(3y7)+(2y2)+3x+(x+y)+(6x2y+3)+(y+1)=14x+5y4.\begin{align*} &(4x+1)+(3y-7)+(2y-2)+3x+(x+y)\\ &\quad +(6x-2y+3)+(y+1)\\ =&\,14x+5y-4. \end{align*}

Since the total weight is 73,

14x+5y4=73,14x+5y=77.\begin{align*} 14x+5y-4=&\,73,\\ 14x+5y=&\,77. \end{align*}

Testing the integer pairs from part (d), only (3,7)(3,7) satisfies this equation, since

14(3)+5(7)=42+35=77.14(3)+5(7)=42+35=77.

Hence

x=3,y=7.\boxed{x=3,\qquad y=7}.