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Chapter 5: Q6E (page 162)

Let G=(V,E) be an undirected graph. Prove that if all its edge weights are distinct, then it has a unique minimum spanning tree

Short Answer

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Here, we get undirected graph G. Given here to prove non - negative real edge weights, and presume that even if you calculated a minimal spanning tree of G and the short distance for that kind of node, it accepts particular nodesV.

Step by step solution

01

Two alternative shortest path trees

If most of the parameters associated in such an undirected graph disagree, the graph has a single minimal spanning tree.

鈥� Assume there are two alternative shortest path trees inside of an undirected graph, including such T1andT2.

鈥� Let e1seems to be the minimal side weight which relates to a few of the trees, for example. e1鈭�T1.

鈥� Including the side role="math" localid="1658904910955" e1into the tree T2making of cycle. The cycle includes the edge of role="math" localid="1658904901130" e2in tree T2, any one is bigger than e1.

鈥� This seems to be an inconsistency for which an indifference curve has two distinct minimum spanning trees.

02

Proof of theoem.

Using the cut property, find the minimal spanning tree based on the graph's topology and edge weight order.

The minimal side value is already on the lowest spanning tree for each and every cut.

鈥� Since all of the respect to the weights inside a graph are different, then all of the cuts are unique.

鈥� If all of the edge weights in a graph are the same, the outcome may be unclear.

If all the edge weights of a graph are distinct, the smallest spanning tree is unique.

For example:

Consider the undirected graph G = (V, E) is given below:

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Most popular questions from this chapter

The basic intuition behind Huffman鈥檚 algorithm, that frequent blocks should have short encodings and infrequent blocks should have long encodings, is also at work in English, where typical words like I, you, is, and, to, from, and so on are short, and rarely used words like velociraptor are longer.

However, words like fire!, help!, and run! are short not because they are frequent, but perhaps because time is precious in situations where they are used.

To make things theoretical, suppose we have a file composed of m different words, with frequencies f1,...,fm. Suppose also that for the ithword, the cost per bit of encoding is ci. Thus, if we find a prefix-free code where the ithword has a codeword of length Ii, then the total cost of the encoding will be localid="1659078764835" 鈭�fi路ci路li.

Show how to modify Huffman鈥檚 algorithm to find the prefix-free encoding of minimum total cost.

A binary counter of unspecified length supports two operations: increment (which increases its value by one) and reset (which sets its value back to zero). Show that, starting from an initially zero counter, any sequence of n increment and reset operations takes time O(n); that is, the amortized time per operation is O(1) .

Prove the following two properties of the Huffman encoding scheme.

(a) If some character occurs with frequency more than 25, then there is guaranteed to be a codeword of length 1 .

(b) If all characters occur with frequency less than13 , then there is guaranteed to be no codeword of length 1 .

Ternary Huffman. Trimedia Disks Inc. has developed 鈥渢ernary鈥� hard disks. Each cell on a disk can now store values 0,1, or 2(instead of just 0 or 1). To take advantage of this new technology, provide a modified Huffman algorithm for compressing sequences of characters from an alphabet of size n, where the characters occur with known frequencies f1, f2,...., fn. Your algorithm should encode each character with a variable-length codeword over the values 0,1,2, such that no codeword is a prefix of another codeword and so as to obtain the maximum possible compression. Prove that your algorithm is correct

Graphs with prescribed degree sequences. Given a list of n positive integers d1,d2,鈥�,dn, we want to efficiently determine whether there exists an undirected graphG=(V,E) whose nodes have degrees preciselyd1,d2,鈥�,dn . That is, if V={v1,鈥�,vn}, then the degree of vi should be exactly di. We call (d1,鈥�,dn) the degree sequence of G. This graph G should not contain self-loops (edges with both endpoints equal to the same node) or multiple edges between the same pair of nodes.

(a) Give an example of d1,d2,d3,d4 where all the di鈮�3 and d1+d2+d3+d4 is even, but for which no graph with degree sequence(d1,d2,d3,d4) exists.

(b) Suppose that d1鈮�d2鈮�d3鈮�dn and that there exists a graph G=(V,E) with degree sequence (d1,鈥�,dn). We want to show that there must exist a graph that has this degree sequence and where in addition the neighbors of v1 are v2,v3,鈥�,vdi+1 . The idea is to gradually transform G into a graph with the desired additional property.

i. Suppose the neighbors ofv1 in Gare not v2,v3,鈥�,vdi+1. Show that there exists i<j鈮�n and u鈭�V and such that {v1,vi},{u,vj}鈭�Eand {v1,vj},{u,vi}鈭�E

ii. Specify the changes you would make to G to obtain a new graph G'=(V,E') with the same degree sequence as G and where (v1,vi)鈭�E'.

iii. Now show that there must be a graph with the given degree sequence but in which v1 has neighbors v2,v3,鈥�,vdi+1.

c) Using the result from part (b), describe an algorithm that on input d1,鈥�,dn (not necessarily sorted) decides whether there exists a graph with this degree sequence. Your algorithm should run in time polynomial in n and in m=鈭�i=1ndi .
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