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By L. Allison

This textbook is an advent to denotational semantics and its functions to programming languages. Dr Allison emphasizes a pragmatic strategy and the scholar is inspired to write down and try out denotational definitions. the 1st part is dedicated to the mathematical foundations of the topic and adequate element is given to demonstrate the basic difficulties. the rest of the booklet covers using denotational semantics to explain sequential programming languages reminiscent of Algol, Pascal and C. all through, a number of workouts, often in Pascal, may help the coed coaching writing definitions and perform easy functions. The publication culminates in discussing an executable semantics of the logic-programming language Prolog. Being an creation, complicated undergraduates in computing device technology and graduates new to the topic will locate this a conveniently available account of 1 of the important subject matters of desktop technological know-how.

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Let G be a bipartite multigraph with bipartition (A, B) such that |A| = |B|. Then the following three statements are equivalent. (i) G is connected, and for each edge e, G has a 1-factor containing e. (ii) For every subset ∅ = X ⊂ A, |NG (X)| > |X|. (iii) For every two vertices a ∈ A and b ∈ B, G − {a, b} has a 1-factor. Proof. (i)⇒(ii) Suppose that |NG (Y )| ≤ |Y | for some subset ∅ = Y ⊂ A. Since G has a 1-factor F , we have |NG (Y )| = |Y | as |NG (Y )| ≥ |NF (Y )| = |Y |. Since G is connected, G has an edge e joining a vertex in A − Y to a vertex in NG (Y ).

19), we have m eG (Ci , S) + eG (Ci , V (G) − S − V (Ci )) rm ≤ i=1 ≤ m eG (Ci , V (G) − S − V (Ci )) degG (x) + x∈S i=1 ≤ r|S| + 2(r − 1). 9). Consequently H has a 1-factor, and the theorem follows. 37, every (r − 1)-edge connected r-regular multigraph of even order has a 1-factor. We can say that this result is the best in the sense that there exist infinitely many (r − 2)-edge connected r-regular multigraphs of even order that have no 1-factor. An example is given below. Example Let r ≥ 3 be an odd integer.

Consequently, the number of factor-critical components of G − S = odd(G − S) > |S|. This contradicts the assumption, and thus the theorem is proved. The next theorem gives a necessary and sufficient condition for a tree to have a 1-factor, and the proof presented here is due to Amahashi [14]. 29 (Chungphaisan ). A tree T of even order has a 1-factor if and only if odd(T − v) = 1 for every vertex v of T . Proof. Suppose that T has a 1-factor F . Then for every vertex v of T , let w be the vertex of T joined to v by an edge of F .

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A Practical Introduction to Denotational Semantics by L. Allison

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