1. Prove: A full m- ary tree with i internal vertices contains n = mi + 1 vertices.

Transcription

1 1. Prove: A full m- ary tree with i internal vertices contains n = mi + 1 vertices. Proof: Every vertex, except the root, is the child of an internal vertex. Since there are i internal vertices, each of which has m children, there are mi vertices other than the root; including the root gives a total of mi + 1 vertices. 2. For a full m- ary tree with n vertices, i internal vertices, and l leaves, prove: (i) i = (n 1)/m and l = [(m 1)n + 1]/m Proofs: Since n= mi + 1 (#2 above), solving for i we get i = (n 1)/m. Next, since n = i + l, we have i = n l. Replacing i with n l in the equation i = (n 1)/m and solving for l, we have n l = n 1 m l = n 1 m n l = n n 1 m l = nm m n 1 m nm n 1 l = m l = nm n+1 m ( l = m 1 )n+1 m (ii) n = mi + 1 vertices and l = (m 1)i + 1 Proofs: We have already proven n = mi + 1 (#2 above). Using the fact that n = i + l we have i+l = mi + 1. Solving for i we have i+l = mi+1 l = mi+1 i l = mi i+1 l = ( m 1)i+1

2 (iii) n = (ml 1)/(m 1) and i = (l 1)/(m 1) Proofs: From (i) above we have l = [(m 1)n + 1]/m Solving for n we have ( l = m 1 )n+1 m ml = ( m 1)n+1 ml 1= ( m 1)n ml 1 m 1 = n From (ii) above we have l = (m 1)i + 1. Solving for i we have l = ( m 1)i+1 l 1= ( m 1)i l 1 m 1 = i 3-8. Refer to this rooted tree: 3. According to universal addressing, what is the address of vertex d? answer: 3 4. According to universal addressing, what is the address of vertex f? answer: According to universal addressing, what is the address of vertex j? answer: Write the preorder traversal. answer: a- b- e- f- c- g- h- d- i- j- k 7. Write the postorder traversal. answer: e- f- b- g- h- c- j- k- i- d- a 8. Write the inorder traversal. answer: e- b- f- a- g- c- h- j- i- k- d

3 9-14. Refer to this rooted tree: 9. According to universal addressing, what is the address of vertex g? answer: According to universal addressing, what is the address of vertex l? answer: According to universal addressing, what is the address of vertex o? answer: Write the preorder traversal. answer: a- b- c- f- g- h- m- d- i- n- o- j- k- e- l- p 13. Write the postorder traversal. answer: b- f- g- m- h- c- n- o- i- j- k- d- p- l- e- a 14. Write the inorder traversal. answer: b- a- f- c- g- m- h- n- i- o- d- j- k- p- l- e 15. Draw the binary tree for the expression (5 (7 + 3)) (8 2) 16. Referring to #15, write the expression in prefix notation. Answer: Prove: On a tree, there is a unique simple path between any two vertices.

4 Proof by contraposition: Recall that a tree is a connected simple graph with no cycles. Suppose G is a connected simple graph such and there are vertices u, v such that there are two distinct paths, P1 and P2, from u to v. Then concatenating these paths produces a cycle from u to u, hence G is not a tree. 18. Prove: A tree with n vertices has exactly n 1 edges. Proof: Each edge of the tree uniquely connects one child to its parent. Of the n vertices, only one vertex (the root) does not have a parent, so the number of edges is n 1. Alternative proof: In discussing connectivity, and graphs in general, we have already proved, for any simple graph G with n vertices: A. If G is connected, then G has at least n 1 edges; and B. If G has more than n 1 edges, then G has a cycle. Combining these two results, we see that any connected acyclic simple graph (i.e. any tree) must have exactly n 1 vertices. 19. Draw the binary tree for the expression (8 3) + ((6 + 2) 5). 20. Referring to #19, write the expression in prefix and postfix notation. prefix: postfix: Draw the binary tree for the expression (5 (3 + (2 4))) (2 + 9). 22. Referring to #21, write the expression prefix and postfix notation.

5 prefix: postfix: Use strong induction to prove: The number of leaves on an m- ary tree of height h is at most m h, for h 1 (the case where h=0 is trivial). Basis: Suppose we have an m- ary tree of height 1. The tree consists only the root and its children, each of which is a leaf. Since this is an m- ary tree, there are at most m of these, so the number of leaves is at most m = m 1. Strong inductive: Suppose, for some k 1, that every m- ary tree of height j has at most m j leaves whenever j k. Let T be an m- ary tree of height k+1, with root r. Consider the graph T r. This graph is a forest of at most m disjoint m- ary subtrees, each of which has height less than or equal to k. Collectively, the leaves of this forest are the same as the leaves of T. Let L be the number of leaves of this forest. Then L (maximum number of subtrees) (maximum number of leaves per subtree) = m m k = m k+1, so T has at most m k+1 leaves. 24. Evaluate the postfix expression: answer: Draw the binary tree for # Referring to #24, write the expression in prefix notation answer: Refer to this graph:

6 A. Suppose we perform a breadth first search rooted at c. If the first two edges are {c, a} and {a, f }, what is the next edge? Answer {a, b} B. Suppose we perform a breadth first search rooted at e. If the first three edges are {e, h}, {e, n} and {h, g}, what is the next edge? Answer {h, i} C. Suppose we perform a depth first search rooted at d, starting with the path d, b, a, f. What edge is added next? Answer: {a, c} D. Suppose we perform a breadth first search rooted at d. If the first two edges are {d, f}, and {d, n}, what is the next edge? Answer {d, b} E. Suppose we perform a depth first search rooted at b, starting with the path b, n, e, h, i. What edge is added next? Answer: {h, g} F. Suppose we perform a depth first search rooted at e, starting with the path e, n, b, d, f, a, c. What edge is added next? Answer: {e, h}

7 28. Refer to this graph: A. Suppose we perform a depth first search rooted at f, starting with the path f, d, b, a. Continuing the search, what will the final edge added to the spanning tree? Answer {e, g} B. Suppose we perform a depth first search rooted at e, employing the path e, b, d, f Continuing the search, what will then next vertex added to the spanning tree? Answer a C. Suppose we perform a breadth first search rooted at c, starting with the edges {c, h}, {c, b}. What will be the next edge? Answer {c, a} D. Suppose we perform a breadth first search rooted at f, starting with the edges {f, d}, {d, e}. What will be the next edge? Answer {d, b} 29. Use Boolean identities, and algebra, to prove x y + y + z = y z + x Justify each step in the proof. Do not skip or combine steps. Here is one correct proof: x y + y + z = x y + yz//demorgan'slaw = yx + yz//commutativelaw = y( x + z)//distributivelaw = y( z + x)//commutativelaw 30. Use Boolean identities, and algebra, to prove x xy + zy = x Justify each step of the proof. Do not skip or combine steps. Here is one proof. It is not the only correct proof.

8 x( xy + zy) = xxy + x yz//distributivelaw = xy + x yz//idempotentlaw = xy + x( y + z)//demorgan'slaw = xy + x y + xz//distributivelaw = x( y + y)+ xz//distributivelaw = x( 1)+ xz//complementslaw =x + xz//identitylaw =x//absorptionlaw 31. T F The set { +, } is functionally complete. Proof: Since our set of operators contains both sum and negation, we only need to prove that product can be expressed in terms of sums and negations: xy = xy //Double negation = x + y //DeMorgan's Law and double negation 32. Write a Boolean expression that is equivalent to xyz + yz, but without any products. Solution: xyz + yz = xyz + yz //Double negation = x + y + z + y + z //DeMorgan's Law, double negation 33. T F The set {, } is functionally complete. Proof: Since our set of operators contains both sum and product, we only need to prove that sum can be expressed used products and negations: x + y = x + y //Double negation = x y //DeMorgan's Law 34. Write a Boolean expression that is equivalent to xyz + yz, but without any sums. Solution: xyz + yz = xyz + yz //Double negation = xyz yz //DeMorgan's Law

9 35. T F If the set, { } is functionally complete, then the set Proof: Since we have the hypothesis that {, }, we can prove that that negation and product can be expressed using only the nor operator. We will also use the definition: x y = x + y. First, x = x x //Idempotent Law = x + x //DeMorgan's Law = x x //Definition of nor Next, xy = x + y //DeMorgan's Law = x y //Definition of nor = ( x x) ( y y) //because a = a a { } is functionally complete. { } is functionally complete by showing This shows that negations and products can be expressed using only the nor operator, so the nor operator is functionally complete. 36. T F If the set +, { } is functionally complete, then the set {} is functionally complete. Proof: Since we have the hypothesis that { +, }, we can prove that {} is functionally complete by showing that sum and negation can be expressed using only the nand operator. We will also use the definition: x y = xy. First, x = x + x //Idempotent Law = x x //DeMorgan's Law = x x //Definition of nand Next, x + y = x y //DeMorgan's Law = x y //Definition of nand = ( x x) ( y y) //because a = a a

10 This shows that negations and sums can be expressed using only the nand operator, so the nand operator is functionally complete. 37. Use algebra (not a truth table) to derive the sum of products expansion (aka disjunctive normal form) y + z = xz + x + yz. You don t have to justify the steps. You may skip or combine steps. for F x, y,z You must write the answer using lexicographic order. The answer is: x yz + x yz + x yz Here is the derivation: y + z yz xz + x + yz = xz + x + yz = xz + x yz + yz yz = xz + x yz + yzz = xz + x yz + y 0 = xz + x yz = xz( y + y)+ x yz = x yz + x yz + x yz = x yz + x yz + x yz 38. Use algebra (not a truth table) to derive the sum of products expansion (aka disjunctive normal form) for F ( x, y,z) = x( y + z)+ yz. You don t have to justify the steps. You may skip or combine steps. You must write the answer using lexicographic order. The answer is: xyz + x yz + x yz + x yz Here is the derivation, with some terse explanations. x( y + z)+ yz = x y + xz + yz //distributive law + xz ( y + y ) + yz ( x + x ) = x y z + z = x yz + x yz + x yz + x yz + xyz + x yz //distribute //multiply various terms by 1 = xyz + x yz + x yz + x yz //place minterms in correct order, remove redundant minterms 39. Use algebra (not a truth table) to derive the sum of products expansion (aka disjunctive normal form) for F ( x, y,z) = ( x + y) yz. You don t have to justify the steps. You may skip or combine steps. You must write the answer using lexicographic order.

11 The answer is: x yz + x yz ( x + y) yz = x y y + z //DeMorgan's Laws = x y + x yz //Distributive and Idempotent Laws = x y z + z + x yz //Multiply a term by 1 = x yz + x yz + x yz //Distributive Law = x yz + x yz //Idempotent Law & check correct ordering of minterms 40. Use Boolean identities, and algebra, to prove ( y + x) x + y =1. Justify each step of the proof. Do not skip or combine steps. Here is one proof. It is not the only correct proof. ( y + x) x + y = ( y + x)+ x + y //DeMorgan's Law, Double Negation = ( y + x)+ [ x + y] //Associative Law = ( x + y)+ [ x + y] //Commutative Law =1 //Complements Law 41. Use algebra to derive the sum of products expansion (aka disjunctive normal form) for = x + z + y x + z F x, y,z You may skip/combine steps. You don t have to write the justifications. Your answer must be written using lexicographic order. Answer: xyz + xyz + x yz + x yz + x yz Solution: Here is one derivation. Some justifications have been provided, for pedagogical reasons, but you would not have to write those justifications on a test. x + z + y( x + z) = x z + yx + yz //DeMorgan's law, Distributive law + yx ( z + z ) + yz ( x + x ) = xz y + y //Multiply terms by 1, in 3 places = x yz + x yz + x yz + x yz + x yz + x yz //Distributive law 3, also Commutative law to arrange literals = x yz + x yz + x yz + x yz //Commutative law to arrange minterms lexicographically, // also Idempotent law to remove redundant minterm 42. Use algebra, not a truth table, to derive the sum of products expansion for F ( x, y, x) = z y + x

12 You may skip/combine steps. You do not have to justify the steps. You must write your answer using lexicographic order. Answer: xyz + xyz + xyz + xyz + xyz Solution: Here is one derivation. Some justifications have been provided, for pedagogical reasons, but you would not have to write those justifications on a test (for this style of problem, only). //DeMorgan's Law z y + x = z + y + x = z + yx //DeMorgan's Law = z( x + x)+ yx z + z //Multiply terms by 1, in two places = zx + zx + yxz + yxz //Distributive law = zx( y + y)+ zx( y + y)+ yxz + yxz //Multiply terms by 1, in two places = zxy + zxy + zxy + zxy + yxz + yxz //Distributive law = xyz + xyz + xyz + xyz + xyz + xyz //Commutative law, repeatedly, to arrange literals within minterms = xyz + xyz + xyz + xyz + xyz + xyz //Commutative law, repeatedly, to arrange minterms lexicographically = xyz + xyz + xyz + xyz + xyz //Idempotent law removes a redundant minterm 43. Find a Boolean function equivalent to ( x + y) ( yz) but without any sums. ( x + y) yz Double negation DeMorgan's law and double negation = x + y yz = xy yz 44. Find a Boolean function equivalent to ( x + y) ( yz) but without any products. ( x + y) yz ( yz) Double negation + ( yz) DeMorgan's law + y + z DeMorgan's law, double negation = x + y = x + y = x + y ( yz) ( xy) + ( y + z) 45. Find the dual to x + y Answer:

13 46. Using, as needed, the Identity, Complements, Commutative, Associative and Distributive laws as axioms, prove this Idempotent law: x x = x. Proof: x = x 1//Identitylaw = x( x + x)//complementslaw = x x + x x//distributivelaw = x x +0//Complementslaw = x x//identitylaw 47. Using, as needed, the Identity, Complements, Commutative, Associative and Distributive laws as axioms, prove this Idempotent law: x + x = x. Proof: x = x +0//Identitylaw //Complementslaw //Distributivelaw = x + x x x + x = x + x == ( x + x) 1//Complementslaw = x + x//identitylaw 48. Having proven the Idempotent laws (Exercises 22 and 23 above), use them as needed, along with the Identity, Complements, Commutative, Associative and Distributive laws as axioms, to prove this Domination law: x 0 = 0 Proof: x 0 = x ( x x)//complementslaw = ( x x) x//associativelaw = x x//idempotentlaw = 0//Complementslaw 49. Having proven the Idempotent laws (Exercises 22 and 23 above), use them as needed, along with the Identity, Complements, Commutative, Associative and Distributive laws as axioms, to prove this Domination law: x +1 = 1

14 Proof: x +1 = x + ( x + x)//complementslaw = ( x + x)+ x//associativelaw = x + x//idempotentlaw = 1//Complementslaw 50. Having proven the Idempotent and Domination laws (Exercises above), use them as needed, along with the Identity, Complements, Commutative, Associative and Distributive laws as axioms, to prove this Absorption law: x + xy = x Proof: x + xy = x 1+ xy//identitylaw = x( 1+ y)//distributivelaw = x 1//Dominationlaw = x//identitylaw 51. Having proven the Idempotent and Domination laws (Exercises above), use them as needed, along with the Identity, Complements, Commutative, Associative and Distributive laws as axioms, to = x prove this Absorption law: x x + y Proof: x + y x x + y = x +0 //Identitylaw = x + ( 0 y)//distributivelaw = x +0//Dominationlaw = x//identitylaw