discrete mathematics for computer science books

Quy tắc quy nạp mạnh suy ra rằng nếu R(0) là đúng và R(n) kéo theo R(n+1) là đúng với mọi n, thì R(m) là đúng với mọi m. Quy tắc này mạnh hơn quy tắc quy nạp thông thường vì nó không yêu cầu chứng minh cơ sở. Quy tắc này được sử dụng để chứng minh rằng số chuỗi nhị phân độ dài n không có 1 nào liên tiếp là Fn+2, trong đó Fn là số Fibonacci thứ n. Bằng cách chứng minh rằng một chuỗi nhị phân độ dài n+1 tốt (không có 1 nào liên tiếp) có thể được tạo thành từ một chuỗi tốt độ dài n kết thúc bằng 0 hoặc một chuỗi tốt độ dài n-1 kết thúc bằng 1, ta có thể chứng minh rằng Cn+1 = Cn + Cn-1, tương đương với Fn+3 = Fn+2 + Fn+1. Do đó, Cn = Fn+2 cho mọi n.

Ngày tải lên: 22/03/2019, 10:53

... David M Keil Framingham State University CSCI 317 Discrete Structures 2/14 Study questions on Discrete Structures for Computer Science The intention in providing these questions is to ... table contains (a) variables; (b) formulas; (c) values of formulas under one interpretation; (d) values of formulas under all interpretations; (e) operations If formulas  and  have the same truth ... (d) variables; (e) quantifiers A formula in logic is valid if (a) it is true for some interpretation; (b) it is true for all interpretations; (c) it is true for no interpretation; (d) it is an

Ngày tải lên: 17/11/2016, 22:23

... steam-curing at 60oC for 24 hours In order to optimize the usage of formwork, the products were cast and steam-cured initially for about hours The steam-curing was then stopped for some time to allow ... used in the case of reinforced Portland cement concrete columns and beams are applicable for reinforced geopolymer concrete columns Mid-span deflection at service load of reinforced geopolymer concrete ... “Bond performance of Reinforcing Bars in Inorganic Polymer Concrete (IPC)”, Journal of Materials Science, Sumajouw, M D J and Rangan, B.V., “Low-Calcium Fly Ash-Based Geopolymer Concrete: Reinforced

Ngày tải lên: 09/12/2016, 07:56

... Functions Nguyen An Khuong, Huynh Tuong Nguyen Chapter Sets and Functions Contents Sets Discrete Structures for Computer Science (CO1007) on Ngày tháng 10 năm 2016 Set Operation Functions One-to-one and ... Functions Sequences and Summation Recursion Nguyen An Khuong, Huynh Tuong Nguyen Faculty of Computer Science and Engineering University of Technology, VNU-HCM 4.1 Contents Sets and Functions Nguyen ... Sets and Functions Nguyen An Khuong, Huynh Tuong Nguyen • Set is a fundamental discrete structure on which all discrete structures are built • Sets are used to group objects, which often have

Ngày tải lên: 29/03/2017, 18:30

... An Khuong, Huynh Tuong Nguyen Chapter Relations Contents Properties of Relations Discrete Structures for Computer Science (CO1007) on Ngày tháng 11 năm 2016 Combining Relations Representing Relations ... Closures of Relations Types of Relations Homeworks Nguyen An Khuong, Huynh Tuong Nguyen Faculty of Computer Science and Engineering University of Technology, VNU-HCM 4.1 Contents Relations Nguyen An ... equivalence classes of an equivalence relation R on a set S form a partition of S Homework Every partition of a set can be used to form an equivalence relation 4.31 Relations Example Nguyen An

Ngày tải lên: 29/03/2017, 18:30

... Counting Nguyen An Khuong, Huynh Tuong Nguyen Chapter Counting Contents Introduction Discrete Structures for Computer Science (CO1007) on Ngày 17 tháng 11 năm 2016 Counting Techniques Pigeonhole Principle ... project from one of three fields: Information system (32 projects), Software Engineering (12 projects) and Computer Science (15 projects) How many ways are there for a student to choose? Solution: ... • Probability Pigeonhole Principle • Statistics Permutations & Combinations • Computer science • Game theory • Information theory • 5.4 Problems Counting Nguyen An Khuong, Huynh Tuong Nguyen

Ngày tải lên: 29/03/2017, 18:30

... ιντεγερκ 2 ⇒ ν = κ , ωηιχη ισεϖεν 1/25/12 More slowly … • Thm For any integer n, n2 is odd if and only if n is odd • To prove a statement of the form “P iff Q,” two separate proofs are needed: – If P ... the assertions together are abbreviated “P iff Q” or “P⇔Q” or “P ≡Q” 1/25/12 More slowly … • Thm For any integer n, n2 is odd if and only if n is odd () “If n2 is odd then n is odd” is equivalent ... (“contrapositive”) which is the same as “if n is even then n2 is even” (since n is an integer) … then n=2k for some k and n2=4k2, which is even 1/25/12 Contrapositive and converse • The contrapositive of

Ngày tải lên: 22/03/2019, 10:35

... bottom pancake 03/22/19 Why does this take 2n-3 flips? • For n≥2, let P(n) := “n pancakes can be sorted using 2n-3 flips” • Suppose this is false for some n • Let C = {n: P(n) is false} • C has a least ... to “lowest terms”  The set of factors of a positive integer is nonempty 03/22/19 To prove P(n) for every nonnegative n: • Let C = {n: P(n) is false} (the set of “counterexamples”) • Assume C ... by WOP Call it m • So m pancakes cannot be sorted using 2m-3 flips and m is the smallest number for which that is the case 03/22/19 Why does this take 2n-3 flips? • m≠2 since one flip sorts pancakes

Ngày tải lên: 22/03/2019, 10:37

... Logic and computers 2/6/12 Binary Arithmetic Only two digits: the bits and (Think: = F, = T) +0 -0 2/6/12 +1 -1 +0 -1 +1 -10 Logic and Computers  A half adder:  Two ... Simpler formulas turn into circuits that use less hardware! • E.g p ⋁ q ⋁ (p⋀q) is equivalent to p ⋁ q but would use more logic gates • But the P=NP? question means that it may be hard to simplify formulas ... possible – Any tautology is equivalent to p ⋁ ¬p so if we could easily simplify formulas we could easily determine whether a formula is a tautology 2/6/12

Ngày tải lên: 22/03/2019, 10:41

... or “A is contained in B” (∀x) (x∈A ⇒ x∈B) N ⊆ Z, {7} ⊆ {7, “Sunday”, π} ∅ ⊆ A for any set A (∀x) (x∈∅ ⇒ x∈A) A ⊆ A for any set A To be clear that A ⊆ B but A ≠ B, write A ⊊ B “Proper subset” (I ... Sets 2/10/12 What is a Set? • Informally, a collection of objects, determined by its members, treated as a single mathematical object

Ngày tải lên: 22/03/2019, 10:45

... → R f(x,y) = x/y Defined for all pairs (x,y) except when y=0! 2/13/12 A Function that is “Partial,” Not Total domain f R×R codomain R f: R ×R → R f(x,y) = x/y Defined for all pairs (x,y) except ... “Size” For finite sets, a bijection exists iff A and B have the same number of elements domain A 2/13/12 f codomain B 10 Cardinality or “Size” Use the same as a definition of “same size” for infinite

Ngày tải lên: 22/03/2019, 10:47

... contradiction: suppose Pf f:A↔P(A) is a bijection Let W::= {a  A|a f(a)}, so for any a, a W iff a f(a) f is a bijection, so W=f(a0), for some a0 A (∀a) a f(a0) iff a f(a ) 2/22/12 12 There is no bijection ... contradiction: suppose Pf f:A↔P(A) is a bijection Let W::= {a  A|a f(a)}, so for any a, a W iff a f(a) f is a bijection, so W=f(a0), for some a0 A 2/22/12 a 0contradiction f(a0) iff a 0f(a ) 13 So ... every row! So cannot be listed: this diagonal sequence will be missing 2/22/12 10 Cantor’s Theorem For every set, A (finite or infinite), there is no bijection A↔P(A) 2/22/12 11 There is no bijection

Ngày tải lên: 22/03/2019, 10:49

... Example Induction Proof Let’s prove: 1+r +r + (for r ≠ 1) n +r = (n+ 1) r -1 r -1 Example Induction Proof Statements in magenta form a template for inductive proofs: • • Proof: (by induction ... +r n = (for r ≠ 1) ( n+ ) r -1 r -1 Example Induction Proof Base Case (n = 0): ? r 0+1 - 1+r +r +L +r = r -1 OK! r -1 = =1 r -1 Example Induction Proof • Inductive Step: Assume P(n) for some

Ngày tải lên: 22/03/2019, 10:51

... n-1 multiplications • Method 2: use successive squaring – How many times can you divide n by before it is reduced to 1? – Repeated squaring requires between log2n and 2∙log2n multiplications ... multiplications • Method 2: use successive squaring – Requires about log2n multiplications • Same idea works for multiplication modulo p • Example: If n is a 500-digit number, we can compute qn (mod p) in

Ngày tải lên: 22/03/2019, 10:55

...  ⎠ For example, n                  Note that n2+1 is being used to name the function f such that f(n) = n2+1 for every n 3/26/12 An example: Stirling’s formula ...   Little-Oh: f = o(g) • Def: f(n) = o(g(n)) if  lim  n   • For example, n2 = o(n3) since   lim      n     3/26/12 = o( ∙ ) is “all one symbol” ...                    So,? ?for? ?example, 3n      3/26/12  Rough Paraphrase • f∼g: f and g grow to be roughly equal •

Ngày tải lên: 22/03/2019, 10:59

... 4-colorable 1850’s: false proof published (was correct for colors) 1970’s: proof with computer 1990’s: much improved 3/16/12 15 Chromatic Number #colors for G is chromatic number, χ(G) lemma: 3/16/12 ... slots How short an exam period? 3/16/12 Harvard’s Solution Different “exam group” for every teaching hour Exams for different groups at different times 3/16/12 10 3/16/12 11 But This May be Suboptimal

Ngày tải lên: 22/03/2019, 11:07

... independent events iff Pr(A|B) = Pr(A) • Proof: That is, knowing whether B is the case gives no information that would help determine the probability of A A and B independent iff Pr(A)∙Pr(B) = Pr(A∩B) ... the probability that Santorum will be the Republican nominee? Total Probability Simple version: For any events A and B whose probability is neither nor 1: Pr(A) = Πρ( Α | Β)⋅ Πρ( Β) + Πρ( Α |

Ngày tải lên: 22/03/2019, 11:14

... one choice for the last So by the product rule, n ∙ (n-1) ∙ (n-2) ∙ … ∙ 2.1 = n! How Many 4-Letter Words Using Each Letter at Most Once? • • • • • • 26 choices for first letter Only 25 for second ... Letter at Most Once? • • • • • • 26 choices for first letter Only 25 for second letter 24 for third letter 23 for fourth letter So 26∙25∙24∙23 or 26!/22! Generalized Product Rule • • Let Q be a set ... |Q| = n1⋅n2⋅⋅⋅nk if n1 possible 1st elements, n2 possible 2nd elements (for each first entry), n3 possible 3rd elements (for each 1st & 2nd entry, ) then, How Many Hands with Cards? • • • • I.e.,

Ngày tải lên: 22/03/2019, 11:18

... transitive if R = G+ for some digraph G 3/22/19 Transitive Closure G+ is the transitive closure of the binary relation G 3/22/19 reflexivity relation R on set A is reflexive if a R a for all a A ≤ ... is reflexive if a R a for all a A ≤ on numbers and ⊆ on sets are reflexive 3/22/19 reflexivity For any digraph G, * G is reflexive 3/22/19 Reflexive Transitive Closure G* is the reflexive transitive ... IMPLIES NOT(v D 3/22/19 * u) 12 antisymmetry antisymmetric relation R: u R v IMPLIES NOT(v R u) for any u ≠ v If D is a DAG then * 3/22/19 13 (weak) partial orders Reflexive, Transitive, and Antisymmetric

Ngày tải lên: 22/03/2019, 11:30

... put the books on the shelves as follows: put all the books before the first piece of wood on shelf 1, all the books between the first and second on shelf 2, and so on until you put all the books ... distinct elements. There are n choices for the first number in the list. For each way of choosing the first element, there are n −1choices for the second. For each choice of the first two elements, ... may we place k distinct books on n shelves of a bookcase (all books pushed to the left as far as possible) if there must be at least one book on each shelf? 8. The formula for the number of multisets...

Ngày tải lên: 21/02/2014, 09:20