CSE 20. Lecture 4: Number System and Boolean Function. CSE 20: Lecture2

Transcription

1 CSE 20 Lecture 4: Number System and Boolean Function

2 Next Weeks Next week we will do Unit:NT, Section 1. There will be an assignment set posted today. It is just for practice.

3 Boolean Functions and Number System

4 Boolean Functions and Number System Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

5 Boolean Functions and Number System Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 What does 217 read like?

6 Boolean Functions and Number System Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 What does 217 read like? Usually we represent our number in decimal representation. Like: 217 =

7 Numbers with base b Usually we represent our number in decimal representation. Like: 217 =

8 Numbers with base b Usually we represent our number in decimal representation. Like: 217 = One can represent a number is any base.

9 Numbers with base b Usually we represent our number in decimal representation. Like: 217 = One can represent a number is any base. Like: 217 =

10 Numbers with base b Usually we represent our number in decimal representation. Like: 217 = One can represent a number is any base. Like: 217 = Thus 217 = [22001] 3.

11 Numbers with base b Usually we represent our number in decimal representation. Like: 217 = One can represent a number is any base. Like: 217 = Thus 217 = [22001] 3. One can represent a number in any base.

12 Base b representation

13 Base b representation Digits: 0, 1,..., b 1

14 Base b representation Digits: 0, 1,..., b 1 Represented as [x] b. (Like [22001] 3 )

15 Base b representation Digits: 0, 1,..., b 1 Represented as [x] b. (Like [22001] 3 ) Base b representation of a number x is the unique way of writing x = x 0 b 0 + x 1 b x k b k, where x 0, x 1,..., x k {0, 1,..., (b 1)}

16 Base b representation Digits: 0, 1,..., b 1 Represented as [x] b. (Like [22001] 3 ) Base b representation of a number x is the unique way of writing x = x 0 b 0 + x 1 b x k b k, where x 0, x 1,..., x k {0, 1,..., (b 1)} So, x = [x k x k 1... x 1 x 0 ] b

17 Questions 1 Why and how can any integer be written in any base?

18 Questions 1 Why and how can any integer be written in any base? Given an method or algorithm to find the representation given x and b.

19 Questions 1 Why and how can any integer be written in any base? Given an method or algorithm to find the representation given x and b. 2 Why does an integer has an unique representation in a base b?

20 Questions 1 Why and how can any integer be written in any base? Given an method or algorithm to find the representation given x and b. 2 Why does an integer has an unique representation in a base b? If an algorithm A 1 write x = [x k x k 1... x 0 ] b and another algorithm A 2 write x = [y k y k 1... y 0 ] b, then is the two representations same.

21 Questions 1 Why and how can any integer be written in any base? Given an method or algorithm to find the representation given x and b. 2 Why does an integer has an unique representation in a base b? If an algorithm A 1 write x = [x k x k 1... x 0 ] b and another algorithm A 2 write x = [y k y k 1... y 0 ] b, then is the two representations same. In other words, is y i = x i for all 0 i k?

22 iclicker question What is the representation of 6102 in binary?

23 iclicker question What is the representation of 6102 in binary? 1 [ ] 2 2 [ ] 2 3 [ ] 2 4 [ ] 2

24 iclicker question What is the sum of [24887] 9 and [13808] 9?

25 iclicker question What is the sum of [24887] 9 and [13808] 9? 1 [38806] 9 2 [38695] 9 3 [38696] 9 4 [38886] 9

26 Addition is base b representation

27 Addition is base b representation Add as numbers and represent in the base b representation.

28 Addition is base b representation Add as numbers and represent in the base b representation. For example: In base 3 [2] 3 + [1] 3 = [10] 3

29 Addition is base b representation Add as numbers and represent in the base b representation. For example: In base 3 [2] 3 + [1] 3 = [10] 3 [11] 3 + [12] 3 = [100] 3

30 Addition is base b representation Add as numbers and represent in the base b representation. For example: In base 3 [2] 3 + [1] 3 = [10] 3 [11] 3 + [12] 3 = [100] 3 [121] 3 + [22] 3 = [220] 3

31 iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b)

32 iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 < Depends on the values of a and b 4 Can t Say

33 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b)

34 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256.

35 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512

36 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512 (2 9 a) + (2 9 b) = ( ) (a + b)

37 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512 (2 9 a) + (2 9 b) = ( ) (a + b) = 2 10 (a + b)

38 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512 (2 9 a) + (2 9 b) = ( ) (a + b) = 2 10 (a + b) Since (a + b) > 0

39 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512 (2 9 a) + (2 9 b) = ( ) (a + b) = 2 10 (a + b) Since (a + b) > 0 and so (2 9 a) + (2 9 b) = 2 10 (a + b) < 2 10

40 iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b)

41 iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) > Depends on the values of a and b

42 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b)

43 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256.

44 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512

45 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512 (2 9 a) + (2 9 b) = (2 9 (a + b))

46 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512 (2 9 a) + (2 9 b) = (2 9 (a + b)) Since (a + b) 512 so 2 9 (a + b) 0

47 Proof of iclicker question If a and b are integers in the range 1 and 256 then is (2 9 a) + (2 9 b) 1 a 256 and 1 b 256. So 2 a + b 512 (2 9 a) + (2 9 b) = (2 9 (a + b)) Since (a + b) 512 so 2 9 (a + b) 0 and so (2 9 a) + (2 9 b) 2 9