Dynamic Programming Part 2: Probability, Combinatorics, and Bitmasks

Transcription

1 Dynamic Programming Part 2: Probability, Combinatorics, and Bitmasks Duke Compsci 309s Siyang Chen

2 Introduction Probability, combinatorics, and bitmasking appear commonly in dynamic programming problems.

3 Kolmogorov s axioms of probability The probability P(A) of an event A is a nonnegative real number. The sum of the probabilities of all atomic events is 1. If A and B are mutually exclusive, then P(A B) = P(A) + P(B).

4 Kolmogorov s axioms of probability The probability P(A) of an event A is a nonnegative real number. The sum of the probabilities of all atomic events is 1. If A and B are mutually exclusive, then P(A B) = P(A) + P(B). I m assuming everyone has a basic understanding of probability, so we won t dwell on these here.

5 Useful properties Independence: Two events A and B are called independent if P(A)P(B) = P(A B)

6 Useful properties Independence: Two events A and B are called independent if P(A)P(B) = P(A B) Linearity of expectation: If a random variable A can take on some values a 1, a 2,..., a n, then its expected value is E(A) = n a i P(A = a i ) i=1

7 Useful properties Independence: Two events A and B are called independent if P(A)P(B) = P(A B) Linearity of expectation: If a random variable A can take on some values a 1, a 2,..., a n, then its expected value is E(A) = n a i P(A = a i ) i=1 Expectation is linear, i.e.: E(A + cb) = E(A) + ce(b) Note that this is true even for events that aren t independent!

8 Example problem: CoinReversing Look at the problem CoinReversing on the syllabus.

9 Example problem: CoinReversing Represent a coin s end state by { 1 if heads C i = 0 if tails

10 Example problem: CoinReversing Represent a coin s end state by { 1 if heads C i = 0 if tails We want to compute E(C 1 + C C N )

11 Example problem: CoinReversing Represent a coin s end state by { 1 if heads C i = 0 if tails We want to compute E(C 1 + C C N ) By linearity of expectation, this is E(C 1 ) + E(C 2 ) E(C N )

12 Example problem: CoinReversing Represent a coin s end state by { 1 if heads C i = 0 if tails We want to compute E(C 1 + C C N ) By linearity of expectation, this is E(C 1 ) + E(C 2 ) E(C N ) Since the coins are the same, this is N E(C)

13 Example problem: CoinReversing Now we want to compute E(C), where C is 1 if the coin is heads after the K flips and 0 otherwise.

14 Example problem: CoinReversing Now we want to compute E(C), where C is 1 if the coin is heads after the K flips and 0 otherwise. Let p i be the probability that the coin is heads up after i flips. We are given that p 0 = 1; E(C) = p K.

15 Example problem: CoinReversing Now we want to compute E(C), where C is 1 if the coin is heads after the K flips and 0 otherwise. Let p i be the probability that the coin is heads up after i flips. We are given that p 0 = 1; E(C) = p K. Let f i be the probability that some coin is flipped on the i-th step:

16 Example problem: CoinReversing Now we want to compute E(C), where C is 1 if the coin is heads after the K flips and 0 otherwise. Let p i be the probability that the coin is heads up after i flips. We are given that p 0 = 1; E(C) = p K. Let f i be the probability that some coin is flipped on the i-th step: f i = a i /N.

17 Example problem: CoinReversing Now we want to compute E(C), where C is 1 if the coin is heads after the K flips and 0 otherwise. Let p i be the probability that the coin is heads up after i flips. We are given that p 0 = 1; E(C) = p K. Let f i be the probability that some coin is flipped on the i-th step: f i = a i /N. Since the probability a coin gets flipped at each point in time is independent, then p i+1 = p i (1 f i ) + (1 p i ) f i

18 Example problem: CoinReversing Now we want to compute E(C), where C is 1 if the coin is heads after the K flips and 0 otherwise. Let p i be the probability that the coin is heads up after i flips. We are given that p 0 = 1; E(C) = p K. Let f i be the probability that some coin is flipped on the i-th step: f i = a i /N. Since the probability a coin gets flipped at each point in time is independent, then p i+1 = p i (1 f i ) + (1 p i ) f i Java solution: master/topcoder/srm/518/coinreversing.java

19 Bitmasking Bitmasking is a compact and efficient way to represent sets as the bits in an integer.

20 Bitmasking Bitmasking is a compact and efficient way to represent sets as the bits in an integer. The i-th bit of S is on if and only if element i is included in S.

21 Bitmasking Bitmasking is a compact and efficient way to represent sets as the bits in an integer. The i-th bit of S is on if and only if element i is included in S. E.g.:

22 Bitmasking Bitmasking is a compact and efficient way to represent sets as the bits in an integer. The i-th bit of S is on if and only if element i is included in S. E.g.: 8 = = {0, 1, 2}

23 Bitmasking Bitmasking is a compact and efficient way to represent sets as the bits in an integer. The i-th bit of S is on if and only if element i is included in S. E.g.: 8 = = {0, 1, 2} 41 = = {0, 3, 5}

24 Bitmasking Bitmasking is a compact and efficient way to represent sets as the bits in an integer. The i-th bit of S is on if and only if element i is included in S. E.g.: 8 = = {0, 1, 2} 41 = = {0, 3, 5} 28 = = {2, 3, 4}

25 Bitmasking Bitmasking is a compact and efficient way to represent sets as the bits in an integer. The i-th bit of S is on if and only if element i is included in S. E.g.: 8 = = {0, 1, 2} 41 = = {0, 3, 5} 28 = = {2, 3, 4} Etc...

26 Bitmasking To iterate over all sets of size N, simply iterate from 0 to 2 N 1: S = 000 S = 001 S = 010 S = S = 110 S = 111

27 Bit operations To shift x left by i bits: x << i

28 Bit operations To shift x left by i bits: x << i... or right: x >> i

29 Bit operations To shift x left by i bits: x << i... or right: x >> i To check if the i-th bit is on: (x & (1 << i)) > 0

30 Bit operations To shift x left by i bits: x << i... or right: x >> i To check if the i-th bit is on: (x & (1 << i)) > 0 To flip the i-th bit: x ^ (1 << i)

31 Bit operations To shift x left by i bits: x << i... or right: x >> i To check if the i-th bit is on: (x & (1 << i)) > 0 To flip the i-th bit: x ^ (1 << i) To turn the i-th bit on: x (1 << i)

32 Example problem: Fish Look at the problem Fish on the syllabus.

33 Combinatorics