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# 07 Problem Solving

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### 07 Problem Solving

1. 1. Stat405 Problem solving Hadley Wickham Wednesday, 16 September 2009
2. 2. Homework Great improvement. A few tips: Headings should be in sentence case, and be a single sentence summary of the section. Include ggsave calls in your code. Don’t use " and " to get “ and ”. Use ` ` and ' '. (Also, I think I’ve been grading quite hard. Think of a 5 as an A+, a 4 as an A, and a 3 as an A-/B+) Wednesday, 16 September 2009
3. 3. Feedback Speed seems about right. Frustration is a natural part of the learning experience. You will get better! In particularly, it will get easier to ﬁnd interesting plots (helps to start homework early!) Split homework help session? Also, feel free to email me to set up time to talk. Wednesday, 16 September 2009
4. 4. 1. Recap of problem 2. Task 3. Basic strategy 4. Turning ideas into code Wednesday, 16 September 2009
5. 5. Slots Casino claims that slot machines have prize payout of 92%. Is this claim true? mean(slots\$payoff) t.test(slots\$prize, mu = 0.92) qplot(prize, data = slots, binwidth = 1) Can we do better? Wednesday, 16 September 2009
6. 6. Bootstrapping Slot machine is basically a random number generator - three numbers between 1 and 7. We can generate a new number by sampling from the data. Doing that lots of times, and calculating the payoff should give us a better idea of the payoff. But ﬁrst we need some way to calculate the prize from the windows Wednesday, 16 September 2009
7. 7. DD DD DD 800 windows <- c("DD", "C", "C") 7 7 7 80 # How can we calculate the BBB BBB BBB 40 # payoff? B B B 10 C C C 10 Any bar Any bar Any bar 5 C C * 5 C * C 5 C * * 2 * C * 2 DD doubles any winning combination. Two DD * * C 2 quadruples. Wednesday, 16 September 2009
8. 8. Your turn There are three basic cases of prizes. What are they? Take 3 minutes to brainstorm with a partner. Wednesday, 16 September 2009
9. 9. Cases 1. All windows have same value 2. A bar (B, BB, or BBB) in every window 3. Cherries and diamonds Wednesday, 16 September 2009
10. 10. Same values Wednesday, 16 September 2009
11. 11. Same values 1. Determine that all windows are the same. How? Wednesday, 16 September 2009
12. 12. Same values 1. Determine that all windows are the same. How? 2. Given single window, look up prize value. How? Wednesday, 16 September 2009
13. 13. Same values 1. Determine that all windows are the same. How? 2. Given single window, look up prize value. How? With a partner, brainstorm for 2 minutes on how to solve one of these problems Wednesday, 16 September 2009
14. 14. # Same value same <- length(unique(windows)) == 1 # OR same <- windows[1] == windows[2] && windows[2] == windows[3] if (same) { # Lookup value } Wednesday, 16 September 2009
15. 15. If if (cond) expression Cond should be a logical vector of length 1 Use && and || to combine sub-conditions Not & and | - these work on vectors && and || are “short-circuiting”: they do the minimum amount of work Wednesday, 16 September 2009
16. 16. if (TRUE) { # This will be run } if (FALSE) { # This won't be run } else { # This will be } # Single line form: (not recommended) if (TRUE) print("True!) if (FALSE) print("True!) Wednesday, 16 September 2009
17. 17. if (TRUE) { # This will be run } if (FALSE) { # This won't be run } else { # This will be } Note indenting. Very important! # Single line form: (not recommended) if (TRUE) print("True!) if (FALSE) print("True!) Wednesday, 16 September 2009
18. 18. x <- 5 if (x < 5) print("x < 5") if (x == 5) print("x == 5") x <- 1:5 if (x < 3) print("What should happen here?") if (x[1] < x[2]) print("x1 < x2") if (x[1] < x[2] && x[2] < x[3]) print("Asc") if (x[1] < x[2] || x[2] < x[3]) print("Asc") Wednesday, 16 September 2009
19. 19. if (window[1] == "DD") { prize <- 800 } else if (windows[1] == "7") { prize <- 80 } else if (windows[1] == "BBB") ... # Or use subsetting c("DD" = 800, "7" = 80, "BBB" = 40) c("DD" = 800, "7" = 80, "BBB" = 40)["BBB"] c("DD" = 800, "7" = 80, "BBB" = 40)["0"] c("DD" = 800, "7" = 80, "BBB" = 40)[window[1]] Wednesday, 16 September 2009
20. 20. Your turn Complete the previous code so that if all the values in win are the same, then prize variable will be set to the correct amount. Wednesday, 16 September 2009
21. 21. All bars How can we determine if all of the windows are B, BB, or BBB? (windows[1] == "B" || windows[1] == "BB" || windows[1] === "BBB") && ... ? Wednesday, 16 September 2009
22. 22. All bars How can we determine if all of the windows are B, BB, or BBB? (windows[1] == "B" || windows[1] == "BB" || windows[1] === "BBB") && ... ? Take 2 minutes to brainstorm possible solutions Wednesday, 16 September 2009
23. 23. windows[1] %in% c("B", "BB", "BBB") windows %in% c("B", "BB", "BBB") allbars <- windows %in% c("B", "BB", "BBB") allbars[1] & allbars[2] & allbars[3] all(allbars) # See also ?any for the complement Wednesday, 16 September 2009
24. 24. Your turn Complete the previous code so that the correct value of prize is set if all the windows are the same, or they are all bars Wednesday, 16 September 2009
25. 25. payoffs <- c("DD" = 800, "7" = 80, "BBB" = 40, "BB" = 25, "B" = 10, "C" = 10, "0" = 0) same <- length(unique(windows)) == 1 allbars <- all(windows %in% c("B", "BB", "BBB")) if (same) { prize <- payoffs[windows[1]] } else if (allbars) { prize <- 5 } Wednesday, 16 September 2009
26. 26. Cherries Need numbers of cherries, and numbers of diamonds (hint: use sum) Then need to look up values (like for the ﬁrst case) and multiply together Wednesday, 16 September 2009
27. 27. cherries <- sum(windows == "C") diamonds <- sum(windows == "DD") c(0, 2, 5)[cherries + 1] * c(1, 2, 4)[diamonds + 1] Wednesday, 16 September 2009
28. 28. payoffs <- c("DD" = 800, "7" = 80, "BBB" = 40, "BB" = 25, "B" = 10, "C" = 10, "0" = 0) same <- length(unique(windows)) == 1 allbars <- all(windows %in% c("B", "BB", "BBB")) if (same) { prize <- payoffs[windows[1]] } else if (allbars) { prize <- 5 } else { cherries <- sum(windows == "C") diamonds <- sum(windows == "DD") prize <- c(0, 2, 5)[cherries + 1] * c(1, 2, 4)[diamonds + 1] } Wednesday, 16 September 2009
29. 29. Writing a function Now we need to wrap up this code in to a reusable fashion. We need a function Have used functions a lot, and this will be our ﬁrst time writing one Wednesday, 16 September 2009
30. 30. Basic structure • Name • Input arguments • Names/positions • Defaults • Body • Output value Wednesday, 16 September 2009
31. 31. calculate_prize <- function(windows) { payoffs <- c("DD" = 800, "7" = 80, "BBB" = 40, "BB" = 25, "B" = 10, "C" = 10, "0" = 0) same <- length(unique(windows)) == 1 allbars <- all(windows %in% c("B", "BB", "BBB")) if (same) { prize <- payoffs[windows[1]] } else if (allbars) { prize <- 5 } else { cherries <- sum(windows == "C") diamonds <- sum(windows == "DD") prize <- c(0, 2, 5)[cherries + 1] * c(1, 2, 4)[diamonds + 1] } prize } Wednesday, 16 September 2009
32. 32. Homework What did we miss? Find out the problem with our scoring function and ﬁx it. Also, some readings related to group work. Wednesday, 16 September 2009