Riddler: Can You Simply Maintain Turning?

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FiveThirtyEight’s Riddler Specific

hyperlink

In Riddler Metropolis, the town streets observe a grid format, working
north-south and east-west. You’re driving north whenever you resolve to
play a bit of recreation. Each time you attain an intersection, you randomly
flip left or proper, every with a 50 % likelihood.

After driving via 10 intersections, what’s the likelihood that
you’re nonetheless driving north?

Further credit score: Now suppose that at each intersection, there’s a
one-third likelihood you flip left, a one-third likelihood you flip proper and
a one-third likelihood you drive straight. After driving via 10
intersections, now what’s the likelihood that you’re nonetheless driving
north?

Plan

This puzzle might be solved analytically, however that might require quite a bit
extra thought than simply simulating it. I’ll attempt to make the algorithm
generalizable to additionally be capable to remedy the additional credit score downside with out
too many adjustments.

Setup

knitr::opts_chunk$set(echo = TRUE, remark = "#>", cache = TRUE, dpi = 400)
library(tidyverse)
library(conflicted)
# Deal with any namespace conflicts.
conflict_prefer("filter", "dplyr")
conflict_prefer("choose", "dplyr")
# Default 'ggplot2' theme.
theme_set(theme_minimal())
# For reproducibility.
set.seed(0)

A single simulation

The abstraction

The code for the simulation itself could be very easy and primarily contained
inside two capabilities, simulate_one_drive() that coordinates every little thing
and adjust_direction() that turns the participant based mostly on the present
course and random flip. Extra on these in a second.

The tough half for this simulation was selecting an abstraction. My
first few makes an attempt relied on conserving monitor of the cardinal course
(i.e. north, south, east, and west), randomly deciding the flip, and
then updating the cardinal course based mostly on the flip. However this was
annoyingly sophisticated as a result of the impact of left or proper on the
cardinal course is determined by the course, itself. Subsequently, it
regarded like I would want to put in writing a large (learn “error-prone”) if-else
assertion.

After fairly a little bit of thought and diagramming, I spotted I might use
angles to unravel the issue. If I set north as $frac{pi}{2}$, then a
left flip could be equal to including $frac{pi}{2}$ and turning
proper could be equal to subtracting $frac{pi}{2}$. And this
could be true whatever the present course!

Utilizing this strategy, I made a decision to simply maintain monitor of the angle of the
present course and ignore truly touring via the town. I
might simply calculate this afterwards, if wanted.

The primary course of

Lastly, we are able to get to the code. The simulate_one_drive() operate
takes chances for turning left (l), proper (r), or persevering with
straight (s) and an argument for the variety of steps within the simulation
(n_steps).

Immediately, the beginning course is outlined as $frac{1}{2}$. I left
out $pi$ from the simulation as a result of I by no means really need radians,
only a relative unit for the angle. Subsequently, as a substitute of starting from 0
to $2pi$, the “angle” ranges from Zero to 2. Multiplying by $pi$
later can return the radians.

Earlier than the for-loop, there’s a fast examine to verify the
chances sum to 1.

Lastly, the tracker is instantiated as a knowledge body with the
interplay, present course, and the selection of flip (“S” for straight
to start with).

In every step of the for-loop

  1. a flip is randomly chosen in accordance with their predetermined
    likelihoods,
  2. the course is modified in accordance with the results of the random
    choice utilizing the adjust_direction() operate (extra in a
    second),
  3. the tracker is up to date with the present step.

The tracker is returned as the results of the simulation.

# Simulate one drive.
simulate_one_drive <- operate(l, r, s, n_steps = 10) {
# Begin dealing with "North".
dir <- 1/2
# Verify that the overall likelihood of turning decisions is 1.
stopifnot(sum(c(l, r, s)) == 1)
# Begin the tracker.
tracker <- update_tracker(tibble(), 0, dir, "S")
# Take `n_steps` for the simulation.
for (i in seq(1, n_steps)) {
next_turn <- pattern(c("L", "R", "S"), 1, prob = c(l, r, s))
dir <- adjust_direction(dir, next_turn)
tracker <- update_tracker(tracker, i, dir, next_turn)
}
return(tracker)
}

The replace tracker operate is only a comfort operate for including
rows to an information body of the present state of the simulation at every
step.

# Replace the tracker knowledge body.
update_tracker <- operate(tracker, i, dir, flip) {
bind_rows(
tracker,
tibble(i = i, course = dir, flip = flip)
)
}

The adjust_direction() operate takes a present course (curr_dir)
and which approach to flip (flip). It then provides $frac{1}{2}$ to show
left ("L") or subtracts $frac{1}{2}$ to show proper ("R"). The
unique course is returned to proceed straight ("S").

Notice that the rationale this operate is so easy shouldn’t be as a result of I did
something intelligent with the code, however as a substitute it’s as a result of the abstraction
is so pure to the issue.

# Regulate the present course `curr_dir` based mostly off of the `flip`.
adjust_direction <- operate(curr_dir, flip) {
new_dir <- curr_dir
if (flip == "L") {
new_dir <- curr_dir + 0.5
} else if (flip == "R") {
new_dir <- curr_dir - 0.5
} else if (flip == "S") {
new_dir <- curr_dir
} else {
cease(paste0("The change in course '", flip, "' shouldn't be acknowledged."))
}
return(new_dir)
}

An instance simulation

Beneath I run a single instance simulation and obtain again the tracker.

set.seed(0)
example_sim <- simulate_one_drive(0.5, 0.5, 0, n_steps = 10)
example_sim

#> # A tibble: 11 x 3
#> i course flip
#> #> 1 Zero 0.5 S
#> 2 1 1 L
#> Three 2 0.5 R
#> Four Three Zero R
#> 5 Four 0.5 L
#> 6 5 1 L
#> 7 6 0.5 R
#> Eight 7 1 L
#> 9 Eight 1.5 L
#> 10 9 2 L
#> 11 10 2.5 L

From the course column, we are able to calculate the change in $x$ and
$y$ by changing from polar coordinates $(r, theta)$ to cartesian
coordinates $(x,y)$:

$x = r occasions cos(theta)$
$y = r occasions sin(theta)$

example_sim2 <- example_sim %>%
mutate(dx = spherical(1 * cos(course * pi)),
dy = spherical(1 * sin(course * pi)))
example_sim2

#> # A tibble: 11 x 5
#> i course flip dx dy
#> #> 1 Zero 0.5 S Zero 1
#> 2 1 1 L -1 0
#> Three 2 0.5 R Zero 1
#> Four Three Zero R 1 0
#> 5 Four 0.5 L Zero 1
#> 6 5 1 L -1 0
#> 7 6 0.5 R Zero 1
#> Eight 7 1 L -1 0
#> 9 Eight 1.5 L 0 -1
#> 10 9 2 L 1 0
#> 11 10 2.5 L Zero 1

With the change in $x$ and $y$ after every flip within the simulation,
the precise place of the automotive on the grid will be calculated. That is
completed by utilizing the
accumulate2()
operate from the
‘purrr’
bundle.

calculate_position <- operate(pos, dx, dy) {
new_pos <- pos
new_pos$x <- pos$x + dx
new_pos$y <- pos$y + dy
return(new_pos)
}
example_sim3 <- example_sim2 %>%
mutate(pos = accumulate2(dx, dy,
calculate_position,
.init = listing(x = 0, y = 0))[-1],
x = map_dbl(pos, ~ .x$x),
y = map_dbl(pos, ~ .x$y))

Lastly, to have a extra satisfying visualization of the simulation, we
can plot the $x$ and $y$ positions for every flip.

example_sim3 %>%
ggplot(aes(x, y)) +
geom_path(group = "a",
arrow = arrow(size = unit(4, "mm"), ends = "final", sort = "closed")) +
geom_point() +
ggrepel::geom_text_repel(aes(label = i))

To facilitate additional evaluation of the outcomes of simulations, I packaged
the above steps right into a single operate
simulation_results_to_cartesian_positions().

calculate_position <- operate(pos, dx, dy) {
new_pos <- pos
new_pos$x <- pos$x + dx
new_pos$y <- pos$y + dy
return(new_pos)
}
simulation_results_to_cartesian_positions <- operate(df) {
df %>%
mutate(dx = spherical(1 * cos(course * pi)),
dy = spherical(1 * sin(course * pi)),
pos = accumulate2(dx, dy,
calculate_position,
.init = listing(x = 0, y = 0))[-1],
x = map_dbl(pos, ~ .x$x),
y = map_dbl(pos, ~ .x$y))
}
simulation_results_to_cartesian_positions(example_sim)

#> # A tibble: 11 x 8
#> i course flip dx dy pos x y
#> #> 1 Zero 0.5 S Zero 1 Zero 1
#> 2 1 1 L -1 0 -1 1
#> Three 2 0.5 R Zero 1 -1 2
#> Four Three Zero R 1 Zero Zero 2
#> 5 Four 0.5 L Zero 1 Zero 3
#> 6 5 1 L -1 0 -1 3
#> 7 6 0.5 R Zero 1 -1 4
#> Eight 7 1 L -1 0 -2 4
#> 9 Eight 1.5 L 0 -1 -2 3
#> 10 9 2 L 1 0 -1 3
#> 11 10 2.5 L Zero 1 -1 4

I additionally made a extra expressive plotting operate plot_simulation() that
exhibits the course at every step.

plot_simulation <- operate(df) {
df %>%
group_by(sim) %>%
mutate(x_start = dplyr::lag(x, default = 0),
y_start = dplyr::lag(y, default = 0)) %>%
ungroup() %>%
ggplot() +
geom_segment(aes(x = x_start, y = y_start, xend = x, yend = y,
coloration = i, group = sim),
arrow = arrow(size = unit(3, "mm"), sort = "closed"),
alpha = 1.0, measurement = 1) +
geom_label(aes((x_start + x) / 2, (y_start + y) / 2, label = i, fill = i),
coloration = "white", label.measurement = 0, fontface = "daring") +
scale_color_gradient(low = "gray70", excessive = "gray15", information = FALSE) +
scale_fill_gradient(low = "gray75", excessive = "gray25", information = FALSE) +
labs(x = "W <-- lateral course --> E",
y = "S <-- longitudinal course --> N")
}
example_sim %>%
simulation_results_to_cartesian_positions() %>%
mutate(sim = 1) %>%
plot_simulation()

The simulation

Lastly, we are able to run a bunch of simulations and answeer the unique
query:

After driving via 10 intersections, what’s the likelihood that
you’re nonetheless driving north?

First, lets plot the outcomes of 5 simulations to make sure that the
simulation is working as anticipated over a number of runs.

set.seed(0)
tibble(sim = 1:5) %>%
mutate(res = map(sim, ~ simulate_one_drive(0.5, 0.5, 0, n_steps = 10)),
res = map(res, simulation_results_to_cartesian_positions)) %>%
unnest(res) %>%
plot_simulation()

With that examine performed, we’re able to run a couple of thousand simulations.

set.seed(0)
N_sims <- 1e4
simulation_results <- tibble(sim = 1:N_sims) %>%
mutate(res = map(sim, ~ simulate_one_drive(0.5, 0.5, 0, n_steps = 10)))
simulation_results

#> # A tibble: 10,000 x 2
#> sim res
#> #> 1 1 #> 2 2 #> Three 3 #> Four 4 #> 5 5 #> 6 6 #> 7 7 #> Eight 8 #> 9 9 #> 10 10 #> # … with 9,990 extra rows

Now now we have a protracted knowledge body with nested knowledge frames, every one
respresenting the outcomes of a single simulation. We now need to inform if
the ultimate course was pointing north. Nevertheless, there’s one refined
downside: $frac{pi}{2} = frac{5pi}{2} = frac{9pi}{2} = …$.
There are a lot of (infinite) potential angles that each one level north.
Subsequently, I wrote the reduce_angle() operate to cut back any angle to
lie inside Zero and a pair of (as a result of we eliminated the fixed $pi$ from the
angle of course).

# Scale back the angle from an worth to between Zero and a pair of.
reduce_angle <- operate(theta) {
theta - (2 * trunc(theta / 2))
}

Now, we are able to unnest the simulation outcomes, take the final course, and
see whether it is pointing north.

simulation_results <- simulation_results %>%
unnest(res) %>%
filter(i == 10) %>%
mutate(reduced_direction = reduce_angle(course))
prob_north <- sum(simulation_results$reduced_direction == 0.5) / N_sims

The likelihood of nonetheless dealing with north after randomly turning left and
proper at every intersection is 0.369.

Since I allowed for a likelihood of going straight within the
simulate_one_drive() operate, fixing the additional credit score downside
requires no change to the code apart from a single argument worth.

Further credit score: Now suppose that at each intersection, there’s a
one-third likelihood you flip left, a one-third likelihood you flip proper and
a one-third likelihood you drive straight. After driving via 10
intersections, now what’s the likelihood that you’re nonetheless driving
north?

set.seed(0)
tibble(sim = 1:5) %>%
mutate(res = map(sim, ~ simulate_one_drive(1/3, 1/3, 1/3, n_steps = 10)),
res = map(res, simulation_results_to_cartesian_positions)) %>%
unnest(res) %>%
plot_simulation()

set.seed(0)
simulation_results <- tibble(sim = 1:N_sims) %>%
mutate(res = map(sim, ~ simulate_one_drive(1/3, 1/3, 1/3, n_steps = 10))) %>%
unnest(res) %>%
filter(i == 10) %>%
mutate(reduced_direction = reduce_angle(course))
prob_north <- sum(simulation_results$reduced_direction == 0.5) / N_sims

The likelihood of nonetheless dealing with north after randomly turning left,
proper, or persevering with straight at every intersection is 0.205.



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