@dewittpe's solution and write up for @standupmaths's frog_problem

dewittpe-solution/frog_jump.R

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+#'---
+#'title: Frog Jump Problem
+#'author: Peter DeWitt
+#'output: pdf_document
+#'header-includes:
+#'  - \usepackage{blkarray}
+#'  - \usepackage{amsmath}
+#'---
+#'
+#+ label = "setup", include = FALSE, cache = FALSE
+knitr::opts_chunk$set(cache = TRUE)
+library(parallel)
+
+#'
+#' # The Problem:
+#'
+#' A frog is on one bank of a pond.  There are $p$ lily pads between the frog
+#' and the opposite bank of the pond.  This frog has the ability to jump from
+#' the bank it starts on to any of the $p$ lily pads or all the way to the
+#' opposite bank.  The frog will randomly jump, with equal probability, forward,
+#' at least one pad.  The question is, what is the expected number of jumps
+#' the frog will make to transverse the pond?
+#'
+#' \begin{center}
+#' \begin{tabular}{cccccccc}
+#' \includegraphics[width=0.10\textwidth]{grass} &
+#' \includegraphics[width=0.10\textwidth]{lilypad}&
+#' \includegraphics[width=0.10\textwidth]{lilypad}&
+#' \ldots &
+#' \includegraphics[width=0.10\textwidth]{lilypad}&
+#' \includegraphics[width=0.10\textwidth]{lilypad}&
+#' \includegraphics[width=0.10\textwidth]{lilypad}&
+#' \includegraphics[width=0.10\textwidth]{pictures-of-green-frogs-7} \\
+#' Pad $0$ & Pad $1$ & Pad $2$ & \ldots & Pad $p - 2$ & Pad $p - 1$ & Pad $p$ & Pad $p+1$ \\
+#' GOAL: Pond bank & & & & & & & Start
+#' \end{tabular}
+#' \end{center}
+#'
+#'
+#' Goals for the solution:
+#'
+#' 1. Solution is to be general for any number of pads.
+#' 2. Solution should not rely on recursion.
+#'
+#'
+#' # Solution
+#'
+#' ## Simulation
+#'
+#' The simulation does rely on recursion.  This simulation is used to test the
+#' other solution.  The code is writen in [R](https://www.r-project.org).
+#'
+# frog_jumper
+# Arguments:
+#   pads    : the number of lily pads between the frog and the bank of the pond.
+#   jumps   : number of jumps taken so far
+#   verbose : if TRUE print status.
+#
+# Return:
+#   the number of jumps needed to transverse the pond for the given simulation.
+#
+frog_jumper <- function(pads, jumps = 0, verbose = FALSE) {
+  if (verbose) {cat("jump:", jumps, "frog is on pad:", pads + 1, "\n")}
+  distance <- sample(x = seq(1, pads + 1, by = 1), size = 1)
+  if (distance > pads) {
+    if (verbose) {cat("jump:", jumps + 1, "frog Has Reached the bank.\n")}
+    return(invisible(jumps + 1))
+  } else {
+    frog_jumper(pads = pads - distance, jumps + 1, verbose = verbose)
+  }
+}
+
+#'
+#' A couple quick checks of the simulation code.  If there are zero pads between
+#' the frog and the pond bank then the function should return 1.
+set.seed(42)
+frog_jumper(0, verbose = TRUE)
+
+#'
+#' If there is one pad beween the frog and the bank then the function should
+#' return the value 1 with probability 1/2 and the value 2 with probability 1/2.
+#' Test this with 1,000,000 simulations.
+x <- replicate(n = 1e6, expr = {frog_jumper(1)})
+prop.table(table(x))
+
+#'
+#' The expected number of jumps the frog will take if there is only one pad
+#' between the frog an the bank is
+simulation <- function(pads, iterations = 1e6) {
+  mean(
+       do.call(c,
+               mclapply(X        = seq(1, iterations),
+                        FUN      = function(i, ...) {frog_jumper(...) },
+                        pads     = pads,
+                        mc.cores = 12))
+      )
+}
+simulation(1)
+
+#'
+#' For 10 pads between the frog and the opposite bank, the expected number of
+#' jumps the frog will take is
+simulation(10)
+
+#'
+#' ## A Maths Solution
+#'
+#' Define the transtion of the frog via a discrete time Markov chain with the
+#' transition probability matrix $\boldsymbol{T}$ with the current state defined
+#' on the rows and next state defined on columns.
+#'
+#' $$ \boldsymbol{T} =
+#' \begin{pmatrix} \boldsymbol{Q} & \boldsymbol{B} \\ \boldsymbol{0} & 1 \end{pmatrix}.$$
+#'
+#' The bottom row of $\boldsymbol{T}$ is a row vector of zeros with $p + 1$
+#' elements and the bottom right entry in $\boldsymbol{T}$ is 1.  This row
+#' represents the transition probability for jumping form the goal bank to the
+#' goal bank.
+#' $\boldsymbol{B}$ is a column vector with $p+1$ entries.  This vector gives
+#' the probabilities of jumping for the current pad to the goal bank.  Finally,
+#' the matrix $\boldsymbol{Q}$ is the $(p+1) \times (p+1)$ matrix of transition
+#' probabilities from pad to pad.
+#'
+#' $$ \boldsymbol{Q} =
+#'\begin{blockarray}{cccccc}
+#' & \text{Pad } p + 1 & \text{Pad } p & \cdots & \text{Pad } 2 & \text{Pad } 1 \\
+#'\begin{block}{c(ccccc)}
+#' \text{Pad } p + 1 & 0      & \frac{1}{p+1} & \cdots & \frac{1}{p+1} & \frac{1}{p+1} \\
+#' \text{Pad } p     & 0      & 0             & \cdots & \frac{1}{p}   & \frac{1}{p}   \\
+#' \vdots            & \vdots & \vdots        & \ddots & \vdots        & \vdots        \\
+#' \text{Pad } 2     & 0      & 0             & 0      & 0             & \frac{1}{2} \\
+#' \text{Pad } 1     & 0      & 0             & 0      & 0             & 0 \\
+#'\end{block}
+#'\end{blockarray}.
+#'$$
+#'
+#' The expected number of transitions, i.e., jumps, the frog will take to
+#' transverse the pond via lily pad can be found via the rowsums of the matrix
+#' $\boldsymbol{J}:$
+#' $$ \boldsymbol{J} = \left( \boldsymbol{I} - \boldsymbol{Q} \right)^{-1}, $$
+#' where $\boldsymbol{I}$ is the identity matrix.
+#'
+#' The form of $\boldsymbol{J}$ is
+#' $$ \boldsymbol{J} =
+#'\begin{blockarray}{ccccccc}
+#' & \text{Pad } p + 1 & \text{Pad } p & \cdots & \text{Pad } 3 & \text{Pad } 2 & \text{Pad } 1 \\
+#'\begin{block}{c(cccccc)}
+#' \text{Pad } p + 1 & 1      & \frac{1}{p+1} & \cdots & \frac{1}{4} & \frac{1}{3} & \frac{1}{2} \\
+#' \text{Pad } p     & 0      & 1             & \cdots & \frac{1}{4} & \frac{1}{3} & \frac{1}{2} \\
+#' \vdots            & \vdots & \vdots        & \ddots & \vdots      & \vdots      & \vdots      \\
+#' \text{Pad } 3     & 0      & 0             & 0      & 1           & \frac{1}{3} & \frac{1}{2} \\
+#' \text{Pad } 2     & 0      & 0             & 0      & 0           & 1           & \frac{1}{2} \\
+#' \text{Pad } 1     & 0      & 0             & 0      & 0           & 0           & 1 \\
+#'\end{block}
+#'\end{blockarray}.
+#'$$
+#'
+#' The expected number of jumps the frog will take from pad $p$ to get to the
+#' bank is
+#'
+#' $$ \text{E}\left(\text{jumps}\right) = \sum_{j = 1}^{p + 1} \frac{1}{j}.$$
+#'
+#' The expected number of jumps if there are 10 pads between the frog and the
+#' bank is
+PADS <- 10
+sum(seq(1, PADS + 1, by = 1)^-1)
+
+#'
+#' For 20 pads between the frog and the bank:
+PADS <- 20
+sum(seq(1, PADS + 1, by = 1)^-1)
+simulation(PADS)
+
+#'
+#' For 30 pads between the frog and the bank:
+PADS <- 30
+sum(seq(1, PADS + 1, by = 1)^-1)
+simulation(PADS)
+
+#'
+#' The simulations and the closed form soluion see to match up.
+#'
+#' **Disclaimer** Part of the goal for this project was to avoid recursion in
+#' the solution.  The derivation of $\boldsymbol{J}$ could be considered
+#' recursive.  However, I would argue the final solution does not rely on
+#' recursion.

dewittpe-solution/frog_jump.Rmd

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+---
+title: Frog Jump Problem
+author: Peter DeWitt
+output: pdf_document
+header-includes:
+ - \usepackage{blkarray}
+ - \usepackage{amsmath}
+---
+
+
+```{r label = "setup", include = FALSE, cache = FALSE}
+knitr::opts_chunk$set(cache = TRUE)
+library(parallel)
+```
+
+
+# The Problem:
+
+A frog is on one bank of a pond.  There are $p$ lily pads between the frog
+and the opposite bank of the pond.  This frog has the ability to jump from
+the bank it starts on to any of the $p$ lily pads or all the way to the
+opposite bank.  The frog will randomly jump, with equal probability, forward,
+at least one pad.  The question is, what is the expected number of jumps
+the frog will make to transverse the pond?
+
+\begin{center}
+\begin{tabular}{cccccccc}
+\includegraphics[width=0.10\textwidth]{grass} &
+\includegraphics[width=0.10\textwidth]{lilypad}&
+\includegraphics[width=0.10\textwidth]{lilypad}&
+\ldots &
+\includegraphics[width=0.10\textwidth]{lilypad}&
+\includegraphics[width=0.10\textwidth]{lilypad}&
+\includegraphics[width=0.10\textwidth]{lilypad}&
+\includegraphics[width=0.10\textwidth]{pictures-of-green-frogs-7} \\
+Pad $0$ & Pad $1$ & Pad $2$ & \ldots & Pad $p - 2$ & Pad $p - 1$ & Pad $p$ & Pad $p+1$ \\
+GOAL: Pond bank & & & & & & & Start
+\end{tabular}
+\end{center}
+
+
+Goals for the solution:
+
+1. Solution is to be general for any number of pads.
+2. Solution should not rely on recursion.
+
+
+# Solution
+
+## Simulation
+
+The simulation does rely on recursion.  This simulation is used to test the
+other solution.  The code is writen in [R](https://www.r-project.org).
+
+
+```{r }
+# frog_jumper
+# Arguments:
+#   pads    : the number of lily pads between the frog and the bank of the pond.
+#   jumps   : number of jumps taken so far
+#   verbose : if TRUE print status.
+#
+# Return:
+#   the number of jumps needed to transverse the pond for the given simulation.
+#
+frog_jumper <- function(pads, jumps = 0, verbose = FALSE) {
+  if (verbose) {cat("jump:", jumps, "frog is on pad:", pads + 1, "\n")}
+  distance <- sample(x = seq(1, pads + 1, by = 1), size = 1)
+  if (distance > pads) {
+    if (verbose) {cat("jump:", jumps + 1, "frog Has Reached the bank.\n")}
+    return(invisible(jumps + 1))
+  } else {
+    frog_jumper(pads = pads - distance, jumps + 1, verbose = verbose)
+  }
+}
+```
+
+
+A couple quick checks of the simulation code.  If there are zero pads between
+the frog and the pond bank then the function should return 1.
+
+```{r }
+set.seed(42)
+frog_jumper(0, verbose = TRUE)
+```
+
+
+If there is one pad beween the frog and the bank then the function should
+return the value 1 with probability 1/2 and the value 2 with probability 1/2.
+Test this with 1,000,000 simulations.
+
+```{r }
+x <- replicate(n = 1e6, expr = {frog_jumper(1)})
+prop.table(table(x))
+```
+
+
+The expected number of jumps the frog will take if there is only one pad
+between the frog an the bank is
+
+```{r }
+simulation <- function(pads, iterations = 1e6) {
+  mean(
+       do.call(c,
+               mclapply(X        = seq(1, iterations),
+                        FUN      = function(i, ...) {frog_jumper(...) },
+                        pads     = pads,
+                        mc.cores = 12))
+      )
+}
+simulation(1)
+```
+
+
+For 10 pads between the frog and the opposite bank, the expected number of
+jumps the frog will take is
+
+```{r }
+simulation(10)
+```
+
+
+## A Maths Solution
+
+Define the transtion of the frog via a discrete time Markov chain with the
+transition probability matrix $\boldsymbol{T}$ with the current state defined
+on the rows and next state defined on columns.
+
+$$ \boldsymbol{T} =
+\begin{pmatrix} \boldsymbol{Q} & \boldsymbol{B} \\ \boldsymbol{0} & 1 \end{pmatrix}.$$
+
+The bottom row of $\boldsymbol{T}$ is a row vector of zeros with $p + 1$
+elements and the bottom right entry in $\boldsymbol{T}$ is 1.  This row
+represents the transition probability for jumping form the goal bank to the
+goal bank.
+$\boldsymbol{B}$ is a column vector with $p+1$ entries.  This vector gives
+the probabilities of jumping for the current pad to the goal bank.  Finally,
+the matrix $\boldsymbol{Q}$ is the $(p+1) \times (p+1)$ matrix of transition
+probabilities from pad to pad.
+
+$$ \boldsymbol{Q} =
+\begin{blockarray}{cccccc}
+& \text{Pad } p + 1 & \text{Pad } p & \cdots & \text{Pad } 2 & \text{Pad } 1 \\
+\begin{block}{c(ccccc)}
+\text{Pad } p + 1 & 0      & \frac{1}{p+1} & \cdots & \frac{1}{p+1} & \frac{1}{p+1} \\
+\text{Pad } p     & 0      & 0             & \cdots & \frac{1}{p}   & \frac{1}{p}   \\
+\vdots            & \vdots & \vdots        & \ddots & \vdots        & \vdots        \\
+\text{Pad } 2     & 0      & 0             & 0      & 0             & \frac{1}{2} \\
+\text{Pad } 1     & 0      & 0             & 0      & 0             & 0 \\
+\end{block}
+\end{blockarray}.
+$$
+
+The expected number of transitions, i.e., jumps, the frog will take to
+transverse the pond via lily pad can be found via the rowsums of the matrix
+$\boldsymbol{J}:$
+$$ \boldsymbol{J} = \left( \boldsymbol{I} - \boldsymbol{Q} \right)^{-1}, $$
+where $\boldsymbol{I}$ is the identity matrix.
+
+The form of $\boldsymbol{J}$ is
+$$ \boldsymbol{J} =
+\begin{blockarray}{ccccccc}
+& \text{Pad } p + 1 & \text{Pad } p & \cdots & \text{Pad } 3 & \text{Pad } 2 & \text{Pad } 1 \\
+\begin{block}{c(cccccc)}
+\text{Pad } p + 1 & 1      & \frac{1}{p+1} & \cdots & \frac{1}{4} & \frac{1}{3} & \frac{1}{2} \\
+\text{Pad } p     & 0      & 1             & \cdots & \frac{1}{4} & \frac{1}{3} & \frac{1}{2} \\
+\vdots            & \vdots & \vdots        & \ddots & \vdots      & \vdots      & \vdots      \\
+\text{Pad } 3     & 0      & 0             & 0      & 1           & \frac{1}{3} & \frac{1}{2} \\
+\text{Pad } 2     & 0      & 0             & 0      & 0           & 1           & \frac{1}{2} \\
+\text{Pad } 1     & 0      & 0             & 0      & 0           & 0           & 1 \\
+\end{block}
+\end{blockarray}.
+$$
+
+The expected number of jumps the frog will take from pad $p$ to get to the
+bank is
+
+$$ \text{E}\left(\text{jumps}\right) = \sum_{j = 1}^{p + 1} \frac{1}{j}.$$
+
+The expected number of jumps if there are 10 pads between the frog and the
+bank is
+
+```{r }
+PADS <- 10
+sum(seq(1, PADS + 1, by = 1)^-1)
+```
+
+
+For 20 pads between the frog and the bank:
+
+```{r }
+PADS <- 20
+sum(seq(1, PADS + 1, by = 1)^-1)
+simulation(PADS)
+```
+
+
+For 30 pads between the frog and the bank:
+
+```{r }
+PADS <- 30
+sum(seq(1, PADS + 1, by = 1)^-1)
+simulation(PADS)
+```
+
+
+The simulations and the closed form soluion see to match up.
+
+**Disclaimer** Part of the goal for this project was to avoid recursion in
+the solution.  The derivation of $\boldsymbol{J}$ could be considered
+recursive.  However, I would argue the final solution does not rely on
+recursion.