We now have all the pieces necessary to build a walk$-$on$-$spheres estimate of temperature on a square domain. Using functions from Tasks $1,2$ and $3,$ write a function, wos$\_$temperature$\_$estimate$($x$0,$ y$0,$ length, cnr$\_$temps, epsilon, number$\_$walks$),$ which returns a tuple with two items: $1)$ the average temperature estimate as a float, and $2)$ a numpy.array with the edge temperatures from each individual walk. The parameter epsilon is a float that indicates when

the distance of the walk is close enough to the edge that we can stop. The parameter number$\_$walks is an integer indicating how many walks to take. Remember that each walk starts afresh from point x$0,$ y$0.$ The other parameters are as described in the earlier tasks. Refer to the algorithm descriptiont

in the figure for what to implement.

Sample

$>>>$ x $=0\mathrm{.}6$; y $=0\mathrm{.}3$

$>>>$ length $=1\mathrm{.}0$; epsilon $=0\mathrm{.}01$

$>>>$ cnr$\_$temps $=\{\text{'}$sw$\text{'}$:$600\mathrm{.}0,\text{'}$se$\text{'}$:$700\mathrm{.}0,\text{'}$ne$\text{'}$:$1000\mathrm{.}0,\text{'}$nw$\text{'}$:$850\mathrm{.}0\}$

$>>>$ number$\_$walks $=1$

$>>>$ random.seed$(1)$

$>>>($T$\_$avg, Ts$)=$ wos$\_$temperature$\_$estimate$($x$,$ y$,$ length, cnr$\_$temps, epsilon,

$,->$ number$\_$walks$)$

$>>>$ T$\_$avg

$635\mathrm{.}4021168323254$

$>>>$ Ts

array$([635\mathrm{.}40211683])$

$>>>$ number$\_$walks $=100$

$>>>$ random.seed$(1)$

$>>>($T$\_$avg, Ts$)=$ wos$\_$temperature$\_$estimate$($x$,$ y$,$ length, cnr$\_$temps, epsilon,

$,->$ number$\_$walks$)$

$>>>$ T$\_$avg

$749\mathrm{.}4923764337983$

It is possible to compute the temperature value analytically for this square domain with linearly varying boundary conditions. In this case, the value at $(0\mathrm{.}6,0\mathrm{.}3)$ is the area$-$weighted average of the corner

temperature values, as follows:

T$(0\mathrm{.}6,0\mathrm{.}3)=$ TSW$\backslash$times $(1\mathrm{.}00\mathrm{.}6)\backslash$times $(1\mathrm{.}00\mathrm{.}3)+$TSE$\backslash$times $0\mathrm{.}6\backslash$times $(1\mathrm{.}00\mathrm{.}3)+$TNE$\backslash$times $0\mathrm{.}6\backslash$times $0\mathrm{.}3+$TNW$\backslash$times $(1\mathrm{.}00\mathrm{.}6)\backslash$times $0\mathrm{.}3$

This results in T$(0\mathrm{.}6,0\mathrm{.}3)=744$ K$.$ What do we notice from the sample usage above? The result from one walk is a terrible estimate: $635$ K$.$ This is not surprising. Remember the value gets more accurate as we take more walks. When using $100$ walks, the result is $749$ K$,$ which only differs $0\mathrm{.}6\%$ from the analytical value of $744$ K$.$