# Calculate the probability of seeing a landmark when starting at a given point and walking straight in a random direction

Input 5 integers: Landmark X & Y, Starting Point X & Y, and View Distance Radius D

Output 1 double: % chance of coming within D of Landmark (X,Y) ("seeing") when walking straight in a random direction from Starting Point (X,Y).

Assume that "random direction" shall be a direction uniform on [0,2π], the land is flat, and of course the 'hiker' never turns after he begins moving.

Example: If landmark is at (10,10), starting point is (5,5), and view distance is 20 then the output would be 100% as no matter which direction you walk the landmark will be immediately visible.

• This kinda lost its appeal since you already linked to the solution. Jul 24, 2014 at 18:23
• @MartinBüttner I hadn't considered that since it is hardly implemented as code... but I suppose I can remove it if people also want to solve the math. Jul 24, 2014 at 19:08
• @ioSamurai just wanted to let you know that people can still see the math if they choose to look at the revisions - although I don't see any problem with leaving the math there.. people would be able to copy the math from the first answer, and would be challenged with optimizing the space anyways... Aug 6, 2014 at 15:18
• @user2813274 Yeah I didn't think showing the math was an issue anyways because it was really about the coding. Aug 6, 2014 at 15:24
• If you want to make this challenging/interesting, i suggest you ban trigonometric functions. Accuracy of probability should also be specified (I suggest to the nearest 1%, given that probability can be from 0 to 100%.) Aug 6, 2014 at 15:39

import Data.Complex
(s,l)%d=100*min(asin(d/magnitude(s-l))/pi)1


Use like (xstart:+ystart, xlandmark:+ylandmark) % distance. Gives result in percent. Why is it so expensive to load modules in Haskell?!?

Note that there is no if/then/else, pattern matching, etc. in this code, min does the magic.

• I didn't expect Haskell to do so well here, nor the Spanish Inquisition. :D Aug 7, 2014 at 20:12
• what is the undocumented feature, if i may ask? Aug 7, 2014 at 20:56
• is it the fixity of min? Aug 7, 2014 at 21:04
• @proud haskeller: the way min deals with NaN. Since NaN compares to a non-NaN to False, the exact implementation of min is relevant (whether or not it uses the inverse of a result of a comparison). As far as I can tell, this behaviour is not specified anywhere, but I'd be glad to be proven wrong. Aug 7, 2014 at 21:05
• when you write NaN, you mean undefined? or is there some other NaN I'm not aware about? Aug 7, 2014 at 21:13

## CJam -41-40-39-38/3531/26

This seems to work. It is my first attempt at CJam and/or codegolf. Run the code at http://cjam.aditsu.net/. In the section called input just place the variables as integers delimited with spaces in the input block in this order: distance D, landmark x, starting x, landmark y, starting y (for example 20 10 5 10 5). I had posted a previous one based on a misunderstanding of the equation that has been resolved. I also had been returning answers as probability ratios rather than percentages. Note: the second code has only 31 characters but combines alot from another user's CJam code.

ri{riri- 2#}:U;UU+.5#/mSP/:A.5<A1?e2

r{~riri-_*}_~+mq/mSP/:A.5<A1?e2


Accuratish one without arcsine being directly called (44 characters):

ririri- 2#riri- 2#+ .5#/:A1<AA3#6/+P/1?100*


Even more accurate (52 characters):

ririri- 2#riri- 2#+ .5#/:A1<AA3#6/+A5#40/3*+P/1?100


Update:

The absolute best I've written is 26 characters. I've still learned alot by watching professorfish's attempts but the crux is mine. I assumes (potentially incorrectly) that if you can see the landmark if it is closer (not closer than or equal to) your sight radius.

1r{~riri-_*}_~+mq/mSP/e<e2


# CJam, 44 40 38 37

First CJam script! Uses the method on the Math.SE answer here.

Supports non-integer inputs as well, at no extra character cost.

rd{rdrd-_*}:U~U+mq/_1<{mSP/}{;1}?100*


## Order of inputs

The inputs are given in this order, on STDIN, separated by spaces:

2. Landmark X
3. Starting Point X
4. Landmark Y
5. Starting Point Y
• Why not use P instead of 3.14? P starts off defined as pi This will give you the same length as mine. Aug 6, 2014 at 19:24
• @kaine Fortunately I've thought of another saving, I'm now beating you by one character
– user16402
Aug 6, 2014 at 19:26
• How is that fortunate? Aug 6, 2014 at 19:26
• There is, what I believe, to be a mistake in this. If you start exactly as far as you can see away, you get 50% on yours 100% on mine. Easy change. An it is debatable which is correct. Aug 6, 2014 at 19:33
• ri{riri-_}:U;UU+mq/_1>{;1}{mSP/}?100 you are welcome dammit Aug 6, 2014 at 20:05

# Haskell — 72 70 69 68

w s y l z d|r<d=100|0<1=asin(d/r)/pi*100where r=sqrt$(l-s)^2+(z-y)^2  • you can replace True with 0<1, and use$ in the definition of r. Aug 6, 2014 at 19:53
• Thanks, I can't believe I missed that \$ opportunity. Aug 6, 2014 at 20:17
• You could also squish where up against 100. Aug 6, 2014 at 20:48
• It breaks vim's syntax highlighting (why I didn't delete the space in the first place), but it does indeed compile. Aug 6, 2014 at 20:54
• Also, asinh should really be asin. The character reduction is a bonus! Aug 6, 2014 at 21:33

# Python - 73

from math import*
f=lambda l,s,d:(abs(l-s)<=d or asin(d/abs(l-s))/pi)*100


l is a complex number (e.g. 5 + 5j) describing the landmark position, s describes the start position and d is the view distance, for the example from the question call f as follows: f(5 + 5j, 10 + 10j, 20)

• I guess you are missing a bracket (. Aug 9, 2014 at 13:49
• Now you are missing two brackets (...). ;) Otherwise you might return True rather than 100 in case the landmark is closer than the distance d. Aug 9, 2014 at 20:29
• Oh, that's what you mean...
– hlt
Aug 9, 2014 at 21:43