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

Question

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.

Answer 1

Haskell, 65 63

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.

Answer 2

CJam -41 -40 -39 -38/35 31/26

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

Answer 3

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*

Interpreter at cjam.aditsu.net.

Order of inputs

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

1. View Distance Radius
2. Landmark X
3. Starting Point X
4. Landmark Y
5. Starting Point Y

Answer 4

Haskell — 72 70 69 68

I think this satisfies the question, but if not please leave a comment and I'll do my best to fix it:

w s y l z d|r<d=100|0<1=asin(d/r)/pi*100where r=sqrt$(l-s)^2+(z-y)^2

Answer 5

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)