27
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The unnecessary and convoluted story

I am walking around manhattan, block by block and my feet have gotten tired and want to go back home.

The traffic is pretty bad, but fortunately I'm very rich and I have a helicopter on standby at the hotel. But I need them to know how much fuel to pack for the flight and for that they need to know my direct distance from the hotel. I did remember which blocks I walked and can tell them what route I took. This distance needs to be precise though, if they are too short we won't make it back, too long and I've bought fuel I can't use.

Can you write me a program to convert that into the distance they will have to travel on their flight to fetch me?

Specification:

Write me a function that:

  1. Accepts a list or string of blocks walked relative to an arbitrary grid:
    • U p, D own, L eft and R ight.
    • Can be either upper or lower case - eg. if its shorter to use u instead of U go ahead.
    • An invalid direction has undefined behaviour - eg. a direction of X can cause a failure.
  2. Returns a float/decimal/double that is twice the straight line distance from the point of origin.

For illustration and clarification:

My trip

My trip could have just as easily been recorded as "luluu..." or ['l','u','l'...] but it must be recorded as Up, Down, Left, Right.

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  • 15
    \$\begingroup\$ You are enough rich to have an helicopter but you care if some extra fuel is bought? :O \$\endgroup\$ – Fez Vrasta Feb 6 '14 at 7:28
  • 8
    \$\begingroup\$ @fezvrasta because I'm stingy. \$\endgroup\$ – user8777 Feb 6 '14 at 7:33
  • 7
    \$\begingroup\$ Way to mess with my head by not making this about Manhattan distance. \$\endgroup\$ – Kendall Frey Feb 6 '14 at 13:33
  • 25
    \$\begingroup\$ The correct answer is "It doesn't matter. You're a rich guy, so you reach into your pocket, pull out a wad of $20's, and wave it in the air to attract the attention of a cabby; you are then set upon by a group of kindergarten thugs who rob you and beat you to a bloody pulp. You are then arrested for littering and public vagrancy, charged with terrorism for attempting to cause mass panic and a pandemic by spreading your bodily ooze across a public sidewalk, convicted, sent to prison, and locked up with a cellmate nicknamed Brutus who takes a real strong liking to you. Welcome to New York!" \$\endgroup\$ – Bob Jarvis - Reinstate Monica Feb 6 '14 at 13:35
  • 2
    \$\begingroup\$ @McKay I interpret it as directions on a map, anyway (otherwise it'd probably be "forward" and "back"), and the distance measure is rather unambiguous "twice the straight line distance from the point of origin", so no manhattan distance). \$\endgroup\$ – FireFly Feb 6 '14 at 16:44

23 Answers 23

32
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J, 17 characters

2*|+/0j1^'urdl'i.

Uses the fact, that the powers of j represent the proper directions.

  • 'urdl'i. take string and calculate indices (0 for 'u', 1 for 'r', ...)
  • 0j1^ transforms into the direction in the complex plane using the corresponding power of j.
  • +/ sums up the single steps
  • 2*| two times the modulus

Example:

> 2*|+/0j1^'urdl'i.'uuuudrrrl'
7.2111
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  • 5
    \$\begingroup\$ Nice job. Maths knowledge for the win. :-) \$\endgroup\$ – Gareth Feb 6 '14 at 10:35
  • \$\begingroup\$ Make this "non-extended" ASCII and then it's just 15 bytes (because you don't use the eighth bit). \$\endgroup\$ – Timtech Feb 8 '14 at 0:40
11
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Python 2.7 56 58 56 51 48

With the stolen Number One Dime from Scrooge McDuck, I made my fortune and now have more wealth than Scrooge.

y=lambda s:2*abs(sum(1j**(ord(i)%15)for i in s))

Python 2.7 - 61 53 50 (case insensitive)

y=lambda s:2*abs(sum(1j**(ord(i)%16%9)for i in s))

Implementation

>>> from random import sample
>>> y=lambda s:2*abs(sum((-1j)**(ord(i)%15)for i in s))
>>> path=sample('RLUD'*1000, 100)
>>> y(path)
20.0
>>> path=sample('RLUD'*1000, 100)
>>> y(path)
34.058772731852805
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  • \$\begingroup\$ I am getting IndexError: list index out of range. What form should the input have? \$\endgroup\$ – plannapus Feb 6 '14 at 9:01
  • \$\begingroup\$ @plannapus: I have added an implementation section \$\endgroup\$ – Abhijit Feb 6 '14 at 9:02
  • \$\begingroup\$ Ah and it was %5 not %8. Ok it makes more sense now :) \$\endgroup\$ – plannapus Feb 6 '14 at 9:04
5
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APL (29)

{|+/2 0j2×-⌿2 2⍴+/'URDL'∘.=⍵}

e.g.

     {|+/2 0j2×-⌿2 2⍴+/'URDL'∘.=⍵} 'UUUUDRRRL'
7.211102551

Explanation:

  • +/'URDL'∘.=⍵: see how often the characters URDL occur in the argument
  • -⌿2 2⍴: subtract the U value from the D value, and the R value from the L value
  • 2 0j2×: multiply the vertical value by 2 and the horizontal value by 2i
  • +/: sum
  • |: magnitude
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4
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Ruby 1.9+ (67)

f=->s{2*(((g=s.method :count)[?U]-g[?D])**2+(g[?R]-g[?L])**2)**0.5}

Example

f["DRUULULLULL"] => 10.0
f["UUUUDRRRL"] => 7.211102550927978
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3
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perl6: 44 chars

2*abs [+] i <<**>>%(<U R D L>Z ^4){get.comb}
  • get.comb gets one line of input and splits into characters
  • <U R L D> is a list of words, chars in this case
  • (1,2,3) Z (4,5,6) == (1,2), (2,5), (3,6), so it zips 2 lists into each other, making a list of parcels that %() turns into a hash
  • <<**>> does pairwise **, extending the shorter list to fit the longer. Shorter list happens to only be i
  • [+] sums all elements of a list, abs takes the modulus for complex numbers

Yes, I removed all possible spaces.

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2
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Python 2.7 - 65

Nice and short one, this uses complex numbers to step through the plane:

x=lambda s:2*abs(sum([[1,-1,1j,-1j]['RLUD'.index(i)]for i in s]))

Props to DSM and Abhijit in other questions that showed me the use of 1j to calculate this.

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  • \$\begingroup\$ Can 1j be written as j, -1j as -j? Also, does this handle upper and lower input, or only upper? \$\endgroup\$ – DavidC Feb 6 '14 at 4:13
  • 1
    \$\begingroup\$ Uncle Scrooze, I hate you. You should at least leave some money for your nephews. \$\endgroup\$ – Abhijit Feb 6 '14 at 8:38
  • 1
    \$\begingroup\$ @DavidCarraher: No you can't. It would be impossible to differentiate between the variable j and the imaginary unit j \$\endgroup\$ – Abhijit Feb 6 '14 at 8:53
  • \$\begingroup\$ Didn't you say it was supposed to output twice the distance? when trying with UUUUDRRRL i'm getting 3.606 with this function instead of 7.21. \$\endgroup\$ – plannapus Feb 6 '14 at 8:59
  • 4
    \$\begingroup\$ You can save 2 more characters, by multiplying the constants by 2 instead of multiplying the final result. \$\endgroup\$ – Abhijit Feb 6 '14 at 9:16
2
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Mathematica 92 49

Calle deserves full credit for streamlining the code.

f@l_:=2 N@Norm[Tr[l/.{"r"→1,"l"→-1,"u"→I,"d"→-I}]]

Example

f[{"u", "u", "u", "u", "d", "r", "r", "r", "l"}]

7.2111

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  • 1
    \$\begingroup\$ You're doing a lot of work that isn't required by the OP, f@l_ := 2 N@Norm[Tr[l /. {"r" -> 1, "l" -> -1, "u" -> I, "d" -> -I}]] will suffice. \$\endgroup\$ – user11030 Feb 6 '14 at 17:25
  • \$\begingroup\$ I get 2 Norm[(2. + 2. I) + "U" + "X"] as the output for your code. \$\endgroup\$ – DavidC Feb 6 '14 at 17:38
  • 1
    \$\begingroup\$ Yes, but the OP says it's OK to fail with such input. That is how I and everyone else interpret it. I can't read these other languages, but you will see that they often hardcode for u, r, l and d. \$\endgroup\$ – user11030 Feb 6 '14 at 17:40
  • \$\begingroup\$ Ok. Got it. Thanks for pointing that out. \$\endgroup\$ – DavidC Feb 6 '14 at 17:41
  • \$\begingroup\$ If you replace two remaining pairs of parentheses with @s you get another two characters less. \$\endgroup\$ – shrx Feb 7 '14 at 0:50
2
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PHP, 67

function f($a){foreach($a as$d)@$$d++;return 2*hypot($U-$D,$L-$R);}

Example:

<?php
var_dump(f(array('U', 'U', 'U', 'U', 'D', 'R', 'R', 'R', 'L')));

>float(7.211102550928)
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2
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Julia, 45

f(l)=2*abs(sum([im^(c=='d'?3:c) for c in l]))

Stole the i to powers trick. Also all the characters except d have values that work as acceptable powers for i.

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1
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J, 29 characters

+:+&.*:/-/_2[\#/.~/:~'ruld'i.

Only works with lower case directions and any characters other than r, u, l, and d will cause it to give a wrong answer.

Usage:

   +:+&.*:/-/_2[\#/.~/:~'ruld'i.'uuuudrrrl'
7.2111

Explanation:

'ruld'i.'uuuudrrrl' The dyadic form of i. finds the index of items from the right argument in the left argument. In this case:

   'ruld'i.'uuuudrrrl'
1 1 1 1 3 0 0 0 2

/:~ sorts this list into ascending order:

   /:~'ruld'i.'uuuudrrrl'
0 0 0 1 1 1 1 2 3

#/.~ counts the number of occurrences of each number:

   #/.~/:~'ruld'i.'uuuudrrrl'
3 4 1 1

_2[\ chops it into 2 rows:

   _2[\#/.~/:~'ruld'i.'uuuudrrrl'
3 4
1 1

-/ subtracts the bottom from the top

   -/_2[\#/.~/:~'ruld'i.'uuuudrrrl'
2 3

+&.*: borrows a trick from another J answer I saw this morning, and squares the items, then sums them, then performs a square root. See under &. documentation:

   +&.*:/-/_2[\#/.~/:~'ruld'i.'uuuudrrrl'
3.60555

+: doubles the result:

   +:+&.*:/-/_2[\#/.~/:~'ruld'i.'uuuudrrrl'
7.2111
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1
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R, 86 74 56 characters

Ok it's actually way shorter with imaginary numbers indeed:

2*Mod(sum(sapply(scan(,""),switch,u=1i,d=-1i,l=-1,r=1)))

Usage:

> 2*Mod(sum(sapply(scan(,""),switch,u=1i,d=-1i,l=-1,r=1)))
1: u u u u d r r r l
10: 
Read 9 items
[1] 7.211103

Old solution at 74 characters with xy coords:

2*sqrt(sum(rowSums(sapply(scan(,""),switch,u=0:1,d=0:-1,l=-1:0,r=1:0))^2))

Usage:

> 2*sqrt(sum(rowSums(sapply(scan(,""),switch,u=0:1,d=0:-1,l=-1:0,r=1:0))^2))
1: u u u u d r r r l
10: 
Read 9 items
[1] 7.211103

Takes input as stdin, need to be lower-case and space-separated. Use x-y coordinates starting from (0,0).

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1
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k (50 49)

{2*sqrt x$x:0 0f+/("udlr"!(1 0;-1 0;0 -1;0 1))@x}

Example

{2*sqrt x$x:0 0f+/("udlr"!(1 0;-1 0;0 -1;0 1))@x}"uuuudrrrl"
7.211103
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1
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Java, 185, 203, 204, 217, 226

class A{public static void main(String[] a){int x=0,y=0;for(int i=0;i<a[0].length();i++) switch(a[0].charAt(i)){case'U':y++;break;case'D':y--;break;case'L':x++;break;case'R':x--;}System.out.print(Math.hypot(x,y)*2);}}

I did assume that each "U" was "1 up", so two units up would be "UU"

Edit: swapped out switch for ifs

class A{public static void main(String[]a){int x=0,y=0;for(int i=0;i<a[0].length();i++){int c=a[0].charAt(i);if(c=='U')y++;if(c=='D')y--;if(c=='L')x++;if(c=='R')x--;}System.out.print(Math.hypot(x,y)*2);}}

Moved for iterator

class A{public static void main(String[]a){int x=0,y=0;for(int i=0;i<a[0].length();){int c=a[0].charAt(i++);if(c=='U')y++;if(c=='D')y--;if(c=='L')x++;if(c=='R')x--;}System.out.print(Math.hypot(x,y)*2);}}

No longer takes input as string, rather array of directions

class A{public static void main(String[]a){int x=0,y=0;for(String s:a){char c=s.charAt(0);if(c=='U')y++;if(c=='D')y--;if(c=='L')x++;if(c=='R')x--;}System.out.print(Math.hypot(x,y)*2);}}
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  • \$\begingroup\$ My understanding of the brief was that you only need a function, not a whole program. \$\endgroup\$ – Boann Feb 6 '14 at 18:32
1
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T-SQL, 158

IF PATINDEX('%[^UDLR]%', @s)=0 select 2*sqrt(power(LEN(REPLACE(@s,'U',''))-LEN(REPLACE(@s,'D','')),2)+power(LEN(REPLACE(@s,'L',''))-LEN(REPLACE(@s,'R','')),2))

The @s is the input string of varchar(max) type

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1
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ES6, 77 69

Definition:

f=s=>{u=d=l=r=0;for(c of s)eval(c+'++');return 2*Math.hypot(u-d,l-r)}

Usage:

>>> f('uuuudrrrl')
7.211102550927979
>>> f( 'uuuudrrrl'.split('') )
7.211102550927979
  • Accepts string OR array (lowercase)
  • Doesn't use imaginary numbers
  • Would not have been possible just 3 days before OP posted the question; that is, it only runs in Firefox 27+ (and maybe also Chrome with experimental stuff enabled, haven't tested :)!!

(Inspired partially by Boann's answer.)

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  • \$\begingroup\$ I really wanna do something tricky to get rid of the return, like turn the whole thing into a boolean expression that just gets evaluated and returned automagically, but I'm not sure there's a way to do this unless I can replace the for statement with some expression (an arrow function body containing statements require the brackets and the explicit return, bodies that are just expressions don't).. \$\endgroup\$ – Noyo Feb 7 '14 at 17:54
1
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JavaScript - 142 characters - no eval()

function r(a){return Math.sqrt(Math.pow(a.match(/u/g).length-a.match(/d/g).length,2)+Math.pow(a.match(/l/g).length-a.match(/r/g).length,2))*2}

where a is a string like 'uudrrl'

use like this -

a='uudrrl'
r(a)

Test in browser console.

var x = "luluurrrrurd"
r(x)
8.48528137423857
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1
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C# - 90 characters

Fresh from LINQPad.

int x=0,y=0;input.Max(i=>i==85?y++:i==82?x++:i==68?y--:x--);(Math.Sqrt(x*x+y*y)*2).Dump();

Where input is a valid string.

>string input = "LULUURRRRURD";

>8.48528137423857
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0
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Befunge-93 (65)

It has 65 non-whitespace characters (217 with whitespace, though that can be reduced by a more compact layout (for 69/176 chars)). It takes some liberality with the output format, but is undeniably accurate. Doesn't seem worth the effort to implement/steal a square root implementation.

v                  >$:*\:*+88*4*5-2.,.@
               >3-:|
           >6-:|
       >8-:|
>~"D"-:|
       $   $   $   $
           \   \
       1   1   1   1
       -   -   +   +
           \   \
^      <   <   <   <

echo 'UUDLLUU' | ./befungee.py ../man outputs 2√13 (actually implementation seems to have issue with the extended ASCII though).

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0
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Matlab, 51 characters

My Matlab submission, works only with captial letters. This was a fun one! The hardest part was converting the string into an array of complex numbers to be summed.

Function:

f=@(s)abs(sum(fix((s-76.5)/8.5)+((s-79)*i/3).^-99))

Usage:

>> f=@(s)abs(sum(fix((s-76.5)/8.5)+((s-79)*i/3).^-99))
>> f('UURDL')
ans =

     1
>>
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0
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Javascript, 136

function z(a){var x=a.split('u').length-a.split('d').length;var y=a.split('r').length-a.split('l').length;return Math.sqrt(x*x+y*y)*2;};
document.write(z('uuuudrrrwl'));
7.211102550927978
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0
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JavaScript, 89

function f(a){U=D=L=R=0;for(d in a)eval(a[d]+'++');return 2*Math.sqrt((U-=D)*U+(L-=R)*L)}

Example:

<script>
document.write(f(['U', 'U', 'U', 'U', 'D', 'R', 'R', 'R', 'L']));
</script>

>7.211102550927978
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0
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C, 120

float d(char *p){int v=0,h=0;while(*p){v+=*p=='U'?1:*p=='D'?-1:0,h+=*p=='R'?1:*p=='L'?-1:0,++p;}return 2*sqrt(v*v+h*h);}

d("LULUURRRRURD") -> 8.485281

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0
\$\begingroup\$

JavaScript (no ES6, no eval) - 131

f=function(h){for(i=0,a=[0,,0,0,0];i<h.length;++i)++a[(h.charCodeAt(i)>>2)-25];x=a[0]-a[4];y=a[2]-a[3];return Math.sqrt(x*x+y*y)*2}

Test:

console.log(f('uuuudrrrl'));     // 7.211102550927978 
console.log(f('luluurrrrurd'));  // 8.48528137423857
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