# How many lights are on?

You are little speck on a Cartesian plane. Pondering your existence, you decide to roam around your plane and touch things. You notice that every coordinate pair you visit has a light switch. To entertain yourself, you decide to toggle every switch that you come across.

The Challenge: Create a program that will take an input of random length, using only the characters v,^,<, and >. These characters correspond to movements.

• < moves you one unit to the left
• ^ moves you one unit up
• > moves you one unit to the right
• v moves you one unit down

Using this randomly generated input, calculate how many switches will be left in the ON position, assuming you start at (0,0), all lights begin in the OFF state, and you don't turn on any lights until you make your first move. (Meaning, if you start at (0,0), and move up 1 unit to (0,1), there will now be 1 light on). If you come across a switch that is on the ON position, it will be turned OFF.

Test cases:

v>v<^^<v<<^<^><<>^^>>><v>vv^v>v^><><<^>><<<vvv^vvv>v>>v><vv^^<<>vv^^>v^<>>^^<^^>^^v<^>^<vvv^v^v><^<<v<<>><<>v>>^><^>^^<>>>>vv>^<<^<<><vvv<v^>>vvv>v^>>><<v^>^^^^v>>^>^v>v<vv^<>^<<v>vv>><^^<^><vv^^v<v<v^^^>v<^vv^v<><vv^^^>v^>v>vv<<^><v<^v><<v^^v>>v<vv<><^^^v<^v><><<^^<>^>><^^^>vv^<>>>>^<^<<<<>>>v<<v<v<vv><<vv<vv<^v^^^>>vv<>v>><<<v^>vv><v<<<v<<>^vvv^<v^v>^^v^v><<v^>>>v<v<v^>>>v><>>>^<><<<<>vv>v><v>v><^v<>v>>>vv<<>^>^>v<^><^<^vv^><^^>v<^^v>v^v<^^^^vv<>v<>><v^^><>v<<<>v^<v^^><>^<>^<>><>^^<>^v><>>><v<^^>>v>^^^<v


yields 125 lights on.

>>><^>>>^vv><^^v^<<<>>^<>>^<^>^<<^^v^<v^>>^<<><<>^v<^^^vv>v><^>>^<^<v><^v<^<<^^v><>>^v<^>^v<vvv<<<v^vv>>^vv>^^<>>vv>^<^>vv<>v^<^v<^^<v^^^<<^><><>^v>vvv<^vvv<vv><vv<^^v^^<^^>>><<<>^<>>><^<<<>><>^v><^^vv<>>><^^^<^>>>>v<v^><<>v<v<v^<<^><v^^><>v>^<>^^^vvv<v^>^^^^v>v<v>>>v^^<v<vv>><<>^vv><<vv<<>^^v>>v<^^v>><v<v<<>^^vv>>^v>v>v<>^>^v<>>><>v>v<<v<^<>>>vv>>v>^<>vv^v><><^v^>v<^^>v<^v>>v^>^>>v>v>^>^<^^>vv>>^vv<^><>^><<v^<><<^<^>^^vv^<<^^<^^v<v<>^>v>>>>>>vv<<v>^>vv^>^><^<^^><<vvvv>vvv<><<><v^vv><v^^<>^>><^vv<><>^>vv>>>vv^vv<<^v^^<<v^^>^>vvv<><<v>^>^>v<v>^<^^^^<^>^>><>>^^vv<>v<^>v><v<v^>>v<^v<vv>v^>v<v^<^^^^v<^<^<<v<<<v<v^^>vv^>><<<v>>^^^>^<^>>>v^v><^^vv^>><^^vv<vv<v^><>>^><<<>>^^v^v<<v^<vv^^^>><>>^v^^^>>^>^<<^>v>^^v>><>v>^>vv^^<vvvv<><^v>^><<>>><^<^v^<<vvv^v<<<<<><><<><><<v>v<v>v><^v^^vvv>><>^>^>^v<<vv^^^v^vv><v><><v<v<<>>>v<^<^v<<>^v<<^v<><>v>>^^>^<v^<<^v^^^vv>><v^<v>^v><^<>>>>^^<vv<>^^>^>v^v^^>><>^^<^v^<v<<v<^<<^^vv>v>^<vv<<^^v^vv^>^^<>v>^>^<>vv><v>>><<><<vv^^<vv<>>^v>^<<vv>^><<>^<v>v><<v^<v<><v>^<^<^><^^^^>>>^<>^^><>>v^<vv^<^<<vvvv>>>v^v>>^>v^><<>>v<>>^>><vvvvv<<vvvv<v>^v<vv^<>><<><v^^<^<v>^v<^<<>^v<v^<>v<<>^<<vvv><^><^^^<>>v^<<>vv><^^^>><^>v^v>v<v^>>v>>v>vv<<v<<vvv^^^>^<v^^<^<v<^<>>v^<<v>>v^><v<vvvvv^^^<v^^<><v<<^>>^><^<v^v^^>><v><>v>^<vvvv><<v^^v^>^>v>><>^^v>v^>vv^>v<^<<^vv^>vv^<v>^<v^<>^v>v^>v^<<>^^<^>^^>vv^>>vv>v>vvv><>^v<<<<v^>v^^v<><v<v>^<^><^><<v<>><<>v^^>>><<><>>><<>><v^^>><^>><^<>v^^vvv>v<^<<vv^>vv^><<>v><>^<>v^^v>^>v^<>^><v>^><<^v<v^^<v>><^^>>^<^<^v<v>^>^<^<v><><>>>><>^<^<v>v<v^>v><>v^>v^<<><^<>>v<^vv^^^>^<<<<>^>^<><^vvv>^^<v^v>v>v^v>v>>vv>^><>^vv>^<v<v^<>vv^<v<><>^>>vvv><>>^<<v^<<>^<v^>v<^^^<^<^^<>^>>v>^<v>vv<v^^>><<<<<>>v>^v>^>>>>v>>^^>^<<<^<<<v>>^><<<<^vv<<>^v<^>v^<v<<>v<>^<^<^<^<>>^^^vvv<v>^vv>^><^<v^>^v<v>><><vvv<^^>>v<^v>^>>>><v^<v^^<^^v<vvv<v>^^<<>><^<v^v<^vv>v>vv>^^<>^^^^>>^v><vv<<<v>^v^>>v^>><<<^v^v<<>><<vvvvv<v^vv>vvvv><v^v<^^^><vv^^<>><>><^>^^^^v>v><><v^<>^v<>^^<^^>^^^vvv>>vv^v^<v<vv^v>v>>>^v^^vv^<^v>v^v>>^v>v<v^<^^><vvv>><<>><>><v>v<^<v>>>>v^^v^^>^><<v><^<<>>v<>^^<<>vv^>>vv^^^v>>>^v^><v<<^>v<v><>>>^>>^<<>>^><<vv<^^>^^^v^^<>>>vv><<>v<><<<>v<<>>>v<>v<>^<vv^v<^^<<<v>^v>>^^>^><<^vv<><><>v>^v>^<>v>>^^^<^^>>vv>v<<<v^><<v><^v><>v<^<<^<>^vv>^><^^^^^<<v^^<>v>><^<v^^^vv>^v<>^<v<v>v>^><^<<^<>><^^>vv^<>^<>vv<>>v<vv^>><^^<^>v<><>vv<v<>>v><v^^^>^^^<<vv^><^^v>v>^<^>v^><<vvv>v^><vv<><^<><^>^v<><<v^<<><>^^^^<<^>>^>^v^>v>^<<^>vv^vv^v<>^<<^>v<^^<<v<v<<<^>vv^>><>v>><><v>v><v^><vvv>vv<<vvv>v^<<^v<^<><<^v>v<>>vv<<v^>v>v<<>>^vv^<^^^<^v>>^<vv>^<v>><>v>^^<<v^<>>>>>v^v>><>v^>>^<>>^<^vvv^^^<^v<><vvv^>^>v><<v>>^v>v<v>v^<v>v>^<>vvv>vvv^^<>vv>^^^^^>v<<^v<>>>><<<>>><vv>>^v^^v<v^>>>^>^v<^v>^v<>vv<><vvv^v<<<<v<vv>vv^<^vvv<^v>v^^vv<^>>>^^>>^^><>^>>v<>>>^^<<v<^<<<<<^^<v^^^<<>><<<^>^v^>vv<>>>^^v><<>^^^^<vvv><^^<>>>^><<^>><v>^<>^v^<vvvv^>>^><<>><^<v^>>vv^vv<^>>>><^>^v<^<v<^^<^<^><^<>>^<>v^v<<>v>v<>><<v<^v<<<^v<v<>><v<^<^>>v>v>><v^<v><>>>>>v^v>><^<^<v>><v^v>v<>v<v><<<>^^><>^^<^vv^^<>v><><><^>^^v^vv^<><>>>>v><>>^>^<<^<v<v^>v^^<v>>><<^^vv^^>><<^<vvvvv>><^>^>>^vv<><<>v>v^<<<^v<^^<<^vv>v<vvv><^v>vv^vvvv<^>^v^<<<<^v<<<>^vvv>^v><<>>v<v<^v^<>v>>^^v^vv>>>^v^^>>^<><><<<<^vv>>>>>v>v^>v<>><<<>^vv>^^^^<^^^>^^^^>^^^v^v><^>^>>>v<v<^<^^<<^v<<^<>vvv^^^^v^<<>vv>^^>>><^^v<^<<<v<>v<<><>v<><>^<v<<^>^^>><<v>^^><^^v<^<v^<^^<>^<>^v^>>^^v^v^<>v<>^<<<>^v^v>^<vvvv<>v<<>vv^<<>vv>>>>^<v><>>>v^><<>^v>><<>>^^v><^<>>vv^^^>vv^<^^v><v>>vvv^v<^v>v<<^^<>v^^^v^^>><v^>>v^v^vv<^>v^<^>^>v<v^><vvv^>^<>v<<^><^^<vv>v>>vv>v^>>^vvv>>v^>^^>vvv>>v><<>>^^v>v<v<><<<<^^v<^<>^v>><v^^^<>>vvv>v><<v>^^<^vvvv^v>v>^vv>^vv^^v><<>>^^>>v>>>^v><^>v<^^<>vv>v>v^^^>>^^^><<<<>>^>>^<^v<^<^<>^><v<<v>v<>^>>^<<v^^<v^vvvvv>>v^>>^^^<^^<><<><><>^v>vvv^>^^>v<^^>^<<^v^^^><>><<v<^^^<<<<>><>><<^^v><v^<<^v<v><<>^<v>^>^v>vv>><v^<^<v<v<v><^^^^>>><^<><^v^v<<<^>vv^<v^>^^v^>>><<<<^<>>><v>>>vv^>^^<v^v>>>v^<^<>vv>^v^^><<<v>v>v>><>>>v<^>^<>>^^<v^<<^<v^>><^v^><><v><><v^vvv<<>v>>><<><v>v<>>><^<^^v<v>^<<>^v>^>^>^^<^^><^>>>>><^^>vv>^<^^<><^>^<^^><^<v>v^>><^>^^^>>v^v<^>>^<v^<>^><><v>>^v<v^^^^v<^vv><^v>>^^^<>^<^<^>vv^v<<>vv>^<>v>^>^>vv^v<vv<^^^v<v>v>v^<^^^v><v<<<^^<><^^>>>><<^^v<<^>v<<vv^^^vv^vv^<v><>^v<v>>><vv^v<v^>>>>^<<<vv^>>v>^><<><<^<^><<vv^>v^>><>v^<<<>v^><>><<>>v><>v^<v><>^v>>><><>>>^vvv^v>vv>>>^^v^><<v<>>^^^v^^><<^v<><><v<<v<v<><<<v^<^^^<>v^^v<^^<<^>v<<v><^<<<<>><>^v>^<>^<^^v^vvv>^^<>^>><v^^vv^<>^<><<^^^v<^^^>>^^v>^>^<^>v><<^<>^v<><vv^vv<><<<<<<v<<v<<vv^<<^<^vvvv><v^v^v<>>>vvvvv^<vv^<^<>vv>^<><<v><>v^^<v<>>>vvv^><^<^>v^^<v>^<>>>^^v^<vv<<<<^><v<<<>v<<<v<>>^^^>^><>v>^v^>^<v^^><^v^^<^v^^>^v>>^^^<<><><<<>v>><^><>>>vvvv>v>>v>^^^^v<><vv<^<v^v>>^^vv<^>vvv>^v>>><v<v<v^<^>^^<vvv<vv<v>>^vv>^<<^<^<v>v^<vv^^^v>vv<v><v><v>^<v>>vv<>v>^^>v^^^<>v<>v^v<>^<><v><^>^<v^v><<^v^v^v<<v><<^^<^vv>^<^v><>v>><v^v^>><><><<<v<>v<^vv>v<v<<>^vvvvv^<<<^<vv><<><>v^<^v<<<^>v>v<v^<<^>v<<^<v><<<^>vv>v>^<^^v>>>><>vv>>vv>vvv<>^^<>^v^<>vvv<^^^vv>v><<<<vv^v><v^<^<<<><v<>^><<>^>v<^^<<>v>>v<<>><^^<<<^<^vv^^>v>v<>^^>>^v^vvv>^v^>v>>v>v>v>>vv^<><<<<>v^^>vv<^^v>>v<vv<^>>^<>^^v<><vv^<><v><v<<v^v<^^<^v^v<>v<<><vvv><<<^<^^<<>>^v>>>^v>>>v<>>^><<<><><<<v<vv<^<>v^^v^^>^<<^^^v^>^<<^>^>^>>>>v<v<v<>vv<<vv^<<^<vv>^^<^<<>><^v><><>^<v><v^>^v>^<^>^^><v><<^<v^^<<^><><v>v<>>><<^><v<^vvv^<<<>><<>^v^^><vv>vv<>^>^>vv<>v^<^<>vv><<>^<v<vv<^<^<><^vv<<^>>>v<>><<>>>^^^^<<^v>>v<vv>^^>v<v<vv^><<><>>>v>>^^v<^v^^>>v^<>>v^>><^<^^v<v<><<><>>^<>><^v<^^^^><>^>vv>>^vv<<>v<<<<<<><<<><<>><v><^^^<>>v<^><^vvv<>^>^^v>^<v><^v^vv^<<>v<<<<v>^vv>>v>vv<<^>^<>>vvv^<v<><>><>^^^^vvvvvvv<<>v<^><^^>vv^^<v<<^^<vvv<v<v<<>><<><v^^>><^<>^v^vv<<v<v<>><<>>>>>^vv<><v<>v><v>v>><v<v^vvvvv<><>v>>v<><<<^^<>^<^^<v>v^<vv>^vv^<>^<<^<vv><v<v>>v>^<>v^<<v^<v>^v<>><v>>>>^<<^^^v<^<>><^<><v>>vv^>^<^<^>>v^>^^^^>vvvvv>^v<^><^^<^^>^<^^^^^^^>v>>vv>v^^^v^^^<>v><^>>>v>^>^>^>vv<vv<^^>>^>>>v<>v><<^<<v^>^>>>>^^><^^<v<<<<>>v>v^v^^<>><v<^<<<<v^^^^<v<<<^>v>^^<vv<^^^^^v>^v^<v><>>^^>^v>^>^vv^v>v>v^>v>^>>^^^^>>^>>^><>><v>v>>><<^v^v^>^>^>>vv><<^>v<v<v^<<>>^v<<^v<<^><^>>^<v>^>vv>v>^^^>v^^<^<^^>vv>^^><v>>^v>^v<<^^^<<^v^>^<<^>vv^>>^<^v><<>v><^^^<^^>>vv>^vv>><^<<<^>vv^v>v<^<<<^<^<<><^^>>>v^<^^^>^<><^v>>^<<v<^v>>v^<^<^<^^^<v^><<vvv^<^v^vv^vv<v<<v<^<>^v>^^^<^^v<v<v><<<^<>^^^^v>v^v^v^v<v><v>>^v><vv^^^v>><<v^vvvv<<<^v<<><^>^<v^^v<>vvvv^vv<>^v<><>^^<>>vvv<^>><v^<<>v>v<>^v^>v^>><<>>^^<^v<>>^>^><>>^<v<v^^<^v><v^<v<><><^<<><v^v<<>vv<v<v<^>>><>vv^^<><<v<^^<<^<><^^^>^>>>^<^>>>^>><^^^<^v^^^v^v^v>v>v><vv>><vvv<<v><><^^>^v<v>><v><^><^<<>v^vv^v><^vv>^>>v<vv><^<^^v<^^vv<vv<v<v>v><v<vv<<>^^v^^v<<<^<>v^^^<><>>><^>v^^^v^vv<<<^>>v><^>v^<>>>>^<>^^vvv^^<><^>^^<><>^vvv^^<vv^>vv^^^^v<>vv<^^^v<<>><<vvvvv>v>^^^vv>><v><v<>vvvv<v^><^<>^>^<>v>v>v^vvvv<><^v>>>^^>><vvv<>^>^v^<vvv>v^vv^vv><>><>v^^v^vv<^v>vv>>v<v><^<<^v<>>^vv^<v>v><v>v>^v>^<v>^<<^>vv>v<^<^vv^<^><<<v<<^^vv<vvv><>v>v<vv^<><><^vvv>>vv<^^^v><^v><<^>^^v>^<>><v<>>^^<<<v><>^>^><vvvv<>^<<<><<<^<>>v^vv^>><^vv^^>^<v^<v>><^^>>>^v>^v<>^v<><^><vv>v^^^<^>>^<<^<^><<<^^<v<<^vv<^<>v<^<<^^<v<vv<<><v<v^<>^<>v>>v<^v>v<>^^vvv<>vv^v^<><v^vv^<^v^v><>^><v^<>>^^^<>>vv^<v>^^v><v<^>^^^^^^><>>vvv<<><><v<^>v<>v^v<<<<>v^>>>>^v>^^<v^>v><v^<^^v<<<<v<<<>^v<^>^v>v^^>v^^vvv>vv<>^>v><v<>^<vv><>>><<^>>><<v>v^^<^<<<<v^<>>>v<<<^v^vv<>^v>v<<<<>^^>><v><>v<v><^^>><>^>^>v>>><v>^vvvv<><><^>>^v^><<>^v<><><^><<<>v^^>v>^>v^<v^vv<>><^vv^^>^^><vv<<>v>v^^>><v^>^<^<>>>vv<>^>v>v^<>v<^<<v>>>^<>v^>v>>vv^^<>>v<v<<^<>v>v^<^^^>v^^>v>v>vv<^<v>v^^><<<v<><>^^<>v>v>^^v>v>v^v>>^<v^v>><>^^>^<>>>^vv^><v^<^>v^>^v><^>^^^vv^^v<>vv<>>^><<^v>^v^>>v>^v^<<^^^vv<<vvv>^vv^v<<<v^^<<><vv<>>^^vv>^^^vv>><><v>v<^v^>>>vv^><>><v<^v<>^><v<^^^^>><^<>v>^v<^vv>v>v<^<>v>v>^<vv>v<^>vvv<v^<vv<vv<>v>^><v^v<>>>>>v>><^v<>v>^v><v^v^vv<>^<vvv^>><v^<vvv^^<^vvv^v^<>><v>v^^v<><>v^^^v<<<^><v<<<>><<vv<<><vvv^v>>v^v<v^>>><<v<>^v><>vv<<v>v^vv>v^v<^<vv<><><^v>^<vv>v^^>>^^^><vv<><^>>>^<v^<<^^>^>vv^><v<vvv>^^>>>^><<vv>vv>^<>>^^><^v><<>^<<<v^>^


yields 1408 lights on.

To generate input, you may use the following Python script:

import random
length = random.randint(15, 10000)
walk = ''
chars = ['v', '<', '>', '^']
for i in range(length):
walk += random.choice(chars)
print(walk)


Standard scoring, winner is the smallest number of bytes.

• 4. 4 lights are on. Aug 15 '16 at 20:36
• No, there are five. Aug 15 '16 at 20:39
• Oh, people, don't be so harsh on a newcomer. Aug 15 '16 at 21:33
• The challenge looks pretty clear (and interesting) to me. Voting to reopen Aug 15 '16 at 21:51
• I don't see any harshness, just advice on what needs improving, and now the challenge is ready. Nice first challenge! Aug 15 '16 at 22:08

# MATL, 191817 13 bytes

5 bytes off thanks to @LeakyNun's idea (see his answer) of using the imaginary unit as a base for exponentiation.

Jj11\^Ys8#uos


Try it online! Test cases: 1, 2.

### Explanation

The code traces the path using unit steps in the complex plane. Then it counts how many times each position was visited, and outputs how many positions were visited an odd number of times.

J         % Push the imaginary unit, 1j
j         % Input string
11\       % Modulo 11. This gives 7 6 5 8 for > ^ < v
^         % 1j raised to those numbers, element-wise. This gives -1j for >, -1 for ^,
% 1j for < and 1 for v. So it gives the displacements in a "transposed"
% complex plane
Ys        % Cumulative sum. This transforms displacements into positions
8#u       % Count of occurrences of each unique position
o         % 1 if odd, 0 if even
s         % Sum. Implicitly display

• This looks curious Aug 15 '16 at 23:38
• If you mean the spacing, yes, it's ugly. On Matlab it does look better: 0+1i 0-1i -1+0i 1+0i Aug 15 '16 at 23:41
• No, that was golfing you 3 bytes Aug 15 '16 at 23:45
• @LeakyNun Yes, I just realized :-) 13 bytes now! Aug 15 '16 at 23:49
• Why use i instead of j? Aug 15 '16 at 23:54

# Python, 68 bytes

25 bytes thanks to Sp3000.

2 bytes thanks to Luis Mendo's idea of taking modulus with 11.

17 bytes thanks to xnor.

d=set()
p=0
for c in input():p+=1j**(ord(c)%11);d^={p}
print(len(d))


Ideone it!

• It's shorter to use a set doing d^={p} and len(d).
– xnor
Aug 16 '16 at 1:38

# Java 8, 169130 127 bytes

Thanks to Leaky Nun for saving 29 32 bytes.

s->{int l=s.length,d=2*l+1,p=l*d+d,r=0,m[]=new int[d*d];for(char c:s){p+=c<61?-1:c<63?1:c<95?-l:l;m[p]^=2;r+=m[p]-1;}return r;}


Ungolfed (sorta):

public static int f(char[] s) {
int l = s.length, d = 2*l+1, p=l*d+d, r = 0;
int[] m = new int[d*d];
for(char c : s) {
p += c<61 ? -1 : c<63 ? 1 : c<95 ? -l : l;
m[p] ^= 2;
r += m[p] - 1;
}
return r;
}

• for(char c:s){p+=c<61?-1:c<63?1:c<95?-l:l;m[p]^=2;r+=m[p]-1;} can be for(int c:s)r+=(m[p+=c<61?-1:c<63?1:c<95?-l:l]^=2)-1; to save 8 bytes: Try it online. Jun 14 '19 at 8:42

# Jelly, 13 bytes

A port of Luis Mendo's answer in MATL

O%11*@ı+\œ^/L


Try it online!

O%11*@ı+\œ^/L
O             convert to codepoint
%11          modulo 11
*@ı       power of i (imaginary unit)
+\     cumulative sum
œ^/  reduce by multiset symmetric difference
L length


Credits of the œ^/ trick to Dennis

# Perl, 49 bytes

Includes +1 for -p Run with the control string on STDIN

./lights.pl <<< "<<>"


lights.pl:

#!/usr/bin/perl -p
$\-=1-($a{$x+=/</-/>/,$y+=/\^/-/v/}^=2)for/./g}{


If input is restricted to 10000 bytes this reduces to 44+1:

$\+=$#{$x+=/</-/>/.e5+/\^/-/v/}*=-1for/./g}{  # TSQL, 238235203 191 bytes Creating a table in memory using recursive SQL. Selecting and calculating from that table(one line code). Golfed: DECLARE @ varchar(max)= 'v>v<'; WITH C as(SELECT 0 a,b=1UNION ALL SELECT a+POWER(CHARINDEX(SUBSTRING(@,b,1),'> <v')-2,15),b+1FROM C WHERE b<=LEN(@))SELECT top 1sum(sum(1-1/b)%2)over()FROM c GROUP BY a OPTION(MAXRECURSION 0)  Ungolfed: DECLARE @ varchar(max)= '>>>><^v^v'; WITH C as ( SELECT 0 a,b=1 UNION ALL SELECT a+POWER(CHARINDEX(SUBSTRING(@,b,1),'> <v')-2,15),b+1 FROM C WHERE b<=LEN(@) ) SELECT top 1sum(sum(1-1/b)%2)over() FROM c GROUP BY a OPTION(MAXRECURSION 0)  Fiddle # Mathematica, 64 bytes A port of Luis Mendo's MATL answer. Count[{_,_?OddQ}]@Tally@Accumulate[I^Mod[ToCharacterCode@#,11]]&  # SQF, 160 bytes Using the function-as-a-file format: x=0;y=0;a=[];{switch(_x){case"v":{y=y-1};case"^":{y=y+1};case"<":{x=x-1};case">":{x=x+1};};if([x,y]in a)then{a=a-[[x,y]]}else{a=a+[[x,y]]}}forEach _this;count a  Call as: "STRING" call NAME_OF_COMPILED_FUNCTION Ungolfed: //position tracker variables x = 0; y = 0; //a keeps track of which coords are switched on a = []; { //switch based on the magic variable _x //which is the current element of the forEach //and adjust coord tracker variables switch(_x) { case "v": { y = y - 1 }; case "^": { y = y + 1 }; case "<": { x = x - 1 }; case ">": { x = x + 1 }; }; //check if the coord is already turned on if ([x, y] in a) then { //remove it from a using array exclusion a = a - [[x, y]] } else { //append it to a a = a + [[x, y]] } //do this for each character in _this (the argument) } forEach _this; //return the length of a count a  • easily followed, but how is the performance? Aug 16 '16 at 9:35 • @zinking That depends on how you execute it. Really fast with a static compile at engine load time, much slower with compile "CODE". Aug 16 '16 at 21:35 ## Python: 207189 181 Bytes a = input() l=[] for i in range(0,len(a)): if a[i]=='v':y-=1 if a[i]=='^':y+=1 if a[i]=='<':x-=1 if a[i]=='>':x+=1 if [x,y] in l:l.remove([x,y]) else:l.append([x,y]) print(l)  Input is a string. Thanks to @LuisMendo for pointing out that I can actually only indent by 1 space. Darn my Python teachers for telling me to always do it increments of four. Fixing it so that input was obtained rather than assumed stored actually saved me 8 bytes. • I don't know Python, but I think 1-space indenting is enough, and it will save you a few bytes Aug 15 '16 at 22:34 • But not in code golf :-) Aug 15 '16 at 22:37 • You can't assume that the input is pre-stored. Just use input() and store it somewhere Aug 15 '16 at 23:02 • If you ever need more than one level of nested indentation, there's a great tip about mixing spaces and tabs Aug 15 '16 at 23:10 • Um... Maybe removing the whitespace around the first = and l+=[[x,y]] is also shorter then l.append([x,y]). Aug 18 '16 at 18:22 # Ruby, 64 + 6 = 70 bytes +6 bytes for -rset flag. A straight port of Leaky Nun's Python answer. ->s{d=Set.new q=0 s.each_char{|c|d^=[q+=1i**(c.ord%11)]} d.size}  See it on repl.it: https://repl.it/Cnjy • I believe the -rset flag only counts as 4 bytes. Aug 18 '16 at 20:17 • @ValueInk Oof. I don’t like that at all. Thanks for the heads-up, though. Jun 15 '19 at 0:48 • I think they made some pretty good points as to why they want to do this (like how -e being "free" in Ruby means -n is 1 byte instead of 3 but people who don't know about that could report different bytecounts) but to each their own. BTW, Set[] saves 2 bytes over Set.new. Jun 15 '19 at 2:27 • Another thought I had while thinking up a non-rset answer: bytes.map saves 4 bytes over each_char because you can take out the ord call. Jun 18 '19 at 22:25 ## Javascript (ES6), 9793 85 bytes i=>[...i].reduce((c,d)=>c-1+(v[p+=v['^<>v'.search(d)]]^=2),0,p=1E6,v=[-1E3,-1,1,1E3])  EDIT: -8 bytes, thanks to Neil Test: var test0 = 'v>v<^^<v<<^<^><<>^^>>><v>vv^v>v^><><<^>><<<vvv^vvv>v>>v><vv^^<<>vv^^>v^<>>^^<^^>^^v<^>^<vvv^v^v><^<<v<<>><<>v>>^><^>^^<>>>>vv>^<<^<<><vvv<v^>>vvv>v^>>><<v^>^^^^v>>^>^v>v<vv^<>^<<v>vv>><^^<^><vv^^v<v<v^^^>v<^vv^v<><vv^^^>v^>v>vv<<^><v<^v><<v^^v>>v<vv<><^^^v<^v><><<^^<>^>><^^^>vv^<>>>>^<^<<<<>>>v<<v<v<vv><<vv<vv<^v^^^>>vv<>v>><<<v^>vv><v<<<v<<>^vvv^<v^v>^^v^v><<v^>>>v<v<v^>>>v><>>>^<><<<<>vv>v><v>v><^v<>v>>>vv<<>^>^>v<^><^<^vv^><^^>v<^^v>v^v<^^^^vv<>v<>><v^^><>v<<<>v^<v^^><>^<>^<>><>^^<>^v><>>><v<^^>>v>^^^<v'; var test1 = '>>><^>>>^vv><^^v^<<<>>^<>>^<^>^<<^^v^<v^>>^<<><<>^v<^^^vv>v><^>>^<^<v><^v<^<<^^v><>>^v<^>^v<vvv<<<v^vv>>^vv>^^<>>vv>^<^>vv<>v^<^v<^^<v^^^<<^><><>^v>vvv<^vvv<vv><vv<^^v^^<^^>>><<<>^<>>><^<<<>><>^v><^^vv<>>><^^^<^>>>>v<v^><<>v<v<v^<<^><v^^><>v>^<>^^^vvv<v^>^^^^v>v<v>>>v^^<v<vv>><<>^vv><<vv<<>^^v>>v<^^v>><v<v<<>^^vv>>^v>v>v<>^>^v<>>><>v>v<<v<^<>>>vv>>v>^<>vv^v><><^v^>v<^^>v<^v>>v^>^>>v>v>^>^<^^>vv>>^vv<^><>^><<v^<><<^<^>^^vv^<<^^<^^v<v<>^>v>>>>>>vv<<v>^>vv^>^><^<^^><<vvvv>vvv<><<><v^vv><v^^<>^>><^vv<><>^>vv>>>vv^vv<<^v^^<<v^^>^>vvv<><<v>^>^>v<v>^<^^^^<^>^>><>>^^vv<>v<^>v><v<v^>>v<^v<vv>v^>v<v^<^^^^v<^<^<<v<<<v<v^^>vv^>><<<v>>^^^>^<^>>>v^v><^^vv^>><^^vv<vv<v^><>>^><<<>>^^v^v<<v^<vv^^^>><>>^v^^^>>^>^<<^>v>^^v>><>v>^>vv^^<vvvv<><^v>^><<>>><^<^v^<<vvv^v<<<<<><><<><><<v>v<v>v><^v^^vvv>><>^>^>^v<<vv^^^v^vv><v><><v<v<<>>>v<^<^v<<>^v<<^v<><>v>>^^>^<v^<<^v^^^vv>><v^<v>^v><^<>>>>^^<vv<>^^>^>v^v^^>><>^^<^v^<v<<v<^<<^^vv>v>^<vv<<^^v^vv^>^^<>v>^>^<>vv><v>>><<><<vv^^<vv<>>^v>^<<vv>^><<>^<v>v><<v^<v<><v>^<^<^><^^^^>>>^<>^^><>>v^<vv^<^<<vvvv>>>v^v>>^>v^><<>>v<>>^>><vvvvv<<vvvv<v>^v<vv^<>><<><v^^<^<v>^v<^<<>^v<v^<>v<<>^<<vvv><^><^^^<>>v^<<>vv><^^^>><^>v^v>v<v^>>v>>v>vv<<v<<vvv^^^>^<v^^<^<v<^<>>v^<<v>>v^><v<vvvvv^^^<v^^<><v<<^>>^><^<v^v^^>><v><>v>^<vvvv><<v^^v^>^>v>><>^^v>v^>vv^>v<^<<^vv^>vv^<v>^<v^<>^v>v^>v^<<>^^<^>^^>vv^>>vv>v>vvv><>^v<<<<v^>v^^v<><v<v>^<^><^><<v<>><<>v^^>>><<><>>><<>><v^^>><^>><^<>v^^vvv>v<^<<vv^>vv^><<>v><>^<>v^^v>^>v^<>^><v>^><<^v<v^^<v>><^^>>^<^<^v<v>^>^<^<v><><>>>><>^<^<v>v<v^>v><>v^>v^<<><^<>>v<^vv^^^>^<<<<>^>^<><^vvv>^^<v^v>v>v^v>v>>vv>^><>^vv>^<v<v^<>vv^<v<><>^>>vvv><>>^<<v^<<>^<v^>v<^^^<^<^^<>^>>v>^<v>vv<v^^>><<<<<>>v>^v>^>>>>v>>^^>^<<<^<<<v>>^><<<<^vv<<>^v<^>v^<v<<>v<>^<^<^<^<>>^^^vvv<v>^vv>^><^<v^>^v<v>><><vvv<^^>>v<^v>^>>>><v^<v^^<^^v<vvv<v>^^<<>><^<v^v<^vv>v>vv>^^<>^^^^>>^v><vv<<<v>^v^>>v^>><<<^v^v<<>><<vvvvv<v^vv>vvvv><v^v<^^^><vv^^<>><>><^>^^^^v>v><><v^<>^v<>^^<^^>^^^vvv>>vv^v^<v<vv^v>v>>>^v^^vv^<^v>v^v>>^v>v<v^<^^><vvv>><<>><>><v>v<^<v>>>>v^^v^^>^><<v><^<<>>v<>^^<<>vv^>>vv^^^v>>>^v^><v<<^>v<v><>>>^>>^<<>>^><<vv<^^>^^^v^^<>>>vv><<>v<><<<>v<<>>>v<>v<>^<vv^v<^^<<<v>^v>>^^>^><<^vv<><><>v>^v>^<>v>>^^^<^^>>vv>v<<<v^><<v><^v><>v<^<<^<>^vv>^><^^^^^<<v^^<>v>><^<v^^^vv>^v<>^<v<v>v>^><^<<^<>><^^>vv^<>^<>vv<>>v<vv^>><^^<^>v<><>vv<v<>>v><v^^^>^^^<<vv^><^^v>v>^<^>v^><<vvv>v^><vv<><^<><^>^v<><<v^<<><>^^^^<<^>>^>^v^>v>^<<^>vv^vv^v<>^<<^>v<^^<<v<v<<<^>vv^>><>v>><><v>v><v^><vvv>vv<<vvv>v^<<^v<^<><<^v>v<>>vv<<v^>v>v<<>>^vv^<^^^<^v>>^<vv>^<v>><>v>^^<<v^<>>>>>v^v>><>v^>>^<>>^<^vvv^^^<^v<><vvv^>^>v><<v>>^v>v<v>v^<v>v>^<>vvv>vvv^^<>vv>^^^^^>v<<^v<>>>><<<>>><vv>>^v^^v<v^>>>^>^v<^v>^v<>vv<><vvv^v<<<<v<vv>vv^<^vvv<^v>v^^vv<^>>>^^>>^^><>^>>v<>>>^^<<v<^<<<<<^^<v^^^<<>><<<^>^v^>vv<>>>^^v><<>^^^^<vvv><^^<>>>^><<^>><v>^<>^v^<vvvv^>>^><<>><^<v^>>vv^vv<^>>>><^>^v<^<v<^^<^<^><^<>>^<>v^v<<>v>v<>><<v<^v<<<^v<v<>><v<^<^>>v>v>><v^<v><>>>>>v^v>><^<^<v>><v^v>v<>v<v><<<>^^><>^^<^vv^^<>v><><><^>^^v^vv^<><>>>>v><>>^>^<<^<v<v^>v^^<v>>><<^^vv^^>><<^<vvvvv>><^>^>>^vv<><<>v>v^<<<^v<^^<<^vv>v<vvv><^v>vv^vvvv<^>^v^<<<<^v<<<>^vvv>^v><<>>v<v<^v^<>v>>^^v^vv>>>^v^^>>^<><><<<<^vv>>>>>v>v^>v<>><<<>^vv>^^^^<^^^>^^^^>^^^v^v><^>^>>>v<v<^<^^<<^v<<^<>vvv^^^^v^<<>vv>^^>>><^^v<^<<<v<>v<<><>v<><>^<v<<^>^^>><<v>^^><^^v<^<v^<^^<>^<>^v^>>^^v^v^<>v<>^<<<>^v^v>^<vvvv<>v<<>vv^<<>vv>>>>^<v><>>>v^><<>^v>><<>>^^v><^<>>vv^^^>vv^<^^v><v>>vvv^v<^v>v<<^^<>v^^^v^^>><v^>>v^v^vv<^>v^<^>^>v<v^><vvv^>^<>v<<^><^^<vv>v>>vv>v^>>^vvv>>v^>^^>vvv>>v><<>>^^v>v<v<><<<<^^v<^<>^v>><v^^^<>>vvv>v><<v>^^<^vvvv^v>v>^vv>^vv^^v><<>>^^>>v>>>^v><^>v<^^<>vv>v>v^^^>>^^^><<<<>>^>>^<^v<^<^<>^><v<<v>v<>^>>^<<v^^<v^vvvvv>>v^>>^^^<^^<><<><><>^v>vvv^>^^>v<^^>^<<^v^^^><>><<v<^^^<<<<>><>><<^^v><v^<<^v<v><<>^<v>^>^v>vv>><v^<^<v<v<v><^^^^>>><^<><^v^v<<<^>vv^<v^>^^v^>>><<<<^<>>><v>>>vv^>^^<v^v>>>v^<^<>vv>^v^^><<<v>v>v>><>>>v<^>^<>>^^<v^<<^<v^>><^v^><><v><><v^vvv<<>v>>><<><v>v<>>><^<^^v<v>^<<>^v>^>^>^^<^^><^>>>>><^^>vv>^<^^<><^>^<^^><^<v>v^>><^>^^^>>v^v<^>>^<v^<>^><><v>>^v<v^^^^v<^vv><^v>>^^^<>^<^<^>vv^v<<>vv>^<>v>^>^>vv^v<vv<^^^v<v>v>v^<^^^v><v<<<^^<><^^>>>><<^^v<<^>v<<vv^^^vv^vv^<v><>^v<v>>><vv^v<v^>>>>^<<<vv^>>v>^><<><<^<^><<vv^>v^>><>v^<<<>v^><>><<>>v><>v^<v><>^v>>><><>>>^vvv^v>vv>>>^^v^><<v<>>^^^v^^><<^v<><><v<<v<v<><<<v^<^^^<>v^^v<^^<<^>v<<v><^<<<<>><>^v>^<>^<^^v^vvv>^^<>^>><v^^vv^<>^<><<^^^v<^^^>>^^v>^>^<^>v><<^<>^v<><vv^vv<><<<<<<v<<v<<vv^<<^<^vvvv><v^v^v<>>>vvvvv^<vv^<^<>vv>^<><<v><>v^^<v<>>>vvv^><^<^>v^^<v>^<>>>^^v^<vv<<<<^><v<<<>v<<<v<>>^^^>^><>v>^v^>^<v^^><^v^^<^v^^>^v>>^^^<<><><<<>v>><^><>>>vvvv>v>>v>^^^^v<><vv<^<v^v>>^^vv<^>vvv>^v>>><v<v<v^<^>^^<vvv<vv<v>>^vv>^<<^<^<v>v^<vv^^^v>vv<v><v><v>^<v>>vv<>v>^^>v^^^<>v<>v^v<>^<><v><^>^<v^v><<^v^v^v<<v><<^^<^vv>^<^v><>v>><v^v^>><><><<<v<>v<^vv>v<v<<>^vvvvv^<<<^<vv><<><>v^<^v<<<^>v>v<v^<<^>v<<^<v><<<^>vv>v>^<^^v>>>><>vv>>vv>vvv<>^^<>^v^<>vvv<^^^vv>v><<<<vv^v><v^<^<<<><v<>^><<>^>v<^^<<>v>>v<<>><^^<<<^<^vv^^>v>v<>^^>>^v^vvv>^v^>v>>v>v>v>>vv^<><<<<>v^^>vv<^^v>>v<vv<^>>^<>^^v<><vv^<><v><v<<v^v<^^<^v^v<>v<<><vvv><<<^<^^<<>>^v>>>^v>>>v<>>^><<<><><<<v<vv<^<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S=i=>[...i].reduce((c,d)=>c-1+(v[p+=v['^<>v'.search(d)]]^=2),0,p=1E6,v=[-1E3,-1,1,1E3]) console.log(S(test0)) console.log(S(test1)) • Use [...i] instead of i.split(''). Also, I think you can write (p=1E6,v=[-1,1,-1E3,1E3],0) as 0,p=1E6,v=[-1,1,-1E3,1E3]. – Neil Aug 16 '16 at 18:35 • If you can arrange for the ^ to be first, search saves a byte over indexOf. (You need the ^ to be first because search treats ^ as a regexp metacharacter.) – Neil Aug 16 '16 at 18:38 • You don't have to count the S= part. Aug 17 '16 at 12:50 • @Arnauld No, the argument name(s) itself is/are part of the function and should remain. However, assigning the function to a variable ("S=") is optional (as you can just call the function like («function code here»)(«argument»)). Aug 17 '16 at 13:56 # R, 172156 137101 96 bytes n=function(a)sum(table(cumsum(sapply(strsplit(a,"")[[1]],switch,v=-1,"^"=1,">"=1i,"<"=-1i)))%%2)  Using @LeakyNun's complex plane idea. ungolfed n=function(a){ b=strsplit(a,"")[[1]] #Splits String up p=sapply(b,switch,v=-1,"^"=1,">"=1i,"<"=-1i) #Replaces characters with directions q=cumsum(p) #Finds each location visitied t=table(q)%%2 #Determines if spots were visited an odd # of times sum(t) #counts odd visited spots }  # PowerShell, 136 bytes $a=@();switch -w($args|% t*y){'^'{$y++}'>'{$x++}'v'{$y--}'<'{$x--}*{if("$x,$y"-in$a){$a=$a|?{$_-ne"$x,$y"}}else{$a+="$x,$y"}}};$a.length  Try it online! I think there's more to optimize here, but this is the lowest I got for now. Easier to read: $a=@();switch -w($args|% t*y){ '^'{$y++}
'>'{$x++} 'v'{$y--}
'<'{$x--} *{if("$x,$y"-in$a){$a=$a|?{$_-ne"$x,$y"}}else{$a+="$x,$y"}}
};$a.length  Explanation: The input is changed to a character array and each piece is matched to one of the first 4 switch cases to increment the x or y coord. Everything matches the 5th case where we decide if a light is turned on or off. Length at the end determines total number of switches turned on. Note: The comma in the coordinates is necessary so we don't get collisions like (10,10) and (101,0) • Strip some apostrophes and pull$x,$y into a variable for 127bytes Jun 14 '19 at 3:40 • -ne works with array 115 bytes Jun 14 '19 at 6:37 • Length -> Count for 114 Jun 26 '19 at 18:42 # PowerShell, 96 95 bytes $(switch -w($args|% t*y){^{$y++}'>'{$x++}v{$y--}'<'{$x--}*{"$x,$y"}})|group|% c*|%{$r+=$_%2}$r


Try it online!

Unrolled:

$plane=$(switch -w($args|% toCharArray){ ^ {$y++}
'>'{$x++} v {$y--}
'<'{$x--} * {"$x,$y"} })$plane|group|% count|{$result+=$_%2}
$result  # PowerShell, 112 109 bytes A port of Luis Mendo's MATL answer. $args|% t*y|%{($p+=($c=[numerics.complex])::Pow($c::new(0,1),$_%11))}|group{'{0:N2}'-f$_}|% c*|?{$r+=$_%2};$r


Try it online!

# Actually, 24 23 bytes

This is a port of Luis Mendo's MATL answer. Golfing suggestions welcome. Try it online!

Edit: One byte thanks to Leaky Nun.

O⌠4P@%ïⁿ⌡Mσ;╗╔⌠╜c2@%⌡MΣ


Ungolfing:

O        ord(c) of the input string (implicit input)
⌠...⌡M   Start a function, and map over the list of ord(). Call the variable i.
4P@%     i mod 11 (the 4-th prime).
ïⁿ       1i to the power of (i%11).

σ        Push a list of cumulative sums of the new list of complex numbers.
This is a list of the coordinate visited.
;╗       Duplicate coord_list, and push to register 0.
╔        uniquify(coord_list)

⌠...⌡M   Map over the uniquified list. Call the variable j.
╜c       Push coord_list.count(j)
2@%      Push count mod 2

Σ        Return the sum of this last list.

• Other than writing :11 as 4P, I am unable to come up with any golf advice Aug 30 '16 at 4:12

# C, 203190 189 bytes

#define S(s)1-(*s%11<7)*2<<*s%11%2*16
char*t,*u;P,p,r,L,e;
f(char*s){for(P=L=0,t=s;*t;++t){P+=S(t);e=r=1;for(p=0,u=s;u<t;u++)e*=(p+=S(u))!=P;for(p=P;*++u;r^=p==P)p+=S(u);L+=e*r;}return L;}


O(n^2) with O(1) memory algorithm.

It loops over every position, checks if this position occurs earlier, and if not checks if this position occurs an even/odd amount of time afterwards, adding it to the total.

## Perl, 102 98 91 90 (89 + -n) bytes

perl -nE '$i=ord()%5,$x+=(-1..1)[$i],$y+=((0)x3,1,-1)[$i],$p{$x,$y}^=1 for/./g;say~~grep$_,values%p'  Readable: $_ = <>; # -n
for (/./g) {
$i = ord() % 5;$x += (-1 .. 1)[$i];$y += ((0) x 3, 1, -1)[$i];$p{$x,$y} ^= 1;
}

say 0 + grep $_, values %p;  • thanks Dada for -4 and -7 • it's the first time I see "readable" and "perl" in the same post I think! – Dada Aug 16 '16 at 16:23 • More seriously, great answer. You can still win a few byte if you use for/./g instead of while, which allows you to use ord() instead of ord($&).
– Dada
Aug 16 '16 at 16:25
• A few more things that can win a few bytes on PPCG : use say instead of print (since -M5.010 or -E are free) and use -n flag (cost 1 byte) instead of doing $_=<>. – Dada Aug 16 '16 at 16:55 # C# 210205139138135128 113 Bytes using System.Linq;int f(string s,int x=0,int y=0)=>s.GroupBy(c=>(x+=c%6%3-1)+s+(y-=c/3%6-2)).Sum(g=>g.Count()%2);  ## Explanation //Have X and Y as default params allows us to use a expression bodied member //and then remove the return statement int f(string s, int x = 0, int y = 0) => //treat the string as a char array s.GroupBy(c => //Using the unerlying int of the char this math reduces down to either x - 1, x + 1, x + 0, y - 1 or y + 1, or y + 0 //Concat the resulting values with the original string to ensure uniqueness ie x=1 and y=11 vs x=11 and y=1 for the grouping (x += c % 6 % 3 - 1) + s + (y -= c / 3 % 6 - 2)) //Mod 2 will return either 1 or 0 so we can use sum to count the groups with odd counts .Sum(g => g.Count() % 2);  • Thanks to VisualMelon for 5 bytes switching Count to Sum • Thanks again to VisualMelon, this time for 66 bytes by suggesting some fancy math to get the translations. • VisualMelon to the rescue again with combining the select and group by for 13 bytes and just using the input string for the x/y separator for another 2 • Nice... I like this a lot. Can you not use Distinct() rather than GroupBy(z=>z)? You can also use Sum(g=>g.Count()%2), since it's always 0 or 1, I should think. Aug 16 '16 at 22:38 • I also recon you can (sadly, I rather like it...) replace the func indexing with something like (x+=c%6%3-1)+"."+(y-=c%18/5-1) (just add a number we compute from the char with some stupid modulus and such). I've not tested, I'll let you fix it if it's broken ;) This ought to save a lot of bytes if I've not messed up! (And there is surly a cheaper pair of expressions, but these are just what I could come up with in the last 20minutes) Aug 16 '16 at 23:01 • @VisualMelon Distinct() doesn't work (unless i'm missing something) as it returns one of each string, rather than giving me a count of the occurences. – user19547 Aug 16 '16 at 23:12 • @VisualMelon good idea with using the sum, saved 5 bytes. As for your suggestion regarding the funcs, i think i need to understand why that works (it does) before i post it. is the choosing of the numbers based on the underlying values of the chars? – user19547 Aug 16 '16 at 23:20 • Yes, don't know what I was thinking about Distinct... I'm far too used to using Group as a lazy man's distinct I suppose... Aye, the stupid expression c%6%3-1 simply maps ASCII values of <>v^ to -1,1,0,0 (change in x, note the zero values) and the other expression produces a similar result for y. There are probably better expressions (I didn't look very hard), a drop of ternary might pay off. This was my first effort, which perhaps makes it more obvious what is going on: (c<61?--x:c<63?++x:x)+"."+(c>95?--y:c>93?++y:y) (but obviously it's quite a bit longer!) Aug 17 '16 at 6:24 # Perl 6, 47 bytes {sum bag([\+] (-i,*i...*)[.ords X%11]){*}X%2}  Try it online! • (-i, *i, ... *) is the infinite repeating sequence -i, 1, i, -1, ... • [.ords X% 11] slices into that sequence with the ordinal values of the characters in the input string, modulo 11. • [\+] performs a triangular reduction (or "scan") on those values, producing a list of the coordinates visited on the complex plane. • bag() creates a Bag containing those coordinates, each of which has an associated multiplicity (the number of times it appeared in the list). • {*} fetches all of the multiplicities from the bag. • X% 2 crosses those multiplicities with the number 2 using the modulus operator %. Odd multiplicities result in a 1, even multiplicities result in a 0. • sum sums those remainders. # Common Lisp, 198 bytes 186 bytes Solution: (let((p 0)(c 0)(h(make-hash-table)))(progn(loop for s across i do(incf(gethash(incf p(nth(position s "^>v<")'(1e3 1 -1e3 -1)))h 0)))(maphash #'(lambda(k v)(incf c(mod v 2)))h)(write c)))  Run it! Explanation: (let ((p 0)(c 0)(h (make-hash-table))) (progn (loop for s across i do(incf (gethash (incf p (nth (position s "^>v<")'(1e3 1 -1e3 -1))) h 0))) (maphash #'(lambda (k v) (incf c (mod v 2))) h) (write c)))  • p is a hashtable map key, generated as an index of the character within the string "^>v<", remapped to values 1000, -1000, 1 -1 representing the x,y. (x axis is +-1, y axis is +-1000) • the hash value at that index is incremented by 1 (switch counter) • the hashtable is than looped trough, and the map function increment the final count as modulo 2 of the value on a given key Note: i (input) is defined it test as: (defparameter i 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>v^^v^^>^<<^^^v^>^<<^>^>^>>>>v<v<v<>vv<<vv^<<^<vv>^^<^<<>><^v><><>^<v><v^>^v>^<^>^^><v><<^<v^^<<^><><v>v<>>><<^><v<^vvv^<<<>><<>^v^^><vv>vv<>^>^>vv<>v^<^<>vv><<>^<v<vv<^<^<><^vv<<^>>>v<>><<>>>^^^^<<^v>>v<vv>^^>v<v<vv^><<><>>>v>>^^v<^v^^>>v^<>>v^>><^<^^v<v<><<><>>^<>><^v<^^^^><>^>vv>>^vv<<>v<<<<<<><<<><<>><v><^^^<>>v<^><^vvv<>^>^^v>^<v><^v^vv^<<>v<<<<v>^vv>>v>vv<<^>^<>>vvv^<v<><>><>^^^^vvvvvvv<<>v<^><^^>vv^^<v<<^^<vvv<v<v<<>><<><v^^>><^<>^v^vv<<v<v<>><<>>>>>^vv<><v<>v><v>v>><v<v^vvvvv<><>v>>v<><<<^^<>^<^^<v>v^<vv>^vv^<>^<<^<vv><v<v>>v>^<>v^<<v^<v>^v<>><v>>>>^<<^^^v<^<>><^<><v>>vv^>^<^<^>>v^>^^^^>vvvvv>^v<^><^^<^^>^<^^^^^^^>v>>vv>v^^^v^^^<>v><^>>>v>^>^>^>vv<vv<^^>>^>>>v<>v><<^<<v^>^>>>>^^><^^<v<<<<>>v>v^v^^<>><v<^<<<<v^^^^<v<<<^>v>^^<vv<^^^^^v>^v^<v><>>^^>^v>^>^vv^v>v>v^>v>^>>^^^^>>^>>^><>><v>v>>><<^v^v^>^>^>>vv><<^>v<v<v^<<>>^v<<^v<<^><^>>^<v>^>vv>v>^^^>v^^<^<^^>vv>^^><v>>^v>^v<<^^^<<^v^>^<<^>vv^>>^<^v><<>v><^^^<^^>>vv>^vv>><^<<<^>vv^v>v<^<<<^<^<<><^^>>>v^<^^^>^<><^v>>^<<v<^v>>v^<^<^<^^^<v^><<vvv^<^v^vv^vv<v<<v<^<>^v>^^^<^^v<v<v><<<^<>^^^^v>v^v^v^v<v><v>>^v><vv^^^v>><<v^vvvv<<<^v<<><^>^<v^^v<>vvvv^vv<>^v<><>^^<>>vvv<^>><v^<<>v>v<>^v^>v^>><<>>^^<^v<>>^>^><>>^<v<v^^<^v><v^<v<><><^<<><v^v<<>vv<v<v<^>>><>vv^^<><<v<^^<<^<><^^^>^>>>^<^>>>^>><^^^<^v^^^v^v^v>v>v><vv>><vvv<<v><><^^>^v<v>><v><^><^<<>v^vv^v><^vv>^>>v<vv><^<^^v<^^vv<vv<v<v>v><v<vv<<>^^v^^v<<<^<>v^^^<><>>><^>v^^^v^vv<<<^>>v><^>v^<>>>>^<>^^vvv^^<><^>^^<><>^vvv^^<vv^>vv^^^^v<>vv<^^^v<<>><<vvvvv>v>^^^vv>><v><v<>vvvv<v^><^<>^>^<>v>v>v^vvvv<><^v>>>^^>><vvv<>^>^v^<vvv>v^vv^vv><>><>v^^v^vv<^v>vv>>v<v><^<<^v<>>^vv^<v>v><v>v>^v>^<v>^<<^>vv>v<^<^vv^<^><<<v<<^^vv<vvv><>v>v<vv^<><><^vvv>>vv<^^^v><^v><<^>^^v>^<>><v<>>^^<<<v><>^>^><vvvv<>^<<<><<<^<>>v^vv^>><^vv^^>^<v^<v>><^^>>>^v>^v<>^v<><^><vv>v^^^<^>>^<<^<^><<<^^<v<<^vv<^<>v<^<<^^<v<vv<<><v<v^<>^<>v>>v<^v>v<>^^vvv<>vv^v^<><v^vv^<^v^v><>^><v^<>>^^^<>>vv^<v>^^v><v<^>^^^^^^><>>vvv<<><><v<^>v<>v^v<<<<>v^>>>>^v>^^<v^>v><v^<^^v<<<<v<<<>^v<^>^v>v^^>v^^vvv>vv<>^>v><v<>^<vv><>>><<^>>><<v>v^^<^<<<<v^<>>>v<<<^v^vv<>^v>v<<<<>^^>><v><>v<v><^^>><>^>^>v>>><v>^vvvv<><><^>>^v^><<>^v<><><^><<<>v^^>v>^>v^<v^vv<>><^vv^^>^^><vv<<>v>v^^>><v^>^<^<>>>vv<>^>v>v^<>v<^<<v>>>^<>v^>v>>vv^^<>>v<v<<^<>v>v^<^^^>v^^>v>v>vv<^<v>v^^><<<v<><>^^<>v>v>^^v>v>v^v>>^<v^v>><>^^>^<>>>^vv^><v^<^>v^>^v><^>^^^vv^^v<>vv<>>^><<^v>^v^>>v>^v^<<^^^vv<<vvv>^vv^v<<<v^^<<><vv<>>^^vv>^^^vv>><><v>v<^v^>>>vv^><>><v<^v<>^><v<^^^^>><^<>v>^v<^vv>v>v<^<>v>v>^<vv>v<^>vvv<v^<vv<vv<>v>^><v^v<>>>>>v>><^v<>v>^v><v^v^vv<>^<vvv^>><v^<vvv^^<^vvv^v^<>><v>v^^v<><>v^^^v<<<^><v<<<>><<vv<<><vvv^v>>v^v<v^>>><<v<>^v><>vv<<v>v^vv>v^v<^<vv<><><^v>^<vv>v^^>>^^^><vv<><^>>>^<v^<<^^>^>vv^><v<vvv>^^>>>^><<vv>vv>^<>>^^><^v><<>^<<<v^>^")  • Does this mean that going up 1 is the equivalent of going right 1000 times - and you could clash? I like the idea of compressing x/y into a single dimension though so I'm going to nab it :) PS - you can use (99 1 -99 -1) to shave 2 bytes.. – mkst Jun 18 '19 at 21:29 • @streetster correct, that's the idea. To compress the x/y into 1d, where y must have some buffer to not interfere with the x. (and yes it might actually clash - therefore bigger gab, smaller the chance - thus I am using actually 1000 (1e3), so 999 would be the same byte length). Jun 19 '19 at 14:40 # K (oK), 3729 27 bytes Solution: +/2!.#:'=+\-99 0 99 -1 1@5!  Try it online! Explanation: Create list of steps, create path, group, count each repeat, if mod 2 is 0 then off, else on. Felt very AdventOfCode-ish. +/2!.#:'=+\-99 0 99 -1 1@5! / the solution 5! / input mod 5, "^v<>" yields 4 3 0 2 -99 0 99 -1 1 / left, null, right, down, up +\ / sum along = / group into key=>value #:' / count each value . / take the value 2! / modulo 2 +/ / sum up  Notes: • -8 bytes by stealing David Horák's method of squashing x-y into x (I know it has limitations, but it works for the examples) • -2 bytes thanks to ngn • 1 -1 -99 99"^v<>"? -> -99 0 99 -1 1@5! – ngn Jun 20 '19 at 17:43 • +/2!.#:' -> +/~=/'^ – ngn Jun 20 '19 at 17:57 • First comment works for -2, second one throws an error... what are you trying to do with the ^? – mkst Jun 21 '19 at 18:08 • my mistake. i wanted to use it as "is null?" on the dict's values (to make them all 0s) but apparently oK doesn't work that way :( – ngn Jun 21 '19 at 18:34 • Yeah, sadly it uses ^ as 'except' rather than null :( – mkst Jun 21 '19 at 20:37 # krrp, 137 bytes ^>:\L\T[length],^v>xy:!tTxy!V?[elem]tv[without]vLtELtv?#?E>V!c#!f>?=c$62.@V#!r>+x1y?=c$60.@V#!r>-x1y?=c$94.@V#!r>x-y1@V#!r>x+y1.LT00E>00.


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# Explanation

^>:\L\T                ~ lambda expression, import list and tuple module
[length],^v>xy:       ~  the answer is the number of lattice points
~  which were visited an odd number of times
!tTxy                ~   current lattice point
!V ?[elem]tv         ~   if the lattice point is already present,
[without]vLtE    ~    remove it, else
Ltv              ~    add it
?#?E> V              ~   no further moves
!c#!f>               ~   move character
?=c$62. @V#!r>+x1y ~ move right ?=c$60. @V#!r>-x1y  ~    move left
?=c\$94. @V#!r>x-y1 ~     move up
@V#!r>x+y1        ~      move down
.LT00E>00.            ~ initialize at point (0, 0)


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Unfortunately, krrp only has one rather slow implementation, making the long test cases difficult to verify.
krrp String conversion.

# APL (Dyalog Unicode), 21 20 bytesSBCS

+/≠/×⊢⌸+\0j1*'^<v'⍳⎕


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uses ⎕io←1

⎕ input

'^<v'⍳ find the index of each input char among '^<v', i.e. ^ becomes 1, < 2, v 3, and anything else 4

0j1* i to the power (the imaginary constant)

+\ partials sums

⊢⌸ matrix in which each row is the list of occurrences (indices) of a unique partial sum; padded with 0s to make it rectangular

× signum - all indices become 1s, the padding remains 0s

≠/ sum mod 2 for each row

+/ sum

# Ruby, 67 bytes

Inspired by orlp's C solution. Assembles a list of all visited positions, and then for each unique one (using setwise intersection (d&d) since it saves a byte over d.uniq), count the ones with an odd number of occurrences (technically, map it to the occurances%2, and sum them together, since that does the same thing)

->s,q=0{d=s.bytes.map{|c|q+=1i**(c%11)}
(d&d).sum{|c|d.count(c)%2}}


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