# Real Chebyshev Rotation

This is a challenge inspired by Chebyshev Rotation. I suggest looking at answers there to get inspiration for this challenge.

Given a point on the plane there is a unique square (a rectangle with equal sides) that is centered on the origin and intersects that point (interactive demo):

Given a point p and a distance d, return the point obtained by moving distance d from p, counter-clockwise (and clockwise for negative d), along the perimeter of the square centered on the origin that intersects p. Your answer must be accurate to at least 4 decimal digits.

### Testcases:

(0, 0), 100 -> (0, 0)
(1, 1), 81.42 -> (-0.4200, 1.0000)
(42.234, 234.12), 2303.34 -> (-234.1200, 80.0940)
(-23, -39.234), -234.3 -> (39.2340, -21.8960)


The following test cases are from the original challenge by Martin Ender, and all are with d = 1:

(0, 0)       -> (0, 0)
(1, 0)       -> (1, 1)
(1, 1)       -> (0, 1)
(0, 1)       -> (-1, 1)
(-1, 1)      -> (-1, 0)
(-1, 0)      -> (-1, -1)
(-1, -1)     -> (0, -1)
(0, -1)      -> (1, -1)
(1, -1)      -> (1, 0)
(95, -12)    -> (95, -11)
(127, 127)   -> (126, 127)
(-2, 101)    -> (-3, 101)
(-65, 65)    -> (-65, 64)
(-127, 42)   -> (-127, 41)
(-9, -9)     -> (-8, -9)
(126, -127)  -> (127, -127)
(105, -105)  -> (105, -104)

• Couldn't almost all of these be only slightly changed from the other challenge? This seems like a needless addition. – ATaco Oct 17 '16 at 2:41
• @ATaco No, it's quite a bit more complicated. – orlp Oct 17 '16 at 2:45
• Should the distance be calculated along the perimeter starting at p? – Gábor Fekete Oct 17 '16 at 16:33
• @GáborFekete What else? – orlp Oct 17 '16 at 16:34
• Yeah I see, the test cases imply this but it isn't stated explicitly. I thought at first that it would start at the positive intersection at the x-axis. – Gábor Fekete Oct 17 '16 at 16:39

# Python 2, 363335296266262258256 233 Bytes

Woo, 130 bytes lost! Thanks to Neil for saving 4 bytes, Nathan Merrill for saving 2 bytes, and xnor for saving a ridiculous 23 bytes!

The general idea is this: We can reduce the distance traveled by taking it's the modulus of it against the perimeter of the square. The perimeter is defined as 8 times the largest of the two coordinates, since the point must rest on it. Then, after modulus is taken we are guaranteed to have no overlap. It also guarantees we only ever have to move counter-clockwise, since modulus gives a positive result.

From there, I simply use what we know from the given x and y-coordinates to figure out where we are: top, bottom, left, right, or in a corner, and determine the direction, which can be one of 0, 1, 2, 3:

0 --> we are on the 'top', moving 'left'
1 --> we are on the 'left', moving 'down'
2 --> we are on the 'bottom', moving 'right'
3 --> we are on the 'right', moving 'up'


After this, it is as simple as looping while we still have distance to travel, and based on the direction we subtract or add to the appropriate coordinate, and tell the loop which direction we are going next.

p,d=input()
x,y=p
s=max(x,y,-x,-y)
d=d%(s*8or 1)
r=[(y<s)*[2,[3,x>-s][x<s]][y>-s],[2*(y<0),3*(y<=0)][x>0]][y*y==x*x]
while s>0<d:f=1-2*(r<2);m=abs(f*s-p[r%2]);j=d>m;p[r%2]=[p[r%2]+f*d,f*s][j];r=-~r%4;d=(d-m)*j
print"%.4f "*2%tuple(p)


While quite long, it certainly works. Here's some example I/O:

[0, 0], 100 --> 0.0000 0.0000
[1, 1], 81.42 --> -0.4200 1.0000
[42.234, 234.12], 2303.34 --> -234.1200 80.0940
[-23, -39.234], -234.3 --> 39.2340 -21.8960

• Does s=max(x,y,-x,-y) work? – Neil Oct 18 '16 at 8:56
• @Neil It does, thanks a bunch! – Kade Oct 18 '16 at 13:06
• (s>0)*(d>0) is s>0<d. The output can be "%.4f "*2%tuple(p). if s:d=d%(8*s) can be d%(s*8or 1). (r+1) can be ~-r. 1*(x>-s) can just be (x>-s). abs(y)==abs(x) can be y*y==x*x – xnor Oct 18 '16 at 16:15
• @xnor Wow, thanks! Only thins I changed are that (x>-s) didn't need the parenthesis, and ~-r decrements, so I used -~r. – Kade Oct 18 '16 at 16:51

## JavaScript (ES6), 147 bytes

f=(x,y,d,s=Math.max(x,y,-x,-y),c=(d/8%s+s)%s*8,v=0,w=x+y>0?1:-1,b=(v?x:y)*w+c-s)=>c?b>0?f(v?s*w:x,v?y:s*w,d,s,b,!v,v?w:-w):[x+c*w*v,y+c*w*!v]:[x,y]


Explanation: Works by trying to add the direction vector while keeping within the bounds of the square. Any overshoot is recursively passed back with the direction rotated anticlockwise by 90°. The direction is actually encoded using a vertical flag v and a unit w so that the vectors (1, 0), (0, 1), (-1, 0) and (0, -1) are encoded with v of 0, 1, 0, 1 and w of 1, 1, -1, -1 respectively. The direction vector may not initially point in a suitable direction but it never points backwards so it will eventually rotate into a usable direction.

f=(x,y,d,                   Input parameters
s=Math.max(x,y,-x,-y),     Calculate half the side of the square
c=(d/8%s+s)%s*8,           Reduce the distance modulo the perimeter
v=0,                       Initial vertical flag
w=x+y>0?1:-1,              Initial direction
b=(v?x:y)*w+c-s)=>         Will we overshoot the corner?
c?b>0?f(v?s*w:x,v?y:s*w,  Advance to the next corner
d,s,b,!v,v?w:-w): Rotate the direction

• This could be because my browser (Opera 40.0.2308.81), but it looks like it has a bit of rounding error for f(42.234, 234.12, 2303.34) -> [-234.12, 80.09399999999988] meaning it doesn't have 4 digit precision. Maybe adding some output formatting would fix this? Nice answer though! :) – Kade Oct 19 '16 at 18:32