Linear Function Solver from Two Points (x,y)

Question

The Task

Write a function L() that takes two Tuple arguments of coordinates in the form (x, y), and returns their respective linear function in the form (a, c), where a is the co-efficent of the x term and c is the y-intercept.

You can assume that the input will not be a line perpendicular the the x axis, and that the two inputs are separate points.

Scoring 

This is Code Golf: shortest program wins.

Please Note: No use of any mathematical functions apart from basic operators (+,-,/,*).

Example

Here is my un-golfed solution in Python.

def L(Point1, Point2):
    x = 0
    y = 1
    Gradient = (float(Point1[y]) - float(Point2[y])) / (float(Point1[x]) - float(Point2[x]))
    YIntercept = Point1[y] - Gradient * Point1[x] 
    return (Gradient, YIntercept)

Output:

>>> L( (0,0) , (1,1) )
(1.0, 0.0)

>>> L( (0,0) , (2,1) )
(0.5, 0.0)

>>> L( (0,0) , (7,1) )
(0.14285714285714285, 0.0)

>>> L( (10,22.5) , (5,12.5) )
(2.0, 2.5)

Answer 1

J - 23 char

Fairly straightforward. Defines a dyadic verb L to be used as (x1,y1) L (x2,y2).

L=:%~/@:-,-/@(*|.)%-&{.

Explanation:

L=:%~/@:-,-/@(*|.)%-&{.  NB. the function L
                    &{.  NB. x coord of both points
                   -     NB. left x minus right x
             ( |.)       NB. flip right argument: (y2,x2)
              *          NB. pointwise multiplication of (x1,y1) and (y2,x2)
          -/@            NB. subtract the two results: (x1*y2)-(y1*x2)
                  %      NB. divide: (x1*y2 - y1*x2)/(x1-x2)
        -                NB. pointwise subtraction
   %~/@:                 NB. divide y difference by x diff: (y1-y2)/(x1-x2)
         ,               NB. append results together
L=:                      NB. assign function to L

Examples:

   L=:%~/@:-,-/@(*|.)%-&{.
   0 0 L 1 1
1 0
   0 0 L 2 1
0.5 0
   0 0 L 7 1
0.142857 0
   10 22.5 L 5 12.5
2 2.5
   0 0 L 0 1  NB. __ is negative infinity
__ 0

Answer 2

GNU dc, 30 24 bytes

[sysxly-rlx-/dlx*lyr-]sL

Defines a macro L such that (x₁, y₁, x₂, y₂) should be pushed to the stack in that order before calling, and after calling L, (a, c) may be popped from the stack (in reverse order of course - it is a stack).

Testcase (save as "linear.dc" and run dc linear.dc):

[sysxly-rlx-/dlx*lyr-]sL   # Define L macro

10                         # Push x1 to the stack
22.5                       # Push y1 to the stack
5                          # Push x2 to the stack
12.5                       # Push y2 to the stack

lLx                        # Call L macro
f                          # Dump the stack

Output is:

$ dc linear.dc 
2.5
2
$ 

Explanation of L macro:

• sy pop y₂ to y register
• sx pop x₂ to x register
• ly push y register (y₂)
• - subtract y₂ from y₁
• r swap (y₁ - y₂) and x₁ on stack
• lx push x register (x₂)
• - subtract x₂ from x₁
• / divide (y₁ - y₂) by (x₁ - x₂) to get gradient
• d duplicate gradient
• lx push x register (x₂)
• * multiply (x₂) by gradient
• ly push y register (y₂)
• r swap (y₂) and (x₂ * gradient) on stack
• - subtract (x₂ * gradient) from (y₂)

Answer 3

Haskell, 41 characters

f(x,y)(u,v)=(a,y-a*x)where a=(y-v)/(x-u)

Not a lot to golf here. It's pretty much what you'd write normally minus whitespace.

Answer 4

Mathematica, 55 38 bytes

This was surprisingly long (those pesky long function names...) EDIT: Changed the approach for the axis intercept (taking some inspiration from the OP's own answer). It turns out calculating it directly wasn't the most clever idea.

L={g=1/Divide@@(#2-#),#[[2]]-g#[[1]]}&

Use like

L[{10,22.5},{5,12.5}]
> {2., 2.5}

Thanks to Mathematica you can also obtain the general result:

L[{r,s},{p,q}]
> {(p - r)/(q - s), (q r - p s)/(q - s)}

(This last example shows how I had originally implemented this.)

Just for the record

L[{0,0},{0,1}]
> {ComplexInfinity, Indeterminate}

which is technically correct.

Answer 5

JavaScript, 62 48

Thanks to @Michael for golfing it down with ES 6.

L=(a,b)=>[s=(b[1]-a[1])/(b[0]-a[0]),a[1]-s*a[0]]

Old version:

function L(a,b){return[s=(b[1]-a[1])/(b[0]-a[0]),a[1]-s*a[0]]}

Sample input:

L([0,0],[7,1])

Sample output:

[0.14285714285714285, 0]

For the record:

L([0,0],[0,1])
[Infinity, NaN]

Answer 6

C# 105 bytes

This is isn't just the function and will compile completely on it's own. I had put L in the System namespace to shorting the using, but it's better to fully qualify and save on using a namespace. Saved the brackets. Also a saving from return new z[] into return new[]

using z=System.Single;class P{z[] L(z[]a,z[]b){z c=(a[1]-b[1])/(a[0]-b[0]);return new[]{c,a[1]-c*a[0]};}}

Answer 7

Python3 (51)

def L(p,q):x,y=p;X,Y=q;m=(Y-y)/(X-x);return m,y-x*m

Answer 8

Lua 5.1.4: 66 64 bytes

function L(q,w)a=(q[2]-w[2])/(q[1]-w[1])return a,q[2]-a*q[1];end

Example Usage:

> print(L( {0,0}, {1,0} ))
-0   0
> print(L( {0,0}, {1,1} ))
1    0
> print(L( {0,0}, {7,1} ))
0.14285714285714    0
> print(L( {0,0}, {0,1} ))
-inf   -nan
> print(L( {0,0}, {0,0} ))
-nan   -nan

Answer 9

C++ 88 (was 106)

Improved: thanks for your comments.

struct t{double x,y;};
t L(t u, t v){u.x=(v.y-u.y)/(v.x-u.x);u.y=v.y-u.x*v.x;return u;}

Golfed:

typedef struct T{double x,y;}t;
t line(t u, t v){t z;z.x=(v.y-u.y)/(v.x-u.x);z.y=v.y-(z.x*v.x);return z;}

Source

typedef struct T{
    double x,y;
} t;

t line(t u, t v)
{
t z;
z.x=(v.y-u.y)/(v.x-u.x);
z.y=v.y-(z.x*v.x);
return z;
}

Answer 10

Apple Swift 95 86

This may be the first Swift entry on PCG.SE??

func L(x:Float...)->(Float,Float){var a=(x[3]-x[1])/(x[2]-x[0]);return(a,x[1]-a*x[0])}

I don't see this language being a huge hit to the Code Golf community.

Answer 11

Python3 - 64 57 Bytes

def L(q,w):a=(q[1]-w[1])/(q[0]-w[0]);return a,q[1]-a*q[0]

You can get it down to 43 if you don't use Tuple, which many people are doing...

def L(x,y,q,w):a=(x-q)/(y-w);return a,y-a*x

Answer 12

Golfscript: 25 bytes

~:y\:x;-\x--1?*.x-1**y+\p

Since the function needs to be named 'L', I saved it as 'L.gs' locally.

The catch, as explained by @Dennis in this post, is that we need to trick Golfscript into using rational numbers instead of integers. So this works if you're willing to accept input X1 Y1 X2 Y2 in golfscript notation

# L( (0,0) , (1,1) )
echo "0 0 1 1" | golfscript L.gs
> 1/1
> 0/1
#L( (10,22.5) , (5,12.5) )
echo "10 22 2-1?+ 5 12 2-1?+" | golfscript L.gs
> 2/1
> 5/2

Answer 13

Ruby – 48 characters

Nearly identical to the JavaScript answer:

L=->u,v{a,b,c,d=*u,*v;[s=(d-b).fdiv(c-a),b-s*a]}

Answer 14

PHP (75 chars)

function L($u,$v){return[$s=($v[1]-$u[1])/($v[0]-$u[0]),$v[1]-($s*$v[0])];}

test : print_r(L([0,0],[7,1]));

output :

Array
(
    [0] => 0.14285714285714
    [1] => 0
)

(thanks @ace)