# Mapping Passing Through Point

(Inspired by this challenge.)

Given six real values in three pairs: $$\(x_1, x_2), (y_1, y_2),\$$ and $$\(x_0, y_0)\$$, where $$\x_1 < x_0 < x_2\$$ and $$\y_1 < y_0 < y_2\$$, create a function which maps between $$\(x_1, x_2)\$$ and $$\(y_1, y_2)\$$ which also passes through $$\(x_0,y_0)\$$. In other words, make some function

$$f: \mathbb{R} \to \mathbb{R},\; f((x_1, x_2)) = (y_1, y_2),\; f(x_0) = y_0$$

Note that the image of the function has to be over the whole interval -- in other words, for every $$\y \in (y_1,y_2)\$$, there must be some $$\x \in (x_1,x_2)\$$, such that $$\f(x)=y\$$.

• The function does not need to be bijective.
• The value of the function cannot be outside of the given $$\y\$$ range.

For example, for $$\(x_1, x_2) = (0, 10), (y_1, y_2) = (0, 10), x_0 = 4, y_0 = 2\$$, a possible function is

$$\begin{cases} \frac12 x & 0 < x \leq 4 \\ -\frac43 x + \frac{46}{3} & 4 < x < 10 \end{cases}$$ On the edge points (i.e. when $$\x = x_1\$$ or $$\x = x_2\$$) or outside of the interval, the function can have whatever value you want, or it can be undefined.

Note: You don't have to return a function -- you could also write your code so that it takes in $$\x_1, x_2, y_1, y_2, x_0, y_0, x\$$ and returns $$\y = f(x)\$$ following the constraints above.

Standard loopholes are forbidden. Since this is , the shortest function wins.

• Is the second line in your example flipped? You say 10 should map to 10, and it would make more sense for 4+Δx to map to 2+Δy.
– Neil
Mar 12 at 19:09
• Is the image of $f$ required to be contained within $(y_1,y_2)$ over the interval $x\in(x_1,x_2)$, or can there be some value within the domain that maps to a value outside of $(y_1,y_2)$? Mar 13 at 15:33
• @AndersKaseorg It does not need to be bijective. Mar 13 at 16:12
• @Neil I don't think I understand your question. Mar 13 at 16:12
• I didn't understand your formulae - what you're trying to say is that the function needs to map the open interval (x₁, x₂) to the open interval (y₁, y₂) and specifically the value x₀ needs to map to y₀. The confusion arises because (x₀, y₀) is a coordinate but the others are open intervals.
– Neil
Mar 13 at 19:38

# R, 6 bytes

approx


Input is x1,x2,x0, y1,y2,y0, x; output is x,y.

This seems like a fairly straightforward task for which built-in solution ought to exit in R, and indeed it does.

# R, 48 38 bytes

\(a,b,c,x)(a+(x*(b+c-2*a)/c)%%b)


Input is x1,y1, x2,y2, x0,y0, x; output is y.

Non-builtin solution. Avoids any kind of curve-fitting tomfoolery by using a deliberately non-bijective function. # MATL, 4 bytes

&1Yn


The implemented mapping consists of two straight segments: the first from the point $$\(x_1,y_1)\$$ to $$\(x_0,y_0)\$$, and the second from $$\(x_0,y_0)\$$ to $$\(x_2,y_2)\$$.

The code inputs two numerical vectors and a number: [x1 x0 x2], [y1 y0 y2], x. Then it simply calls the interp1 function, which linearly interpolates the data defined by [x1 x0 x2] and [y1 y0 y2] at the abscissa x. The ouput is the value y corresponding to x.

Try it online! You can also see the graph of the mapping here.

# JavaScript (ES6), 71 bytes

Expects $$\(x_0,x_1,x_2,y_0,y_1,y_2,x)\$$.

Just uses quadratic interpolation, which is most certainly not the shortest approach.

(a,b,c,d,e,f,x)=>(x-b)*((f-e)/(c-b)-(e-=d)/(b-=a))/(c-a)*(x-=a)+e/b*x+d


Try it online!

# Charcoal, 31 bytes

ＮθＮηＮζＦ⊕›θη«ＮεＮδ»Ｉ⁺ζ∕×⁻θη⁻δζ⁻εη


Try it online! Link is to verbose version of code. Takes input as seven numbers x, x₀, y₀, x₁, y₁, x₂ and y₂. Explanation: Performs linear interpolation between x₀ and y₀ and either x₁ and y₁ or x₂ and y₂ depending on whether x is greater than x₀ or not.

Ｎθ


Input x.

ＮηＮζ


Input x₀ and y₀.

Ｆ⊕›θη«ＮεＮδ»


Input x₁ and y₁ or x₂ and y₂ depending on whether x is greater than x₀.

Ｉ⁺ζ∕×⁻θη⁻δζ⁻εη


Perform linear interpolation.

# Pyt, 33 bytes

Đ←=?ĉ←:←ŕŕ←←-Đ↔⇹ᵮ₄%⇹/4*←Đ←⇹-⇹↔*+;


Try it online!

Uses $$\f(x)=\left\{\begin{array}{ c l }y_0 &\quad\textrm{if }x=x_0 \\y_1+(y_2-y_1)*{4\left(x\!\!\!\!\mod\!{x_2-x_1\over 4}\right)\over x_2-x_1}& \quad\textrm{otherwise}\end{array}\right. \$$

Takes input as the following, each on a new line: x x0 y0 x2 x1 y1 y2

Code Stack Action
Đ $$\x\ x\$$ implicit input; Đuplicate
$$\x\ x\ x_0\$$ get $$\x_0\$$
=? $$\x\ \{x\!=\!x_0\}\$$ if $$\x=x_0\$$:
ĉ← $$\y_0\$$ ĉlear the stack and get $$\y_0\$$
:←ŕŕ $$\x\$$ otherwise, get $$\y_0\$$ and then ŕemove $$\\{x\!=\!x_0\}\$$ and $$\y_0\$$
←←- $$\x\ z\$$ get $$\x_2\$$ and $$\x_1\$$ and then subtract (call it $$\z\$$)
Đ $$\x\ z\ z\$$ Đuplicate
↔⇹ $$\z\ x\ z\$$ manipulate the stack
ᵮ₄ $$\z\ x\ {z\over4}\$$ cast to ᵮloat, then divide by 4
% $$\z\ q\$$ $$\x\!\!\!\!\mod\!\! {z\over4}\$$ (call it $$\q\$$)
$$\q\ z\$$ swap top two items on stack
/ $$\{q\over z}\$$ divide
4* $$\{4q\over z}\$$ multiply by 4
←Đ $$\{4q\over z}\ y_1\ y_1\$$ get $$\y_1\$$ and Đuplicate
$$\{4q\over z}\ y_1\ y_1\ y_2\$$ get $$\y_2\$$
⇹- $$\{4q\over z}\ y_1\ m\$$ swap top two on stack and then subtract (call it $$\m\$$)
⇹↔ $$\y_1\ m\ {4q\over z}\$$ manipulate the stack
* $$\y_1\ {4mq\over z}\$$ multiply
+ $$\y_1+{4mq\over z}\$$ add
; either way, implicit print

The function is onto $$\(y_1,y_2)\$$ as follows:

It can be trivially shown that if the discontinuity at $$\x_0\$$ did not exist, then $$\\forall y\in(y_1,y_2)\$$, $$\\exists\$$ at least one $$\x\in(x_1,x_2)\$$ s.t. $$\f(x)=y\$$. In fact, there would be between two and four such points for all values.

Some of you may object here and say that $$\\nexists x\in(x_1,x_2)\$$ s.t. $$\f(x)=y_2\$$, and you'd be right. However, the problem lists the intervals as open, and so the endpoints are not necessary.

Since without the discontinuity, there would be at least two distinct $$\x\in(x_1,x_2)\$$ such that $$\f(x)=y\$$, with the discontinuity, every point must have at least one such $$\x\$$. Therefore, $$\f(x)\$$ is onto $$\(y_1,y_2)\$$.

# Python, 81 bytes

lambda a,b,c,d,e,f,x:(x-b)*((f-e)/(c-b)-(e:=e-d)/(b:=b-a))/(c-a)*(x:=x-a)+e/b*x+d


Attempt This Online!