Mapping Passing Through Point

Question

(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

$$\begin{cases} \frac12 x & 0 < x \leq 4 \\ -\frac43 x + \frac{46}{3} & 4 < x < 10 \end{cases}$$

Which looks like:

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.

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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 code-golf, the shortest function wins.

Answer 1

R, 6 bytes

approx

Try it online! or Try it with graphical output at rdrr.io

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.

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R, 48 38 bytes

\(a,b,c,x)(a+(x*(b+c-2*a)/c[1])%%b)[2]

Attempt This Online! or Try it with graphical output at rdrr.io

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.

Answer 2

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.

Answer 3

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!

Answer 4

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.

Answer 5

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)\$.

Answer 6

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

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Port of Arnauld's JS answer.