Recovering Range Parameters

Question

Description

You are given the results of a range function where every element has been rounded down to the nearest whole number. Your goal is to recover the original list.

For example, the following function (in Python3) will produce an input for your program:

from numpy import arange, floor
def floored_range(A, B, C):
    return list(floor(arange(A, B, C)))

The output of your program should be a valid guess of the original data. Here, valid guess means that it must exactly match the input when floored and it must be a possible output of a range function (ie, when graphed it must form a perfectly straight line).

Examples

Input: [1,2,3,4]  
Output: [1,2,3,4]  

Input: [1,2,3,4]  
Output: [1.9,2.7,3.5,4.3]  

Input: [1,2,3,4,5,5]  
Output: [1.9,2.7,3.5,4.3,5.1,5.9]  

Input: [1,1,2,2,3,3,4,4]  
Output: [1,1.5,2,2.5,3,3.5,4,4.5]  

Input: [1,1,2,3,3,4]  
Output: [1,1.7,2.4,3.1,3.8,4.5]

Input: [56, 54, 52, 50, 48, 45, 43, 41, 39, 37, 35, 32, 30, 28, 26, 24, 22, 19, 17, 15, 13, 11]
Output: [56.7  , 54.541, 52.382, 50.223, 48.064, 45.905, 43.746, 41.587,
   39.428, 37.269, 35.11 , 32.951, 30.792, 28.633, 26.474, 24.315,
   22.156, 19.997, 17.838, 15.679, 13.52 , 11.361]

Answer 1

Octave, 82 bytes

function y=f(x)
while any(floor(y=linspace(x(1)+rand,x(end)+rand,numel(x)))-x),end

Running time is non-deterministic, but the code ends in finite time with probability 1.

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Explanation

The code defines a function of x that outputs y. The function consists of a while loop.

In each iteration, the right amount (numel(x)) of linearly spaced values are generated (linspace), starting at x(1)+rand and ending at x(end)+rand. These two calls to the rand function give random offsets between 0 and 1, which are applied to the initial and final values of x.

The loop is repeated for as long as any of the floored results differs (-) from the corresponding entry in x.

Answer 2

Python 3, 189 bytes

def f(l):
 R=range(len(l));e=1-1e-9
 for j in R:
  for I in range(j*4):
   i=I//4;L=[((l[i]+I//2%2*e)*(x-j)-(l[j]+I%2*e)*(x-i))/(i-j)for x in R]
   if[x//1 for x in L]==l:return L
 return l

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Cubic time.

Has some numerical issues.

Answer 3

R, 86 bytes

function(n){while(any(n-(x=seq(n[1]+runif(1),tail(n,1)+runif(1),l=sum(n|1)))%/%1))0;x}

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R port of Luis Mendo's answer; it does issue a number of warnings because of any coercing to logical but these can be ignored.

Answer 4

Python 3, 168 bytes

def f(l):r=range(len(l));g=lambda n:[(l[b]+n-l[a])/(b-a)for a in r for b in r if b>a]or[0];s=(max(g(-1))+min(g(1)))/2;m=min(a*s-l[a]for a in r);return[a*s-m for a in r]

Try it online! Explanation: g calculates the limiting values for C that lie just outside the range for A and B to exist. The average is then taken to give a usable value for C, and then the lowest possible range is then generated.

Answer 5

Jelly, 31 bytes

ṾṚ”.;V×LḶ$}+©1ị$}IEȧḞ⁼¥ʋ
0ç1#®ḷ

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Warning: Floating-point inaccuracies.

Answer 6

JavaScript (Node.js), 94 bytes, assuming input length > 1

f=x=>(t=x.map(_=>a+=b,b=x[1]+(c=Math.random)(a=x[0]+c())-a,a-=b)).map(Math.floor)+''==x?t:f(x)

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97 bytes

f=x=>(t=x.map(_=>a+=b,b=x[1]+(c=Math.random)(a=x[0]+c())-a||0,a-=b)).map(Math.floor)+''==x?t:f(x)

Answer 7

Python 2, 212 bytes

def f(l):
 C=[(0,1.,0,1.)]
 for a,A,b,B in C:
  x,y=(A+a)/2,(B+b)/2;r=[l[0]+x+i*(l[-1]+y-l[0]-x)/(~-len(l)or 1)for i in range(len(l))];C+=[(x,A,y,B),(a,x,y,B),(x,A,b,y),(a,x,b,y)]
  if[n//1for n in r]==l:return r

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