Oscillation equality

Question

We have objects that oscillate between two integer points, [l, r], at the speed of one unit per time unit, starting at l on t=0. You may assume l < r. For example, if an object oscillates on [3, 6], then we have:

t=0 -> 3
t=1 -> 4
t=2 -> 5
t=3 -> 6
t=4 -> 5
t=6 -> 4
t=7 -> 3
t=8 -> 4

Etc. But objects oscillate continuously, so we also have t=0.5 -> 3.5 and t=3.7 -> 5.3.

Given two objects oscillating between [l1, r1], [l2, r2], determine if there is ever a time t such that the two objects share the same position. You make take l1, r1, l2, r2 in any convenient format, and output any truthy/falsy values.

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Truthy inputs:

[[3, 6], [3, 6]]
[[3, 6], [4, 8]]
[[0, 2], [2, 3]]
[[0, 3], [2, 4]]
[[7, 9], [8, 9]]

Falsy inputs:

[[0, 3], [3, 5]] 
[[0, 2], [2, 4]]
[[5, 8], [9, 10]]
[[6, 9], [1, 2]]
[[1, 3], [2, 6]]

Answer 1

Python 2, 69 bytes

l,r,L,R=input()
d=r-l;k=1
while(R-L)*k%d:k+=1
print r-L>=k%2*d/k<=R-l

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Answer 2

Husk, 13 bytes

VEΣUẊeTmȯ…¢mD

Takes input in format [[l,r],[L,R]]. Returns 0 for falsy instances and a positive integer for truthy instances. Try it online!

Explanation

The main ideas are

1. A collision can only happen at an integer or half-integer coordinate.
2. It's enough to simulate the system until a repetition of two consecutive states is encountered.

Here's the code annotated.

VEΣUẊeTmȯ…¢mD  Implicit input, say [[0,2],[2,3]]
       mȯ      For both pairs do:
           mD   Double each: [[0,4],[4,6]]
          ¢     Cycle: [[0,4,0,4..],[4,6,4,6..]]
         …      Rangify: [[0,1,2,3,4,3,2,1,0,1,2..],[4,5,6,5,4,5,6..]]
      T        Transpose: [[0,4],[1,5],[2,6],[3,5],[4,4],[3,5],[2,6]..
    Ẋe         Adjacent pairs: [[[0,4],[1,5]],[[1,5],[2,6]],[[2,6],[3,5]],[[3,5],[4,4]]..
   U           Prefix of unique elements: [[[0,4],[1,5]],[[1,5],[2,6]],[[2,6],[3,5]],[[3,5],[4,4]]..[[1,5],[0,4]]]
  Σ            Concatenate: [[0,4],[1,5],[1,5],[2,6],[2,6],[3,5],[3,5],[4,4]..[1,5],[0,4]]
VE             Index of first pair whose elements are equal (or 0 if not found): 8

Answer 3

JavaScript (ES6), 104 100 bytes

A naive implementation that just runs the simulation. Takes (a, b, c, d) as 4 distinct variables.

(a,b,c,d)=>(g=(X,Y)=>x==y|x+X==y&(y+=Y)==x||(x+=X)-a|y-c&&g(x>a&x<b?X:-X,y>c&y<d?Y:-Y))(1,1,x=a,y=c)

Test cases

let f =

(a,b,c,d)=>(g=(X,Y)=>x==y|x+X==y&(y+=Y)==x||(x+=X)-a|y-c&&g(x>a&x<b?X:-X,y>c&y<d?Y:-Y))(1,1,x=a,y=c)

console.log('Truthy')
console.log(f(3, 6, 3, 6))
console.log(f(3, 6, 4, 8))
console.log(f(0, 2, 2, 3))
console.log(f(0, 3, 2, 4))
console.log(f(7, 9, 8, 9))

console.log('Falsy')
console.log(f(0, 3, 3, 5))
console.log(f(0, 2, 2, 4))
console.log(f(5, 8, 9, 10))
console.log(f(6, 9, 1, 2))

Answer 4

Wolfram Language (Mathematica), 77 69 61 bytes

If[#>#3,#0[##3,#,#2],(z=GCD[x=#-#2,#3-#4])Mod[x/z,2]<=#2-#3]&

A pure function taking the four arguments l1, r1, l2, r2 as input: e.g., [0,3,2,4] when the intervals are [0,3] and [2,4].

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How it works

To get a point in [a,b] close to a point in [c,d], assuming a<c<b<d, we want an odd multiple of b-a within b-c of an even multiple of d-c. If b-a has more factors of 2 than d-c, we can make this happen exactly: there will be a time when the first point is at b and the second point is at c, and then we're in good shape. If not, then the best we can do is the GCD of b-a and d-c.

Answer 5

JavaScript (ES6), 89 bytes

(a,b,c,d)=>[...Array((b-=a)*(d-=c)*4)].some((g=e=>i/e&2?e-i/2%e:i/2%e,i)=>a+g(b)==c+g(d))

Takes l1,r1,l2,r2 as separate arguments. Explanation: The simulation is guaranteed to repeat after (r1-l1)*(r2-l2)*2 time units (or a factor thereof); g calculates the offset of the appropriate object after i/2 time units, so i needs to range up to (r1-l1)*(r2-l2)*4.

Answer 6

05AB1E, 12 10 14 bytes

+4 bytes to handle negative ranges

Return 0 if falsy, or a positive integer otherwise

Use Zgarb's idea of doubling values to make same position detection easier

Thanks to @Zacharý for pointing out my mistakes

ÄZU\·εXиŸ}øüQO

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Explanations:

ÄZU\·εXиŸ}øüQO 
ÄZU\            Store in X the largest absolute number in the lists
    ·           Double lists ([3,6],[4,8] => [6,12],[8,16])
     ε   }      For each...
      X             Push X
       и            List-repeat that much times ([6,12]*12 => [6,12,6,12,6,...])
        Ÿ           Rangify ([6,12,6,...] => [6,7,8,9,10,11,12,11,...])
          ø     Zip lists ([6,7,8,...],[8,9,10,...] => [6,8],[7,9],[8,10],...)
           üQ   1 if both elements of a pair are equal, 0 otherwise
             O  Sum result list (=0 if the same position is never shared)
                Implicit output

Answer 7

Java (OpenJDK 8), 81 bytes

(l,r,L,R)->{int d=r-l,k=0;for(;(R-L)*++k%d>0;);return(r-L<R-l?r-L:R-l)>=k%2*d/k;}

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Reusing xnor's Python algorithm.