How to use a given list of values for a parameter in FindInstance?

Question

How should I specify a discrete range for a parameter as one of the conditions in FindInstance? MemberQ/Element functions don't seem to work.

For example,

FindInstance[{x^2 + y^2 < 10, MemberQ[{1, 2, 5, 10}, x], MemberQ[{1/8, 1/4, 1/2, 1}, y]}, {x, y}]

FindInstance[{x^2 + y^2 < 10, x ∈ {1, 2, 5, 10}, y ∈ {1/8, 1/4, 1/2, 1} }, {x, y}] 

Answer 1

You can use region objects for this purpose:

xvalues = {1, 2, 5, 10};
yvalues = {1/8, 1/4, 1/2, 1};

FindInstance[
    {
    x^2+y^2 < 10,
    {x} ∈ Point[List/@xvalues],
    {y} ∈ Point[List/@yvalues]
    },
    {x,y},
    10
]

{{x -> 1, y -> 1/8}, {x -> 1, y -> 1/4}, {x -> 1, y -> 1/2}, {x -> 1, y -> 1}, {x -> 2, y -> 1/8}, {x -> 2, y -> 1/4}, {x -> 2, y -> 1/2}, {x -> 2, y -> 1}}

Another possibility:

FindInstance[
    pt ∈ RegionIntersection[{
        Disk[{0, 0}, Sqrt[10]],
        RegionProduct[Point[List/@xvalues], Point[List/@yvalues]]
    }],
    pt,
    10
]

{{pt -> {1, 1/8}}, {pt -> {1, 1/4}}, {pt -> {1, 1/2}}, {pt -> {1, 1}}, {pt -> {2, 1/8}}, {pt -> {2, 1/4}}, {pt -> {2, 1/2}}, {pt -> {2, 1}}}

The problem with using MemberQ is that it always evaluates to True/False, while Element expects a domain specification or a region.

Answer 2

xvalues = {1, 2, 5, 10};

yvalues = {1/8, 1/4, 1/2, 1};

FindInstance[
 x^2 + y^2 < 10 && Or @@ Thread[x == xvalues] && 
  Or @@ Thread[y == yvalues], {x, y}, 8]

(* {{x -> 1, y -> 1/8}, {x -> 1, y -> 1/4}, {x -> 1, y -> 1/2}, 
    {x -> 1, y -> 1}, {x -> 2, y -> 1/8}, {x -> 2, y -> 1/4}, 
    {x -> 2, y -> 1/2}, {x -> 2, y -> 1}} *)

EDIT: Based on comparative timings, FindInstance is not the preferred approach.

$HistoryLength = 0;

hanlon = FindInstance[
    x^2 + y^2 < 10 && Or @@ Thread[x == xvalues] &&
     Or @@ Thread[y == yvalues], {x, y}, 8] //
   AbsoluteTiming;

woll = FindInstance[
    {x^2 + y^2 < 10,
     {x} ∈ Point[List /@ xvalues],
     {y} ∈ Point[List /@ yvalues]},
    {x, y}, 8] //
   AbsoluteTiming;

outer = Outer[
     If[#1^2 + #2^2 < 10, {x -> #1, y -> #2}, Nothing] &,
     xvalues, yvalues] //
    Flatten[#, 1] & //
   AbsoluteTiming;

The results are identical

Equal @@ Last /@ {hanlon, woll, outer}

(* True *)

However, using Outer is 300 times faster

(First /@ {hanlon, woll, outer})/outer[[1]]

(* {333.446, 308.769, 1.} *)