How to generate random points without duplication?

Question

In generating a list of random 2D integral coordinates e.g. {{u1,v1}, {u2,v2}, {u3,v3},...}, how can one write a condition specifying that no point is repeated, i.e no {ui,vi} == {uj,vj}?

Answer 1

If you only have 20 integers, you can proceed as in the other answers, and generate all possible tuples, and then sample randomly. On the other hand, if you have many more integers, you will not want to generate all possible tuples. For example, suppose you have 10^6 integers, and you want to guarantee that there are no duplicates. In this case, creating the list of 10^12 tuples will obviously not work.

In this case, you can create a function that converts between integers and tuples. For example, with integers from 0 to 999999, that function is QuotientRemainder. Since there are (10^6) * (10^6) tuples, we can construct a random sample of tuples with:

QuotientRemainder[RandomSample[0 ;; 10^12-1, 20], 10^6]

{{907983, 133497}, {517855, 551768}, {630894, 992111}, {729528, 990425}, {908605, 700641}, {109298, 886800}, {419959, 75716}, {829838, 303614}, {629060, 96075}, {302111, 741921}, {173138, 477425}, {89844, 799405}, {738388, 606520}, {419960, 677229}, {708126, 327841}, {420295, 191655}, {32408, 973949}, {181724, 12744}, {203054, 506577}, {288186, 436908}}

Here's a comparison with the Tuples approach:

SeedRandom[1]
r1 = QuotientRemainder[RandomSample[0 ;; 440, 20], 21]

SeedRandom[1]
r2 = RandomSample[Tuples[Range[0,20],2],20]

r1 === r2

{{19, 19}, {12, 5}, {18, 6}, {6, 8}, {1, 10}, {16, 0}, {10, 18}, {4, 11}, {5, 19}, {6, 11}, {14, 6}, {15, 8}, {19, 9}, {0, 19}, {3, 12}, {8, 1}, {18, 18}, {0, 9}, {9, 3}, {19, 6}}

{{19, 19}, {12, 5}, {18, 6}, {6, 8}, {1, 10}, {16, 0}, {10, 18}, {4, 11}, {5, 19}, {6, 11}, {14, 6}, {15, 8}, {19, 9}, {0, 19}, {3, 12}, {8, 1}, {18, 18}, {0, 9}, {9, 3}, {19, 6}}

True

And, a timing comparison for integers from 0 to 1000:

SeedRandom[1]
r1 = QuotientRemainder[RandomSample[0 ;; 1001^2-1, 20], 1001]; //AbsoluteTiming

SeedRandom[1]
r2 = RandomSample[Tuples[Range[0, 1000], 2], 20]; //AbsoluteTiming

r1 === r2

{0.000042, Null}

{0.003316, Null}

True

Answer 2

RandomSample[Tuples[Range[0, 20], 2], 10]

{{14, 13}, {2, 17}, {12, 7}, {1, 11}, {6, 6}, {12, 4}, {13, 14}, {2, 16}, {15, 6}, {1, 0}}

Note: The length of the sample cannot exceed 21^2 (441).

Answer 3

For 10 points, oversample and select unique ones using RandomSample.

points = RandomInteger[20, {50, 2}];
points = RandomSample[points, 10]

Actually, RandomInteger[20, {50, 2}] appear to produce 50 unique 2D points.

E.g.

tries = 0;
points = RandomInteger[20, {50, 2}];
While[(sample = Quiet[Check[RandomSample[points, 50], True]]),
 tries++;]
tries

0

Answer 4

Since you mentioned that your coordinates are drawn from integers in the regions 20 >= u >= 0 and 20 >= v >= 0, you can first generate a one-dimensional vector containing all lattice points:

V = Flatten[Table[x[i,j],{i,0,20},{j,0,20}]];

Then you can select a random sample from this vector, e.g.:

RandomSample[V,5]/.x->List

{{16, 8}, {10, 17}, {8, 5}, {0, 17}, {16, 9}}

Since V has no duplicates, the random sample will also contain no duplicates.

Answer 5

It kind of depends on what you want to do next. You only need to generate one such list, and only once? Then efficiency is not important, and so you can proceed very naively as follows: generate random lists until one works:

Module[{l = RandomInteger[{0, 20}, {10, 2}]},
       While[Length[l] != Length[DeleteDuplicates[l]], 
       l = RandomInteger[{0, 20}, {10, 2}]];
       l
]

On the other hand, if you need many of these lists, then a more efficient approach is as follows: first generate all possible pairs, and pick a random sample ever time you need one such list:

lists = Tuples[Range[0, 20], 2];
sample := RandomSample[lists,10];

Now every time you execute sample you will get one such list, picked at random.