Placing seven point-sized pawns

Question

Can you place seven (point-sized) pawns on a $7\times7$ checkerboard, so that

• every pawn is placed precisely in the middle of one of the little checkerboard squares, and
• all distances between pairs are different?

Answer 1

Aight, second try. This time was computer generated, so let's see if my code is dummy or what:

Pos{x=0, y=0}, Pos{x=6, y=6}, Pos{x=5, y=5}, Pos{x=1, y=2}, Pos{x=0, y=2}, Pos{x=2, y=6}, Pos{x=3, y=0}

and these are the used distances:

2.0, 5.830951894845301, 7.211102550927978, 7.0710678118654755, 5.385164807134504, 6.324555320336759, 3.0, 2.23606797749979, 6.4031242374328485, 2.8284271247461903, 8.48528137423857, 4.0, 1.0, 5.0, 3.605551275463989, 6.708203932499369, 4.123105625617661, 3.1622776601683795, 4.47213595499958, 1.4142135623730951, 6.082762530298219

For the curious, here is the code (minus syntactic sugar for brevity):

class Pos {
    final int x;
    final int y;
}

class State {
    final Set<Double> usedDistances;
    final Set<Pos> usedCells;
}

void run(String[] args) {
    State initial = new State(new HashSet<>(), new HashSet<>());
    Set<State> visited = new HashSet<>();
    Deque<State> queue = new LinkedList<>();
    queue.add(initial);
    while (!queue.isEmpty()) {
        State state = queue.remove();
        if (!visited.contains(state)) {
            visited.add(state);
            if (isFinal(state)) {
                System.out.println(state);
                return;
            }
            else {
                Set<State> next = nextOf(state);
                queue.addAll(next);
            }
        }
    }
    System.out.println("nothing");
}

private Set<State> nextOf(State state) {
    Set<State> res = new HashSet<>();
    for (int i = 0; i<7; i++) {
        for (int j = 0; j<7; j++) {
            Pos pos = new Pos(i, j);
            if (!state.usedCells.contains(pos)) {
                Set<Double> nextDistances = new HashSet<>(state.usedDistances);
                Set<Pos> nextPos = new HashSet<>(state.usedCells);
                boolean valid = true;
                for (Pos used: state.usedCells) {
                    double distance = computeDistance(pos, used);
                    if (nextDistances.contains(distance)) {
                        valid = false;
                        break;
                    }
                    else {
                        nextDistances.add(distance);
                    }
                }
                if (valid) {
                    nextPos.add(pos);
                    res.add(new State(nextDistances, nextPos));
                }
            }
        }
    }
    return res;
}

Finally, here is a picture of the solution:

       -----------------------------
       | X |   |   | X |   |   |   |
       -----------------------------
       |   |   |   |   |   |   |   |
       -----------------------------
       | X | X |   |   |   |   |   |
       -----------------------------
       |   |   |   |   |   |   |   |
       -----------------------------
       |   |   |   |   |   |   |   |
       -----------------------------
       |   |   |   |   |   | X |   |
       -----------------------------
       |   |   | X |   |   |   | X |
       -----------------------------

Answer 2

Not really an answer but I'd like to add my thought process to this

First I checked: How many possible distances are there really on a 7x7 checkerboard? To do this you can draw lines from (1,1) to (1,7) till (7,7), from (1,1) to (1,6) till (6,6) and so on. This mean $7+6+5+4+3+2 = 27$, However because 3,4,5 is a pythagorean triple it is one less because (1,1) to (4,5) = (1,1) to (6,1). There are 21 connections between 7 points. My intuition therefore says it is very unlikely that there are 21 connections that are all different because there are only 26 possibilities to begin with.

This does tell us that if there is a solution that at least 3 points are in a square along the border. Why? There are 7 distances that go from border to border and only a maximum of 5 are not used of these. The two distances could share a point, therefore at least 3 points are on the border squares

Answer 3

I would say:

No this isn't possible. You should only have 26 different options of distance between pawns and the total number of relationships to take into account is 5040. Still thinking on it, but if I'm right feel free to tell me.

This is unless:

The board itself isn't square ;)