How many $\mathbb R$s must a Mathematician walk down?

Question

A mathematician is lost on the complex plane. He knows neither his position nor the direction he is facing. He wants to return to the main road, a strip of width $1$ around the real axis (that is, from $\mathbb R+\frac{i}{2}$ to $\mathbb R-\frac{i}{2}$). Naturally, he will recognize the road once he is on it.

The only instrument in his possession is a compass-like device that can show him the absolute value $|z|$ of his position $z$. The problem: He bought the cheap "finite precision" version, which does not display any digits after the decimal point (it just rounds to the nearest integer).

Looking at his device, he sees the number $1000$. Can he get to his destination in a finite number of steps, where a step consists of consulting the device, turning an angle $\phi$ of his choice and then walking straight for a distance $d$ of his choice (both choices can be different in each step)? If yes, how can he minimize the number of steps he must take? Or is it all hopeless and in the end the answer is blowing in the wind?

Clarification: He will recognize the road when a step takes him directly on to it. He will not recognize the road simply by crossing it in the course of a step.

Answer 1

He doesn't even need his special device to find the road in a finite time. He just needs to walk in an expanding spiral, in straight-line steps shorter than two units, until he hits the road.

The spiral can be any scale, but if he knows that he is 1000 units from the origin, it is enough to go forward 1000 units, left 1000 units, left 2000 units, left 2000 units, and left again 2000 units, which takes at most 501 + 501 + 1001 + 1001 + 1001 = 4005 steps. (Of course this can be improved on, for instance by following a circle more closely.)

Answer 2

I can do it in three. Walk $10$ in the direction he is facing. Turn $\frac \pi 2$ and walk $10$. At least one of the two will show a change of at least $7$ units. Use the arctangent of the two observations to determine the direction to the origin. You can't be too far off, so walk that way $1500$ and he will hit the road.

Answer 3

It seems to me that he can at least estimate his angle to the origin pretty effectively with just one step: take a step of length 500 in whatever angle he happens to be facing, and consult the device for the new distance $d$. If O,A,B denote origin, old position, new position respectively, then he knows all the distances OA, AB, OB to reasonable precision (all the lengths have to be at least 500, so the relative error introduced by rounding should be small). Using the law of cosines he can estimate all the angles of this triangle, which would give him a decent estimate of his angle to the origin.

I don't know whether this estimate is actually good enough that just turning to face the estimated origin and taking a step of length OB will get him close to the origin, though -- seems that OB being large gives the error a chance to get multiplied as well.

edit: I guess this is basically the same as Ross Millikan's answer, but it saves a step

Answer 4

Since the problem statement says "he," may we assume the mathematician is not female? That rules out one obvious answer that would get the mathematician back on the road in one step.

The initial reading tells the mathematician that the origin of the plane is somewhere in an annulus with inner radius $999.5,$ outer radius $1000.5,$ and center at his current location.

The question is how efficiently he can subdivide the possible locations of the origin until either he happens to land on the road while searching, or reduces the possible locations of the origin to a figure that fits within a disk of radius $\frac12,$ at which point his next step is to travel to the center of that disk (and he will then be on the road).

One method is to travel $10000$ units (or other suitably long distance) in the initial direction. This narrows the location of the origin either to a long, thin sliver between two arcs, to a figure like a double-headed "spear" (with a bent or straight shaft), or to a disconnected figure composed two regions each bounded by four arcs. In any case, there are two extremal points of the figure that are as far apart as any two points in that figure can be.

The mathematician's second step is $\frac12$ unit in the same direction. This effectively draws a curve "lengthwise" through the locus of the origin, and depending on whether the distance reading goes up by one or stays the same, the locus is reduced to the part on one side or the other of that curve. The remaining locus has a maximum width of about $\frac12,$ possibly more than $\frac12$ but certainly much less than $1.$ In particular, it is possible to construct a line (relative to the points already visited) that passes within less than $\frac12$ unit of all points in the remaining locus of the origin.

The mathematician's third step is to a point on that line a great distance from the nearest point of the locus of the origin, and his fourth step is a distance $\frac12$ to another point on that line.

The locus of the origin is now bounded between one pair if arcs approximately $\frac12$ unit apart, and also between a pair of arcs approximately $\frac12$ unit apart and approximately "perpendicular" to the first pair. The locus easily fits within a disk of radius $\frac12,$ so the fifth (and final) step is to any point in that locus.

I'm not entirely certain it can't be done in four or fewer steps.