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mathfloating-pointprecisionangle

Find angle between hour and minute hands in an analog clock

I was given this interview question recently:

Given a 12-hour analog clock, compute in degree the smaller angle between the hour and minute hands. Be as precise as you can.

I'm wondering what's the simplest, most readable, most precise algorithm is. Solution in any language is welcome (but do explain it a bit if you think it's necessary).

Solution

  • It turns out that Wikipedia does have the best answer:

    // h = 1..12, m = 0..59
static double angle(int h, int m) {
    double hAngle = 0.5D * (h * 60 + m);
    double mAngle = 6 * m;
    double angle = Math.abs(hAngle - mAngle);
    angle = Math.min(angle, 360 - angle);
    return angle;
}

    Basically:

    • The hour hand moves at the rate of 0.5 degrees per minute
    • The minute hand moves at the rate of of 6 degrees per minute

    Problem solved.

    And precision isn't a concern because the fractional part is either .0 or .5, and in the range of 0..360, all of these values are exactly representable in double.