/**
     * this method finds the vector similarity between two arrays between 0 and 1
     * where 1 is the highest value of similarity
     * returns to within a margin of error because even double precision floats
     * in java are not perfect.
     * @param in1 The first vector
     * @param in2 the second vector
     * @return the cosine of the angle between the two vectors, which is a measure of similarity.
     */
    public static double findSimilarity(ArrayList<Double> in1, ArrayList<Double> in2) {
        //calculate dot product (in1[0]*in2[0] + in1[1]*in[1] ... in1[n]*in2[n])
        double dot = 0;
        //since when calculating we are only multiplying and adding same numbered indicies, we can stop at whatever length is the smallest
        for(int i = 0; i<in1.size() && i<in2.size(); i++) {
            dot += in1.get(i)*in2.get(i);
        }

        //calculate the magnitude of each (sqrt(in1[0]^2+in1[1]^2...in[n]^2))
        double squareofmagofin1 = 0;
        for(int i = 0; i < in1.size(); i++) {
            squareofmagofin1 += in1.get(i)*in1.get(i);
        }

        double squareofmagofin2 = 0;
        for(int i = 0; i < in1.size(); i++) {
            squareofmagofin2 += in2.get(i)*in2.get(i);
        }

        //calculate the square roots:
        double sqrtofmagofin2 = Math.sqrt(squareofmagofin2);
        double sqrtofmagofin1 = Math.sqrt(squareofmagofin2);

        /* Now doe this:
         * model the inputs as vectors, take cosine of both vectors as a measure of similarity
         * using   item . args
         * cosx = _______________
         *       ||item||||args||
         */

        double result = dot/(sqrtofmagofin1*sqrtofmagofin2);
        return result;
    }