“…Suppose the two adjacent symbols 12 and 2 are deleted, then π = (2, 6,12,1,9,5,13,4,8,10,3,7,11) is received. By checking the positions of values from [1,4] and [10,13] in π, we have R = [4,7] and R = [3,6]. That is a = k j = 4, b = 8 in Algorithm 2, and there is no h j in [a, a + s − 1] = [4,6].…”