the longest dependency path is +1 operations long. Examples Take Pn(x) to mean the nth order polynomial of the form: Pn(x) = C0 + C1x + C2x2 + C3x3 + ⋯ + Cnxn Written with Estrin's scheme we have: P3(x) = (C0 + C1x) + (C2 + C3x) x2 P4(x) = (C0 + C1x) + (C2 + C3x) x2 + C4x4 P5(x) = (C0 + C1x) + (C2 + C3x) x2 + (C4 + C5x) x4 P6(x) = (C0 + C1x) + (C2 + C3x) x2 + ((C4 + C5x) + C6x2)x4 P7(x) = (C0 + C1x) + (C2 + C3x) x2 + ((C4 + C5x) + (C6 + C7x) x2)x4 P8(x) = (C0 + C1x) + (C2 + C3x) x2 + ((C4 + C5x) + (C6 + C7x) x2)x4 + C8x8 P9(x) = (C0 + C1x) + (C2 + C3x) x2 + ((C4 + C5x) + (C6 + C7x) x2)x4 + (C8 + C9x) x8 … In full detail, consider the evaluation of P15(x): Inputs: x, C0, C1, C2, C3, C4, C5 C6, C7, C8, C9 C10, C11, C12, C13 C14, C15 Step 1: x2, C0+C1x, C2+C3x, C4+C5x, C6+C7x, C8+C9x, C10+C11x, C12+C13x, C14+C15x Step 2: x4, (C0+C1x) + (C2+C3x)x2, (C4+C5x) + (C6+C7x)x2, (C8+C9x) + (C10+C11x)x2, (C12+C13x) + (C14+C15x)x2 Step 3: x8, ((C0+C1x) + (C2+C3x)x2) + ((C4+C5x) + (C6+C7x)x2)x4, ((C8+C9x) + (C10+C11x)x2) + ((C12+C13x) + (C14+C15x)x2)x4 Step 4: (((C0+C1x) + (C2+C3x)x2) + ((C4+C5x) + (C6+C7x)x2)x4) + (((C8+C9x) + (C10+C11x)x2) + ((C12+C13x) + (C14+C15x)x2)x4)x8 References Further reading fast_polynomial, a Sage library using an improved scheme (extended abstract). Numerical analysis