一维搜索的二次插值法

# 搜索区间，分别是x[0]（为了对齐下标，无意义）,x1,x2,x3
x = [0, 0, 1, 3]
# 结果x
xb = 0
# 精确度
e = 0.2


def f(x2):
    return 3 * (x2 ** 3) - 4 * x2 + 2


def c1():
    res = (f(x[1]) - f(x[3])) / (x[1] - x[3])

    return res


def c2():
    res = (((f(x[1]) - f(x[2])) / (x[1] - x[2])) - c1()) / (x[2] - x[3])
    print("c2 =", res)
    return res


def XB():
    c_1 = c1()
    print("c1 =", c_1)
    res = 0.5 * (x[1] + x[3] - (c_1 / c2()))
    print("xb =", res)
    return res


def getNewx1x3(newx2):
    a = {
        "1": x[1],
        "new2": newx2,
        "old2": x[2],
        "3": x[3]
    }

    sorted(a.items(), key=lambda t: t[1])
    l = list(a.keys())
    i = 0
    for k in l:
        if k == "new2":
            break
        i += 1

    return a[l[i - 1]], a[l[i + 1]]


for i in range(1000):
    xb = XB()
    fxb = f(xb)
    print("fxb =", fxb)
    if abs(x[2] - xb) < e:
        break
    aa = {
        "fxb": [fxb, xb],
        "f(x[1])": [f(x[1]), x[1]],
        "f(x[2])": [f(x[2]), x[2]],
        "f(x[3])": [f(x[3]), x[3]],
    }

    sorted(aa.items(), key=lambda t: t[1][0])

    newx2 = list(aa.values())[0][1]
    x[1], x[3] = getNewx1x3(newx2)
    x[2] = newx2
    print("new x1, x2, x3 =", x[1], x[2], x[3], end="

")

print("xb, fxb =", xb, fxb)