姚氏百萬富翁問題數值解法「非電路解法」詳解（python代碼實現，原理分析）

寫在前面：

任何大師所發明的都是通用的規律，而不是賦予表象的技巧：

高斯的「正十七邊形尺規做圖法」是表象，而尺規做圖的加法器、乘法器、開根器是通用規律，更核心的是如何將正多邊形的角度轉換成開根式。

 

廢話很少，進入正題：

運行結果：

****************************************************
J_bob = 5 ; I_alice = 5
transfer_to_alice 1375
range_list       [    1,    2,    3,    4,    5,    6,    7,    8,    9,   10]
Yu_list          [ 1355, 1340, 1967,   32,  473,  758, 2051,   26,  883,  537]
Zu_list          [   71,   56,   41,   32,   45,    9,   18,   26,   27,    2]
list_back_to_bob [   71,   56,   41,   32,   45,   10,   19,   27,   28,    3]
         list[J_bob - 1] = list[4] = 45
G = x_random_bob % p_alice_for_bob = 45
result: Bob <= Alice

****************************************************
J_bob = 7 ; I_alice = 5
transfer_to_alice 1373
range_list       [    1,    2,    3,    4,    5,    6,    7,    8,    9,   10]
Yu_list          [  125, 2110, 1355, 1340, 1967,   32,  473,  758, 2051,   26]
Zu_list          [   18,   77,   71,   56,   41,   32,   45,    9,   18,   26]
list_back_to_bob [   18,   77,   71,   56,   41,   33,   46,   10,   19,   27]
         list[J_bob - 1] = list[6] = 46
G = x_random_bob % p_alice_for_bob = 45
result: Bob > Alice

****************************************************
J_bob = 3 ; I_alice = 5
transfer_to_alice 1377
range_list       [    1,    2,    3,    4,    5,    6,    7,    8,    9,   10]
Yu_list          [ 1967,   32,  473,  758, 2051,   26,  883,  537, 1110, 1428]
Zu_list          [   41,   32,   45,    9,   18,   26,   27,    2,   40,   37]
list_back_to_bob [   41,   32,   45,    9,   18,   27,   28,    3,   41,   38]
         list[J_bob - 1] = list[2] = 45
G = x_random_bob % p_alice_for_bob = 45
result: Bob <= Alice

Process finished with exit code 0

實現代碼：

def map_list_5_str(list_in):
    ready_list = list(map(lambda x: '%5s' % str(x), list_in))
    r_str = ','.join(ready_list)
    return '[' + r_str + ']'


# public key of alice as e, and private key as d, total encode space is N
e_alice = 79
d_alice = 1019
N_alice = 3337

o1 = 1
o2 = 10

I_alice = 5
for J_bob in [5, 7, 3]:
    print "\n****************************************************"
    print "J_bob =", J_bob, "; I_alice =", I_alice

    assert o1 < I_alice < o2
    assert o1 < J_bob < o2

    # ------------------------------
    x_random_bob = 473  # bob's random number
    c = (x_random_bob ** e_alice) % N_alice
    transfer_to_alice = c - J_bob
    print 'transfer_to_alice', transfer_to_alice

    range_alice = [1, o2 - o1 + 1]
    range_list = range(range_alice[0], range_alice[1] + 1)
    print 'range_list      ', map_list_5_str(range_list)
    
    Yu_list = [((transfer_to_alice + u) ** d_alice) % N_alice for u in range_list]
    print 'Yu_list         ', map_list_5_str(Yu_list)

    p_alice_for_bob = 107
    assert p_alice_for_bob < N_alice
    Zu_list = list(map(lambda x: x % p_alice_for_bob, Yu_list))
    print 'Zu_list         ', map_list_5_str(Zu_list)

    list_back_to_bob = Zu_list[0:I_alice]
    list_back_to_bob.extend(list(map(lambda x: x + 1, Zu_list[I_alice:])))
    print 'list_back_to_bob', map_list_5_str(list_back_to_bob)
    g = x_random_bob % p_alice_for_bob
    g_ = list_back_to_bob[J_bob - 1]
    print '         list[J_bob - 1] = list['+str(J_bob - 1)+'] =', g_
    print 'G = x_random_bob % p_alice_for_bob =', g
    print 'result:', 'Bob <= Alice' if g >= g_ else 'Bob > Alice'

 

參考：

Andrew Yao’s Millionaire’s Problem

Explain it like I’m Five: The Socialist Millionaire Problem and Secure Multi-Party Computation

YAO’S MILLIONAIRE PROBLEM

YAO’S MILLIONAIRES’ PROBLEM AND PUBLIC-KEY ENCRYPTION WITHOUT COMPUTATIONAL ASSUMPTIONS

一種有意思的想法

1. 先在一不透光試管中加入未知量的水
2. A女士量取她年齡那麼多毫升的1mol/L的HCl 加入試管
3. B女士量取她年齡那麼多毫升的1mol/L的NaOH 加入試管
4. 用Ph試紙就可判斷誰的年齡大了

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百萬富翁問題