假设rand(n)在这里表示“从0到n-1均匀分布的随机整数”，下面是使用Python的randint的代码示例，它具有这种效果。它只使用randint(5)和常量来产生randint(7)的效果。其实有点傻

from random import randint
sum = 7
while sum >= 7:
    first = randint(0,5)   
    toadd = 9999
    while toadd>1:
        toadd = randint(0,5)
    if toadd:
        sum = first+5
    else:
        sum = first

assert 7>sum>=0 
print sum

#!/usr/bin/env ruby
class Integer
  def rand7
    rand(6)+1
  end
end

def rand5
  rand(4)+1
end

x = rand5() # x => int between 1 and 5

y = x.rand7() # y => int between 1 and 7

．．尽管这可能被认为是作弊。

通过使用滚动总数，您可以同时

保持平均分配;而且 不需要牺牲随机序列中的任何元素。

这两个问题都是简单的rand(5)+rand(5)…类型的解决方案。下面的Python代码展示了如何实现它(其中大部分是证明发行版)。

import random
x = []
for i in range (0,7):
    x.append (0)
t = 0
tt = 0
for i in range (0,700000):
    ########################################
    #####            qq.py             #####
    r = int (random.random () * 5)
    t = (t + r) % 7
    ########################################
    #####       qq_notsogood.py        #####
    #r = 20
    #while r > 6:
        #r =     int (random.random () * 5)
        #r = r + int (random.random () * 5)
    #t = r
    ########################################
    x[t] = x[t] + 1
    tt = tt + 1
high = x[0]
low = x[0]
for i in range (0,7):
    print "%d: %7d %.5f" % (i, x[i], 100.0 * x[i] / tt)
    if x[i] < low:
        low = x[i]
    if x[i] > high:
        high = x[i]
diff = high - low
print "Variation = %d (%.5f%%)" % (diff, 100.0 * diff / tt)

这个输出显示了结果:

pax$ python qq.py
0:   99908 14.27257
1:  100029 14.28986
2:  100327 14.33243
3:  100395 14.34214
4:   99104 14.15771
5:   99829 14.26129
6:  100408 14.34400
Variation = 1304 (0.18629%)

pax$ python qq.py
0:   99547 14.22100
1:  100229 14.31843
2:  100078 14.29686
3:   99451 14.20729
4:  100284 14.32629
5:  100038 14.29114
6:  100373 14.33900
Variation = 922 (0.13171%)

pax$ python qq.py
0:  100481 14.35443
1:   99188 14.16971
2:  100284 14.32629
3:  100222 14.31743
4:   99960 14.28000
5:   99426 14.20371
6:  100439 14.34843
Variation = 1293 (0.18471%)

一个简单的rand(5)+rand(5)，忽略那些返回大于6的情况，其典型变化为18%，是上面所示方法的100倍:

pax$ python qq_notsogood.py
0:   31756 4.53657
1:   63304 9.04343
2:   95507 13.64386
3:  127825 18.26071
4:  158851 22.69300
5:  127567 18.22386
6:   95190 13.59857
Variation = 127095 (18.15643%)

pax$ python qq_notsogood.py
0:   31792 4.54171
1:   63637 9.09100
2:   95641 13.66300
3:  127627 18.23243
4:  158751 22.67871
5:  126782 18.11171
6:   95770 13.68143
Variation = 126959 (18.13700%)

pax$ python qq_notsogood.py
0:   31955 4.56500
1:   63485 9.06929
2:   94849 13.54986
3:  127737 18.24814
4:  159687 22.81243
5:  127391 18.19871
6:   94896 13.55657
Variation = 127732 (18.24743%)

并且，根据Nixuz的建议，我已经清理了脚本，所以您可以提取并使用rand7…材料:

import random

# rand5() returns 0 through 4 inclusive.

def rand5():
    return int (random.random () * 5)

# rand7() generator returns 0 through 6 inclusive (using rand5()).

def rand7():
    rand7ret = 0
    while True:
        rand7ret = (rand7ret + rand5()) % 7
        yield rand7ret

# Number of test runs.

count = 700000

# Work out distribution.

distrib = [0,0,0,0,0,0,0]
rgen =rand7()
for i in range (0,count):
    r = rgen.next()
    distrib[r] = distrib[r] + 1

# Print distributions and calculate variation.

high = distrib[0]
low = distrib[0]
for i in range (0,7):
    print "%d: %7d %.5f" % (i, distrib[i], 100.0 * distrib[i] / count)
    if distrib[i] < low:
        low = distrib[i]
    if distrib[i] > high:
        high = distrib[i]
diff = high - low
print "Variation = %d (%.5f%%)" % (diff, 100.0 * diff / count)

int rand7() {
    int value = rand5()
              + rand5() * 2
              + rand5() * 3
              + rand5() * 4
              + rand5() * 5
              + rand5() * 6;
    return value%7;
}

与选定的解决方案不同，该算法将在常数时间内运行。然而，它对rand5的调用比所选解决方案的平均运行时间多2次。

请注意，这个生成器并不完美(数字0比任何其他数字都有0.0064%的可能性)，但对于大多数实际目的，保证恒定的时间可能比这种不准确性更重要。

解释

这个解源于数字15624能被7整除的事实，因此，如果我们可以随机且均匀地生成从0到15624的数字，然后对7取余，我们就可以得到一个近乎均匀的rand7生成器。将rand5滚动6次，将0到15624之间的数字统一生成，并使用这些数字组成以5为基数的数字，如下所示:

rand5 * 5^5 + rand5 * 5^4 + rand5 * 5^3 + rand5 * 5^2 + rand5 * 5 + rand5

mod 7的属性允许我们稍微简化一下方程:

5^5 = 3 mod 7
5^4 = 2 mod 7
5^3 = 6 mod 7
5^2 = 4 mod 7
5^1 = 5 mod 7

So

rand5 * 5^5 + rand5 * 5^4 + rand5 * 5^3 + rand5 * 5^2 + rand5 * 5 + rand5

就变成了

rand5 * 3 + rand5 * 2 + rand5 * 6 + rand5 * 4 + rand5 * 5 + rand5

理论

15624这个数字不是随机选择的，而是可以用费马小定理来发现的，该定理指出，如果p是质数，那么

a^(p-1) = 1 mod p

这就得到，

(5^6)-1 = 0 mod 7

(5^6)-1等于

4 * 5^5 + 4 * 5^4 + 4 * 5^3 + 4 * 5^2 + 4 * 5 + 4

这是一个以5为底的数，因此我们可以看到，这种方法可以用于从任何随机数发生器到任何其他随机数发生器。尽管在使用指数p-1时总是会引入对0的小偏差。

为了更准确地推广这种方法，我们可以有这样一个函数:

def getRandomconverted(frm, to):
    s = 0
    for i in range(to):
        s += getRandomUniform(frm)*frm**i
    mx = 0
    for i in range(to):
        mx = (to-1)*frm**i 
    mx = int(mx/to)*to # maximum value till which we can take mod
    if s < mx:
        return s%to
    else:
        return getRandomconverted(frm, to)

function rand7() {
    while (true) { //lowest base 5 random number > 7 reduces memory
        int num = (rand5()-1)*5 + rand5()-1;
    if (num < 21)  // improves performance
        return 1 + num%7;
    }
}

Python代码:

from random import randint
def rand7():
    while(True):
        num = (randint(1, 5)-1)*5 + randint(1, 5)-1
        if num < 21:
                return 1 + num%7

100000次运行的测试分布:

>>> rnums = []
>>> for _ in range(100000):
    rnums.append(rand7())
>>> {n:rnums.count(n) for n in set(rnums)}
{1: 15648, 2: 15741, 3: 15681, 4: 15847, 5: 15642, 6: 15806, 7: 15635}

你需要的函数是rand1_7()，我写了rand1_5()，这样你就可以测试它并绘制它。

import numpy
def rand1_5():
    return numpy.random.randint(5)+1

def rand1_7():
    q = 0
    for i in xrange(7):  q+= rand1_5()
    return q%7 + 1