Generating four complementary intervals between 0 and 1-CodePudding

How do I generate a list that contains four tuples and each tuple is compose by two numbers between 0.0 and 1.0 following this rules:

The first and the second number of each tuple cannot be the samecov
The second number must be bigger than the first one
The range cover by all four tuples must add 1
The range covered by each tuple cannot have a intersection

(0.01, 0.1), (0.1,0.2), (0.3,1), (0.2,0.29) -> acceptable

(0.1,0.2), (0.2,0.3) ,(0.2,0.3) ,(0.4,0.5) -> repeated interval

(0.1, 0.2), (0.30000000000000004, 0.5), (0.4, 0.8), (0.6, 0.9) -> the range between the four tuples not add to 1 or 100%

So far I code this logic but I'm not respecting the condition for the ranges to be complementary

interval_range = range(0, 100, 10)
interval = [
    (lowerlimit, upperlimit)
    for lowerlimit, upperlimit in product(interval_range, interval_range)
    if lowerlimit < upperlimit
]

interval_final = []
for one, two, three, four in product(interval, interval, interval, interval):
    size = one   two   three   four
    size_no_dup = set(size)
    if sum([one[1]-one[0],two[1]-two[0],three[1]-three[0],four[1]-four[0]]) == 100 and len(size) == len(size_no_dup):
        interval_final.append((one, two, three, four))
        break #just to see if the logic is working

CodePudding user response：

Based on the discussion in the comments, this solution uses ints from 0 to 100 rather than floats, and gives non-overlapping ranges whose widths sum to 100.

This uses itertools.pairwise which is new in Python 3.10.

import random
import itertools

cuts = random.sample(range(1, 100), k=3)  # 3 guaranteed different random numbers from 1 to 99 inclusive
ranges = list(itertools.pairwise([0]   sorted(cuts)   [100]))
print(ranges)

Example output:

[(0, 11), (11, 63), (63, 72), (72, 100)]

If you want multiple combinations, you can wrap this in a for loop. If you want to reproducibly generate the same series of solutions, set random.seed to a fixed value.

For earlier versions of Python 3, an alternative is:

import random

cuts = random.sample(range(1, 100), k=3)
points = [0]   sorted(cuts)   [100]
ranges = [(points[i], points[i 1]) for i in range(len(points)-1)]
print(ranges)

To enumerate every valid combination:

valid = []
for x in range(1, 98):
    for y in range(x 1, 99):
        for z in range(y 1, 100):
            valid.append([(0, x), (x, y), (y, z), (z, 100)])
print(len(valid))
print(valid[0])
print(valid[-1])

prints:

156849
[(0, 1), (1, 2), (2, 3), (3, 100)]
[(0, 97), (97, 98), (98, 99), (99, 100)]