Python，递归：给出满足布尔表达式的所有可能的元组组合

Question

So, for school, I got an exercise on recursion, which goes as follows: I'm given a string and a random int value 'N'. The string is a boolean expression, for example '3*x - 2* y <0' . The result has to be a list of tuples(x, y), (-N < x < N and -N < y < N), from all the possible tuple combinations who meet the expression. I did this exercise first with for loops etc and that was not so difficult, but then I had to do it recursively, and here's where I got stuck: As you can see, I just add up x and y by '1' at the end of the code, which will give me all tuple combinations where X and Y are the same. For example, if N = 5, my code only evaluates the combinations (-4,-4), (-3,-3) ... (4,4) but not (-2,1) or (1,3) for example. So my question is: can anyone help me writing a recursive code which evaluates the boolean expression in all the possible tuple combinations? My code has to be written recursively and I can't use functions as 'itertools' etc, it's not allowed in our school.

**MY CODE:**
def solution(expression, N,x=None,y=None):
  if x is None: x = -N + 1
  if y is None: y = -N + 1
  res = []
  if x >= N and y >= N:
      return []

  if eval(expression) == True:
      res.append((x, y))

  return res + solution(expression, N, x+1, y+1)

Answer 1

I have modified your code and I think it works now:

UPDATE: I have corrected this expression if x < N - 1 : to that one if x < N - 1 or y < N - 1:

def solution(expression, N, x=None, y=None):
  if x is None: x = -N + 1
  if y is None: y = -N + 1
  res = []

  if eval(expression) == True:
    res.append((x, y))

  # if x < N - 1 :
  if x < N - 1 or y < N - 1:
    if y < N - 1:
      y += 1
    else:
      x += 1
      y = - N + 1
    return res + solution(expression, N, x, y)
  else:
    return res

print(solution('3*x - 2* y <0', 4))

Answer 2

Slightly different approach building on getting permutations for a list and in the end checking the expression. Not the prettiest code but does the job.

results = []


def check(expression, items):
    x = y = None

    if len(items) == 1:
        x = y = items[0]
        if eval(expression) and (x, y) not in results:
            results.append((x, y))

    if len(items) == 2:
        x = items[0]
        y = items[1]
        if eval(expression) and (x, y) not in results:
            results.append((x, y))
        x = items[1]
        y = items[0]
        if eval(expression) and (x, y) not in results:
            results.append((x, y))

    if len(items) > 2:
        for i in items:
            remaining_elements = [x for x in items if x != i]
            check(expression, remaining_elements)


expression = "3*x - 2*y < 0"
N = 4
items = range(-N + 1, N)
check(expression, items)
print(results)