Initialisation du repository de Beta

venv/lib/python3.12/site-packages/sympy/combinatorics/schur_number.py

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+"""
+The Schur number S(k) is the largest integer n for which the interval [1,n]
+can be partitioned into k sum-free sets.(https://mathworld.wolfram.com/SchurNumber.html)
+"""
+import math
+from sympy.core import S
+from sympy.core.basic import Basic
+from sympy.core.function import Function
+from sympy.core.numbers import Integer
+
+
+class SchurNumber(Function):
+    r"""
+    This function creates a SchurNumber object
+    which is evaluated for `k \le 5` otherwise only
+    the lower bound information can be retrieved.
+
+    Examples
+    ========
+
+    >>> from sympy.combinatorics.schur_number import SchurNumber
+
+    Since S(3) = 13, hence the output is a number
+    >>> SchurNumber(3)
+    13
+
+    We do not know the Schur number for values greater than 5, hence
+    only the object is returned
+    >>> SchurNumber(6)
+    SchurNumber(6)
+
+    Now, the lower bound information can be retrieved using lower_bound()
+    method
+    >>> SchurNumber(6).lower_bound()
+    536
+
+    """
+
+    @classmethod
+    def eval(cls, k):
+        if k.is_Number:
+            if k is S.Infinity:
+                return S.Infinity
+            if k.is_zero:
+                return S.Zero
+            if not k.is_integer or k.is_negative:
+                raise ValueError("k should be a positive integer")
+            first_known_schur_numbers = {1: 1, 2: 4, 3: 13, 4: 44, 5: 160}
+            if k <= 5:
+                return Integer(first_known_schur_numbers[k])
+
+    def lower_bound(self):
+        f_ = self.args[0]
+        # Improved lower bounds known for S(6) and S(7)
+        if f_ == 6:
+            return Integer(536)
+        if f_ == 7:
+            return Integer(1680)
+        # For other cases, use general expression
+        if f_.is_Integer:
+            return 3*self.func(f_ - 1).lower_bound() - 1
+        return (3**f_ - 1)/2
+
+
+def _schur_subsets_number(n):
+
+    if n is S.Infinity:
+        raise ValueError("Input must be finite")
+    if n <= 0:
+        raise ValueError("n must be a non-zero positive integer.")
+    elif n <= 3:
+        min_k = 1
+    else:
+        min_k = math.ceil(math.log(2*n + 1, 3))
+
+    return Integer(min_k)
+
+
+def schur_partition(n):
+    """
+
+    This function returns the partition in the minimum number of sum-free subsets
+    according to the lower bound given by the Schur Number.
+
+    Parameters
+    ==========
+
+    n: a number
+        n is the upper limit of the range [1, n] for which we need to find and
+        return the minimum number of free subsets according to the lower bound
+        of schur number
+
+    Returns
+    =======
+
+    List of lists
+        List of the minimum number of sum-free subsets
+
+    Notes
+    =====
+
+    It is possible for some n to make the partition into less
+    subsets since the only known Schur numbers are:
+    S(1) = 1, S(2) = 4, S(3) = 13, S(4) = 44.
+    e.g for n = 44 the lower bound from the function above is 5 subsets but it has been proven
+    that can be done with 4 subsets.
+
+    Examples
+    ========
+
+    For n = 1, 2, 3 the answer is the set itself
+
+    >>> from sympy.combinatorics.schur_number import schur_partition
+    >>> schur_partition(2)
+    [[1, 2]]
+
+    For n > 3, the answer is the minimum number of sum-free subsets:
+
+    >>> schur_partition(5)
+    [[3, 2], [5], [1, 4]]
+
+    >>> schur_partition(8)
+    [[3, 2], [6, 5, 8], [1, 4, 7]]
+    """
+
+    if isinstance(n, Basic) and not n.is_Number:
+        raise ValueError("Input value must be a number")
+
+    number_of_subsets = _schur_subsets_number(n)
+    if n == 1:
+        sum_free_subsets = [[1]]
+    elif n == 2:
+        sum_free_subsets = [[1, 2]]
+    elif n == 3:
+        sum_free_subsets = [[1, 2, 3]]
+    else:
+        sum_free_subsets = [[1, 4], [2, 3]]
+
+    while len(sum_free_subsets) < number_of_subsets:
+        sum_free_subsets = _generate_next_list(sum_free_subsets, n)
+        missed_elements = [3*k + 1 for k in range(len(sum_free_subsets), (n-1)//3 + 1)]
+        sum_free_subsets[-1] += missed_elements
+
+    return sum_free_subsets
+
+
+def _generate_next_list(current_list, n):
+    new_list = []
+
+    for item in current_list:
+        temp_1 = [number*3 for number in item if number*3 <= n]
+        temp_2 = [number*3 - 1 for number in item if number*3 - 1 <= n]
+        new_item = temp_1 + temp_2
+        new_list.append(new_item)
+
+    last_list = [3*k + 1 for k in range(len(current_list)+1) if 3*k + 1 <= n]
+    new_list.append(last_list)
+    current_list = new_list
+
+    return current_list