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diff --git a/2025/10/b.py b/2025/10/b.py
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+++ b/2025/10/b.py
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+import fileinput
+from math import gcd
+
+inf = 999999999
+
+# Read an input line and return a triple A, b, c where:
+#   A is the matrix of coefficient (A[i][j] is 1 if pressing the j-th button
+#     increases the i-th counter, 0 otherwise).
+#   b is the list of required final values of each counter.
+#   c is a list of bounds for the number of times each button can be pressed:
+#     c[i] is the minimum required value of a counter increased by button i.
+def readline(l):
+        b = [int(x) for x in l[l.index('{')+1:l.index('}')].split(',')]
+
+        A = [[] for j in b]
+        c = []
+
+        end = l.index(']')
+        while '(' in l[end:]:
+                begin = l[end:].index('(') + end + 1
+                end = l[begin:].index(')') + begin
+                bu = [int(x) for x in l[begin:end].split(',')]
+                c.append(min(b[j] for j in bu))
+                for i in range(len(b)):
+                        A[i].append(1 if i in bu else 0)
+
+        return A, b, c
+
+# swaprow, swapcol and reducerow are utility functions for the matrix
+# row reduction procedure.
+def swaprow(A, b, i, j):
+        if i != j:
+                A[i], A[j] = A[j], A[i]
+                b[i], b[j] = b[j], b[i]
+
+def swapcol(A, c, i, j):
+        if i != j:
+                for k in range(len(A)):
+                        A[k][i], A[k][j] = A[k][j], A[k][i]
+                c[i], c[j] = c[j], c[i]
+
+def reducerow(A, b, i, j):
+        if A[i][i] != 0:
+                x = A[i][i]
+                y = -A[j][i]
+                d = gcd(x, y)
+                A[j] = [(y*A[i][k]+x*A[j][k])//d for k in range(len(A[i]))]
+                b[j] = (y*b[i]+x*b[j])//d
+
+# Reduce the matrix A by row, including back subsitution. Returns a triple
+# A, b, c where:
+#   A is the reduced matrix in the form [D|p] with D diagonal and p a matrix
+#     whose number of columns is the number of free parameters for Ax = b.
+#   b and c are the adjusted versions of the input paramters with the same
+#     names. In particular, b is adjusted every time a row is reduced and
+#     when rows are swapped, while c (the list of bounds) is adjusted when
+#     two columns are swapped (which is equivalent to changing the order of
+#     two buttons).
+def reduce(A, b, c):
+        for i in range(len(A[0])):
+                # Swap columns until there is one in position i with at least
+                # one non-zero element.
+                I = []
+                k = i
+                while len(I) == 0 and k < len(A[0]):
+                        swapcol(A, c, i, k)
+                        I = [j for j in range(i, len(A)) if A[j][i] != 0]
+                        k += 1
+
+                # If no such column is found, we are done
+                if len(I) == 0:
+                        break
+
+                # Swap rows so that A[i][i] is non-zero
+                swaprow(A, b, i, I[0])
+
+                # Reduce all other rows
+                for j in range(i+1, len(A)):
+                        reducerow(A, b, i, j)
+
+        # Remove all rows of zero and check if the system is solvable
+        I = [i for i in range(len(A)) if any(a != 0 for a in A[i])]
+        A = [A[i] for i in I]
+        b = [b[i] for i in I]
+
+        # Back substitution
+        for i in range(len(A)-1, -1, -1):
+                for j in range(i):
+                        reducerow(A, b, i, j)
+
+        return A, b, c
+
+# Find all combinations of parameters respecting the bounds in c. The
+# free parameters are always the last n-rank(A) columns of the matrix A
+# (where n is the total number of columns).
+def paramcomb(nparam, c):
+        if nparam == 0:
+                return [[]]
+
+        ret = []
+        for i in range(c[-nparam]+1):
+                ret += [[i, *l] for l in paramcomb(nparam-1, c)]
+        return ret
+
+# Solve the integer linear system Ax = b, minimizing the sum of the
+# coordinates of x.
+def solve_system_min_sum(A, b, c):
+        k = len(A[0]) - len(A)
+        mins = inf
+        for c in paramcomb(k, c):
+                sol = sum(c)
+                for i in range(len(A)):
+                        cc = (c[j]*A[i][len(A[0])-k+j] for j in range(len(c)))
+                        s = b[i]-sum(cc)
+                        a = s // A[i][i]
+                        # If a is negative or not integer, we skip this solution
+                        if a < 0 or s % A[i][i] != 0:
+                                sol = inf
+                                break
+                        sol += a
+                mins = min(mins, sol)
+        return mins
+
+with fileinput.input() as lines:
+        print(sum(solve_system_min_sum(*reduce(*readline(l))) for l in lines))