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adventofcode2024
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day 13 actually numpy is fine

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magical 2024-12-13 05:49:42 +00:00
parent 1c7035d433
commit 7cc13becde
1 changed files with 13 additions and 46 deletions
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day13/sol.py
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@ -32,64 +32,31 @@ def solve(file):
# Solve the linear equation:
# k_a ax + k_b * bx = px
# k_a ay + k_b * by = py
M = numpy.array([[ax, bx], [ay, by]])
V = numpy.array([px, py])
sol = numpy.linalg.solve(M,V)
k1 = int(round(sol[0]))
k2 = int(round(sol[1]))
if k1*ax + k2*bx == px and k1*ay + k2*by == py:
print("yes", k1, k2)
#print("yes", k1, k2)
total += 3*k1 + k2
for record in records:
ax, ay, bx, by, px, py = record
# Part 2
#
# Can't use numpy because the values are too large for floats
# So here's an ad-hoc solver
qx = px + 10000000000000
qy = py + 10000000000000
px += 10000000000000
py += 10000000000000
V2 = numpy.array([qx, qy])
sol = numpy.linalg.solve(M,V2)
k1 = int(round(sol[0]))
k2 = int(round(sol[1]))
m00, m01 = ax, bx
m10, m11 = ay, by
v1 = px
v2 = py
# subtract c*eq1 from eq2 to cancel first coefficient
c = F(m10,m00)
m10 -= c*m00
m11 -= c*m01
v2 -= c*v1
if k1*ax + k2*bx == qx and k1*ay + k2*by == qy:
#print("yes2", k1, k2)
total2 += 3*k1 + k2
# scale eq2
c = F(1,m11)
v2 *= c
m11 *= c
# subtract c*eq2 from eq1 to cancel second coefficient
c = F(m01,m11)
m01 -= c*m11
v1 -= c*v2
# scale eq1
c = F(1,m00)
v1 *= c
m00 *= c
# we now have
# 1*k_a + 0 = v1
# 0 + 1*k_b = v2
print(m00, m01, m10, m11, v1, v2)
if v1.denominator == 1 and v2.denominator == 1 and v1 > 0 and v2 > 0:
if v1*ax + v2*bx == px and v1*ay + v2*by == py:
print("yes2", v1, v2)
total2 += 3*v1 + v2
print(total)
print(total2)