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Reduce performance impact on cost function

master
Forest Belton 2 years ago
parent
commit
22f5e10812
2 changed files with 10 additions and 8 deletions
  1. +2
    -2
      ex.py
  2. +8
    -6
      gbso/optimize.py

+ 2
- 2
ex.py View File

@ -29,7 +29,7 @@ initial_cycles = prgm.perf()
print("Program to optimize:")
prgm.display()
print(f"Cost: {initial_cost}")
print(f"Cost: {initial_cost[0]}")
print(f"Cycles: {initial_cycles}")
start_time = time()
@ -51,7 +51,7 @@ optimized_prgm.display()
optimized_cost = cost(prgm, test_cases, outputs, optimized_prgm)
optimized_cycles = optimized_prgm.perf()
print(f"Cost: {optimized_cost}")
print(f"Cost: {optimized_cost[0]}")
print(f"Cycles: {optimized_cycles}")
print(f"Took {round(end_time - start_time, 3)} seconds")

+ 8
- 6
gbso/optimize.py View File

@ -19,15 +19,17 @@ DEFAULT_PROB_INSN_UNUSED = 0.1
def cost(orig_prgm, test_cases, outputs, prgm) -> Tuple[int, bool]:
c = prgm.perf() - orig_prgm.perf()
eq = c == 0
# print(f"init cost: {c}")
# Since each instruction executes in 4*k cycles (for some k), this can have
# the undesirable effect of performance improvements being weighted much
# higher than correctness. This hurts convergence pretty badly, so we scale
# by 1/4 to compensate.
perf = (prgm.perf() - orig_prgm.perf()) / 4.0
eq = 0
for test_case in test_cases:
c += eq_on_testcase(orig_prgm, prgm, test_case, outputs)
# print(f"cost after testcase: {c}")
eq += eq_on_testcase(orig_prgm, prgm, test_case, outputs)
return c, eq
return perf + eq, eq == 0
def optimize(

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