I asked LLM and what it did was try DP solution, couldnt find DP solution, then launched brute-force solution. Everything in Python. so what LLM really did, was execute the SoTA human knowledge including solving using code.
here are LLM "thoughts" for example in the article:
________
I need to distribute 11 tracks across four programs on an 8-track cartridge. Every program must play for the exact same length of tape. Shorter programs waste the remaining silence. The goal is to minimize the longest program length.
________
The search found a best maximum program length of 613 seconds.
The dynamic programming solver claimed feasibility for 601, yet the search returned zero solutions.
A discrepancy appeared between the binary search result and the partition check.
The binary search lower bound was set to the longest track instead of the average length. This caused the search to miss the true minimum of 613.
Reconstruction failed — running exhaustive search for the true optimum.
________
The optimal assignment has been confirmed.
Program 3 is the longest at 613 seconds, which sets the tape length. Program 4 is the shortest at 573 seconds, leaving 40 seconds of wasted silence. The remaining three programs are 607, 610, and 613 seconds. A perfect 601-second maximum is impossible because the tracks cannot be split.
this was code used by LLM:
python3 << 'EOF'
tracks = {1:174,2:238,3:195,4:194,5:220,6:203,7:258,8:135,9:213,10:271,11:302}
def fmt(s): return f"{s//60}:{s%60:02d}"
items = list(tracks.items())
all_best = []
def search(i, bins, progs, best_max):
if i == len(items):
if max(bins) == best_max:
all_best.append([sorted(p) for p in progs])
return
if max(bins) > best_max:
return
tid, dur = items[i]
tried = set()
for pi in range(4):
if bins[pi] in tried: continue
tried.add(bins[pi])
bins[pi] += dur
progs[pi].append(tid)
search(i+1, bins, progs, best_max)
progs[pi].pop(); bins[pi] -= dur
search(0, [0,0,0,0], [[],[],[],[]], 613)
# dedupe
seen = set()
unique = []
for sol in all_best:
key = tuple(sorted(tuple(p) for p in sol))
if key not in seen:
seen.add(key)
unique.append(sol)
print(f"All {len(unique)} distinct optimal solutions at 10:13:")
for sol in unique[:5]:
sums = [sum(tracks[t] for t in p) for p in sol]
print(f" {sol} -> {[fmt(s) for s in sums]}")
EOF