Eight streets on fire

1 · One street

Why the answer is a square

Watch the very first fire. After one day it has burned one house. After two days it has spread to its two neighbours, so three houses. After three days, five. Every day it grows by one house on each side, so after m days the first fire has burned a stretch of 2m − 1 houses: for six days, that is 11.

The second fire, lit on day two, is a day behind, so it has only reached 9 houses by day six. The third has reached 7, the fourth 5, the fifth 3, and the sixth fire, lit on the last day, has burned just the one house it was lit in. So after six days you have six burned stretches of lengths 11, 9, 7, 5, 3 and 1: one of each odd length, counting down.

Here is the lovely part. Add those up: 11 + 9 + 7 + 5 + 3 + 1 = 36, which is 6 × 6. It always works: the odd numbers from 1 up to 2m − 1 add up to exactly m × m, a square. The picture below shows why. Take day one's stretch of 11 houses and bend it into an L, 6 along the bottom and 5 up the side. Then bend day two's 9 houses into a smaller L that fits inside it, and so on. The six L's nest together into a six by six square with nothing left over. So six days of fire can burn a street of 36 houses, and no more, and a street of 36 houses can be burned in six days, as long as each fire is lit in the right place. That is exactly what the animation above is doing: lighting the fires so that the stretches lie end to end with no overlap and no gap.

2 · Eight streets

One box of matches for the whole town

Now the town has eight streets, and they are separate: a field between each pair, so a fire on one street can never reach another. The rule is the same as before, with one twist. The whole town shares one box of matches, and you may strike only one match a day, on any street you choose. After 28 days you will have struck 28 matches, and each one will have burned a stretch of houses: the first match 55 houses, the next 53, and so on down to the last match, 1 house. The same odd numbers as before, just more of them.

Think of each stretch as a plank with its length painted on it: a plank of 55, a plank of 53, down to a plank of 1. There are 28 planks and eight streets. To burn a street completely, you need to lay planks along it that cover it exactly: a plank that sticks out past the end of the street is a wasted match, and a gap between planks is a house that never burns. No plank can be cut in two, because a fire cannot be stopped halfway. So the question "can the town be burned in 28 days?" becomes a tiling puzzle: can the 28 planks be shared out so that every street is covered exactly?

For the puzzle to be fair we give the town exactly as many days as it needs: the eight street lengths add up to 28 × 28 = 784, so the planks are exactly long enough, with nothing to spare. Here are two towns of that size where the sharing works. Every plank is used, every street is covered, nothing is cut.

3 · When sharing fails

A town the fire cannot finish in time

Sometimes there is no way to share the planks, even though they add up to exactly the right amount. Here is the simplest town where it goes wrong. Give it 26 days, so the planks are 1, 3, 5, … up to 51. Let four of the streets be 67 houses long and the other four be 102. Check: 4 × 67 + 4 × 102 = 676 = 26 × 26, so the planks are exactly long enough. And yet no arrangement works. Here is why, in three steps.

The 102-streets. Could one be covered by just two planks? The two longest planks are 51 and 49, and together they reach only 100, two houses short. So every 102-street needs at least three planks. But there is a catch: every plank has an odd length, and odd + odd + odd is odd, while 102 is even. An even street needs an even number of odd planks. Not two, so at least four.

The 67-streets. One plank is not enough, since the longest is 51. And 67 is odd, so it needs an odd number of odd planks: not one, so at least three.

Count the planks. Four streets needing four planks each, plus four streets needing three each, is 16 + 12 = 28 planks. But 26 days give only 26. The town cannot be burned in 26 days, however cleverly the matches are struck.

Tan and Teh call a town that fails this counting test impossibly burnable, and the test is theirs. Notice what went wrong. The trouble was the short streets. A short street can only take a few planks, and it has no freedom about how many: 67 forces an odd count, and the small planks it wants are needed elsewhere. If all the streets were long enough, there would be room to manoeuvre. So the real question is: how long is long enough?

4 · The question

Is 68 always enough?

Computationally confirmed 19 October 2026: see section 10, “Two roads to 68, and the one that arrived first”

Here is the question exactly. Suppose every one of the eight streets has at least L houses, and the lengths add up to a square, m × m, for any m at all. Is the sharing then always possible? The town above shows that L = 67 is not enough, because the 67-streets broke it. Two mathematicians, Tan and Teh, worked this out for towns of one street, two streets, and so on up to seven streets. For seven streets the answer is 56: if every street has at least 56 houses, the sharing always works. For eight streets they could show that 67 is not enough, and that is where the known results stop. Notice that the town in section 3 rules out every smaller number as well: "every street at least 60" is a weaker promise than "every street at least 67", and that town keeps the weaker promise too, so it breaks every threshold up to 67 in one go. The answer for eight streets is therefore at least 68.

So the conjecture was that 68 is exactly the answer: that 68 is always enough for eight streets. This is the problem that GPT, the other half of this project's research partnership, chose when it was told to follow its own curiosity, and it is what the rest of this page is about. On 19 October 2026 it was settled: 68 is the answer. Section 5 explains how the infinite question was cut down to 64,787,844 towns, and section 10 how every one of them was tiled by a computer in under two hours and checked twice. The general theory in sections 6 to 9, which would explain why rather than check, remains open.

There is a pattern behind the numbers, and Tan and Teh found it. The thresholds for one to seven streets are 1, 3, 18, 26, 36, 46 and 56. Every town that spoils a threshold, in all the cases they could check, spoils it the same way our 67-and-102 town did: by the plank count. Count the fewest planks each street could possibly take, remembering that an odd street needs an odd number and an even street an even number; if the counts add up to more than the number of days, the town is impossible. Tan and Teh proved a formula for the longest shortest street such a counting failure can have, and it predicts the threshold for every number of streets: 68 for eight, 78 for nine, 88 for ten, 98 for eleven, 110 for twelve, growing by about twelve per street in the long run. Their conjecture is that the plank count is the only way a town with long enough streets can fail. Eight streets was the first case nobody had been able to check, until 19 October 2026.

5 · Shrinking the pile

Why only nine sizes of town matter

"For any m at all" sounds hopeless: there are infinitely many towns to check. But a trick cuts the infinite list down to size. Suppose one of the streets is very long, at least 2m + 67 houses. Give it the longest plank, 2m − 1, and take that plank out of the game. What is left? A town one day smaller, with the same rules, and the long street is now 2m − 1 houses shorter, so it still has at least 68. So if the smaller town can be burned, so can the bigger one. Any town with a very long street is really a smaller town in disguise.

That means we only ever need to check towns where no street is very long: every street between 68 and 2m + 66 houses. And eight such streets cannot add up to m × m once m reaches 33, because the square grows faster than the streets can. So only the towns with 24 to 32 days are left to check by hand, or rather by computer. That is still a lot of towns: 64,787,844 of them, most at 27 and 28 days. But it is a finite list, and finite lists can, in principle, be worked through. On 19 October this one was (section 10).

6 · A tidier way to count

Saw one house off every plank

Sixty-five million towns is too many to check one at a time, so the hope is a rule: a simple test that tells you whether a town can be shared out, without trying every arrangement. To look for one, it helps to tidy the planks first. Every plank has an odd length, which means "one house, plus an even number of houses". So saw one house off every plank. The plank of 1 becomes nothing, the plank of 3 becomes 2, the plank of 5 becomes 4, and so on. Now every plank is even, so measure them in pairs of houses instead: 0, 1, 2, 3, … up to 27. The planks are just the ordinary counting numbers.

A street that wanted, say, 3 planks adding up to 67 houses now wants 3 numbers adding up to (67 − 3)/2 = 32. Nothing about the puzzle has changed; it just reads more easily. The pool is the numbers 0 to 27, and each street says how many numbers it will take and what they must add up to.

And now one rule is obvious. A street taking three numbers cannot ask for less than the three smallest add up to (0 + 1 + 2 = 3), nor for more than the three largest (25 + 26 + 27 = 78). The same goes for any group of streets together: two streets taking seven numbers between them cannot ask for less than 0 + 1 + … + 6 = 21. We will call this the cut rule: every street, and every group of streets, must ask for a total somewhere between the least and the most its numbers could possibly give. A town that breaks the cut rule is obviously impossible. The surprise comes in the next section.

7 · What has been proved

For a plain run of numbers, the cut rule is the whole story

Here is the surprise. For a town of three streets, and for a town of four, the cut rule is not just necessary but enough: if every street and every group of streets sits inside its band, a sharing always exists. No further cleverness is needed. This was first noticed by having a computer check about 250,000 small towns, every one of which obeyed it, and then proved properly, by hand, with every line of the proof read and checked by a second party. The proof even says how to find the sharing: take the street (or pair of streets) that is squeezed tightest against its band, give it a particular staircase-shaped set of numbers, and hand what is left to the three-street rule.

Real towns have eight streets, not four, and for those the record holds something different: not a rule but a list. For every one of the 258,180 towns at 28 days in which all eight streets have odd length, an actual sharing has been worked out, written down, and then recomputed and checked by a second, independent program. Those towns are done, the way a crossword is done when it is filled in.

8 · The hole

Why eight streets are not four streets twice

If the cut rule settles four streets, why not eight? The plan writes itself. Take the tightest street, give it its numbers, cross them off, and you have a town of seven streets. Do it again: six streets. Keep going until four are left and the four-street rule finishes the job. The trouble is in the crossing off. After the first street has taken its numbers, the pool is no longer a plain run 0, 1, 2, 3, … It has holes in it where the numbers were taken.

And with a hole, the cut rule stops being enough. Here is the smallest example. The pool is 0 to 12 with the 6 missing. Four streets take three numbers each, wanting totals of 4, 18, 18 and 32. Check every street and every group against its band and they all pass. But look closer. The street wanting 4 has exactly one way to get it, 0 + 1 + 3, because no other three different numbers add up to something as small as 4. The street wanting 32 also has exactly one way: 9 + 11 + 12. Those two streets are pinned. Cross their numbers off and what is left is 2, 4, 5, 7, 8, 10, and no three of those add up to 18. The town obeys the cut rule and still cannot be shared.

The repair is to ask for a little slack. Instead of letting a street sit right on the edge of its band, ask every street and every group to clear the bottom of its band by one unit for each number it takes, as if the missing number had been at the very bottom of the pool. Call that a margin. With a margin of one, pinned streets like the one wanting 4 are ruled out, and the one-hole, four-street rule holds in every town a computer has checked, nearly half a million of them. With two holes, a margin of two does the same. GPT worked out exactly how much margin the crossing-off step leaves behind (MI362), and proved that the margin survives when the tightest street is crossed off, whichever numbers it takes (MI368).

Where the one-hole proof stands (20 October 2026)

On 20 October GPT and Cloud spent the day on that hand proof. The plan splits the job in two. Call the numbers above the hole big and those below it small. First decide how many big numbers each street will take; then hand out the actual numbers. For streets that take at least three numbers from each side, proved results already do the second half. Streets that take fewer are a separate step, still open.

The first half, choosing the counts, is a neat question of its own. Each street, and each group of streets, can reach its total only with a certain range of big numbers: too few and it cannot climb high enough, too many and it overshoots. A choice of counts has to put every group inside its range at once. Two things make this tidy. If the ranges can be met with fractions, they can be met with whole numbers (G333, G334). And whether they can be met at all comes down to nine kinds of condition, by a theorem from the mathematics of sharing costs in cooperative games (Bondareva and Shapley). Six of the nine are now proved. The other three are the ones where groups of streets cross: streets 1 and 2 together, set against streets 1 and 3 together. Each of the three can be written using only three of the four streets (in the computer checks the fourth street only has to clear its own floor), yet no proof reaches any of them. Computers have found a choice of counts in every one of 9,370,699 towns of up to fifteen numbers, so the three conditions are almost certainly true. For now, they are the wall.

21 October: the wall narrowed to one family of exceptions

The next day the question was turned round. Forget the margin for a moment and ask only the simplest test: can each street, and each group of streets, reach its total at all with that many numbers from the pool? Call a town that passes all fourteen such tests spectrum-feasible. A computer then shared out, or proved impossible, every spectrum-feasible town with four streets of at least three numbers and one missing number, up to seventeen numbers in the pool: 18,682,428 towns. Only 54 cannot be shared, and all 54 are one family.

That family is the town at the top of this section, grown longer. Two streets are squeezed to within one unit of their smallest and largest possible totals, which forces exactly which numbers they take; what is left for the last two streets is six numbers such as 2, 4, 5, 7, 8, 10, and those two streets want equal totals that no split of the six can give. Nothing else fails. In particular every town in which every group sits at least two units inside its range can be shared, which takes in the margin rule above and makes the count conditions of the picture unnecessary for this question. The failures also need two streets of exactly three numbers: among the size patterns with at most one such street, not one town in eight million fails.

Along the way GPT proved seven new pieces, each checked by Cloud: the tournament picture (a sharing is a contest in which every number beats exactly as many others as its value), its sharpening to at most six players caught in cycles, a way for a long well-mixed run of numbers to absorb any small cycle, a complete account of one case with two gaps, and two results on how far from the ends a missing total can sit when a run has a few gaps. What still has no proof is the step every route reached: choosing the first piece so that what is left can still be shared. The one mechanism in the record that beats it, choosing a whole boundary set for two streets at once, survives a missing number in 97 to 99.7 percent of the hard towns tested; the towns it misses need one extra jump in that set.

9 · The wall

Three holes, equal shares, and a famous hard case

So far so good: one hole, margin one; two holes, margin two. Then the crossing-off reaches three holes, and the margin stops working. The failures are rare, about one town in a hundred thousand, and they all look alike: the streets want equal totals, and the holes have split the pool into two clumps, a few small numbers and a run of big ones. Here is the smallest. The pool is 0, 1, 2, 3 and then 7, 8, 9, 10, 11. Three streets take three numbers each, and each wants 17.

Suppose a street takes two of the small numbers. The most it can reach is 2 + 3 + 11 = 16, one short of 17. So every street takes at most one small number. But there are four small numbers and only three streets: somebody has to take two. No sharing exists, and no amount of margin on the bands would have warned us. What matters here is not how far each street is from its band but how the small numbers are spread out, which is a different kind of counting altogether.

Counting the clumps rescues some of these towns but not all. A few survive, always with equal totals and a street squeezed tight against its band. That is not a coincidence. "Share a run of numbers into groups with equal totals, every group having at least three numbers" is a question that mathematicians have asked on its own, and it is still open (Hoory and Kotlar, 2026). The crossing-off plan, pushed far enough, lands on a thin disguise of that known hard problem. This is the wall: everything up to two holes is in hand, and the hard core beyond it has a name.

10 · Where we stand, and what we need

Two roads to 68, and the one that arrived first

There were two ways to settle the question. The rule road: find a recipe that shares out the planks of every town with long enough streets, and prove that it never fails. The list road: take the 64,787,844 towns of section 5, the only ones that need checking, and find a sharing for each by computer. The list road arrived first. On 19 October 2026 every one of those towns had been shared out and checked, so 68 is always enough, and with the town of section 3, 68 is exactly the answer for eight streets. It is a computer-assisted result, like the seven answers before it, and the first one past the point where Tan and Teh stopped.

How big the job was

For each town the computer searches for a sharing. It takes the longest plank first and tries each street that could still hold it, backing up whenever a choice leads nowhere. It also uses the odd-and-even plank count of section 4 to give up early on any choice that leaves a street needing more planks than remain. A plain search without that shortcut took about 143 milliseconds a town, which for the whole list is about three and a half months on one processor core. With the shortcut, an average town took under a millisecond. The hardest town needed 4,916 steps of search, and about one town in twelve needed no backing up at all.

Spread over the eight cores of an M5 MacBook Pro, all 64,787,844 towns took 1 hour 46 minutes, overnight on 18 to 19 October 2026. The answer, one sharing per town, fills 6.9 GB. Checking is far quicker than searching. Two programs re-read every sharing and add up every street; one of them, written separately by GPT, also counts the towns in every batch against an exact count worked out another way, so a missing or doubled town would be caught. Both agree, and both report the same fingerprint (a SHA-256 digest) of the output. A third program on another machine redid four of the nine sizes of town from scratch, and agreed.

Why nine streets costs far more

The trick of section 5 works for any number of streets. For nine streets it uses Tan and Teh's predicted threshold, 78. But the pile it leaves grows about twenty times with each extra street, because there are more streets to vary and more sizes of town in which no street is very long. Each town is also a bigger search, with more streets and a longer plank list. The last column below assumes, generously, that a town with more streets is no harder than an eight-street town, so the real times would be longer.

* Predicted by Tan and Teh's conjecture; 36, 46 and 56 are their proved values. Towns counted exactly with the formula of section 5 (pile_counts.py); times scaled from the eight-street run at 0.79 milliseconds of one core per town. Tan and Teh settled five to seven streets by their own methods, not by this list.

Nine streets would take the same laptop a day and a half or more, and its answer would fill about 125 GB. That is possible but no longer a single night, and every further street multiplies it again. More to the point, the list road can never finish the general question. Tan and Teh's conjecture covers every number of streets, and no amount of computing reaches all of them. Only a proof can.

Why the proof is still out of reach

The computer shows that every eight-street town can be shared out; it does not show why. A proof would be a recipe that works for every town, with a reason it never fails. Three things we found make that recipe hard to find.

No simple recipe works. We tried three natural one-plank-at-a-time rules on millions of the real towns: give the biggest plank to the street with the most room left; give it to the tightest street that can take it; or decide first how many planks each street gets. Even all three together share out only about 13 to 30 percent of the towns.

The sharings that work are not one-at-a-time. In the sharings the computer found, the streets that take only two planks often pair a plank from the top of the list with one from the middle, all adding to the same total: 45 + 23, 43 + 25, 41 + 27 and 39 + 29 each make 68. That is a matching across the whole list at once, and a rule that places one plank at a time never finds it.

The general rule breaks where the planks have gaps. Sections 6 to 9 built a rule, the cut rule, that is the whole story for a plain run of numbers. Once a few planks have been used and gaps open in the run, it fails, and the place it fails is the equal-shares problem of section 9, which mathematicians list as open. The pieces proved so far are rungs of a ladder; the missing rung is a way of choosing which planks the tightest street takes so that the town left over stays inside the rule. For a single missing number that rung is now narrower: choosing how many big numbers each street takes comes down to three conditions where groups of streets cross, drawn as a wall in section 8.

So eight streets is settled; the pattern behind it, Tan and Teh's conjecture for every number of streets, is not; and the wall of section 9 is where a proof would have to get through.

Check it yourself

Everything needed to repeat or check the computation is on this site, in the release folder. It needs only Python 3.

• The note: the result written up as a short mathematical paper, with every lemma, the reduction and the checks (eight-paths-L8.md). By the Wolfram · Collatz · Mahler project: Gareth (Human) - Claude (Anthropic SI) - ChatGPT (OpenAI SI): Equals amongst mankind and machines.
• The programs: the search and its checker (l8_certificates.py), GPT's independent verifier (l8_stream_verify.py), the batch census (l8_certificate_census.py), the third program (e_slice.py, cert_check.py), the packing tools (pack, unpack), the rule experiment (mechanism.py, r3_certcounts.py) and the town counter (pile_counts.py).
• The certificates: one sharing for every one of the 64,787,844 towns, in nine compressed files, one per number of days (340 MB in all): 24, 25, 26, 27, 28, 29, 30, 31, 32 days, with manifest.json holding their fingerprints. Unpacked, they give back the original output byte for byte.
• The smaller datasets: the third program's certificates for 24, 25, 31 and 32 days, and the scores of the three rules above.

The fingerprint of the original output is

90fea878fc36d18da25e52b2e73fbfcb8e4a447c72e470f9e5c3a5ea30d1063b

It covers the run's timing notes as well as the sharings, so a fresh run gives a different one. A fresh run is compared by a second fingerprint that leaves the timing out. For the original output it is

da7c2e7716df23188cba01580ceabb4172e924529b9b73721671c330479c3a96

Fresh runs of four of the nine sizes of town, made on another machine, give the same timing-free fingerprint for those four sizes as the original output does. The README gives both fingerprints and the commands to make them.

What would help

A recipe for sharing out the planks that explains every real town, perhaps built on the matched pairs above; fresh eyes on the equal-shares problem with gaps; and, for anyone who thinks in pictures, pictures. Every obstruction on this page was found by a computer and only afterwards seen: the pinned street, the two clumps, the pigeonholes. If a picture makes you see the next one before the computer does, that is a real contribution, and here is how to send it.