Twelve cables survive a slack one

A cable is half a brace found the rule for a square grid braced with wires or rods that pull and do not push. A cell’s cable forbids its cell’s shear in one sense and allows the other. So it is one direction of an edge between the cell’s column and its row, and a grid braced with cables is rigid exactly when the directed graph on its columns and rows can reach every node from every other. A three-by-three grid needs six cables where it needs five struts, and only twelve of the 5,376 ways of placing six do it.

It also found the flaw in every one of those bracings. A cable carries load only while it is taut, and a cable that stretches, creeps, loses its pretension or is simply cut is an arrow taken out of the graph. With one cable per cell, no bracing of a three-by-three grid survived that, not even with a cable in all nine cells. The degree count said why. Surviving the loss of any one arrow needs every node to keep an arrow in and an arrow out afterwards, so every node needs two of each before, twelve arrows at least, and nine cells with one cable each cannot hold twelve.

So the cells have to hold more than one. This essay lets a cell hold nothing, one cable rising, one falling, or both crossed, tries every filling, and asks three things: how few cables survive a slack one, where they go, and whether the count at each node that ruled out nine is also the rule that picks out the survivors.

The degree count, one level up #

The count that set the old minimum carries over directly. Each node of the directed graph, a column or a row of cells, must be reachable and must reach, so it needs an arrow in and an arrow out. To survive the loss of any single arrow, it must still have one of each after the loss, so it needs two in and two out before. On a three-by-three grid there are six nodes, so at least twelve arrows, and every arrow is one cable.

That is a necessary condition and nothing more. A graph can give every node two arrows each way and still fall into two halves joined by a single arrow in each direction, and then losing either of those two arrows disconnects it. The seventh contact found that a constraint added to an exactly constrained body could fail to touch and so constrain nothing. A slack cable is a contact that has stopped touching. The count says how many a grid must carry to afford one such failure. It does not say that carrying that many is enough.

The hero figure is one of the answers. Its nine cells, read row by row, hold rising, falling, crossed; falling, crossed, rising; crossed, rising, falling. Every row of cells and every column of cells holds exactly one of each. Its graph gives every column node and every row node two arrows out and two in. The slider takes each cable away in turn. Whichever goes, the eleven arrows left still reach every node from every other, and the grid stays rigid.

Every filling of nine cells #

Four states per cell and nine cells make 262,144 fillings, few enough to try every one. For each, the census builds the directed graph, asks whether it is strongly connected, and then removes each arrow in turn and asks again. It also records whether every node has at least two arrows in and two out, which is the count’s condition, so that the count can be compared with survival rather than assumed to predict it.

Below twelve cables nothing survives, as the count requires. At eleven cables 14,328 placements are rigid, more than at any other count, and not one of them can lose a cable. At twelve, 12,216 placements are rigid and thirty-six survive. Survivors then grow in number to 432 at fourteen cables and fall back as the grid fills. There are 336 survivors at fifteen, 117 at sixteen and eighteen at seventeen, and the single filling with every cell crossed survives at eighteen.

The column that matters most is the one beside the survivors. At every count from twelve to eighteen, the placements that meet the count and the placements that survive are the same placements. There are 36 of each at twelve, 216 at thirteen, 432 at fourteen, and so on to the one at eighteen. Across all 262,144 fillings of this grid, a bracing survives a slack cable exactly when every row and every column of cells carries at least two cables into it and two out of it. The count that looked like a necessary condition is, on this grid, the whole rule.

The census cannot say why from its own numbers, and a rule found by exhausting one grid might be a coincidence of that grid’s size. The reason is short, and it also says exactly where the rule must fail.

Why the count is enough, and where it cannot be #

Suppose a bracing meets the count and still fails when some cable goes slack. Then its graph, with that arrow gone, has a set of nodes that nothing outside can get into, or nothing inside can get out of. Call that set $S$. It must hold $a$ columns and $b$ rows. Before the loss, at most one arrow left $S$, the slack one.

Each of $S$’s columns sends out at least two arrows, and a column can send at most one arrow to any one row, since a cell holds one cable each way at most. So $S$’s columns send at most $ab$ arrows to $S$’s own rows, and at least $2a - ab$ of their arrows leave $S$. The same holds for its rows, which gives at least $2b - ab$ more. At most one arrow leaves, so

$$2(a + b - ab) \le 1, \quad\text{that is}\quad (a-1)(b-1) \ge \tfrac12.$$

Both $a$ and $b$ must be at least two. $S$ needs two columns and two rows of its own. The same argument run with arrows coming in, on the nodes outside $S$, needs two more columns and two more rows there. A bracing can meet the count and fail only on a grid with at least four columns and four rows.

That settles every grid with a side of three or fewer: on it the count is the whole rule, at every number of cables. The census was not a coincidence of three by three. It also predicts where to look for the first failure, a four-by-four grid split into two blocks of two by two, each block meeting the count on its own, with at most one arrow each way between them.

Thirty-six, in three families #

The thirty-six bracings of twelve cables are the fewest that survive a slack one on this grid, and they are not a list to memorise. At twelve cables the count is met with equality, every node exactly two in and two out, so every row of cells and every column of cells carries exactly four cables. A row has three cells and needs four cables, so at least one of its cells is crossed. How the rest are spent sorts them into three families.

Twelve cross exactly three cells and leave none empty. In every one of them each row and each column holds one rising cell, one falling cell and one crossed cell. That is a Latin square on three symbols, and there are exactly twelve Latin squares of order three, so every Latin square is a survivor and nothing else in this family is. The hero figure is one. Six cross six cells and leave three empty, one in each row and column. They are the struts’ six-cycle from which diagonal rigidifies a grid with each strut replaced by a crossed pair of cables, and there are six such cycles. The remaining eighteen cross four cells, leave one empty, and sit between the two.

That second family says something about the first answer. A crossed pair of cables in one cell forbids its shear both ways, which is exactly what a strut does — the second diagonal that the cell that repeats for ever counted as pure redundancy in a strut grid is, for cables, the other half of the brace — and six struts on a cycle survive the loss of any one strut. So the struts’ version of this question has the answer six struts, arranged round a cycle. Six struts drawn as crossed cables is twelve cables, the same twelve the count asked for. Nothing is gained by crossing more cells than that, and the Latin squares show that nothing is lost by crossing fewer. A bracing with three crossed cells and six single cables is exactly as tolerant as one with six crossed cells, and uses the same material.

Two routes on the same thirty-six #

Everything above is a statement about a graph, and the graph is the model of a cable that a cable is half a brace derived rather than the grid itself. The check that does not trust the model is the one that essay used, Roth and Whiteley’s. A framework with cables is rigid to first order when it is rigid with every cable read as a bar and a self-stress pulls on every cable. That route builds the grid from coordinates, takes the rank of its rigidity matrix, and solves a small linear program for the stress. It shares nothing with the directed graph except the list of cables.

It was run on all thirty-six survivors with each of their twelve cables removed in turn, 432 frameworks. Every one is rigid with a stress pulling on the eleven cables left. It was run again on the four-by-four bracings drawn further down that the graph says fail, and there it names the same fatal cables the graph names, by index.

A spare costs struts one more and cables double #

The count generalises at once, and the census can be run on other grids to check it. Every arrow joins a column to a row. The side of the grid with more nodes needs two arrows in at every node and two out, and each arrow supplies one end at that side in one direction. So a grid needs at least twice its longer side in arrows each way, $4\max(m, n)$ cables in all. That is exactly twice the $2\max(m, n)$ a bare cable bracing needs.

The table bears the law out on seven sizes, and in each case one fewer cable leaves no survivor at all. It also puts the struts beside the cables, and the comparison is the design point. A strut bracing with a spare needs a graph that stays connected after losing any edge, which on these grids takes $2\max(m, n)$ struts, one more than the bare $m + n - 1$ on a square grid. With cables, the spare doubles the whole bracing. The fewest struts that survive a loss are, on every size here, the fewest cables that are rigid at all. Redundancy costs a strut bracing one member and a cable bracing half of everything.

The narrow grids are the extreme. A grid two cells wide has only two cells in each row and needs four cables in each, so every cell must be crossed, and there is exactly one survivor at each length, the fully crossed grid. The same arithmetic, applied to each line of the longer side, says how many cells must be crossed in general. A line of $k$ cells that must carry four cables needs $4 - k$ of them crossed, so a grid needs its long side times four less its short side, and none at all once the short side reaches four. The census matches that on all seven sizes: three crossed cells at three by three, four at three by four, five at three by five, all of them on the two-wide grids, and none required at four by four.

Four by four, where the count stops being enough #

The argument above said the count could fail only on a grid with four columns and four rows, and four by four is the first. The census at sixteen cables there does not enumerate all 4¹⁶ fillings. It walks only the placements that meet the count, pruning any cell choice that would leave a node unable to reach two arrows each way, which finds all 8,100 of them.

Of the 8,100, 7,506 survive and 594 do not, and the failures take exactly the shape the argument predicted. Eighteen are two blocks of two columns and two rows, each braced on its own, with no arrow between them at all. They meet the count everywhere and are not even rigid. The other 576 are the same two blocks joined by one arrow each way, through a single rising and a single falling cable. They are rigid, and each of those two cables is fatal. The left and centre panels of the figure are one of each. The count cannot see either, because every node in both is locally perfect. What is wrong with them is global: the bracing has a waist.

The right-hand panel is the other thing a fourth column buys. Its sixteen cables sit one to a cell, none crossed, and it survives any one going slack. A column of four cells can carry four cables without doubling any, so the crossed cells that every narrower grid needed are no longer needed at all. Ninety of the 7,506 survivors are like it. The price is that the rule for placing them is no longer local. The designer of a four-by-four bracing has to check the count at every row and column and then check that no split into two blocks crosses only one arrow each way.

What the grid falls back on #

A cable bracing must be pretensioned, because the self-stress that makes it rigid is what keeps every cable taut under load. A bracing that survives a slack cable therefore has two self-stresses to think about, the one it is built with and the one it is left with.

With all twelve cables pulling, the linear program finds a stress with every cable at the same tension. With the first cable slack, it finds one with eight of the eleven at that tension and three at twice it. The second is also a self-stress of the complete grid, the one that happens to be nought in the slack cable — a constraint that has been said already read a self-stress as a combination of constraints that sums to nothing, and this one simply leaves one constraint out of the sum — since a stress that balances every joint without a member still balances with that member present and carrying nothing. That is the physical content of surviving a slack cable. The grid is not rescued by a new mechanism appearing when a cable goes. It already contains, among its self-stresses, one that does not use that cable, and the loss simply moves the pretension onto it.

The price is one self-stress found a single self-stress deciding what a framework’s play must absorb. A cable bracing with a spare has a family of them. Survival means that for every cable there is a member of the family that leaves it out, and the designer’s real choice is which of those the pretension should sit near.

The rule a builder can carry #

The rule for a cable bracing that tolerates one slack cable, on a grid of cells with a side of three or fewer, is a count that can be checked with a pencil.

Give every row of cells and every column of cells at least two cables rising and two falling. On such a grid that is necessary and sufficient, and the fewest cables that meet it are four times the longer side. The cheapest way to meet it on a square three-by-three grid is a Latin square, one rising, one falling and one crossed cell in every row and column. It uses twelve cables, three cells crossed.

On a grid four cells or more in both directions, the count is still necessary and no longer sufficient. The extra check is that no split of the grid’s columns and rows into two blocks is crossed by only one cable each way. That is the same test the loops are in the graph made on an undirected drawing, asked this time of a directed one with a cut in each direction.

What is not measured #

One slack cable, not two. The same counting gives three arrows each way for two losses, eighteen cables on a three-by-three grid. Only the fully crossed grid has eighteen, and it survives any two slack cables. What the rule is on larger grids, where three each way can be met in many ways, is not computed.

Cables that go slack, not cables that stretch. A slack cable here carries nothing and the rest are taut. A real cable that creeps loses tension gradually, and the grid passes through a range where the stress is shifting from one member of the family to another. The analysis is first-order throughout. A first-order rigid framework is rigid, so the verdicts stand, but the first-order picture is not the whole of a framework’s behaviour and it says nothing about stiffness. How large a load a pretensioned grid carries before the next cable goes slack is not measured.

Square cells and one cable per diagonal. As for every grid in this field, the rule depends on each cell being a parallelogram, so that a cable compares a column’s lean with a row’s.

Still open: how hard to pull the spare #

Every survivor has a family of self-stresses, and the census answered only whether a member that leaves out each cable exists. The pretension a builder puts in is one member of the family, and when a cable goes slack under load the grid does not jump to the linear program’s answer. It carries whatever share of the load the remaining members give it.

The distinct argument there would be the pretension that makes a spare worth having. It would choose the self-stress of a Latin-square bracing that maximises the smallest load, in any direction of shear, that the grid can take before any cable goes slack, and then do the same after each single cable is lost. The measurement would compare the best pretension of the Latin square with that of the six-cycle of crossed pairs, which uses the same twelve cables. It would say whether two bracings that are equal under the count are equal under load, or whether one of the three families is the one to build.