Cycles and Repeating Schedules

Everything in a repeating schedule follows from the cycle length, so find that first. Here a tour visits seven cities and then returns to the first, so the cycle length is seven. Drawing it as a ring rather than a straight line is worth the few seconds it takes, because a ring makes the wrap-around obvious — after the seventh stop you are back at the start, and there is nowhere else to go.

A ten-day tour running a seven-day cycle repeats: days eight, nine and ten are days one, two and three again. To find the city on day nine, subtract the cycle length — nine take seven is two — so day nine matches day two. Keep subtracting seven until you land on a day between one and seven. And notice the consequence: a ten-day tour visits three of its cities twice and the other four once. Any schedule longer than its own cycle must repeat something, and spotting that immediately answers a whole class of question.

For a day a long way out, subtracting the cycle length over and over is slow and easy to miscount, so divide instead. The one subtlety is that days are counted from one rather than nought, so take one off first: forty-five take one is forty-four, and forty-four divided by seven is six remainder two. A remainder of two means two steps past day one, which is day three. You can sanity-check it the short way — six whole cycles is forty-two days, and forty-five take forty-two is three. Same answer, far fewer steps.

When several schedules share a cycle they differ only in where they join it. Tour two is tour one shifted along by one step, tour three by two, and so on. That means you never need to memorise the whole table: learn the ring once and note where each tour joins it. A table like this looks like a great deal of information and is really one cycle plus six starting offsets, which is why reading the first row carefully is worth more than skimming all six.

A harder version asks about one day across all the tours at once. Day ten is day three of the cycle, so on day ten each tour is at its own third city. Six tours give six different third cities, and since there are seven cities in the ring, exactly one city is never a third city — that is the one left out. The way in is always the same: reduce the day to its place in the cycle first, then ask the question about that position rather than about day ten. Cycle length, wrap around, remainders, offsets. Quiz and exercises now.