Picture in your mind some platonic ideal of a set of traffic lights in operation. Take a bird’s-eye view.<\/p>\n
Chances are you are conjuring a simple image of a simple cross-intersection with traffic travelling North and South for a bit, and then East and West for a bit, before repeating the two-state cycle.<\/p>\n
In reality, traffic lights tend to be more complicated. It is common for one direction, e.g. North, to be given a green right-arrows, meaning South must be stopped.<\/p>\n
So, commonly, rather than a length-2 cycle, it is actually a length-4 cycle.<\/p>\n
There are more elaborate systems, including, but not limited to; more than 4 roads intersecting; only 3 roads intersecting; pedestrian crossings; pedestrians being permitted to cross diagonally; and systems that react to cars pulling up in a lane or a lane getting busy.<\/p>\n
However, I am going to focus on the four-way intersection with the length-4 cycle, because they are quite common on busy city roads.<\/p>\n
Imagine a law-abiding pedestrian that is at the NW<\/abbr> corner, and needs to get to the SE corner.<\/p>\n The pedestrian has two choices: Head East then South, or head South then East. Which way is preferable?<\/p>\n The obvious answer is to take whichever green walk signal appears first.<\/p>\n In a length-2 cycle, that is the simple solution to the dilemma: both ways have the same expected time.<\/p>\n Unexpectedly, in a length-4 cycle, the answer is not that simple. It is sometimes desirable to ignore a green light. It is this mildly-surprising result that I analyze here.<\/p>\n When a traffic light is showing green for cars coming from the West, only the NW↔NE crosswalk is open to foot traffic. In particular, the SW↔SE crosswalk is closed, and cars turning right (from the West to the South) get a green arrow.<\/p>\n Under such restrictions, there are six possible length-4 cycles. I’ve listed them below in a canonical form – each row, represents one cycle:<\/p>\n The following assumptions are fairly arbitrary, but I think they illustrate the point.<\/p>\n For our pedestrian trying to travel from the NW to SE corner, Cycle #2 is the best. Follow the obvious rule and take whichever light is green first to get to the other side. I compute the average time to get across the intersection is 2.0 time units (with a range of 1.0 to 3.0 time units.)<\/p>\n Cycles #3, #4 and #5 are also best approached with the same, obvious, rule, although the lights are less favourable. <\/p>\n Cycle #3: Average time: 3.25 time units; Range: 2.0 – 5.0 time units. So far, no real surprises.<\/p>\n Cycles #1 and #6 represent clockwise and anti-clockwise patterns in the green lights.<\/p>\n They, similarly, exhibit a “handedness” in optimal behaviour. The pedestrian should always choose to travel in one direction around the intersection, even if it means ignoring a green light<\/strong> if it would take you in the wrong direction. The average time to cross the intersection is 3.0 time units, with a range from 1.0 to 5.0 time units.<\/p>\n Ignoring the rule, and instead adopting the simple “first green light” approach, is sub-optimal. It increases the average travel time by 0.75 time units to 3.75, and the worst-case travel time to 6.0 time units.<\/p>\n This preference for travel in one direction is very strong; it sometimes applies in the case where you simply want to cross one<\/em> road, instead of two. In the extreme case, where you are just miss the walk signal, it can drop your travel time from 4.0 time units to 3.0 time units, to cross three roads to get to your destination, instead of waiting for the single green.<\/p>\n For some traffic lights, it is possible to negotiate a diagonal street-crossing at a cross intersection faster by selectively ignoring green lights. <\/p>\n To make this decision, you need some local knowledge about the green-light cycles at that particular intersection. The difficulty is in giving pedestrians clues as to the optimal direction.<\/p>\n [Update: Corrected the time taken to do three-crossings instead of one.]<\/p>\n","protected":false},"excerpt":{"rendered":" I give a theoretical analysis of the time taken for a pedestrian to negotiate diagonally across an intersection, depending on the cycles of the traffic lights… although my name’s not Bamber.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_s2mail":"","footnotes":""},"categories":[33],"tags":[],"class_list":["post-140","post","type-post","status-publish","format-standard","hentry","category-puzzle-solving"],"_links":{"self":[{"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/posts\/140","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/comments?post=140"}],"version-history":[{"count":0,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/posts\/140\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/media?parent=140"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/categories?post=140"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/tags?post=140"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Valid Length-4 Cycles<\/h2>\n
\n
Evaluating Travel Costs<\/h2>\n
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Impact of the Cycle Pattern on the Decision<\/h2>\n
\nCycle #4: Average time: 4.0 time units; Range: 3.0 – 5.0 time units.
\nCycle #5: Average time: 3.75 time units; Range: 2.0 – 5.0 time units.<\/p>\nConclusion<\/h2>\n