Saturday, January 17, 2015

Counting steno strokes

How many different strokes are possible on a steno keyboard, using the conventional fingering?

Background

The world's fastest typists can type more than 200 words per minute, and they all use steno keyboards. On these keyboards, it's usual to press more than one key at the same time. The act of pressing one or more keys simultaneously is called a stroke. Each stroke is automatically converted to text by a computer, using a digital dictionary. A particular stroke could represent a letter, a symbol, a word, a phrase, or even an instruction such as "capitalize the next word." To learn more about how steno works, see How court reporting is done from the U.S. Bureau of Labor, or watch Plover: Thought to Text at 240 WPM by Mirabai Knight.

Counting strokes

How many different strokes are possible? To answer the question, I'll assume that each finger operates independently, count the number of different places each finger can be during a stroke, and multiply those numbers together to get the answer. This way of counting includes an "empty stroke" in which every finger is off the keyboard, but this doesn't count as stroke, so we'll subtract 1 at the end to account for that.

Steno keyboards and fingers

A steno keyboard is arranged as in the diagram below. In any particular stroke, each finger can press zero or more keys. Except for the number bar, which we'll deal with below, the conventional fingering assigns each key to one finger. (Some unconventional fingering techniques like the Philadelphia Shift allow the same key to be pressed by different fingers, but I won't consider them here.) For some reason, steno keyboards have two keys (both labelled '*') in the 5th column, but in terms of the text that comes out on the screen, it never matters which one of those two is pressed. So in this note, I'll consider the 5th column to have only one key.

The diagram shows the 23 keys of a steno keyboard, with an oval for each finger. A solid red disc in the middle of a key represents the possibility that a finger is pressing only that key. A solid red disc between two keys indicates that a finger can press both keys at once. The right pinky can press 4 keys at once; this is indicated by a solid red disc in the middle of those 4 keys. There is also an empty circle for each finger to represent the possibility that the finger isn't pressing a key.

Counting the number bar and the left pinky

The long key at the top of the steno keyboard is called the number bar. For the purposes of counting steno strokes, it doesn't matter which finger presses it. To simplify the counting, and without sacrificing accuracy, we'll assume that if the number bar is pressed, it's pressed by the left pinky. How do we know that it's always possible for the left pinky to press the number bar? Not counting the number bar, the left pinky can be doing one of 2 things: either pressing the key in the first column, or doing nothing. No matter which of those it's doing, it could also be pressing the number key at the same time.

Counting the other fingers

Moving on, the left 3rd finger can be doing one of 4 things: (1) pressing the top key in the 2nd column, (2) pressing both keys in the 2nd column, (3) pressing the bottom key in the 2nd column, or (4) not pressing any key. Continuing on like this, we get to the right 1st finger, which can be doing any of 8 things, including pressing the key in the 5th column in various combinations with the top and bottom keys in the 6th column, or not pressing any key. Finally, we get to the right pinky, which can be in any of 10 possible places during a stroke, and the two thumbs, which can each be in any of 4 places.

And the answer is...

To count the number of possible strokes, multiply the number of possibilities for each individual finger together, and subtract 1 for the "empty stroke" when no keys are pressed: 4•4•4•4•8•4•4•10•4•4 - 1. That equals 5242879, or around 5.25 million.

Or is it?

A Google search for ["5242879" steno OR stenotype] turns up no results. Did I made a mistake somewhere? Let me know.

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