# Finding the optimal (non-)T9 cellphone keyboard layout

This idea was born before we all had smartphones and swipey keyboards and the like: how can we arrange the 26 letters of the alphabet on a cellphone keyboard to make T9-like predictive typing as efficient as possible? In other words, which assignment of 26 letters to eight keys (0 and 1 are non-letter keys) minimizes the number of textonyms – such as, with the standard layout, home, gone, and good?

(Sidenode: I solved this several years ago with IBM CPLEX, which took all but a few seconds to find the optimal key assignment. Now, with a slower machine, access to only open source tools, and an interest in the Minizinc ecosystem, I’m revisiting the problem.)

The answer to the problem will, of course, depend on the word list from which potential textonyms are drawn. I’ve made my list of 1000 common English words, along with their frequency, available here. Unfortunately, I don’t remember where I found it or what corpus I compiled it from.

## Constraints

First, we’ll assemble the constraints we can feed to our solver engine.

Each of the 26 letters will be assigned a key index between 0 and 7 – that’s the output we’re looking for. So we need a 1×26 matrix of integer variables over `{0..7}`

, named `letter_keys`

.

In the Minizinc language, we can express this as:

array[1..26] of var 0..7: letter_keys;

We channel this into a 26×26 matrix of booleans, `letter_keys_match`

, which indicate whether or not any two letters occupy the same key. For instance, if a and b both sit on key number 3, then `letter_keys_match[1, 2] == true`

.

Wait – there’s a symmetry here: `letter_keys_match[i, j] == letter_keys_match[j, i]`

! Indeed, we can make do with just one triangle of the matrix (let’s use the upper right). Instead of 26×26 = 676 variables, that’s just (26×26 – 26)/2 = 325. It’s a bit annoying having to convert between the two running letter indices of the loop and the one-dimensional index of `letter_keys_match`

, but once we work out the arithmetic it’s simple enough:

array[1..325] of var bool: letter_keys_match; constraint forall (letter1 in 1..25, letter2 in (letter1+1)..26) ( let { % convert from 2d alphabet index (upper right triangle) to 1d index int: index = 26 * (letter1 - 1) + letter2 - letter1 - (letter1*letter1 - letter1) div 2 } in letter_keys_match[index] = (letter_keys[letter1] == letter_keys[letter2]) );

Let’s now get to the optimization part. To quantify how bad a solution is, we need to look at all word pairs where:

- both words have the same length, and
- for each letter of the first word, the corresponding letter of the second word sits on the same key (given whatever
`letter_keys`

will be).

For each one of these word pairs, let’s come up with a number – a “badness value” – that represents how inconvenient it would be to have the pair be a textonym. Something like `min(10000/frequency1, 10000/frequency2)`

works well: the frequency of the less-used word determines the degree of annoyance. (There’s nothing wrong with `1.0/frequency`

, but most finite domain solvers work with integers only, so we need bigger numbers.) If, for example, we end up with a layout in which *the* is a textonym with *fix*, then that’s not too terrible: having to manually choose *fix* over *the* will happen relatively rarely, because *fix* is itself a relatively rare word. A conflict between *the* and *and*, however, would be seriously annoying, because *and* comes up so often.

Our word list contains 76436 pairs of equal-length words (finding this number shall be an exercise for the reader). So let’s create:

array[1..76436] of var float: pair_badness;

along with the 76436 Minizinc constraints, one per pair. They should all look somewhat like this:

constraint pair_badness[43] = if letter_keys_match[14,14]+letter_keys_match[5,7] == 2 then 11 else 0 endif;

## More constraints

There is more we can do. First off, each key should accommodate three or four letters:

constraint global_cardinality_low_up(letter_keys, [0, 1, 2, 3, 4, 5, 6, 7], [3, 3, 3, 3, 3, 3, 3, 3], [4, 4, 4, 4, 4, 4, 4, 4]);

Then, we can observe that the positions of any two keys can be swapped, without affecting the quality of the solution at all. To exclude those symmetries, we’ll require that the keys be sorted by the first of their respective letters. In other words: letter a must be on key 0; letter b must be on key 0 or 1 (but not 2), etc.:

constraint forall (letter in 2..26) ( let { var int: maxkey = max([letter_keys[k] | k in 1..(letter-1)]), } in letter_keys[letter] <= maxkey + 1 );

And to explicitly strengthen this constraint, we can write for the first eight letters:

constraint forall (key in 1..8) ( letter_keys[key] <= key - 1 );

Finally, we can pre-prune our search tree by stating a priori which letters we definitely don’t want on the same key. This information is easily found: just look at the most frequent word pairs with a letter-distance of one. For instance, you certainly won’t want *he* and *me* to be textonyms, so as potential key-mates the letters *h* and *m* are out from the start. Similarly, each vowel *aeiou* should most definitely get its own key. We can write down a bunch of common ones:

constraint all_different([letter_keys[k] | k in {1,5,9,15,21}]); % aeiou constraint all_different([letter_keys[k] | k in {7,12}]); % gl constraint all_different([letter_keys[k] | k in {13,12}]); % ml constraint all_different([letter_keys[k] | k in {13,8}]); % etc... constraint all_different([letter_keys[k] | k in {7,8}]); constraint all_different([letter_keys[k] | k in {13,14}]); constraint all_different([letter_keys[k] | k in {16,19}]); constraint all_different([letter_keys[k] | k in {18,19}]); constraint all_different([letter_keys[k] | k in {14,15}]); constraint all_different([letter_keys[k] | k in {7,9}]); constraint all_different([letter_keys[k] | k in {5,6}]); constraint all_different([letter_keys[k] | k in {5,18}]); constraint all_different([letter_keys[k] | k in {12,20}]); constraint all_different([letter_keys[k] | k in {7,19}]); constraint all_different([letter_keys[k] | k in {6,20}]); constraint all_different([letter_keys[k] | k in {13,18}]);

The more we add, the smaller the search tree, and the faster the solving process. On the other hand, if we exclude too many solutions, we might miss the optimal one – or end up finding none at all.

## Solution

Finally, we’ll add

solve minimize sum(pair_badness); output([show(letter_keys)]);

and we’re on our way!

On my (admittedly slow) laptop, the conversion from Minizinc to Flatzinc takes about two minutes when targeting Google OR-tools, and the resulting Flatzinc file weighs about 62MB.

I started OR-tools and went for a run. When I came back, it had found the optimal key assignment with a total badness of 149:

letter_keys = array1d(1..26, [0, 1, 0, 1, 2, 1, 2, 3, 4, 1, 0, 5, 2, 6, 5, 4, 6, 4, 7, 3, 7, 3, 6, 5, 0, 7]);

which corresponds to

```
acky bdfj egm htv ipr lox nqw suz
```

– so there you go; the least annoying T9 keyboard for Simple English!

Our layout still suffers from nine textonyms:

```
32 its its # LOL, I guess this one's unavoidable
26 told hold
19 along alone
14 wife wide
13 are arm
13 fire firm
11 here term
11 fall ball
10 from film
```

But that’s nothing compared to the traditional alphabetic ordering, which has 46 textonyms for a total badness of 1998:

```
222 he if
222 on no
153 there these
144 them then
121 of me
103 in go
93 or mr
90 up us
62 good home
50 was war
45 might night
34 line kind
33 work york
32 its its
31 have gave
30 any boy
30 seem seen
29 an am
28 go im
28 in im
27 part past
25 say pay
22 but cut
22 case care
21 they view
20 good gone
20 home gone
20 room soon
18 see red
17 move note
16 done food
16 find fine
14 hand game
14 said paid
14 wall walk
14 wife wide
13 dont foot
13 got hot
13 law lay
12 alone blood
12 bed add
11 call ball
11 needs offer
11 purpose suppose
11 reason season
11 run sun
```

Man, remember on/no and of/me? I always hated those two.

The word list, along with a Python file that will generate the complete Minizinc source is available at github.com/skosch/t9_improved.