One way to memorize telephone numbers is to convert the number into a simple phrase in natural language (e.g., English) so that when dialing on a keypad one merely types out that word or phrase. Here are the conversions, taken from a standard telephone keypad:

subs1 = {2 -> {"a", "b", "c"}, 
         3 -> {"d", "e", "f"}, 
         4 -> {"g", "h", "i"}, 
         5 -> {"j", "k", "l"}, 
         6 -> {"m", "n", "o"}, 
         7 -> {"p", "q", "r", "s"}, 
         8 -> {"t", "u", "v"}, 
         9 -> {"w", "x", "y", "z"}};

(We'll return to the issue of 0 and 1.)

Thus the (seven-digit) telephone number 3987228 would be converted into "extract", which the user merely types into the phone, guided by the letters on the keypad. (Note that, of course, a given number might have several valid corresponding words: 228 could be converted to "act," "bat" or "cat".)

Here's simple code that finds (single) valid English words corresponding to a number (here 228):

Select[StringJoin /@ Tuples[IntegerDigits[228] /. subs1], DictionaryWordQ]

(* {"act", "bat", "cat"} *)


Select[StringJoin /@ Tuples[IntegerDigits[5865] /. subs], DictionaryWordQ]

(* {"junk"} *)

The overall problem gets a bit trickier for two reasons. First, one must deal with the digits 0 and 1, which do not have corresponding letters on a keypad. Second, a full 10-digit phone number rarely has a single corresponding word. So one needs to modify the above code to search for sets or phrases of valid English words (of any length) commensurate with the length of the number. (Semantic meaning of the phrase is irrelevant.) Moreover, there are a few heuristics, such as 0 can be read as "oh", 1 can be read as "one", etc.

Thus the phone number 2427793647 could be rendered as "2 happy dogs" (or, given the below substitutions, "two happy dogs" and the user knows to press $2$ rather than spell out "t w o").

Here is an expanded set of substitutions that capture some of these ideas.

subs = {0 -> {"o"}, 
        1 -> {"i", "one"}, 
        2 -> {"a", "b", "c", "two", "too"}, 
        3 -> {"d", "e", "f"}, 
        4 -> {"g", "h", "i", "for"}, 
        5 -> {"j", "k", "l"}, 
        6 -> {"m", "n", "o"}, 
        7 -> {"p", "q", "r", "s"}, 
        8 -> {"t", "u", "v"}, 
        9 -> {"w", "x", "y", "z"}};

What modification of the above code would take an arbitrary 10-digit telephone and generate a phrase in English (of any number of word/components) that could be used to memorize this number?

Surely one will want to use code to detect various splits of strings. For instance:

testphrases = {"catdog", "longit", "hotbag"};

Select[testphrases, AllTrue[StringPartition[#, 3], DictionaryWordQ] &]

(* {"catdog", "hotbag"} *)

This tests for just two valid words generated by splitting the source string after the third character. We need to check all splitting positions, and more simultaneous splitting position (giving three or four or more components).

  • $\begingroup$ why is i in both 4 and 1 lists? $\endgroup$
    – ZaMoC
    Commented Nov 28, 2018 at 23:10
  • $\begingroup$ @J42161217: Because you might want to convert 19268227 to "I want car" and 44779744 to "hippy pig". Admittedly, the user will have to know that in the first case the 1 should be "read" as "I". $\endgroup$ Commented Nov 28, 2018 at 23:12
  • $\begingroup$ then if I want to call "hippy pig" I would have 4 different numbers?? $\endgroup$
    – ZaMoC
    Commented Nov 28, 2018 at 23:14
  • $\begingroup$ @J42161217 Not really. For "hippy" the user knows the "i" is part of a word (so press $4$), but for "I want car" the 1 is a stand-alone character/word (so press $1$). But feel free to modify the substitution rules so long as the code generates interpretable phrases. $\endgroup$ Commented Nov 28, 2018 at 23:16
  • $\begingroup$ Does it have to be keypad-associated? Personally I find the Major System more general and powerfull to encode any number: „Meteor Tail Pink“ (nicely visualized) has you remeber that Pi is approximately 3.1415927. $\endgroup$
    – gwr
    Commented Nov 29, 2018 at 8:41

1 Answer 1


10 digit is difficult...

here is a code that works with 10-digit numbers
(of course there are many limitations)
It only finds words with 1,2,3,4, or 5 letters and in a particular order, but it is fast.
The number-letter associations are based on frequency of letters in english texts (but you can change it and make your own)
Try it by placing a 10-digit number in the beginning

TenDigitNumber = 7599039547
num = IntegerDigits@TenDigitNumber;
subs = {0 -> {"u", "k"}, 1 -> {"d", "v"}, 2 -> {"r", "p", "x"}, 
3 -> {"i", "m", "z"}, 4 -> {"o", "h", "j"}, 5 -> {"t", "g", "q"}, 
6 -> {"n", "b"}, 7 -> {"s", "f"}, 8 -> {"l", "y", "e"}, 
9 -> {"a", "c", "w"}};
"" <> # & /@ Tuples[Select[Tuples[IntegerDigits[#] /. subs], (r = #;
   Length@Flatten[Select[Select[Join[p=Subsets@(def = 
             Flatten[Subsequences[r, {#}] & /@ Range@Length@r, 1];
  Do[def=DeleteCases[def,{(Delete[Alphabet[],{{1},{9}}])[[i]]}],{i,24}]; def), 
Reverse/@p],Flatten@#==r&],And @@ (DictionaryWordQ /@ StringJoin /@ #) &], 2] > 
    0) &] & /@ FromDigits /@ Partition[num, 5]]  


another example

TenDigitNumber = 3896164467     


  • $\begingroup$ The justification for your new substitutions, and the substitutions make little sense. (What does letter frequency have to do with anything?) Also, so many of your "words" or "phrases" are neither: {"mlandnoons", "mlandnoobs", "mlandboons", "mlandboobs", "mlandboobf", "myandboobf", "meandboobf"}. Why only partial phrases with non-phrase elements? How would you eliminate the problem cases? $\endgroup$ Commented Nov 29, 2018 at 3:22
  • 1
    $\begingroup$ DictionaryWordQ["ml"] is True and many other 2 letter words, so it is not my fault. Letter frequency is related because by choosing a random number we must try to have letters that produce words. So the frequent letters must be distributed equally on numbers... $\endgroup$
    – ZaMoC
    Commented Nov 29, 2018 at 3:37
  • 2
    $\begingroup$ Gosh... "ml" as a word! Thanks! $\endgroup$ Commented Nov 29, 2018 at 3:40
  • $\begingroup$ @DavidG.Stork updated. Now "i" and "a" also work as one letter words $\endgroup$
    – ZaMoC
    Commented Nov 29, 2018 at 4:08
  • $\begingroup$ A good approach (+1). Let me wait a bit before accepting this, as I have a feeling someone might have a slightly superior solution. $\endgroup$ Commented Nov 29, 2018 at 7:21

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