579085261 is CRAZY, 725582 is GOLF, 10757494 is ...?

Question

Your task is to translate a 103-smooth number into an English word, using the method described below.

How?

1. Generate the list of prime factors (with repetition) of the input number.
2. Sort the list:
   • If 2 is not one of the prime factors, sort the list in ascending order.
   • If 2 is one of the prime factors, remove it from the list and sort the remaining factors in descending order.

3. Translate each factor into a letter, using the following table:

    3 = S   13 = L   29 = X   43 = O   61 = Z   79 = H  101 = K  
    5 = P   17 = Q   31 = N   47 = R   67 = T   83 = V  103 = Y  
    7 = M   19 = U   37 = C   53 = A   71 = E   89 = D  
   11 = F   23 = I   41 = W   59 = G   73 = J   97 = B  
   

_{Note: This table was built empirically to maximize the number of possible words. For the curious, here is a list of 2,187 words that can be encoded that way (may include rude language). It's definitely not guaranteed to be optimal, but it's good enough for this challenge.}

Examples

Example 1: 579085261 (ascending order)

1. The prime factors are [ 37, 47, 53, 61, 103 ].
2. 2 is not a prime factor, so we keep the list sorted in ascending order.
3. 37 = C, 47 = R, etc. The output is "CRAZY".

Example 2: 725582 (descending order)

1. The prime factors are [ 2, 11, 13, 43, 59 ].
2. 2 is a prime factor, so we remove it and sort the list in descending order, which gives:
   [ 59, 43, 13, 11 ].
3. 59 = G, 43 = O, etc. The output is "GOLF".

Example 3: 10757494 (with a repeated factor)

1. The prime factors are [ 2, 11, 71, 71, 97 ].
2. 2 is a prime factor, so we remove it and sort the list in descending order, which gives:
   [ 97, 71, 71, 11 ].
3. 97 = B, 71 = E, 11 = F. The output is "BEEF".

Clarifications and rules

• The input number is guaranteed to be 103-smooth and divisible by 2 at most once.
• By definition, a smooth-number is a positive integer.
• Input and output can be handled in any reasonable format. The output can be in lowercase or uppercase. Trailing whitespace is acceptable. Leading whitespace is not.
• If your program/function can't support large inputs, please specify it in your answer.
• This is code golf, so the shortest answer in bytes wins.

Test cases

34874          --> ARM
483254         --> BAR
353722         --> EAR
494302         --> EGG
39061          --> FAT
6479           --> FUN
60421          --> ICE
54166          --> JAM
48911474       --> BETA
2510942        --> BOOM
2303854        --> DOOM
844261         --> FIRE
1606801        --> MAZE
1110085        --> PAGE
5212974        --> BALLS
67892046       --> BEANS
885396199      --> CREEK
67401037       --> FUNKY
27762173       --> QUICK
1238440506     --> ARROWS
33045832681    --> CRAGGY
1362714005     --> PIRATE
137302698      --> TROLLS
358310128062   --> BEGGARS
40255151586    --> DETAILS
164633248153   --> FIXATED
621172442227   --> UNRATED
2467812606     --> VACUUMS
86385078330    --> GROWNUPS
26607531423091 --> UNWORTHY

Answer 1

Jelly, 29 27 bytes

ÆEµØA“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’œ?xḊṚḢ}¡

Thanks to @JonathanAllan for golfing off 1 byte!

Try it online!

Background

“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’

is a numeric literal. The characters between the quotes are replaced with their 1-based indices in the Jelly code page, and the resulting array is interpreted as a base-250 number. This yields the integer c := 288824892868083015619552399.

How it works

ÆEµØA“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’œ?xḊṚḢ}¡  Main link. Argument: n

ÆE                           Yield the exponents of n's prime factorization, with.
                             zeroes. This yields an array A.
  µ                          Begin a new monadic chain with argument A.
   ØA                        Set the return value to “ABC...XYZ”.
     “¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’œ?        Select the c-th permutation of the alphabet, yielding
                             s := “SPMFLQUIXNCWORAGZTEJHVDBKY”.
                      Ḋ      Dequeue; yield A without its first element, stripping
                             the exponent of 2.
                     x       Repeat the k-th letter of s r times, where r is the
                             exponent of the k-th odd prime number.
                          ¡  Combine the two links to the left into a quicklink:
                        Ḣ}     - Apply head to the right argument (A), yielding the
                                 exponent of 2. Since n is at most divisible by the
                                 first power of 2, this yields 1 for even numbers
                                 and 0 for odd ones. Call the link to the left that
                                 many times on the previous return value.
                       Ṛ       - Reverse the string to the left.

Answer 2

Jelly, 36 bytes

“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’œ?ØA1;;⁶
×107ÆE¢×UḢ¡t⁶

Try it online!

Explanation

Helper constant 1£ (produces “SPMFLQUIXNCWORAGZTEJHVDBKY ” with a 1 prepended)

“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’œ?ØA1;;⁶
“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’          288824892868083015619552399 (compressed representation)
              œ?ØA      th permutation of the alphabet
                  1;    prepend 1
                    ;⁶  append a space

Main program

×107ÆE¢×UḢ¡t⁶
×107           Multiply {the input} by 107
    ÆE         Convert to a list of frequencies for each factor
      ¢×       {Vectorized} multiply by the return value of 1£
        UḢ¡    Delete the first value, reverse the list that many times
           t⁶  Delete trailing/leading space

I have a feeling that my compression of the list easily beats the other Jelly answer, but that my algorithm for using it could be a lot more efficient. Maybe I'll try to combine them.

Jelly, 31 bytes, inspired by @Leakynun's answer

“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’œ?ØA⁷;
ÆfÆCị¢U⁸¡U

Try it online! (slightly modified to run a lot faster)

Is inconsistent in whether it prints a trailing newline (but PPCG normally allows answers either with or without a trailing newline, so I guess this works too?). Is very slow (O(n) where n is the input, and those numbers aren't exactly small…)

Explanation

Helper constant 1£ (produces “¶SPMFLQUIXNCWORAGZTEJHVDBKY”, where ¶ is newline)

“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’œ?ØA⁷;
“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’          288824892868083015619552399 (compressed representation)
              œ?ØA      th permutation of the alphabet
                  ⁷;    prepend newline

Main program

ÆfÆCị¢U⁸¡U
Æf          Produce list of prime factors (repeating repeated factors)
  ÆC        Map the nth prime to n
    ị¢      Index into the output of 1£
      U     Reverse
        ¡   a number of times
       ⁸    equal to the input
         U  Reverse again

Answer 3

05AB1E, 39 38 bytes

ÒW<iR¨}26LØR•6Ê2"£´õþÕàçŸôëÂÛ*™•36BS:J

Uses the CP-1252 encoding. Try it online!

Answer 4

Ruby, 139 138 134 125 120 115+7 = 146 145 141 132 127 122 bytes

Uses the -rprime flag for +7 bytes.

-1 byte from @daniero. -4 bytes by remembering that I can just do a regular divisibility check instead of checking the prime division for the existence of 2.

-9 bytes from @mbomb007's Python solution reminding me of a shorter way to retrieve the matching letter.

-5 bytes because trailing whitespace is now allowed.

-5 bytes from discovering Enumerable#find_index

->n{x=Prime.prime_division n;x.reverse!if n%2<1;x.map{|i,c|" SPMFLQUIXNCWORAGZTEJHVDBKY"[Prime.find_index i]*c}*''}

Try it online! (all test cases)

Explanation

->n{                                   # Anonymous procedure with one argument n
    x=Prime.prime_division n;          # Get prime factorization of n, sorted
                                       # p0^e0 * p1^e1 ... -> [[p0,e0],[p1,e1],...]
    x.reverse!if n%2<1;                # Reverse if divisible by 2
    x.map{|i,c|                        # For each prime/exponent pair:
        " SPMFLQUIXNCWORAGZTEJHVDBKY"[ # Get corresponding character by obtaining:
            Prime.find_index i]        # Determine index of the current prime
                               *c      # Repeat the letter by the supplied exponent
                                 }*''} # Join all letter sequences together

Answer 5

Python 2, 220 217 bytes

n=input()
i=1
L=[]
exec'i+=1;c=0\nwhile n%i<1:c+=1;n/=i\nif c:L+=[i]*c\n'*n
T='SPMFLQUIXNCWORAGZTEJHVDBKY'
print''.join(T[[p for p in range(3,104)if all(p%k for k in range(2,p))].index(q)]for q in[L,L[:0:-1]][L[0]<3])

Try it online - only runs the smallest test case without running out of memory

Ungolfed:

This version doesn't use exec, so you can actually test all of the test cases without running out of memory.

n=input()
i=1
L=[]
while~-n:
 i+=1;c=0
 while n%i<1:c+=1;n/=i
 if c:L+=[i]*c
if L[0]<3:L=L[:0:-1]
T='SPMFLQUIXNCWORAGZTEJHVDBKY'
print''.join(T[[p for p in range(3,104)if all(p%k for k in range(2,p))].index(q)]for q in L)

Try it online

Answer 6

Bash + GNU utilities + bsd-games package, 170

Seems pretty non-optimal, but it works:

p='printf %03d\n'
a=(`factor $1`)
x=$[a[1]<3]
$p `primes 3 104`|paste - <(fold -1<<<SPMFLQUIXNCWORAGZTEJHVDBKY)|join -o1.2 - <($p ${a[@]:x+1})|(((x))&&tac||cat)|tr -d \\n

Try it online.

Answer 7

Japt, 51 50 bytes

49 bytes of code, +1 for the -P flag.

%2?Uk :Uk Åw)£`yspmflquixncÙgz’jhvdbk`g#ho fj bX

Try it online!

This could be way shorter if only Japt had a couple more features...

Explanation

`yspmflquixncÙgz’jhvdbk` is just the same string everyone else is using compressed as much as Japt can compress it (3 bytes shorter than the original!). Japt's only built-in compression tool at the moment replaces common pairs of lowercase letter with a single-byte char.

So let's examine the actual code:

%2?Uk :Uk Å  w)
%2?Uk :Uk s1 w)
%2?             // If the input mod 2 is non-zero,
   Uk           //   take the prime factors of the input (U).
      :Uk       // Otherwise, take the prime factors of the input,
          s1 w  //   slice off the first one (2), and reverse.

Then £ is used to replace each item X in the result like this:

"string"g#h o fj bX
"string"g104o fj bX

         104o         // Create the range [0...104).
              fj      // Filter to only items Z where Z.j() is truthy (Z is prime).
                      // This results in the list of prime numbers from 2 to 103.
                 bX   // Take the index of X in this list.
"string"g             // Get the char in the compressed string at that index.
                      // For `y`, the index is 26, but since the string is only 26 chars
                      // long, Japt wraps around and grabs the first char in the string.

The result is an array of characters at this point, so the -P flag joins it into a single string, and the result is implicitly sent to output.

Answer 8

Jelly, 33 bytes

ÆfÆCUṖ$¹e@?1’ị“¡3ḅḲ+Ṿɼ¡ẏƙẊƘ’œ?ØA¤

Try it online!

Answer 9

Pyth, 54 47 bytes

7 bytes thanks to isaacg

s__W%Q2@L."AZ❤O❤❤❤❤❤❤Q❤9❤❤×❤❤"xL_MP#r_3_103-PQ2

(❤ represents an unprintable character)

Pyth doesn't have many prime built-ins...

Hexdump:

0000000: 73 5f 5f 57 25 51 32 40 4c 2e 22 41 5a 03 4f f3 s__W%Q2@L."AZ.O.
0000010: 14 af 15 ed f5 51 90 39 d5 18 d7 20 a8 22 78 4c .....Q.9... ."xL
0000020: 5f 4d 50 23 72 5f 33 5f 31 30 33 2d 50 51 32    _MP#r_3_103-PQ2

Try it online!

Answer 10

Jelly, 40 bytes

Æfḟ2ÆCṚ⁸Ḃ¤¡Ṛị“YSPMFLQUIXNCWORAGZTEJHVDBK

Try it online!

Verify all testcases at once! (slightly modified)

Answer 11

J, 59 bytes

'SPMFLQUIXNCWORAGZTEJHVDBKY'{~(p:>:i.26)i.[:|.@}.^:(2={.)q:

Try it online!

Quite simple...

'SPMFLQUIXNCWORAGZTEJHVDBKY'{~(p:>:i.26)i.[:|.@}.^:(2={.)q:
                              (p:>:i.26)                      first 26 primes after 2
                                        i.                    for which the indices is
                                                         q:   prime factors
                                            |.@}.             remove first and reverse
                                                 ^:           if
                                                   (2={.)     the first entry is 2
                            {~                                reverse index
'..........................'                                  hardcoded string

Answer 12

PHP, 173 Bytes

for($i=2;1<$n=&$argn;$n%$i?++$i:$r[]=$i.!$n/=$i)for($t=$i;$i>2&!$w[$i]&&$i%--$t;$t>2?:$w[$i]=SPMFLQUIXNCWORAGZTEJHVDBKY[$p++]);$r[0]>2?:rsort($r);foreach($r as$s)echo$w[$s];

Online Version

Expanded

for($i=2;1<$n=&$argn; # loop till input is 1
$n%$i?++$i:$r[]=$i.!$n/=$i) #after loop add value to result if input is divisible and divide input
  for($t=$i;$i>2&!$w[$i]&&$i%--$t; # loop if number is gt 2 and not in letter array till number is divisible 
  $t>2?:$w[$i]=SPMFLQUIXNCWORAGZTEJHVDBKY[$p++]) # if is prime add to letter array
  ; # make nothing in the loop
$r[0]>2?:rsort($r); # reverse result array if 2 is in result array
foreach($r as$s) # loop result array
  echo$w[$s]; # Output 

PHP, 178 Bytes

for($z=2;$p<26;$t>1?:$w[$z]=SPMFLQUIXNCWORAGZTEJHVDBKY[$p++])for($t=++$z;$z%--$t;);for($i=2;1<$n=&$argn;)$n%$i?++$i:$r[]=$i.!$n/=$i;$r[0]>2?:rsort($r);foreach($r as$s)echo$w[$s];

Online Version

Expanded

for($z=2;$p<26;
$t>1?:$w[$z]=SPMFLQUIXNCWORAGZTEJHVDBKY[$p++]) # after loop if is prime add to letter array
  for($t=++$z;$z%--$t;); 
for($i=2;1<$n=&$argn;)  # loop till input is 1
  $n%$i?++$i:$r[]=$i.!$n/=$i; #add value to result if input is divisible and divide input
$r[0]>2?:rsort($r); # reverse result array if 2 is in result array
foreach($r as$s) # loop result array
  echo$w[$s]; # Output 

Answer 13

Python, 1420 bytes

lambda x:(lambda a,b,e,k,l,m,q,p:len.__name__[:a].join(dict(zip((lambda n:(lambda n,f,g:f(n,e,[],f,g))(n,lambda n,i,r,f,g:g(n,i+b,r,f,g)if i<n else r,lambda n,i,r,f,g:f(n,i,[r,r+[i]][all(i%x!=a for x in[e]+r)],f,g)))(l*e*(k*l+b)),(lambda n,o,t:(lambda n,f,g:f(n, len.__name__[:a],f,g))(n,lambda n,s,f,g:g(n,s,f,g)if n>o else s,lambda n,s,f,g:f(n//t,s+chr(n%t),f,g)))((((((k<<e)-b)<<m)+m)<<((k<<q)+(k<<b)))+(((((k<<e)-b)<<e)+b)<<((k<<q)-(b<<b)))+((((((b<<k)+b))<<l)+b)<<((((k<<e)-b)<<l)+(b<<b)))+(((((k<<e)-b)<<l)-k)<<((m<<m)+p))-(((p<<m)-b)<<((m<<m)-(b<<b)))+(((m<<k)+b)<<((((m<<e)-b)<<k)-(b<<b)))+(((((k<<e)-b)<<l)-m)<<((((b<<l)+b)<<k)+k))-(((((b<<l)-b)<<l)-p)<<((b<<p)+(b<<b)))-(((p<<k)-b)<<((((b<<l)-b)<<k)+k))-(((k<<q)-b)<<((p<<l)))+(((m<<m)+m)<<((((k<<e)+b)<<k)-b))-(((k<<m)+b)<<((k<<m)-b))-(((m<<m)+k)<<((((k<<e)-b)<<k)-(b<<b)))+(((((k<<e)+b)<<e)+b)<<((m<<l)-(b<<e)))-(((((k<<e)+b)<<e)+b)<<((((b<<l)+b)<<e)-b))+((((((b<<k)+b))<<k)+b)<<((p<<k)))+(((((k<<e)-b)<<k)-k)<<((k<<l)))+(((m<<l)+b)<<((m<<k)))+(((m<<e)-b)<<((b<<m)+(b<<b)))+((((((b<<k)+b))<<e)-b)<<((k<<k)+b))+(((m<<k)-b)<<((b<<l)+b))-((((k<<e)-b))<<((k<<e)))+(((m<<e)+b)<<e)-b,b,b<<l*e)))[i]for i in(lambda x: [x.remove(e), x[::-b]][b] if e in x else x)((lambda x:(lambda x,g,h:g(x,b,[],g,h))(x,lambda x,n,r,g,h:h(x,n+b,r,g,h)if x>b else r,lambda x,n,r,g,h:h(x//n,n,r+[n],g,h)if x%n==a else g(x,n,r,g,h)))(x))))(*[x for x in range(True<<len(len.__name__))])

This could definitely by shorted some, but it's my attempt at solving it with no numbers or string literals. I realize this is a code-golf problem and this isn't exactly short but wanted to share anyway, not sure if this breaks any rules or not.

Was a lot of fun to make, I used the algorithm on this blog post to reduce the ASCII numerical representation of "SPMFLQUIXNCWORAGZTEJHVDBKY" in to the bit-shift expression I use. I also generally took a lot of inspiration from this blog, I wanted to try it out myself and this seemed like a good challenge to do it with.

Here is a slightly more readable version, with some more sensible variable names too