24 game/Solve

24 game/Solve
You are encouraged to solve this task according to the task description, using any language you may know.

Write a function that given four digits subject to the rules of the 24 game, computes an expression to solve the game if possible.

Show examples of solutions generated by the function

C.F: Arithmetic Evaluator

ABAP

Will generate all possible solutions of any given four numbers according to the rules of the 24 game.

Note: the permute function was locally from [here] <lang ABAP>data: lv_flag type c,

     lv_number type i,
     lt_numbers type table of i.

constants: c_no_val type i value 9999.

append 1 to lt_numbers. append 1 to lt_numbers. append 2 to lt_numbers. append 7 to lt_numbers.

write 'Evaluating 24 with the following input: '. loop at lt_numbers into lv_number.

 write lv_number.

endloop. perform solve_24 using lt_numbers.

form eval_formula using iv_eval type string changing ev_out type i.

 call function 'EVAL_FORMULA' "analysis of a syntactically correct formula
   exporting
     formula = iv_eval
   importing
     value   = ev_out
   exceptions
  others     = 1.

 if sy-subrc <> 0.
   ev_out = -1.
 endif.

endform.

" Solve a 24 puzzle. form solve_24 using it_numbers like lt_numbers.

 data: lv_flag   type c,
       lv_op1    type c,
       lv_op2    type c,
       lv_op3    type c,
       lv_var1   type c,
       lv_var2   type c,
       lv_var3   type c,
       lv_var4   type c,
       lv_eval   type string,
       lv_result type i,
       lv_var     type i.

 define retrieve_var.
   read table it_numbers index &1 into lv_var.
   &2 = lv_var.
 end-of-definition.

 define retrieve_val.
   perform eval_formula using lv_eval changing lv_result.
   if lv_result = 24.
       write / lv_eval.
   endif.
 end-of-definition.
 " Loop through all the possible number permutations.
 do.
   " Init. the operations table.

   retrieve_var: 1 lv_var1, 2 lv_var2, 3 lv_var3, 4 lv_var4.
   do 4 times.
     case sy-index.
       when 1.
         lv_op1 = '+'.
       when 2.
         lv_op1 = '*'.
       when 3.
         lv_op1 = '-'.
       when 4.
         lv_op1 = '/'.
     endcase.
     do 4 times.
       case sy-index.
       when 1.
         lv_op2 = '+'.
       when 2.
         lv_op2 = '*'.
       when 3.
         lv_op2 = '-'.
       when 4.
         lv_op2 = '/'.
       endcase.
       do 4 times.
         case sy-index.
         when 1.
           lv_op3 = '+'.
         when 2.
           lv_op3 = '*'.
         when 3.
           lv_op3 = '-'.
         when 4.
           lv_op3 = '/'.
         endcase.
         concatenate '(' '(' lv_var1 lv_op1 lv_var2 ')' lv_op2 lv_var3 ')' lv_op3 lv_var4  into lv_eval separated by space.
         retrieve_val.
         concatenate '(' lv_var1 lv_op1 lv_var2 ')' lv_op2 '(' lv_var3 lv_op3 lv_var4 ')'  into lv_eval separated by space.
         retrieve_val.
         concatenate '(' lv_var1 lv_op1 '(' lv_var2 lv_op2 lv_var3 ')' ')' lv_op3 lv_var4  into lv_eval separated by space.
         retrieve_val.
         concatenate lv_var1 lv_op1 '(' '(' lv_var2 lv_op2 lv_var3 ')' lv_op3 lv_var4 ')'  into lv_eval separated by space.
         retrieve_val.
         concatenate lv_var1 lv_op1 '(' lv_var2 lv_op2 '(' lv_var3 lv_op3 lv_var4 ')' ')'  into lv_eval separated by space.
         retrieve_val.
       enddo.
     enddo.
   enddo.

   " Once we've reached the last permutation -> Exit.
   perform permute using it_numbers changing lv_flag.
   if lv_flag = 'X'.
     exit.
   endif.
 enddo.

endform.

" Permutation function - this is used to permute: " A = {A1...AN} -> Set of supplied variables. " B = {B1...BN - 1} -> Set of operators. " Can be used for an unbounded size set. Relies " on lexicographic ordering of the set. form permute using iv_set like lt_numbers

            changing ev_last type c.
 data: lv_len     type i,
       lv_first   type i,
       lv_third   type i,
       lv_count   type i,
       lv_temp    type i,
       lv_temp_2  type i,
       lv_second  type i,
       lv_changed type c,
       lv_perm    type i.
 describe table iv_set lines lv_len.

 lv_perm = lv_len - 1.
 lv_changed = ' '.
 " Loop backwards through the table, attempting to find elements which
 " can be permuted. If we find one, break out of the table and set the
 " flag indicating a switch.
 do.
   if lv_perm <= 0.
     exit.
   endif.
   " Read the elements.
   read table iv_set index lv_perm into lv_first.
   add 1 to lv_perm.
   read table iv_set index lv_perm into lv_second.
   subtract 1 from lv_perm.
   if lv_first < lv_second.
     lv_changed = 'X'.
     exit.
   endif.
   subtract 1 from lv_perm.
 enddo.

 " Last permutation.
 if lv_changed <> 'X'.
   ev_last = 'X'.
   exit.
 endif.

 " Swap tail decresing to get a tail increasing.
 lv_count = lv_perm + 1.
 do.
   lv_first = lv_len + lv_perm - lv_count + 1.
   if lv_count >= lv_first.
     exit.
   endif.

   read table iv_set index lv_count into lv_temp.
   read table iv_set index lv_first into lv_temp_2.
   modify iv_set index lv_count from lv_temp_2.
   modify iv_set index lv_first from lv_temp.
   add 1 to lv_count.
 enddo.

 lv_count = lv_len - 1.
 do.
   if lv_count <= lv_perm.
     exit.
   endif.

   read table iv_set index lv_count into lv_first.
   read table iv_set index lv_perm into lv_second.
   read table iv_set index lv_len into lv_third.
   if ( lv_first < lv_third ) and ( lv_first > lv_second ).
     lv_len = lv_count.
   endif.

   subtract 1 from lv_count.
 enddo.

 read table iv_set index lv_perm into lv_temp.
 read table iv_set index lv_len into lv_temp_2.
 modify iv_set index lv_perm from lv_temp_2.
 modify iv_set index lv_len from lv_temp.

endform.</lang>

Sample Runs:

Evaluating 24 with the following input:  1 1 2 7
( 1 + 2 ) * ( 1 + 7 )
( 1 + 2 ) * ( 7 + 1 )
( 1 + 7 ) * ( 1 + 2 )
( 1 + 7 ) * ( 2 + 1 )
( 2 + 1 ) * ( 1 + 7 )
( 2 + 1 ) * ( 7 + 1 )
( 7 + 1 ) * ( 1 + 2 )
( 7 + 1 ) * ( 2 + 1 )

Evaluating 24 with the following input:  1
( ( 1 + 2 ) + 3 ) * 4
( 1 + ( 2 + 3 ) ) * 4
( ( 1 * 2 ) * 3 ) * 4
( 1 * 2 ) * ( 3 * 4 )
( 1 * ( 2 * 3 ) ) * 4
1 * ( ( 2 * 3 ) * 4 )
1 * ( 2 * ( 3 * 4 ) )
( ( 1 * 2 ) * 4 ) * 3
( 1 * 2 ) * ( 4 * 3 )
( 1 * ( 2 * 4 ) ) * 3
1 * ( ( 2 * 4 ) * 3 )
1 * ( 2 * ( 4 * 3 ) )
( ( 1 + 3 ) + 2 ) * 4
( 1 + ( 3 + 2 ) ) * 4
( 1 + 3 ) * ( 2 + 4 )
( ( 1 * 3 ) * 2 ) * 4
( 1 * 3 ) * ( 2 * 4 )
( 1 * ( 3 * 2 ) ) * 4
1 * ( ( 3 * 2 ) * 4 )
1 * ( 3 * ( 2 * 4 ) )
( 1 + 3 ) * ( 4 + 2 )
( ( 1 * 3 ) * 4 ) * 2
( 1 * 3 ) * ( 4 * 2 )
( 1 * ( 3 * 4 ) ) * 2
1 * ( ( 3 * 4 ) * 2 )
1 * ( 3 * ( 4 * 2 ) )
( ( 1 * 4 ) * 2 ) * 3
( 1 * 4 ) * ( 2 * 3 )
( 1 * ( 4 * 2 ) ) * 3
1 * ( ( 4 * 2 ) * 3 )
1 * ( 4 * ( 2 * 3 ) )
( ( 1 * 4 ) * 3 ) * 2
( 1 * 4 ) * ( 3 * 2 )
( 1 * ( 4 * 3 ) ) * 2
1 * ( ( 4 * 3 ) * 2 )
1 * ( 4 * ( 3 * 2 ) )
( ( 2 + 1 ) + 3 ) * 4
( 2 + ( 1 + 3 ) ) * 4
( ( 2 * 1 ) * 3 ) * 4
( 2 * 1 ) * ( 3 * 4 )
( 2 * ( 1 * 3 ) ) * 4
2 * ( ( 1 * 3 ) * 4 )
2 * ( 1 * ( 3 * 4 ) )
( ( 2 / 1 ) * 3 ) * 4
( 2 / 1 ) * ( 3 * 4 )
( 2 / ( 1 / 3 ) ) * 4
2 / ( 1 / ( 3 * 4 ) )
2 / ( ( 1 / 3 ) / 4 )
( ( 2 * 1 ) * 4 ) * 3
( 2 * 1 ) * ( 4 * 3 )
( 2 * ( 1 * 4 ) ) * 3
2 * ( ( 1 * 4 ) * 3 )
2 * ( 1 * ( 4 * 3 ) )
( ( 2 / 1 ) * 4 ) * 3
( 2 / 1 ) * ( 4 * 3 )
( 2 / ( 1 / 4 ) ) * 3
2 / ( 1 / ( 4 * 3 ) )
2 / ( ( 1 / 4 ) / 3 )
( ( 2 + 3 ) + 1 ) * 4
( 2 + ( 3 + 1 ) ) * 4
( ( 2 * 3 ) * 1 ) * 4
( 2 * 3 ) * ( 1 * 4 )
( 2 * ( 3 * 1 ) ) * 4
2 * ( ( 3 * 1 ) * 4 )
2 * ( 3 * ( 1 * 4 ) )
( ( 2 * 3 ) / 1 ) * 4
( 2 * ( 3 / 1 ) ) * 4
2 * ( ( 3 / 1 ) * 4 )
( 2 * 3 ) / ( 1 / 4 )
2 * ( 3 / ( 1 / 4 ) )
( ( 2 * 3 ) * 4 ) * 1
( 2 * 3 ) * ( 4 * 1 )
( 2 * ( 3 * 4 ) ) * 1
2 * ( ( 3 * 4 ) * 1 )
2 * ( 3 * ( 4 * 1 ) )
( ( 2 * 3 ) * 4 ) / 1
( 2 * 3 ) * ( 4 / 1 )
( 2 * ( 3 * 4 ) ) / 1
2 * ( ( 3 * 4 ) / 1 )
2 * ( 3 * ( 4 / 1 ) )
( 2 + 4 ) * ( 1 + 3 )
( ( 2 * 4 ) * 1 ) * 3
( 2 * 4 ) * ( 1 * 3 )
( 2 * ( 4 * 1 ) ) * 3
2 * ( ( 4 * 1 ) * 3 )
2 * ( 4 * ( 1 * 3 ) )
( ( 2 * 4 ) / 1 ) * 3
( 2 * ( 4 / 1 ) ) * 3
2 * ( ( 4 / 1 ) * 3 )
( 2 * 4 ) / ( 1 / 3 )
2 * ( 4 / ( 1 / 3 ) )
( 2 + 4 ) * ( 3 + 1 )
( ( 2 * 4 ) * 3 ) * 1
( 2 * 4 ) * ( 3 * 1 )
( 2 * ( 4 * 3 ) ) * 1
2 * ( ( 4 * 3 ) * 1 )
2 * ( 4 * ( 3 * 1 ) )
( ( 2 * 4 ) * 3 ) / 1
( 2 * 4 ) * ( 3 / 1 )
( 2 * ( 4 * 3 ) ) / 1
2 * ( ( 4 * 3 ) / 1 )
2 * ( 4 * ( 3 / 1 ) )
( ( 3 + 1 ) + 2 ) * 4
( 3 + ( 1 + 2 ) ) * 4
( 3 + 1 ) * ( 2 + 4 )
( ( 3 * 1 ) * 2 ) * 4
( 3 * 1 ) * ( 2 * 4 )
( 3 * ( 1 * 2 ) ) * 4
3 * ( ( 1 * 2 ) * 4 )
3 * ( 1 * ( 2 * 4 ) )
( ( 3 / 1 ) * 2 ) * 4
( 3 / 1 ) * ( 2 * 4 )
( 3 / ( 1 / 2 ) ) * 4
3 / ( 1 / ( 2 * 4 ) )
3 / ( ( 1 / 2 ) / 4 )
( 3 + 1 ) * ( 4 + 2 )
( ( 3 * 1 ) * 4 ) * 2
( 3 * 1 ) * ( 4 * 2 )
( 3 * ( 1 * 4 ) ) * 2
3 * ( ( 1 * 4 ) * 2 )
3 * ( 1 * ( 4 * 2 ) )
( ( 3 / 1 ) * 4 ) * 2
( 3 / 1 ) * ( 4 * 2 )
( 3 / ( 1 / 4 ) ) * 2
3 / ( 1 / ( 4 * 2 ) )
3 / ( ( 1 / 4 ) / 2 )
( ( 3 + 2 ) + 1 ) * 4
( 3 + ( 2 + 1 ) ) * 4
( ( 3 * 2 ) * 1 ) * 4
( 3 * 2 ) * ( 1 * 4 )
( 3 * ( 2 * 1 ) ) * 4
3 * ( ( 2 * 1 ) * 4 )
3 * ( 2 * ( 1 * 4 ) )
( ( 3 * 2 ) / 1 ) * 4
( 3 * ( 2 / 1 ) ) * 4
3 * ( ( 2 / 1 ) * 4 )
( 3 * 2 ) / ( 1 / 4 )
3 * ( 2 / ( 1 / 4 ) )
( ( 3 * 2 ) * 4 ) * 1
( 3 * 2 ) * ( 4 * 1 )
( 3 * ( 2 * 4 ) ) * 1
3 * ( ( 2 * 4 ) * 1 )
3 * ( 2 * ( 4 * 1 ) )
( ( 3 * 2 ) * 4 ) / 1
( 3 * 2 ) * ( 4 / 1 )
( 3 * ( 2 * 4 ) ) / 1
3 * ( ( 2 * 4 ) / 1 )
3 * ( 2 * ( 4 / 1 ) )
( ( 3 * 4 ) * 1 ) * 2
( 3 * 4 ) * ( 1 * 2 )
( 3 * ( 4 * 1 ) ) * 2
3 * ( ( 4 * 1 ) * 2 )
3 * ( 4 * ( 1 * 2 ) )
( ( 3 * 4 ) / 1 ) * 2
( 3 * ( 4 / 1 ) ) * 2
3 * ( ( 4 / 1 ) * 2 )
( 3 * 4 ) / ( 1 / 2 )
3 * ( 4 / ( 1 / 2 ) )
( ( 3 * 4 ) * 2 ) * 1
( 3 * 4 ) * ( 2 * 1 )
( 3 * ( 4 * 2 ) ) * 1
3 * ( ( 4 * 2 ) * 1 )
3 * ( 4 * ( 2 * 1 ) )
( ( 3 * 4 ) * 2 ) / 1
( 3 * 4 ) * ( 2 / 1 )
( 3 * ( 4 * 2 ) ) / 1
3 * ( ( 4 * 2 ) / 1 )
3 * ( 4 * ( 2 / 1 ) )
4 * ( ( 1 + 2 ) + 3 )
4 * ( 1 + ( 2 + 3 ) )
( ( 4 * 1 ) * 2 ) * 3
( 4 * 1 ) * ( 2 * 3 )
( 4 * ( 1 * 2 ) ) * 3
4 * ( ( 1 * 2 ) * 3 )
4 * ( 1 * ( 2 * 3 ) )
( ( 4 / 1 ) * 2 ) * 3
( 4 / 1 ) * ( 2 * 3 )
( 4 / ( 1 / 2 ) ) * 3
4 / ( 1 / ( 2 * 3 ) )
4 / ( ( 1 / 2 ) / 3 )
4 * ( ( 1 + 3 ) + 2 )
4 * ( 1 + ( 3 + 2 ) )
( ( 4 * 1 ) * 3 ) * 2
( 4 * 1 ) * ( 3 * 2 )
( 4 * ( 1 * 3 ) ) * 2
4 * ( ( 1 * 3 ) * 2 )
4 * ( 1 * ( 3 * 2 ) )
( ( 4 / 1 ) * 3 ) * 2
( 4 / 1 ) * ( 3 * 2 )
( 4 / ( 1 / 3 ) ) * 2
4 / ( 1 / ( 3 * 2 ) )
4 / ( ( 1 / 3 ) / 2 )
( 4 + 2 ) * ( 1 + 3 )
4 * ( ( 2 + 1 ) + 3 )
4 * ( 2 + ( 1 + 3 ) )
( ( 4 * 2 ) * 1 ) * 3
( 4 * 2 ) * ( 1 * 3 )
( 4 * ( 2 * 1 ) ) * 3
4 * ( ( 2 * 1 ) * 3 )
4 * ( 2 * ( 1 * 3 ) )
( ( 4 * 2 ) / 1 ) * 3
( 4 * ( 2 / 1 ) ) * 3
4 * ( ( 2 / 1 ) * 3 )
( 4 * 2 ) / ( 1 / 3 )
4 * ( 2 / ( 1 / 3 ) )
( 4 + 2 ) * ( 3 + 1 )
4 * ( ( 2 + 3 ) + 1 )
4 * ( 2 + ( 3 + 1 ) )
( ( 4 * 2 ) * 3 ) * 1
( 4 * 2 ) * ( 3 * 1 )
( 4 * ( 2 * 3 ) ) * 1
4 * ( ( 2 * 3 ) * 1 )
4 * ( 2 * ( 3 * 1 ) )
( ( 4 * 2 ) * 3 ) / 1
( 4 * 2 ) * ( 3 / 1 )
( 4 * ( 2 * 3 ) ) / 1
4 * ( ( 2 * 3 ) / 1 )
4 * ( 2 * ( 3 / 1 ) )
4 * ( ( 3 + 1 ) + 2 )
4 * ( 3 + ( 1 + 2 ) )
( ( 4 * 3 ) * 1 ) * 2
( 4 * 3 ) * ( 1 * 2 )
( 4 * ( 3 * 1 ) ) * 2
4 * ( ( 3 * 1 ) * 2 )
4 * ( 3 * ( 1 * 2 ) )
( ( 4 * 3 ) / 1 ) * 2
( 4 * ( 3 / 1 ) ) * 2
4 * ( ( 3 / 1 ) * 2 )
( 4 * 3 ) / ( 1 / 2 )
4 * ( 3 / ( 1 / 2 ) )
4 * ( ( 3 + 2 ) + 1 )
4 * ( 3 + ( 2 + 1 ) )
( ( 4 * 3 ) * 2 ) * 1
( 4 * 3 ) * ( 2 * 1 )
( 4 * ( 3 * 2 ) ) * 1
4 * ( ( 3 * 2 ) * 1 )
4 * ( 3 * ( 2 * 1 ) )
( ( 4 * 3 ) * 2 ) / 1
( 4 * 3 ) * ( 2 / 1 )
( 4 * ( 3 * 2 ) ) / 1
4 * ( ( 3 * 2 ) / 1 )
4 * ( 3 * ( 2 / 1 ) )

Evaluating 24 with the following input:  5 6 7 8
5 * ( 6 - ( 8 / 7 ) )
( 5 + 7 ) * ( 8 - 6 )
( ( 5 + 7 ) - 8 ) * 6
( 5 + ( 7 - 8 ) ) * 6
( ( 5 - 8 ) + 7 ) * 6
( 5 - ( 8 - 7 ) ) * 6
6 * ( ( 5 + 7 ) - 8 )
6 * ( 5 + ( 7 - 8 ) )
6 * ( ( 5 - 8 ) + 7 )
6 * ( 5 - ( 8 - 7 ) )
6 * ( ( 7 + 5 ) - 8 )
6 * ( 7 + ( 5 - 8 ) )
( 6 / ( 7 - 5 ) ) * 8
6 / ( ( 7 - 5 ) / 8 )
6 * ( ( 7 - 8 ) + 5 )
6 * ( 7 - ( 8 - 5 ) )
( 6 * 8 ) / ( 7 - 5 )
6 * ( 8 / ( 7 - 5 ) )
( 6 - ( 8 / 7 ) ) * 5
( 7 + 5 ) * ( 8 - 6 )
( ( 7 + 5 ) - 8 ) * 6
( 7 + ( 5 - 8 ) ) * 6
( ( 7 - 8 ) + 5 ) * 6
( 7 - ( 8 - 5 ) ) * 6
( 8 - 6 ) * ( 5 + 7 )
( 8 * 6 ) / ( 7 - 5 )
8 * ( 6 / ( 7 - 5 ) )
( 8 - 6 ) * ( 7 + 5 )
( 8 / ( 7 - 5 ) ) * 6
8 / ( ( 7 - 5 ) / 6 )

Argile

Works with: Argile version 1.0.0

<lang Argile>die "Please give 4 digits as argument 1\n" if argc < 2

print a function that given four digits argv[1] subject to the rules of \ the _24_ game, computes an expression to solve the game if possible.

use std, array

let digits be an array of 4 byte let operators be an array of 4 byte (: reordered arrays :) let (type of digits) rdigits let (type of operators) roperators

.: a function that given four digits <text digits> subject to

  the rules of the _24_ game, computes an expression to solve
  the game if possible.                                       :. -> text
 if #digits != 4 {return "[error: need exactly 4 digits]"}
 operators[0] = '+' ; operators[1] = '-'
 operators[2] = '*' ; operators[3] = '/'
 for each (val int d) from 0 to 3
   if (digits[d] < '1') || (digits[d] > '9')
     return "[error: non-digit character given]"
   (super digits)[d] = digits[d]
 let expr = for each operand order stuff
 return "" if expr is nil
 expr

.:for each operand order stuff:. -> text

 for each (val int a) from 0 to 3
   for each (val int b) from 0 to 3
     next if (b == a)
     for each (val int c) from 0 to 3
       next if (c == b) or (c == a)

for each (val int d) from 0 to 3 next if (d == c) or (d == b) or (d == a) rdigits[0] = digits[a] ; rdigits[1] = digits[b] rdigits[2] = digits[c] ; rdigits[3] = digits[d] let found = for each operator order stuff return found unless found is nil

 nil

.:for each operator order stuff:. -> text

 for each (val int i) from 0 to 3
   for each (val int j) from 0 to 3
     for each (val int k) from 0 to 3
       roperators[0] = operators[i]

roperators[1] = operators[j] roperators[2] = operators[k] let found = for each RPN pattern stuff return found if found isn't nil

 nil

our (raw array of text) RPN_patterns = Cdata

 "xx.x.x."
 "xx.xx.."
 "xxx..x."
 "xxx.x.."
 "xxxx..."

our (raw array of text) formats = Cdata

 "((%c%c%c)%c%c)%c%c"
 "(%c%c%c)%c(%c%c%c)"
 "(%c%c(%c%c%c))%c%c"
 "%c%c((%c%c%c)%c%c)"
 "%c%c(%c%c(%c%c%c))"

our (raw array of array of 3 int) rrop = Cdata

 {0;1;2}; {0;2;1}; {1;0;2}; {2;0;1}; {2;1;0}

.:for each RPN pattern stuff:. -> text

 let RPN_stack be an array of 4 real
 for each (val int rpn) from 0 to 4
   let (nat) sp=0, op=0, dg=0.
   let text p
   for (p = RPN_patterns[rpn]) (*p != 0) (p++)
     if *p == 'x'
       if sp >= 4 {die "RPN stack overflow\n"}

if dg > 3 {die "RPN digits overflow\n"} RPN_stack[sp++] = (rdigits[dg++] - '0') as real

     if *p == '.'
       if sp < 2 {die "RPN stack underflow\n"}

if op > 2 {die "RPN operators overflow\n"} sp -= 2 let x = RPN_stack[sp] let y = RPN_stack[sp + 1] switch roperators[op++] case '+' {x += y} case '-' {x -= y} case '*' {x *= y} case '/' {x /= y} default {die "RPN operator unknown\n"} RPN_stack[sp++] = x

   if RPN_stack[0] == 24.0
     our array of 12 byte buffer (: 4 paren + 3 ops + 4 digits + null :)
     snprintf (buffer as text) (size of buffer) (formats[rpn])		\
        (rdigits[0]) (roperators[(rrop[rpn][0])]) (rdigits[1])		\
                     (roperators[(rrop[rpn][1])]) (rdigits[2])		\
                     (roperators[(rrop[rpn][2])]) (rdigits[3]);
     return buffer as text
 nil</lang>

Examples:

$ arc 24_game_solve.arg -o 24_game_solve.c
$ gcc -Wall 24_game_solve.c -o 24_game_solve
$ ./24_game_solve 1234
((1+2)+3)*4
$ ./24_game_solve 9999

$ ./24_game_solve 5678
((5+7)-8)*6
$ ./24_game_solve 1127
(1+2)*(1+7)

AutoHotkey

Works with: AutoHotkey_L

Output is in RPN. <lang AHK>#NoEnv InputBox, NNNN  ; user input 4 digits NNNN := RegExReplace(NNNN, "(\d)(?=\d)", "$1,") ; separate with commas for the sort command sort NNNN, d`, ; sort in ascending order for the permutations to work StringReplace NNNN, NNNN, `,, , All ; remove comma separators after sorting

ops := "+-*/" patterns := [ "x x.x.x." ,"x x.x x.." ,"x x x..x." ,"x x x.x.." ,"x x x x..." ]

build bruteforce operator list ("+++, ++-, ++* ... ///")

a := b := c := 0 While (++a<5){

While (++b<5){
 While (++c<5){
  l := SubStr(ops, a, 1) . SubStr(ops, b, 1) . SubStr(ops, c, 1)

  ; build bruteforce template ("x x+x+x+, x x+x x++ ... x x x x///")
  For each, pattern in patterns
  {
     Loop 3
        StringReplace, pattern, pattern, ., % SubStr(l, A_Index, 1)
     pat .= pattern "`n"
  }
 }c := 0
}b := 0

} StringTrimRight, pat, pat, 1 ; remove trailing newline

permutate input. As the lexicographic algorithm is used, each permutation generated is unique

While NNNN { StringSplit, N, NNNN ; substitute numbers in for x's and evaluate Loop Parse, pat, `n { eval := A_LoopField ; current line Loop 4 StringReplace, eval, eval, x, % N%A_Index% ; substitute number for "x" If Round(evalRPN(eval), 4) = 24 final .= eval "`n" } NNNN := perm_next(NNNN) ; next lexicographic permutation of user's digits } MsgBox % final ? clipboard := final : "No solution"

simple stack-based evaluation. Integers only. Whitespace is used to push a value.

evalRPN(s){ stack := [] Loop Parse, s If A_LoopField is number t .= A_LoopField else { If t stack.Insert(t), t := "" If InStr("+-/*", l := A_LoopField) { a := stack.Remove(), b := stack.Remove() stack.Insert( l = "+" ? b + a :l = "-" ? b - a :l = "*" ? b * a :l = "/" ? b / a :0 ) } } return stack.Remove() }

perm_Next(str){ p := 0, sLen := StrLen(str) Loop % sLen { If A_Index=1 continue t := SubStr(str, sLen+1-A_Index, 1) n := SubStr(str, sLen+2-A_Index, 1) If ( t < n ) { p := sLen+1-A_Index, pC := SubStr(str, p, 1) break } } If !p return false Loop { t := SubStr(str, sLen+1-A_Index, 1) If ( t > pC ) { n := sLen+1-A_Index, nC := SubStr(str, n, 1) break } } return SubStr(str, 1, p-1) . nC . Reverse(SubStr(str, p+1, n-p-1) . pC . SubStr(str, n+1)) }

Reverse(s){ Loop Parse, s o := A_LoopField o return o }</lang> Sample output for 1127:

1 2+1 7+*
1 2+7 1+*
1 7+1 2+*
1 7+2 1+*
2 1+1 7+*
2 1+7 1+*
7 1+1 2+*
7 1+2 1+*

And for 8338:

8 3 8 3/-/

BBC BASIC

<lang bbcbasic>

     PROCsolve24("1234")
     PROCsolve24("6789")
     PROCsolve24("1127")
     PROCsolve24("5566")
     END
     
     DEF PROCsolve24(s$)
     LOCAL F%, I%, J%, K%, L%, P%, T%, X$, o$(), p$(), t$()
     DIM o$(4), p$(24,4), t$(11)
     o$() = "", "+", "-", "*", "/"
     RESTORE
     FOR T% = 1 TO 11
       READ t$(T%)
     NEXT
     DATA "abcdefg", "(abc)defg", "ab(cde)fg", "abcd(efg)", "(abc)d(efg)", "(abcde)fg"
     DATA "ab(cdefg)", "((abc)de)fg", "(ab(cde))fg", "ab((cde)fg)", "ab(cd(efg))"
     
     FOR I% = 1 TO 4
       FOR J% = 1 TO 4
         FOR K% = 1 TO 4
           FOR L% = 1 TO 4
             IF I%<>J% IF J%<>K% IF K%<>L% IF I%<>K% IF J%<>L% IF I%<>L% THEN
               P% += 1
               p$(P%,1) = MID$(s$,I%,1)
               p$(P%,2) = MID$(s$,J%,1)
               p$(P%,3) = MID$(s$,K%,1)
               p$(P%,4) = MID$(s$,L%,1)
             ENDIF
           NEXT
         NEXT
       NEXT
     NEXT
     
     FOR I% = 1 TO 4
       FOR J% = 1 TO 4
         FOR K% = 1 TO 4
           FOR T% = 1 TO 11
             FOR P% = 1 TO 24
               X$ = t$(T%)
               MID$(X$, INSTR(X$,"a"), 1) = p$(P%,1)
               MID$(X$, INSTR(X$,"b"), 1) = o$(I%)
               MID$(X$, INSTR(X$,"c"), 1) = p$(P%,2)
               MID$(X$, INSTR(X$,"d"), 1) = o$(J%)
               MID$(X$, INSTR(X$,"e"), 1) = p$(P%,3)
               MID$(X$, INSTR(X$,"f"), 1) = o$(K%)
               MID$(X$, INSTR(X$,"g"), 1) = p$(P%,4)
               F% = TRUE : ON ERROR LOCAL F% = FALSE
               IF F% IF EVAL(X$) = 24 THEN PRINT X$ : EXIT FOR I%
               RESTORE ERROR
             NEXT
           NEXT
         NEXT
       NEXT
     NEXT
     IF I% > 4 PRINT "No solution found"
     ENDPROC

</lang> Sample output:

(1+2+3)*4
6*8/(9-7)
(1+2)*(1+7)
(5+5-6)*6

C

This is a solver that's generic enough to deal with more than 4 numbers, goals other than 24, or different digit ranges. It guarantees a solution if there is one. Its output format is reasonably good looking, though not necessarily optimal. <lang C>#include <stdio.h>

1. include <stdlib.h>
2. include <time.h>

1. define n_cards 4
2. define solve_goal 24
3. define max_digit 9

typedef struct { int num, denom; } frac_t, *frac; typedef enum { C_NUM = 0, C_ADD, C_SUB, C_MUL, C_DIV, } op_type;

typedef struct expr_t *expr; typedef struct expr_t {

       op_type op;
       expr left, right;
       int value;

} expr_t;

void show_expr(expr e, op_type prec, int is_right) {

       char * op;
       switch(e->op) {
       case C_NUM:     printf("%d", e->value);
                       return;
       case C_ADD:     op = " + "; break;
       case C_SUB:     op = " - "; break;
       case C_MUL:     op = " x "; break;
       case C_DIV:     op = " / "; break;
       }

       if ((e->op == prec && is_right) || e->op < prec) printf("(");
       show_expr(e->left, e->op, 0);
       printf("%s", op);
       show_expr(e->right, e->op, 1);
       if ((e->op == prec && is_right) || e->op < prec) printf(")");

}

void eval_expr(expr e, frac f) {

       frac_t left, right;
       if (e->op == C_NUM) {
               f->num = e->value;
               f->denom = 1;
               return;
       }
       eval_expr(e->left, &left);
       eval_expr(e->right, &right);
       switch (e->op) {
       case C_ADD:
               f->num = left.num * right.denom + left.denom * right.num;
               f->denom = left.denom * right.denom;
               return;
       case C_SUB:
               f->num = left.num * right.denom - left.denom * right.num;
               f->denom = left.denom * right.denom;
               return;
       case C_MUL:
               f->num = left.num * right.num;
               f->denom = left.denom * right.denom;
               return;
       case C_DIV:
               f->num = left.num * right.denom;
               f->denom = left.denom * right.num;
               return;
       default:
               fprintf(stderr, "Unknown op: %d\n", e->op);
               return;
       }

} int solve(expr ex_in[], int len) {

       int i, j;
       expr_t node;
       expr ex[n_cards];
       frac_t final;

       if (len == 1) {
               eval_expr(ex_in[0], &final);
               if (final.num == final.denom * solve_goal && final.denom) {
                       show_expr(ex_in[0], 0, 0);
                       return 1;
               }
               return 0;
       }

       for (i = 0; i < len - 1; i++) {
               for (j = i + 1; j < len; j++)
                       ex[j - 1] = ex_in[j];
               ex[i] = &node;
               for (j = i + 1; j < len; j++) {
                       node.left = ex_in[i];
                       node.right = ex_in[j];
                       for (node.op = C_ADD; node.op <= C_DIV; node.op++)
                               if (solve(ex, len - 1))
                                       return 1;

                       node.left = ex_in[j];
                       node.right = ex_in[i];
                       node.op = C_SUB;
                       if (solve(ex, len - 1)) return 1;
                       node.op = C_DIV;
                       if (solve(ex, len - 1)) return 1;

                       ex[j] = ex_in[j];
               }
               ex[i] = ex_in[i];
       }

       return 0;

}

int solve24(int n[]) {

       int i;
       expr_t ex[n_cards];
       expr   e[n_cards];
       for (i = 0; i < n_cards; i++) {
               e[i] = ex + i;
               ex[i].op = C_NUM;
               ex[i].left = ex[i].right = 0;
               ex[i].value = n[i];
       }
       return solve(e, n_cards);

}

int main() {

       int i, j, n[] = { 3, 3, 8, 8, 9 };
       srand(time(0));

       for (j = 0; j < 10; j++) {
               for (i = 0; i < n_cards; i++) {
                       n[i] = 1 + (double) rand() * max_digit / RAND_MAX;
                       printf(" %d", n[i]);
               }
               printf(":  ");
               printf(solve24(n) ? "\n" : "No solution\n");
       }

       return 0;

}</lang>Sample output:

 1 8 2 1:  1 x 8 x (2 + 1)
 6 8 2 8:  6 + 8 + 2 + 8
 4 2 8 1:  (4 - 2 + 1) x 8
 3 1 9 9:  (9 - 1) / (3 / 9)
 5 7 5 1:  No solution
 5 8 4 1:  (5 + 1) x (8 - 4)
 8 3 4 9:  8 + 3 + 4 + 9
 3 7 4 4:  ((3 + 7) - 4) x 4
 5 6 4 1:  4 / (1 - 5 / 6)
 5 5 9 8:  5 x 5 - 9 + 8

For the heck of it, using seven numbers ranging from 0 to 99, trying to calculate 1:

 54 64 44 67 60 54 97:  (54 + 64 + 44) / 54 + 60 / (67 - 97)
 83 3 52 50 14 48 55:  55 - (((83 + 3 + 52) - 50 + 14) - 48)
 70 14 26 6 4 50 19:  ((70 + 14 + 26) / 4 - 19) x 6 - 50
 75 29 61 95 1 6 73:  6 / (73 - ((75 + 29 + 61) - 95)) - 1
 99 65 59 54 29 3 21:  3 - (99 + 65 + 54) / (59 + 29 + 21)
 88 57 18 72 60 70 22:  (72 - 70) x (60 + 22) - (88 + 57 + 18)
 73 18 76 44 32 3 49:  32 / (49 - (44 + 3)) - ((73 + 18) - 76)
 36 53 68 12 82 30 8:  ((36 + 53 + 68) - 82) / 30 - 12 / 8
 83 35 81 82 99 40 36:  ((83 + 35) x 81 - 82 x 99) / 40 / 36
 29 43 57 18 1 74 89:  (1 + 74) / (((29 + 43) - 57) / 18) - 89

Clojure

The code:

<lang lisp>(use 'clojure.contrib.combinatorics)

(defn nested-replace [l m] (cond (= l '()) '() (m (first l)) (concat (list (m (first l))) (nested-replace (rest l) m)) (seq? (first l)) (concat (list (nested-replace (first l) m)) (nested-replace (rest l) m)) true (concat (list (first l)) (nested-replace (rest l) m))))

(defn format-solution [sol] (cond (number? sol) sol (seq? sol)

   (list (format-solution (second sol)) (first sol) (format-solution (nth sol 2)))))

(defn play24 [& digits] (count (map #(-> % format-solution println) (let [operator-map-list (map (fn [a] {:op1 (nth a 0) :op2 (nth a 1) :op3 (nth a 2)})

      (selections '(* + - /) 3))
    digits-map-list 
      (map (fn [a] {:num1 (nth a 0) :num2 (nth a 1) :num3 (nth a 2) :num4 (nth a 3)}) 
        (permutations digits))
    patterns-list (list 
      '(:op1 (:op2 :num1 :num2) (:op3 :num3 :num4)) 
      '(:op1 :num1 (:op2 :num2 (:op3 :num3 :num4))))
      ;other patterns can be added here, e.g. '(:op1 (:op2 (:op3 :num1 :num2) :num3) :num4)
    op-subbed (reduce concat '() 
      (map (fn [a] (map #(nested-replace a % ) operator-map-list)) patterns-list))
    full-subbed (reduce concat '()
      (map (fn [a] (map #(nested-replace % a) op-subbed)) digits-map-list))] 
    (filter #(= (try (eval %) (catch Exception e nil)) 24) full-subbed)))))</lang>

The function play24 works by substituting the given digits and the four operations into the two binary tree patterns (o (o n n) (o n n)) and (o n (o n (o n n))). The substitution is the complex part of the program: two pairs of nested maps (the function) are used to sub in first the operations and then the digits, which are matched into the tree patterns using maps (the datatype.)

Example use:

user=> (play24 5 6 7 8)
((5 + 7) * (8 - 6))
(6 * (5 + (7 - 8)))
(6 * (5 - (8 - 7)))
(6 * (7 + (5 - 8)))
(6 * (7 - (8 - 5)))
((6 * 8) / (7 - 5))
(6 * (8 / (7 - 5)))
((7 + 5) * (8 - 6))
((8 - 6) * (5 + 7))
((8 - 6) * (7 + 5))
((8 * 6) / (7 - 5))
(8 * (6 / (7 - 5)))
12

CoffeeScript

<lang coffeescript>

1. This program tries to find some way to turn four digits into an arithmetic
2. expression that adds up to 24.
4. Example solution for 5, 7, 8, 8:
5. (((8 + 7) * 8) / 5)

solve_24_game = (digits...) ->

 # Create an array of objects for our helper functions
 arr = for digit in digits
   {
     val: digit
     expr: digit
   }
 combo4 arr...

combo4 = (a, b, c, d) ->

 arr = [a, b, c, d]
 # Reduce this to a three-node problem by combining two
 # nodes from the array.
 permutations = [
   [0, 1, 2, 3]
   [0, 2, 1, 3]
   [0, 3, 1, 2]
   [1, 2, 0, 3]
   [1, 3, 0, 2]
   [2, 3, 0, 1]
 ]
 for permutation in permutations
   [i, j, k, m] = permutation
   for combo in combos arr[i], arr[j]
     answer = combo3 combo, arr[k], arr[m]  
     return answer if answer
 null

combo3 = (a, b, c) ->

 arr = [a, b, c]
 permutations = [
   [0, 1, 2]
   [0, 2, 1]
   [1, 2, 0]
 ]
 for permutation in permutations
   [i, j, k] = permutation
   for combo in combos arr[i], arr[j]
     answer = combo2 combo, arr[k]
     return answer if answer
 null
 

combo2 = (a, b) ->

 for combo in combos a, b
   return combo.expr if combo.val == 24
 null
 

combos = (a, b) ->

 [
   val: a.val + b.val
   expr: "(#{a.expr} + #{b.expr})"
 ,
   val: a.val * b.val
   expr: "(#{a.expr} * #{b.expr})"
 ,
   val: a.val - b.val
   expr: "(#{a.expr} - #{b.expr})"
 ,
   val: b.val - a.val
   expr: "(#{b.expr} - #{a.expr})"
 ,
   val: a.val / b.val
   expr: "(#{a.expr} / #{b.expr})"
 ,
   val: b.val / a.val
   expr: "(#{b.expr} / #{a.expr})"
 ,
 ]
 

1. test

do ->

 rand_digit = -> 1 + Math.floor (9 * Math.random())

 for i in [1..15]
   a = rand_digit()
   b = rand_digit()
   c = rand_digit()
   d = rand_digit()
   solution = solve_24_game a, b, c, d
   console.log "Solution for #{[a,b,c,d]}: #{solution ? 'no solution'}"

</lang> output <lang> > coffee 24_game.coffee Solution for 8,3,1,8: ((1 + 8) * (8 / 3)) Solution for 6,9,5,7: (6 - ((5 - 7) * 9)) Solution for 4,2,1,1: no solution Solution for 3,5,1,3: (((3 + 5) * 1) * 3) Solution for 6,4,1,7: ((7 - (4 - 1)) * 6) Solution for 8,1,3,1: (((8 + 1) - 1) * 3) Solution for 6,1,3,3: (((6 + 1) * 3) + 3) Solution for 7,1,5,6: (((7 - 1) * 5) - 6) Solution for 4,2,3,1: ((3 + 1) * (4 + 2)) Solution for 8,8,5,8: ((5 * 8) - (8 + 8)) Solution for 3,8,4,1: ((1 - (3 - 8)) * 4) Solution for 6,4,3,8: ((8 - (6 / 3)) * 4) Solution for 2,1,8,7: (((2 * 8) + 1) + 7) Solution for 5,2,7,5: ((2 * 7) + (5 + 5)) Solution for 2,4,8,9: ((9 - (2 + 4)) * 8) </lang>

Common Lisp

<lang lisp>(defconstant +ops+ '(* / + -))

(defun digits ()

 (sort (loop repeat 4 collect (1+ (random 9))) #'<))

(defun expr-value (expr)

 (eval expr))

(defun divides-by-zero-p (expr)

 (when (consp expr)
   (destructuring-bind (op &rest args) expr
     (or (divides-by-zero-p (car args))
         (and (eq op '/)
              (or (and (= 1 (length args))
                       (zerop (expr-value (car args))))
                  (some (lambda (arg)
                          (or (divides-by-zero-p arg)
                              (zerop (expr-value arg))))
                        (cdr args))))))))

(defun solvable-p (digits &optional expr)

 (unless (divides-by-zero-p expr)
   (if digits
       (destructuring-bind (next &rest rest) digits
         (if expr
             (some (lambda (op)
                     (solvable-p rest (cons op (list next expr))))
                   +ops+)
           (solvable-p rest (list (car +ops+) next))))
     (when (and expr
                (eql 24 (expr-value expr)))
       (merge-exprs expr)))))

(defun merge-exprs (expr)

 (if (atom expr)
     expr
   (destructuring-bind (op &rest args) expr
     (if (and (member op '(* +))
              (= 1 (length args)))
         (car args)
       (cons op
             (case op
               ((* +)
                (loop for arg in args
                      for merged = (merge-exprs arg)
                      when (and (consp merged)
                                (eq op (car merged)))
                      append (cdr merged)
                      else collect merged))
               (t (mapcar #'merge-exprs args))))))))

(defun solve-24-game (digits)

 "Generate a lisp form using the operators in +ops+ and the given

digits which evaluates to 24. The first form found is returned, or NIL if there is no solution."

 (solvable-p digits))</lang>

Example Output:

CL-USER 138 > (loop repeat 24 for soln = (solve-24-game (digits)) when soln do (pprint soln))

(+ 7 5 (* 4 3))
(* 6 4 (- 3 2))
(+ 9 8 4 3)
(* 8 (- 6 (* 3 1)))
(* 6 4 (/ 2 2))
(* 9 (/ 8 (- 8 5)))
NIL

D

This uses the Rational struct and permutations functions of two other Rosetta Code Tasks.

Translation of: Scala

<lang d>import std.stdio, std.algorithm, std.range, std.typecons, std.conv,

      std.string, rational, perm;

string solve(in int target, in int[] L) {

 static struct computeAllOperations {
   alias Tuple!(Rational,"r", string,"e") T;
   Rational[] L;

   int opApply(int delegate(ref T) dg) {
     int result;

     if (L.length) {
       auto x = L[0];
       auto xs = L[1 .. $];
       if (L.length == 1) {
         T aux = T(x, text(x));
         result = dg(aux);
       } else {
         OUTER: foreach (o; computeAllOperations(xs)) {
           auto y = o.r;
           T[] sub = (y == 0) ?
             [T(x*y,"*"), T(x+y,"+"), T(x-y,"-")] :
             [T(x*y,"*"), T(x/y,"/"), T(x+y,"+"), T(x-y,"-")];
           foreach (e; sub) {
             auto aux = T(e.r, format("(%s%s%s)", x, e.e, o.e));
             result = dg(aux); if (result) break OUTER;
           }
         }
       }
     }

     return result;
   }
 }

 foreach (p; permutations(array(map!Rational(L))))
   foreach (sol; computeAllOperations(p))
     if (sol.r == target)
       return sol.e;
 return "No solution";

}

void main() {

 foreach (p; [[6, 7, 9, 5], [3, 3, 8, 8], [1, 1, 1, 1]])
   writeln(p, ": ", solve(24, p));

}</lang> Output:

[6, 7, 9, 5]: (6+(9*(7-5)))
[3, 3, 8, 8]: (8/(3-(8/3)))
[1, 1, 1, 1]: No solution

Fortran

<lang Fortran>program solve_24

 use helpers
 implicit none
 real                 :: vector(4), reals(4), p, q, r, s
 integer              :: numbers(4), n, i, j, k, a, b, c, d
 character, parameter :: ops(4) = (/ '+', '-', '*', '/' /)
 logical              :: last
 real,parameter       :: eps = epsilon(1.0)

 do n=1,12
   call random_number(vector)
   reals   = 9 * vector + 1
   numbers = int(reals)
   call Insertion_Sort(numbers)
   
   permutations: do
     a = numbers(1); b = numbers(2); c = numbers(3); d = numbers(4)
     reals = real(numbers)
     p = reals(1);   q = reals(2);   r = reals(3);   s = reals(4)
     ! combinations of operators:
     do i=1,4
       do j=1,4
         do k=1,4
           if      ( abs(op(op(op(p,i,q),j,r),k,s)-24.0) < eps ) then
             write (*,*) numbers, ' : ', '((',a,ops(i),b,')',ops(j),c,')',ops(k),d
             exit permutations
           else if ( abs(op(op(p,i,op(q,j,r)),k,s)-24.0) < eps ) then
             write (*,*) numbers, ' : ', '(',a,ops(i),'(',b,ops(j),c,'))',ops(k),d
             exit permutations
           else if ( abs(op(p,i,op(op(q,j,r),k,s))-24.0) < eps ) then
             write (*,*) numbers, ' : ', a,ops(i),'((',b,ops(j),c,')',ops(k),d,')'
             exit permutations
           else if ( abs(op(p,i,op(q,j,op(r,k,s)))-24.0) < eps ) then
             write (*,*) numbers, ' : ', a,ops(i),'(',b,ops(j),'(',c,ops(k),d,'))'
             exit permutations
           else if ( abs(op(op(p,i,q),j,op(r,k,s))-24.0) < eps ) then
             write (*,*) numbers, ' : ', '(',a,ops(i),b,')',ops(j),'(',c,ops(k),d,')'
             exit permutations
           end if
         end do
       end do
     end do
     call nextpermutation(numbers,last)  
     if ( last ) then
       write (*,*) numbers, ' : no solution.'
       exit permutations
     end if
   end do permutations

 end do

contains

 pure real function op(x,c,y)
   integer, intent(in) :: c
   real, intent(in)    :: x,y
   select case ( ops(c) )
     case ('+')
       op = x+y
     case ('-')
       op = x-y
     case ('*')
       op = x*y
     case ('/')
       op = x/y
   end select
 end function op

end program solve_24</lang>

<lang Fortran>module helpers

contains

 pure subroutine Insertion_Sort(a)
   integer, intent(inout) :: a(:)
   integer                :: temp, i, j
   do i=2,size(a)
     j = i-1
     temp = a(i)
     do while ( j>=1 .and. a(j)>temp )
       a(j+1) = a(j)
       j = j - 1
     end do
     a(j+1) = temp
   end do
 end subroutine Insertion_Sort

 subroutine nextpermutation(perm,last)
   integer, intent(inout) :: perm(:)
   logical, intent(out)   :: last
   integer :: k,l
   k = largest1()
   last = k == 0
   if ( .not. last ) then    
     l = largest2(k)
     call swap(l,k)
     call reverse(k)
   end if
 contains
   pure integer function largest1()
     integer :: k, max
     max = 0
     do k=1,size(perm)-1
       if ( perm(k) < perm(k+1) ) then
         max = k
       end if
     end do
     largest1 = max
   end function largest1

   pure integer function largest2(k)
     integer, intent(in) :: k
     integer             :: l, max
     max = k+1
     do l=k+2,size(perm)
       if ( perm(k) < perm(l) ) then
         max = l
       end if
     end do
     largest2 = max
   end function largest2

   subroutine swap(l,k)
     integer, intent(in) :: k,l
     integer             :: temp
     temp    = perm(k)
     perm(k) = perm(l)
     perm(l) = temp
   end subroutine swap
   
   subroutine reverse(k)
     integer, intent(in) :: k
     integer             :: i
     do i=1,(size(perm)-k)/2
       call swap(k+i,size(perm)+1-i)
     end do
   end subroutine reverse
   
 end subroutine nextpermutation

end module helpers</lang>

Sample output (using g95):

 3 6 7 9  :  3 *(( 6 - 7 )+ 9 )
 3 9 5 8  : (( 3 * 9 )+ 5 )- 8
 4 5 6 9  : (( 4 + 5 )+ 6 )+ 9
 2 9 9 8  : ( 2 +( 9 / 9 ))* 8
 1 4 7 5  : ( 1 +( 4 * 7 ))- 5
 8 7 7 6  : no solution.
 3 3 8 9  : ( 3 *( 3 + 8 ))- 9
 1 5 6 7  : ( 1 +( 5 * 6 ))- 7
 2 3 5 3  :  2 *(( 3 * 5 )- 3 )
 4 5 6 9  : (( 4 + 5 )+ 6 )+ 9
 1 1 3 6  : ( 1 +( 1 * 3 ))* 6
 2 4 6 8  : (( 2 / 4 )* 6 )* 8

GAP

<lang gap># Solution in RPN check := function(x, y, z) local r, c, s, i, j, k, a, b, p; i := 0; j := 0; k := 0; s := [ ]; r := ""; for c in z do if c = 'x' then i := i + 1; k := k + 1; s[k] := x[i]; Append(r, String(x[i])); else j := j + 1; b := s[k]; k := k - 1; a := s[k]; p := y[j]; r[Size(r) + 1] := p; if p = '+' then a := a + b; elif p = '-' then a := a - b; elif p = '*' then a := a * b; elif p = '/' then if b = 0 then continue; else a := a / b; fi; else return fail; fi; s[k] := a; fi; od; if s[1] = 24 then return r; else return fail; fi; end;

Player24 := function(digits) local u, v, w, x, y, z, r; u := PermutationsList(digits); v := Tuples("+-*/", 3); w := ["xx*x*x*", "xx*xx**", "xxx**x*", "xxx*x**", "xxxx***"]; for x in u do for y in v do for z in w do r := check(x, y, z); if r <> fail then return r; fi; od; od; od; return fail; end;

Player24([1,2,7,7]);

1. "77*1-2/"

Player24([9,8,7,6]);

1. "68*97-/"

Player24([1,1,7,7]);

1. fail

1. Solutions with only one distinct digit are found only for 3, 4, 5, 6:

Player24([3,3,3,3]);

1. "33*3*3-"

Player24([4,4,4,4]);

1. "44*4+4+"

Player24([5,5,5,5]);

1. "55*55/-"

Player24([6,6,6,6]);

1. "66*66+-"

1. A tricky one:

Player24([3,3,8,8]); "8383/-/"</lang>

Go

<lang go>package main

import ( "fmt" "math/rand" "time" )

const ( op_num = iota op_add op_sub op_mul op_div )

type frac struct { num, denom int }

// Expression: can either be a single number, or a result of binary // operation from left and right node type Expr struct { op int left, right *Expr value frac }

var n_cards = 4 var goal = 24 var digit_range = 9

func (x *Expr) String() string { if x.op == op_num { return fmt.Sprintf("%d", x.value.num) }

var bl1, br1, bl2, br2, opstr string switch { case x.left.op == op_num: case x.left.op >= x.op: case x.left.op == op_add && x.op == op_sub: bl1, br1 = "", "" default: bl1, br1 = "(", ")" }

if x.right.op == op_num || x.op < x.right.op { bl2, br2 = "", "" } else { bl2, br2 = "(", ")" }

switch { case x.op == op_add: opstr = " + " case x.op == op_sub: opstr = " - " case x.op == op_mul: opstr = " * " case x.op == op_div: opstr = " / " }

return bl1 + x.left.String() + br1 + opstr + bl2 + x.right.String() + br2 }

func expr_eval(x *Expr) (f frac) { if x.op == op_num { return x.value }

l, r := expr_eval(x.left), expr_eval(x.right)

switch { case x.op == op_add: f.num = l.num*r.denom + l.denom*r.num f.denom = l.denom * r.denom return

case x.op == op_sub: f.num = l.num*r.denom - l.denom*r.num f.denom = l.denom * r.denom return

case x.op == op_mul: f.num = l.num * r.num f.denom = l.denom * r.denom return

case x.op == op_div: f.num = l.num * r.denom f.denom = l.denom * r.num return } return }

func solve(ex_in []*Expr) bool { // only one expression left, meaning all numbers are arranged into // a binary tree, so evaluate and see if we get 24 if len(ex_in) == 1 { f := expr_eval(ex_in[0]) if f.denom != 0 && f.num == f.denom*goal { fmt.Println(ex_in[0].String()) return true } return false }

var node Expr ex := make([]*Expr, len(ex_in)-1)

// try to combine a pair of expressions into one, thus reduce // the list length by 1, and recurse down for i := range ex { copy(ex[i:len(ex)], ex_in[i+1:len(ex_in)])

ex[i] = &node for j := i + 1; j < len(ex_in); j++ { node.left = ex_in[i] node.right = ex_in[j]

// try all 4 operators for o := op_add; o <= op_div; o++ { node.op = o if solve(ex) { return true } }

// also - and / are not commutative, so swap arguments node.left = ex_in[j] node.right = ex_in[i]

node.op = op_sub if solve(ex) { return true }

node.op = op_div if solve(ex) { return true }

if j < len(ex) { ex[j] = ex_in[j] } } ex[i] = ex_in[i] } return false }

func main() { cards := make([]*Expr, n_cards) rand.Seed(time.Now().Unix())

for k := 0; k < 10; k++ { for i := 0; i < n_cards; i++ { cards[i] = &Expr{op_num, nil, nil, frac{rand.Intn(digit_range-1) + 1, 1}} fmt.Printf(" %d", cards[i].value.num) } fmt.Print(": ") if !solve(cards) { fmt.Println("No solution") } } }</lang> Output:

 8 6 7 6:  No solution
 7 2 6 6:  (7 - 2) * 6 - 6
 4 8 7 3:  4 * (7 - 3) + 8
 3 8 8 7:  3 * 8 * (8 - 7)
 5 7 3 7:  No solution
 5 7 8 3:  5 * 7 - 8 - 3
 3 6 5 2:  ((3 + 5) * 6) / 2
 8 4 5 4:  (8 - 4) * 5 + 4
 2 2 8 8:  (2 + 2) * 8 - 8
 6 8 8 2:  6 + 8 + 8 + 2

Haskell

<lang haskell>import Data.List import Data.Ratio import Control.Monad import System.Environment (getArgs)

data Expr = Constant Rational |

   Expr :+ Expr | Expr :- Expr |
   Expr :* Expr | Expr :/ Expr
   deriving (Eq)

ops = [(:+), (:-), (:*), (:/)]

instance Show Expr where

   show (Constant x) = show $ numerator x
     -- In this program, we need only print integers.
   show (a :+ b)     = strexp "+" a b
   show (a :- b)     = strexp "-" a b
   show (a :* b)     = strexp "*" a b
   show (a :/ b)     = strexp "/" a b

strexp :: String -> Expr -> Expr -> String strexp op a b = "(" ++ show a ++ " " ++ op ++ " " ++ show b ++ ")"

templates :: [[Expr] -> Expr] templates = do

   op1 <- ops
   op2 <- ops
   op3 <- ops
   [\[a, b, c, d] -> op1 a $ op2 b $ op3 c d,
    \[a, b, c, d] -> op1 (op2 a b) $ op3 c d,
    \[a, b, c, d] -> op1 a $ op2 (op3 b c) d,
    \[a, b, c, d] -> op1 (op2 a $ op3 b c) d,
    \[a, b, c, d] -> op1 (op2 (op3 a b) c) d]

eval :: Expr -> Maybe Rational eval (Constant c) = Just c eval (a :+ b) = liftM2 (+) (eval a) (eval b) eval (a :- b) = liftM2 (-) (eval a) (eval b) eval (a :* b) = liftM2 (*) (eval a) (eval b) eval (a :/ b) = do

   denom <- eval b
   guard $ denom /= 0
   liftM (/ denom) $ eval a

solve :: Rational -> [Rational] -> [Expr] solve target r4 = filter (maybe False (== target) . eval) $

   liftM2 ($) templates $
   nub $ permutations $ map Constant r4 

main = getArgs >>= mapM_ print . solve 24 . map (toEnum . read)</lang>

Example use:

$ runghc 24Player.hs 2 3 8 9
(8 * (9 - (3 * 2)))
(8 * (9 - (2 * 3)))
((9 - (2 * 3)) * 8)
((9 - (3 * 2)) * 8)
((9 - 3) * (8 / 2))
((8 / 2) * (9 - 3))
(8 * ((9 - 3) / 2))
(((9 - 3) / 2) * 8)
((9 - 3) / (2 / 8))
((8 * (9 - 3)) / 2)
(((9 - 3) * 8) / 2)
(8 / (2 / (9 - 3)))

Icon and Unicon

This shares code with and solves the 24 game. A series of pattern expressions are built up and then populated with the permutations of the selected digits. Equations are skipped if they have been seen before. The procedure 'eval' was modified to catch zero divides. The solution will find either all occurrences or just the first occurrence of a solution.

<lang Icon>invocable all link strings # for csort, deletec, permutes

procedure main() static eL initial {

  eoP := []  # set-up expression and operator permutation patterns
  every ( e := !["a@b#c$d", "a@(b#c)$d", "a@b#(c$d)", "a@(b#c$d)", "a@(b#(c$d))"] ) & 
        ( o := !(opers := "+-*/") || !opers || !opers ) do
     put( eoP, map(e,"@#$",o) )    # expr+oper perms
  
  eL := []   # all cases
  every ( e := !eoP ) & ( p := permutes("wxyz") ) do
     put(eL, map(e,"abcd",p))

  }

write("This will attempt to find solutions to 24 for sets of numbers by\n",

     "combining 4 single digits between 1 and 9 to make 24 using only + - * / and ( ).\n",
     "All operations have equal precedence and are evaluated left to right.\n",
     "Enter 'use n1 n2 n3 n4' or just hit enter (to use a random set),",
     "'first'/'all' shows the first or all solutions, 'quit' to end.\n\n")

repeat {

  e := trim(read()) | fail
  e ?  case tab(find(" ")|0) of {
     "q"|"quit" : break
     "u"|"use"  : e := tab(0)
     "f"|"first": first := 1 & next
     "a"|"all"  : first := &null & next
     ""         : e := " " ||(1+?8) || " " || (1+?8) ||" " || (1+?8) || " " || (1+?8)
     }
     
  writes("Attempting to solve 24 for",e)
  
  e := deletec(e,' \t') # no whitespace   
  if e ? ( tab(many('123456789')), pos(5), pos(0) ) then 
     write(":")
  else write(" - invalid, only the digits '1..9' are allowed.") & next  
  
  eS := set()
  every ex := map(!eL,"wxyz",e) do {
     if member(eS,ex) then next # skip duplicates of final expression
     insert(eS,ex)
     if ex ? (ans := eval(E()), pos(0)) then # parse and evaluate
        if ans = 24 then {
           write("Success ",image(ex)," evaluates to 24.")
           if \first then break
           }
     }
  }

write("Quiting.") end

procedure eval(X) #: return the evaluated AST

  if type(X) == "list" then {
     x := eval(get(X)) 
     while o := get(X) do 
        if y := get(X) then
           x := o( real(x), (o ~== "/" | fail, eval(y) ))
        else write("Malformed expression.") & fail
  }
  return \x | X

end

procedure E() #: expression

  put(lex := [],T())
  while put(lex,tab(any('+-*/'))) do
     put(lex,T())  
  suspend if *lex = 1 then lex[1] else lex     # strip useless []  

end

procedure T() #: Term

  suspend 2(="(", E(), =")") | # parenthesized subexpression, or ...
      tab(any(&digits))        # just a value

end</lang>

Library: Icon Programming Library

strings.icn provides deletec and permutes

J

<lang J>perm=: (A.&i.~ !) 4 ops=: ' ',.'+-*%' {~ >,{i.each 4 4 4 cmask=: 1 + 0j1 * i.@{:@$@[ e. ] left=: [ #!.'('~"1 cmask right=: [ #!.')'~"1 cmask paren=: 2 :'[: left&m right&n' parens=: ], 0 paren 3, 0 paren 5, 2 paren 5, [: 0 paren 7 (0 paren 3) all=: [: parens [:,/ ops ,@,."1/ perm { [:;":each answer=: ({.@#~ 24 = ".)@all</lang>

This implementation tests all 7680 candidate sentences.

Example use:

   answer 2 3 5 7
 2+7+3*5  
   answer 8 4 7 1
 8*7-4*1 
  answer 1 1 2 7
(1+2)*1+7  

The answer will be either a suitable J sentence or blank if none can be found. "J sentence" means that, for example, the sentence 8*7-4*1 is equivalent to the sentence 8*(7-(4*1)). [Many infix languages use operator precedence to make polynomials easier to express without parenthesis, but J has other mechanisms for expressing polynomials and minimal operator precedence makes the language more regular.]

JavaScript

This is a translation of the C code. <lang javascript>var ar=[],order=[0,1,2],op=[],val=[]; var NOVAL=9999,oper="+-*/",out;

function rnd(n){return Math.floor(Math.random()*n)}

function say(s){

try{document.write(s+"
")}
catch(e){WScript.Echo(s)}

}

function getvalue(x,dir){

var r=NOVAL;
if(dir>0)++x;
while(1){
 if(val[x]!=NOVAL){
  r=val[x];
  val[x]=NOVAL;
  break;
 }
 x+=dir;
}
return r*1;

}

function calc(){

var c=0,l,r,x;
val=ar.join('/').split('/');
while(c<3){
 x=order[c];
 l=getvalue(x,-1);
 r=getvalue(x,1);
 switch(op[x]){
  case 0:val[x]=l+r;break;
  case 1:val[x]=l-r;break;
  case 2:val[x]=l*r;break;
  case 3:
  if(!r||l%r)return 0;
  val[x]=l/r;
 }
 ++c;
}
return getvalue(-1,1);

}

function shuffle(s,n){

var x=n,p=eval(s),r,t;
while(x--){
 r=rnd(n);
 t=p[x];
 p[x]=p[r];
 p[r]=t;
}

}

function parenth(n){

while(n>0)--n,out+='(';
while(n<0)++n,out+=')';

}

function getpriority(x){

for(var z=3;z--;)if(order[z]==x)return 3-z;
return 0;

}

function showsolution(){

var x=0,p=0,lp=0,v=0;
while(x<4){
 if(x<3){
  lp=p;
  p=getpriority(x);
  v=p-lp;
  if(v>0)parenth(v);
 }
 out+=ar[x];
 if(x<3){
  if(v<0)parenth(v);
  out+=oper.charAt(op[x]);
 }
 ++x;
}
parenth(-p);
say(out);

}

function solve24(s){

var z=4,r;
while(z--)ar[z]=s.charCodeAt(z)-48;
out="";
for(z=100000;z--;){
 r=rnd(256);
 op[0]=r&3;
 op[1]=(r>>2)&3;
 op[2]=(r>>4)&3;
 shuffle("ar",4);
 shuffle("order",3);
 if(calc()!=24)continue;
 showsolution();
 break;
}

}

solve24("1234"); solve24("6789"); solve24("1127");</lang>

Examples:

(((3*1)*4)*2)
((6*8)/((9-7)))
(((1+7))*(2+1))

Liberty BASIC

<lang lb>dim d(4) input "Enter 4 digits: "; a$ nD=0 for i =1 to len(a$)

   c$=mid$(a$,i,1)
   if instr("123456789",c$) then
       nD=nD+1
       d(nD)=val(c$)
   end if

next 'for i = 1 to 4 ' print d(i); 'next

'precompute permutations. Dumb way. nPerm = 1*2*3*4 dim perm(nPerm, 4) n = 0 for i = 1 to 4

   for j = 1 to 4
       for k = 1 to 4
           for l = 1 to 4
           'valid permutation (no dupes?)
               if i<>j and i<>k and i<>l _
                   and j<>k and j<>l _
                       and k<>l then
                   n=n+1
                   '

' perm(n,1)=i ' perm(n,2)=j ' perm(n,3)=k ' perm(n,4)=l

                   'actually, we can as well permute given digits
                   perm(n,1)=d(i)
                   perm(n,2)=d(j)
                   perm(n,3)=d(k)
                   perm(n,4)=d(l)
               end if
           next
       next
   next

next 'check if permutations look OK. They are 'for i =1 to n ' print i, ' for j =1 to 4: print perm(i,j);:next ' print 'next

'possible brackets NBrackets = 11 dim Brakets$(NBrackets) DATA "4#4#4#4" DATA "(4#4)#4#4" DATA "4#(4#4)#4" DATA "4#4#(4#4)" DATA "(4#4)#(4#4)" DATA "(4#4#4)#4" DATA "4#(4#4#4)" DATA "((4#4)#4)#4" DATA "(4#(4#4))#4" DATA "4#((4#4)#4)" DATA "4#(4#(4#4))" for i = 1 to NBrackets

   read Tmpl$: Brakets$(i) = Tmpl$

next

'operations: full search count = 0 Ops$="+ - * /" dim Op$(3) For op1=1 to 4

   Op$(1)=word$(Ops$,op1)
   For op2=1 to 4
       Op$(2)=word$(Ops$,op2)
       For op3=1 to 4
           Op$(3)=word$(Ops$,op3)
           'print "*"
           'substitute all brackets
           for t = 1 to NBrackets
               Tmpl$=Brakets$(t)
               'print , Tmpl$
               'now, substitute all digits: permutations.
               for p = 1 to nPerm
                   res$= ""
                   nOp=0
                   nD=0
                   for i = 1 to len(Tmpl$)
                       c$ = mid$(Tmpl$, i, 1)
                       select case c$
                       case "#"                'operations
                           nOp = nOp+1
                           res$ = res$+Op$(nOp)
                       case "4"                'digits
                           nD = nOp+1
                           res$ = res$; perm(p,nD)
                       case else               'brackets goes here
                           res$ = res$+ c$ 
                       end select
                   next
                   'print,, res$
                   'eval here
                   if evalWithErrCheck(res$) = 24 then
                       print "24 = ";res$
                       end 'comment it out if you want to see all versions
                   end if
                   count = count + 1
               next
           next
       Next
   Next

next

print "If you see this, probably task cannot be solved with these digits" 'print count end

function evalWithErrCheck(expr$)

   on error goto [handler]
   evalWithErrCheck=eval(expr$)
   exit function

[handler] end function</lang>

Lua

Generic solver: pass card of any size with 1st argument and target number with second.

<lang lua> local SIZE = #arg[1] local GOAL = tonumber(arg[2]) or 24

local input = {} for v in arg[1]:gmatch("%d") do table.insert(input, v) end assert(#input == SIZE, 'Invalid input')

local operations = {'+', '-', '*', '/'}

local function BinaryTrees(vert) if vert == 0 then return {false} else local buf = {} for leften = 0, vert - 1 do local righten = vert - leften - 1 for _, left in pairs(BinaryTrees(leften)) do for _, right in pairs(BinaryTrees(righten)) do table.insert(buf, {left, right}) end end end return buf end end local trees = BinaryTrees(SIZE-1) local c, opc, oper, str local max = math.pow(#operations, SIZE-1) local function op(a,b) opc = opc + 1 local i = math.floor(oper/math.pow(#operations, opc-1))%#operations+1 return '('.. a .. operations[i] .. b ..')' end

local function EvalTree(tree) if tree == false then c = c + 1 return input[c-1] else return op(EvalTree(tree[1]), EvalTree(tree[2])) end end

local function printResult() for _, v in ipairs(trees) do for i = 0, max do c, opc, oper = 1, 0, i str = EvalTree(v) loadstring('res='..str)() if(res == GOAL) then print(str, '=', res) end end end end

local uniq = {} local function permgen (a, n) if n == 0 then local str = table.concat(a) if not uniq[str] then printResult() uniq[str] = true end else for i = 1, n do a[n], a[i] = a[i], a[n] permgen(a, n - 1) a[n], a[i] = a[i], a[n] end end end

permgen(input, SIZE) </lang>

Output:

$ lua 24game.solve.lua 2389
(8*(9-(3*2)))	=	24
(8*((9-3)/2))	=	24
((8*(9-3))/2)	=	24
((9-3)*(8/2))	=	24
(((9-3)*8)/2)	=	24
(8*(9-(2*3)))	=	24
(8/(2/(9-3)))	=	24
((8/2)*(9-3))	=	24
((9-3)/(2/8))	=	24
((9-(3*2))*8)	=	24
(((9-3)/2)*8)	=	24
((9-(2*3))*8)	=	24
$ lua 24game.solve.lua 1172
((1+7)*(2+1))	=	24
((7+1)*(2+1))	=	24
((1+2)*(7+1))	=	24
((2+1)*(7+1))	=	24
((1+2)*(1+7))	=	24
((2+1)*(1+7))	=	24
((1+7)*(1+2))	=	24
((7+1)*(1+2))	=	24
$ lua 24game.solve.lua 123456789 1000
(2*(3+(4-(5+(6-(7*(8*(9*1))))))))	=	1000
(2*(3+(4-(5+(6-(7*(8*(9/1))))))))	=	1000
(2*(3*(4*(5+(6*(7-(8/(9*1))))))))	=	1000
(2*(3*(4*(5+(6*(7-(8/(9/1))))))))	=	1000
(2*(3+(4-(5+(6-(7*((8*9)*1)))))))	=	1000
(2*(3+(4-(5+(6-(7*((8*9)/1)))))))	=	1000
(2*(3*(4*(5+(6*(7-((8/9)*1)))))))	=	1000
(2*(3*(4*(5+(6*(7-((8/9)/1)))))))	=	1000
.....

Mathematica

The code: <lang Mathematica> treeR[n_] := Table[o[trees[a], trees[n - a]], {a, 1, n - 1}] treeR[1] := n tree[n_] :=

Flatten[treeR[n] //. {o[a_List, b_] :> (o[#, b] & /@ a), 
   o[a_, b_List] :> (o[a, #] & /@ b)}]

game24play[val_List] :=

Union[StringReplace[StringTake[ToString[#, InputForm], {10, -2}], 
    "-1*" ~~ n_ :> "-" <> n] & /@ (HoldForm /@ 
     Select[Union@
       Flatten[Outer[# /. {o[q_Integer] :> #2q, 
            n[q_] :> #3q} &, 
         Block[{O = 1, N = 1}, # /. {o :> o[O++], n :> n[N++]}] & /@ 
          tree[4], Tuples[{Plus, Subtract, Times, Divide}, 3], 
         Permutations[Array[v, 4]], 1]], 
      Quiet[(# /. v[q_] :> valq) == 24] &] /. 
    Table[v[q] -> valq, {q, 4}])]</lang>

The treeR method recursively computes all possible operator trees for a certain number of inputs. It does this by tabling all combinations of distributions of inputs across the possible values. (For example, treeR[4] is allotted 4 inputs, so it returns {o[treeR[3],treeR[1]],o[treeR[2],treeR[2]],o[treeR[1],treeR[3]]}, where o is the operator (generic at this point). The base case treeR[1] returns n (the input). The final output of tree[4] (the 24 game has 4 random inputs) (tree cleans up the output of treeR) is:

{o[n, o[n, o[n, n]]],
 o[n, o[o[n, n], n]],
 o[o[n, n], o[n, n]], 
 o[o[n, o[n, n]], n],
 o[o[o[n, n], n], n]}

game24play takes the four random numbers as input and does the following (the % refers to code output from previous bullets):

• Block[{O = 1, N = 1}, # /. {o :> o[O++], n :> n[N++]}] & /@ tree[4]
  • Assign ascending numbers to the input and operator placeholders.
  • Ex: o[1][o[2][n[1], n[2]], o[3][n[3], n[4]]]
• Tuples[{Plus, Subtract, Times, Divide}, 3]
  • Find all combinations (Tuples allows repeats) of the four allowed operations.
  • Ex: {{Plus, Plus, Plus}, {Plus, Plus, Subtract}, <<60>>, {Divide, Divide, Times}, {Divide, Divide, Divide}}
• Permutations[Array[v, 4]]
  • Find all permutations (Permutations does not allow repeats) of the four given values.
  • Ex: {{v[1],v[2],v[3],v[4]}, {v[1],v[2],v[4],v[3]}, <<20>>, {v[4],v[3],v[1],v[2]}, {v[4],v[3],v[2],v[1]}}
• Outer[# /. {o[q_Integer] :> #2[[q]], n[q_] :> #3[[q]]} &, %%%, %%, %, 1]
  • Perform an outer join on the three above lists (every combination of each element) and with each combination put into the first (the operator tree) the second (the operation at each level) and the third (the value indexes, not actual values).
  • Ex: v[1] + v[2] - v[3] + v[4]
• Union@Flatten[%]
  • Get rid of any sublists caused by Outer and remove any duplicates (Union).
• Select[%, Quiet[(# /. v[q_] :> val[[q]]) == 24] &]
  • Select the elements of the above list where substituting the real values returns 24 (and do it Quietly because of div-0 concerns).
• HoldForm /@ % /. Table[v[q] -> val[[q]], {q, 4}]
  • Apply HoldForm so that substituting numbers will not cause evaluation (otherwise it would only ever return lists like {24, 24, 24}!) and substitute the numbers in.
• Union[StringReplace[StringTake[ToString[#, InputForm], {10, -2}], "-1*" ~~ n_ :> "-" <> n] & /@ %]
  • For each result, turn the expression into a string (for easy manipulation), strip the "HoldForm" wrapper, replace numbers like "-1*7" with "-7" (a idiosyncrasy of the conversion process), and remove any lingering duplicates. Some duplicates will still remain, notably constructs like "3 - 3" vs. "-3 + 3" and trivially similar expressions like "(8*3)*(6-5)" vs "(8*3)/(6-5)". Example run input and outputs:

<lang Mathematica>game24play[RandomInteger[{1, 9}, 4]]</lang>

Example outputs

{7, 2, 9, 5}
{-2 - 9 + 7*5}

{7, 5, 6, 2}
{6*(7 - 5 + 2), (7 - 5)*6*2, 7 + 5 + 6*2}

{7, 6, 7, 7}
{}

{3, 7, 6, 1}
{(-3 + 6)*(7 + 1), ((-3 + 7)*6)/1, (-3 + 7)*6*1,
 6 - 3*(-7 + 1), 6*(-3 + 7*1), 6*(-3 + 7/1),
 6 + 3*(7 - 1), 6*(7 - 3*1), 6*(7 - 3/1), 7 + 3*6 - 1}

Note that although this program is designed to be extensible to higher numbers of inputs, the largest working set in the program (the output of the Outer function can get very large:

• tree[n] returns a list with the length being the (n-1)-th Catalan number.
• Tuples[{Plus, Subtract, Times, Divide}, 3] has fixed length 64 (or p³ for p operations).
• Permutations[Array[v, n]] returns $n!$ permutations.

Therefore, the size of the working set is $64\cdot n!\,C_{n-1}=64\cdot (n-1)!!!!=64{\frac {(2n-2)!}{(n-1)!}}$, where $n!!!!$ is the quadruple factorial. It goes without saying that this number increases very fast. For this game, the total is 7680 elements. For higher numbers of inputs, it is {7 680, 107 520, 1 935 360, 42 577 920, 1 107 025 920, ...}.

OCaml

<lang ocaml>type expression =

 | Const of float
 | Sum  of expression * expression   (* e1 + e2 *)
 | Diff of expression * expression   (* e1 - e2 *)
 | Prod of expression * expression   (* e1 * e2 *)
 | Quot of expression * expression   (* e1 / e2 *)

let rec eval = function

 | Const c -> c
 | Sum (f, g) -> eval f +. eval g
 | Diff(f, g) -> eval f -. eval g
 | Prod(f, g) -> eval f *. eval g
 | Quot(f, g) -> eval f /. eval g

let print_expr expr =

 let open_paren prec op_prec =
   if prec > op_prec then print_string "(" in
 let close_paren prec op_prec =
   if prec > op_prec then print_string ")" in
 let rec print prec = function   (* prec is the current precedence *)
   | Const c -> Printf.printf "%g" c
   | Sum(f, g) ->
       open_paren prec 0;
       print 0 f; print_string " + "; print 0 g;
       close_paren prec 0
   | Diff(f, g) ->
       open_paren prec 0;
       print 0 f; print_string " - "; print 1 g;
       close_paren prec 0
   | Prod(f, g) ->
       open_paren prec 2;
       print 2 f; print_string " * "; print 2 g;
       close_paren prec 2
   | Quot(f, g) ->
       open_paren prec 2;
       print 2 f; print_string " / "; print 3 g;
       close_paren prec 2
 in
 print 0 expr

let rec insert v = function

 | [] -> v
 | x::xs as li -> (v::li) :: (List.map (fun y -> x::y) (insert v xs))

let permutations li =

 List.fold_right (fun x z -> List.concat (List.map (insert x) z)) li [[]]

let rec comp expr = function

 | x::xs ->
     comp (Sum (expr, x)) xs;
     comp (Diff(expr, x)) xs;
     comp (Prod(expr, x)) xs;
     comp (Quot(expr, x)) xs;
 | [] ->
     if (eval expr) = 24.0
     then (print_expr expr; print_newline())

let () =

 Random.self_init();
 let digits = Array.init 4 (fun _ -> 1 + Random.int 9) in
 print_string "Input digits: ";
 Array.iter (Printf.printf " %d") digits; print_newline();
 let digits = Array.to_list(Array.map float_of_int digits) in
 let digits = List.map (fun v -> Const v) digits in
 let all = permutations digits in
 List.iter (function
   | x::xs -> comp x xs
   | [] -> assert false
 ) all</lang>

Input digits: 5 7 4 1
7 * 4 - 5 + 1
7 * 4 + 1 - 5
4 * 7 - 5 + 1
4 * 7 + 1 - 5
(5 - 1) * 7 - 4

(notice that the printer only puts parenthesis when needed)

Perl

Will generate all possible solutions of any given four numbers according to the rules of the 24 game.

Note: the permute function was taken from here <lang Perl># Fischer-Krause ordered permutation generator

1. http://faq.perl.org/perlfaq4.html#How_do_I_permute_N_e

sub permute (&@) { my $code = shift; my @idx = 0..$#_; while ( $code->(@_[@idx]) ) { my $p = $#idx; --$p while $idx[$p-1] > $idx[$p]; my $q = $p or return; push @idx, reverse splice @idx, $p; ++$q while $idx[$p-1] > $idx[$q]; @idx[$p-1,$q]=@idx[$q,$p-1]; } }

@formats = ( '((%d %s %d) %s %d) %s %d', '(%d %s (%d %s %d)) %s %d', '(%d %s %d) %s (%d %s %d)', '%d %s ((%d %s %d) %s %d)', '%d %s (%d %s (%d %s %d))', );

1. generate all possible combinations of operators

@op = qw( + - * / ); @operators = map{ $a=$_; map{ $b=$_; map{ "$a $b $_" }@op }@op }@op;

while(1) { print "Enter four integers or 'q' to exit: "; chomp($ent = <>); last if $ent eq 'q';

if($ent !~ /^[1-9] [1-9] [1-9] [1-9]$/){ print "invalid input\n"; next }

@n = split / /,$ent; permute { push @numbers,join ' ',@_ }@n;

for $format (@formats) { for(@numbers) { @n = split; for(@operators) { @o = split; $str = sprintf $format,$n[0],$o[0],$n[1],$o[1],$n[2],$o[2],$n[3]; $r = eval($str); print "$str\n" if $r == 24; } } } }</lang>

PicoLisp

We use Pilog (PicoLisp Prolog) to solve this task <lang PicoLisp>(be play24 (@Lst @Expr) # Define Pilog rule

  (permute @Lst (@A @B @C @D))
  (member @Op1 (+ - * /))
  (member @Op2 (+ - * /))
  (member @Op3 (+ - * /))
  (or
     ((equal @Expr (@Op1 (@Op2 @A @B) (@Op3 @C @D))))
     ((equal @Expr (@Op1 @A (@Op2 @B (@Op3 @C @D))))) )
  (@ = 24 (catch '("Div/0") (eval (-> @Expr)))) )

(de play24 (A B C D) # Define PicoLisp function

  (pilog
     (quote
        @L (list A B C D)
        (play24 @L @X) )
     (println @X) ) )

(play24 5 6 7 8) # Call 'play24' function</lang> Output:

(* (+ 5 7) (- 8 6))
(* 6 (+ 5 (- 7 8)))
(* 6 (- 5 (- 8 7)))
(* 6 (- 5 (/ 8 7)))
(* 6 (+ 7 (- 5 8)))
(* 6 (- 7 (- 8 5)))
(* 6 (/ 8 (- 7 5)))
(/ (* 6 8) (- 7 5))
(* (+ 7 5) (- 8 6))
(* (- 8 6) (+ 5 7))
(* (- 8 6) (+ 7 5))
(* 8 (/ 6 (- 7 5)))
(/ (* 8 6) (- 7 5))

ProDOS

Note This example uses the math module: <lang ProDOS>editvar /modify -random- = <10

a

editvar /newvar /withothervar /value=-random- /title=1 editvar /newvar /withothervar /value=-random- /title=2 editvar /newvar /withothervar /value=-random- /title=3 editvar /newvar /withothervar /value=-random- /title=4 printline These are your four digits: -1- -2- -3- -4- printline Use an algorithm to make the number 24. editvar /newvar /value=a /userinput=1 /title=Algorithm: do -a- if -a- /hasvalue 24 printline Your algorithm worked! & goto :b ( ) else printline Your algorithm did not work. editvar /newvar /value=b /userinput=1 /title=Do you want to see how you could have done it? if -b- /hasvalue y goto :c else goto :b

b

editvar /newvar /value=c /userinput=1 /title=Do you want to play again? if -c- /hasvalue y goto :a else exitcurrentprogram

c

editvar /newvar /value=do -1- + -2- + -3- + -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- - -2- + -3- + -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- / -2- + -3- + -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- * -2- + -3- + -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- + -2- - -3- + -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- + -2- / -3- + -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- + -2- * -3- + -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- + -2- + -3- - -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- + -2- + -3- / -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- + -2- + -3- * -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- - -2- - -3- - -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- / -2- / -3- / -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve editvar /newvar /value=do -1- * -2- * -3- * -4- /title=c & do -c- >d & if -d- /hasvalue 24 goto :solve

solve

printline you could have done it by doing -c- stoptask goto :b</lang>

Sample Output: <lang ProDOS>These are your four digits: 1 4 5 2 Use an algorithm to make the number 24. Algorithm: 4 + 2 - 5 + 1 Your algorithm did not work. Do you want to play again? y

These are your four digits: 1 8 9 6 Use an algorithm to make the number 24. Algorithm: 1 + 8 + 9 + 6 Your algorithm worked! Do you want to play again? n</lang>

Prolog

Works with SWI-Prolog.
The game is generic, you can choose to play with a goal different of 24, any number of numbers in other ranges than 1 .. 9 !
rdiv/2 is use instead of //2 to enable the program to solve difficult cases as [3 3 8 8].

<lang Prolog>play24(Len, Range, Goal) :- game(Len, Range, Goal, L, S), maplist(my_write, L), format(': ~w~n', [S]).

game(Len, Range, Value, L, S) :- length(L, Len), maplist(choose(Range), L), compute(L, Value, [], S).

choose(Range, V) :- V is random(Range) + 1.

write_tree([M], [M]).

write_tree([+, M, N], S) :- write_tree(M, MS), write_tree(N, NS), append(MS, [+ | NS], S).

write_tree([-, M, N], S) :- write_tree(M, MS), write_tree(N, NS), ( is_add(N) -> append(MS, [-, '(' | NS], Temp), append(Temp, ')', S) ; append(MS, [- | NS], S)).

write_tree([Op, M, N], S) :- member(Op, [*, /]), write_tree(M, MS), write_tree(N, NS), ( is_add(M) -> append(['(' | MS], [')'], TempM) ; TempM = MS), ( is_add(N) -> append(['(' | NS], [')'], TempN) ; TempN = NS), append(TempM, [Op | TempN], S).

is_add([Op, _, _]) :- member(Op, [+, -]).

compute([Value], Value, _R-S1, S) :- write_tree(S1, S2), with_output_to(atom(S), maplist(write, S2)).

compute(L, Value, CS, S) :- select(M, L, L1), select(N, L1, L2), next_value(M, N, R, CS, Expr), compute([R|L2], Value, Expr, S).

next_value(M, N, R, CS,[[R - [+, M1, N1]] | CS2]) :- R is M+N, ( member([M-ExprM], CS) -> select([M-ExprM], CS, CS1), M1 = ExprM ; M1 = [M], CS1 = CS ), ( member([N-ExprN], CS1) -> select([N-ExprN], CS1, CS2), N1 = ExprN ; N1 = [N], CS2 = CS1 ).

next_value(M, N, R, CS,[[R - [-, M1, N1]] | CS2]) :- R is M-N, ( member([M-ExprM], CS) -> select([M-ExprM], CS, CS1), M1 = ExprM ; M1 = [M], CS1 = CS ), ( member([N-ExprN], CS1) -> select([N-ExprN], CS1, CS2), N1 = ExprN ; N1 = [N], CS2 = CS1 ).

next_value(M, N, R, CS,[[R - [*, M1, N1]] | CS2]) :- R is M*N, ( member([M-ExprM], CS) -> select([M-ExprM], CS, CS1), M1 = ExprM ; M1 = [M], CS1 = CS ), ( member([N-ExprN], CS1) -> select([N-ExprN], CS1, CS2), N1 = ExprN ; N1 = [N], CS2 = CS1 ).

next_value(M, N, R, CS,[[R - [/, M1, N1]] | CS2]) :- N \= 0, R is rdiv(M,N), ( member([M-ExprM], CS) -> select([M-ExprM], CS, CS1), M1 = ExprM ; M1 = [M], CS1 = CS ), ( member([N-ExprN], CS1) -> select([N-ExprN], CS1, CS2), N1 = ExprN ; N1 = [N], CS2 = CS1 ).

my_write(V) :- format('~w ', [V]).</lang> Example of output :

?- play24(4,9, 24).
6 2 3 4 : (6-2+4)*3
true ;
6 2 3 4 : 3*(6-2+4)
true ;
6 2 3 4 : (6-2+4)*3
true ;
6 2 3 4 : 3*(6-2+4)
true ;
6 2 3 4 : (6*2-4)*3
true ;
6 2 3 4 : 3*(6*2-4)
true ;
6 2 3 4 : 3*4+6*2
true ;
6 2 3 4 : 3*4+6*2
true ;
6 2 3 4 : 4*3+6*2
true ;
6 2 3 4 : 4*3+6*2
true ;
6 2 3 4 : (6/2+3)*4
true ;
6 2 3 4 : 4*(6/2+3)
true ;
6 2 3 4 : (6/2+3)*4
true ;
6 2 3 4 : 4*(6/2+3)
true ;
6 2 3 4 : (6-3)*2*4
true ;
6 2 3 4 : 4*(6-3)*2
true ;
6 2 3 4 : (6-3)*4*2
...

?- play24(7,99, 1).
66 40 2 76 95 59 12 : (66+40)/2-76+95-59-12
true ;
66 40 2 76 95 59 12 : (66+40)/2-76+95-12-59
true ;
66 40 2 76 95 59 12 : (66+40)/2-76-59+95-12
true ;
66 40 2 76 95 59 12 : (66+40)/2-76-59-12+95
true ;
66 40 2 76 95 59 12 : 95+(66+40)/2-76-59-12
true ;
66 40 2 76 95 59 12 : 95+(66+40)/2-76-59-12
true ;
66 40 2 76 95 59 12 : 95-12+(66+40)/2-76-59
true ;
66 40 2 76 95 59 12 : (66+40)/2-76-59+95-12
....

Python

The function is called solve, and is integrated into the game player. The docstring of the solve function shows examples of its use when isolated at the Python command line. <lang Python>

The 24 Game Player

Given any four digits in the range 1 to 9, which may have repetitions,
Using just the +, -, *, and / operators; and the possible use of
brackets, (), show how to make an answer of 24.

An answer of "q"  will quit the game.
An answer of "!"  will generate a new set of four digits.
An answer of "!!" will ask you for a new set of four digits.
An answer of "?"  will compute an expression for the current digits.

Otherwise you are repeatedly asked for an expression until it evaluates to 24

Note: you cannot form multiple digit numbers from the supplied digits,
so an answer of 12+12 when given 1, 2, 2, and 1 would not be allowed.

from __future__ import division, print_function from itertools import permutations, combinations, product, \

                        chain

from pprint import pprint as pp from fractions import Fraction as F import random, ast, re import sys

if sys.version_info[0] < 3:

   input = raw_input
   from itertools import izip_longest as zip_longest

else:

   from itertools import zip_longest

def choose4():

   'four random digits >0 as characters'
   return [str(random.randint(1,9)) for i in range(4)]

def ask4():

   'get four random digits >0 from the plaayer'
   digits = 
   while len(digits) != 4 or not all(d in '123456789' for d in digits):
       digits = input('Enter the digits to solve for: ')
       digits = .join(digits.strip().split())
   return list(digits)

def welcome(digits):

   print (__doc__)
   print ("Your four digits: " + ' '.join(digits))

def check(answer, digits):

   allowed = set('() +-*/\t'+.join(digits))
   ok = all(ch in allowed for ch in answer) and \
        all(digits.count(dig) == answer.count(dig) for dig in set(digits)) \
        and not re.search('\d\d', answer)
   if ok:
       try:
           ast.parse(answer)
       except:
           ok = False
   return ok

def solve(digits):

   """\
   >>> for digits in '3246 4788 1111 123456 1127 3838'.split():
           solve(list(digits))


   Solution found: 2 + 3 * 6 + 4
   '2 + 3 * 6 + 4'
   Solution found: ( 4 + 7 - 8 ) * 8
   '( 4 + 7 - 8 ) * 8'
   No solution found for: 1 1 1 1
   '!'
   Solution found: 1 + 2 + 3 * ( 4 + 5 ) - 6
   '1 + 2 + 3 * ( 4 + 5 ) - 6'
   Solution found: ( 1 + 2 ) * ( 1 + 7 )
   '( 1 + 2 ) * ( 1 + 7 )'
   Solution found: 8 / ( 3 - 8 / 3 )
   '8 / ( 3 - 8 / 3 )'
   >>> """
   digilen = len(digits)
   # length of an exp without brackets 
   exprlen = 2 * digilen - 1
   # permute all the digits
   digiperm = sorted(set(permutations(digits)))
   # All the possible operator combinations
   opcomb   = list(product('+-*/', repeat=digilen-1))
   # All the bracket insertion points:
   brackets = ( [()] + [(x,y)
                        for x in range(0, exprlen, 2)
                        for y in range(x+4, exprlen+2, 2)
                        if (x,y) != (0,exprlen+1)]
                + [(0, 3+1, 4+2, 7+3)] ) # double brackets case
   for d in digiperm:
       for ops in opcomb:
           if '/' in ops:
               d2 = [('F(%s)' % i) for i in d] # Use Fractions for accuracy
           else:
               d2 = d
           ex = list(chain.from_iterable(zip_longest(d2, ops, fillvalue=)))
           for b in brackets:
               exp = ex[::]
               for insertpoint, bracket in zip(b, '()'*(len(b)//2)):
                   exp.insert(insertpoint, bracket)
               txt = .join(exp)
               try:
                   num = eval(txt)
               except ZeroDivisionError:
                   continue
               if num == 24:
                   if '/' in ops:
                       exp = [ (term if not term.startswith('F(') else term[2])
                              for term in exp ]
                   ans = ' '.join(exp).rstrip()
                   print ("Solution found:",ans)
                   return ans
   print ("No solution found for:", ' '.join(digits))            
   return '!'

def main():

   digits = choose4()
   welcome(digits)
   trial = 0
   answer = 
   chk = ans = False
   while not (chk and ans == 24):
       trial +=1
       answer = input("Expression %i: " % trial)
       chk = check(answer, digits)
       if answer == '?':
           solve(digits)
           answer = '!'
       if answer.lower() == 'q':
           break
       if answer == '!':
           digits = choose4()
           trial = 0
           print ("\nNew digits:", ' '.join(digits))
           continue
       if answer == '!!':
           digits = ask4()
           trial = 0
           print ("\nNew digits:", ' '.join(digits))
           continue
       if not chk:
           print ("The input '%s' was wonky!" % answer)
       else:
           if '/' in answer:
               # Use Fractions for accuracy in divisions
               answer = .join( (('F(%s)' % char) if char in '123456789' else char)
                                 for char in answer )
           ans = eval(answer)
           print (" = ", ans)
           if ans == 24:
               print ("Thats right!")
   print ("Thank you and goodbye")   

main()</lang>

Sample Output

 The 24 Game Player

 Given any four digits in the range 1 to 9, which may have repetitions,
 Using just the +, -, *, and / operators; and the possible use of
 brackets, (), show how to make an answer of 24.

 An answer of "q" will quit the game.
 An answer of "!" will generate a new set of four digits.
 An answer of "?" will compute an expression for the current digits.
 
 Otherwise you are repeatedly asked for an expression until it evaluates to 24

 Note: you cannot form multiple digit numbers from the supplied digits,
 so an answer of 12+12 when given 1, 2, 2, and 1 would not be allowed.


Your four digits: 6 7 9 5
Expression 1: ?
Solution found: 6 - ( 5 - 7 ) * 9
Thank you and goodbye

Difficult case requiring precise division

The digits 3,3,8 and 8 have a solution that is not equal to 24 when using Pythons double-precision floating point because of a division in all answers. The solver above switches to precise fractional arithmetic when division is involved and so can both recognise and solve for cases like this, (rather than allowing some range of closeness to 24).

Evaluation needing precise division

...
Expression 1: !!
Enter the digits to solve for: 3388

New digits: 3 3 8 8
Expression 1: 8/(3-(8/3))
 =  24
Thats right!
Thank you and goodbye

Solving needing precise division

...
Expression 1: !!
Enter the digits to solve for: 3388

New digits: 3 3 8 8
Expression 1: ?
Solution found: 8 / ( 3 - 8 / 3 )

R

This uses exhaustive search and makes use of R's ability to work with expressions as data. It is in principle general for any set of operands and binary operators. <lang r> library(gtools)

solve24 <- function(vals=c(8, 4, 2, 1),

                   goal=24,
                   ops=c("+", "-", "*", "/")) {
 
 val.perms <- as.data.frame(t(
                 permutations(length(vals), length(vals))))

 nop <- length(vals)-1
 op.perms <- as.data.frame(t(
                 do.call(expand.grid,
                         replicate(nop, list(ops)))))
 
 ord.perms <- as.data.frame(t(
                  do.call(expand.grid,
                          replicate(n <- nop, 1:((n <<- n-1)+1)))))

 for (val.perm in val.perms)
   for (op.perm in op.perms)
     for (ord.perm in ord.perms)
       {
         expr <- as.list(vals[val.perm])
         for (i in 1:nop) {
           expr[[ ord.perm[i] ]] <- call(as.character(op.perm[i]),
                                         expr[[ ord.perm[i]   ]],
                                         expr[[ ord.perm[i]+1 ]])
           expr <- expr[ -(ord.perm[i]+1) ]
         }
         if (identical(eval(expr1), goal)) return(expr1)
       }

 return(NA)

} </lang> Example Output: <lang r> > solve24() 8 * (4 - 2 + 1) > solve24(c(6,7,9,5)) 6 + (7 - 5) * 9 > solve24(c(8,8,8,8)) [1] NA > solve24(goal=49) #different goal value 8 * (4 + 2) + 1 > solve24(goal=52) #no solution [1] NA > solve24(ops=c('-', '/')) #restricted set of operators (8 - 2)/(1/4) </lang>

REXX

<lang rexx>/*REXX program to help the user find solutions to the game of 24. */ /*------------------------------------------------------------------+

| Argument is either of two forms:     ssss    ==or==   ssss-ffff  |
|                                                                  |
| where one or both strings must be exactly four numerals (digits) |
| comprised soley of the numerals (digits)  1 --> 9   (no zeroes). |
|                                                                  |
| In    SSSS-FFFF           SSSS  is the start,                    |
|                           FFFF  is the start.                    |
+------------------------------------------------------------------*/

parse arg orig /*get the guess from the argument. */ parse var orig start '-' finish /*get the start & finish (maybe).*/ start=space(start,0) /*remove any blanks from the START. */ finish=space(finish,0) /*remove any blanks from the FINISH. */ finish=word(finish start,1) /*if no FINISH specified, use START.*/ call validate start call validate finish opers='+-*/' /*define the legal arithmetic operators*/ ops=length(opers) /* ... and the count of them (length). */

        do j=1 for ops      /*define a version for fast execution. */
        o.j=substr(opers,j,1)
        end

finds=0 /*number of found solutions (so far). */ x.=0 /*a method to hold unique expressions. */ indent=left(,30) /*used to indent display of solutions. */

                            /*alternative:  indent=copies(' ',30)  */

Lpar='(' /*a string to make REXX code prettier. */ Rpar=')' /*ditto. */

 do g=start to finish       /*process a (possible) range of values.*/
 if pos(0,g)\==0 then iterate   /*ignore values with zero in them. */

     do j=1 for 4           /*define a version for fast execution. */
     g.j=substr(g,j,1)
     end

   do i=1 for ops           /*insert an operator after 1st number. */
     do j=1 for ops         /*insert an operator after 2nd number. */
       do k=1 for ops       /*insert an operator after 2nd number. */
         do m=0 to 4-1
         L.=              /*assume no left parenthesis so far.   */
           do n=m+1 to 4    /*match left paren with a right paren. */
           L.m=Lpar         /*define a left paren, m=0 means ignore*/
           R.=            /*un-define all right parenthesis.     */
           if m==1 & n==2 then L.=   /*special case:  (n)+ ...   */
                          else if m\==0 then R.n=Rpar  /*no (, no )*/
           e=L.1 g.1 o.i L.2 g.2 o.j L.3 g.3 R.3 o.k g.4 R.4
           e=space(e,0)     /*remove all blanks from the expression*/

                            /*(below) change expression:           */
                            /*       /(yyy)   ===>   /div(yyy)     */
                            /*Enables to check for division by zero*/
           origE=e          /*keep old version for the display.    */
           if pos('/(',e)\==0 then e=changestr('/(',e,"/div(")
                            /*The above could be replaced by:      */
                            /*   e=changestr('/(',e,"/div(")       */

                                /*INTERPRET stresses REXX's groin, */
                                /*so try to avoid repeated lifting.*/
           if x.e then iterate  /*was the expression already used? */
           x.e=1                /*mark this expression as unique.  */
                                /*have REXX do the heavy lifting.  */
           interpret 'x='e
           x=x/1                /*remove trailing decimal points.  */
           if x\==24 then iterate        /*Not correct?  Try again.*/
           finds=finds+1        /*bump number of found solutions.  */
           _=translate(origE,'][',")(")       /*show  [],  not  ().*/
           say indent 'a solution:' _         /*display a solution.*/
           end   /*n*/
         end     /*m*/
       end       /*k*/
     end         /*j*/
   end           /*i*/
 end             /*g*/

sols=finds if sols==0 then sols='No' /*make the sentence not so geek-like. */ say say sols 'unique solution's(finds) "found for" orig /*pluralize.*/ exit /*---------------------------DIV subroutine-------------------------*/ div: procedure; parse arg q /*tests if dividing by 0 (zero). */ if q=0 then q=1e9 /*if dividing by zero, change divisor. */ return q /*changing Q invalidates the expression*/ /*---------------------------GER subroutine-------------------------*/ ger: say; say '*** error! ***'; if _\== then say 'guess=' _ say arg(1) say exit 13 /*---------------------------validate subroutine--------------------*/ validate: parse arg _ digs=123456789 /*numerals (digits) that can be used. */ if _== then call ger 'no digits entered.' if length(_)<4 then call ger 'not enough digits entered, must be 4' if length(_)>4 then call ger 'too many digits entered, must be 4' if pos(0,_)\==0 then call ger "can't use the digit 0 (zero)" _=verify(_,digs) return /*---------------------------S subroutine---------------------------*/ s:if arg(1)=1 then return ; return 's' /*simple pluralizer.*/</lang> output when the following input is used: 1111-1234

                               a solution: [1+1+1]*8
                               a solution: [1+1+2]*6
                               a solution: [1+1*2]*8
                               a solution: [1*1+2]*8
                               a solution: 1*[1+2]*8
                               a solution: [1/1+2]*8
                               a solution: [1+1*3]*6
                               a solution: [1*1+3]*6
                               a solution: 1*[1+3]*6
                               a solution: [1/1+3]*6
                               a solution: 1-1+3*8
                               a solution: [1-1+3]*8
                               a solution: [1-1+3*8]
                               a solution: 1-1+[3*8]
                               a solution: 1-[1-3*8]
                               a solution: 1*1*3*8
                               a solution: [1*1*3]*8
                               a solution: [1*1*3*8]
                               a solution: 1*[1*3]*8
                               a solution: 1*[1*3*8]
                               a solution: 1*1*[3*8]
                               a solution: 1/1*3*8
                               a solution: [1/1*3]*8
                               a solution: [1/1*3*8]
                               a solution: 1/1*[3*8]
                               a solution: 1/[1/3]*8
                               a solution: 1/[1/3/8]
                               a solution: [1+1+4]*4
                               a solution: 1-1+4*6
                               a solution: [1-1+4]*6
                               a solution: [1-1+4*6]
                               a solution: 1-1+[4*6]
                               a solution: 1-[1-4*6]
                               a solution: 1*1*4*6
                               a solution: [1*1*4]*6
                               a solution: [1*1*4*6]
                               a solution: 1*[1*4]*6
                               a solution: 1*[1*4*6]
                               a solution: 1*1*[4*6]
                               a solution: 1/1*4*6
                               a solution: [1/1*4]*6
                               a solution: [1/1*4*6]
                               a solution: 1/1*[4*6]
                               a solution: 1/[1/4]*6
                               a solution: 1/[1/4/6]
                               a solution: [1+1*5]*4
                               a solution: [1*1+5]*4
                               a solution: 1*[1+5]*4
                               a solution: [1/1+5]*4
                               a solution: [1+1+6]*3
                               a solution: 1-1+6*4
                               a solution: [1-1+6]*4
                               a solution: [1-1+6*4]
                               a solution: 1-1+[6*4]
                               a solution: 1-[1-6*4]
                               a solution: 1*1*6*4
                               a solution: [1*1*6]*4
                               a solution: [1*1*6*4]
                               a solution: 1*[1*6]*4
                               a solution: 1*[1*6*4]
                               a solution: 1*1*[6*4]
                               a solution: 1/1*6*4
                               a solution: [1/1*6]*4
                               a solution: [1/1*6*4]
                               a solution: 1/1*[6*4]
                               a solution: 1/[1/6]*4
                               a solution: [1+1*7]*3
                               a solution: [1*1+7]*3
                               a solution: 1*[1+7]*3
                               a solution: [1/1+7]*3
                               a solution: 1-1+8*3
                               a solution: [1-1+8]*3
                               a solution: [1-1+8*3]
                               a solution: 1-1+[8*3]
                               a solution: 1-[1-8*3]
                               a solution: 1*1*8*3
                               a solution: [1*1*8]*3
                               a solution: [1*1*8*3]
                               a solution: 1*[1*8]*3
                               a solution: 1*[1*8*3]
                               a solution: 1*1*[8*3]
                               a solution: 1/1*8*3
                               a solution: [1/1*8]*3
                               a solution: [1/1*8*3]
                               a solution: 1/1*[8*3]
                               a solution: 1/[1/8]*3
                               a solution: 1/[1/8/3]
                               a solution: [1+2+1]*6
                               a solution: [1+2*1]*8
                               a solution: [1+2/1]*8
                               a solution: [1*2+1]*8
                               a solution: 1*[2+1]*8
                               a solution: [1*2+2]*6
                               a solution: 1*[2+2]*6
                               a solution: 1*2*2*6
                               a solution: [1*2*2]*6
                               a solution: [1*2*2*6]
                               a solution: 1*[2*2]*6
                               a solution: 1*[2*2*6]
                               a solution: 1*2*[2*6]
                               a solution: [1+2+3]*4
                               a solution: 1*2*3*4
                               a solution: [1*2*3]*4
                               a solution: [1*2*3*4]
                               a solution: 1*[2*3]*4
                               a solution: 1*[2*3*4]
                               a solution: 1*2*[3*4]

107 unique solutions found for 1111-1234

Ruby

Translation of: Tcl

<lang ruby>require 'rational'

class TwentyFourGamePlayer

 EXPRESSIONS = [
   '((%d %s %d) %s %d) %s %d',
   '(%d %s (%d %s %d)) %s %d',
   '(%d %s %d) %s (%d %s %d)',
   '%d %s ((%d %s %d) %s %d)',
   '%d %s (%d %s (%d %s %d))',
 ]
 OPERATORS = [:+, :-, :*, :/]

 @@objective = Rational(24,1)

 def initialize(digits)
   @digits = digits
   @solutions = []
   solve
 end

 attr_reader :digits, :solutions

 def solve
   digits.permutation.to_a.uniq.each do |a,b,c,d|
     OPERATORS.each   do |op1| 
     OPERATORS.each   do |op2| 
     OPERATORS.each   do |op3|
     EXPRESSIONS.each do |expr|
       # evaluate using rational arithmetic
       test = expr.gsub('%d', 'Rational(%d,1)') % [a, op1, b, op2, c, op3, d]
       value = eval(test) rescue -1  # catch division by zero
       if value == @@objective
         @solutions << expr % [a, op1, b, op2, c, op3, d]
       end
     end;end;end;end
   end
 end

end

1. validate user input

digits = ARGV.map do |arg|

 begin
   Integer(arg)
 rescue ArgumentError
   raise "error: not an integer: '#{arg}'"
 end

end digits.size == 4 or raise "error: need 4 digits, only have #{digits.size}"

player = TwentyFourGamePlayer.new(digits) if player.solutions.empty?

 puts "no solutions"

else

 puts "found #{player.solutions.size} solutions, including #{player.solutions.first}"
 puts player.solutions.sort.join("\n")

end</lang>

Sample output:

$ ruby 24game.player.rb 1 1 1 1
no solutions

$ ruby 24game.player.rb 1 1 2 7
found 8 solutions, including (1 + 2) * (1 + 7)
(1 + 2) * (1 + 7)
(1 + 2) * (7 + 1)
(1 + 7) * (1 + 2)
(1 + 7) * (2 + 1)
(2 + 1) * (1 + 7)
(2 + 1) * (7 + 1)
(7 + 1) * (1 + 2)
(7 + 1) * (2 + 1)

$ ruby 24game.player.rb 2 3 8 9
found 12 solutions, including (8 / 2) * (9 - 3)
((9 - 3) * 8) / 2
((9 - 3) / 2) * 8
(8 * (9 - 3)) / 2
(8 / 2) * (9 - 3)
(9 - (2 * 3)) * 8
(9 - (3 * 2)) * 8
(9 - 3) * (8 / 2)
(9 - 3) / (2 / 8)
8 * ((9 - 3) / 2)
8 * (9 - (2 * 3))
8 * (9 - (3 * 2))
8 / (2 / (9 - 3))

Scala

A non-interactive player.

<lang scala>def permute(l: List[Double]): List[List[Double]] = l match {

 case Nil => List(Nil)
 case x :: xs =>
   for {
     ys <- permute(xs)
     position <- 0 to ys.length
     (left, right) = ys splitAt position
   } yield left ::: (x :: right)

}

def computeAllOperations(l: List[Double]): List[(Double,String)] = l match {

 case Nil => Nil
 case x :: Nil => List((x, "%1.0f" format x))
 case x :: xs =>
   for {
     (y, ops) <- computeAllOperations(xs)
     (z, op) <- 
       if (y == 0) 
         List((x*y, "*"), (x+y, "+"), (x-y, "-")) 
       else 
         List((x*y, "*"), (x/y, "/"), (x+y, "+"), (x-y, "-"))
   } yield (z, "(%1.0f%s%s)" format (x,op,ops))

}

def hasSolution(l: List[Double]) = permute(l) flatMap computeAllOperations filter (_._1 == 24) map (_._2)</lang>

Example:

val problemsIterator = (
    Iterator
    continually List.fill(4)(scala.util.Random.nextInt(9) + 1 toDouble)
    filter (!hasSolution(_).isEmpty)
)

val solutionIterator = problemsIterator map hasSolution

scala> solutionIterator.next
res8: List[String] = List((3*(5-(3-6))), (3*(5-(3-6))), (3*(5+(6-3))), (3+(6+(3*5))), (3*(6-(3-5))), (3+(6+(5*3))), (3*(
6+(5-3))), (3*(5+(6-3))), (3+(6+(5*3))), (3*(6+(5-3))), (6+(3+(5*3))), (6*(5-(3/3))), (6*(5-(3/3))), (3+(6+(3*5))), (3*(
6-(3-5))), (6+(3+(3*5))), (6+(3+(3*5))), (6+(3+(5*3))))

scala> solutionIterator.next
res9: List[String] = List((4-(5*(5-9))), (4-(5*(5-9))), (4+(5*(9-5))), (4+(5*(9-5))), (9-(5-(4*5))), (9-(5-(5*4))), (9-(
5-(4*5))), (9-(5-(5*4))))

scala> solutionIterator.next
res10: List[String] = List((2*(4+(3+5))), (2*(3+(4+5))), (2*(3+(5+4))), (4*(3-(2-5))), (4*(3+(5-2))), (2*(4+(5+3))), (2*
(5+(4+3))), (2*(5+(3+4))), (4*(5-(2-3))), (4*(5+(3-2))))

scala> solutionIterator.next
res11: List[String] = List((4*(5-(2-3))), (2*(4+(5+3))), (2*(5+(4+3))), (2*(5+(3+4))), (2*(4+(3+5))), (2*(3+(4+5))), (2*
(3+(5+4))), (4*(5+(3-2))), (4*(3+(5-2))), (4*(3-(2-5))))

Tcl

This is a complete Tcl script, intended to be invoked from the command line.

Library: Tcllib (Package: struct::list)

<lang tcl>package require struct::list

1. Encoding the various expression trees that are possible

set patterns {

   {((A x B) y C) z D}
   {(A x (B y C)) z D}
   {(A x B) y (C z D)}
   {A x ((B y C) z D)}
   {A x (B y (C z D))}

}

1. Encoding the various permutations of digits

set permutations [struct::list map [struct::list permutations {a b c d}] \

       {apply {v {lassign $v a b c d; list A $a B $b C $c D $d}}}]

1. The permitted operations

set operations {+ - * /}

1. Given a list of four integers (precondition not checked!) return a list of
2. solutions to the 24 game using those four integers.

proc find24GameSolutions {values} {

   global operations patterns permutations
   set found {}
   # For each possible structure with numbers at the leaves...
   foreach pattern $patterns {

foreach permutation $permutations { set p [string map [subst { a [lindex $values 0].0 b [lindex $values 1].0 c [lindex $values 2].0 d [lindex $values 3].0 }] [string map $permutation $pattern]]

           # For each possible structure with operators at the branches...

foreach x $operations { foreach y $operations { foreach z $operations { set e [string map [subst {x $x y $y z $z}] $p]

# Try to evaluate (div-zero is an issue!) and add it to # the result if it is 24 catch { if {[expr $e] == 24.0} { lappend found [string map {.0 {}} $e] } } } } } }

   }
   return $found

}

1. Wrap the solution finder into a player

proc print24GameSolutionFor {values} {

   set found [lsort -unique [find24GameSolutions $values]]
   if {![llength $found]} {

puts "No solution possible"

   } else {

puts "Total [llength $found] solutions (may include logical duplicates)"

       puts "First solution: [lindex $found 0]"
   }

} print24GameSolutionFor $argv</lang> Demonstrating it in use:

bash$ tclsh8.4 24player.tcl 3 2 8 9
Total 12 solutions (may include logical duplicates)
First solution: ((9 - 3) * 8) / 2
bash$ tclsh8.4 24player.tcl 1 1 2 7
Total 8 solutions (may include logical duplicates)
First solution: (1 + 2) * (1 + 7)
bash$ tclsh8.4 24player.tcl 1 1 1 1
No solution possible

Ursala

This uses exhaustive search and exact rational arithmetic to enumerate all solutions. The algorithms accommodate data sets with any number of digits and any target value, but will be limited in practice by combinatorial explosion as noted elsewhere. (Rationals are stored as pairs of integers, hence ("n",1) for n/1, etc..)

The tree_shapes function generates a list of binary trees of all possible shapes for a given number of leaves. The with_leaves function substitutes a list of numbers into the leaves of a tree in every possible way. The with_roots function substitutes a list of operators into the non-terminal nodes of a tree in every possible way. The value function evaluates a tree and the format function displays it in a readable form. <lang Ursala>#import std

1. import nat
2. import rat

tree_shapes = "n". (@vLPiYo //eql iota "n")*~ (rep"n" ~&iiiK0NlrNCCVSPTs) {0^:<>} with_leaves = ^|DrlDrlK34SPSL/permutations ~& with_roots = ^DrlDrlK35dlPvVoPSPSL\~&r @lrhvdNCBvLPTo2DlS @hiNCSPtCx ~&K0=> value = *^ ~&v?\(@d ~&\1) ^|H\~&hthPX '+-*/'-$<sum,difference,product,quotient> format = *^ ~&v?\-+~&h,%zP@d+- ^H/mat@d *v ~&t?\~& :/`(+ --')'

game"n" "d" = format* value==("n",1)*~ with_roots/'+-*/' with_leaves/"d"*-1 tree_shapes length "d"</lang> test program: <lang Ursala>#show+

test_games = mat` * pad` *K7 pad0 game24* <<2,3,8,9>,<5,7,4,1>,<5,6,7,8>></lang> output:

8/(2/(9-3)) 1-(5-(7*4)) 6*(5+(7-8))
8*(9-(2*3)) 1-(5-(4*7)) 6*(7+(5-8))
8*(9-(3*2)) 1+((7*4)-5) 6*(7-(8-5))
8*((9-3)/2) 1+((4*7)-5) 6*(5-(8-7))
(8/2)*(9-3) (7*4)-(5-1) 6*(8/(7-5))
(9-3)/(2/8) (7*4)+(1-5) 8*(6/(7-5))
(9-3)*(8/2) (4*7)-(5-1) 6*((5+7)-8)
(8*(9-3))/2 (4*7)+(1-5) 6*((7+5)-8)
(9-(2*3))*8 (1-5)+(7*4) 6/((7-5)/8)
(9-(3*2))*8 (1-5)+(4*7) 6*((7-8)+5)
((9-3)/2)*8 (7*(5-1))-4 6*((5-8)+7)
((9-3)*8)/2 (1+(7*4))-5 8/((7-5)/6)
            (1+(4*7))-5 (5+7)*(8-6)
            ((7*4)-5)+1 (7+5)*(8-6)
            ((7*4)+1)-5 (6*8)/(7-5)
            ((4*7)-5)+1 (8-6)*(5+7)
            ((4*7)+1)-5 (8-6)*(7+5)
            ((5-1)*7)-4 (8*6)/(7-5)
                        (6/(7-5))*8
                        (5+(7-8))*6
                        (7+(5-8))*6
                        (7-(8-5))*6
                        (5-(8-7))*6
                        (8/(7-5))*6
                        ((5+7)-8)*6
                        ((7+5)-8)*6
                        ((7-8)+5)*6
                        ((5-8)+7)*6