24 game/Solve
From Rosetta Code
< 24 game
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
[edit] 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
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.
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 )
[edit] 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
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)
[edit] AutoHotkey
Output is in RPN.
#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
}
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/-/
[edit] BBC BASIC
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
Sample output:
(1+2+3)*4 6*8/(9-7) (1+2)*(1+7) (5+5-6)*6
[edit] 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.
#include <stdio.h>Sample output:
#include <stdlib.h>
#include <time.h>
#define n_cards 4
#define solve_goal 24
#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)
{
const 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;
}
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
[edit] Clojure
The code:
(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)))))
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
[edit] CoffeeScript
# This program tries to find some way to turn four digits into an arithmetic
# expression that adds up to 24.
#
# Example solution for 5, 7, 8, 8:
# (((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})"
,
]
# 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'}"
output
> 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)
[edit] Common 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))
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
[edit] D
This uses the Rational struct and permutations functions of two other Rosetta Code Tasks.
import std.stdio, std.algorithm, std.range, std.typecons, std.conv,
std.string, permutations2, arithmetic_rational;
string solve(in int target, in int[] problem) {
static struct ComputeAllOperations {
//static struct T { Rational r; string e; }
alias T = Tuple!(Rational,"r", string,"e");
Rational[] L;
int opApply(in int delegate(ref T) dg) {
int result;
if (!L.empty) {
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;
auto sub = [T(x * y, "*"), T(x + y, "+"), T(x - y, "-")];
if (y) sub ~= [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; problem.map!Rational.array.permutations)
foreach (sol; ComputeAllOperations(p))
if (sol.r == target)
return sol.e;
return "No solution";
}
void main() {
foreach (prob; [[6, 7, 9, 5], [3, 3, 8, 8], [1, 1, 1, 1]])
writeln(prob, ": ", solve(24, prob));
}
- Output:
[6, 7, 9, 5]: (6+(9*(7-5))) [3, 3, 8, 8]: (8/(3-(8/3))) [1, 1, 1, 1]: No solution
[edit] Euler Math Toolbox
Via brute force.
>function try24 (v) ...
$n=cols(v);
$if n==1 and v[1]~=24 then
$ "Solved the problem",
$ return 1;
$endif
$loop 1 to n
$ w=tail(v,2);
$ loop 1 to n-1
$ h=w; a=v[1]; b=w[1];
$ w[1]=a+b; if try24(w); ""+a+"+"+b+"="+(a+b), return 1; endif;
$ w[1]=a-b; if try24(w); ""+a+"-"+b+"="+(a-b), return 1; endif;
$ w[1]=a*b; if try24(w); ""+a+"*"+b+"="+(a*b), return 1; endif;
$ if not b~=0 then
$ w[1]=a/b; if try24(w); ""+a+"/"+b+"="+(a/b), return 1; endif;
$ endif;
$ w=rotright(w);
$ end;
$ v=rotright(v);
$end;
$return 0;
$endfunction
>try24([1,2,3,4]);
Solved the problem
6*4=24
3+3=6
1+2=3
>try24([8,7,7,1]);
Solved the problem
22+2=24
14+8=22
7+7=14
>try24([8,4,7,1]);
Solved the problem
6*4=24
7-1=6
8-4=4
>try24([3,4,5,6]);
Solved the problem
4*6=24
-1+5=4
3-4=-1
[edit] 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
module helpersSample output (using g95):
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
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
[edit] 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]);
# "77*1-2/"
Player24([9,8,7,6]);
# "68*97-/"
Player24([1,1,7,7]);
# fail
# Solutions with only one distinct digit are found only for 3, 4, 5, 6:
Player24([3,3,3,3]);
# "33*3*3-"
Player24([4,4,4,4]);
# "44*4+4+"
Player24([5,5,5,5]);
# "55*55/-"
Player24([6,6,6,6]);
# "66*66+-"
# A tricky one:
Player24([3,3,8,8]);
"8383/-/"
[edit] 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")
}
}
}
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
[edit] Gosu
uses java.lang.Integer
uses java.lang.Double
uses java.lang.System
uses java.util.ArrayList
uses java.util.LinkedList
uses java.util.List
uses java.util.Scanner
uses java.util.Stack
function permutations<T>( lst : List<T> ) : List<List<T>> {
if( lst.size() == 0 ) return {}
if( lst.size() == 1 ) return { lst }
var pivot = lst.get(lst.size()-1)
var sublist = new ArrayList<T>( lst )
sublist.remove( sublist.size() - 1 )
var subPerms = permutations( sublist )
var ret = new ArrayList<List<T>>()
for( x in subPerms ) {
for( e in x index i ) {
var next = new LinkedList<T>( x )
next.add( i, pivot )
ret.add( next )
}
x.add( pivot )
ret.add( x )
}
return ret
}
function readVals() : List<Integer> {
var line = new java.io.BufferedReader( new java.io.InputStreamReader( System.in ) ).readLine()
var scan = new Scanner( line )
var ret = new ArrayList<Integer>()
for( i in 0..3 ) {
var next = scan.nextInt()
if( 0 >= next || next >= 10 ) {
print( "Invalid entry: ${next}" )
return null
}
ret.add( next )
}
return ret
}
function getOp( i : int ) : char[] {
var ret = new char[3]
var ops = { '+', '-', '*', '/' }
ret[0] = ops[i / 16]
ret[1] = ops[(i / 4) % 4 ]
ret[2] = ops[i % 4 ]
return ret
}
function isSoln( nums : List<Integer>, ops : char[] ) : boolean {
var stk = new Stack<Double>()
for( n in nums ) {
stk.push( n )
}
for( c in ops ) {
var r = stk.pop().doubleValue()
var l = stk.pop().doubleValue()
if( c == '+' ) {
stk.push( l + r )
} else if( c == '-' ) {
stk.push( l - r )
} else if( c == '*' ) {
stk.push( l * r )
} else if( c == '/' ) {
// Avoid division by 0
if( r == 0.0 ) {
return false
}
stk.push( l / r )
}
}
return java.lang.Math.abs( stk.pop().doubleValue() - 24.0 ) < 0.001
}
function printSoln( nums : List<Integer>, ops : char[] ) {
// RPN: a b c d + - *
// Infix (a * (b - (c + d)))
print( "Found soln: (${nums.get(0)} ${ops[0]} (${nums.get(1)} ${ops[1]} (${nums.get(2)} ${ops[2]} ${nums.get(3)})))" )
}
System.out.print( "#> " )
var vals = readVals()
var opPerms = 0..63
var solnFound = false
for( i in permutations( vals ) ) {
for( j in opPerms ) {
var opList = getOp( j )
if( isSoln( i, opList ) ) {
printSoln( i, opList )
solnFound = true
}
}
}
if( ! solnFound ) {
print( "No solution!" )
}
[edit] 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)
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)))
[edit] 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.
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
strings.icn provides deletec and permutes
[edit] 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
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.]
[edit] JavaScript
This is a translation of the C code.
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+"<br>")}
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");
Examples:
(((3*1)*4)*2) ((6*8)/((9-7))) (((1+7))*(2+1))
[edit] Julia
function solve24(nums)
length(nums) != 4 && error("Input must be a 4-element Array")
syms = [+,-,*,/]
for x in syms, y in syms, z in syms
for i = 1:24
a,b,c,d = nthperm(nums,i)
if round(x(y(a,b),z(c,d)),5) == 24
return "($a$y$b)$x($c$z$d)"
elseif round(x(a,y(b,z(c,d))),5) == 24
return "$a$x($b$y($c$z$d))"
elseif round(x(y(z(c,d),b),a),5) == 24
return "(($c$z$d)$y$b)$x$a"
elseif round(x(y(b,z(c,d)),a),5) == 24
return "($b$y($c$z$d))$x$a"
end
end
end
return "0"
end
- Output:
julia> solve24([6, 8, 2, 8]) "(6+8)+(2+8)" julia> solve24([4, 2, 8, 1]) "8*(4+(1-2))" julia> solve24([3, 1, 9, 9])# "(9-1)*(9/3)" julia> solve24([5, 7, 5, 1])# "0" julia> solve24([5, 8, 4, 1]) "8+(4*(5-1))" julia> solve24([8, 3, 4, 9]) "(8+3)+(4+9)" julia> solve24([3, 7, 4, 4])# "4*(3+(7-4))" julia> solve24([5, 6, 4, 1]) "4/(1-(5/6))" julia> solve24([5, 5, 9, 8]) "(8-9)+(5*5)"
[edit] Liberty BASIC
dim d(4)
input "Enter 4 digits: "; a$
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
'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
[edit] Lua
Generic solver: pass card of any size with 1st argument and target number with second.
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)
- 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 .....
[edit] Mathematica
The code:
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] :> #2[[q]],
n[q_] :> #3[[q]]} &,
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_] :> val[[q]]) == 24] &] /.
Table[v[q] -> val[[q]], {q, 4}])]
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 (
Tuplesallows repeats) of the four allowed operations. - Ex:
{{Plus, Plus, Plus}, {Plus, Plus, Subtract}, <<60>>, {Divide, Divide, Times}, {Divide, Divide, Divide}}
- Find all combinations (
Permutations[Array[v, 4]]- Find all permutations (
Permutationsdoes 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]}}
- Find all permutations (
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
Outerand remove any duplicates (Union).
- Get rid of any sublists caused by
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).
- Select the elements of the above list where substituting the real values returns 24 (and do it
HoldForm /@ % /. Table[v[q] -> val[[q]], {q, 4}]- Apply
HoldFormso that substituting numbers will not cause evaluation (otherwise it would only ever return lists like{24, 24, 24}!) and substitute the numbers in.
- Apply
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:
- For each result, turn the expression into a string (for easy manipulation), strip the "
game24play[RandomInteger[{1, 9}, 4]]
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 
, 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, ...}.
[edit] 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
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)
[edit] 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
# Fischer-Krause ordered permutation generator
# 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))',
);
# 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;
}
}
}
}
Example output:
E:\Temp>24solve.pl Enter four integers or 'q' to exit: 1 3 3 8 ((1 + 8) * 3) - 3 ((1 + 8) * 3) - 3 ((8 + 1) * 3) - 3 ((8 - 1) * 3) + 3 ((8 + 1) * 3) - 3 ((8 - 1) * 3) + 3 (3 * (1 + 8)) - 3 (3 * (8 + 1)) - 3 (3 * (8 - 1)) + 3 (3 * (1 + 8)) - 3 (3 * (8 + 1)) - 3 (3 * (8 - 1)) + 3 3 - ((1 - 8) * 3) 3 + ((8 - 1) * 3) 3 - ((1 - 8) * 3) 3 + ((8 - 1) * 3) 3 - (3 * (1 - 8)) 3 + (3 * (8 - 1)) 3 - (3 * (1 - 8)) 3 + (3 * (8 - 1)) Enter four integers or 'q' to exit: q E:\Temp>
[edit] Perl 6
A loose translation of the Perl entry. Does not return every possible permutation of the possible solutions. Filters out duplicates (from repeated digits) and only reports the solution for a particular order of digits and operators with the fewest parenthesis (avoids reporting duplicate solutions only differing by unnecessary parenthesis).
Since Perl 6 uses Rational numbers for division (whenever possible) there is no loss of precision as is common with floating point division. So a comparison like (1 + 7) / (1 / 3) == 24 "Just Works"™
my @digits;
my $amount = 4;
# Get $amount digits from the user,
# ask for more if they don't supply enough
while @digits.elems < $amount {
@digits ,= (prompt "Enter {$amount - @digits} digits from 1 to 9, "
~ '(repeats allowed): ').comb(/<[1..9]>/);
}
# Throw away any extras
@digits = @digits[^$amount];
# Generate combinations of operators
my @op = <+ - * />;
my @ops = map {my $a = $_; map {my $b = $_; map {[$a,$b,$_]}, @op}, @op}, @op;
# Enough sprintf formats to cover most precedence orderings
my @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)',
'%d %s (%d %s %d %s %d)',
'%d %s (%d %s (%d %s %d))',
);
# Brute force test the different permutations
for unique permutations @digits -> @p {
for @ops -> @o {
for @formats -> $format {
my $string = sprintf $format, @p[0], @o[0],
@p[1], @o[1], @p[2], @o[2], @p[3];
my $result = try { eval($string) };
say "$string = 24" and last if $result and $result == 24;
}
}
}
# Perl 6 translation of Fischer-Krause ordered permutation algorithm
sub permutations (@array) {
my @index = ^@array;
my $last = @index[*-1];
my (@permutations, $rev, $fwd);
loop {
push @permutations, [@array[@index]];
$rev = $last;
--$rev while $rev and @index[$rev-1] > @index[$rev];
return @permutations unless $rev;
$fwd = $rev;
push @index, @index.splice($rev).reverse;
++$fwd while @index[$rev-1] > @index[$fwd];
@index[$rev-1,$fwd] = @index[$fwd,$rev-1];
}
}
# Only return unique sub-arrays
sub unique (@array) {
my %h = map { $_.Str => $_ }, @array;
%h.values;
}
Sample Output
Enter 4 digits from 1 to 9, (repeats allowed): 3711 3 * (7 + 1 * 1) = 24 3 * (7 + 1 / 1) = 24 3 * (7 * 1 + 1) = 24 3 * (7 / 1 + 1) = 24 (3 + 1) * (7 - 1) = 24 3 * (1 + 7 * 1) = 24 3 * (1 + 7 / 1) = 24 (3 * 1) * (7 + 1) = 24 (3 / 1) * (7 + 1) = 24 3 / (1 / (7 + 1)) = 24 3 * (1 + 1 * 7) = 24 (3 * 1) * (1 + 7) = 24 3 * (1 / 1 + 7) = 24 (3 / 1) * (1 + 7) = 24 3 / (1 / (1 + 7)) = 24 (7 + 1) * 3 * 1 = 24 (7 + 1) * 3 / 1 = 24 (7 - 1) * (3 + 1) = 24 (7 + 1) * 1 * 3 = 24 (7 + 1) / 1 * 3 = 24 (7 + 1) / (1 / 3) = 24 (7 - 1) * (1 + 3) = 24 (7 * 1 + 1) * 3 = 24 (7 / 1 + 1) * 3 = 24 (1 + 3) * (7 - 1) = 24 (1 * 3) * (7 + 1) = 24 (1 * 3) * (1 + 7) = 24 (1 + 7) * 3 * 1 = 24 (1 + 7) * 3 / 1 = 24 (1 + 7) * 1 * 3 = 24 (1 + 7) / 1 * 3 = 24 (1 + 7) / (1 / 3) = 24 (1 * 7 + 1) * 3 = 24 (1 + 1 * 7) * 3 = 24 (1 * 1 + 7) * 3 = 24 (1 / 1 + 7) * 3 = 24 Enter 4 digits from 1 to 9, (repeats allowed): 5 5 5 5 5 * 5 - 5 / 5 = 24 Enter 4 digits from 1 to 9, (repeats allowed): 8833 8 / (3 - 8 / 3) = 24
[edit] PicoLisp
We use Pilog (PicoLisp Prolog) to solve this task
(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
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))
[edit] ProDOS
Note This example uses the math module:
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
Sample Output:
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
[edit] 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].
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]).
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 ....
[edit] 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.
'''
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 player'
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()
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
[edit] 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 )
[edit] 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.
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(expr[[1]]), goal)) return(expr[[1]])
}
return(NA)
}
Example Output:
> 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)
[edit] Racket
The sequence of all possible variants of expressions with given numbers n1, n2, n3, n4 and operations o1, o2, o3.
(define (in-variants n1 o1 n2 o2 n3 o3 n4)
(let ([o1n (object-name o1)]
[o2n (object-name o2)]
[o3n (object-name o3)])
(with-handlers ((exn:fail:contract:divide-by-zero? (λ (_) empty-sequence)))
(in-parallel
(list (o1 (o2 (o3 n1 n2) n3) n4)
(o1 (o2 n1 (o3 n2 n3)) n4)
(o1 (o2 n1 n2) (o3 n3 n4))
(o1 n1 (o2 (o3 n2 n3) n4))
(o1 n1 (o2 n2 (o3 n3 n4))))
(list `(((,n1 ,o3n ,n2) ,o2n ,n3) ,o1n ,n4)
`((,n1 ,o2n (,n2 ,o3n ,n3)) ,o1n ,n4)
`((,n1 ,o2n ,n2) ,o1n (,n3 ,o3n ,n4))
`(,n1 ,o1n ((,n2 ,o3n ,n3) ,o2n ,n4))
`(,n1 ,o1n (,n2 ,o2n (,n3 ,o3n ,n4))))))))
Search for all solutions using brute force:
(define (find-solutions numbers (goal 24))
(define in-operations (list + - * /))
(remove-duplicates
(for*/list ([n1 numbers]
[n2 (remove-from numbers n1)]
[n3 (remove-from numbers n1 n2)]
[n4 (remove-from numbers n1 n2 n3)]
[o1 in-operations]
[o2 in-operations]
[o3 in-operations]
[(res expr) (in-variants n1 o1 n2 o2 n3 o3 n4)]
#:when (= res goal))
expr)))
(define (remove-from numbers . n) (foldr remq numbers n))
Examples:
> (find-solutions '(3 8 3 8)) '((8 / (3 - (8 / 3)))) > (find-solutions '(3 8 2 9)) '(((8 / 2) * (9 - 3)) (8 / (2 / (9 - 3))) (8 * (9 - (3 * 2))) (8 * ((9 - 3) / 2)) ((8 * (9 - 3)) / 2) (8 * (9 - (2 * 3))) ((9 - 3) * (8 / 2)) (((9 - 3) * 8) / 2) ((9 - (3 * 2)) * 8) (((9 - 3) / 2) * 8) ((9 - 3) / (2 / 8)) ((9 - (2 * 3)) * 8))
In order to find just one solution effectively one needs to change for*/list to for*/first in the function find-solutions.
[edit] 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 and 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.*/
digs=123456789 /*numerals (digits) that can be used. */
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 /*j*/
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 _=1 for 4 /*define versions for faster execution.*/
g._=substr(g,_,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 3; 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 heavy 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 (ugh).*/
interpret 'x=' e /*... strain... */
x=x/1 /*remove trailing decimal points(maybe)*/
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
/*───────────────────────────S subroutine───────────────────────────────*/
s: if arg(1)==1 then return ''; return 's' /*simple pluralizer.*/
/*───────────────────────────validate subroutine────────────────────────*/
validate: parse arg y; errCode=0; _v=verify(y,digs)
select
when y=='' then call ger 'no digits entered.'
when length(y)<4 then call ger 'not enough digits entered, must be 4'
when length(y)>4 then call ger 'too many digits entered, must be 4'
when pos(0,y)\==0 then call ger "can't use the digit 0 (zero)"
when _v\==0 then call ger 'illegal character:' substr(y,_v,1)
otherwise nop
end /*select*/
return \errCode
Some older REXXes don't have a changestr bif, so one is included here ──► CHANGESTR.REX.
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
[edit] Ruby
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))',
].map{|expr| [expr, expr.gsub('%d', 'Rational(%d,1)')]}
OPERATORS = [:+, :-, :*, :/].repeated_permutation(3)
OBJECTIVE = Rational(24,1)
def self.solve(digits)
solutions = []
digits.permutation.to_a.uniq.each do |a,b,c,d|
OPERATORS.each do |op1,op2,op3|
EXPRESSIONS.each do |expr,expr_rat|
# evaluate using rational arithmetic
test = expr_rat % [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
solutions
end
end
# 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}"
solutions = TwentyFourGamePlayer.solve(digits)
if solutions.empty?
puts "no solutions"
else
puts "found #{solutions.size} solutions, including #{solutions.first}"
puts solutions.sort
end
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))
[edit] Scala
A non-interactive player.
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)
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))))
[edit] Tcl
This is a complete Tcl script, intended to be invoked from the command line.
package require struct::list
# 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))}
}
# 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}}}]
# The permitted operations
set operations {+ - * /}
# Given a list of four integers (precondition not checked!) return a list of
# 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
}
# 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 $argv
Demonstrating it in use:
bash$ 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
[edit] 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.
#import std
#import nat
#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"
test program:
#show+
test_games = mat` * pad` *K7 pad0 game24* <<2,3,8,9>,<5,7,4,1>,<5,6,7,8>>
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
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