Sum to 100
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Find solutions to the sum to one hundred puzzle.
Add (insert) the mathematical
operators + or - (plus
or minus) before any of the digits in the
decimal numeric string 123456789 such that the
resulting mathematical expression adds up to a
particular sum (in this iconic case, 100).
Example:
123 + 4 - 5 + 67 - 89 = 100
Show all output here.
- Show all solutions that sum to 100
- Show the sum that has the maximum number of solutions (from zero to infinity‡)
- Show the lowest positive sum that can't be expressed (has no solutions), using the rules for this task
- Show the ten highest numbers that can be expressed using the rules for this task (extra credit)
‡ (where infinity would be a relatively small 123,456,789)
An example of a sum that can't be expressed (within the rules of this task) is: 5074
(which, of course, isn't the lowest positive sum that can't be expressed).
Ada
The Package Sum_To
Between any two consecutive digits, there can be a "+", a "-", or no operator. E.g., the digits "4" and "5" occur in the string as either of the following three substrings: "4+5", "4-5", or "45". For the first digit, we only have two choices: "+1" (written as "1"), and "-1". This makes 2*3^8 (two times (three to the power of eight)) different strings. Essential is the generic function Eval in the package Sum_To calls the procedure Callback for each such string Str, with the number Int holding the sum corresponding to the evaluation of Str. The second generic procedure Print is for convenience. If the Sum fits the condition, i.e., if Print_If(Sum, Number), then Print writes Sum = Str to the output.
<lang Ada>package Sum_To is
generic
with procedure Callback(Str: String; Int: Integer);
procedure Eval;
generic
Number: Integer;
with function Print_If(Sum, Number: Integer) return Boolean;
procedure Print(S: String; Sum: Integer);
end Sum_To;</lang>
The implementation of Eval follows the observation above: Eval calls Rec_Eval with the initial string "1" and "-1". For each call, Rec_Eval recursively evaluates a ternary tree with 3^8 leafs. At each leaf, Rec_Eval calls Callback. The implementation of Print is straightforward.
<lang Ada>with Ada.Text_IO, Ada.Containers.Ordered_Maps;
package body Sum_To is
procedure Eval is
procedure Rec_Eval(Str: String; Previous, Current, Next: Integer) is
Next_Image: String := Integer'Image(Next); -- Next_Image(1) holds a blank, Next_Image(2) a digit
function Sign(N: Integer) return Integer is (if N<0 then -1 elsif N>0 then 1 else 0);
begin
if Next = 10 then -- end of recursion Callback(Str, Previous+Current); else -- Next < 10 Rec_Eval(Str & Next_Image(2), -- concatenate current and Next Previous, Sign(Current)*(10*abs(Current)+Next), Next+1); Rec_Eval(Str & "+" & Next_Image(2), -- add Next Previous+Current, Next, Next+1); Rec_Eval(Str & "-" & Next_Image(2), -- subtract Next Previous+Current, -Next, Next+1); end if;
end Rec_Eval;
begin -- Eval
Rec_Eval("1", 0, 1, 2); -- unary "+", followed by "1"
Rec_Eval("-1", 0, -1, 2); -- unary "-", followed by "1"
end Eval;
procedure Print(S: String; Sum: Integer) is
-- print solution (S,N), if N=Number
begin
if Print_If(Sum, Number) then
Ada.Text_IO.Put_Line(Integer'Image(Sum) & " = " & S & ";");
end if; end Print;
end Sum_To;</lang>
The First Subtask
Given the package Sum_To, the solution to the first subtask (print all solution for the sum 100) is trivial: Eval_100 calls Print_100 for all 2*3^8 strings, and Print_100 writes the output if the sum is equal to 100.
<lang Ada>with Sum_To;
procedure Sum_To_100 is
procedure Print_100 is new Sum_To.Print(100, "="); procedure Eval_100 is new Sum_To.Eval(Print_100);
begin
Eval_100;
end Sum_To_100;</lang>
- Output:
100 = 123+45-67+8-9; 100 = 123+4-5+67-89; 100 = 123-45-67+89; 100 = 123-4-5-6-7+8-9; 100 = 12+3+4+5-6-7+89; 100 = 12+3-4+5+67+8+9; 100 = 12-3-4+5-6+7+89; 100 = 1+23-4+56+7+8+9; 100 = 1+23-4+5+6+78-9; 100 = 1+2+34-5+67-8+9; 100 = 1+2+3-4+5+6+78+9; 100 = -1+2-3+4+5+6+78+9;
The other subtasks (including the extra credit)
For the three other subtasks, we maintain an ordered map of sums (as the keys) and counters for the number of solutions (as the elements). The procedure Generate_Map generates the Map by calling the procedure Insert_Solution for all 2*3^8 solutions. Finding (1) the sum with the maximal number of solutions, (2) the first sum>=0 without a solution and (3) the ten largest sums with a solution (extra credit) are done by iterating this map.
<lang Ada>with Sum_To, Ada.Containers.Ordered_Maps, Ada.Text_IO; use Ada.Text_IO;
procedure Three_Others is
package Num_Maps is new Ada.Containers.Ordered_Maps
(Key_Type => Integer, Element_Type => Positive);
use Num_Maps;
Map: Num_Maps.Map;
-- global Map stores how often a sum did occur
procedure Insert_Solution(S: String; Sum: Integer) is
-- inserts a solution into global Map
use Num_Maps;
-- use type Num_Maps.Cursor;
Position: Cursor := Map.Find(Sum);
begin
if Position = No_Element then -- first solutions for Sum
Map.Insert(Key => Sum, New_Item => 1); -- counter is 1
else -- increase counter for Sum
Map.Replace_Element(Position => Position, New_Item => (Element(Position))+1);
end if; end Insert_Solution; procedure Generate_Map is new Sum_To.Eval(Insert_Solution); Current: Cursor; -- Points into Map Sum: Integer; -- current Sum of interest Max: Natural;
begin
Generate_Map;
-- find Sum >= 0 with maximum number of solutions
Max := 0; -- number of solutions for Sum (so far, none)
Current := Map.Ceiling(0); -- first element in Map with Sum >= 0
while Has_Element(Current) loop
if Element(Current) > Max then
Max := Element(Current); -- the maximum of solutions, so far Sum := Key(Current); -- the Sum with Max solutions
end if;
Next(Current);
end loop;
Put_Line("Most frequent result:" & Integer'Image(Sum));
Put_Line("Frequency of" & Integer'Image(Sum) & ":" &
Integer'Image(Max));
New_Line;
-- find smallest Sum >= 0 with no solution
Sum := 0;
while Map.Find(Sum) /= No_Element loop
Sum := Sum + 1;
end loop;
Put_Line("Smallest nonnegative impossible sum:" & Integer'Image(Sum));
New_Line;
-- find ten highest numbers with a solution
Current := Map.Last; -- highest element in Map with a solution
Put_Line("Highest sum:" & Integer'Image(Key(Current)));
Put("Next nine:");
for I in 1 .. 9 loop -- 9 steps backward
Previous(Current);
Put(Integer'Image(Key(Current)));
end loop;
New_Line;
end Three_others;</lang>
- Output:
Most frequent result: 9 Frequency of 9: 46 Smallest nonnegative impossible sum: 211 Highest sum: 123456789 Next nine: 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
Aime
<lang aime>integer b, i, j, k, l, p, s, z; index r, w;
i = 0; while (i < 512) {
b = i.bcount;
j = 0;
while (j < 1 << b) {
data e;
j += 1;
k = s = p = 0;
l = j;
z = 1;
while (k < 9) {
if (i & 1 << k) {
e.append("-+"[l & 1]);
s += p * z;
z = (l & 1) * 2 - 1;
l >>= 1;
p = 0;
}
e.append('1' + k);
p = p * 10 + 1 + k;
k += 1;
}
s += p * z;
if (e[0] != '+') {
if (s == 100) {
o_(e, "\n");
}
w[s] += 1;
}
}
i += 1;
}
w.wcall(i_fix, 1, 1, r);
o_(r.back, "\n");
k = 0; for (+k in w) {
if (!w.key(k + 1)) {
o_(k + 1, "\n");
break;
}
}
i = 10; for (k of w) {
o_(k, "\n");
if (!(i -= 1)) {
break;
}
}</lang>
- Output:
123-45-67+89 123+4-5+67-89 12+3+4+5-6-7+89 12-3-4+5-6+7+89 1+23-4+5+6+78-9 1+2+3-4+5+6+78+9 -1+2-3+4+5+6+78+9 123+45-67+8-9 1+2+34-5+67-8+9 12+3-4+5+67+8+9 1+23-4+56+7+8+9 123-4-5-6-7+8-9 9 211 123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
ALGOL 68
<lang algol68>BEGIN
# find the numbers the string 123456789 ( with "+/-" optionally inserted # # before each digit ) can generate #
# experimentation shows that the largest hundred numbers that can be # # generated are are greater than or equal to 56795 # # as we can't declare an array with bounds -123456789 : 123456789 in # # Algol 68G, we use -60000 : 60000 and keep counts for the top hundred #
INT max number = 60 000; [ - max number : max number ]STRING solutions; [ - max number : max number ]INT count; FOR i FROM LWB solutions TO UPB solutions DO solutions[ i ] := ""; count[ i ] := 0 OD;
# calculate the numbers ( up to max number ) we can generate and the strings leading to them #
# also determine the largest numbers we can generate #
[ 100 ]INT largest;
[ 100 ]INT largest count;
INT impossible number = - 999 999 999;
FOR i FROM LWB largest TO UPB largest DO
largest [ i ] := impossible number;
largest count[ i ] := 0
OD;
[ 1 : 18 ]CHAR sum string := ".1.2.3.4.5.6.7.8.9";
[]CHAR sign char = []CHAR( "-", " ", "+" )[ AT -1 ];
# we don't distinguish between strings starting "+1" and starting " 1" #
FOR s1 FROM -1 TO 0 DO
sum string[ 1 ] := sign char[ s1 ];
FOR s2 FROM -1 TO 1 DO
sum string[ 3 ] := sign char[ s2 ];
FOR s3 FROM -1 TO 1 DO
sum string[ 5 ] := sign char[ s3 ];
FOR s4 FROM -1 TO 1 DO
sum string[ 7 ] := sign char[ s4 ];
FOR s5 FROM -1 TO 1 DO
sum string[ 9 ] := sign char[ s5 ];
FOR s6 FROM -1 TO 1 DO
sum string[ 11 ] := sign char[ s6 ];
FOR s7 FROM -1 TO 1 DO
sum string[ 13 ] := sign char[ s7 ];
FOR s8 FROM -1 TO 1 DO
sum string[ 15 ] := sign char[ s8 ];
FOR s9 FROM -1 TO 1 DO
sum string[ 17 ] := sign char[ s9 ];
INT number := 0;
INT part := IF s1 < 0 THEN -1 ELSE 1 FI;
IF s2 = 0 THEN part *:= 10 +:= 2 * SIGN part ELSE number +:= part; part := 2 * s2 FI;
IF s3 = 0 THEN part *:= 10 +:= 3 * SIGN part ELSE number +:= part; part := 3 * s3 FI;
IF s4 = 0 THEN part *:= 10 +:= 4 * SIGN part ELSE number +:= part; part := 4 * s4 FI;
IF s5 = 0 THEN part *:= 10 +:= 5 * SIGN part ELSE number +:= part; part := 5 * s5 FI;
IF s6 = 0 THEN part *:= 10 +:= 6 * SIGN part ELSE number +:= part; part := 6 * s6 FI;
IF s7 = 0 THEN part *:= 10 +:= 7 * SIGN part ELSE number +:= part; part := 7 * s7 FI;
IF s8 = 0 THEN part *:= 10 +:= 8 * SIGN part ELSE number +:= part; part := 8 * s8 FI;
IF s9 = 0 THEN part *:= 10 +:= 9 * SIGN part ELSE number +:= part; part := 9 * s9 FI;
number +:= part;
IF number >= LWB solutions
AND number <= UPB solutions
THEN
solutions[ number ] +:= ";" + sum string;
count [ number ] +:= 1
FI;
BOOL inserted := FALSE;
FOR l pos FROM LWB largest TO UPB largest WHILE NOT inserted DO
IF number > largest[ l pos ] THEN
# found a new larger number #
FOR m pos FROM UPB largest BY -1 TO l pos + 1 DO
largest [ m pos ] := largest [ m pos - 1 ];
largest count[ m pos ] := largest count[ m pos - 1 ]
OD;
largest [ l pos ] := number;
largest count[ l pos ] := 1;
inserted := TRUE
ELIF number = largest[ l pos ] THEN
# have another way of generating this number #
largest count[ l pos ] +:= 1;
inserted := TRUE
FI
OD
OD
OD
OD
OD
OD
OD
OD
OD
OD;
# show the solutions for 100 #
print( ( "100 has ", whole( count[ 100 ], 0 ), " solutions:" ) );
STRING s := solutions[ 100 ];
FOR s pos FROM LWB s TO UPB s DO
IF s[ s pos ] = ";" THEN print( ( newline, " " ) )
ELIF s[ s pos ] /= " " THEN print( ( s[ s pos ] ) )
FI
OD;
print( ( newline ) );
# find the number with the most solutions #
INT max solutions := 0;
INT number with max := LWB count - 1;
FOR n FROM 0 TO max number DO
IF count[ n ] > max solutions THEN
max solutions := count[ n ];
number with max := n
FI
OD;
FOR n FROM LWB largest count TO UPB largest count DO
IF largest count[ n ] > max solutions THEN
max solutions := largest count[ n ];
number with max := largest[ n ]
FI
OD;
print( ( whole( number with max, 0 ), " has the maximum number of solutions: ", whole( max solutions, 0 ), newline ) );
# find the smallest positive number that has no solutions #
BOOL have solutions := TRUE;
FOR n FROM 0 TO max number
WHILE IF NOT ( have solutions := count[ n ] > 0 )
THEN print( ( whole( n, 0 ), " is the lowest positive number with no solutions", newline ) )
FI;
have solutions
DO SKIP OD;
IF have solutions
THEN print( ( "All positive numbers up to ", whole( max number, 0 ), " have solutions", newline ) )
FI;
print( ( "The 10 largest numbers that can be generated are:", newline ) );
FOR t pos FROM 1 TO 10 DO
print( ( " ", whole( largest[ t pos ], 0 ) ) )
OD;
print( ( newline ) )
END</lang>
- Output:
100 has 12 solutions:
-1+2-3+4+5+6+78+9
12-3-4+5-6+7+89
123-4-5-6-7+8-9
123-45-67+89
123+4-5+67-89
123+45-67+8-9
12+3-4+5+67+8+9
12+3+4+5-6-7+89
1+23-4+56+7+8+9
1+23-4+5+6+78-9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
9 has the maximum number of solutions: 46
211 is the lowest positive number with no solutions
The 10 largest numbers that can be generated are:
123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
AppleScript
AppleScript is essentially out of its depth at this scale. The first task (number of distinct paths to 100) is accessible within a few seconds. Subsequent tasks, however, terminate only (if at all) after impractical amounts of time. Note the contrast with the lighter and more optimised JavaScript interpreter, which takes less than half a second to return full results for all the listed tasks. <lang AppleScript>use framework "Foundation" -- for basic NSArray sort
property pSigns : {1, 0, -1} --> ( + | unsigned | - ) property plst100 : {"Sums to 100:", ""} property plstSums : {} property plstSumsSorted : missing value property plstSumGroups : missing value
-- data Sign :: [ 1 | 0 | -1 ] = ( Plus | Unsigned | Minus ) -- asSum :: [Sign] -> Int on asSum(xs)
script
on |λ|(a, sign, i)
if sign ≠ 0 then
{digits:{}, n:(n of a) + (sign * ((i & digits of a) as string as integer))}
else
{digits:{i} & (digits of a), n:n of a}
end if
end |λ|
end script
set rec to foldr(result, {digits:{}, n:0}, xs)
set ds to digits of rec
if length of ds > 0 then
(n of rec) + (ds as string as integer)
else
n of rec
end if
end asSum
-- data Sign :: [ 1 | 0 | -1 ] = ( Plus | Unisigned | Minus ) -- asString :: [Sign] -> String on asString(xs)
script
on |λ|(a, sign, i)
set d to i as string
if sign ≠ 0 then
if sign > 0 then
a & " +" & d
else
a & " -" & d
end if
else
a & d
end if
end |λ|
end script
foldl(result, "", xs)
end asString
-- sumsTo100 :: () -> String on sumsTo100()
-- From first permutation without leading '+' (3 ^ 8) to end of universe (3 ^ 9)
repeat with i from 6561 to 19683
set xs to nthPermutationWithRepn(pSigns, 9, i)
if asSum(xs) = 100 then set end of plst100 to asString(xs)
end repeat
intercalate(linefeed, plst100)
end sumsTo100
-- mostCommonSum :: () -> String
on mostCommonSum()
-- From first permutation without leading '+' (3 ^ 8) to end of universe (3 ^ 9)
repeat with i from 6561 to 19683
set intSum to asSum(nthPermutationWithRepn(pSigns, 9, i))
if intSum ≥ 0 then set end of plstSums to intSum
end repeat
set plstSumsSorted to sort(plstSums)
set plstSumGroups to group(plstSumsSorted)
script groupLength
on |λ|(a, b)
set intA to length of a
set intB to length of b
if intA < intB then
-1
else if intA > intB then
1
else
0
end if
end |λ|
end script
set lstMaxSum to maximumBy(groupLength, plstSumGroups)
intercalate(linefeed, ¬
{"Most common sum: " & item 1 of lstMaxSum, ¬
"Number of instances: " & length of lstMaxSum})
end mostCommonSum
-- TEST ----------------------------------------------------------------------
on run
return sumsTo100() -- Also returns a value, but slow: -- mostCommonSum()
end run
-- GENERIC FUNCTIONS ---------------------------------------------------------
-- nthPermutationWithRepn :: [a] -> Int -> Int -> [a] on nthPermutationWithRepn(xs, groupSize, iIndex)
set intBase to length of xs
set intSetSize to intBase ^ groupSize
if intBase < 1 or iIndex > intSetSize then
{}
else
set baseElems to inBaseElements(xs, iIndex)
set intZeros to groupSize - (length of baseElems)
if intZeros > 0 then
replicate(intZeros, item 1 of xs) & baseElems
else
baseElems
end if
end if
end nthPermutationWithRepn
-- inBaseElements :: [a] -> Int -> [String] on inBaseElements(xs, n)
set intBase to length of xs
script nextDigit
on |λ|(residue)
set {divided, remainder} to quotRem(residue, intBase)
{valid:divided > 0, value:(item (remainder + 1) of xs), new:divided}
end |λ|
end script
reverse of unfoldr(nextDigit, n)
end inBaseElements
-- sort :: [a] -> [a] on sort(lst)
((current application's NSArray's arrayWithArray:lst)'s ¬
sortedArrayUsingSelector:"compare:") as list
end sort
-- maximumBy :: (a -> a -> Ordering) -> [a] -> a on maximumBy(f, xs)
set cmp to mReturn(f)
script max
on |λ|(a, b)
if a is missing value or cmp's |λ|(a, b) < 0 then
b
else
a
end if
end |λ|
end script
foldl(max, missing value, xs)
end maximumBy
-- group :: Eq a => [a] -> a on group(xs)
script eq
on |λ|(a, b)
a = b
end |λ|
end script
groupBy(eq, xs)
end group
-- groupBy :: (a -> a -> Bool) -> [a] -> a on groupBy(f, xs)
set mf to mReturn(f)
script enGroup
on |λ|(a, x)
if length of (active of a) > 0 then
set h to item 1 of active of a
else
set h to missing value
end if
if h is not missing value and mf's |λ|(h, x) then
{active:(active of a) & x, sofar:sofar of a}
else
{active:{x}, sofar:(sofar of a) & {active of a}}
end if
end |λ|
end script
if length of xs > 0 then
set dct to foldl(enGroup, {active:{item 1 of xs}, sofar:{}}, tail(xs))
if length of (active of dct) > 0 then
sofar of dct & {active of dct}
else
sofar of dct
end if
else
{}
end if
end groupBy
-- tail :: [a] -> [a] on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- quotRem :: Integral a => a -> a -> (a, a) on quotRem(m, n)
{m div n, m mod n}
end quotRem
-- replicate :: Int -> a -> [a] on replicate(n, a)
set out to {}
if n < 1 then return out
set dbl to {a}
repeat while (n > 1)
if (n mod 2) > 0 then set out to out & dbl
set n to (n div 2)
set dbl to (dbl & dbl)
end repeat
return out & dbl
end replicate
-- foldr :: (a -> b -> a) -> a -> [b] -> a on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldr
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- unfoldr :: (b -> Maybe (a, b)) -> b -> [a] on unfoldr(f, v)
set mf to mReturn(f)
set lst to {}
set recM to mf's |λ|(v)
repeat while (valid of recM) is true
set end of lst to value of recM
set recM to mf's |λ|(new of recM)
end repeat
lst & value of recM
end unfoldr
-- until :: (a -> Bool) -> (a -> a) -> a -> a on |until|(p, f, x)
set mp to mReturn(p)
set v to x
tell mReturn(f)
repeat until mp's |λ|(v)
set v to |λ|(v)
end repeat
end tell
return v
end |until|
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn</lang>
- Output:
Sums to 100: 1 +2 +34 -5 +67 -8 +9 1 +2 +3 -4 +5 +6 +78 +9 1 +23 -4 +5 +6 +78 -9 1 +23 -4 +56 +7 +8 +9 12 +3 +4 +5 -6 -7 +89 12 +3 -4 +5 +67 +8 +9 123 +45 -67 +8 -9 123 +4 -5 +67 -89 123 -45 -67 +89 123 -4 -5 -6 -7 +8 -9 12 -3 -4 +5 -6 +7 +89 -1 +2 -3 +4 +5 +6 +78 +9
AutoHotkey
<lang AutoHotkey>output:="" for k, v in (sum2num(100))
output .= k "`n"
MsgBox, 262144, , % output
mx := [] loop 123456789{
x := sum2num(A_Index) mx[x.Count()] := mx[x.Count()] ? mx[x.Count()] ", " A_Index : A_Index
} MsgBox, 262144, , % mx[mx.MaxIndex()] " has " mx.MaxIndex() " solutions"
loop {
if !sum2num(A_Index).Count(){
MsgBox, 262144, , % "Lowest positive sum that can't be expressed is " A_Index
break
}
} return
sum2num(num){
output := []
loop % 6561
{
oper := SubStr("00000000" ConvertBase(10, 3, A_Index-1), -7)
oper := StrReplace(oper, 0, "+")
oper := StrReplace(oper, 1, "-")
oper := StrReplace(oper, 2, ".")
str := ""
loop 9
str .= A_Index . SubStr(oper, A_Index, 1)
str := StrReplace(str, ".")
loop 2
{
val := 0
for i, v in StrSplit(str, "+")
for j, m in StrSplit(v, "-")
val += A_Index=1 ? m : 0-m
if (val = num)
output[str] := true
str := "-" str
}
}
Sort, output
return output
}
ConvertBase(InputBase, OutputBase, nptr){
static u := A_IsUnicode ? "_wcstoui64" : "_strtoui64"
static v := A_IsUnicode ? "_i64tow" : "_i64toa"
VarSetCapacity(s, 66, 0)
value := DllCall("msvcrt.dll\" u, "Str", nptr, "UInt", 0, "UInt", InputBase, "CDECL Int64")
DllCall("msvcrt.dll\" v, "Int64", value, "Str", s, "UInt", OutputBase, "CDECL")
return s
}</lang>
- Output:
--------------------------- -1+2-3+4+5+6+78+9 1+2+3-4+5+6+78+9 1+2+34-5+67-8+9 1+23-4+5+6+78-9 1+23-4+56+7+8+9 12+3+4+5-6-7+89 12+3-4+5+67+8+9 12-3-4+5-6+7+89 123+4-5+67-89 123+45-67+8-9 123-4-5-6-7+8-9 123-45-67+89 --------------------------- 9 has 46 solutions --------------------------- Lowest positive sum that can't be expressed is 211 ---------------------------
AWK
Awk is a weird language: there are no integers, no switch-case (in the standard language version), programs are controlled by data flow, the interpreter speed is moderate. The advantage of Awk are associative arrays, used here for counting how many times we get the same sum as the result of calculations. <lang AWK>#
- RossetaCode: Sum to 100, AWK.
- Find solutions to the "sum to one hundred" puzzle.
function evaluate(code) {
value = 0
number = 0
power = 1
for ( k = 9; k >= 1; k-- )
{
number = power*k + number
op = code % 3
if ( op == 0 ) {
value = value + number
number = 0
power = 1
} else if (op == 1 ) {
value = value - number
number = 0
power = 1
} else if ( op == 2) {
power = power * 10
} else {
}
code = int(code / 3);
}
return value;
}
function show(code) {
s = ""
a = 19683
b = 6561
for ( k = 1; k <= 9; k++ )
{
op = int( (code % a) / b )
if ( op == 0 && k > 1 )
s = s "+"
else if ( op == 1 )
s = s "-"
else {
}
a = b
b = int(b / 3)
s = s k
}
printf "%9d = %s\n", evaluate(code), s;
}
BEGIN {
nexpr = 13122
print print "Show all solutions that sum to 100" print for ( i = 0; i < nexpr; i++ ) if ( evaluate(i) == 100 ) show(i);
print
print "Show the sum that has the maximum number of solutions"
print
for ( i = 0; i < nexpr; i++ ) {
sum = evaluate(i);
if ( sum >= 0 )
stat[sum]++;
}
best = (-1);
for ( sum in stat )
if ( best < stat[sum] ) {
best = stat[sum]
bestSum = sum
}
delete stat
printf "%d has %d solutions\n", bestSum, best
print
print "Show the lowest positive number that can't be expressed"
print
for ( i = 0; i <= 123456789; i++ ){
for ( j = 0; j < nexpr; j++ )
if ( i == evaluate(j) )
break;
if ( i != evaluate(j) )
break;
}
printf "%d\n",i
print
print "Show the ten highest numbers that can be expressed"
print
limit = 123456789 + 1;
for ( i = 1; i <= 10; i++ )
{
best = 0;
for ( j = 0; j < nexpr; j++ )
{
test = evaluate(j);
if ( test < limit && test > best ) best = test;
}
for ( j = 0; j < nexpr; j++ ) if ( evaluate(j) == best ) show(j)
limit = best
}
}</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
C
Optimized for speed
Warning: this version requires at least four byte integers. <lang C>/*
* RossetaCode: Sum to 100, C99, an algorithm using ternary numbers. * * Find solutions to the "sum to one hundred" puzzle. */
- include <stdio.h>
- include <stdlib.h>
/*
* There are only 13122 (i.e. 2*3**8) different possible expressions, * thus we can encode them as positive integer numbers from 0 to 13121. */
- define NUMBER_OF_EXPRESSIONS (2 * 3*3*3*3 * 3*3*3*3 )
enum OP { ADD, SUB, JOIN }; typedef int (*cmp)(const void*, const void*);
// Replacing struct Expression and struct CountSum by a tuple like // struct Pair { int first; int last; } is possible but would make the source // code less readable.
struct Expression{
int sum; int code;
}expressions[NUMBER_OF_EXPRESSIONS]; int expressionsLength = 0; int compareExpressionBySum(const struct Expression* a, const struct Expression* b){
return a->sum - b->sum;
}
struct CountSum{
int counts; int sum;
}countSums[NUMBER_OF_EXPRESSIONS]; int countSumsLength = 0; int compareCountSumsByCount(const struct CountSum* a, const struct CountSum* b){
return a->counts - b->counts;
}
int evaluate(int code){
int value = 0, number = 0, power = 1;
for ( int k = 9; k >= 1; k-- ){
number = power*k + number;
switch( code % 3 ){
case ADD: value = value + number; number = 0; power = 1; break;
case SUB: value = value - number; number = 0; power = 1; break;
case JOIN: power = power * 10 ; break;
}
code /= 3;
}
return value;
}
void print(int code){
static char s[19]; char* p = s;
int a = 19683, b = 6561;
for ( int k = 1; k <= 9; k++ ){
switch((code % a) / b){
case ADD: if ( k > 1 ) *p++ = '+'; break;
case SUB: *p++ = '-'; break;
}
a = b;
b = b / 3;
*p++ = '0' + k;
}
*p = 0;
printf("%9d = %s\n", evaluate(code), s);
}
void comment(char* string){
printf("\n\n%s\n\n", string);
}
void init(void){
for ( int i = 0; i < NUMBER_OF_EXPRESSIONS; i++ ){
expressions[i].sum = evaluate(i);
expressions[i].code = i;
}
expressionsLength = NUMBER_OF_EXPRESSIONS;
qsort(expressions,expressionsLength,sizeof(struct Expression),(cmp)compareExpressionBySum);
int j = 0;
countSums[0].counts = 1;
countSums[0].sum = expressions[0].sum;
for ( int i = 0; i < expressionsLength; i++ ){
if ( countSums[j].sum != expressions[i].sum ){
j++;
countSums[j].counts = 1;
countSums[j].sum = expressions[i].sum;
}
else
countSums[j].counts++;
}
countSumsLength = j + 1;
qsort(countSums,countSumsLength,sizeof(struct CountSum),(cmp)compareCountSumsByCount);
}
int main(void){
init();
comment("Show all solutions that sum to 100");
const int givenSum = 100;
struct Expression ex = { givenSum, 0 };
struct Expression* found;
if ( found = bsearch(&ex,expressions,expressionsLength,
sizeof(struct Expression),(cmp)compareExpressionBySum) ){
while ( found != expressions && (found-1)->sum == givenSum )
found--;
while ( found != &expressions[expressionsLength] && found->sum == givenSum )
print(found++->code);
}
comment("Show the positve sum that has the maximum number of solutions");
int maxSumIndex = countSumsLength - 1;
while( countSums[maxSumIndex].sum < 0 )
maxSumIndex--;
printf("%d has %d solutions\n",
countSums[maxSumIndex].sum, countSums[maxSumIndex].counts);
comment("Show the lowest positive number that can't be expressed");
for ( int value = 0; ; value++ ){
struct Expression ex = { value, 0 };
if (!bsearch(&ex,expressions,expressionsLength,
sizeof(struct Expression),(cmp)compareExpressionBySum)){
printf("%d\n", value);
break;
}
}
comment("Show the ten highest numbers that can be expressed");
for ( int i = expressionsLength-1; i >= expressionsLength-10; i-- )
print(expressions[i].code);
return 0;
}</lang>
- Output:
Show all solutions that sum to 100
100 = 123+4-5+67-89
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = 1+2+34-5+67-8+9
100 = 123+45-67+8-9
100 = 1+2+3-4+5+6+78+9
100 = 1+23-4+5+6+78-9
100 = 12-3-4+5-6+7+89
100 = 12+3+4+5-6-7+89
100 = -1+2-3+4+5+6+78+9
100 = 12+3-4+5+67+8+9
100 = 1+23-4+56+7+8+9
Show the positve sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
Optimized for memory consumption
Warning: this program needs at least four byte integers. <lang C>/*
* RossetaCode: Sum to 100, C11, MCU friendly. * * Find solutions to the "sum to one hundred" puzzle. * * We optimize algorithms for size. Therefore we don't use arrays, but recompute * all values again and again. It is a little surprise that the time efficiency * is quite acceptable. */
- include <stdio.h>
enum OP { ADD, SUB, JOIN };
int evaluate(int code){
int value = 0, number = 0, power = 1;
for ( int k = 9; k >= 1; k-- ){
number = power*k + number;
switch( code % 3 ){
case ADD: value = value + number; number = 0; power = 1; break;
case SUB: value = value - number; number = 0; power = 1; break;
case JOIN: power = power * 10 ; break;
}
code /= 3;
}
return value;
}
void print(int code){
static char s[19]; char* p = s;
int a = 19683, b = 6561;
for ( int k = 1; k <= 9; k++ ){
switch((code % a) / b){
case ADD: if ( k > 1 ) *p++ = '+'; break;
case SUB: *p++ = '-'; break;
}
a = b;
b = b / 3;
*p++ = '0' + k;
}
*p = 0;
printf("%9d = %s\n", evaluate(code), s);
}
int main(void){
int i,j; const int nexpr = 13122;
- define LOOP(K) for (K = 0; K < nexpr; K++)
puts("\nShow all solutions that sum to 100\n");
LOOP(i) if ( evaluate(i) == 100 ) print(i);
puts("\nShow the sum that has the maximum number of solutions\n");
int best, nbest = (-1);
LOOP(i){
int test = evaluate(i);
if ( test > 0 ){
int ntest = 0;
LOOP(j) if ( evaluate(j) == test ) ntest++;
if ( ntest > nbest ){ best = test; nbest = ntest; }
}
}
printf("%d has %d solutions\n", best,nbest);
puts("\nShow the lowest positive number that can't be expressed\n");
for ( i = 0; i <= 123456789; i++ ){
LOOP(j) if ( i == evaluate(j) ) break;
if ( i != evaluate(j) ) break;
}
printf("%d\n",i);
puts("\nShow the ten highest numbers that can be expressed\n");
int limit = 123456789 + 1;
for ( i = 1; i <= 10; i++ ) {
int best = 0;
LOOP(j){
int test = evaluate(j);
if ( test < limit && test > best ) best = test;
}
LOOP(j) if ( evaluate(j) == best ) print(j);
limit = best;
}
return 0;
}</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq;
class Program {
static void Main(string[] args)
{
// All unique expressions that have a plus sign in front of the 1; calculated in parallel
var expressionsPlus = Enumerable.Range(0, (int)Math.Pow(3, 8)).AsParallel().Select(i => new Expression(i, 1));
// All unique expressions that have a minus sign in front of the 1; calculated in parallel
var expressionsMinus = Enumerable.Range(0, (int)Math.Pow(3, 8)).AsParallel().Select(i => new Expression(i, -1));
var expressions = expressionsPlus.Concat(expressionsMinus);
var results = new Dictionary<int, List<Expression>>();
foreach (var e in expressions)
{
if (results.Keys.Contains(e.Value))
results[e.Value].Add(e);
else
results[e.Value] = new List<Expression>() { e };
}
Console.WriteLine("Show all solutions that sum to 100");
foreach (Expression e in results[100])
Console.WriteLine(" " + e);
Console.WriteLine("Show the sum that has the maximum number of solutions (from zero to infinity)");
var summary = results.Keys.Select(k => new Tuple<int, int>(k, results[k].Count));
var maxSols = summary.Aggregate((a, b) => a.Item2 > b.Item2 ? a : b);
Console.WriteLine(" The sum " + maxSols.Item1 + " has " + maxSols.Item2 + " solutions.");
Console.WriteLine("Show the lowest positive sum that can't be expressed (has no solutions), using the rules for this task");
var lowestPositive = Enumerable.Range(1, int.MaxValue).First(x => !results.Keys.Contains(x));
Console.WriteLine(" " + lowestPositive);
Console.WriteLine("Show the ten highest numbers that can be expressed using the rules for this task (extra credit)");
var highest = from k in results.Keys
orderby k descending
select k;
foreach (var x in highest.Take(10))
Console.WriteLine(" " + x);
}
} public enum Operations { Plus, Minus, Join }; public class Expression {
protected Operations[] Gaps;
// 123456789 => there are 8 "gaps" between each number
/// with 3 possibilities for each gap: plus, minus, or join
public int Value; // What this expression sums up to
protected int _one;
public Expression(int serial, int one)
{
_one = one;
Gaps = new Operations[8];
// This represents "serial" as a base 3 number, each Gap expression being a base-three digit
int divisor = 2187; // == Math.Pow(3,7)
int times;
for (int i = 0; i < 8; i++)
{
times = Math.DivRem(serial, divisor, out serial);
divisor /= 3;
if (times == 0)
Gaps[i] = Operations.Join;
else if (times == 1)
Gaps[i] = Operations.Minus;
else
Gaps[i] = Operations.Plus;
}
// go ahead and calculate the value of this expression
// because this is going to be done in a parallel thread (save time)
Value = Evaluate();
}
public override string ToString()
{
string ret = _one.ToString();
for (int i = 0; i < 8; i++)
{
switch (Gaps[i])
{
case Operations.Plus:
ret += "+";
break;
case Operations.Minus:
ret += "-";
break;
}
ret += (i + 2);
}
return ret;
}
private int Evaluate()
/* Calculate what this expression equals */
{
var numbers = new int[9];
int nc = 0;
var operations = new List<Operations>();
int a = 1;
for (int i = 0; i < 8; i++)
{
if (Gaps[i] == Operations.Join)
a = a * 10 + (i + 2);
else
{
if (a > 0)
{
if (nc == 0)
a *= _one;
numbers[nc++] = a;
a = i + 2;
}
operations.Add(Gaps[i]);
}
}
if (nc == 0)
a *= _one;
numbers[nc++] = a;
int ni = 0;
int left = numbers[ni++];
foreach (var operation in operations)
{
int right = numbers[ni++];
if (operation == Operations.Plus)
left = left + right;
else
left = left - right;
}
return left;
}
}</lang>
- Output:
Show all solutions that sum to 100 123-45-67+89 123-4-5-6-7+8-9 123+45-67+8-9 123+4-5+67-89 12-3-4+5-6+7+89 12+3-4+5+67+8+9 12+3+4+5-6-7+89 1+23-4+5+6+78-9 1+23-4+56+7+8+9 1+2+34-5+67-8+9 1+2+3-4+5+6+78+9 -1+2-3+4+5+6+78+9 Show the sum that has the maximum number of solutions (from zero to infinity) The sum 9 has 46 solutions. Show the lowest positive sum that can't be expressed (has no solutions), using the rules for this task 211 Show the ten highest numbers that can be expressed using the rules for this task (extra credit) 123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
C++
For each expression of sum s, there is at least one expression whose sum is -s. If the sum s can be represented by n expressions, the sum -s can also be represented by n expressions. The change of all signs in an expression change the sign of the sum of this expression. For example, -1+23-456+789 has the opposite sign than +1-23+456-789. Therefore only the positive sum with the maximum number of solutions is shown. The program does not check uniqueness of this sum. We can easily check (modifying the program) that: sum 9 has 46 solutions; sum -9 has 46 solutions; any other sum has less than 46 solutions. <lang Cpp>/*
* RossetaCode: Sum to 100, C++, STL, OOP. * Works with: MSC 16.0 (MSVS2010); GCC 5.1 (use -std=c++11 or -std=c++14 etc.). * * Find solutions to the "sum to one hundred" puzzle. */
- include <iostream>
- include <iomanip>
- include <algorithm>
- include <string>
- include <set>
- include <map>
using namespace std;
class Expression{
private:
enum { NUMBER_OF_DIGITS = 9 }; // hack for C++98, use const int in C++11
enum Op { ADD, SUB, JOIN };
int code[NUMBER_OF_DIGITS];
public:
static const int NUMBER_OF_EXPRESSIONS;
Expression(){
for ( int i = 0; i < NUMBER_OF_DIGITS; i++ )
code[i] = ADD;
}
Expression& operator++(int){ // post incrementation
for ( int i = 0; i < NUMBER_OF_DIGITS; i++ )
if ( ++code[i] > JOIN ) code[i] = ADD;
else break;
return *this;
}
operator int() const{
int value = 0, number = 0, sign = (+1);
for ( int digit = 1; digit <= 9; digit++ )
switch ( code[NUMBER_OF_DIGITS - digit] ){
case ADD: value += sign*number; number = digit; sign = (+1); break;
case SUB: value += sign*number; number = digit; sign = (-1); break;
case JOIN: number = 10*number + digit; break;
}
return value + sign*number;
}
operator string() const{
string s;
for ( int digit = 1; digit <= NUMBER_OF_DIGITS; digit++ ){
switch( code[NUMBER_OF_DIGITS - digit] ){
case ADD: if ( digit > 1 ) s.push_back('+'); break;
case SUB: s.push_back('-'); break;
}
s.push_back('0' + digit);
}
return s;
}
}; const int Expression::NUMBER_OF_EXPRESSIONS = 2 * 3*3*3*3 * 3*3*3*3;
ostream& operator<< (ostream& os, Expression& ex){
ios::fmtflags oldFlags(os.flags());
os << setw(9) << right << static_cast<int>(ex) << " = "
<< setw(0) << left << static_cast<string>(ex) << endl;
os.flags(oldFlags);
return os;
}
struct Stat{
map<int,int> countSum;
map<int, set<int> > sumCount;
Stat(){
Expression expression;
for ( int i = 0; i < Expression::NUMBER_OF_EXPRESSIONS; i++, expression++ )
countSum[expression]++;
for ( auto it = countSum.begin(); it != countSum.end(); it++ )
sumCount[it->second].insert(it->first);
}
};
void print(int givenSum){
Expression expression;
for ( int i = 0; i < Expression::NUMBER_OF_EXPRESSIONS; i++, expression++ )
if ( expression == givenSum )
cout << expression;
}
void comment(string commentString){
cout << endl << commentString << endl << endl;
}
int main(){
Stat stat;
comment( "Show all solutions that sum to 100" ); const int givenSum = 100; print(givenSum);
comment( "Show the sum that has the maximum number of solutions" ); auto maxi = max_element(stat.sumCount.begin(),stat.sumCount.end()); auto it = maxi->second.begin(); while ( *it < 0 ) it++; cout << static_cast<int>(*it) << " has " << maxi->first << " solutions" << endl;
comment( "Show the lowest positive number that can't be expressed" ); int value = 0; while(stat.countSum.count(value) != 0) value++; cout << value << endl;
comment( "Show the ten highest numbers that can be expressed" ); auto rit = stat.countSum.rbegin(); for ( int i = 0; i < 10; i++, rit++ ) print(rit->first);
return 0;
}</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
Common Lisp
<lang lisp>(defun f (lst &optional (sum 100) (so-far nil))
"Takes a list of digits as argument"
(if (null lst)
(cond ((= sum 0) (format t "~d = ~{~@d~}~%" (apply #'+ so-far) (reverse so-far)) 1)
(t 0) )
(let ((total 0)
(len (length lst)) )
(dotimes (i len total)
(let* ((str1 (butlast lst i))
(num1 (or (numlist-to-string str1) 0))
(rem (nthcdr (- len i) lst)) )
(incf total
(+ (f rem (- sum num1) (cons num1 so-far))
(f rem (+ sum num1) (cons (- num1) so-far)) )))))))
(defun numlist-to-string (lst)
"Convert a list of digits into an integer"
(when lst
(parse-integer (format nil "~{~d~}" lst)) ))
</lang>
- Output:
>(f '(1 2 3 4 5 6 7 8 9)) 100 = +123+45-67+8-9 100 = +123-45-67+89 100 = +123+4-5+67-89 100 = +123-4-5-6-7+8-9 100 = +12+3+4+5-6-7+89 100 = +12+3-4+5+67+8+9 100 = +12-3-4+5-6+7+89 100 = +1+23-4+56+7+8+9 100 = +1+23-4+5+6+78-9 100 = +1+2+34-5+67-8+9 100 = +1+2+3-4+5+6+78+9 100 = -1+2-3+4+5+6+78+9 12
The other subtasks are not yet implemented.
D
<lang D>import std.stdio;
void main() {
import std.algorithm : each, max, reduce, sort; import std.range : take;
Stat stat = new Stat();
comment("Show all solutions that sum to 100");
immutable givenSum = 100;
print(givenSum);
comment("Show the sum that has the maximum number of solutions");
const int maxCount = reduce!max(stat.sumCount.keys);
int maxSum;
foreach(key, entry; stat.sumCount[maxCount]) {
if (key >= 0) {
maxSum = key;
break;
}
}
writeln(maxSum, " has ", maxCount, " solutions");
comment("Show the lowest positive number that can't be expressed");
int value = 0;
while (value in stat.countSum) {
value++;
}
writeln(value);
comment("Show the ten highest numbers that can be expressed");
const int n = stat.countSum.keys.length;
auto sums = stat.countSum.keys;
sums.sort!"a>b"
.take(10)
.each!print;
}
void comment(string commentString) {
writeln(); writeln(commentString); writeln();
}
void print(int givenSum) {
Expression expression = new Expression();
for (int i=0; i<Expression.NUMBER_OF_EXPRESSIONS; i++, expression.next()) {
if (expression.toInt() == givenSum) {
expression.print();
}
}
}
class Expression {
private enum NUMBER_OF_DIGITS = 9; private enum ADD = 0; private enum SUB = 1; private enum JOIN = 2;
enum NUMBER_OF_EXPRESSIONS = 2 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3; byte[NUMBER_OF_DIGITS] code;
Expression next() {
for (int i=0; i<NUMBER_OF_DIGITS; i++) {
if (++code[i] > JOIN) {
code[i] = ADD;
} else {
break;
}
}
return this;
}
int toInt() {
int value = 0;
int number = 0;
int sign = (+1);
for (int digit=1; digit<=9; digit++) {
switch (code[NUMBER_OF_DIGITS - digit]) {
case ADD:
value += sign * number;
number = digit;
sign = (+1);
break;
case SUB:
value += sign * number;
number = digit;
sign = (-1);
break;
case JOIN:
number = 10 * number + digit;
break;
default:
assert(false);
}
}
return value + sign * number;
}
void toString(scope void delegate(const(char)[]) sink) const {
import std.conv : to;
import std.format : FormatSpec, formatValue;
import std.range : put;
auto fmt = FormatSpec!char("s");
for (int digit=1; digit<=NUMBER_OF_DIGITS; digit++) {
switch (code[NUMBER_OF_DIGITS - digit]) {
case ADD:
if (digit > 1) {
put(sink, '+');
}
break;
case SUB:
put(sink, '-');
break;
default:
break;
}
formatValue(sink, digit, fmt);
}
}
void print() {
print(stdout);
}
void print(File printStream) {
printStream.writefln("%9d = %s", toInt(), this);
}
}
class Stat {
int[int] countSum; bool[int][int] sumCount;
this() {
Expression expression = new Expression();
for (int i=0; i<Expression.NUMBER_OF_EXPRESSIONS; i++, expression.next()) {
int sum = expression.toInt();
countSum[sum]++;
}
foreach (key, entry; countSum) {
bool[int] set;
if (entry in sumCount) {
set = sumCount[entry];
} else {
set.clear();
}
set[key] = true;
sumCount[entry] = set;
}
}
}</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
Delphi
See Pascal.
Elixir
<lang elixir>defmodule Sum do
def to(val) do
generate
|> Enum.map(&{eval(&1), &1})
|> Enum.filter(fn {v, _s} -> v==val end)
|> Enum.each(&IO.inspect &1)
end
def max_solve do
generate
|> Enum.group_by(&eval &1)
|> Enum.filter_map(fn {k,_} -> k>=0 end, fn {k,v} -> {length(v),k} end)
|> Enum.max
|> fn {len,sum} -> IO.puts "sum of #{sum} has the maximum number of solutions : #{len}" end.()
end
def min_solve do
solve = generate |> Enum.group_by(&eval &1)
Stream.iterate(1, &(&1+1))
|> Enum.find(fn n -> solve[n]==nil end)
|> fn sum -> IO.puts "lowest positive sum that can't be expressed : #{sum}" end.()
end
def highest_sums(n\\10) do
IO.puts "highest sums :"
generate
|> Enum.map(&eval &1)
|> Enum.uniq
|> Enum.sort_by(fn sum -> -sum end)
|> Enum.take(n)
|> IO.inspect
end
defp generate do
x = ["+", "-", ""]
for a <- ["-", ""], b <- x, c <- x, d <- x, e <- x, f <- x, g <- x, h <- x, i <- x,
do: "#{a}1#{b}2#{c}3#{d}4#{e}5#{f}6#{g}7#{h}8#{i}9"
end
defp eval(str), do: Code.eval_string(str) |> elem(0)
end
Sum.to(100) Sum.max_solve Sum.min_solve Sum.highest_sums</lang>
- Output:
{100, "-1+2-3+4+5+6+78+9"}
{100, "1+2+3-4+5+6+78+9"}
{100, "1+2+34-5+67-8+9"}
{100, "1+23-4+5+6+78-9"}
{100, "1+23-4+56+7+8+9"}
{100, "12+3+4+5-6-7+89"}
{100, "12+3-4+5+67+8+9"}
{100, "12-3-4+5-6+7+89"}
{100, "123+4-5+67-89"}
{100, "123+45-67+8-9"}
{100, "123-4-5-6-7+8-9"}
{100, "123-45-67+89"}
sum of 9 has the maximum number of solutions : 46
lowest positive sum that can't be expressed : 211
highest sums :
[123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790,
3456788, 3456786]
F#
<lang fsharp> (* Generate the data set Nigel Galloway February 22nd., 2017
- )
type N = {n:string; g:int} let N = seq {
let rec fn n i g e l = seq {
match i with
|9 -> yield {n=l + "-9"; g=g+e-9}
yield {n=l + "+9"; g=g+e+9}
yield {n=l + "9"; g=g+e*10+9*n}
|_ -> yield! fn -1 (i+1) (g+e) -i (l + string -i)
yield! fn 1 (i+1) (g+e) i (l + "+" + string i)
yield! fn n (i+1) g (e*10+i*n) (l + string i)
}
yield! fn 1 2 0 1 "1"
yield! fn -1 2 0 -1 "-1"
} </lang>
- Output:
<lang fsharp> N |> Seq.filter(fun n->n.g=100) |> Seq.iter(fun n->printfn "%s" n.n) </lang>
1+2+3-4+5+6+78+9 1+2+34-5+67-8+9 1+23-4+5+6+78-9 1+23-4+56+7+8+9 12-3-4+5-6+7+89 12+3-4+5+67+8+9 12+3+4+5-6-7+89 123-4-5-6-7+8-9 123-45-67+89 123+4-5+67-89 123+45-67+8-9 -1+2-3+4+5+6+78+9
<lang fsharp> let n,g = N |> Seq.filter(fun n->n.g>=0) |> Seq.countBy(fun n->n.g) |> Seq.maxBy(snd) printfn "%d has %d solutions" n g </lang>
9 has 46 solutions
<lang fsharp> match N |> Seq.filter(fun n->n.g>=0) |> Seq.distinctBy(fun n->n.g) |> Seq.sortBy(fun n->n.g) |> Seq.pairwise |> Seq.tryFind(fun n->(snd n).g-(fst n).g > 1) with
|Some(n) -> printfn "least non-value is %d" ((fst n).g+1) |None -> printfn "No non-values found"
</lang>
least non-value is 211
<lang fsharp> N |> Seq.filter(fun n->n.g>=0) |> Seq.distinctBy(fun n->n.g) |> Seq.sortBy(fun n->(-n.g)) |> Seq.take 10 |> Seq.iter(fun n->printfn "%d" n.g ) </lang>
123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
Forth
This solution uses EVALUATE on a string buffer to compute the sum. Given the copious string manipulations, EVALUATE, and the large byte-array used to keep sum counts, this implementation is optimized neither for speed nor for memory. On my machine it runs in about 3.8 seconds, compared to the speed-optimized C solution which runs in about 0.005 seconds.
<lang Forth>CREATE *OPS CHAR + C, CHAR - C, CHAR # C,
CREATE 0OPS CHAR - C, CHAR # C,
CREATE BUFF 43 C, 43 CHARS ALLOT
CREATE PTR CELL ALLOT
CREATE LIMITS 2 C, 3 C, 3 C, 3 C, 3 C, 3 C, 3 C, 3 C, 3 C,
CREATE INDX 0 C, 0 C, 0 C, 0 C, 0 C, 0 C, 0 C, 0 C, 0 C,
CREATE OPS 0OPS , *OPS , *OPS , *OPS , *OPS , *OPS , *OPS , *OPS , *OPS ,
- B0 BUFF 1+ dup PTR ! 43 blank ;
- B, ( c --) PTR @ C! 1 PTR +! ;
CREATE STATS 123456790 ALLOT STATS 123456790 ERASE
- inc ( c-addr c-lim u -- t|f)
1- tuck + >r swap dup rot + ( addr a-addr) ( R: l-addr)
BEGIN dup C@ 1+ dup r@ C@ =
IF drop 2dup =
IF 2drop FALSE rdrop EXIT \ no inc, contents invalid
ELSE 0 over C! 1- r> 1- >r \ reset and carry
THEN
ELSE swap C! drop TRUE rdrop EXIT
THEN
AGAIN ;
- INDX+ INDX LIMITS 9 inc 0= ;
- SYNTH B0 [CHAR] 0 B, 9 0 DO
INDX I + C@ OPS I CELLS + @ + C@
dup [CHAR] # <> IF BL B, B, BL B, ELSE drop THEN
I [CHAR] 1 + B,
LOOP BUFF COUNT ;
- .MOST cr ." Sum that has the maximum number of solutions" cr 4 spaces
STATS 0 STATS 1+ 123456789 bounds DO
dup I c@ < IF drop drop I I c@ THEN
LOOP swap STATS - . ." has " . ." solutions" ;
- .CANT cr ." Lowest positive sum that can't be expressed" cr 4 spaces
STATS 1+ ( 0 not positive) BEGIN dup c@ WHILE 1+ REPEAT STATS - . ;
- .BEST cr ." Ten highest numbers that can be expressed" cr 4 spaces
0 >r [ STATS 123456789 + ]L BEGIN r@ 10 < over STATS >= and WHILE dup c@ IF dup STATS - . r> 1+ >r THEN 1- REPEAT r> drop ;
- . 0 <# #S #> TYPE ;
- .INFX cr 4 spaces 9 0 DO
INDX I + C@ OPS I cells + @ + C@
dup [char] # <> IF emit ELSE drop THEN I 1+ .
LOOP ;
- REPORT ( n) dup 100 = IF .INFX THEN
dup 0> IF STATS + dup c@ 1+ swap c! ELSE drop THEN ;
- >NUM 0. bl word count >number 2drop d>s ;
- # 10 * + ; \ numeric concatenation
- + >NUM + ; \ infix +
- - >NUM - ; \ infix -
- .SOLUTIONS cr ." Solutions that sum to 100:"
BEGIN SYNTH EVALUATE REPORT INDX+ UNTIL ;
- SUM100 .SOLUTIONS .MOST .CANT .BEST cr ;</lang>
- Output:
Note: must start Gforth with a larger-than-default dictionary size:
gforth -m 124M sum100.fs -e SUM100
Solutions that sum to 100:
-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89
Sum that has the maximum number of solutions
9 has 46 solutions
Lowest positive sum that can't be expressed
211
Ten highest numbers that can be expressed
123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
Fortran
Fortran IV
The program below is written in Fortran IV. Nevertheless, Fortran IV had a variety of dialects. It did not work the same on every type of computer. The source code below is compiled without problems using today's compilers. It have not been checked on any old mainframe. Sorry, I have no access to CDC6000. The algorithm used is not very fast, but uses little memory. In practice, this program took about 15 seconds to complete the task on PC (in 2017). For comparison: the program written in C ++ (using maps and STL collections) took about 1 second on the same machine.
C ROSSETACODE: SUM TO 100, FORTRAN IV
C FIND SOLUTIONS TO THE "SUM TO ONE HUNDRED" PUZZLE
C =================================================
PROGRAM SUMTO100
DATA NEXPRM1/13121/
WRITE(6,110)
110 FORMAT(1X/1X,34HSHOW ALL SOLUTIONS THAT SUM TO 100/)
DO 10 I = 0,NEXPRM1
10 IF ( IEVAL(I) .EQ. 100 ) CALL PREXPR(I)
WRITE(6,120)
120 FORMAT(1X/1X,
153HSHOW THE SUM THAT HAS THE MAXIMUM NUMBER OF SOLUTIONS/)
NBEST = -1
DO 30 I = 0, NEXPRM1
ITEST = IEVAL(I)
IF ( ITEST .LT. 0 ) GOTO 30
NTEST = 0
DO 20 J = 0, NEXPRM1
20 IF ( IEVAL(J) .EQ. ITEST ) NTEST = NTEST + 1
IF ( NTEST .LE. NBEST ) GOTO 30
IBEST = ITEST
NBEST = NTEST
30 CONTINUE
WRITE(6,121) IBEST, NBEST
121 FORMAT(1X,I8,5H HAS ,I8,10H SOLUTIONS/)
WRITE(6,130)
130 FORMAT(1X/1X,
155HSHOW THE LOWEST POSITIVE NUMBER THAT CAN'T BE EXPRESSED/)
DO 50 I = 0,123456789
DO 40 J = 0,NEXPRM1
40 IF ( I .EQ. IEVAL(J) ) GOTO 50
GOTO 60
50 CONTINUE
60 WRITE(6,131) I
131 FORMAT(1X,I8)
WRITE(6,140)
140 FORMAT(1X/1X,
150HSHOW THE TEN HIGHEST NUMBERS THAT CAN BE EXPRESSED/)
ILIMIT = 123456789
DO 90 I = 1,10
IBEST = 0
DO 70 J = 0, NEXPRM1
ITEST = IEVAL(J)
70 IF( (ITEST .LE. ILIMIT) .AND. (ITEST .GT. IBEST)) IBEST = ITEST
DO 80 J = 0, NEXPRM1
80 IF ( IEVAL(J) .EQ. IBEST ) CALL PREXPR(J)
90 ILIMIT = IBEST - 1
END
C EVALUATE THE VALUE OF THE GIVEN ENCODED EXPRESSION
C --------------------------------------------------
FUNCTION IEVAL(ICODE)
IC = ICODE
IEVAL = 0
N = 0
IP = 1
DO 50 K = 9,1,-1
N = IP*K + N
GOTO (10,20,40,30) MOD(IC,3)+1
10 IEVAL = IEVAL + N
GOTO 30
20 IEVAL = IEVAL - N
30 N = 0
IP = 1
GOTO 50
40 IP = IP * 10
50 IC = IC / 3
END
C PRINT THE ENCODED EXPRESSION IN THE READABLE FORMAT
C ---------------------------------------------------
SUBROUTINE PREXPR(ICODE)
DIMENSION IH(9),IHPMJ(4)
DATA IHPMJ/1H+,1H-,1H ,1H?/
IA = 19683
IB = 6561
DO 10 K = 1,9
IH(K) = IHPMJ(MOD(ICODE,IA) / IB+1)
IA = IB
10 IB = IB / 3
IVALUE = IEVAL(ICODE)
WRITE(6,110) IVALUE, IH
110 FORMAT(I9,3H = 1A1,1H1,1A1,1H2,1A1,1H3,1A1,1H4,1A1,1H5,1A1,1H6,1A1
1,1H7,1A1,1H8,1A1,1H9)
END
- Output:
SHOW ALL SOLUTIONS THAT SUM TO 100
100 = +1+2+3-4+5+6+7 8+9
100 = +1+2+3 4-5+6 7-8+9
100 = +1+2 3-4+5+6+7 8-9
100 = +1+2 3-4+5 6+7+8+9
100 = +1 2+3+4+5-6-7+8 9
100 = +1 2+3-4+5+6 7+8+9
100 = +1 2-3-4+5-6+7+8 9
100 = +1 2 3+4-5+6 7-8 9
100 = +1 2 3+4 5-6 7+8-9
100 = +1 2 3-4-5-6-7+8-9
100 = +1 2 3-4 5-6 7+8 9
100 = -1+2-3+4+5+6+7 8+9
SHOW THE SUM THAT HAS THE MAXIMUM NUMBER OF SOLUTIONS
9 HAS 46 SOLUTIONS
SHOW THE LOWEST POSITIVE NUMBER THAT CAN'T BE EXPRESSED
211
SHOW THE TEN HIGHEST NUMBERS THAT CAN BE EXPRESSED
123456789 = +1 2 3 4 5 6 7 8 9
23456790 = +1+2 3 4 5 6 7 8 9
23456788 = -1+2 3 4 5 6 7 8 9
12345687 = +1 2 3 4 5 6 7 8+9
12345669 = +1 2 3 4 5 6 7 8-9
3456801 = +1 2+3 4 5 6 7 8 9
3456792 = +1+2+3 4 5 6 7 8 9
3456790 = -1+2+3 4 5 6 7 8 9
3456788 = +1-2+3 4 5 6 7 8 9
3456786 = -1-2+3 4 5 6 7 8 9
Fortran 95
<lang Fortran>! RossetaCode: Sum to 100, Fortran 95, an algorithm using ternary numbers. ! ! Find solutions to the 'sum to one hundred' puzzle. ! ! We optimize algorithms for size. Therefore we don't use arrays, but recompute ! all values again and again. It is a little surprise that the time efficiency ! is quite acceptable. Actually the code is more compact than the implementation ! in C++ (STL maps and sets). We purposely break DRY and use magic values. ! Nevertheless, it is Fortran 95, free form lines, do-endo etc.
program sumto100
parameter (nexpr = 13122)
print *
print *, 'Show all solutions that sum to 100'
print *
do i = 0, nexpr-1
if ( ievaluate(i) .eq. 100 ) then
call printexpr(i)
endif
enddo
print *
print *, 'Show the sum that has the maximum number of solutions'
print *
ibest = -1
nbest = -1
do i = 0, nexpr-1
itest = ievaluate(i)
if ( itest .ge. 0 ) then
ntest = 0
do j = 0, nexpr-1
if ( ievaluate(j) .eq. itest ) then
ntest = ntest + 1
endif
enddo
if ( (ntest .gt. nbest) ) then
ibest = itest
nbest = ntest
endif
endif
enddo
print *, ibest, ' has ', nbest, ' solutions'
print *
! do i = 0, nexpr-1 ! if ( ievaluate(i) .eq. ibest ) then ! call printexpr(i) ! endif ! enddo
print *
print *, 'Show the lowest positive number that cant be expressed'
print *
loop: do i = 0,123456789
do j = 0,nexpr-1
if ( i .eq. ievaluate(j) ) then
cycle loop
endif
enddo
exit
enddo loop
print *, i
print *
print *, 'Show the ten highest numbers that can be expressed'
print *
ilimit = 123456789
do i = 1,10
ibest = 0
do j = 0, nexpr-1
itest = ievaluate(j)
if ( (itest .le. ilimit) .and. (itest .gt. ibest ) ) then
ibest = itest
endif
enddo
do j = 0, nexpr-1
if ( ievaluate(j) .eq. ibest ) then
call printexpr(j)
endif
enddo
ilimit = ibest - 1;
enddo
end
function ievaluate(icode)
ic = icode
ievaluate = 0
n = 0
ip = 1
do k = 9,1,-1
n = ip*k + n
select case(mod(ic,3))
case ( 0 )
ievaluate = ievaluate + n
n = 0
ip = 1
case ( 1 )
ievaluate = ievaluate - n
n = 0
ip = 1
case ( 2 )
ip = ip * 10
end select
ic = ic / 3
enddo
end
subroutine printexpr(icode)
character(len=32) s
ia = 19683
ib = 6561
s = ""
do k = 1,9
ic = mod(icode,ia) / ib
ia = ib
ib = ib / 3
select case(mod(ic,3))
case ( 0 )
if ( k .gt. 1 ) then
s = trim(s) // '+'
endif
case ( 1 )
s = trim(s) // '-'
end select
s = trim(s) // char(ichar('0')+k)
end do
ivalue = ievaluate(icode)
print *, ivalue, ' = ', s
end</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
Batch processing
By the simple expedient of storing all evaluations in an array (which is not so large) and then sorting the array, the required results appear in a blink. The source is essentially F77 except for the usage of an array assignment of OP = -1, writing out the highest ten results via an array expression instead of a DO-loop, array OPNAME extending from -1 to +1, a CYCLE statement rather than a GO TO, and the use of the I0 format code. Subroutine DEBLANK is straightforward, and omitted. It was only to remove spaces from the text of the expression. Reading the expression from right to left is about as picky as left-to-right. <lang Fortran> INTEGER NDIGITS,TARGET !Document the shape.
PARAMETER (NDIGITS = 9, TARGET = 100)
INTEGER*1 OP(NDIGITS) !A set of operation codes, associated with each digit.
INTEGER N,D,P !Number, digit, power.
CHARACTER*1 OPNAME(-1:+1) !Encodement of the operations.
PARAMETER (OPNAME = (/"-"," ","+"/)) !These will look nice.
CHARACTER*20 TEXT !A scratchpad for the expression. Single digits only.
INTEGER I,L,H,ME !Assistants.
LOGICAL CURSE !Needed for a Comb Sort.
INTEGER LOOP,NHIT !Some counters.
INTEGER ENUFF !Collect the results.
PARAMETER (ENUFF = 20000) !Surely big enough...
INTEGER VALUE(ENUFF) !A table.
INTEGER V,VV,PV,VE !For scanning the table.
INTEGER MSG !I/O unit number.
MSG = 6 !Standard output.
WRITE(MSG,1) NDIGITS,TARGET !Announce.
1 FORMAT ("To find expressions of ",I0," digits in order, "
1 "interspersed with + or -, adding to ",I0,/)
NHIT = 0 !No matches to TARGET.
LOOP = 0 !Because none have been made.
OP = -1 !Start the expression sequence.
Calculate the value of the expression given by OP(i) i pairs.
100 LOOP = LOOP + 1 !Here we go again.
N = 0 !Clear the number.
D = 0 !No previous digits have been seen.
P = 1 !The power for the first digit.
DO I = NDIGITS,1,-1 !Going backwards sees the digits before the sign.
D = D + I*P !Assimilate the digit string backwards.
IF (OP(I).EQ.0) THEN !A no-operation?
P = P*10 !Yes. Prepare the power for the next digit leftwards.
ELSE !Otherwise, add or subtract the digit string's value.
N = N + SIGN(D,OP(I)) !By transferring the sign to D..
D = 0 !Clear, ready for the next digit string.
P = 1 !The starting power, again.
END IF !So much for that step.
END DO !On to the next.
IF (OP(1).EQ.0) N = N + D !Provide an implicit add for an unsigned start.
VALUE(LOOP) = N !Save the value for later...
IF (N.EQ.TARGET) THEN !Well then?
NHIT = NHIT + 1 !Yay!
WRITE (TEXT,101) (OPNAME(OP(I)),I, I = 1,NDIGITS) !Translate the expression.
101 FORMAT (10(A1,I1)) !Single-character operation codes, single-digit number parts.
CALL DEBLANK(TEXT,L) !Squeeze out the no-operations, so numbers are together.
WRITE (MSG,102) N,TEXT(1:L) !Result!
102 FORMAT (I5,": ",A) !This should do.
END IF !So much for that.
Concoct the next expression, working as if with a bignumber in base three, though offset.
200 P = NDIGITS !Start with the low-order digit.
201 OP(P) = OP(P) + 1 !Add one to it.
IF (OP(P).GT.1) THEN !Is a carry needed?
OP(P) = -1 !Yes. Set the digit back to the start.
P = P - 1 !Go up a power.
IF (P.GT.0) GO TO 201 !And augment the next digit up.
END IF !Once the carry fizzles, the increment is complete.
IF (OP(1).LE.0) GO TO 100 !A leading + is equivalent to a leading no-op.
Contemplate the collection.
300 WRITE (6,301) LOOP,NHIT 301 FORMAT (/,I0," evaluations, ",I0," hit the target.")
Crank up a comb sort.
H = LOOP - 1 !Last - First, and not +1.
IF (H.LE.0) STOP "Huh?" !Ha ha.
310 H = MAX(1,H*10/13) !The special feature.
IF (H.EQ.9 .OR. H.EQ.10) H = 11 !A twiddle.
CURSE = .FALSE. !So far, so good.
DO I = LOOP - H,1,-1 !If H = 1, this is a BubbleSort.
IF (VALUE(I) .GT. VALUE(I + H)) THEN !One compare.
N = VALUE(I);VALUE(I)=VALUE(I+H);VALUE(I+H)=N !One swap.
CURSE = .TRUE. !One curse.
END IF !One test.
END DO !One loop.
IF (CURSE .OR. H.GT.1) GO TO 310 !Work remains?
Chase after some results.
H = 0 !Hunt the first omitted positive number.
VE = 0 !No equal values have been seen.
ME = 0 !So, their maximum run length is short.
PV = VALUE(1) !Grab the first value,
DO I = 2,LOOP !And scan the successors.
V = VALUE(I) !The value of the moment.
IF (V.LE.0) CYCLE !Only positive numbers are of interest.
IF (V.GT.PV + 1) THEN !Is there a gap?
IF (H.LE.0) H = PV + 1 !Recall the first such.
END IF !Perhaps a list of the first dew?
IF (V.EQ.PV) THEN !Is it the same as the one before?
VE = VE + 1 !Yes. Count up the length of the run.
IF (VE.GT.ME) THEN !Is this a longer run?
ME = VE !Yes. Remember its length.
VV = V !And its value.
END IF !So much for runs of equal values.
ELSE !But if it is not the same,
VE = 0 !A fresh count awaits.
END IF !So much for comparing one value to its predecessor.
PV = V !Be ready for the next time around.
END DO !On to the next.
Cast forth the results.
IF (ME.GT.1) WRITE (MSG,320) VV,ME + 1 !Counting started with the second occurrence.
320 FORMAT (I0," has the maximum number of attainments:",I0)
IF (H.GT.0) WRITE (MSG,321) H !Surely there will be one.
321 FORMAT ("The lowest positive sum that can't be expressed is ",I0)
WRITE (MSG,322) VALUE(LOOP - 9:LOOP) !Surely LOOP > 9.
322 FORMAT ("The ten highest sums: ",10(I0:","))
END !That was fun!</lang>
Results:
To find expressions of 9 digits in order, interspersed with + or -, adding to 100
1: -1+2-3+4+5+6+78+9
2: 12-3-4+5-6+7+89
3: 123-4-5-6-7+8-9
4: 123-45-67+89
5: 123+4-5+67-89
6: 123+45-67+8-9
7: 12+3-4+5+67+8+9
8: 12+3+4+5-6-7+89
9: 1+23-4+56+7+8+9
10: 1+23-4+5+6+78-9
11: 1+2+3-4+5+6+78+9
12: 1+2+34-5+67-8+9
13122 evaluations, 12 hit the target.
9 has the maximum number of attainments:46
The lowest positive sum that can't be expressed is 211
The ten highest sums: 3456786,3456788,3456790,3456792,3456801,12345669,12345687,23456788,23456790,123456789
Frink
<lang frink> digits = array[1 to 9] opList = makeArray[[8], ["", " + ", " - "]] opList.pushFirst"", "-" countDict = new dict
multifor ops = opList {
str = ""
for d = rangeOf[digits]
str = str + ops@d + digits@d
e = eval[str]
countDict.increment[e, 1]
if e == 100
println[str]
} println[]
// Find the sum that has the maximum number of solutions freq = toArray[countDict] sort[freq, {|a,b| -(a@1 <=> b@1)}] max = freq@0@1 print["Maximum count is $max at: "] n = 0 while freq@n@1 == max {
print[freq@n@0 + " "] n = n + 1
} println[]
// Find the smallest non-representable positive sum sort[freq, {|a,b| a@0 <=> b@0}] last = 0 for [num, count] = freq {
if num > 0 and last+1 != num
{
println["Lowest non-representable positive sum is " + (last+1)]
break
}
last = num
}
// Find highest 10 representable numbers println["\nHighest representable numbers:"] size = length[freq] for i = size-10 to size-1
println[freq@i@0]
</lang>
- Output:
123 + 45 - 67 + 8 - 9 123 + 4 - 5 + 67 - 89 123 - 45 - 67 + 89 123 - 4 - 5 - 6 - 7 + 8 - 9 12 + 3 + 4 + 5 - 6 - 7 + 89 12 + 3 - 4 + 5 + 67 + 8 + 9 12 - 3 - 4 + 5 - 6 + 7 + 89 1 + 23 - 4 + 56 + 7 + 8 + 9 1 + 23 - 4 + 5 + 6 + 78 - 9 1 + 2 + 34 - 5 + 67 - 8 + 9 1 + 2 + 3 - 4 + 5 + 6 + 78 + 9 -1 + 2 - 3 + 4 + 5 + 6 + 78 + 9 Maximum count is 46 at: 9 -9 Lowest non-representable positive sum is 211 Highest representable numbers: 3456786 3456788 3456790 3456792 3456801 12345669 12345687 23456788 23456790 123456789
Go
<lang Go>package main
import ( "fmt" "sort" )
const pow3_8 = 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 // 3^8 const pow3_9 = 3 * pow3_8 // 3^9 const maxExprs = 2 * pow3_8 // not 3^9 since first op can't be Join
type op uint8
const ( Add op = iota // insert a "+" Sub // or a "-" Join // or just join together )
// code is an encoding of [9]op, the nine "operations" // we do on each each digit. The op for 1 is in // the highest bits, the op for 9 in the lowest. type code uint16
// evaluate 123456789 with + - or "" prepended to each as indicated by `c`. func (c code) evaluate() (sum int) { num, pow := 0, 1 for k := 9; k >= 1; k-- { num += pow * k switch op(c % 3) { case Add: sum += num num, pow = 0, 1 case Sub: sum -= num num, pow = 0, 1 case Join: pow *= 10 } c /= 3 } return sum }
func (c code) String() string { buf := make([]byte, 0, 18) a, b := code(pow3_9), code(pow3_8) for k := 1; k <= 9; k++ { switch op((c % a) / b) { case Add: if k > 1 { buf = append(buf, '+') } case Sub: buf = append(buf, '-') } buf = append(buf, '0'+byte(k)) a, b = b, b/3 } return string(buf) }
type sumCode struct { sum int code code } type sumCodes []sumCode
type sumCount struct { sum int count int } type sumCounts []sumCount
// For sorting (could also use sort.Slice with just Less). func (p sumCodes) Len() int { return len(p) } func (p sumCodes) Swap(i, j int) { p[i], p[j] = p[j], p[i] } func (p sumCodes) Less(i, j int) bool { return p[i].sum < p[j].sum } func (p sumCounts) Len() int { return len(p) } func (p sumCounts) Swap(i, j int) { p[i], p[j] = p[j], p[i] } func (p sumCounts) Less(i, j int) bool { return p[i].count > p[j].count }
// For printing. func (sc sumCode) String() string { return fmt.Sprintf("% 10d = %v", sc.sum, sc.code) } func (sc sumCount) String() string { return fmt.Sprintf("% 10d has %d solutions", sc.sum, sc.count) }
func main() { // Evaluate all expressions. expressions := make(sumCodes, 0, maxExprs/2) counts := make(sumCounts, 0, 1715) for c := code(0); c < maxExprs; c++ { // All negative sums are exactly like their positive // counterpart with all +/- switched, we don't need to // keep track of them. sum := c.evaluate() if sum >= 0 { expressions = append(expressions, sumCode{sum, c}) } } sort.Sort(expressions)
// Count all unique sums sc := sumCount{expressions[0].sum, 1} for _, e := range expressions[1:] { if e.sum == sc.sum { sc.count++ } else { counts = append(counts, sc) sc = sumCount{e.sum, 1} } } counts = append(counts, sc) sort.Sort(counts)
// Extract required results
fmt.Println("All solutions that sum to 100:") i := sort.Search(len(expressions), func(i int) bool { return expressions[i].sum >= 100 }) for _, e := range expressions[i:] { if e.sum != 100 { break } fmt.Println(e) }
fmt.Println("\nThe positive sum with maximum number of solutions:") fmt.Println(counts[0])
fmt.Println("\nThe lowest positive number that can't be expressed:") s := 1 for _, e := range expressions { if e.sum == s { s++ } else if e.sum > s { fmt.Printf("% 10d\n", s) break } }
fmt.Println("\nThe ten highest numbers that can be expressed:") for _, e := range expressions[len(expressions)-10:] { fmt.Println(e) } }</lang>
- Output:
All solutions that sum to 100:
100 = -1+2-3+4+5+6+78+9
100 = 1+23-4+5+6+78-9
100 = 123+45-67+8-9
100 = 123+4-5+67-89
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 12+3-4+5+67+8+9
100 = 1+23-4+56+7+8+9
100 = 123-45-67+89
100 = 12-3-4+5-6+7+89
100 = 12+3+4+5-6-7+89
100 = 123-4-5-6-7+8-9
The positive sum with maximum number of solutions:
9 has 46 solutions
The lowest positive number that can't be expressed:
211
The ten highest numbers that can be expressed:
3456786 = -1-2+3456789
3456788 = 1-2+3456789
3456790 = -1+2+3456789
3456792 = 1+2+3456789
3456801 = 12+3456789
12345669 = 12345678-9
12345687 = 12345678+9
23456788 = -1+23456789
23456790 = 1+23456789
123456789 = 123456789
Haskell
<lang Haskell>import Control.Monad (replicateM) import Data.Char (intToDigit) import Data.List
( find, group, intercalate, nub, sort, sortBy, )
import Data.Monoid ((<>)) import Data.Ord (comparing)
data Sign
= Unsigned | Plus | Minus deriving (Eq, Show)
SUM TO 100 ----------------------
universe :: (Int, Sign) universe =
zip [1 .. 9]
<$> filter
((/= Plus) . head)
(replicateM 9 [Unsigned, Plus, Minus])
allNonNegativeSums :: [Int] allNonNegativeSums =
sort $
filter
(>= 0)
(asSum <$> universe)
uniqueNonNegativeSums :: [Int] uniqueNonNegativeSums = nub allNonNegativeSums
asSum :: [(Int, Sign)] -> Int asSum xs =
n
+ ( case s of
[] -> 0
_ -> read s :: Int
)
where
(n, s) = foldr readSign (0, []) xs
readSign ::
(Int, Sign) ->
(Int, String) ->
(Int, String)
readSign (i, x) (n, s)
| x == Unsigned = (n, intToDigit i : s)
| otherwise =
( ( case x of
Plus -> (+)
_ -> (-)
)
n
(read (show i <> s) :: Int),
[]
)
asString :: [(Int, Sign)] -> String asString = foldr signedDigit []
where
signedDigit (i, x) s
| x == Unsigned = intToDigit i : s
| otherwise =
( case x of
Plus -> " +"
_ -> " -"
)
<> [intToDigit i]
<> s
TEST -------------------------
main :: IO () main =
putStrLn $
unlines
[ "Sums to 100:",
unlines
(asString <$> filter ((100 ==) . asSum) universe),
"\n10 commonest sums (sum, number of routes to it):",
show
( ((,) <$> head <*> length)
<$> take
10
( sortBy
(flip (comparing length))
(group allNonNegativeSums)
)
),
"\nFirst positive integer not expressible "
<> "as a sum of this kind:",
maybeReport
( find
(uncurry (/=))
(zip [0 ..] uniqueNonNegativeSums)
),
"\n10 largest sums:",
show
( take
10
( sortBy
(flip compare)
uniqueNonNegativeSums
)
)
]
where
maybeReport ::
Show a =>
Maybe (a, b) ->
String
maybeReport (Just (x, _)) = show x
maybeReport _ = "No gaps found"</lang>
- Output:
(Run in Atom editor, through Script package)
Sums to 100: 123 +45 -67 +8 -9 123 +4 -5 +67 -89 123 -45 -67 +89 123 -4 -5 -6 -7 +8 -9 12 +3 +4 +5 -6 -7 +89 12 +3 -4 +5 +67 +8 +9 12 -3 -4 +5 -6 +7 +89 1 +23 -4 +56 +7 +8 +9 1 +23 -4 +5 +6 +78 -9 1 +2 +34 -5 +67 -8 +9 1 +2 +3 -4 +5 +6 +78 +9 -1 +2 -3 +4 +5 +6 +78 +9 10 commonest sums [sum, number of routes to it]: [(9,46),(27,44),(1,43),(15,43),(21,43),(45,42),(3,41),(5,40),(7,39),(17,39)] First positive integer not expressible as a sum of this kind: 211 10 largest sums: [123456789,23456790,23456788,12345687,12345669,3456801,3456792,3456790,3456788,3456786] [Finished in 1.204s]
J
Since J has no verb precedence, -1-2 would evaluate to 1 and not to -3. That's why I decided to multiply each of the partitions of '123456789' (like '123','45', '6', '78', '9') with each possible +1/-1 vectors of length 9 (like 1 1 -1 1 -1 -1 1 1 -1) and to add up the results. This leads to 512*256 results, that of course include a lot of duplicates. To use directly ~. (nub) on the 512x256x9 vector is very slow and that's why I computed a sort of a hash to use it to get only the unique expressions. The rest is trivial - I check which expressions add up to 100; sort the sum vector and find the longest sequence ot repeating sums; get the 10 largest sums and finnaly check which sum differs with more then 1 from the previous one.
<lang J> p =: ,"2".>(#: (+ i.)2^8) <;.1 '123456789' m =. (9$_1x)^"1#:i.2^9 s =. 131072 9 $ ,m *"1/ p s2 =: (~: (10x^i._9)#.s)#s ss =: +/"1 s2 '100=';<'bp<+>' 8!:2 (I.100=ss){s2 pos =: (0<ss)#ss =: /:~ss ({.;'times';{:)>{.\:~(#,{.) each </.~ ss 'Ten largest:';,.(->:i.10){ss 'First not expressible:';>:pos{~ 1 i.~ 1<|2-/\pos </lang>
- Output:
┌───┬────────────────────────┐ │100│+12 +3 +4 +5 -6 -7 +89 │ │ │+1 +2 +3 -4 +5 +6 +78+9│ │ │+1 +2 +34-5 +67-8 +9 │ │ │+12 +3 -4 +5 +67+8 +9 │ │ │+1 +23-4 +56+7 +8 +9 │ │ │+1 +23-4 +5 +6 +78-9 │ │ │+123+45-67+8 -9 │ │ │+123+4 -5 +67-89 │ │ │+123-45-67+89 │ │ │+12 -3 -4 +5 -6 +7 +89 │ │ │+123-4 -5 -6 -7 +8 -9 │ │ │-1 +2 -3 +4 +5 +6 +78+9│ └───┴────────────────────────┘ ┌──┬─────┬─┐ │46│times│9│ └──┴─────┴─┘ ┌────────────┬─────────┐ │Ten largest:│123456789│ │ │ 23456790│ │ │ 23456788│ │ │ 12345687│ │ │ 12345669│ │ │ 3456801│ │ │ 3456792│ │ │ 3456790│ │ │ 3456788│ │ │ 3456786│ └────────────┴─────────┘ ┌───────────────────────┬───┐ │First not expressible :│211│ └───────────────────────┴───┘
Java
For each expression of sum s, there is at least one expression whose sum is -s. If the sum s can be represented by n expressions, the sum -s can also be represented by n expressions. The change of all signs in an expression change the sign of the sum of this expression. For example, -1+23-456+789 has the opposite sign than +1-23+456-789. Therefore only the positive sum with the maximum number of solutions is shown. The program does not check uniqueness of this sum. We can easily check (modifying the program) that: sum 9 has 46 solutions; sum -9 has 46 solutions; any other sum has less than 46 solutions. <lang Java>/*
* RossetaCode: Sum to 100, Java 8. * * Find solutions to the "sum to one hundred" puzzle. */
package rosettacode;
import java.io.PrintStream; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.HashSet; import java.util.Iterator; import java.util.Map; import java.util.Set;
public class SumTo100 implements Runnable {
public static void main(String[] args) {
new SumTo100().run();
}
void print(int givenSum) {
Expression expression = new Expression();
for (int i = 0; i < Expression.NUMBER_OF_EXPRESSIONS; i++, expression.next()) {
if (expression.toInt() == givenSum) {
expression.print();
}
}
}
void comment(String commentString) {
System.out.println();
System.out.println(commentString);
System.out.println();
}
@Override
public void run() {
final Stat stat = new Stat();
comment("Show all solutions that sum to 100");
final int givenSum = 100;
print(givenSum);
comment("Show the sum that has the maximum number of solutions");
final int maxCount = Collections.max(stat.sumCount.keySet());
int maxSum;
Iterator<Integer> it = stat.sumCount.get(maxCount).iterator();
do {
maxSum = it.next();
} while (maxSum < 0);
System.out.println(maxSum + " has " + maxCount + " solutions");
comment("Show the lowest positive number that can't be expressed");
int value = 0;
while (stat.countSum.containsKey(value)) {
value++;
}
System.out.println(value);
comment("Show the ten highest numbers that can be expressed");
final int n = stat.countSum.keySet().size();
final Integer[] sums = stat.countSum.keySet().toArray(new Integer[n]);
Arrays.sort(sums);
for (int i = n - 1; i >= n - 10; i--) {
print(sums[i]);
}
}
private static class Expression {
private final static int NUMBER_OF_DIGITS = 9;
private final static byte ADD = 0;
private final static byte SUB = 1;
private final static byte JOIN = 2;
final byte[] code = new byte[NUMBER_OF_DIGITS];
final static int NUMBER_OF_EXPRESSIONS = 2 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3;
Expression next() {
for (int i = 0; i < NUMBER_OF_DIGITS; i++) {
if (++code[i] > JOIN) {
code[i] = ADD;
} else {
break;
}
}
return this;
}
int toInt() {
int value = 0;
int number = 0;
int sign = (+1);
for (int digit = 1; digit <= 9; digit++) {
switch (code[NUMBER_OF_DIGITS - digit]) {
case ADD:
value += sign * number;
number = digit;
sign = (+1);
break;
case SUB:
value += sign * number;
number = digit;
sign = (-1);
break;
case JOIN:
number = 10 * number + digit;
break;
}
}
return value + sign * number;
}
@Override
public String toString() {
StringBuilder s = new StringBuilder(2 * NUMBER_OF_DIGITS + 1);
for (int digit = 1; digit <= NUMBER_OF_DIGITS; digit++) {
switch (code[NUMBER_OF_DIGITS - digit]) {
case ADD:
if (digit > 1) {
s.append('+');
}
break;
case SUB:
s.append('-');
break;
}
s.append(digit);
}
return s.toString();
}
void print() {
print(System.out);
}
void print(PrintStream printStream) {
printStream.format("%9d", this.toInt());
printStream.println(" = " + this);
}
}
private static class Stat {
final Map<Integer, Integer> countSum = new HashMap<>();
final Map<Integer, Set<Integer>> sumCount = new HashMap<>();
Stat() {
Expression expression = new Expression();
for (int i = 0; i < Expression.NUMBER_OF_EXPRESSIONS; i++, expression.next()) {
int sum = expression.toInt();
countSum.put(sum, countSum.getOrDefault(sum, 0) + 1);
}
for (Map.Entry<Integer, Integer> entry : countSum.entrySet()) {
Set<Integer> set;
if (sumCount.containsKey(entry.getValue())) {
set = sumCount.get(entry.getValue());
} else {
set = new HashSet<>();
}
set.add(entry.getKey());
sumCount.put(entry.getValue(), set);
}
}
}
}</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
JavaScript
ES5
<lang JavaScript>(function () {
'use strict';
// GENERIC FUNCTIONS ----------------------------------------------------
// permutationsWithRepetition :: Int -> [a] -> a var permutationsWithRepetition = function (n, as) { return as.length > 0 ? foldl1(curry(cartesianProduct)(as), replicate(n, as)) : []; };
// cartesianProduct :: [a] -> [b] -> a, b var cartesianProduct = function (xs, ys) { return [].concat.apply([], xs.map(function (x) { return [].concat.apply([], ys.map(function (y) { return [ [x].concat(y) ]; })); })); };
// curry :: ((a, b) -> c) -> a -> b -> c
var curry = function (f) {
return function (a) {
return function (b) {
return f(a, b);
};
};
};
// flip :: (a -> b -> c) -> b -> a -> c
var flip = function (f) {
return function (a, b) {
return f.apply(null, [b, a]);
};
};
// foldl1 :: (a -> a -> a) -> [a] -> a
var foldl1 = function (f, xs) {
return xs.length > 0 ? xs.slice(1)
.reduce(f, xs[0]) : [];
};
// replicate :: Int -> a -> [a]
var replicate = function (n, a) {
var v = [a],
o = [];
if (n < 1) return o;
while (n > 1) {
if (n & 1) o = o.concat(v);
n >>= 1;
v = v.concat(v);
}
return o.concat(v);
};
// group :: Eq a => [a] -> a var group = function (xs) { return groupBy(function (a, b) { return a === b; }, xs); };
// groupBy :: (a -> a -> Bool) -> [a] -> a var groupBy = function (f, xs) { var dct = xs.slice(1) .reduce(function (a, x) { var h = a.active.length > 0 ? a.active[0] : undefined, blnGroup = h !== undefined && f(h, x);
return {
active: blnGroup ? a.active.concat(x) : [x],
sofar: blnGroup ? a.sofar : a.sofar.concat([a.active])
};
}, {
active: xs.length > 0 ? [xs[0]] : [],
sofar: []
});
return dct.sofar.concat(dct.active.length > 0 ? [dct.active] : []);
};
// compare :: a -> a -> Ordering
var compare = function (a, b) {
return a < b ? -1 : a > b ? 1 : 0;
};
// on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
var on = function (f, g) {
return function (a, b) {
return f(g(a), g(b));
};
};
// nub :: [a] -> [a]
var nub = function (xs) {
return nubBy(function (a, b) {
return a === b;
}, xs);
};
// nubBy :: (a -> a -> Bool) -> [a] -> [a]
var nubBy = function (p, xs) {
var x = xs.length ? xs[0] : undefined;
return x !== undefined ? [x].concat(nubBy(p, xs.slice(1)
.filter(function (y) {
return !p(x, y);
}))) : [];
};
// find :: (a -> Bool) -> [a] -> Maybe a
var find = function (f, xs) {
for (var i = 0, lng = xs.length; i < lng; i++) {
if (f(xs[i], i)) return xs[i];
}
return undefined;
};
// Int -> [a] -> [a]
var take = function (n, xs) {
return xs.slice(0, n);
};
// unlines :: [String] -> String
var unlines = function (xs) {
return xs.join('\n');
};
// show :: a -> String
var show = function (x) {
return JSON.stringify(x);
}; //, null, 2);
// head :: [a] -> a
var head = function (xs) {
return xs.length ? xs[0] : undefined;
};
// tail :: [a] -> [a]
var tail = function (xs) {
return xs.length ? xs.slice(1) : undefined;
};
// length :: [a] -> Int
var length = function (xs) {
return xs.length;
};
// SIGNED DIGIT SEQUENCES (mapped to sums and to strings)
// data Sign :: [ 0 | 1 | -1 ] = ( Unsigned | Plus | Minus )
// asSum :: [Sign] -> Int
var asSum = function (xs) {
var dct = xs.reduceRight(function (a, sign, i) {
var d = i + 1; // zero-based index to [1-9] positions
if (sign !== 0) {
// Sum increased, digits cleared
return {
digits: [],
n: a.n + sign * parseInt([d].concat(a.digits)
.join(), 10)
};
} else return { // Digits extended, sum unchanged
digits: [d].concat(a.digits),
n: a.n
};
}, {
digits: [],
n: 0
});
return dct.n + (
dct.digits.length > 0 ? parseInt(dct.digits.join(), 10) : 0
);
};
// data Sign :: [ 0 | 1 | -1 ] = ( Unsigned | Plus | Minus )
// asString :: [Sign] -> String
var asString = function (xs) {
var ns = xs.reduce(function (a, sign, i) {
var d = (i + 1)
.toString();
return sign === 0 ? a + d : a + (sign > 0 ? ' +' : ' -') + d;
}, );
return ns[0] === '+' ? tail(ns) : ns; };
// SUM T0 100 ------------------------------------------------------------
// universe :: Sign var universe = permutationsWithRepetition(9, [0, 1, -1]) .filter(function (x) { return x[0] !== 1; });
// allNonNegativeSums :: [Int]
var allNonNegativeSums = universe.map(asSum)
.filter(function (x) {
return x >= 0;
})
.sort();
// uniqueNonNegativeSums :: [Int] var uniqueNonNegativeSums = nub(allNonNegativeSums);
return ["Sums to 100:\n", unlines(universe.filter(function (x) {
return asSum(x) === 100;
})
.map(asString)),
"\n\n10 commonest sums (sum, followed by number of routes to it):\n",
show(take(10, group(allNonNegativeSums)
.sort(on(flip(compare), length))
.map(function (xs) {
return [xs[0], xs.length];
}))),
"\n\nFirst positive integer not expressible as a sum of this kind:\n",
show(find(function (x, i) {
return x !== i;
}, uniqueNonNegativeSums.sort(compare)) - 1), // zero-based index
"\n10 largest sums:\n",
show(take(10, uniqueNonNegativeSums.sort(flip(compare))))
].join('\n') + '\n';
})();</lang>
- Output:
(Run in Atom editor, through Script package)
Sums to 100: 123 +45 -67 +8 -9 123 +4 -5 +67 -89 123 -45 -67 +89 123 -4 -5 -6 -7 +8 -9 12 +3 +4 +5 -6 -7 +89 12 +3 -4 +5 +67 +8 +9 12 -3 -4 +5 -6 +7 +89 1 +23 -4 +56 +7 +8 +9 1 +23 -4 +5 +6 +78 -9 1 +2 +34 -5 +67 -8 +9 1 +2 +3 -4 +5 +6 +78 +9 -1 +2 -3 +4 +5 +6 +78 +9 10 commonest sums (sum, followed by number of routes to it): [[9,46],[27,44],[1,43],[15,43],[21,43],[45,42],[3,41],[5,40],[17,39],[7,39]] First positive integer not expressible as a sum of this kind: 211 10 largest sums: [123456789,23456790,23456788,12345687,12345669,3456801,3456792,3456790,3456788,3456786] [Finished in 0.381s]
ES6
<lang JavaScript>(() => {
'use strict';
// GENERIC FUNCTIONS ----------------------------------------------------
// permutationsWithRepetition :: Int -> [a] -> a const permutationsWithRepetition = (n, as) => as.length > 0 ? ( foldl1(curry(cartesianProduct)(as), replicate(n, as)) ) : [];
// cartesianProduct :: [a] -> [b] -> a, b const cartesianProduct = (xs, ys) => [].concat.apply([], xs.map(x => [].concat.apply([], ys.map(y => [[x].concat(y)]))));
// curry :: ((a, b) -> c) -> a -> b -> c const curry = f => a => b => f(a, b);
// flip :: (a -> b -> c) -> b -> a -> c const flip = f => (a, b) => f.apply(null, [b, a]);
// foldl1 :: (a -> a -> a) -> [a] -> a
const foldl1 = (f, xs) =>
xs.length > 0 ? xs.slice(1)
.reduce(f, xs[0]) : [];
// replicate :: Int -> a -> [a]
const replicate = (n, a) => {
let v = [a],
o = [];
if (n < 1) return o;
while (n > 1) {
if (n & 1) o = o.concat(v);
n >>= 1;
v = v.concat(v);
}
return o.concat(v);
};
// group :: Eq a => [a] -> a const group = xs => groupBy((a, b) => a === b, xs);
// groupBy :: (a -> a -> Bool) -> [a] -> a const groupBy = (f, xs) => { const dct = xs.slice(1) .reduce((a, x) => { const h = a.active.length > 0 ? a.active[0] : undefined, blnGroup = h !== undefined && f(h, x);
return {
active: blnGroup ? a.active.concat(x) : [x],
sofar: blnGroup ? a.sofar : a.sofar.concat([a.active])
};
}, {
active: xs.length > 0 ? [xs[0]] : [],
sofar: []
});
return dct.sofar.concat(dct.active.length > 0 ? [dct.active] : []);
};
// compare :: a -> a -> Ordering const compare = (a, b) => a < b ? -1 : (a > b ? 1 : 0);
// on :: (b -> b -> c) -> (a -> b) -> a -> a -> c const on = (f, g) => (a, b) => f(g(a), g(b));
// nub :: [a] -> [a] const nub = xs => nubBy((a, b) => a === b, xs);
// nubBy :: (a -> a -> Bool) -> [a] -> [a]
const nubBy = (p, xs) => {
const x = xs.length ? xs[0] : undefined;
return x !== undefined ? [x].concat(
nubBy(p, xs.slice(1)
.filter(y => !p(x, y)))
) : [];
};
// find :: (a -> Bool) -> [a] -> Maybe a
const find = (f, xs) => {
for (var i = 0, lng = xs.length; i < lng; i++) {
if (f(xs[i], i)) return xs[i];
}
return undefined;
}
// Int -> [a] -> [a] const take = (n, xs) => xs.slice(0, n);
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// show :: a -> String const show = x => JSON.stringify(x); //, null, 2);
// head :: [a] -> a const head = xs => xs.length ? xs[0] : undefined;
// tail :: [a] -> [a] const tail = xs => xs.length ? xs.slice(1) : undefined;
// length :: [a] -> Int const length = xs => xs.length;
// SIGNED DIGIT SEQUENCES (mapped to sums and to strings)
// data Sign :: [ 0 | 1 | -1 ] = ( Unsigned | Plus | Minus )
// asSum :: [Sign] -> Int
const asSum = xs => {
const dct = xs.reduceRight((a, sign, i) => {
const d = i + 1; // zero-based index to [1-9] positions
if (sign !== 0) { // Sum increased, digits cleared
return {
digits: [],
n: a.n + (sign * parseInt([d].concat(a.digits)
.join(), 10))
};
} else return { // Digits extended, sum unchanged
digits: [d].concat(a.digits),
n: a.n
};
}, {
digits: [],
n: 0
});
return dct.n + (dct.digits.length > 0 ? (
parseInt(dct.digits.join(), 10)
) : 0);
};
// data Sign :: [ 0 | 1 | -1 ] = ( Unsigned | Plus | Minus )
// asString :: [Sign] -> String
const asString = xs => {
const ns = xs.reduce((a, sign, i) => {
const d = (i + 1)
.toString();
return (sign === 0 ? (
a + d
) : (a + (sign > 0 ? ' +' : ' -') + d));
}, );
return ns[0] === '+' ? tail(ns) : ns; };
// SUM T0 100 ------------------------------------------------------------
// universe :: Sign const universe = permutationsWithRepetition(9, [0, 1, -1]) .filter(x => x[0] !== 1);
// allNonNegativeSums :: [Int]
const allNonNegativeSums = universe.map(asSum)
.filter(x => x >= 0)
.sort();
// uniqueNonNegativeSums :: [Int] const uniqueNonNegativeSums = nub(allNonNegativeSums);
return [
"Sums to 100:\n",
unlines(universe.filter(x => asSum(x) === 100)
.map(asString)),
"\n\n10 commonest sums (sum, followed by number of routes to it):\n",
show(take(10, group(allNonNegativeSums)
.sort(on(flip(compare), length))
.map(xs => [xs[0], xs.length]))),
"\n\nFirst positive integer not expressible as a sum of this kind:\n",
show(find(
(x, i) => x !== i,
uniqueNonNegativeSums.sort(compare)
) - 1), // i is the the zero-based Array index.
"\n10 largest sums:\n",
show(take(10, uniqueNonNegativeSums.sort(flip(compare))))
].join('\n') + '\n';
})();</lang>
- Output:
(Run in Atom editor, through Script package)
Sums to 100: 123 +45 -67 +8 -9 123 +4 -5 +67 -89 123 -45 -67 +89 123 -4 -5 -6 -7 +8 -9 12 +3 +4 +5 -6 -7 +89 12 +3 -4 +5 +67 +8 +9 12 -3 -4 +5 -6 +7 +89 1 +23 -4 +56 +7 +8 +9 1 +23 -4 +5 +6 +78 -9 1 +2 +34 -5 +67 -8 +9 1 +2 +3 -4 +5 +6 +78 +9 -1 +2 -3 +4 +5 +6 +78 +9 10 commonest sums (sum, followed by number of routes to it): [[9,46],[27,44],[1,43],[15,43],[21,43],[45,42],[3,41],[5,40],[17,39],[7,39]] First positive integer not expressible as a sum of this kind: 211 10 largest sums: [123456789,23456790,23456788,12345687,12345669,3456801,3456792,3456790,3456788,3456786] [Finished in 0.382s]
ES3 (JScript)
<lang javascript>SumTo100();
function SumTo100() {
var
ADD = 0,
SUB = 1,
JOIN = 2;
var
nexpr = 13122;
function out(something)
{
WScript.Echo(something);
}
function evaluate(code)
{
var
value = 0,
number = 0,
power = 1;
for ( var k = 9; k >= 1; k-- )
{
number = power*k + number;
switch( code % 3 )
{
case ADD: value = value + number; number = 0; power = 1; break;
case SUB: value = value - number; number = 0; power = 1; break;
case JOIN: power = power * 10 ; break;
}
code = Math.floor(code/3);
}
return value;
}
function print(code)
{
var
s = "";
var
a = 19683,
b = 6561;
for ( var k = 1; k <= 9; k++ )
{
switch( Math.floor( (code % a) / b ) ){
case ADD: if ( k > 1 ) s = s + '+'; break;
case SUB: s = s + '-'; break;
}
a = b;
b = Math.floor(b/3);
s = s + String.fromCharCode(0x30+k);
}
out(evaluate(code) + " = " + s);
}
function comment(commentString)
{
out("");
out(commentString);
out("");
}
comment("Show all solutions that sum to 100");
for ( var i = 0; i < nexpr; i++)
if ( evaluate(i) == 100 )
print(i);
comment("Show the sum that has the maximum number of solutions");
var stat = {};
for ( var i = 0; i < nexpr; i++ )
{
var sum = evaluate(i);
if (stat[sum])
stat[sum]++;
else
stat[sum] = 1;
}
var best = 0;
var nbest = -1;
for ( var i = 0; i < nexpr; i++ )
{
var sum = evaluate(i);
if ( sum > 0 )
if ( stat[sum] > nbest )
{
best = i;
nbest = stat[sum];
}
}
out("" + evaluate(best) + " has " + nbest + " solutions");
comment("Show the lowest positive number that can't be expressed");
for ( var i = 0; i <= 123456789; i++ )
{
for ( var j = 0; j < nexpr; j++)
if ( i == evaluate(j) ) break;
if ( i != evaluate(j) ) break;
}
out(i);
comment("Show the ten highest numbers that can be expressed");
var limit = 123456789 + 1;
for ( i = 1; i <= 10; i++ )
{
var best = 0;
for ( var j = 0; j < nexpr; j++)
{
var test = evaluate(j);
if ( test < limit && test > best )
best = test;
}
for ( var j = 0; j < nexpr; j++)
if ( evaluate(j) == best ) print(j);
limit = best;
}
} </lang>
- Output:
Show all solutions that sum to 100 100 = 1+2+3-4+5+6+78+9 100 = 1+2+34-5+67-8+9 100 = 1+23-4+5+6+78-9 100 = 1+23-4+56+7+8+9 100 = 12+3+4+5-6-7+89 100 = 12+3-4+5+67+8+9 100 = 12-3-4+5-6+7+89 100 = 123+4-5+67-89 100 = 123+45-67+8-9 100 = 123-4-5-6-7+8-9 100 = 123-45-67+89 100 = -1+2-3+4+5+6+78+9 Show the sum that has the maximum number of solutions 9 has 46 solutions Show the lowest positive number that can't be expressed 211 Show the ten highest numbers that can be expressed 123456789 = 123456789 23456790 = 1+23456789 23456788 = -1+23456789 12345687 = 12345678+9 12345669 = 12345678-9 3456801 = 12+3456789 3456792 = 1+2+3456789 3456790 = -1+2+3456789 3456788 = 1-2+3456789 3456786 = -1-2+3456789
jq
For ease of understanding, the problems will be solved separately, using the machinery defined in the following section.
All possible sums <lang jq># Generate a "sum" in the form: [I, 1, X, 2, X, 3, ..., X, n] where I is "-" or "", and X is "+", "-", or "" def generate(n):
def pm: ["+"], ["-"], [""]; if n == 1 then (["-"], [""]) + [1] else generate(n-1) + pm + [n] end;
- The numerical value of a "sum"
def addup:
reduce .[] as $x ({sum:0, previous: "0"};
if $x == "+" then .sum += (.previous|tonumber) | .previous = ""
elif $x == "-" then .sum += (.previous|tonumber) | .previous = "-"
elif $x == "" then .
else .previous += ($x|tostring)
end)
| .sum + (.previous | tonumber) ;
- Pretty-print a "sum", e.g. ["",1,"+", 2] => 1 + 2
def pp: map(if . == "+" or . == "-" then " " + . else tostring end) | join(""); </lang> Solutions to "Sum to 100" problem <lang jq>generate(9) | select(addup == 100) | pp</lang>
- Output:
1 +23 -4 +56 +7 +8 +9 12 +3 -4 +5 +67 +8 +9 1 +2 +34 -5 +67 -8 +9 -1 +2 -3 +4 +5 +6 +78 +9 1 +2 +3 -4 +5 +6 +78 +9 123 -4 -5 -6 -7 +8 -9 123 +45 -67 +8 -9 1 +23 -4 +5 +6 +78 -9 12 -3 -4 +5 -6 +7 +89 12 +3 +4 +5 -6 -7 +89 123 -45 -67 +89 123 +4 -5 +67 -89
Helper Functions
For brevity, we define an efficient function for computing a histogram in the form of a JSON object, and a helper function for identifying the values with the n highest frequencies. <lang jq>def histogram(s): reduce s as $x ({}; ($x|tostring) as $k | .[$k] += 1);
- Emit an array of [ value, frequency ] pairs
def greatest(n):
to_entries | map( [.key, .value] ) | sort_by(.[1]) | .[(length-n):] | reverse ; </lang>
Maximum number of solutions <lang jq>histogram(generate(9) | addup | select(.>0)) | greatest(1)</lang>
- Output:
[["9",46]]
Ten most frequent sums <lang jq>histogram(generate(9) | addup | select(.>0)) | greatest(1)</lang>
- Output:
[["9",46],["27",44],["1",43],["21",43],["15",43],["45",42],["3",41],["5",40],["7",39],["17",39]]
First unsolvable <lang jq>def first_missing(s):
first( foreach s as $i (null;
if . == null or $i == . or $i == .+1 then $i else [.+1] end;
select(type == "array") | .[0]));
first_missing( [generate(9) | addup | select(.>0) ] | unique[])</lang>
- Output:
211
Ten largest sums <lang jq>[generate(9) | addup | select(.>0)] | unique | .[(length-10):]</lang>
- Output:
[3456786,3456788,3456790,3456792,3456801,12345669,12345687,23456788,23456790,123456789]
Julia
<lang julia>using Printf, IterTools, DataStructures
expr(p::String...)::String = @sprintf("%s1%s2%s3%s4%s5%s6%s7%s8%s9", p...) function genexpr()::Vector{String}
op = ["+", "-", ""] return collect(expr(p...) for (p) in Iterators.product(op, op, op, op, op, op, op, op, op) if p[1] != "+")
end
using DataStructures
function allexpr()::Dict{Int,Int}
rst = DefaultDict{Int,Int}(0)
for e in genexpr()
val = eval(Meta.parse(e))
rst[val] += 1
end
return rst
end
sumto(val::Int)::Vector{String} = filter(e -> eval(Meta.parse(e)) == val, genexpr()) function maxsolve()::Dict{Int,Int}
ae = allexpr()
vmax = maximum(values(ae))
smax = filter(ae) do (v, f)
f == vmax
end
return smax
end function minsolve()::Int
ae = keys(allexpr())
for i in 1:typemax(Int)
if i ∉ ae
return i
end
end
end function highestsums(n::Int)::Vector{Int}
sums = collect(keys(allexpr())) return sort!(sums; rev=true)[1:n]
end
const solutions = sumto(100) const smax = maxsolve() const smin = minsolve() const hsums = highestsums(10)
println("100 =") foreach(println, solutions)
println("\nMax number of solutions:") for (v, f) in smax
@printf("%3i -> %2i\n", v, f)
end
println("\nMin number with no solutions: $smin")
println("\nHighest sums representable:") foreach(println, hsums)
</lang>
- Output:
100 = 1+23-4+56+7+8+9 12+3-4+5+67+8+9 1+2+34-5+67-8+9 -1+2-3+4+5+6+78+9 1+2+3-4+5+6+78+9 123-4-5-6-7+8-9 123+45-67+8-9 1+23-4+5+6+78-9 12-3-4+5-6+7+89 12+3+4+5-6-7+89 123-45-67+89 123+4-5+67-89 Max number of solutions: 9 -> 46 -9 -> 46 Min number with no solutions: 211 Highest sums representable: 123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
Kotlin
<lang scala>// version 1.1.51
class Expression {
private enum class Op { ADD, SUB, JOIN }
private val code = Array<Op>(NUMBER_OF_DIGITS) { Op.ADD }
companion object {
private const val NUMBER_OF_DIGITS = 9
private const val THREE_POW_4 = 3 * 3 * 3 * 3
private const val FMT = "%9d"
const val NUMBER_OF_EXPRESSIONS = 2 * THREE_POW_4 * THREE_POW_4
fun print(givenSum: Int) {
var expression = Expression()
repeat(Expression.NUMBER_OF_EXPRESSIONS) {
if (expression.toInt() == givenSum) println("${FMT.format(givenSum)} = $expression")
expression++
}
}
}
operator fun inc(): Expression {
for (i in 0 until code.size) {
code[i] = when (code[i]) {
Op.ADD -> Op.SUB
Op.SUB -> Op.JOIN
Op.JOIN -> Op.ADD
}
if (code[i] != Op.ADD) break
}
return this
}
fun toInt(): Int {
var value = 0
var number = 0
var sign = +1
for (digit in 1..9) {
when (code[NUMBER_OF_DIGITS - digit]) {
Op.ADD -> { value += sign * number; number = digit; sign = +1 }
Op.SUB -> { value += sign * number; number = digit; sign = -1 }
Op.JOIN -> { number = 10 * number + digit }
}
}
return value + sign * number
}
override fun toString(): String {
val sb = StringBuilder()
for (digit in 1..NUMBER_OF_DIGITS) {
when (code[NUMBER_OF_DIGITS - digit]) {
Op.ADD -> if (digit > 1) sb.append(" + ")
Op.SUB -> sb.append(" - ")
Op.JOIN -> {}
}
sb.append(digit)
}
return sb.toString().trimStart()
}
}
class Stat {
val countSum = mutableMapOf<Int, Int>() val sumCount = mutableMapOf<Int, MutableSet<Int>>()
init {
var expression = Expression()
repeat (Expression.NUMBER_OF_EXPRESSIONS) {
val sum = expression.toInt()
countSum.put(sum, 1 + (countSum[sum] ?: 0))
expression++
}
for ((k, v) in countSum) {
val set = if (sumCount.containsKey(v))
sumCount[v]!!
else
mutableSetOf<Int>()
set.add(k)
sumCount.put(v, set)
}
}
}
fun main(args: Array<String>) {
println("100 has the following solutions:\n")
Expression.print(100)
val stat = Stat()
val maxCount = stat.sumCount.keys.max()
val maxSum = stat.sumCount[maxCount]!!.max()
println("\n$maxSum has the maximum number of solutions, namely $maxCount")
var value = 0
while (stat.countSum.containsKey(value)) value++
println("\n$value is the lowest positive number with no solutions")
println("\nThe ten highest numbers that do have solutions are:\n")
stat.countSum.keys.toIntArray().sorted().reversed().take(10).forEach { Expression.print(it) }
}</lang>
- Output:
100 has the following solutions:
100 = 1 + 2 + 3 - 4 + 5 + 6 + 78 + 9
100 = 1 + 2 + 34 - 5 + 67 - 8 + 9
100 = 1 + 23 - 4 + 5 + 6 + 78 - 9
100 = 1 + 23 - 4 + 56 + 7 + 8 + 9
100 = 12 + 3 + 4 + 5 - 6 - 7 + 89
100 = 12 + 3 - 4 + 5 + 67 + 8 + 9
100 = 12 - 3 - 4 + 5 - 6 + 7 + 89
100 = 123 + 4 - 5 + 67 - 89
100 = 123 + 45 - 67 + 8 - 9
100 = 123 - 4 - 5 - 6 - 7 + 8 - 9
100 = 123 - 45 - 67 + 89
100 = - 1 + 2 - 3 + 4 + 5 + 6 + 78 + 9
9 has the maximum number of solutions, namely 46
211 is the lowest positive number with no solutions
The ten highest numbers that do have solutions are:
123456789 = 123456789
23456790 = 1 + 23456789
23456788 = - 1 + 23456789
12345687 = 12345678 + 9
12345669 = 12345678 - 9
3456801 = 12 + 3456789
3456792 = 1 + 2 + 3456789
3456790 = - 1 + 2 + 3456789
3456788 = 1 - 2 + 3456789
3456786 = - 1 - 2 + 3456789
Lua
<lang lua>local expressionsLength = 0 function compareExpressionBySum(a, b)
return a.sum - b.sum
end
local countSumsLength = 0 function compareCountSumsByCount(a, b)
return a.counts - b.counts
end
function evaluate(code)
local value = 0
local number = 0
local power = 1
for k=9,1,-1 do
number = power*k + number
local mod = code % 3
if mod == 0 then
-- ADD
value = value + number
number = 0
power = 1
elseif mod == 1 then
-- SUB
value = value - number
number = 0
power = 1
elseif mod == 2 then
-- JOIN
power = 10 * power
else
print("This should not happen.")
end
code = math.floor(code / 3)
end
return value
end
function printCode(code)
local a = 19683
local b = 6561
local s = ""
for k=1,9 do
local temp = math.floor((code % a) / b)
if temp == 0 then
-- ADD
if k>1 then
s = s .. '+'
end
elseif temp == 1 then
-- SUB
s = s .. '-'
end
a = b
b = math.floor(b/3)
s = s .. tostring(k)
end
print("\t"..evaluate(code).." = "..s)
end
-- Main local nexpr = 13122
print("Show all solutions that sum to 100") for i=0,nexpr-1 do
if evaluate(i) == 100 then
printCode(i)
end
end print()
print("Show the sum that has the maximum number of solutions") local nbest = -1 for i=0,nexpr-1 do
local test = evaluate(i)
if test>0 then
local ntest = 0
for j=0,nexpr-1 do
if evaluate(j) == test then
ntest = ntest + 1
end
if ntest > nbest then
best = test
nbest = ntest
end
end
end
end print(best.." has "..nbest.." solutions\n")
print("Show the lowest positive number that can't be expressed") local code = -1 for i=0,123456789 do
for j=0,nexpr-1 do
if evaluate(j) == i then
code = j
break
end
end
if evaluate(code) ~= i then
code = i
break
end
end print(code.."\n")
print("Show the ten highest numbers that can be expressed") local limit = 123456789 + 1 for i=1,10 do
local best=0
for j=0,nexpr-1 do
local test = evaluate(j)
if (test<limit) and (test>best) then
best = test
end
end
for j=0,nexpr-1 do
if evaluate(j) == best then
printCode(j)
end
end
limit = best
end</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
Mathematica
Defining all possible sums:
<lang Mathematica>operations =
DeleteCases[Tuples[{"+", "-", ""}, 9], {x_, y__} /; x == "+"];
sums =
Map[StringJoin[Riffle[#, CharacterRange["1", "9"]]] &, operations];</lang>
Sums to 100:
<lang Mathematica> TableForm@Select[sums, ToExpression@# == 100 &] </lang>
- Output:
-1+2-3+4+5+6+78+9 1+2+3-4+5+6+78+9 1+2+34-5+67-8+9 1+23-4+5+6+78-9 1+23-4+56+7+8+9 12+3+4+5-6-7+89 12+3-4+5+67+8+9 12-3-4+5-6+7+89 123+4-5+67-89 123+45-67+8-9 123-4-5-6-7+8-9 123-45-67+89
Maximum number of solutions: <lang Mathematica> MaximalBy[Counts@ToExpression@sums, Identity] </lang>
- Output:
<|9 -> 46, -9 -> 46|>
First unsolvable: <lang Mathematica> pos = Cases[ToExpression@sums, _?Positive]; n = 1; While[MemberQ[pos, n], ++n]; </lang>
- Output:
211
Ten largest sums: <lang Mathematica> {#, ToExpression@#}&/@TakeLargestBy[sums, ToExpression, 10]//TableForm </lang>
- Output:
123456789 123456789 1+23456789 23456790 -1+23456789 23456788 12345678+9 12345687 12345678-9 12345669 12+3456789 3456801 1+2+3456789 3456792 -1+2+3456789 3456790 1-2+3456789 3456788 -1-2+3456789 3456786
Modula-2
<lang modula2>MODULE SumTo100; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE Evaluate(code : INTEGER) : INTEGER; VAR
value,number,power,k : INTEGER;
BEGIN
value := 0; number := 0; power := 1;
FOR k:=9 TO 1 BY -1 DO
number := power * k + number;
IF code MOD 3 = 0 THEN
(* ADD *)
value := value + number;
number := 0;
power := 1
ELSIF code MOD 3 = 1 THEN
(* SUB *)
value := value - number;
number := 0;
power := 1
ELSE
(* CAT *)
power := power * 10
END;
code := code / 3
END;
RETURN value
END Evaluate;
PROCEDURE Print(code : INTEGER); VAR
expr,buf : ARRAY[0..63] OF CHAR; a,b,k,p : INTEGER;
BEGIN
a := 19683; b := 6561; p := 0;
FOR k:=1 TO 9 DO
IF (code MOD a) / b = 0 THEN
IF k > 1 THEN
expr[p] := '+';
INC(p)
END
ELSIF (code MOD a) / b = 1 THEN
expr[p] := '-';
INC(p)
END;
a := b;
b := b / 3;
expr[p] := CHR(k + 30H);
INC(p)
END;
expr[p] := 0C;
FormatString("%9i = %s\n", buf, Evaluate(code), expr);
WriteString(buf)
END Print;
(* Main *) CONST nexpr = 13122; VAR
i,j : INTEGER; best,nbest,test,ntest,limit : INTEGER; buf : ARRAY[0..63] OF CHAR;
BEGIN
WriteString("Show all solution that sum to 100");
WriteLn;
FOR i:=0 TO nexpr-1 DO
IF Evaluate(i) = 100 THEN
Print(i)
END
END;
WriteLn;
WriteString("Show the sum that has the maximum number of solutions");
WriteLn;
nbest := -1;
FOR i:=0 TO nexpr-1 DO
test := Evaluate(i);
IF test > 0 THEN
ntest := 0;
FOR j:=0 TO nexpr-1 DO
IF Evaluate(j) = test THEN
INC(ntest)
END;
IF ntest > nbest THEN
best := test;
nbest := ntest
END
END
END
END;
FormatString("%i has %i solutions\n\n", buf, best, nbest);
WriteString(buf);
WriteString("Show the lowest positive number that can't be expressed");
WriteLn;
FOR i:=0 TO 123456789 DO
FOR j:=0 TO nexpr-1 DO
IF i = Evaluate(j) THEN
BREAK
END
END;
IF i # Evaluate(j) THEN
BREAK
END
END;
FormatString("%i\n\n", buf, i);
WriteString(buf);
WriteString("Show the ten highest numbers that can be expressed");
WriteLn;
limit := 123456789 + 1;
FOR i:=1 TO 10 DO
best := 0;
FOR j:=0 TO nexpr-1 DO
test := Evaluate(j);
IF (test < limit) AND (test > best) THEN
best := test
END
END;
FOR j:=0 TO nexpr-1 DO
IF Evaluate(j) = best THEN
Print(j)
END
END;
limit := best
END;
ReadChar
END SumTo100.</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
Nim
Recursive solution.
<lang Nim>import algorithm, parseutils, sequtils, strutils, tables
type Expression = string
proc buildExprs(start: Natural = 0): seq[Expression] =
let item = if start == 0: "" else: $start if start == 9: return @[item] for expr in buildExprs(start + 1): result.add item & expr result.add item & '-' & expr if start != 0: result.add item & '+' & expr
proc evaluate(expr: Expression): int =
var idx = 0 var val: int while idx < expr.len: let n = expr.parseInt(val, idx) inc idx, n result += val
let exprs = buildExprs() var counts: CountTable[int]
echo "The solutions for 100 are:" for expr in exprs:
let sum = evaluate(expr) if sum == 100: echo expr if sum > 0: counts.inc(sum)
let (n, count) = counts.largest() echo "\nThe maximum count of positive solutions is $1 for number $2.".format(count, n)
var s = 1 while true:
if s notin counts: echo "\nThe smallest number than cannot be expressed is: $1.".format(s) break inc s
echo "\nThe ten highest numbers than can be expressed are:" let numbers = sorted(toSeq(counts.keys), Descending) echo numbers[0..9].join(", ")</lang>
- Output:
The solutions for 100 are: 123+4-5+67-89 123-45-67+89 12+3+4+5-6-7+89 12-3-4+5-6+7+89 1+23-4+5+6+78-9 123+45-67+8-9 123-4-5-6-7+8-9 1+2+3-4+5+6+78+9 -1+2-3+4+5+6+78+9 1+2+34-5+67-8+9 12+3-4+5+67+8+9 1+23-4+56+7+8+9 The maximum count of positive solutions is 46 for number 9. The smallest number than cannot be expressed is: 211. The ten highest numbers than can be expressed are: 123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790, 3456788, 3456786
Iterative previous solution written in French (updated).
<lang Nim>import strutils
var
ligne: string = "" sum: int opera: array[0..9, int] = [0,0,1,1,1,1,1,1,1,1] curseur: int = 9 boucle: bool tot: array[1..123456789, int] pG: int plusGrandes: array[1..10, string]
let
ope: array[0..3, string] = ["-",""," +"," -"] aAtteindre = 100
proc calcul(li: string): int =
var liS: seq[string] liS = split(li," ") for i in liS: if i.len > 0: result += parseInt(i)
echo "Valeur à atteindre : ",aAtteindre
while opera[1]<2:
ligne.add(ope[opera[1]])
ligne.add("1")
for i in 2..9:
ligne.add(ope[opera[i]])
ligne.add($i)
sum = calcul(ligne)
if sum == aAtteindre:
stdout.write(ligne)
echo " = ",sum
if sum>0:
tot[sum] += 1
pG = 1
while pG<10:
if sum>calcul(plusGrandes[pG]):
for k in countdown(10,pG+1):
plusGrandes[k]=plusGrandes[k-1]
plusGrandes[pG]=ligne
pG = 11
pG += 1
ligne = ""
boucle = true
while boucle:
opera[curseur] += 1
if opera[curseur] == 4:
opera[curseur]=1
curseur -= 1
else:
curseur = 9
boucle = false
echo "Valeur atteinte ",tot[aAtteindre]," fois." echo ""
var
min0: int = 0 max: int = 0 valmax: int = 0
for i in 1..123456789:
if tot[i]==0 and min0 == 0: min0 = i if tot[i]>max: max = tot[i] valmax = i
echo "Plus petite valeur ne pouvant pas être atteinte : ",min0 echo "Valeur atteinte le plus souvent : ",valmax,", atteinte ",max," fois." echo "" echo "Plus grandes valeurs pouvant être atteintes :" for i in 1..10:
echo calcul(plusGrandes[i])," = ",plusGrandes[i]</lang>
- Output:
Valeur à atteindre : 100 -1 +2 -3 +4 +5 +6 +78 +9 = 100 123 +45 -67 +8 -9 = 100 123 +4 -5 +67 -89 = 100 123 -45 -67 +89 = 100 123 -4 -5 -6 -7 +8 -9 = 100 12 +3 +4 +5 -6 -7 +89 = 100 12 +3 -4 +5 +67 +8 +9 = 100 12 -3 -4 +5 -6 +7 +89 = 100 1 +23 -4 +56 +7 +8 +9 = 100 1 +23 -4 +5 +6 +78 -9 = 100 1 +2 +34 -5 +67 -8 +9 = 100 1 +2 +3 -4 +5 +6 +78 +9 = 100 Valeur atteinte 12 fois. Plus petite valeur ne pouvant pas être atteinte : 211 Valeur atteinte le plus souvent : 9, atteinte 46 fois. Plus grandes valeurs pouvant être atteintes : 123456789 = 123456789 23456790 = 1 +23456789 23456788 = -1 +23456789 12345687 = 12345678 +9 12345669 = 12345678 -9 3456801 = 12 +3456789 3456792 = 1 +2 +3456789 3456790 = -1 +2 +3456789 3456788 = 1 -2 +3456789 3456786 = -1 -2 +3456789
Pascal
<lang Pascal>{ RossetaCode: Sum to 100, Pascal.
Find solutions to the "sum to one hundred" puzzle.
We don't use arrays, but recompute all values again and again. It is a little surprise that the time efficiency is quite acceptable. }
program sumto100;
const
ADD = 0; SUB = 1; JOIN = 2; { opcodes inserted between digits }
NEXPR = 13122; { the total number of expressions }
var
i, j: integer; loop: boolean; test, ntest, best, nbest, limit: integer;
function evaluate(code: integer): integer;
var
k: integer;
value, number, power: integer;
begin
value := 0;
number := 0;
power := 1;
for k := 9 downto 1 do
begin
number := power * k + number;
case code mod 3 of
ADD: begin value := value + number; number := 0; power := 1; end;
SUB: begin value := value - number; number := 0; power := 1; end;
JOIN: power := power * 10
end;
code := code div 3
end;
evaluate := value
end;
procedure print(code: integer);
var
k: integer;
a, b: integer;
begin
a := 19683;
b := 6561;
write( evaluate(code):9 );
write(' = ');
for k := 1 to 9 do
begin
case ((code mod a) div b) of
ADD: if k > 1 then write('+');
SUB: { always } write('-');
end;
a := b;
b := b div 3;
write( k:1 )
end;
writeln
end;
begin
writeln;
writeln('Show all solutions that sum to 100');
writeln;
for i := 0 to NEXPR - 1 do
if evaluate(i) = 100 then
print(i);
writeln;
writeln('Show the sum that has the maximum number of solutions');
writeln;
nbest := (-1);
for i := 0 to NEXPR - 1 do
begin
test := evaluate(i);
if test > 0 then
begin
ntest := 0;
for j := 0 to NEXPR - 1 do
if evaluate(j) = test then
ntest := ntest + 1;
if ntest > nbest then
begin
best := test;
nbest := ntest;
end
end
end;
writeln(best, ' has ', nbest, ' solutions');
writeln;
writeln('Show the lowest positive number that cant be expressed');
writeln;
i := 0;
loop := TRUE;
while (i <= 123456789) and loop do
begin
j := 0;
while (j < NEXPR - 1) and (i <> evaluate(j)) do
j := j + 1;
if i <> evaluate(j) then
loop := FALSE
else
i := i + 1;
end;
writeln(i);
writeln;
writeln('Show the ten highest numbers that can be expressed');
writeln;
limit := 123456789 + 1;
for i := 1 to 10 do
begin
best := 0;
for j := 0 to NEXPR - 1 do
begin
test := evaluate(j);
if (test < limit) and (test > best) then
best := test;
end;
for j := 0 to NEXPR - 1 do
if evaluate(j) = best then
print(j);
limit := best;
end
end.</lang>
- Output:
Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions
9 has 46 solutions
Show the lowest positive number that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
Perl
<lang perl>#!/usr/bin/perl use warnings; use strict; use feature qw{ say };
my $string = '123456789'; my $length = length $string; my @possible_ops = ("" , '+', '-');
{
my @ops;
sub Next {
return @ops = (0) x ($length) unless @ops;
my $i = 0;
while ($i < $length) {
if ($ops[$i]++ > $#possible_ops - 1) {
$ops[$i++] = 0;
next
}
# + before the first number
next if 0 == $i && '+' eq $possible_ops[ $ops[0] ];
return @ops
}
return
}
}
sub evaluate {
my ($expression) = @_; my $sum; $sum += $_ for $expression =~ /([-+]?[0-9]+)/g; return $sum
}
my %count = ( my $max_count = 0 => 0 );
say 'Show all solutions that sum to 100';
while (my @ops = Next()) {
my $expression = "";
for my $i (0 .. $length - 1) {
$expression .= $possible_ops[ $ops[$i] ];
$expression .= substr $string, $i, 1;
}
my $sum = evaluate($expression);
++$count{$sum};
$max_count = $sum if $count{$sum} > $count{$max_count};
say $expression if 100 == $sum;
}
say 'Show the sum that has the maximum number of solutions'; say "sum: $max_count; solutions: $count{$max_count}";
my $n = 1; ++$n until ! exists $count{$n}; say "Show the lowest positive sum that can't be expressed"; say $n;
say 'Show the ten highest numbers that can be expressed'; say for (sort { $b <=> $a } keys %count)[0 .. 9];</lang>
- Output:
Show all solutions that sum to 100 123-45-67+89 12-3-4+5-6+7+89 12+3+4+5-6-7+89 123+4-5+67-89 -1+2-3+4+5+6+78+9 1+2+3-4+5+6+78+9 12+3-4+5+67+8+9 1+23-4+56+7+8+9 1+2+34-5+67-8+9 1+23-4+5+6+78-9 123+45-67+8-9 123-4-5-6-7+8-9 Show the sum that has the maximum number of solutions sum: 9; solutions: 46 Show the lowest positive sum that can't be expressed 211 Show the ten highest numbers that can be expressed 123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
oneliner version
The first task posed can be solved simply with (pay attention to doublequotes around the program: adjust for you OS): <lang perl> perl -E "say for grep{eval $_ == 100} glob '{-,}'.join '{+,-,}',1..9" </lang>
While the whole task can be solved by: <lang perl> perl -MList::Util="first" -E "@c[0..10**6]=(0..10**6);say for grep{$e=eval;$c[$e]=undef if $e>=0;$h{$e}++;eval $_==100}glob'{-,}'.join'{+,-,}',1..9;END{say for(sort{$h{$b}<=>$h{$a}}grep{$_>=0}keys %h)[0],first{defined $_}@c;say for(sort{$b<=>$a}grep{$_>0}keys %h)[0..9]}" </lang> which outputs
-1+2-3+4+5+6+78+9 1+2+3-4+5+6+78+9 1+2+34-5+67-8+9 1+23-4+5+6+78-9 1+23-4+56+7+8+9 12+3+4+5-6-7+89 12+3-4+5+67+8+9 12-3-4+5-6+7+89 123+4-5+67-89 123+45-67+8-9 123-4-5-6-7+8-9 123-45-67+89 9 211 123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
Phix
This is just a trivial count in base 3, with a leading '+' being irrelevant, so from 0(3)000_000_000 to 0(3)122_222_222 which is only (in decimal) 13,122 ...
Admittedly, categorising them into 3429 bins is slightly more effort, but otherwise I am somewhat bemused by all the applescript/javascript/Haskell shenanegins.
enum SUB=-1, NOP=0, ADD=1 function eval(sequence s) integer res = 0, tmp = 0, op = ADD for i=1 to length(s) do if s[i]=NOP then tmp = tmp*10+i else res += op*tmp tmp = i op = s[i] end if end for return res + op*tmp end function procedure show(sequence s) string res = "" for i=1 to length(s) do if s[i]!=NOP then res &= ','-s[i] end if res &= '0'+i end for printf(1,"%s = %d\n",{res,eval(s)}) end procedure -- Logically this intersperses -/nop/+ between each digit, but you do not actually need the digit. sequence s = repeat(SUB,9) -- (==> ..nop+add*8) bool done = false integer maxl = 0, maxr integer count = 0 while not done do count += 1 integer r = eval(s), k = getd_index(r) sequence solns = iff(k=0?{s}:append(getd_by_index(k),s)) setd(r,solns) if r>0 and maxl<length(solns) then maxl = length(solns) maxr = r end if for i=length(s) to 1 by -1 do if i=1 and s[i]=NOP then done = true exit elsif s[i]!=ADD then s[i] += 1 exit end if s[i] = SUB end for end while printf(1,"%d solutions considered (dictionary size: %d)\n",{count,dict_size()}) sequence s100 = getd(100) printf(1,"There are %d sums to 100:\n",{length(s100)}) papply(s100,show) printf(1,"The positive sum of %d has the maximum number of solutions: %d\n",{maxr,maxl}) integer prev = 0 function missing(integer key, sequence /*data*/, integer /*pkey*/, object /*user_data=-2*/) if key!=prev+1 then return 0 end if prev = key return 1 end function traverse_dict_partial_key(missing,1) printf(1,"The lowest positive sum that cannot be expressed: %d\n",{prev+1}) sequence highest = {} function top10(integer key, sequence /*data*/, object /*user_data*/) highest &= key return length(highest)<10 end function traverse_dict(top10,rev:=1) printf(1,"The 10 highest sums: %v\n",{highest})
- Output:
13122 solutions considered (dictionary size: 3429)
There are 12 sums to 100:
-1+2-3+4+5+6+78+9 = 100
12-3-4+5-6+7+89 = 100
123-4-5-6-7+8-9 = 100
123-45-67+89 = 100
123+4-5+67-89 = 100
123+45-67+8-9 = 100
12+3-4+5+67+8+9 = 100
12+3+4+5-6-7+89 = 100
1+23-4+56+7+8+9 = 100
1+23-4+5+6+78-9 = 100
1+2+3-4+5+6+78+9 = 100
1+2+34-5+67-8+9 = 100
The positive sum of 9 has the maximum number of solutions: 46
The lowest positive sum that cannot be expressed: 211
The 10 highest sums: {123456789,23456790,23456788,12345687,12345669,3456801,3456792,3456790,3456788,3456786}
PureBasic
<lang PureBasic>#START=6561
- STOPP=19682
- SUMME=100
- BASIS="123456789"
Structure TSumTerm
sum.i ter.s
EndStructure
NewList Solutions.TSumTerm() NewMap SolCount.i() Dim op.s{1}(8) Dim b.s{1}(8) PokeS(@b(),#BASIS)
Procedure StripTerm(*p_Term)
If PeekS(*p_Term,1)="+" : PokeC(*p_Term,' ') : EndIf
EndProcedure
Procedure.s Triadisch(v)
While v : r$=Str(v%3)+r$ : v/3 : Wend ProcedureReturn r$
EndProcedure
Procedure.i Calc(t$)
While Len(t$) x=Val(t$) : r+x If x<0 : s$=Str(x) : Else : s$="+"+Str(x) : EndIf t$=RemoveString(t$,s$,#PB_String_NoCase,1,1) Wend ProcedureReturn r
EndProcedure
For n=#START To #STOPP
PokeS(@op(),Triadisch(n))
Term$=""
For i=0 To 8
Select op(i)
Case "0" : Term$+ b(i)
Case "1" : Term$+"+"+b(i)
Case "2" : Term$+"-"+b(i)
EndSelect
Next
AddElement(Solutions()) : Solutions()\sum=Calc(Term$) : StripTerm(@Term$) : Solutions()\ter=Term$
Next SortStructuredList(Solutions(),#PB_Sort_Ascending,OffsetOf(TSumTerm\sum),TypeOf(TSumTerm\sum))
If OpenConsole()
PrintN("Show all solutions that sum to 100:")
ForEach Solutions()
If Solutions()\sum=#SUMME : PrintN(#TAB$+Solutions()\ter) : EndIf
SolCount(Str(Solutions()\sum))+1
Next
ForEach SolCount()
If SolCount()>MaxCount : MaxCount=SolCount() : MaxVal$=MapKey(SolCount()) : EndIf
Next
PrintN("Show the positve sum that has the maximum number of solutions:")
PrintN(#TAB$+MaxVal$+" has "+Str(MaxCount)+" solutions")
If LastElement(Solutions())
MaxVal=Solutions()\sum
PrintN("Show the lowest positive number that can't be expressed:")
For i=1 To MaxVal
If SolCount(Str(i))=0 : PrintN(#TAB$+Str(i)) : Break : EndIf
Next
PrintN("Show the 10 highest numbers that can be expressed:")
For i=1 To 10
PrintN(#TAB$+LSet(Str(Solutions()\sum),9)+" = "+Solutions()\ter)
If Not PreviousElement(Solutions()) : Break : EndIf
Next
EndIf
Input()
EndIf</lang>
- Output:
Show all solutions that sum to 100: 123+45-67+8-9 123+4-5+67-89 123-45-67+89 123-4-5-6-7+8-9 12+3+4+5-6-7+89 12+3-4+5+67+8+9 12-3-4+5-6+7+89 1+23-4+56+7+8+9 1+23-4+5+6+78-9 1+2+34-5+67-8+9 1+2+3-4+5+6+78+9 -1+2-3+4+5+6+78+9 Show the positve sum that has the maximum number of solutions: 9 has 46 solutions Show the lowest positive number that can't be expressed: 211 Show the 10 highest numbers that can be expressed: 123456789 = 123456789 23456790 = 1+23456789 23456788 = -1+23456789 12345687 = 12345678+9 12345669 = 12345678-9 3456801 = 12+3456789 3456792 = 1+2+3456789 3456790 = -1+2+3456789 3456788 = 1-2+3456789 3456786 = -1-2+3456789
Python
<lang python>from itertools import product, islice
def expr(p):
return "{}1{}2{}3{}4{}5{}6{}7{}8{}9".format(*p)
def gen_expr():
op = ['+', '-', ] return [expr(p) for p in product(op, repeat=9) if p[0] != '+']
def all_exprs():
values = {}
for expr in gen_expr():
val = eval(expr)
if val not in values:
values[val] = 1
else:
values[val] += 1
return values
def sum_to(val):
for s in filter(lambda x: x[0] == val, map(lambda x: (eval(x), x), gen_expr())):
print(s)
def max_solve():
print("Sum {} has the maximum number of solutions: {}".
format(*max(all_exprs().items(), key=lambda x: x[1])))
def min_solve():
values = all_exprs()
for i in range(123456789):
if i not in values:
print("Lowest positive sum that can't be expressed: {}".format(i))
return
def highest_sums(n=10):
sums = map(lambda x: x[0],
islice(sorted(all_exprs().items(), key=lambda x: x[0], reverse=True), n))
print("Highest Sums: {}".format(list(sums)))
sum_to(100)
max_solve()
min_solve()
highest_sums()</lang>
- Output:
(100, '-1+2-3+4+5+6+78+9') (100, '1+2+3-4+5+6+78+9') (100, '1+2+34-5+67-8+9') (100, '1+23-4+5+6+78-9') (100, '1+23-4+56+7+8+9') (100, '12+3+4+5-6-7+89') (100, '12+3-4+5+67+8+9') (100, '12-3-4+5-6+7+89') (100, '123+4-5+67-89') (100, '123+45-67+8-9') (100, '123-4-5-6-7+8-9') (100, '123-45-67+89') Sum 9 has the maximum number of solutions: 46 Lowest positive sum that can't be expressed: 211 Highest Sums: [123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790, 3456788, 3456786]
Alternate solution
Mostly the same algorithm, but both shorter and faster.
<lang python>import itertools from collections import defaultdict, Counter
s = "123456789" h = defaultdict(list) for v in itertools.product(["+", "-", ""], repeat=9):
if v[0] != "+":
e = "".join("".join(u) for u in zip(v, s))
h[eval(e)].append(e)
print("Solutions for 100") for e in h[100]:
print(e)
c = Counter({k: len(v) for k, v in h.items() if k >= 0})
k, m = c.most_common(1)[0] print("Maximum number of solutions for %d (%d solutions)" % (k, m))
v = sorted(c.keys())
for i in range(v[-1]):
if i not in c:
print("Lowest impossible sum: %d" % i)
break
print("Ten highest sums") for k in reversed(v[-10:]):
print(k)</lang>
- Output:
Solutions for 100 -1+2-3+4+5+6+78+9 1+2+3-4+5+6+78+9 1+2+34-5+67-8+9 1+23-4+5+6+78-9 1+23-4+56+7+8+9 12+3+4+5-6-7+89 12+3-4+5+67+8+9 12-3-4+5-6+7+89 123+4-5+67-89 123+45-67+8-9 123-4-5-6-7+8-9 123-45-67+89 Maximum number of solutions for 9 (46 solutions) Lowest impossible sum: 211 Ten highest sums 123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
Racket
<lang racket>#lang racket
(define list-partitions
(match-lambda
[(list) (list null)]
[(and L (list _)) (list (list L))]
[(list L ...)
(for*/list
((i (in-range 1 (add1 (length L))))
(r (in-list (list-partitions (drop L i)))))
(cons (take L i) r))]))
(define digits->number (curry foldl (λ (dgt acc) (+ (* 10 acc) dgt)) 0))
(define partition-digits-to-numbers
(let ((memo (make-hash)))
(λ (dgts)
(hash-ref! memo dgts
(λ ()
(map (λ (p) (map digits->number p))
(list-partitions dgts)))))))
(define (fold-sum-to-ns digits kons k0)
(define (get-solutions nmbrs acc chain k)
(match nmbrs
[(list)
(kons (cons acc (let ((niahc (reverse chain)))
(if (eq? '+ (car niahc)) (cdr niahc) niahc)))
k)]
[(cons a d)
(get-solutions d (- acc a) (list* a '- chain)
(get-solutions d (+ acc a) (list* a '+ chain) k))]))
(foldl (λ (nmbrs k) (get-solutions nmbrs 0 null k)) k0 (partition-digits-to-numbers digits)))
(define sum-to-ns/hash-promise
(delay (fold-sum-to-ns
'(1 2 3 4 5 6 7 8 9)
(λ (a.s d) (hash-update d (car a.s) (λ (x) (cons (cdr a.s) x)) list))
(hash))))
(module+ main
(define S (force sum-to-ns/hash-promise))
(displayln "Show all solutions that sum to 100")
(pretty-print (hash-ref S 100))
(displayln "Show the sum that has the maximum number of solutions (from zero to infinity*)")
(let-values (([k-max v-max]
(for/fold ((k-max #f) (v-max 0))
(([k v] (in-hash S)) #:when (> (length v) v-max))
(values k (length v)))))
(printf "~a has ~a solutions~%" k-max v-max))
(displayln "Show the lowest positive sum that can't be expressed (has no solutions),
using the rules for this task")
(for/first ((n (in-range 1 (add1 123456789))) #:unless (hash-has-key? S n)) n)
(displayln "Show the ten highest numbers that can be expressed using the rules for this task")
(take (sort (hash-keys S) >) 10))
(module+ test
(require rackunit) (check-equal? (list-partitions null) '(())) (check-equal? (list-partitions '(1)) '(((1)))) (check-equal? (list-partitions '(1 2)) '(((1) (2)) ((1 2)))) (check-equal? (partition-digits-to-numbers '()) '(())) (check-equal? (partition-digits-to-numbers '(1)) '((1))) (check-equal? (partition-digits-to-numbers '(1 2)) '((1 2) (12))))</lang>
- Output:
Show all solutions that sum to 100 '((123 - 45 - 67 + 89) (123 + 45 - 67 + 8 - 9) (123 + 4 - 5 + 67 - 89) (123 - 4 - 5 - 6 - 7 + 8 - 9) (12 + 3 - 4 + 5 + 67 + 8 + 9) (12 - 3 - 4 + 5 - 6 + 7 + 89) (12 + 3 + 4 + 5 - 6 - 7 + 89) (1 + 23 - 4 + 56 + 7 + 8 + 9) (1 + 23 - 4 + 5 + 6 + 78 - 9) (1 + 2 + 34 - 5 + 67 - 8 + 9) (- 1 + 2 - 3 + 4 + 5 + 6 + 78 + 9) (1 + 2 + 3 - 4 + 5 + 6 + 78 + 9)) Show the sum that has the maximum number of solutions (from zero to infinity*) 9 has 46 solutions Show the lowest positive sum that can't be expressed (has no solutions), using the rules for this task 211 Show the ten highest numbers that can be expressed using the rules for this task '(123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786)
Raku
(formerly Perl 6)
<lang perl6>my $sum = 100; my $N = 10; my @ops = ['-', ], |( [' + ', ' - ', ] xx 8 ); my @str = [X~] map { .Slip }, ( @ops Z 1..9 ); my %sol = @str.classify: *.subst( ' - ', ' -', :g )\
.subst( ' + ', ' ', :g ).words.sum;
my %count.push: %sol.map({ .value.elems => .key });
my $max-solutions = %count.max( + *.key ); my $first-unsolvable = first { %sol{$_} :!exists }, 1..*; sub n-largest-sums (Int $n) { %sol.sort(-*.key)[^$n].fmt: "%8s => %s\n" }
given %sol{$sum}:p {
say "{.value.elems} solutions for sum {.key}:";
say " $_" for .value.list;
}
.say for :$max-solutions, :$first-unsolvable, "$N largest sums:", n-largest-sums($N);</lang>
- Output:
12 solutions for sum 100:
-1 + 2 - 3 + 4 + 5 + 6 + 78 + 9
1 + 2 + 3 - 4 + 5 + 6 + 78 + 9
1 + 2 + 34 - 5 + 67 - 8 + 9
1 + 23 - 4 + 5 + 6 + 78 - 9
1 + 23 - 4 + 56 + 7 + 8 + 9
12 + 3 + 4 + 5 - 6 - 7 + 89
12 + 3 - 4 + 5 + 67 + 8 + 9
12 - 3 - 4 + 5 - 6 + 7 + 89
123 + 4 - 5 + 67 - 89
123 + 45 - 67 + 8 - 9
123 - 4 - 5 - 6 - 7 + 8 - 9
123 - 45 - 67 + 89
max-solutions => 46 => [-9 9]
first-unsolvable => 211
10 largest sums:
123456789 => 123456789
23456790 => 1 + 23456789
23456788 => -1 + 23456789
12345687 => 12345678 + 9
12345669 => 12345678 - 9
3456801 => 12 + 3456789
3456792 => 1 + 2 + 3456789
3456790 => -1 + 2 + 3456789
3456788 => 1 - 2 + 3456789
3456786 => -1 - 2 + 3456789
REXX
<lang rexx>/*REXX pgm solves a puzzle: using the string 123456789, insert - or + to sum to 100*/ parse arg LO HI . /*obtain optional arguments from the CL*/ if LO== | LO=="," then LO= 100 /*Not specified? Then use the default.*/ if HI== | HI=="," then HI= LO /* " " " " " " */ if LO==00 then HI= 123456789 /*LOW specified as zero with leading 0.*/ ops= '+-'; L= length(ops) + 1 /*define operators (and their length). */ @.=; do i=1 for L-1; @.i= substr(ops,i,1) /* " some handy-dandy REXX literals*/
end /*i*/ /* " individual operators for speed*/
mx= 0; mn= 999999 /*initialize the minimums and maximums.*/ mxL=; mnL=; do j=LO to HI until LO==00 & mn==0 /*solve with range of sums*/
z= ???(j) /*find # solutions for J. */
if z> mx then mxL= /*is this a new maximum ? */
if z>=mx then do; mxL=mxL j; mx=z; end /*remember this new max. */
if z< mn then mnL= /*is this a new minimum ? */
if z<=mn then do; mnL=mnL j; mn=z; end /*remember this new min. */
end /*j*/
if LO==HI then exit 0 /*don't display max&min ? */ @@= 'number of solutions: '; say _= words(mxL); say 'sum's(_) "of" mxL ' 's(_,"have",'has') 'the maximum' @@ mx _= words(mnL); say 'sum's(_) "of" mnL ' 's(_,"have",'has') 'the minimum' @@ mn exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ s: if arg(1)==1 then return arg(3); return word( arg(2) "s",1) /*simple pluralizer*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ ???: parse arg answer; #= 0 /*obtain the answer (sum) to the puzzle*/
do a=L-1 for 2; aa= @.a'1' /*choose one of - or nothing. */
do b=1 for L; bb= aa || @.b'2' /* " " " - +, or abutment.*/
do c=1 for L; cc= bb || @.c'3' /* " " " " " " " */
do d=1 for L; dd= cc || @.d'4' /* " " " " " " " */
do e=1 for L; ee= dd || @.e'5' /* " " " " " " " */
do f=1 for L; ff= ee || @.f'6' /* " " " " " " " */
do g=1 for L; gg= ff || @.g'7' /* " " " " " " " */
do h=1 for L; hh= gg || @.h'8' /* " " " " " " " */
do i=1 for L; ii= hh || @.i'9' /* " " " " " " " */
interpret '$=' ii /*calculate the sum of modified string.*/
if $\==answer then iterate /*Is sum not equal to answer? Then skip*/
#= # + 1; if LO==HI then say 'solution: ' $ " ◄───► " ii
end /*i*/ /* */
end /*h*/ /* d */
end /*g*/ /* d */
end /*f*/ /* eeeee n nnnn dddddd sssss */
end /*e*/ /* e e nn n d d s */
end /*d*/ /* eeeeeee n n d d sssss */
end /*c*/ /* e n n d d s */
end /*b*/ /* eeeee n n ddddd sssss */
end /*a*/ /* */
y= # /* [↓] adjust the number of solutions?*/
if y==0 then y= 'no' /* [↓] left justify plural of solution*/
if LO\==00 then say right(y, 9) 'solution's(#, , " ") 'found for' ,
right(j, length(HI) ) left(, #, "─")
return # /*return the number of solutions found.*/</lang>
- output when using the default input:
solution: 100 ◄───► -1+2-3+4+5+6+78+9
solution: 100 ◄───► 1+2+3-4+5+6+78+9
solution: 100 ◄───► 1+2+34-5+67-8+9
solution: 100 ◄───► 1+23-4+5+6+78-9
solution: 100 ◄───► 1+23-4+56+7+8+9
solution: 100 ◄───► 12+3+4+5-6-7+89
solution: 100 ◄───► 12+3-4+5+67+8+9
solution: 100 ◄───► 12-3-4+5-6+7+89
solution: 100 ◄───► 123+4-5+67-89
solution: 100 ◄───► 123+45-67+8-9
solution: 100 ◄───► 123-4-5-6-7+8-9
solution: 100 ◄───► 123-45-67+89
12 solutions found for 100
- output when the following input is used: 00
sum of 9 has the maximum number of solutions: 46 sum of 211 has the minimum number of solutions: 0
Ruby
<lang ruby>def gen_expr
x = ['-', ]
y = ['+', '-', ]
x.product(y,y,y,y,y,y,y,y)
.map do |a,b,c,d,e,f,g,h,i|
"#{a}1#{b}2#{c}3#{d}4#{e}5#{f}6#{g}7#{h}8#{i}9"
end
end
def sum_to(val)
gen_expr.map{|expr| [eval(expr), expr]}.select{|v,expr| v==val}.each{|x| p x}
end
def max_solve
n,size = gen_expr.group_by{|expr| eval(expr)}
.select{|val,_| val>=0}
.map{|val,exprs| [val, exprs.size]}
.max_by{|_,size| size}
puts "sum of #{n} has the maximum number of solutions : #{size}"
end
def min_solve
solves = gen_expr.group_by{|expr| eval(expr)}
n = 0.step{|i| break i unless solves[i]}
puts "lowest positive sum that can't be expressed : #{n}"
end
def highest_sums(n=10)
n = gen_expr.map{|expr| eval(expr)}.uniq.sort.reverse.take(n)
puts "highest sums : #{n}"
end
sum_to(100) max_solve min_solve highest_sums</lang>
- Output:
[100, "-1+2-3+4+5+6+78+9"] [100, "1+2+3-4+5+6+78+9"] [100, "1+2+34-5+67-8+9"] [100, "1+23-4+5+6+78-9"] [100, "1+23-4+56+7+8+9"] [100, "12+3+4+5-6-7+89"] [100, "12+3-4+5+67+8+9"] [100, "12-3-4+5-6+7+89"] [100, "123+4-5+67-89"] [100, "123+45-67+8-9"] [100, "123-4-5-6-7+8-9"] [100, "123-45-67+89"] sum of 9 has the maximum number of solutions : 46 lowest positive sum that can't be expressed : 211 highest sums : [123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790, 3456788, 3456786]
Scala
<lang scala>object SumTo100 {
def main(args: Array[String]): Unit = {
val exps = expressions(9).map(str => (str, eval(str)))
val sums = exps.map(_._2).sortWith(_>_)
val s1 = exps.filter(_._2 == 100)
val s2 = sums.distinct.map(s => (s, sums.count(_ == s))).maxBy(_._2)
val s3 = sums.distinct.reverse.filter(_>0).zipWithIndex.dropWhile{case (n, i) => n == i + 1}.head._2 + 1
val s4 = sums.distinct.take(10)
println(s"""All ${s1.size} solutions that sum to 100:
|${s1.sortBy(_._1.length).map(p => s"${p._2} = ${p._1.tail}").mkString("\n")}
|
|Most common sum: ${s2._1} (${s2._2})
|Lowest unreachable sum: $s3
|Highest 10 sums: ${s4.mkString(", ")}""".stripMargin)
}
def expressions(l: Int): LazyList[String] = configurations(l).map(p => p.zipWithIndex.map{case (op, n) => s"${opChar(op)}${n + 1}"}.mkString)
def configurations(l: Int): LazyList[Vector[Int]] = LazyList.range(0, math.pow(3, l).toInt).map(config(l)).filter(_.head != 0)
def config(l: Int)(num: Int): Vector[Int] = Iterator.iterate((num%3, num/3)){case (_, n) => (n%3, n/3)}.map(_._1 - 1).take(l).toVector
def eval(exp: String): Int = (exp.headOption, exp.tail.takeWhile(_.isDigit), exp.tail.dropWhile(_.isDigit)) match{
case (Some(op), n, str) => doOp(op, n.toInt) + eval(str)
case _ => 0
}
def doOp(sel: Char, n: Int): Int = if(sel == '-') -n else n
def opChar(sel: Int): String = sel match{
case -1 => "-"
case 1 => "+"
case _ => ""
}
}</lang>
- Output:
All 12 solutions that sum to 100: 100 = 123-45-67+89 100 = 123+45-67+8-9 100 = 123+4-5+67-89 100 = 1+23-4+5+6+78-9 100 = 123-4-5-6-7+8-9 100 = 12+3+4+5-6-7+89 100 = 12-3-4+5-6+7+89 100 = 1+2+34-5+67-8+9 100 = 12+3-4+5+67+8+9 100 = 1+23-4+56+7+8+9 100 = 1+2+3-4+5+6+78+9 100 = 1+2-3+4+5+6+78+9 Most common sum: 9 (46) Lowest unreachable sum: 211 Highest 10 sums: 123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790, 3456788, 3456786
Sidef
<lang ruby>func gen_expr() is cached {
var x = ['-', ] var y = ['+', '-', ]
gather {
cartesian([x,y,y,y,y,y,y,y,y], {|a,b,c,d,e,f,g,h,i|
take("#{a}1#{b}2#{c}3#{d}4#{e}5#{f}6#{g}7#{h}8#{i}9")
})
}
}
func eval_expr(expr) is cached {
expr.scan(/([-+]?\d+)/).sum_by { Num(_) }
}
func sum_to(val) {
gen_expr().grep { eval_expr(_) == val }
}
func max_solve() {
gen_expr().grep { eval_expr(_) >= 0 } \
.group_by { eval_expr(_) } \
.max_by {|_,v| v.len }
}
func min_solve() {
var h = gen_expr().group_by { eval_expr(_) }
for i in (0..Inf) { h.exists(i) || return i }
}
func highest_sums(n=10) {
gen_expr().map { eval_expr(_) }.uniq.sort.reverse.first(n)
}
sum_to(100).each { say "100 = #{_}" }
var (n, solutions) = max_solve()... say "Sum of #{n} has the maximum number of solutions: #{solutions.len}" say "Lowest positive sum that can't be expressed : #{min_solve()}" say "Highest sums: #{highest_sums()}"</lang>
- Output:
100 = -1+2-3+4+5+6+78+9 100 = 1+2+3-4+5+6+78+9 100 = 1+2+34-5+67-8+9 100 = 1+23-4+5+6+78-9 100 = 1+23-4+56+7+8+9 100 = 12+3+4+5-6-7+89 100 = 12+3-4+5+67+8+9 100 = 12-3-4+5-6+7+89 100 = 123+4-5+67-89 100 = 123+45-67+8-9 100 = 123-4-5-6-7+8-9 100 = 123-45-67+89 Sum of 9 has the maximum number of solutions: 46 Lowest positive sum that can't be expressed : 211 Highest sums: [123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790, 3456788, 3456786]
Tcl
<lang Tcl>proc sum_to_100 {} {
for {set i 0} {$i <= 13121} {incr i} {
set i3 [format %09d [dec2base 3 $i]] set form "" set subs {"" - +} foreach a [split $i3 ""] b [split 123456789 ""] { append form [lindex $subs $a] $b } lappend R([expr $form]) $form
}
puts "solutions for sum=100:\n[join [lsort $R(100)] \n]"
set max -1
foreach key [array names R] {
if {[llength $R($key)] > $max} { set max [llength $R($key)] set maxkey $key }
}
puts "max solutions: $max for $maxkey"
for {set i 0} {$i <= 123456789} {incr i} {
if ![info exists R($i)] { puts "first unsolvable: $i" break }
} puts "highest 10:\n[lrange [lsort -integer -decr [array names R]] 0 9]"
} proc dec2base {base dec} {
set res ""
while {$dec > 0} {
set res [expr $dec%$base]$res set dec [expr $dec/$base]
}
if {$res eq ""} {set res 0}
return $res
} sum_to_100</lang>
~ $ ./sum_to_100.tcl solutions for sum=100: -1+2-3+4+5+6+78+9 1+2+3-4+5+6+78+9 1+2+34-5+67-8+9 1+23-4+5+6+78-9 1+23-4+56+7+8+9 12+3+4+5-6-7+89 12+3-4+5+67+8+9 12-3-4+5-6+7+89 123+4-5+67-89 123+45-67+8-9 123-4-5-6-7+8-9 123-45-67+89 max solutions: 46 for 9 first unsolvable: 211 highest 10: 123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
Visual Basic .NET
Of course, one could just code-convert the existing C# example, but I thought this could be written with some simpler constructs. The point of doing this is to make the code more compatible with other BASIC languages. Not every language has something similar to the Enumerable Range construct. I also found the Dictionary construct could be implemented with something more primitive.
Another interesting thing this program can do is solve for other sets of numbers easily, as neither the number of digits, nor the digit sequence itself, is hard-coded. You could solve for the digits 1 through 8, for example, or the digits starting at 9 and going down to 1. One can even override the target sum (of 100) parameter, if you happen to be interested in another number. <lang vbnet>' Recursively iterates (increments) iteration array, returns -1 when out of "digits". Function plusOne(iAry() As Integer, spot As Integer) As Integer
Dim spotLim As Integer = If(spot = 0, 1, 2) ' The first "digit" has a lower limit.
If iAry(spot) = spotLim Then ' Check if spot has reached limit
If spot = 0 Then Return -1 ' No previous spot to increment, so indicate completed.
iAry(spot) = 0 ' Reset current spot, and
Return plusOne(iAry, spot - 1) ' Increment previous spot.
Else
iAry(spot) += 1 ' Increment current spot.
End If
Return spot
End Function
' Returns string sequence of operations from iAry and terms string Function generate(iAry() As Integer, terms As String) As String
Dim operations As String() = {"", "-", "+"} ' Possible operations.
generate = ""
For i As Integer = 0 To iAry.Count - 1
generate &= operations(iAry(i)) & Mid(terms, i + 1, 1).ToString()
Next
End Function
' Returns evaluation of string sequence Function eval(sequence As String) As Integer
eval = 0
Dim term As Integer = 0, operation As Integer = 1
For Each ch As Char In sequence
Select Case ch
Case "-", "+" ' New operation detected, apply previous operation to term,
eval += If(operation = 0, -term, term) : term = 0 ' and reset term.
operation = If(ch = "-", 0, 1) ' Note next operation.
Case Else ' Digit detected, increase term.
term = term * 10 + Val(ch)
End Select
Next
eval += If(operation = 0, -term, term) ' Apply final term.
End Function
' Sorts a pair of List(Of Integer) by the first Sub reSort(ByRef first As List(Of Integer), ByRef second As List(Of Integer))
Dim lou As New List(Of ULong) ' Temporary list of ULong for sorting.
For i As Integer = 0 To first.Count - 1
lou.Add((CULng(first(i)) << 32) + second(i)) ' "Pack" list items.
Next
lou.Sort()
For k As Integer = 0 To first.Count - 1
first(k) = lou(k) >> 32 ' "Unpack" first list item.
second(k) = lou(k) And &H7FFFFFFF ' "Unpack" second list item.
Next
End Sub
' Returns first result not in sequence, assumes passed list is sorted before call, ' uses binary search algo. Function firstMiss(loi As List(Of Integer))
Dim low As Integer = 0, high As Integer = loi.Count - 1, middle = (low + high) \ 2
Do
If loi(middle) = middle Then low = middle + 1 Else high = middle - 1
middle = (low + high) \ 2
Loop Until high <= low
Return middle + If(loi(middle) = middle, 1, 0)
End Function
' Iterates through all possible operations, ' uses a pair of List (of Integer) to tabulate solutions. Sub Solve100(Optional terms As String = "123456789",
Optional targSum As Integer = 100,
Optional highNums As Integer = 10)
Dim lastDig As Integer = Len(terms) - 1 ' The final "digit".
Dim iAry() As Integer = New Integer(lastDig) {} ' Iterations array.
Dim seq As String ' Sequence of numbers and operations.
Dim sVal As Integer ' Sequence value.
Dim sCnt As Integer = 1 ' Solution count (targSum).
Dim res As New List(Of Integer) ' List of results.
Dim tally As New List(Of Integer) ' Tally of results.
Console.WriteLine("List of solutions that evaluate to 100:")
Do ' Tabulate results until digits are exhausted.
seq = generate(iAry, terms) ' Obtain next expression.
sVal = eval(seq) ' Obtain next evaluation.
If sVal >= 0 Then ' Don't bother saving the negative results.
If res.Contains(sVal) Then tally(res.IndexOf(sVal)) += 1 _
Else res.Add(sVal) : tally.Add(1)
If sVal = targSum Then _
Console.WriteLine(" {0,2} {1}", sCnt, seq) : sCnt += 1
End If
Loop Until plusOne(iAry, lastDig) < 0
reSort(tally, res) ' Sort by tally to find result with the most solutions.
Console.WriteLine("The sum that has the the most solutions is {0}, (at {1}).",
res.Last, tally.Last)
reSort(res, tally) ' Sort by result to find first missing result and top results.
Console.WriteLine("The lowest positive sum that can't be expressed is {0}.",
firstMiss(res))
Console.WriteLine("The ten highest numbers that can be expressed are:")
res.Reverse() ' To let us take the last items for output.
sCnt = 0 ' Keep track of items displayed (for formatting).
For Each item As Integer In res.Take(highNums)
Console.Write("{0, -11}", item)
sCnt = (sCnt + 1) Mod 5 : If sCnt = 0 Then Console.WriteLine()
Next
End Sub
Sub Main()
Solve100() ' if interested, try this: Solve100("987654321")
End Sub</lang>
- Output:
List of solutions that evaluate to 100: 1 123-45-67+89 2 123-4-5-6-7+8-9 3 123+45-67+8-9 4 123+4-5+67-89 5 12-3-4+5-6+7+89 6 12+3-4+5+67+8+9 7 12+3+4+5-6-7+89 8 1+23-4+56+7+8+9 9 1+23-4+5+6+78-9 10 1+2+34-5+67-8+9 11 1+2+3-4+5+6+78+9 12 -1+2-3+4+5+6+78+9 The sum that has the the most solutions is 9, (at 46). The lowest positive sum that can't be expressed is 211. The ten highest numbers that can be expressed are: 123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
Wren
<lang ecmascript>import "/dynamic" for Enum import "/fmt" for Fmt import "/set" for Set import "/math" for Nums import "/sort" for Sort
var Op = Enum.create("Op", ["ADD", "SUB", "JOIN"])
var NUMBER_OF_DIGITS = 9 var THREE_POW_4 = 3 * 3 * 3 * 3 var NUMBER_OF_EXPRESSIONS = 2 * THREE_POW_4 * THREE_POW_4
class Expression {
static print(givenSum) {
var expr = Expression.new()
for (i in 0...NUMBER_OF_EXPRESSIONS) {
if (expr.toInt == givenSum) Fmt.print("$9d = $s", givenSum, expr)
expr.inc
}
}
construct new() {
_code = List.filled(NUMBER_OF_DIGITS, Op.ADD)
}
inc {
for (i in 0..._code.count) {
_code[i] = (_code[i] == Op.ADD) ? Op.SUB : (_code[i] == Op.SUB) ? Op.JOIN : Op.ADD
if (_code[i] != Op.ADD) break
}
return this
}
toInt {
var value = 0
var number = 0
var sign = 1
for (digit in 1..9) {
var c = _code[NUMBER_OF_DIGITS - digit]
if (c == Op.ADD) {
value = value + sign * number
number = digit
sign = 1
} else if (c == Op.SUB) {
value = value + sign * number
number = digit
sign = -1
} else {
number = 10 * number + digit
}
}
return value + sign * number
}
toString {
var sb = ""
for (digit in 1..NUMBER_OF_DIGITS) {
var c = _code[NUMBER_OF_DIGITS - digit]
if (c == Op.ADD) {
if (digit > 1) sb = sb + " + "
} else if (c == Op.SUB) {
sb = sb + " - "
}
sb = sb + digit.toString
}
return sb.trimStart()
}
}
class Stat {
construct new() {
_countSum = {}
_sumCount = {}
var expr = Expression.new()
for (i in 0...NUMBER_OF_EXPRESSIONS) {
var sum = expr.toInt
_countSum[sum] = _countSum[sum] ? 1 + _countSum[sum] : 1
expr.inc
}
for (me in _countSum) {
var set = _sumCount.containsKey(me.value) ? _sumCount[me.value] : Set.new()
set.add(me.key)
_sumCount[me.value] = set
}
}
countSum { _countSum }
sumCount { _sumCount }
}
System.print("100 has the following solutions:\n") Expression.print(100)
var stat = Stat.new() var maxCount = Nums.max(stat.sumCount.keys) var maxSum = Nums.max(stat.sumCount[maxCount]) System.print("\n%(maxSum) has the maximum number of solutions, namely %(maxCount)")
var value = 0 while (stat.countSum.containsKey(value)) value = value + 1 System.print("\n%(value) is the lowest positive number with no solutions")
System.print("\nThe ten highest numbers that do have solutions are:\n") var res = stat.countSum.keys.toList Sort.quick(res) res[-1..0].take(10).each { |e| Expression.print(e) }</lang>
- Output:
100 has the following solutions:
100 = 1 + 2 + 3 - 4 + 5 + 6 + 78 + 9
100 = 1 + 2 + 34 - 5 + 67 - 8 + 9
100 = 1 + 23 - 4 + 5 + 6 + 78 - 9
100 = 1 + 23 - 4 + 56 + 7 + 8 + 9
100 = 12 + 3 + 4 + 5 - 6 - 7 + 89
100 = 12 + 3 - 4 + 5 + 67 + 8 + 9
100 = 12 - 3 - 4 + 5 - 6 + 7 + 89
100 = 123 + 4 - 5 + 67 - 89
100 = 123 + 45 - 67 + 8 - 9
100 = 123 - 4 - 5 - 6 - 7 + 8 - 9
100 = 123 - 45 - 67 + 89
100 = - 1 + 2 - 3 + 4 + 5 + 6 + 78 + 9
9 has the maximum number of solutions, namely 46
211 is the lowest positive number with no solutions
The ten highest numbers that do have solutions are:
123456789 = 123456789
23456790 = 1 + 23456789
23456788 = - 1 + 23456789
12345687 = 12345678 + 9
12345669 = 12345678 - 9
3456801 = 12 + 3456789
3456792 = 1 + 2 + 3456789
3456790 = - 1 + 2 + 3456789
3456788 = 1 - 2 + 3456789
3456786 = - 1 - 2 + 3456789
zkl
Taking a big clue from Haskell and just calculate the world. <lang zkl>var all = // ( (1,12,123...-1,-12,...), (2,23,...) ...)
(9).pump(List,fcn(n){ split("123456789"[n,*]) }) // 45
.apply(fcn(ns){ ns.extend(ns.copy().apply('*(-1))) }); // 90
fcn calcAllSums{ // calculate all 6572 sums (1715 unique)
fcn(n,sum,soFar,r){
if(n==9) return();
foreach b in (all[n]){
if(sum+b>=0 and b.abs()%10==9) r.appendV(sum+b,"%s%+d".fmt(soFar,b)); self.fcn(b.abs()%10,sum + b,"%s%+d".fmt(soFar,b),r);
} }(0,0,"",r:=Dictionary()); r
}
// "123" --> (1,12,123)
fcn split(nstr){ (1).pump(nstr.len(),List,nstr.get.fp(0),"toInt") }</lang> <lang zkl>fcn showSums(allSums,N=100,printSolutions=2){
slns:=allSums.find(N,T);
if(printSolutions) println("%d solutions for N=%d".fmt(slns.len(),N));
if(printSolutions==2) println(slns.concat("\n"));
println();
}
allSums:=calcAllSums(); showSums(allSums); showSums(allSums,0,1);
println("Smallest postive integer with no solution: ",
[1..].filter1('wrap(n){ Void==allSums.find(n) }));
println("5 commonest sums (sum, number of ways to calculate to it):"); ms:=allSums.values.apply("len").sort()[-5,*]; // 5 mostest sums allSums.pump(List, // get those pairs
'wrap([(k,v)]){ v=v.len(); ms.holds(v) and T(k.toInt(),v) or Void.Skip })
.sort(fcn(kv1,kv2){ kv1[1]>kv2[1] }) // and sort .println();</lang>
- Output:
12 solutions for N=100 +1+2+3-4+5+6+78+9 +1+2+34-5+67-8+9 +1+23-4+5+6+78-9 +1+23-4+56+7+8+9 +12+3+4+5-6-7+89 +12+3-4+5+67+8+9 +12-3-4+5-6+7+89 +123+4-5+67-89 +123+45-67+8-9 +123-4-5-6-7+8-9 +123-45-67+89 -1+2-3+4+5+6+78+9 22 solutions for N=0 Smallest postive integer with no solution: 211 5 commonest sums (sum, number of ways to calculate to it): L(L(9,46),L(27,44),L(15,43),L(1,43),L(21,43))
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