# Prelude:

This task is taken from Five Problems Every Software Engineer Should Be Able to Solve in Less Than an Hour, which I really recommend.

Write a program that outputs all possibilities to put + or - or nothing between the numbers 1, 2, ..., 9 (in this order) such that the result is always 100.

# Winning Criteria:

Shortest code in bytes wins.

• Are spaces allowed between the numbers, like 1 2 3+4+5 rather than 123+4+5? – Tim May 10 '15 at 16:27
• Pretty similar to codegolf.stackexchange.com/questions/3019/… (this link allows for * & / and does not specify the target 100) – Kyle Kanos May 10 '15 at 17:44

# Perl 5.10.0 - 5046 45B

say for grep{eval==100}glob join'{,+,-}',1..9


Usage: perl -M5.10.0 scratch.pl (I think I get the version for free)

• A very nice solution; both concise and fairly readable even without much knowledge of Perl. – JohnE May 10 '15 at 18:15

# CJam, 43 40 bytes

"-"SL]8m*{A,2>.+1\+s}%{~]:+100=},N*S/'+*


Here is how it works:

"-"SL]                     e# This is an array containing -, space and empty character
e# This represents -, + and joints correspondingly
8m*                  e# Get all cartesian products of these 3 strings of length 8
e# For instance, all cartesian products of length 2 are:
e# [["-" " "] ["-" "-"] ["-" ""] [" " " "] [" " "-"] [" " ""]
e# ["" " "] ["" "-"] ["" ""]]
{          }%     e# Map these cartesian products over this loop
A,2>             e# Get array [2, 3, 4, 5, 6, 7, 8, 9]
.+           e# In-order concatenation of the above two arrays
1\+s       e# Prepend 1 to the result from above and convert it to string
{        },                e# Now we have all possible strings of 1 to 9 with all possible
e# operators as well as joins (things like 567). Its time to
e# filter out what add up to 100
~                         e# Evaluate the string. This leaves some numbers on stack.
]:+                      e# Wrap those numbers in an array and take their sum
100=                  e# Check if the sum is equal to 100
N*              e# Join all filtered strings by newline
S/'+*         e# We used space instead of +. So now we replace space with +


Try it online here

My shot at this:

# Python 2, 93 bytes

def a(e,i):
if i<=9:[a(e+x+i,i+1)for x in("+","-","")]
elif eval(e)==100:print e
a("1",2)

• You can use single space indents and also str(i) -> i. Also, you might want to mention that this is Python 2 specific :) – Sp3000 May 10 '15 at 16:52
• @Sp3000 Thanks! I'm trying to improve my code golf skills and I'd never heard of the backtick thing. – James Williams May 10 '15 at 17:44
• One of the best code I have ever seen... – Arun Joshla Apr 1 '19 at 11:32

# Mathematica, 93 92 bytes

Print/@Select[""<>ToString/@Range@9~Riffle~#&/@{"+","-",""}~Tuples~{8},ToExpression@#==100&]


Fairly straightforward: generates all possible strings using Tuples and then filters them.

For reference, I get:

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


# Pyth, 26 bytes

fqvT100m+ssC,r1Td9^c3"+-"8


First, we generate all 8 element combinations of ["+", "-", ""]. c3"+-" chops "+-" into 3 rougly equal sied pieces, giving the appropriate list. ^c3"+-"8 gives the 8 element combinations.

Then, we greate the arithmetic sequences. C,r1Td zips together the operators with the numbers 1 through 8, which are then summed twice to give the string, and + ... 9 puts a 9 at the end.

Finally, fqvT100 filters for those expressions which evaluate to 100.

The update to the interpreter which allowed this code to work was this one, which was made about a day before this question was asked, so this code would have only worked on the command line compiler, not the online one, when the question was asked.

Demonstration.

# JavaScript, 86 92

Head to head with the best. That must be an anomalous problem.

Edit -6 thx @nderscore

for(i=6562;k=--i;eval(o)-100||alert(o))for(j=o=1;j++<9;k=k/3|0)o+=k%3?' +-'[k%3]+j:[j]


Test in any browser console, it pops up just 11 times.

Or this snippet:

function out(x) { OUT.innerHTML = OUT.innerHTML + '\n' + x; }

// 6562 is 3 powered to 8 +1
for(i=6562;k=--i;eval(o)-100||out(o))
for(j=o=1;j++<9;k=k/3|0)
o+=k%3?' +-'[k%3]+j:[j]

<pre id=OUT></pre>

• Here's -6: for(i=6562;k=--i;eval(o)^100||alert(o))for(j=o=1;9>j++;k=k/3|0)o+=k%3?' +-'[k%3]+j:[j] – nderscore May 11 '15 at 1:56
• @nderscore thanks, some obvious changes against my sloppy golfing, but the last trick with the ternary is very clever – edc65 May 11 '15 at 3:58

# Python (176)

39 minutes and counting (for all problems that is)

from itertools import*
for i in product(('+','-',''),repeat=8):
s=''.join(''.join(j)for j in chain(izip_longest(map(str,range(1,10)),i,fillvalue='')))
if eval(s)==100:print s


This could probably be shorter, but this was my actual solution for this problem, just golfed down a bit.

# K, 61 50 bytes

t@&100=.:'t:{:[x>9;,y;,/_f[x+1]'(y,"+";y,"-";y),\:$x]}[2;"1"]  Tested with Kona. The general approach is to recursively generate all the expressions and then filter out those which eval (.) to 100. This is reasonably competitive as written, but I think there's still room for improvement. ## edit: t@&100=.:'t:(,"1"){,/x,/:\:("+";"-";""),\:$y}/2+!8


Altering my approach a bit to be iterative rather than recursive, which removes the need for explicit base cases as well as saving me some brackets.