Electrical outlet
Retina, 3 bytes
1
The trailing linefeed is significant.
Input is a space-separated list of unary numbers.
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Explanation
The code simply removes all spaces as well as the 1
after them from the string. Here is why that works:
Addition in unary is simple: just concatenate the numbers which is the same as removing the delimiters. Decrementing by 1 is also simple: just remove a 1
from each number. We want 1 more than the sum of the decremented inputs though, so we simply only remove the 1
s we find after spaces, thereby decrementing all but the first input.
Jelly, 3 bytes
’S‘
Decrement (all), sum, increment. Try it here.
Hexagony, 18 14 bytes
.?<_(@'")>{+.!
Unfolded:
. ? <
_ ( @ '
" ) > { +
. ! . .
. . .
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I don't think side-length 2 is possible, but there must might be a more efficient side-length 3 solution that this.
This is the usual "decrement all, sum, increment" approach, but I'll have to add diagrams later to show how exactly it works in Hexagony.