Independence Day weekend puzzler

Inspired by an assignment in my son's Java programming class:

Write a one-liner that takes as input an array of numbers (e.g. x = [1 2 3]) and which outputs an array of integers that is "incremented" properly, (in this case, y = [1 2 4]).

Examples of proper input/output:

x = [1 9 1 9] ----> y = [1 9 2 0]

and

x = [9 9 9] ----> y = [1 0 0 0]

No semicolons allowed in your one line!

[Fangjun Jiang]

I need a little clarification. As if the output is 124=123+1, 1920=1919+1 and 1000=999+1? What if your input is [1 2 34], should the output be [1 2 35] or [1 2 3 5]?

[the cyclist]

You can assume the elements in the input array are single digits, 0-9.

[David Young]

How big can the array be? For example, should the answer work correctly for x=repmat(9, 1, 100)?

[David Young]

Sorry, the example isn't stringent enough: I mean x=repmat(9,1,1000).

[the cyclist]

I only intended it to be for "smallish" vectors of numbers, but feel free to create whatever conditions you want, for maximum enjoyment.

[Paulo Silva]

+1 vote for the interesting puzzler, it's the first vote!

Please vote on it if you found it interesting and you want more of them.

[Fangjun Jiang]

Sure. +1

Answer 1

[Fangjun Jiang]

I like this Golf challenge. Inspired by Paulo's entry.

num2str(str2num(sprintf('%d',x))+1)-'0'

is shorter.

Answer 2

[Jan]

My submission is neither a one-liner nor free of semicolons. But the total number of lines and semicolons is less than in STR2NUM and NUM2STR, which have 86 and 217 lines and call INT2STR in addtion.

n = length(x);
q = find(x ~= 9, 1, 'last');
if isempty(q)    % [9, 9, 9, ...]
  x = 1;
  x(n + 1) = 0;  % or x = [1, zeros(1, n)]
else             % Any non-9 is found
  x = [x(1:q - 1), x(q) + 1, zeros(1, n - q)];
end

[David Young]

This works for long vectors (thousands of elements).

Answer 3

[bym]

kind of a hybrid:

sprintf('%d',sum(x.*10.^(numel(x)-1:-1:0))+1)-'0'

[Jan]

+1: This does not call other M-functions, therefore it is the one-linest one-liner yet.

[Andrei Bobrov]

num2str(10.^(numel(x)-1:-1:0)*x'+1)-'0'

[David Young]

Jan: my earlier answer uses only built-in functions.

[Jan]

@David: Which one is your earlier answer?

[David Young]

@Jan: The one that starts with a call to diff

Answer 4

[Andrei Bobrov]

str2num(num2str(10.^(length(x)-1:-1:0)*x'+1)')'

ADD

z = 10.^(numel(x)-1:-1:0)*x'+1
y = round(rem(fix(z.*10.^-(fix(log10(z)):-1:0))*.1,1)*10)

[bym]

+1 vote for the 'z' solution; compact & elegant

[Andrei Bobrov]

@proecsm, thanks!

Answer 5

[David Young]

One-liner, avoiding string operations:

diff([0 (floor((sum(x.*10.^(length(x)-1:-1:0))+1) ./ 10.^(floor(log10(sum(x.*10.^(length(x)-1:-1:0))+1)):-1:0))) .* 10.^(floor(log10(sum(x.*10.^(length(x)-1:-1:0))+1)):-1:0)]) ./ 10.^(floor(log10(sum(x.*10.^(length(x)-1:-1:0))+1)):-1:0)

EDIT: This is only one line of code, even though formatting on the Answers web page makes it look like 4 lines.

[David Young]

This fails if there are more than about 16 elements in the vector, because you only get about 16 significant figures in a double.

[Jan]

+1: A one-liner without calls to other M-files and therefore even no hidden semicolons.

[David Young]

Thanks Jan!

Answer 6

[Paulo Silva]

x=[1 2 3]; %example input
xr=num2str(str2num(strrep(num2str(x),' ',''))+1)-'0'
%xr =[1 2 3 4]

Answer 7

[the cyclist]

Here's one just a bit shorter than andrei bobrov's [using numel() to trim one character from length()]:

str2num(num2str(10.^(numel(x)-1:-1:0)*x'+1)')'

Anything shorter?!

Answer 8

[David Young]

Also one line of code (formatting for the web page will display it over more than one line of text):

double(regexprep(char(x), {['([' char(0:8) ']?)(' char(9) '*)$'] ['^(' char(0) '+)$']}, {'${char($1+1)}${regexprep($2,char(9),char(0))}' [char(1) '$1']}))

[David Young]

This one works for long vectors with thousdands of elements (my arithmetic-based solution doesn't).

Answer 9

[David Young]

I'm sorry about this one - it's somewhat over the top, and I promise I won't do any more. However, since I think it's a different approach to the others (arithmetic operations but no powers of 10!), here it is:

[ones(1, sum(x)==9*length(x)) x(1:length(x)-sum(cumsum(fliplr(x)) == 9*(1:length(x)))-1) repmat(x(max(1, length(x)-sum(cumsum(fliplr(x)) == 9*(1:length(x)))))+1, 1, sum(x)~=9*length(x)) zeros(1, sum(cumsum(fliplr(x)) == 9*(1:length(x))))]