Independence Day weekend puzzler

2011 7 3
9 个回答
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5 个投票

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!

7 个评论
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Fangjun Jiang
Fangjun Jiang 2011-7-3
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
the cyclist 2011-7-3
You can assume the elements in the input array are single digits, 0-9.
David Young
David Young 2011-7-4
How big can the array be? For example, should the answer work correctly for x=repmat(9, 1, 100)?
David Young
David Young 2011-7-4
Sorry, the example isn't stringent enough: I mean x=repmat(9,1,1000).
the cyclist
the cyclist 2011-7-4
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
Paulo Silva 2011-7-4
+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.

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 采纳的回答

Fangjun Jiang
Fangjun Jiang 2011-7-3

3 个投票

I like this Golf challenge. Inspired by Paulo's entry.
num2str(str2num(sprintf('%d',x))+1)-'0'
is shorter.

更多回答（8 个）

Jan
Jan 2011-7-3

2 个投票

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
bym
bym 2011-7-3

2 个投票

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

5 个评论
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Jan
Jan 2011-7-3
+1: This does not call other M-functions, therefore it is the one-linest one-liner yet.
Andrei Bobrov
Andrei Bobrov 2011-7-4
num2str(10.^(numel(x)-1:-1:0)*x'+1)-'0'
David Young
David Young 2011-7-4
Jan: my earlier answer uses only built-in functions.
Jan
Jan 2011-7-4
@David: Which one is your earlier answer?
David Young
David Young 2011-7-4
@Jan: The one that starts with a call to diff

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Andrei Bobrov
Andrei Bobrov 2011-7-3

1 个投票

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)

2 个评论
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bym
bym 2011-7-4
+1 vote for the 'z' solution; compact & elegant
Andrei Bobrov
Andrei Bobrov 2011-7-4
@proecsm, thanks!

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David Young
David Young 2011-7-3

1 个投票

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.

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David Young
David Young 2011-7-4
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
Jan 2011-7-4
+1: A one-liner without calls to other M-files and therefore even no hidden semicolons.

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Paulo Silva
Paulo Silva 2011-7-3

0 个投票

x=[1 2 3]; %example input
xr=num2str(str2num(strrep(num2str(x),' ',''))+1)-'0'
%xr =[1 2 3 4]
the cyclist
the cyclist 2011-7-3

0 个投票

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?!
David Young
David Young 2011-7-4

0 个投票

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']}))

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David Young
David Young 2011-7-4
This one works for long vectors with thousdands of elements (my arithmetic-based solution doesn't).

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David Young
David Young 2011-7-4

0 个投票

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))))]

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