Problem 68. Kaprekar Steps
Solution Stats
Problem Comments

16 Comments
The behavior at x=6174 is artifically set to 0, which taints the purerecursive solution, but hey it was cool problem!
The K constant is 495 for 3 digits.
So the test with 691 is wrong.
Why tests with only one digit ?
Did I miss something ?
I don't understand the Test Suite for x = 3 and x = 1 too. what should I do in this case?
For x=3, the steps are 30000003=2997, 99722799=7173, etc.
Problem description is confusing as there are different Kaprekar constants depending on the number of digits. [0 9 495 6174 for 1, 2,3 4 digits respectively.
getting this error:
Internal Server Error  Read
The server encountered an internal error or misconfiguration and was unable to complete your request.
Reference #3.c2c1ab8.1412090777.18623697
any ideas?
The problem should specify that any number with less than four digits should be filled up to four digits with leading zeros. (e.g. 3 > 0003)
Very nice and interesting problem!
I like the recursion aspect of this problem.
There is a small correction needed in the problem statement. Not all natural numbers, but 4 digit numbers can be reduced to Kaprekar number by the mentioned method. Similarly 3 digit numbers can be reduced to 495
https://en.wikipedia.org/wiki/D._R._Kaprekar
How it works x = 1？？？？？
For those confused with test cases 2,3 and 5, like myself before, do conversion to 4digit integer. Here is an example:
x = 1:
10000001 = 999
99900999 = 8991
99811899 = 8082
88200288 = 8532
85322358 = 6174
Therefore, y_correct = 5
love it!!!
Very nice. Took some few minutes to crack this.
What I did is to convert x into string and then use sort function.
nice problem
ATTENTION
This problem statement is poorly given and leaves out a crucial element the input number must always have 4 digits, so if 1 is the input, the first iteration should be 10000001. the second iteration should be 99900999. for the problem as stated, the test suite gives solutions that are incorrect.
Solution Comments

1 Comment
I am used to program with C, could someone have a vectorization solution for this problem in MATLAB way?
function y = KaprekarSteps(x)
y = 0;
while ~(x == 6174  x == 0)
x = step(x);
y = y + 1;
end
if x == 0
y = inf;
end
end
function x = step(x)
digits = getDigits(x);
gain = [1000 100 10 1]';
x_d = sort(digits,"descend")*gain;
x_a = sort(digits,"ascend")*gain;
x = x_d  x_a;
end
function digits = getDigits(x)
digits = [0 0 0 0];
for i = 1:4
digits(i) = fix(x/(10^(4i)));
x = mod(x,10^(4i));
end
end

1 Comment
The test suite doesn't match the problem description  how can the answer to test two be 5?

1 Comment
Cheated with 1...:
1000  0001 = 999.
999  999 = 0
y = inf;
No?

1 Comment
How it works with x = 3, x = 691, x = 1?

2 Comments
This works for all but test 3 where it gives, in my opinion, the correct answer 8.
But zero is not sorted in this solution. You should get same answer for input x = 691 and 6910. (And then no need for abs())
1) 96100169=9441
2) 94411449=7992
3) 99722799=7173
4) 77311377=6354
5) 65433456=3087
6) 87300378=8352
7) 85322358=6174

1 Comment
This has been the lamest test so far. You need to pad the numbers with zeros to build it up to be a 4 digit number. Not explained in the rules.

1 Comment
Not all test cases seem to be correct. For example x=3 would imply that the next value should be x=33=0, so y_correct=Inf and not 6

1 Comment
An efficient lookup table solution

1 Comment
Interesting  a recursive approach

1 Comment
Could you stop doing this kind of thing? I think it would be a lot more fun if we could see the _actual_ best solution..
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