Hunting high and low

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Here is a nice little challenge, Benjamin came up with some months ago: “From a list of numbers find the smallest and the largest one – but do not use more than 1,5 comparisons per number.”

Dear reader: Try it!

The solution is simple, straight forward and very nice. And I didn’t find it.

I found something else. My approach was to look at the largest-number-so-far (max) and the smallest-number-so-far (min) as the boundaries of a region. A number outside this region must be larger than max or smaller than min and causes that boundary to change.

To check if a number (a) was outside that region a little computation came to my mind:

check = (max - a)*(min - a);

You would expect n to be smaller than max, so (max-n) should be positive.

You would expect n to be larger than min, so (n-min) should be positive, too.

Two positive numbers multiplied result in a positive number again.

So, if n was outside the boundaries, check would become negative. I only had to check which boundary was exceeded and that’s it:

if (check < 0)
    if (a > max)
        max = a;
        min = a;

Using this approach the number of comparisons  converges to 1 when the length of the list grows. So I found a solution that is way better than 1,5 comparisons per number, didn’t I?


No, I didn’t. I cheated. In fact  (max-a) and (min-a) are comparisons, they just don’t use > or <.

(Actually it’s the other way around: To compute < or > most processors do a subtraction and compare the result to 0.)

So – if you count the subtractions as well you get 3+ comparisons per number…

The intended solution to the challenge (and the code you should provide in your exam) is:

a = list[count++];
b = list[count++];

if (a<b)
    if (a<min) min = a;
    if (b>max) max = b;
    if (b < min) min = b;
    if (a > max) max = a;                    


So – what’s the point of this post?

The point is: My silly approach can be faster than the standard solution. Depending on the type and range of the numbers in the list calculating and probing check needs less time than the comparisons in the standard solution.

It may not be much, but sometimes small advantages matter.  For example: For an integer-list of length 10^8 with values from -10.000 to 10.000 my approach is 0.02 seconds faster (on my current laptop). And has less code.

So if you feel like using it – I won’t charge you.

Permutations and the number 9 (proof)

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Ten days ago I promised a proof why the differences of any permutation of the same digits is always a multiple of 9. Here it is

Let’s say x and y are digits and ‘xy’ is not ‘x*y’, but the number that consist of the digits x and y.

xy – yx = multiple of 9.

Is it always true? Yes, it is:

(10x + 1y) – (10y + 1x) = multiple of 9?
9x – 9y = multiple of 9?
9(x-y) = multiple of 9? Yes, for sure.

I calculated the numeric values of xy and yx by "10 times the higher position + 1 time the lower position", just like we all do every day in our beloved decimal-system. More formally we describe the factor for the position (1, 10, 100…) by

10^position, (position is zero-based, counting from right to left, of course)

Now, when we change the position of a single digit within a number we change its numeric value from

x*10^(old position) to x*10^(new position).

We can neglect x here, because it’s a factor that occurs in both values, so we can get rid of it by division.
The remaining part for building the difference is

10^p1 – 10^p2, a formula that always produces multiples of 9: 10-1, 1000-100, 10-1000000.


Since the difference of a number and one of it’s permutations is a sum of (10^p1-10^p2)-parts, which are all multiples of 9, the total also can be divided by 9.

Permutations and the number 9

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Lately I stumbled over permutations and noticed a funny fact: When you take any integer, produce a permutation and subtract one form the other, you always get a multiple of 9.

It works with a any length. Some examples:

Length 2
23 – 32 = -9 
84 – 48 = 36
60 – 06 = 54

Length 3
123 – 132 = -9
and so on…

It still works for changing the front position:
123 – 213 = -90

In fact it works for changing any position:
123 – 321 = -198 (-22*9)

This leads to the conclusion I stated in the beginning. The difference of any permutation of the same number is a multiple of 9. Feel free to check it with you favorite numbers.

It’s not a great mathematical invention (if it is, please let me know), but I didn’t know it, didn’t expect it und find it kind of fascinating. I asked a couple of friends and no one was aware of this.

I still try to find a way to use these fact, maybe for spell-check-like for numbers or a fast calculation of permutations, but so far it seems just useless 🙂

Instead I can provide a proof for numbers of any length and I’ll post it here in ten days from today. Meanwhile I want you to try it yourself. The proof is really, really simple, almost trivial and it won’t cost you more than five minutes to understand it all.

Please email or comment your solution!

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