Dear TomGrundy,
You can sort any numbered items by sorting by each digit separately as
long as you start with the least significant digit, i.e. the
right-hand, or units digit.
Let's imagine that your items have three-digit numbers. First sort
them by the last digit into ten piles and reassemble the ten piles
into one, in order. You may not be very impressed by what you have
achieved so far, as very different numbers are mixed together, with
numbers as different as 0 and 990 being mixed in the first part of the
pile. But persevere. Now sort the pile that you have created by the
penultimate, or tens digit. It is important at this stage that you do
not split or shuffle or otherwise mix the pile that you have created
in the first step. Once again you wil create ten piles, which you
again reassemble into a single pile. Now you have all the numbers
with zero in the tens digit first, and then the ones, and so on. But
the important thing is that within those sections, the last digits
will still be in order. In other words, within the twos, for example,
numbers ending 20, 21, 22, and so on will come in order (even though
20, 120, 320, and so on are mixed randomly).
Now all will be coming clear: all you have left is to sort and
reassemble the whole pile again, this time by the first digit. Not
only will the pile start with the numbers below 100, then the 100s,
then the 200s, and so on, but the numbers within those groups will
have remained sorted from the earlier passes. You've done it! You
can imagine that this will work with any-length numbers, simply
needing an appropriate number of passes.
You may find your first attempt at this technique reverses the sort at
each stage - so that, unhelpfully, 9 comes first, then 8, and so on.
To prvent this, you need to turn the whole pile over at each stage
before sorting or else turn each item over into the piles as you sort.
I trust this helps.
Carnegie |
