Revisting the good 'ol days- hand collating a set......

[bigdcardinals7]

I may have that card I'll look thru some of the boxes I have if I find it i'll send it your way pwe

Sent from my SPH-L710 using Freedom Card Board mobile app

 

[rsmath]

I'm currently hand collating 30+ 3200 count boxes into sets and dupes...it's brutal

the magical sorting system should help if they are all the same card set (or if you have separated out the boxes by card set).

Through the magical sort system, all the card number 1's will be together, 2's, 300's, etc. so at the very end you just lay them out into stacks to make your h/c sets with the stragglers as dups.

 

[fordman]

I'm trying to get a grasp on how this sorting method works. I'm not fully understanding the explanation so if i may, try to use examples:

1) Sort the cards by the ones place into the 10 stacks.

So you sort by cards numbers right to left? (Random card numbers I pulled out of the air by the way!)
stack #1 :80, 20, 200, 350, 430, 100, 550, 730.....?
stack #2 :81, 21, 201, 351, 431, 101, 551, 731.....?
stack #3 :82, 22, 202, 352, 432, 102, 552, 732.....?
stack(s) 4-10 the same way?

2) You will have to take the 10 stacks and put them back into one big stack again for the next sort. How you reassemble the 10 stacks back into one main stack matters or the sort will be messed up and you have to start over. Pick up the 9 "ones" stack and place it in a cardboard box card-number side up. Pick up the 8 "ones" stack and place it in the box card-number side up, then the "7" ones, and so on. When done, the "0" ones stack should be card-number side up at the top of the master stack.

Then take a stack, sort left to right again, but using the second digit from the right?
(using illustration from above)
Stack #1 : 100, 101, 102, 200, 201, 202
Stack #2 : (any card with the number 1 in the tens spot from right to left)
Stack #3 : 20, 21, 22
Stack #4 : 430, 431, 432, 730, 731, 732
Stack #5 : (any card with the number 4 in the tens spot from right to left)
Stack #6 : 350, 351, 351, 550, 551, 552
Stack #7 : (any card with the number 6 in the tens spot from right to left)
Stack #8 : (any card with the number 7 in the tens spot from right to left)
Stack #9 : 80, 81, 82
stack #1 0 : (any card with the number 9 in the tens spot from right to left)

3) Do your next sort of the master stack by the "tens" column starting from the very top of the master stack. Single digits like "1" are "zero" in the tens column so they go in the zero "tens" column stack.

Then take stack #1 from above to create 10 more stacks?
Stack 1a : 100, 101, 102
Stack 1b : 200, 201, 202
Stack 1c : (any card with a three in the hundreds spot from right to left)
Stack 1d-1j : (any card with a four - nine in the hundreds spot from right to left)

than take stack #2 from above to and pile onto stack 1a-1j

Im assuming eventually, the numbers will fall into place?

(my brain hurts now!)

I dont have any cards in from of me right now to try this but if Im wrong, please correct me!

Fordman

 

[rsmath]

I think the first two look more or less correct (but don't forget to add a leading "zero" when doing the sort by tens or hundreds column since people always write "1" instead of "001" with the leading zeros). Also I'm not sure if you got the point that after the ones column sort, all the cards go back together into one big stack before doing the tens column sort (and the 10 piles have to be reassembled back into that one big stack in a certain order in order not to mess up the sorting method). I know when I first learned the sorting method, that was my biggest flub in figuring out the method as I was taking my 0 in the ones column stack and doing the tens column sort in another area of the table without recombining the 10 stacks back into one and my sort didn't work right until I figured out the recombining step.

The final step seems a little wrong because you are ignoring the card numbers 1-99 in the first stack (or "001-099" if you put back in the assumed zeros).

I would say take a stack of 50 cards and sort them to test the method. 50 cards shouldn't take too long to sort which is why I recommend that size for learning the method rather than going full-blown on a 330 card set. With a sorting size of 50 cards, they'll be in number order if the sort is done right, even though the number order may look like 1, 8, 12, 17, 21, 32 since you are sorting a small sample instead of a complete or nearly completed set where you will see 1, 2, 3, 4, 5, 6, 7, etc. without any or much gaps.