Lobotomy642
EuropeMember Posts: **1**

in Collecting

...or more probably 6, unless you go for 4, but you'll have to go for the TL;DR to know why.

But in any case, you certainly won't need 2000 copies, you scumbag!

Hello dear Brickset fellows,

I was wondering how many Christmas build-up 2017 sets would be needed if we wanted to build all the models without disassembling the previous ones, so I made a little study which I share with the curious amongst you.

*Disclaimer:* building all the models will deprive you from the irreplaceable pleasure of the LEGO rebuilding concept (and putting to good use the offered brick separator). Please reserve this kind of practice to the LEGO display fetishist like me. You've been warned.

*Disclaimer:* Despite being computer aided, this study contains traces of human work, so it is subject to the unavoidable Errare Humanum Est concept. Use at your own risk.

*Disclaimer:* Lending yourself to this exercise will leave you with a huge pile of unused bricks. But can a serious LEGO builder really complain about "having too many bricks" ?

So, why would you need 7 times the set? The cause is part 6029156, Flat tile 1x1 round with opened eye pattern, provided in two specimen but used in 7 different model.

This one is actually the easiest limit to work around, because the set also provides two 6097402, Flat tile 1x1 round with closed eye pattern, which is used in only one model, so it is very easy to swap one for the other.

The next limit is harder to work around: you will get four 6175968 (Brick 1x1 with 2 knobs corner), but you will need 22 of them (4 on dec 6th, 2 on dec 9th, 4 on dec 13th, 4 on dec 18th, 4 on dec 20th and 4 on dec 23rd). The only trade off at your disposal are to find 2 spare ones in your tremendous collection, or to sacrifice one of the models. Your decision.

This case solved, the next limits are easy to work around: 614101 and 614121, 1x1 round plates in red and white, which you'll need 5 sets for. I have assumed that considering LEGO's politics on spare parts it is very likely that you will have extra of them already in your collection, if not even already in these sets (Bricklink tells me you will probably have them).

Then you'll reach the last limit I consider, the 4 instances of the set: this time it concern 10 parts.

I believe the blocking ones will be 4243812 (4x4 plate in white needed on dec 11, dec 15, dec 23 and dec 24) which never comes as a spare; then 4121972 (Roof tile 1x2/45 in brick yellow); 6004990 (Angular Place 1.5 bot. 1x2 1/2 in medium stone grey); 4529239 (Nose cone small 1x1 in Dark green); 6092587 (The famous Jumper, Plate 1x2 with 1 knob in Brick yellow) and 625401 (Ice Cream in White).

There is also the case of 6015344 (Brick 1x2 with 2 knobs in medium stone grey) although you may use the extra 4558955 (Brick 1x1 with 1 knob) that you may end up with (2 instances used in only 1 model).

The other parts are not worth spending time as you'll likely have spares on them in the set (3005744, 6168647, 4523159).

As a final note, I will add that if you really go for 4 sets, you will be able to build 1 more instance of some of the subsets, I'm thinking about dec 1st (the boat) and dec 17th (the wreath) and dec 14th (the lighthouse), but I do not have the statistics for these so it is pure conjecture from me.

Hope you liked this information! Feel free to share your feelings about all this...

*Behind the scene:*

For the very curious amongst you, the "computer aided" part is because amongst my fetishes I like to do a little bit of programming, so it is possible that all this was just an excuse to write a little program to help in the study.

For those who are in the field: I have been using Python+Tkinter to make a gui, the algorithm part was to write a function to isolate the part from the part-list image taken from the PDF instruction (I did not trust brickset part list at that time because sometimes it is not complete); then I manually enter the part count (I tried to use Tesseract OCR but it did not work in this case and I did not investigate further); then I follow the building instruction and pick the bricks per model (the error prone part); Finally I click on the (basic) statistics buttons which list me how many time each part is needed (All that code just for that ???).

But in any case, you certainly won't need 2000 copies, you scumbag!

Hello dear Brickset fellows,

I was wondering how many Christmas build-up 2017 sets would be needed if we wanted to build all the models without disassembling the previous ones, so I made a little study which I share with the curious amongst you.

So, why would you need 7 times the set? The cause is part 6029156, Flat tile 1x1 round with opened eye pattern, provided in two specimen but used in 7 different model.

This one is actually the easiest limit to work around, because the set also provides two 6097402, Flat tile 1x1 round with closed eye pattern, which is used in only one model, so it is very easy to swap one for the other.

The next limit is harder to work around: you will get four 6175968 (Brick 1x1 with 2 knobs corner), but you will need 22 of them (4 on dec 6th, 2 on dec 9th, 4 on dec 13th, 4 on dec 18th, 4 on dec 20th and 4 on dec 23rd). The only trade off at your disposal are to find 2 spare ones in your tremendous collection, or to sacrifice one of the models. Your decision.

This case solved, the next limits are easy to work around: 614101 and 614121, 1x1 round plates in red and white, which you'll need 5 sets for. I have assumed that considering LEGO's politics on spare parts it is very likely that you will have extra of them already in your collection, if not even already in these sets (Bricklink tells me you will probably have them).

Then you'll reach the last limit I consider, the 4 instances of the set: this time it concern 10 parts.

I believe the blocking ones will be 4243812 (4x4 plate in white needed on dec 11, dec 15, dec 23 and dec 24) which never comes as a spare; then 4121972 (Roof tile 1x2/45 in brick yellow); 6004990 (Angular Place 1.5 bot. 1x2 1/2 in medium stone grey); 4529239 (Nose cone small 1x1 in Dark green); 6092587 (The famous Jumper, Plate 1x2 with 1 knob in Brick yellow) and 625401 (Ice Cream in White).

There is also the case of 6015344 (Brick 1x2 with 2 knobs in medium stone grey) although you may use the extra 4558955 (Brick 1x1 with 1 knob) that you may end up with (2 instances used in only 1 model).

The other parts are not worth spending time as you'll likely have spares on them in the set (3005744, 6168647, 4523159).

As a final note, I will add that if you really go for 4 sets, you will be able to build 1 more instance of some of the subsets, I'm thinking about dec 1st (the boat) and dec 17th (the wreath) and dec 14th (the lighthouse), but I do not have the statistics for these so it is pure conjecture from me.

Hope you liked this information! Feel free to share your feelings about all this...

For the very curious amongst you, the "computer aided" part is because amongst my fetishes I like to do a little bit of programming, so it is possible that all this was just an excuse to write a little program to help in the study.

For those who are in the field: I have been using Python+Tkinter to make a gui, the algorithm part was to write a function to isolate the part from the part-list image taken from the PDF instruction (I did not trust brickset part list at that time because sometimes it is not complete); then I manually enter the part count (I tried to use Tesseract OCR but it did not work in this case and I did not investigate further); then I follow the building instruction and pick the bricks per model (the error prone part); Finally I click on the (basic) statistics buttons which list me how many time each part is needed (All that code just for that ???).

11

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## Comments

6,44017,788778I also feel like the computer part of what you talked about could be the start of an amazing tool for multi-build set analysis into all kinds of things. Definitely a great post!