# Following the official rules of Jenga, what is the highest possible level that can be placed?

I can't seem to find the answer to this question anywhere. Also, to preempt a couple possible answers:

I have read about the current record. However, I want to know if this is the highest level possible following the official rules. That is, the preceding level must be completed before beginning the next highest level.

I'm sure there is a mathematical method to solve this but my brain isn't working sufficiently right now to figure it out.

Also, I saw an answer posted somewhere of 54. Though it is possible to stack all 54 blocks atop one another individually, this would violate the rules of the game.

• Does this answer your question? What is the tallest Jenga tower ever created? – Joe W Oct 30 '20 at 1:04
• nb: this is not a duplicate. The suggested dup asks about the largest one created in the wild and not about the theoretical max (and none of the answers there address the theory). – L. Scott Johnson Oct 30 '20 at 11:33

51 levels. Achieving this requires 97 moves.

Let's start with a fresh tower, and let's try to minimize the number of blocks in the top levels. The rest of the tower will consist of one-block levels.

I shall denote the state by the number of blocks on the top level, followed by the number of blocks on the next level down, etc. So a fresh tower is 3,3,3,...

One can't remove a block from the uppermost complete level, so our attempt goes as follows:

1. 1,3,2,... ⇒ 6 in top 3.
2. 2,3,1,... ⇒ 6 in top 3.
3. 3,3,1,... ⇒ 7 in top 3.

And from there it loops.

Jenga has 54 blocks, so the tallest tower can have at most 51 level.

• 1,3,2,<48 levels with just 1> after 97 moves.
• 2,3,1,<48 levels with just 1> after 98 moves.

From the above, we can also determine that the tallest tower with a complete top level will have 50 levels. This takes 96 moves.

• Technically, you should also prove that achieving this limit is (in principle) possible without the tower ever falling over, i.e. that it's possible to do the moves in such an order that the center of mass of the N highest levels is always above the blocks on level N + 1 from the top. But I do think this is in fact possible, and it should even be possible to prove a stronger property, namely that, with reasonable play, the center of mass of the N highest levels (for all N ≥ 2) will always stay within half a block's width from the original central axis of the tower. – Ilmari Karonen Oct 30 '20 at 16:05
• @Ilmari Karonen, I did consider mentioning "physics aside", but it seemed too obvious. If you start considering physics, the question really becomes "What's the highest tower that's been achieved", but that wasn't what was asked. – ikegami Oct 30 '20 at 16:07

The key rule is that you can't remove a block from the uppermost complete level. We can prove (see below) that this means the top 3 levels will always have at least 6 blocks between them. Underneath that, the remaining blocks (up to 48 of them) could theoretically be just a single block per level, so this gives you an upper bound of 3 + 48 = 51 levels.

(I'm counting the uppermost row of blocks as a level even if it only ever contained one or two blocks; if you don't want to count the top level until it's complete then the answer is 50.)

Proofs:

First, we prove that the 2nd level is always a complete level of 3 blocks, by induction on the number of turns taken. At the beginning of the game (0 turns), it is true. Then after any turn on which it is true, there is one of two cases:

1. The top level is incomplete. In this case, the 2nd level can't be modified and the top level will remain the top, so it will still be true after one more turn.
2. The top level is complete. In this case, after one more turn that top level will become the (complete) 2nd level.

Now, we prove that the top 3 levels will always have at least 6 blocks. Assume towards contradiction that at some point there are fewer. Then since the 2nd level must have 3 by the lemma above, the 1st and 3rd level must each have only 1 block. But then that 3rd level with one block must, one turn earlier, have been an incomplete 2nd level, contradiction.

• Can you please explain why the top 3 levels will always have at least 6 blocks between them? – Cohensius Oct 30 '20 at 9:06
• "You can prove …". Your answer would be greatly improved if you proved it. – Ray Butterworth Oct 30 '20 at 13:51
• Ok, proof added. – Benjamin Cosman Oct 30 '20 at 15:32
• @Benjamin Cosman. You are mistaken. The top two layers are necessarily full before you can start a new one. so the only way to get fewer than a full layer is to remove from them. My proof shows the only way to do that, thus giving the lower bound on the number of blocks in the upper layers (6) and thus the upper bound on the number of levels in the tower (51). – ikegami Oct 30 '20 at 16:05
• @Cohensius when the top level has 2 blocks on it, the one below it is the uppermost complete and can't be touched, top has 2, second 3, third at least 1 = 6. When you complete the new top it has 3, 2nd has 3 and 3rd has at least 1 = 7. top two now have 6 between them and if you then start to remove from what used to be the uppermost complete level, those go to become the new top level, making those three levels always 6 or more. – Andrew Nov 1 '20 at 19:34