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By: John Wed, 01 Sep 2010 22:04:37 +0000 Nathan’s comment has a couple errors…

When you have:

Total Boards = (20^0 * 6^16) + (20^1 * 6^15) + … + (20^16 + 6^0)

It should be:

Total Boards = (20 choose 0)(20^0 * 6^16) + (20 choose 1)(20^1 * 6^15) + (20 choose 2)(20^2 * 6^14) … + (20 choose 20)(20^16 + 6^0)

since you need to CHOOSE which of the “n” spots you put your “n” consonants. The total is actually a lot higher.

Also, the eight-fold symmetry comes from reflection, since reflected boards have the same score.

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By: » Solving Boggle by Taking Option Three Wed, 05 Aug 2009 01:19:09 +0000 [...] board, we’d need to show that every other board scores fewer points. In a previous post, I estimated that there were 2^69 possible boggle boards. At 10,000 boards/second, this would take 1.9 billion [...]

By: Nathan Buch Sun, 08 Feb 2009 20:10:04 +0000 I had a couple thoughts about finding the upper limit of boggle boards. Please let me know if there’s something wrong with this line of thinking!

You could divide the alphabet into 20 consonants and 6 vowels, and for a 4×4 board the total number of possibilities would be the sum of every combination of consonants and vowels, namely:

Total Boards = (20^0 * 6^16) + (20^1 * 6^15) + … + (20^16 + 6^0)

Of course, in English every word must have at least one vowel, so the last term drops out since it represents the all-consonant boards. Therefore, the total is:

Total Boards = 2.8 x 10^20 ~ e^47

At 10,000 boards/second, that would still take almost 900 million years to evaluate. However, at this point one could employ a couple heuristics. For example, by examining the frequency distribution of letters in the English language, it is reasonable to assume that the letters J, X, Q, and Z will not be part of the best Boggle board. If those letters are excluded, then the number of consonants falls to 16, and the new number of boards decreases by a factor of 25, to 1.1 x 10^19 boards (35 million years).

From here, one could go a step further, and remove B, V, and K from the running, to decrease the total to 5.7 x 10^17 boards (1.8 million years). Or, one could examine the letter frequencies again and surmise that the best Boggle board will in all likelihood contain an E. If one E is guaranteed, the total boards changes to:

Total Boards = (20^0 * 6^15) + (20^1 * 6^14) + … + (20^14 + 6^1)

Or, 5.1 x 10^19 (162 million years)

With one E guaranteed and J, X, Q, Z removed, you would have 6.9 x 10^17 boards (2.2 million years), and with one E guaranteed and J, X, Q, Z, B, V and K removed, you would have 4.4 x 10^16 boards (139,000 years).

Of course, rotating a Boggle board 90 degrees will produce a different board with the same score, so by symmetry you could divide all of these totals by 4 (I saw in your posts that you found eight-fold symmetry – how is that?).

Removing or guaranteeing certain letters is pretty unscientific, but taking about a Z or putting in an E is also pretty reasonable. I guess it depends on how thorough one wants to be. Also, even with the most ‘flexibility’ employed (one E, no J, X, Q, Z, B, V or K, and four-fold symmetry), the solution would still take about 35,000 years to solve. But I imagine that’s a bit more palatable than 2 billion years.

– Nathan

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