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Which has more entries, BDB or Strong's Hebrew?

Is there any known mapping between the two?

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I found a mapping here, e.g. concordances.org/hebrew/543.htm – Walrus the Cat May 11 '12 at 19:03
Also, the first sentence here seems to indicate that the BDB's entries are at least an improper subset of Strong's: bible-discovery.com/dictionary-license-bdb.php – Walrus the Cat May 11 '12 at 19:08
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Walrus the Cat, welcome to Mi Yodeya. I hope you stick around and enjoy it. Your question would be more valuable if it would include some (brief) background info on what BDB and Strong's are, perhaps linking to Wikipedia or somewhere, that explains why you thought of the possible existence of a mapping. Also, if you discovered an answer to your question within ten minutes of asking it by Googling (as it seems), then the question is rather weak; but the answer might be good: why not write it up as an answer, below? – msh210♦ May 11 '12 at 19:25
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@WalrustheCat Welcome to the site! While your point is valid, in this community we often encourage people to dejargonify so that even if someone doesn't know the answer and isn't familiar with the territory, they may find the Q and A interesting. There's a lot of fancy lingo that gets thrown around here, believe me, but we try to make it so that everyone can understand. :D – HodofHod May 13 '12 at 1:29
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@yoel, are there better lexicons out there that I don't know about? – Walrus the Cat May 22 '12 at 18:49
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1 Answer

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OK, I'm going to answer it myself, based on the comments above.

  1. Mapping

1 Strongs entry -> possibly multiple BDB entries

Apparently, because Hebrew words can be browsed by Strong's entry at concordances.org/hebrew, and each entry is linked to a BDB entry, then there is at most 1 BDB entry per Strong's entry, and the two are mapped (from Strongs to BDB) at that URL.

  1. Number of Entries

Strongs > BDB

Shemmy's reasoning seems good. Strong's is exhaustive. I believe BDB is also exhaustive, but it is organized by one entry per tri-radical root. Being that Strong's contains multiple entries per root, and BDB contains only one entry (containing multiple instances of that root in the text), Strong's is bound to have more entries.

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Just reading this, it doesn't make any sense. How can there be more Strongs entries, yet each maps to possibly multiple BDB entries? Does redundancy account for this apparent discrepancy? – Walrus the Cat May 23 '12 at 6:46
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@Warlus the Cat, if you think this doesn't answer the question, you might want to remove "accepted" mark from this answer, in order to avoid confusion of occasional googlers – jutky May 23 '12 at 11:46
Thanks, but I do think it answers the question. Redundancy can, and probably does account for the oddity. – Walrus the Cat May 24 '12 at 21:04
Of course, I'm open opinions on that. – Walrus the Cat Jun 6 '12 at 11:20

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