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Just want to check when you parallize Strassen's algorithm you get a time complexity of $O(\log n )$ because you divide and conquer $\log n$ times in parallel and a space complexity of $ O( n^{\log_2 7} ) $ because the space complexity in the parallel setting satisfies the same recursive identity as time complexity in the sequential setting.

Should be obvious but I couldn't find this by googling.

Also looking for a reference on parallel matrix algorithms.

asked Aug 11 at 22:53
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Well, it calculates 7 products of matrices, so you can just hand each product to its own thread.

Or if you had eight cores, you could split a 8n x 8n product into 343 = 8 x 43 - 1 nxn products.

answered Aug 12 at 16:26
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