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a PRAM is not MIMD

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QUOTING FROM: Algorithms and Theory of Computation Handbook, CRC Press LLC, 1999, "parallel random-access machine", in Dictionary of Algorithms and Data Structures [online], Paul E. Black, ed., U.S. National Institute of Standards and Technology. 27 February 2004. (accessed 6/4/06) Available from: http://www.nist.gov/dads/HTML/parallelRandomAccessMachine.html:

parallel random-access machine (definition)

Definition: A shared memory model of computation, where typically the processors all execute the same instruction synchronously, and access to any memory location occurs in unit time.

Also known as PRAM.

See also work-depth model, multiprocessor model.

PLEASE OBSERVE: that the definition clearly indicates use of the SIMD machine and ***not*** an MIMD machine as this Wiki entry states. Vonkje 16:10, 3 June 2006 (UTC)[reply]

The definition you prodivded is wrong. If that definition was true, PRAMs would be virtualy useless for examination of any serious (i.e. the one which has conditional statement in it) paralel algorithm. And PRAMs are used for examination of those. --Dijxtra 17:59, 27 June 2006 (UTC)[reply]
In SIMD, conditional statements are executed using predication. In other words, the fact that an algorithm has a conditional statement does not mean that it cannot be executed on a SIMD machine. -- 69.11.220.126 (talk) 21:52, 7 November 2013 (UTC)[reply]

That is a definition of SIMD, indeed. Nevertheless, a wrong definition does not prove the otherwise, namely that PRAM is MIMD. The "Advanced Computer Architecture and Parallel Processing, 2005" also states that "Active processors must execute the same instruction, generally, on different data. Hence, this model is sometimes called the shared memory, single instruction, multiple data (SMSIMD) machine." The Google is full of presentations stating "the abstract PRAM model can be SIMD or MIMD". There is no ground for the MIMD-thesis, therefore. --Javalenok (talk) 17:52, 20 March 2009 (UTC)[reply]

There is a commercial implementation of PRAM!

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NVIDIA's CUDA programming model provides a parallel data cache within a single "cooperative thread array" (CTA, aka thread block). All threads within the CTA have CRCW access to the parallel data cache, so one CTA is effectively a PRAM machine (with an installed base of millions!). However, CTAs are currently limited to 512 threads. Skadron 17:11, 15 November 2007 (UTC)[reply]

I don't agree that CUDA is a PRAM implementation. This is because: 1. The PRAM model is a generalization of the Random Access Machine (RAM) model, the basic sequential computing model exposed to programmers in traditional programming languages. 2. The NVIDIA CUDA model cannot support the PRAM since it does not support the RAM.

Move?

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The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.


Parallel Random Access MachineParallel random access machine

The hyphenated redirect that you linked is so that the random access article can be easily linked from contexts in which the adjective form, as in random-access machine, is used. Dicklyon (talk) 04:43, 14 September 2012 (UTC)[reply]
  • Support with hyphen. With the hyphen I could understand much more easily; WP is for non-experts where possible and practical. And as a card-carrying non-expert in this field, I don't care that only a minority of sources are kind to us. Tony (talk) 01:28, 14 September 2012 (UTC)[reply]
  • Comment – I just noticed that my comments on the technical move request got moved to start this RM, and that I moved the article already, as part of a batch of related fixes. Sorry if anyone thinks this was potentially controversial. It shouldn't be. Dicklyon (talk) 02:10, 14 September 2012 (UTC)[reply]
I just moved the page back since it makes this RM discussion confusing. You know I did not support your move so I am surprised you did it. It definitely makes this RM confusing. Jojalozzo 02:17, 14 September 2012 (UTC)[reply]
Why at that article? It's not relevant except where the compound is used as an adjective. Dicklyon (talk) 03:09, 14 September 2012 (UTC)[reply]
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.