COMPUTER FUNDAMENTALS TRAINING - School of …
Notes about this manual: It was my intent to make this useful and easy to use by everyone—yes, even from those who have used a computer to those who have never touched one.
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Class Overview and General Introduction to Machine Learning
www.cs.utah.eduUnderstand how various machine learning algorithms work Implement them (and, hopefully, their variants/improvements) on your own Look at a real-world problem and identify if …
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DeepLog: Anomaly Detection and Diagnosis from System …
www.cs.utah.eduand anomaly detection. Existing approaches that leverage system log data for anomaly detection can be broadly classi•ed into three groups: PCA based approaches over log message counters [39], invariant mining based methods to capture co-occurrence pa−erns between di‡erent log keys [21], and work…ow based methods to identify execution anom-
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www.cs.utah.edufor integer overflows, and used it to conduct the first detailed ... and there are code idioms that deliberately use it. On the other hand, C and C++ have ... Although it is commonly known that C and C++ programs contain numerical errors and also benign, deliberate use
ISAAC: A Convolutional Neural Network Accelerator with In ...
www.cs.utah.eduISAAC: A Convolutional Neural Network Accelerator with In-Situ Analog Arithmetic in Crossbars Ali Shafiee ∗, Anirban Nag , Naveen Muralimanohar†, Rajeev Balasubramonian∗, John Paul Strachan †, Miao Hu , R. Stanley Williams†, Vivek Srikumar∗ ∗School of Computing, University of Utah, Salt Lake City, Utah, USA Email: {shafiee, anirban, rajeev, svivek}@cs.utah.edu
Highlights of the High- Bandwidth Memory (HBM) Standard
www.cs.utah.eduJun 14, 2014 · The Memory Forum – June 14, 2014 HBM RAS Challenges Stacked Memory has some challenges with respect to RAS requirements Traditional DRAM DIMMs get only a subset of bits (e.g. 4) from each burst from a single DRAM device HBM gives you all the bits of a burst from a single row of a single bank of a single DRAM device
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www.cs.utah.eduThey key to this theorem is again the Chernoff-Hoeffding bound. Fix some q2R, and for each point s i in S, let X ibe a random event describing the effect on q(S) of s i. That is X i= 1 if s i2R qand X i= 0 if s i2=R q, so i= 1 for all i2[k]. Let M= P i X i= q(S), and note that E[M] = jSjq(P)=jPj. Multiplying Mby k= jSjwe can now apply Theorem 2 ...
Lecture 3: MIPS Instruction Set - University of Utah
www.cs.utah.eduit seems to disappear every time we switch procedures – a procedure’s values are therefore backed up in memory on a stack Proc A’s values Proc B’s values Proc C’s values … High address Low address Stack grows this way Proc A call Proc B … call Proc C … return return return
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