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diff --git a/candle-examples/examples/phi/README.md b/candle-examples/examples/phi/README.md index 566411d1..70af6650 100644 --- a/candle-examples/examples/phi/README.md +++ b/candle-examples/examples/phi/README.md @@ -1,14 +1,33 @@ -# candle-phi: 1.3b LLM with state of the art performance for <10b models. +# candle-phi: 1.3b and 2.7b LLM with state of the art performance for <10b models. -[Phi-1.5](https://huggingface.co/microsoft/phi-1_5) is a language model using -only 1.3 billion parameters but with state of the art performance compared to +[Phi-1.5](https://huggingface.co/microsoft/phi-1_5) and +[Phi-2](https://huggingface.co/microsoft/phi-2) are language models using +only 1.3 and 2.7 billion parameters but with state of the art performance compared to models with up to 10 billion parameters. The candle implementation provides both the standard version as well as a quantized variant. -## Running some example +## Running some examples +For the v2 version. +```bash +$ cargo run --example phi --release -- --model 2 \ + --prompt "A skier slides down a frictionless slope of height 40m and length 80m. What's the skier speed at the bottom?" + +A skier slides down a frictionless slope of height 40m and length 80m. What's the skier speed at the bottom? + +Solution: +The potential energy of the skier is converted into kinetic energy as it slides down the slope. The formula for potential energy is mgh, where m is mass, g is acceleration due to gravity (9.8 m/s^2), and h is height. Since there's no friction, all the potential energy is converted into kinetic energy at the bottom of the slope. The formula for kinetic energy is 1/2mv^2, where v is velocity. We can equate these two formulas: +mgh = 1/2mv^2 +Solving for v, we get: +v = sqrt(2gh) +Substituting the given values, we get: +v = sqrt(2*9.8*40) = 28 m/s +Therefore, the skier speed at the bottom of the slope is 28 m/s. +``` + +For the v1.5 version. ```bash $ cargo run --example phi --release -- --prompt "def print_prime(n): " |