Update the readme to mention mixtral. (huggingface#1443)

README.md

@@ -173,6 +173,7 @@ If you have an addition to this list, please submit a pull request.
         - StarCoder.
         - Phi 1, 1.5, and 2.
         - Mistral 7b v0.1.
+        - Mixtral 8x7b v0.1.
         - StableLM-3B-4E1T.
         - Replit-code-v1.5-3B.
         - Bert.
@@ -182,6 +183,7 @@ If you have an addition to this list, please submit a pull request.
         - Mistral 7b, and 7b instruct.
         - Zephyr 7b a and b (Mistral based).
         - OpenChat 3.5 (Mistral based).
+        - Mixtral 8x7b.
     - Text to text.
         - T5 and its variants: FlanT5, UL2, MADLAD400 (translation), CoEdit (Grammar correction).
         - Marian MT (Machine Translation).

candle-examples/examples/quantized/README.md

@@ -26,6 +26,19 @@ cargo run --example quantized --release -- --prompt "The best thing about coding
 > The best thing about coding in rust is 1.) that I don’t need to worry about memory leaks, 2.) speed and 3.) my program will compile even on old machines.
 ```
 
+Using the mixtral sparse mixture of expert model:
+```bash
+
+$ cargo run --example quantized --release -- --which mixtral --prompt "Lebesgue's integral is superior to Riemann's because "
+> avx: true, neon: false, simd128: false, f16c: true
+> temp: 0.80 repeat-penalty: 1.10 repeat-last-n: 64
+> loaded 995 tensors (26.44GB) in 0.03s
+Lebesgue's integral is superior to Riemann's because 1. it is defined for a wider class of functions, those which are absolutely integrable; 2. the definition does not involve limits in two variables---one being computed before the other (which makes some computations more difficult); and 3. interchange of order of integration is easier to establish than with Riemann's integral. On the other hand, Lebesgue's integral applies only for bounded functions defined on finite intervals; it does not provide numerical values for improper integrals. The latter are best evaluated using Cauchy's limit definition.
+
+The reason $f(x) = x^2$ is discontinuous at the ends of its interval of definition, and Riemann's integral requires continuity on the whole of an open interval containing it (see our earlier post), sine no such function exists with this property, is that the endpoints are infinite in measure for Lebesgue's integral.
+ ```
+
+
 ## Command-line flags
 
 Run with `--help` to see all options.