---
title: "91 - Using the 'SymPy' object directly"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{91 - Using the 'SymPy' object directly}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

```{r, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)
```

```{r, message=FALSE}
library(caracas)
```

```{r, include = FALSE}
inline_code <- function(x) {
  x
}

if (!has_sympy()) {
  # SymPy not available, so the chunks shall not be evaluated
  knitr::opts_chunk$set(eval = FALSE)
  
  inline_code <- function(x) {
    deparse(substitute(x))
  }
}
```

## Using `SymPy` directly

First we get the `SymPy` object:

```{r}
sympy <- get_sympy()
```

```{r}
sympy$diff("2*a*x", "x")
sympy$solve("x**2 - 1", "x")
```

## Elaborate example

How can we minimise the amount of material used to produce a cylindric
tin can that contains 1 litre.  The cylinder has diameter $d$ and
height $h$. The question is therefore: What is $d$ and $h$?

We introduce the variables `d` (diameter) and `h` (height):

```{r}
d <- sympy$symbols('d')
h <- sympy$symbols('h')
```

The problem is a constrained optimisation problem, and we solve it by
a Lagrange multiplier, and therefore we introduce `lam` (the Lagrange
multiplier):

```{r}
lam <- sympy$symbols('lam')
```

We now set up the problem:

```{r}
area_str <- "Pi/2 * d**2 + Pi * h * d"
vol_str <- "Pi/4 * d**2 * h"
lap_str <- paste0("(", area_str, ") - lam*((", vol_str, ") - 1)")
lap <- sympy$parsing$sympy_parser$parse_expr(
  lap_str,
  local_dict = list('d' = d, 'h' = h, 'lam' = lam))
```

We can now find the gradient:

```{r}
grad <- sympy$derive_by_array(lap, list(d, h, lam))
grad
```

And find the critical points:

```{r}
sol <- sympy$solve(grad, list(d, h, lam), dict = TRUE)
sol
```

We take the one with the real solution:

```{r}
sol[[1]]
```

We now have a short helper function to help getting appropriate `R`
expressions (such a function will be included in later versions of
this package):

```{r}
to_r <- function(x) {
  x <- as.character(x)
  x <- gsub("Pi", "pi", x, fixed = TRUE)
  x <- gsub("**", "^", x, fixed = TRUE)
  x <- parse(text = x)
  return(x)
}

sol_d <- to_r(sol[[1]]$d)
sol_d
eval(sol_d)
sol_h <- to_r(sol[[1]]$h)
sol_h
eval(sol_h)
```

(It is left as an exercise to the reader to show that the critical point indeed is a minimum.)

## Simple example with assumptions

```{r}
x <- sympy$symbols('x')
x$assumptions0
x <- sympy$symbols('x', positive = TRUE)
x$assumptions0
eq <- sympy$parsing$sympy_parser$parse_expr("x**2 - 1",
                                            local_dict = list('x' = x))
sympy$solve(eq, x, dict = TRUE)
```

## Another example with assumptions

```{r}
x <- sympy$symbols('x', positive = TRUE)
eq <- sympy$parsing$sympy_parser$parse_expr("x**3/3 - x",
                                            local_dict = list('x' = x))
eq
grad <- sympy$derive_by_array(eq, x)
grad
sympy$solve(grad, x, dict = TRUE)
```