# How much pizza and how much frozen yogurt? …with Gröbner bases

**R-bloggers on Mikkel Meyer Andersen**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

In a recent blog post I tried to get `yacas`

to solve a system of polynomial equations.

Unfortunately it could not do that, so I solved it numerically instead.

Now it is possible – together with many other systems of polynomial equations thanks to

fixing a small error in `yacas`

.

It has now been fixed, also in `Ryacas`

(development version), so

hurry up and update `Ryacas`

to the latest version `0.9.0.9122`

.

First I will show how to do it directly in `Ryacas`

(and afterwards show what happens behind the curtains).

library(Ryacas)

## Solving a system of equations

We define the system from the recent blog post in `R`

:

L <- "x^2 * (y/4) - a*(3*x + 3*y/2 - 45)" eq1 <- paste0("D(x) ", L) %>% yac_str() eq2 <- paste0("D(y) ", L) %>% yac_str() eq3 <- paste0("D(a) ", L) %>% yac_str() eqs <- c(eq1, eq2, eq3) eqs ## [1] "(x*y)/2-3*a" "x^2/4-(3*a)/2" "45-(3*x+(3*y)/2)"

As known from the theory of Gröbner bases,

instead of finding roots in this entangled system,

we can find the Gröbner bases and find roots for those instead:

eq_str <- paste0("{", paste0(eqs, " == 0", collapse = ", "), "}") %>% y_fn("Solve", "{x, y, a}") eq_str ## [1] "Solve({(x*y)/2-3*a == 0, x^2/4-(3*a)/2 == 0, 45-(3*x+(3*y)/2) == 0}, {x, y, a})" sol_all <- eq_str %>% yac_str() sol_all ## [1] "{{x==0,y==30,a==0},{x==10,y==10,a==50/3}}"

As seen, we get two solutions. The solution \(a = 0\) is not feasible in our case (read the recent blog post) so we just pick the second:

sol_elible_vars <- sol_all %>% y_fn("Nth", 2) %>% yac_str() sol_elible_vars ## [1] "{x==10,y==10,a==50/3}"

We have now finally arrived at the answer:

\[

(x, y, a) = (10, 10, 50/3) .

\]

It is also possible to postprocess the answer a bit if that is needed:

# Remove variable names if wanted: sol_elible <- sol_elible_vars %>% y_rmvars() %>% yac_str() sol_elible ## [1] "{10,10,50/3}" # And maybe convert to numbers: sol_elible %>% yac_expr() ## expression(c(10, 10, 50/3)) # ...maybe even evaluated: sol_elible %>% yac_expr() %>% eval() ## [1] 10.00000 10.00000 16.66667

## Enter Gröbner bases

We could actually have solved the system manually earlier using Gröbner bases.

As known from the theory of Gröbner bases (but I will not start trying to explain it because

I do not know them nearly as well as I should),

instead of finding roots in this entangled system,

we can find the Gröbner bases and find roots for those instead:

fs <- paste0("{", paste0(eqs, collapse = ", "), "}") fs ## [1] "{(x*y)/2-3*a, x^2/4-(3*a)/2, 45-(3*x+(3*y)/2)}" gs <- fs %>% y_fn("Groebner") %>% yac_str() %>% as_r() gs ## [1] "(x*y)/2-3*a" "x^2/4+((-3)*a)/2" ## [3] "(-3)*x+((-3)*y)/2+45" "9*y*a-90*a" ## [5] "15*y-y^2/2-6*a" "(-6)*a^2+100*a" ## [7] "((-15)*y)/2+(-9)*a+225"

As seen the system is now “triangulated”. So we proceed:

yac_str("Solve((-6)*a^2+100*a == 0, a)") ## [1] "{a==0,a==50/3}"

As mentioned, the solution \(a = 0\) is not feasible in our case (read the recent blog post), so we use \(a = 50/3\) and obtain:

yac_str("WithValue(a, 50/3, Solve(9*y*a-90*a == 0, y))") ## [1] "{y==10}"

So \(y = 10\). And we get

yac_str("WithValue({a, y}, {50/3, 10}, Solve((x*y)/2-3*a == 0, x))") ## [1] "{x==10}"

We have now finally arrived at the answer:

\[

(x, y, a) = (10, 10, 50/3) .

\]

**leave a comment**for the author, please follow the link and comment on their blog:

**R-bloggers on Mikkel Meyer Andersen**.

R-bloggers.com offers

**daily e-mail updates**about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.