# exercici 1 Parcial Q2 20-21

# variables

# m      : matriu donada
# k      : valor del radi de l'ordre
# iniSub : index inferior mínim de la submatriu en estudi
# fiSub  : index superior màxim de la submatriu en estudi

# La submatriu associada a un k-centre d'ordre es
# de (2k+1)x(2k+1). Per tant, perque els indexs que la recorren
# siguin sempre dins de la matriu donada de nxn, cal que
# 2k+1 <= n. Es a dir, k <= (n - 1) %/% 2.


matOrdenada <- function(m)
{
  ant <- m[1, 1]
  for(i in 1 : nrow(m)){
    for (j in 1 : ncol(m)){
      if (ant > m[i, j]) return (FALSE)
      ant <- m[i, j]
    }
  }
  return (TRUE)
}


nombreCentres <- function(m, k)
{
  n <- nrow(m)
  if (k <= (n - 1) %/% 2){
    iniSub <- 1 + k    
    fiSub  <- n - k    

    for (i in iniSub : fiSub){
      for (j in iniSub : fiSub){
        subMatriu <- m[(i-k):(i+k), (j-k):(j+k)]
        if( matOrdenada(subMatriu)){
           cat("(",i, ",", j, ")", "\n")
        }
      }
    }
  }
}


# programa principal

# A definida per columnes

A <- matrix( c(11, 14, 17,  2,  3, 90, 12, 11, 77,
               12, 15, 18,  1, 11, 65, 74, 81, 96,
	       13, 16, 19, 27, 47, 70, 75, 90, 96,
	        2, 17, 20, 30, 50, 71, 76, 95, 97,
	        1, 18, 21, 40, 52, 74, 12, 13, 15,
	        3, 19, 25, 45, 63, 76, 10, 11, 11,
	        1, 19, 27, 46, 65, 80, 10, 10, 14,
	        1,  1,  1,  1,  1,  1,  1,  1,  1,
	        1,  1,  1,  1,  1,  1,  1,  1,  1
	   ), nrow = 9)

k <- 1
cat("\n k = ", k, "\n")
nombreCentres(A, k)

k <- 2
cat("\n k = ", k, "\n")
nombreCentres(A, k)

k <- 3
cat("\n k = ", k, "\n")
nombreCentres(A, k)

k <- 7
cat("\n k = ", k, "\n")
nombreCentres(A, k)