The challenge here is interesting but also very clear: for a given un-ordered vector of numbers, select from it a longest increasing (or decreasing) subsequence. This subsequence may not be unique.

For example, in the following sequence,

62 49 42 54 59 39 35 48 57 60


One obvious longest increasing subsequence is 48, 57, 60, which has length 3;

The (unique) longest decreasing subsequence is 62, 49, 42, 39, 35, which has length 5.

For a very long vector, say having 10000 numbers, is it the case that all increasing and decreasing sequences are very short? This question was actually answered 80 years ago, the longest increasing or decreasing subsequence has ( O(\sqrt{n}) ) length.

Interested readers can find the proof in Erdos P. and Szekeres G., 1935 for following theorem: any sequence of ( n^2+1 ) distinct integers has either an increasing or a decreasing subsequence of length ( n+1 ).

That means for a general vector of length ( m ), there is either an increasing or decreasing subsequence of length at least ( \sqrt{m-1} ).

Once we know that there is always a longest increasing or decreasing subsequence of at least a certain size, how do we find it? At a glance this may seem difficult. Actually there is a good dynamic programming algorithm using only ( O(n\log(n)) ) time.

## R version

The following code has been implemented following Wiki page.

library(compiler)
longest_subseq.R = cmpfun(function(x) {
P = integer(length(x))
M = integer(length(x) + 1)
L = newL = 0
for (i in seq_along(x) - 1) {
lo = 1
hi = L
while (lo <= hi) {
mid = (lo + hi)%/%2
if (x[M[mid + 1] + 1] < x[i + 1]) {
lo = mid + 1
} else {
hi = mid - 1
}
}
newL = lo
P[i + 1] = M[newL]
if (newL > L) {
M[newL + 1] = i
L = newL
} else if (x[i + 1] < x[M[newL + 1] + 1]) {
M[newL + 1] = i
}
}
k = M[L + 1]
re = integer(L)
for (i in L:1) {
re[i] = k + 1
k = P[k + 1]
}
re
})
longest_subseq.R(c(5, 1, 3, 2, 4))

## [1] 2 4 5


Now let’s play with a larger vector.

x = rnorm(1000)
plot(x, main = "Longest Increasing (blue) and Decreasing (red) Subsequences\nfor 1000 random numbers",
ylab = "Random Value", pch = 16, col = "#0000004D")
# Longest Increasing Sequence
ind = longest_subseq.R(x)
points(ind, x[ind], pch = 16, col = "blue")
# Longest Decreasing Sequence
rind = longest_subseq.R(-x)
points(rind, x[rind], pch = 16, col = "red")


y = rnorm(250000)
system.time(ind1 <- longest_subseq.R(y))

## user system elapsed
## 3.36 0.01 3.48


A bit slow, right?

But 250,000 is typically not long for a time series.

## Rcpp version

library(Rcpp)

## Warning: package 'Rcpp' was built under R version 3.0.3

cppcode <- "\n#include <vector>\n#include <Rcpp.h>\nusing namespace std;\ntypedef int index_type;\n\n// [[Rcpp::export]]\nvector<index_type> longest_subseq(SEXP X){\n Rcpp::NumericVector x(X);\n\tvector<index_type> P(x.size(), 0);\n\tvector<index_type> M(x.size()+1, 0);\n\tindex_type L(0), newL;\n\tfor(index_type i=0; i < x.size(); ++i) {\n\t\tindex_type lo(1), hi(L), mid;\n\t\twhile( lo <= hi) {\n\t\t\tmid = (lo + hi) / 2;\n\t\t\tif (x[M[mid]] < x[i] ) {\n\t\t\t\tlo = mid + 1;\n\t\t\t} else {\n\t\t\t\thi = mid - 1;\n\t\t\t}\n\t\t}\n\t\tnewL = lo;\n\t\tP[i] = M[newL - 1];\n\t\tif (newL > L) {\n\t\t\tM[newL] = i;\n\t\t\tL = newL;\n\t\t} else if (x[i] < x[M[newL]]) {\n\t\t\tM[newL] = i;\n\t\t}\n\t}\n vector<index_type> re(L);\n\tindex_type k(M[L]);\n\tfor(index_type i=L-1; i>=0; --i){\n\t\tre[i] = k + 1;\n\t\tk = P[k];\n\t}\n return(re);\n}"

sourceCpp(code = cppcode)


The sourceCpp will export the cpp function longest_subseq as R function longest_subseq, now we can test it.

system.time(ind2 <- longest_subseq(y))

## user system elapsed
## 0.03 0.00 0.03

all.equal(ind1, ind2)

## [1] TRUE


That’s a big improvement.

Now we can play with some time series data sets.

data(EuStockMarkets)
matplot(EuStockMarkets, main = "Longest Increasing Subsequence\nEU Stock Market Data (1991–1998)",
col = 2:5, pch = "o", cex = 0.3)
for (i in 1:4) {
ind = longest_subseq(EuStockMarkets[, i])
points(ind, EuStockMarkets[ind, i], type = "l", col = i + 1)
}
i = 3
ind = longest_subseq(EuStockMarkets[, i])
for (j in ind) lines(c(j, j), c(0, EuStockMarkets[j, i]), col = i + 1, lty = 3)
legend("topleft", legend = colnames(EuStockMarkets), col = 2:5, pch = "o")