adagio (version 0.9.2)

subsetsum: Subset Sum Problem

Description

Subset sum routine for positive integers.

Usage

subsetsum(S, t, method = "greedy")
sss_test(S, t)

Value

List with the target value, if reached, and vector of indices of elements in S that sum up to t.

If no solution is found, the dynamic method will return indices for the largest value below the target, the greedy method witll return NULL.

sss_test will simply return maximum sum value found.

Arguments

S

vector of positive integers.

t

target value, bigger than all items in S.

method

can be ``greedy'' or ``dynamic'', where ``dynamic'' stands for the dynamic programming approach.

Author

HwB email: <[e-mail]>

Details

subsetsum is searching for a set of elements in S that sum up to t by continuously adding more elements of S.

It is not required that S is decreasingly sorted. But for reasons of efficiency and smaller execution times it is urgently recommended to sort the item set in decreasing order. See the examples to find out how to handle your data.

The first components will be preferred, i.e., if S is decreasing, the sum with larger elements will be found, if increasing, the sum with smaller elements. Because of timing considerations, the default is to sort decreasingly before processing.

The dynamic method may be faster for large sets, but will also require much more memory if the target value is large.

sss_test will find the biggest number below or equal to t that can be expressed as a sum of items in S. It will not return any indices. It can be quite fast, though it preprocesses the set S to be sorted decreasingly, too.

References

Horowitz, E., and S. Sahni (1978). Fundamentals of Computer Algorithms. Computer Science Press, Rockville, ML.

See Also

maxsub

Examples

t <- 5842
S <- c(267, 493, 869, 961, 1000, 1153, 1246, 1598, 1766, 1922)

# S is not decreasingly sorted, so ...
o  <- order(S, decreasing = TRUE)
So <- S[o]                          # So is decreasingly sorted

sol <- subsetsum(So, t)             # $inds:  2 4 6 7 8  w.r.t.  So
is  <- o[sol$inds]                  # is:     9 7 5 4 3  w.r.t.  S
sum(S[is])                          # 5842

if (FALSE) {
amount <- 4748652
products <- 
c(30500,30500,30500,30500,42000,42000,42000,42000,
  42000,42000,42000,42000,42000,42000,71040,90900,
  76950,35100,71190,53730,456000,70740,70740,533600,
  83800,59500,27465,28000,28000,28000,28000,28000,
  26140,49600,77000,123289,27000,27000,27000,27000,
  27000,27000,80000,33000,33000,55000,77382,48048,
  51186,40000,35000,21716,63051,15025,15025,15025,
  15025,800000,1110000,59700,25908,829350,1198000,1031655)

# prepare set
prods <- products[products <= amount]  # no elements > amount
prods <- sort(prods, decreasing=TRUE)  # decreasing order

# now find one solution
system.time(is <- subsetsum(prods, amount))
#  user  system elapsed 
# 0.030   0.000   0.029 

prods[is]
#  [1]   70740   70740   71190   76950   77382   80000   83800
#  [8]   90900  456000  533600  829350 1110000 1198000

sum(prods[is]) == amount
# [1] TRUE

# Timings:
#             unsorted   decr.sorted
# "greedy"      22.930         0.030    (therefore the default settings)
# "dynamic"      2.515         0.860    (overhead for smaller sets)
# sss_test       8.450         0.040    (no indices returned)
}