Find non-first match in R - r

I know that match(x,y) returns the first match of all elements of x in y.
Assuming that x may contain the same value multiple time, I am looking for a concise way to match the nth occurrence in x with the nth occurrence in y.
For example: `
x <- c(3,4,4,3,2,4)
y <- c(1,2,3,4,1,2,3,4)
my.match(x, y)
## 3,4,8,7,2,NA


Using a for loop to match, store and overwrite a match with NA.
idx <- c()
for (i in x) {
k <- match(i, y)
idx <- c(idx, k)
y[k] <- NA
}
idx
#[1] 3 4 8 7 2 NA


The following function is much faster when vectors are large because it does not iterate over the whole vector
my.match <- function(x,y){
fidx <- rep(FALSE,length(x))
fidy <- rep(FALSE,length(y))
ret <- rep(NA,length(x))
repeat{
nidx <- which(!fidx)
nidy <- which(!fidy)
idx <- match(x[nidx],y[nidy])
idy <- match(y[nidy],x[nidx])
ret[nidx] <- nidy[idx]
fidx[nidx[unique(idy)]] <- TRUE
fidy[nidy[unique(idx)]] <- TRUE
if(sum(!is.na(idx))==0 | sum(!is.na(idy))==0){
break
}
}
return(ret)
}
Benchmarking with the other proposed method yields:
my.match1 <- function(x,y){
idx <- c()
for (i in x) {
k <- match(i, y)
idx <- c(idx, k)
y[k] <- NA
}
return(idx)
}
x <- sample.int(100,10000,replace=T)
y <- sample.int(100,10000,replace=T)
system.time(my.match1(x,y))
## user system elapsed
## 1.016 0.003 1.020
system.time(my.match(x,y))
## user system elapsed
## 0.049 0.000 0.049

Related

What's wrong with the logic of this function in R?

I'm tyring to build a function in R which calculates the percentage change between rows based on any arbitrary index, this is, between any given row and the preceding one or any given row and n preceding ones.
perc_change <- function(x,n) {
y <- c()
z <- c()
for (i in 1:length(x)) {
z[i] <- (x[i]/(x[i-n])-1)*100
}
y <- c(rep(NA,n),z[(n+1):length(z)])
y
}
When n is one the function works properly:
x <- c(2,3.5,4,6)
perc_change(x,1)
[1] NA 75.00000 14.28571 50.00000
But when I change to 2 or other n, I receive this error:
Error in z[i] <- (x[i]/(x[i - n]) - 1) * 100 :
replacement has length zero
I just can't find why and where the logic of my function is wrong so I appreciate any comment or suggestion.

In the loop, when n is greater than 1, the i starting at 1 can result in negative or zero index (i.e. when n =2, 1 - 2). To avoid, that an if/else condition can be added
perc_change <- function(x,n) {
y <- c()
z <- c()
for (i in 1:length(x)) {
if(i > n) {
z[i] <- (x[i]/(x[i-n])-1)*100
} else z[i] <- NA
}
y <- c(rep(NA,n),z[(n+1):length(z)])
y
}
perc_change(x,1)
#[1] NA 75.00000 14.28571 50.00000
perc_change(x, 2)
#[1] NA NA 100.00000 71.42857
perc_change(x, 3)
#[1] NA NA NA 200

The following function lags the input vector and then computes the percent change with vectorized operations, no need for for loops. The lag function is a copy&paste of the last code lines of dplyr::lag.
perc_change <- function(x, n = 1) {
lag <- function(x, n = 1){
if(n == 0)
return(x)
xlen <- length(x)
n <- pmin(n, xlen)
out <- c(rep(NA, n), x[seq_len(xlen - n)])
attributes(out) <- attributes(x)
out
}
y <- lag(x, n)
(x/y - 1)*100
}
x <- c(2, 3.5, 4, 6)
perc_change(x,1)
#[1] NA 75.00000 14.28571 50.00000
perc_change(x, 2)
#[1] NA NA 100.00000 71.42857

compare the information between two matrices R

I have two matrices, one is generated out of the other by deleting some rows. For example:
m = matrix(1:18, 6, 3)
m1 = m[c(-1, -3, -6),]
Suppose I do not know which rows in m were eliminated to create m1, how should I find it out by comparing the two matrices? The result I want looks like this:
1, 3, 6
The actual matrix I am dealing with is very big. I was wondering if there is any efficient way of conducting it.

Here are some approaches:
1) If we can assume that there are no duplicated rows in m -- this is the case in the example in the question -- then:
which(tail(!duplicated(rbind(m1, m)), nrow(m)))
## [1] 1 3 6
2) Transpose m and m1 giving tm and tm1 since it is more efficient to work on columns than rows.
Define match_indexes(i) which returns a vector r such that each row in m[r, ] matches m1[i, ].
Apply that to each i in 1:n1 and remove the result from 1:n.
n <- nrow(m); n1 <- nrow(m1)
tm <- t(m); tm1 <- t(m1)
match_indexes <- function(i) which(colSums(tm1[, i] == tm) == n1)
setdiff(1:n, unlist(lapply(1:n1, match_indexes)))
## [1] 1 3 6
3) Calculate an interaction vector for each matrix and then use setdiff and finally match to get the indexes:
i <- interaction(as.data.frame(m))
i1 <- interaction(as.data.frame(m1))
match(setdiff(i, i1), i)
## [1] 1 3 6
Added If there can be duplicates in m then (1) and (3) will only return the first of any multiply occurring row in m not in m1.
m <- matrix(1:18, 6, 3)
m1 <- m[c(2, 4, 5),]
m <- rbind(m, m[1:2, ])
# 1
which(tail(!duplicated(rbind(m1, m)), nrow(m)))
## 1 3 6
# 2
n <- nrow(m); n1 <- nrow(m1)
tm <- t(m); tm1 <- t(m1)
match_indexes <- function(i) which(colSums(tm1[, i] == tm) == n1)
setdiff(1:n, unlist(lapply(1:n1, match_indexes)))
## 1 3 6 7
# 3
i <- interaction(as.data.frame(m))
i1 <- interaction(as.data.frame(m1))
match(setdiff(i, i1), i)
## 1 3 6

A possible way is to represent each row as a string:
x1 <- apply(m, 1, paste0, collapse = ';')
x2 <- apply(m1, 1, paste0, collapse = ';')
which(!x1 %in% x2)
# [1] 1 3 6
Some benchmark with a large matrix using my solution and G. Grothendieck's solutions:
set.seed(123)
m <- matrix(rnorm(20000 * 5000), nrow = 20000)
m1 <- m[-sample.int(20000, 1000), ]
system.time({
which(tail(!duplicated(rbind(m1, m)), nrow(m)))
})
# user system elapsed
# 339.888 2.368 342.204
system.time({
x1 <- apply(m, 1, paste0, collapse = ';')
x2 <- apply(m1, 1, paste0, collapse = ';')
which(!x1 %in% x2)
})
# user system elapsed
# 395.428 0.568 395.955
system({
n <- nrow(m); n1 <- nrow(m1)
tm <- t(m); tm1 <- t(m1)
match_indexes <- function(i) which(colSums(tm1[, i] == tm) == n1)
setdiff(1:n, unlist(lapply(1:n1, match_indexes)))
})
# > 15 min, not finish
system({
i <- interaction(as.data.frame(m))
i1 <- interaction(as.data.frame(m1))
match(setdiff(i, i1), i)
})
# run out of memory. My 32G RAM machine crashed.

We can also use do.call
which(!do.call(paste, as.data.frame(m)) %in% do.call(paste, as.data.frame(m1)))
#[1] 1 3 6

R, for loop, scalable solutions

I have data that looks like this :
char_column date_column1 date_column2 integer_column
415 18JT9R6EKV 2014-08-28 2014-09-06 1
26 18JT9R6EKV 2014-12-08 2014-12-11 2
374 18JT9R6EKV 2015-03-03 2015-03-09 1
139 1PEGXAVCN5 2014-05-06 2014-05-10 3
969 1PEGXAVCN5 2014-06-11 2014-06-15 2
649 1PEGXAVCN5 2014-08-12 2014-08-16 3
I want to perform a loop that would check every row against the preceding row, and given certain conditions assign them the same number (so I can group them later) , the point is that if the date segments are close enough I would collapse them into one segment.
my attempt is the following :
i <- 1
z <- 1
v <- 1
for (i in 2:nrow(df)){
z[i] <- ifelse(df[i,'char_column'] == df[i-1,'char_column'],
ifelse((df[i,'date_column1'] - df[i-1,'date_column2']) <= 5,
ifelse(df[i,'integer_column'] == df[i-1,'integer_column'],
v, v<- v+1),
v <- v+1),
v <- v+1)}
df$grouping <- z
then I would just group using min(date_column1) and max(date_column2).
this method works perfectly for say 100,000 rows (22.86 seconds)
but for a million rows : 33.18 minutes!! I have over 60m rows to process,
is there a way I can make the process more efficient ?
PS: to generate a similar table you can use the following code :
x <- NULL
for (i in 1:200) { x[i] <- paste(sample(c(LETTERS, 1:9), 10), collapse = '')}
y <- sample((as.Date('2014-01-01')):as.Date('2015-05-01'), 1000, replace = T)
y2 <- y + sample(1:10)
df <- data.frame(char_column = sample(x, 1000, rep = T),
date_column1 = as.Date(y, origin = '1970-01-01'),
date_column2 = as.Date(y2,origin = '1970-01-01'),
integer_column = sample(1:3,1000, replace = T),
row.names = NULL)
df <- df[order(df$char_column, df$date_column1),]

Since data.table::rleid does not work, I post another (hopefully) fast solution
1. Get rid of nested ifelse
ifelse is often slow, especially for scalar evaluation, use if.
Nested ifelse should be avoided whenever possible: observe that ifelse(A, ifelse(B, x, y), y) can be suitably replaced by if (A&B) x else y
f1 <- function(df){
z <- rep(NA, nrow(df))
z[1] <- 1
char_col <- df[, 'char_column']
date_col1 <- df[, 'date_column1']
date_col2 <- df[, 'date_column2']
int_col <- df[, 'integer_column']
for (i in 2:nrow(df)){
if((char_col[i] == char_col[i-1])&((date_col1[i] - date_col2[i-1]) <= 5)&(int_col[i] == int_col[i-1]))
{
z[i] <- z[i-1]
}
else
{
z[i] <- z[i-1]+1
}
}
z
}
f1 is about 40% faster than the original solution for 10.000 rows.
system.time(f1(df))
user system elapsed
2.72 0.00 2.79
2. Vectorize
Upon closer inspection the conditions inside if can be vectorized
library(data.table)
f2 <- function(df){
z <- rep(NA, nrow(df))
z[1] <- 1
char_col <- df[, 'char_column']
date_col1 <- df[, 'date_column1']
date_col2 <- df[, 'date_column2']
int_col <- df[, 'integer_column']
cond <- (char_col==shift(char_col))&(date_col1 - shift(date_col2) <= 5)&(int_col==shift(int_col))
for (i in 2:nrow(df)){
if(cond[i])
{
z[i] <- z[i-1]
}
else
{
z[i] <- z[i-1]+1
}
}
z
}
# for 10000 rows
system.time(f2(df))
# user system elapsed
# 0.01 0.00 0.02
3. Vectorize, Vectorize
While f2 is already quite fast, a further vectorization is possible. Observe how z is calculated: cond is a logical vector, and z[i] = z[i-1] + 1 when cond is FALSE. This is none other than cumsum(!cond).
f3 <- function(df){
setDT(df)
df[, cond := (char_column==shift(char_column))&(date_column1 - shift(date_column2) <= 5)&(integer_column==shift(integer_column)),]
df[, group := cumsum(!c(FALSE, cond[-1L])),]
}
For 1M rows
system.time(f3(df))
# user system elapsed
# 0.05 0.05 0.09
system.time(f2(df))
# user system elapsed
# 1.83 0.05 1.87

Implement table() function as a user defined function

x <- c(1,2,3,2,1)
table(x)
# x
# 1 2 3
# 2 2 1
Outputs how many times each element occur in the vector.
I am trying to imitate the above function using function()
Below is my code:
TotalTimes = function(x){
times = 0
y = unique(x)
for (i in 1:length(y)) {
for (i in 1:length(x)) {
if(y[i] == x[i])
times = times + 1
}
return(times)
}
}
What would be the right approach?

Here's a one-liner, using rle():
f <- function(x) {
with(rle(sort(x)), setNames(lengths, values))
}
f(c(1,2,3,2,1))
# 1 2 3
# 2 2 1
Alternatively, here's an option that's less "tricky", and is probably a better model for learning to code in an R-ish way:
f2 <- function(x) {
ss <- sort(x)
uu <- unique(ss)
names(uu) <- uu
sapply(uu, function(u) sum(ss == u))
}
f2(c(1,2,3,2,1))
# 1 2 3
# 2 2 1

function(x) {
q = sapply(unique(x), function(i) sum(x == i))
names(q) = unique(x)
return(q)
}

Here is one method using base R:
# data
x <- c(1,2,3,2,1)
# set up
y <- sort(unique(x))
counts <- rep_len(0, length.out=length(y))
names(counts) <- y
for(i in seq_along(x)) {
counts[x[i] == y] <- counts[x[i] == y] + 1
}
Wrapping it in a function:
table2 <- function(x) {
# transform x into character vector to reduce search cost in loop
x <- as.character(x)
y <- sort(unique(x))
counts <- rep_len(0, length.out=length(y))
names(counts) <- y
for(i in seq_along(x)) {
counts[x[i]] <- counts[x[i]] + 1L
}
return(counts)
}
This version only accepts a single vector, of course. At #Frank's suggestion, the function version is slightly different, and possibly faster, in that it transforms the input x into a character. The potential speed up is in the search in counts[x[i]] where the name in counts is referred to (as x[i]), rather than performing a search using "==."

R Function for returning ALL factors

My normal search foo is failing me. I'm trying to find an R function that returns ALL of the factors of an integer. There are at least 2 packages with factorize() functions: gmp and conf.design, however these functions return only prime factors. I'd like a function that returns all factors.
Obviously searching for this is made difficult since R has a construct called factors which puts a lot of noise in the search.

To follow up on my comment (thanks to #Ramnath for my typo), the brute force method seems to work reasonably well here on my 64 bit 8 gig machine:
FUN <- function(x) {
x <- as.integer(x)
div <- seq_len(abs(x))
factors <- div[x %% div == 0L]
factors <- list(neg = -factors, pos = factors)
return(factors)
}
A few examples:
> FUN(100)
$neg
[1] -1 -2 -4 -5 -10 -20 -25 -50 -100
$pos
[1] 1 2 4 5 10 20 25 50 100
> FUN(-42)
$neg
[1] -1 -2 -3 -6 -7 -14 -21 -42
$pos
[1] 1 2 3 6 7 14 21 42
#and big number
> system.time(FUN(1e8))
user system elapsed
1.95 0.18 2.14

You can get all factors from the prime factors. gmp calculates these very quickly.
library(gmp)
library(plyr)
get_all_factors <- function(n)
{
prime_factor_tables <- lapply(
setNames(n, n),
function(i)
{
if(i == 1) return(data.frame(x = 1L, freq = 1L))
plyr::count(as.integer(gmp::factorize(i)))
}
)
lapply(
prime_factor_tables,
function(pft)
{
powers <- plyr::alply(pft, 1, function(row) row$x ^ seq.int(0L, row$freq))
power_grid <- do.call(expand.grid, powers)
sort(unique(apply(power_grid, 1, prod)))
}
)
}
get_all_factors(c(1, 7, 60, 663, 2520, 75600, 15876000, 174636000, 403409160000))

Update
This is now implemented in the package RcppBigIntAlgos. See this answer for more details.
Original Post
The algorithm has been fully updated and now implements multiple polynomials as well as some clever sieving techniques that eliminates millions of checks. In addition to the original links, this paper along with this post from primo were very helpful for this last stage (many kudos to primo). Primo does a great job of explaining the guts of the QS in a relatively short space and also wrote a pretty amazing algorithm (it will factor the number at the bottom, 38! + 1, in under 2 secs!! Insane!!).
As promised, below is my humble R implementation of the Quadratic Sieve. I have been working on this algorithm sporadically since I promised it in late January. I will not try to explain it fully (unless requested... also, the links below do a very good job) as it is very complicated and hopefully, my function names speak for themselves. This has proved to be one of the most challenging algorithms I have ever attempted to execute as it is demanding both from a programmer's point of view as well as mathematically. I have read countless papers and ultimately, I found these five to be the most helpful (QSieve1, QSieve2, QSieve3, QSieve4, QSieve5).
N.B. This algorithm, as it stands, does not serve very well as a general prime factorization algorithm. If it was optimized further, it would need to be accompanied by a section of code that factors out smaller primes (i.e. less than 10^5 as suggested by this post), then call QuadSieveAll, check to see if these are primes, and if not, call QuadSieveAll on both of these factors, etc. until you are left with all primes (all of these steps are not that difficult). However, the main point of this post is to highlight the heart of the Quadratic Sieve, so the examples below are all semiprimes (even though it will factor most odd numbers not containing a square… Also, I haven’t seen an example of the QS that didn’t demonstrate a non-semiprime). I know the OP was looking for a method to return all factors and not the prime factorization, but this algorithm (if optimized further) coupled with one of the algorithms above would be a force to reckon with as a general factoring algorithm (especially given that the OP was needing something for Project Euler, which usually requires much more than brute force methods). By the way, the MyIntToBit function is a variation of this answer and the PrimeSieve is from a post that #Dontas appeared on a while back (Kudos on that as well).
QuadSieveMultiPolysAll <- function(MyN, fudge1=0L, fudge2=0L, LenB=0L) {
### 'MyN' is the number to be factored; 'fudge1' is an arbitrary number
### that is used to determine the size of your prime base for sieving;
### 'fudge2' is used to set a threshold for sieving;
### 'LenB' is a the size of the sieving interval. The last three
### arguments are optional (they are determined based off of the
### size of MyN if left blank)
### The first 8 functions are helper functions
PrimeSieve <- function(n) {
n <- as.integer(n)
if (n > 1e9) stop("n too large")
primes <- rep(TRUE, n)
primes[1] <- FALSE
last.prime <- 2L
fsqr <- floor(sqrt(n))
while (last.prime <= fsqr) {
primes[seq.int(last.prime^2, n, last.prime)] <- FALSE
sel <- which(primes[(last.prime + 1):(fsqr + 1)])
if (any(sel)) {
last.prime <- last.prime + min(sel)
} else {
last.prime <- fsqr + 1
}
}
MyPs <- which(primes)
rm(primes)
gc()
MyPs
}
MyIntToBit <- function(x, dig) {
i <- 0L
string <- numeric(dig)
while (x > 0) {
string[dig - i] <- x %% 2L
x <- x %/% 2L
i <- i + 1L
}
string
}
ExpBySquaringBig <- function(x, n, p) {
if (n == 1) {
MyAns <- mod.bigz(x,p)
} else if (mod.bigz(n,2)==0) {
MyAns <- ExpBySquaringBig(mod.bigz(pow.bigz(x,2),p),div.bigz(n,2),p)
} else {
MyAns <- mod.bigz(mul.bigz(x,ExpBySquaringBig(mod.bigz(
pow.bigz(x,2),p), div.bigz(sub.bigz(n,1),2),p)),p)
}
MyAns
}
TonelliShanks <- function(a,p) {
P1 <- sub.bigz(p,1); j <- 0L; s <- P1
while (mod.bigz(s,2)==0L) {s <- s/2; j <- j+1L}
if (j==1L) {
MyAns1 <- ExpBySquaringBig(a,(p+1L)/4,p)
MyAns2 <- mod.bigz(-1 * ExpBySquaringBig(a,(p+1L)/4,p),p)
} else {
n <- 2L
Legendre2 <- ExpBySquaringBig(n,P1/2,p)
while (Legendre2==1L) {n <- n+1L; Legendre2 <- ExpBySquaringBig(n,P1/2,p)}
x <- ExpBySquaringBig(a,(s+1L)/2,p)
b <- ExpBySquaringBig(a,s,p)
g <- ExpBySquaringBig(n,s,p)
r <- j; m <- 1L
Test <- mod.bigz(b,p)
while (!(Test==1L) && !(m==0L)) {
m <- 0L
Test <- mod.bigz(b,p)
while (!(Test==1L)) {m <- m+1L; Test <- ExpBySquaringBig(b,pow.bigz(2,m),p)}
if (!m==0) {
x <- mod.bigz(x * ExpBySquaringBig(g,pow.bigz(2,r-m-1L),p),p)
g <- ExpBySquaringBig(g,pow.bigz(2,r-m),p)
b <- mod.bigz(b*g,p); r <- m
}; Test <- 0L
}; MyAns1 <- x; MyAns2 <- mod.bigz(p-x,p)
}
c(MyAns1, MyAns2)
}
SieveLists <- function(facLim, FBase, vecLen, sieveD, MInt) {
vLen <- ceiling(vecLen/2); SecondHalf <- (vLen+1L):vecLen
MInt1 <- MInt[1:vLen]; MInt2 <- MInt[SecondHalf]
tl <- vector("list",length=facLim)
for (m in 3:facLim) {
st1 <- mod.bigz(MInt1[1],FBase[m])
m1 <- 1L+as.integer(mod.bigz(sieveD[[m]][1] - st1,FBase[m]))
m2 <- 1L+as.integer(mod.bigz(sieveD[[m]][2] - st1,FBase[m]))
sl1 <- seq.int(m1,vLen,FBase[m])
sl2 <- seq.int(m2,vLen,FBase[m])
tl1 <- list(sl1,sl2)
st2 <- mod.bigz(MInt2[1],FBase[m])
m3 <- vLen+1L+as.integer(mod.bigz(sieveD[[m]][1] - st2,FBase[m]))
m4 <- vLen+1L+as.integer(mod.bigz(sieveD[[m]][2] - st2,FBase[m]))
sl3 <- seq.int(m3,vecLen,FBase[m])
sl4 <- seq.int(m4,vecLen,FBase[m])
tl2 <- list(sl3,sl4)
tl[[m]] <- list(tl1,tl2)
}
tl
}
SieverMod <- function(facLim, FBase, vecLen, SD, MInt, FList, LogFB, Lim, myCol) {
MyLogs <- rep(0,nrow(SD))
for (m in 3:facLim) {
MyBool <- rep(FALSE,vecLen)
MyBool[c(FList[[m]][[1]][[1]],FList[[m]][[2]][[1]])] <- TRUE
MyBool[c(FList[[m]][[1]][[2]],FList[[m]][[2]][[2]])] <- TRUE
temp <- which(MyBool)
MyLogs[temp] <- MyLogs[temp] + LogFB[m]
}
MySieve <- which(MyLogs > Lim)
MInt <- MInt[MySieve]; NewSD <- SD[MySieve,]
newLen <- length(MySieve); GoForIT <- FALSE
MyMat <- matrix(integer(0),nrow=newLen,ncol=myCol)
MyMat[which(NewSD[,1L] < 0),1L] <- 1L; MyMat[which(NewSD[,1L] > 0),1L] <- 0L
if ((myCol-1L) - (facLim+1L) > 0L) {MyMat[,((facLim+2L):(myCol-1L))] <- 0L}
if (newLen==1L) {MyMat <- matrix(MyMat,nrow=1,byrow=TRUE)}
if (newLen > 0L) {
GoForIT <- TRUE
for (m in 1:facLim) {
vec <- rep(0L,newLen)
temp <- which((NewSD[,1L]%%FBase[m])==0L)
NewSD[temp,] <- NewSD[temp,]/FBase[m]; vec[temp] <- 1L
test <- temp[which((NewSD[temp,]%%FBase[m])==0L)]
while (length(test)>0L) {
NewSD[test,] <- NewSD[test,]/FBase[m]
vec[test] <- (vec[test]+1L)
test <- test[which((NewSD[test,]%%FBase[m])==0L)]
}
MyMat[,m+1L] <- vec
}
}
list(MyMat,NewSD,MInt,GoForIT)
}
reduceMatrix <- function(mat) {
tempMin <- 0L; n1 <- ncol(mat); n2 <- nrow(mat)
mymax <- 1L
for (i in 1:n1) {
temp <- which(mat[,i]==1L)
t <- which(temp >= mymax)
if (length(temp)>0L && length(t)>0L) {
MyMin <- min(temp[t])
if (!(MyMin==mymax)) {
vec <- mat[MyMin,]
mat[MyMin,] <- mat[mymax,]
mat[mymax,] <- vec
}
t <- t[-1]; temp <- temp[t]
for (j in temp) {mat[j,] <- (mat[j,]+mat[mymax,])%%2L}
mymax <- mymax+1L
}
}
if (mymax<n2) {simpMat <- mat[-(mymax:n2),]} else {simpMat <- mat}
lenSimp <- nrow(simpMat)
if (is.null(lenSimp)) {lenSimp <- 0L}
mycols <- 1:n1
if (lenSimp>1L) {
## "Diagonalizing" Matrix
for (i in 1:lenSimp) {
if (all(simpMat[i,]==0L)) {simpMat <- simpMat[-i,]; next}
if (!simpMat[i,i]==1L) {
t <- min(which(simpMat[i,]==1L))
vec <- simpMat[,i]; tempCol <- mycols[i]
simpMat[,i] <- simpMat[,t]; mycols[i] <- mycols[t]
simpMat[,t] <- vec; mycols[t] <- tempCol
}
}
lenSimp <- nrow(simpMat); MyList <- vector("list",length=n1)
MyFree <- mycols[which((1:n1)>lenSimp)]; for (i in MyFree) {MyList[[i]] <- i}
if (is.null(lenSimp)) {lenSimp <- 0L}
if (lenSimp>1L) {
for (i in lenSimp:1L) {
t <- which(simpMat[i,]==1L)
if (length(t)==1L) {
simpMat[ ,t] <- 0L
MyList[[mycols[i]]] <- 0L
} else {
t1 <- t[t>i]
if (all(t1 > lenSimp)) {
MyList[[mycols[i]]] <- MyList[[mycols[t1[1]]]]
if (length(t1)>1) {
for (j in 2:length(t1)) {MyList[[mycols[i]]] <- c(MyList[[mycols[i]]], MyList[[mycols[t1[j]]]])}
}
}
else {
for (j in t1) {
if (length(MyList[[mycols[i]]])==0L) {MyList[[mycols[i]]] <- MyList[[mycols[j]]]}
else {
e1 <- which(MyList[[mycols[i]]]%in%MyList[[mycols[j]]])
if (length(e1)==0) {
MyList[[mycols[i]]] <- c(MyList[[mycols[i]]],MyList[[mycols[j]]])
} else {
e2 <- which(!MyList[[mycols[j]]]%in%MyList[[mycols[i]]])
MyList[[mycols[i]]] <- MyList[[mycols[i]]][-e1]
if (length(e2)>0L) {MyList[[mycols[i]]] <- c(MyList[[mycols[i]]], MyList[[mycols[j]]][e2])}
}
}
}
}
}
}
TheList <- lapply(MyList, function(x) {if (length(x)==0L) {0} else {x}})
list(TheList,MyFree)
} else {
list(NULL,NULL)
}
} else {
list(NULL,NULL)
}
}
GetFacs <- function(vec1, vec2, n) {
x <- mod.bigz(prod.bigz(vec1),n)
y <- mod.bigz(prod.bigz(vec2),n)
MyAns <- c(gcd.bigz(x-y,n),gcd.bigz(x+y,n))
MyAns[sort.list(asNumeric(MyAns))]
}
SolutionSearch <- function(mymat, M2, n, FB) {
colTest <- which(apply(mymat, 2, sum) == 0)
if (length(colTest) > 0) {solmat <- mymat[ ,-colTest]} else {solmat <- mymat}
if (length(nrow(solmat)) > 0) {
nullMat <- reduceMatrix(t(solmat %% 2L))
listSol <- nullMat[[1]]; freeVar <- nullMat[[2]]; LF <- length(freeVar)
} else {LF <- 0L}
if (LF > 0L) {
for (i in 2:min(10^8,(2^LF + 1L))) {
PosAns <- MyIntToBit(i, LF)
posVec <- sapply(listSol, function(x) {
t <- which(freeVar %in% x)
if (length(t)==0L) {
0
} else {
sum(PosAns[t])%%2L
}
})
ansVec <- which(posVec==1L)
if (length(ansVec)>0) {
if (length(ansVec) > 1L) {
myY <- apply(mymat[ansVec,],2,sum)
} else {
myY <- mymat[ansVec,]
}
if (sum(myY %% 2) < 1) {
myY <- as.integer(myY/2)
myY <- pow.bigz(FB,myY[-1])
temp <- GetFacs(M2[ansVec], myY, n)
if (!(1==temp[1]) && !(1==temp[2])) {
return(temp)
}
}
}
}
}
}
### Below is the main portion of the Quadratic Sieve
BegTime <- Sys.time(); MyNum <- as.bigz(MyN); DigCount <- nchar(as.character(MyN))
P <- PrimeSieve(10^5)
SqrtInt <- .mpfr2bigz(trunc(sqrt(mpfr(MyNum,sizeinbase(MyNum,b=2)+5L))))
if (DigCount < 24) {
DigSize <- c(4,10,15,20,23)
f_Pos <- c(0.5,0.25,0.15,0.1,0.05)
MSize <- c(5000,7000,10000,12500,15000)
if (fudge1==0L) {
LM1 <- lm(f_Pos ~ DigSize)
m1 <- summary(LM1)$coefficients[2,1]
b1 <- summary(LM1)$coefficients[1,1]
fudge1 <- DigCount*m1 + b1
}
if (LenB==0L) {
LM2 <- lm(MSize ~ DigSize)
m2 <- summary(LM2)$coefficients[2,1]
b2 <- summary(LM2)$coefficients[1,1]
LenB <- ceiling(DigCount*m2 + b2)
}
LimB <- trunc(exp((.5+fudge1)*sqrt(log(MyNum)*log(log(MyNum)))))
B <- P[P<=LimB]; B <- B[-1]
facBase <- P[which(sapply(B, function(x) ExpBySquaringBig(MyNum,(x-1)/2,x)==1L))+1L]
LenFBase <- length(facBase)+1L
} else if (DigCount < 67) {
## These values were obtained from "The Multiple Polynomial
## Quadratic Sieve" by Robert D. Silverman
DigSize <- c(24,30,36,42,48,54,60,66)
FBSize <- c(100,200,400,900,1200,2000,3000,4500)
MSize <- c(5,25,25,50,100,250,350,500)
LM1 <- loess(FBSize ~ DigSize)
LM2 <- loess(MSize ~ DigSize)
if (fudge1==0L) {
fudge1 <- -0.4
LimB <- trunc(exp((.5+fudge1)*sqrt(log(MyNum)*log(log(MyNum)))))
myTarget <- ceiling(predict(LM1, DigCount))
while (LimB < myTarget) {
LimB <- trunc(exp((.5+fudge1)*sqrt(log(MyNum)*log(log(MyNum)))))
fudge1 <- fudge1+0.001
}
B <- P[P<=LimB]; B <- B[-1]
facBase <- P[which(sapply(B, function(x) ExpBySquaringBig(MyNum,(x-1)/2,x)==1L))+1L]
LenFBase <- length(facBase)+1L
while (LenFBase < myTarget) {
fudge1 <- fudge1+0.005
LimB <- trunc(exp((.5+fudge1)*sqrt(log(MyNum)*log(log(MyNum)))))
myind <- which(P==max(B))+1L
myset <- tempP <- P[myind]
while (tempP < LimB) {
myind <- myind + 1L
tempP <- P[myind]
myset <- c(myset, tempP)
}
for (p in myset) {
t <- ExpBySquaringBig(MyNum,(p-1)/2,p)==1L
if (t) {facBase <- c(facBase,p)}
}
B <- c(B, myset)
LenFBase <- length(facBase)+1L
}
} else {
LimB <- trunc(exp((.5+fudge1)*sqrt(log(MyNum)*log(log(MyNum)))))
B <- P[P<=LimB]; B <- B[-1]
facBase <- P[which(sapply(B, function(x) ExpBySquaringBig(MyNum,(x-1)/2,x)==1L))+1L]
LenFBase <- length(facBase)+1L
}
if (LenB==0L) {LenB <- 1000*ceiling(predict(LM2, DigCount))}
} else {
return("The number you've entered is currently too big for this algorithm!!")
}
SieveDist <- lapply(facBase, function(x) TonelliShanks(MyNum,x))
SieveDist <- c(1L,SieveDist); SieveDist[[1]] <- c(SieveDist[[1]],1L); facBase <- c(2L,facBase)
Lower <- -LenB; Upper <- LenB; LenB2 <- 2*LenB+1L; MyInterval <- Lower:Upper
M <- MyInterval + SqrtInt ## Set that will be tested
SqrDiff <- matrix(sub.bigz(pow.bigz(M,2),MyNum),nrow=length(M),ncol=1L)
maxM <- max(MyInterval)
LnFB <- log(facBase)
## N.B. primo uses 0.735, as his siever
## is more efficient than the one employed here
if (fudge2==0L) {
if (DigCount < 8) {
fudge2 <- 0
} else if (DigCount < 12) {
fudge2 <- .7
} else if (DigCount < 20) {
fudge2 <- 1.3
} else {
fudge2 <- 1.6
}
}
TheCut <- log10(maxM*sqrt(2*asNumeric(MyNum)))*fudge2
myPrimes <- as.bigz(facBase)
CoolList <- SieveLists(LenFBase, facBase, LenB2, SieveDist, MyInterval)
GetMatrix <- SieverMod(LenFBase, facBase, LenB2, SqrDiff, M, CoolList, LnFB, TheCut, LenFBase+1L)
if (GetMatrix[[4]]) {
newmat <- GetMatrix[[1]]; NewSD <- GetMatrix[[2]]; M <- GetMatrix[[3]]
NonSplitFacs <- which(abs(NewSD[,1L])>1L)
newmat <- newmat[-NonSplitFacs, ]
M <- M[-NonSplitFacs]
lenM <- length(M)
if (class(newmat) == "matrix") {
if (nrow(newmat) > 0) {
PosAns <- SolutionSearch(newmat,M,MyNum,myPrimes)
} else {
PosAns <- vector()
}
} else {
newmat <- matrix(newmat, nrow = 1)
PosAns <- vector()
}
} else {
newmat <- matrix(integer(0),ncol=(LenFBase+1L))
PosAns <- vector()
}
Atemp <- .mpfr2bigz(trunc(sqrt(sqrt(mpfr(2*MyNum))/maxM)))
if (Atemp < max(facBase)) {Atemp <- max(facBase)}; myPoly <- 0L
while (length(PosAns)==0L) {LegTest <- TRUE
while (LegTest) {
Atemp <- nextprime(Atemp)
Legendre <- asNumeric(ExpBySquaringBig(MyNum,(Atemp-1L)/2,Atemp))
if (Legendre == 1) {LegTest <- FALSE}
}
A <- Atemp^2
Btemp <- max(TonelliShanks(MyNum, Atemp))
B2 <- (Btemp + (MyNum - Btemp^2) * inv.bigz(2*Btemp,Atemp))%%A
C <- as.bigz((B2^2 - MyNum)/A)
myPoly <- myPoly + 1L
polySieveD <- lapply(1:LenFBase, function(x) {
AInv <- inv.bigz(A,facBase[x])
asNumeric(c(((SieveDist[[x]][1]-B2)*AInv)%%facBase[x],
((SieveDist[[x]][2]-B2)*AInv)%%facBase[x]))
})
M1 <- A*MyInterval + B2
SqrDiff <- matrix(A*pow.bigz(MyInterval,2) + 2*B2*MyInterval + C,nrow=length(M1),ncol=1L)
CoolList <- SieveLists(LenFBase, facBase, LenB2, polySieveD, MyInterval)
myPrimes <- c(myPrimes,Atemp)
LenP <- length(myPrimes)
GetMatrix <- SieverMod(LenFBase, facBase, LenB2, SqrDiff, M1, CoolList, LnFB, TheCut, LenP+1L)
if (GetMatrix[[4]]) {
n2mat <- GetMatrix[[1]]; N2SD <- GetMatrix[[2]]; M1 <- GetMatrix[[3]]
n2mat[,LenP+1L] <- rep(2L,nrow(N2SD))
if (length(N2SD) > 0) {NonSplitFacs <- which(abs(N2SD[,1L])>1L)} else {NonSplitFacs <- LenB2}
if (length(NonSplitFacs)<2*LenB) {
M1 <- M1[-NonSplitFacs]; lenM1 <- length(M1)
n2mat <- n2mat[-NonSplitFacs,]
if (lenM1==1L) {n2mat <- matrix(n2mat,nrow=1)}
if (ncol(newmat) < (LenP+1L)) {
numCol <- (LenP + 1L) - ncol(newmat)
newmat <- cbind(newmat,matrix(rep(0L,numCol*nrow(newmat)),ncol=numCol))
}
newmat <- rbind(newmat,n2mat); lenM <- lenM+lenM1; M <- c(M,M1)
if (class(newmat) == "matrix") {
if (nrow(newmat) > 0) {
PosAns <- SolutionSearch(newmat,M,MyNum,myPrimes)
}
}
}
}
}
EndTime <- Sys.time()
TotTime <- EndTime - BegTime
print(format(TotTime))
return(PosAns)
}
With Old QS algorithm
> library(gmp)
> library(Rmpfr)
> n3 <- prod(nextprime(urand.bigz(2, 40, 17)))
> system.time(t5 <- QuadSieveAll(n3,0.1,myps))
user system elapsed
164.72 0.77 165.63
> system.time(t6 <- factorize(n3))
user system elapsed
0.1 0.0 0.1
> all(t5[sort.list(asNumeric(t5))]==t6[sort.list(asNumeric(t6))])
[1] TRUE
With New Muli-Polynomial QS algorithm
> QuadSieveMultiPolysAll(n3)
[1] "4.952 secs"
Big Integer ('bigz') object of length 2:
[1] 342086446909 483830424611
> n4 <- prod(nextprime(urand.bigz(2,50,5)))
> QuadSieveMultiPolysAll(n4) ## With old algo, it took over 4 hours
[1] "1.131717 mins"
Big Integer ('bigz') object of length 2:
[1] 166543958545561 880194119571287
> n5 <- as.bigz("94968915845307373740134800567566911") ## 35 digits
> QuadSieveMultiPolysAll(n5)
[1] "3.813167 mins"
Big Integer ('bigz') object of length 2:
[1] 216366620575959221 438925910071081891
> system.time(factorize(n5)) ## It appears we are reaching the limits of factorize
user system elapsed
131.97 0.00 131.98
Side note: The number n5 above is a very interesting number. Check it out here
The Breaking Point!!!!
> n6 <- factorialZ(38) + 1L ## 45 digits
> QuadSieveMultiPolysAll(n6)
[1] "22.79092 mins"
Big Integer ('bigz') object of length 2:
[1] 14029308060317546154181 37280713718589679646221
> system.time(factorize(n6)) ## Shut it down after 2 days of running
Latest Triumph (50 digits)
> n9 <- prod(nextprime(urand.bigz(2,82,42)))
> QuadSieveMultiPolysAll(n9)
[1] "12.9297 hours"
Big Integer ('bigz') object of length 2:
[1] 2128750292720207278230259 4721136619794898059404993
## Based off of some crude test, factorize(n9) would take more than a year.
It should be noted that the QS generally doesn't perform as well as the Pollard's rho algorithm on smaller numbers and the power of the QS starts to become apparent as the numbers get larger.

The following approach deliver correct results, even in cases of really big numbers (which should be passed as strings). And it's really fast.
# TEST
# x <- as.bigz("12345678987654321")
# all_divisors(x)
# all_divisors(x*x)
# x <- pow.bigz(2,89)-1
# all_divisors(x)
library(gmp)
options(scipen =30)
sort_listz <- function(z) {
#==========================
z <- z[order(as.numeric(z))] # sort(z)
} # function sort_listz
mult_listz <- function(x,y) {
do.call('c', lapply(y, function(i) i*x))
}
all_divisors <- function(x) {
#==========================
if (abs(x)<=1) return(x)
else {
factorsz <- as.bigz(factorize(as.bigz(x))) # factorize returns up to
# e.g. x= 12345678987654321 factors: 3 3 3 3 37 37 333667 333667
factorsz <- sort_listz(factorsz) # vector of primes, sorted
prime_factorsz <- unique(factorsz)
#prime_ekt <- sapply(prime_factorsz, function(i) length( factorsz [factorsz==i]))
prime_ekt <- vapply(prime_factorsz, function(i) sum(factorsz==i), integer(1), USE.NAMES=FALSE)
spz <- vector() # keep all divisors
all <-1
n <- length(prime_factorsz)
for (i in 1:n) {
pr <- prime_factorsz[i]
pe <- prime_ekt[i]
all <- all*(pe+1) #counts all divisors
prz <- as.bigz(pr)
pse <- vector(mode="raw",length=pe+1)
pse <- c( as.bigz(1), prz)
if (pe>1) {
for (k in 2:pe) {
prz <- prz*pr
pse[k+1] <- prz
} # for k
} # if pe>1
if (i>1) {
spz <- mult_listz (spz, pse)
} else {
spz <- pse;
} # if i>1
} #for n
spz <- sort_listz (spz)
return (spz)
}
} # function factors_all_divisors
#====================================
Refined version, very fast. Code remains simple, readable & clean.
TEST
#Test 4 (big prime factor)
x <- pow.bigz(2,256)+1 # = 1238926361552897 * 93461639715357977769163558199606896584051237541638188580280321
system.time(z2 <- all_divisors(x))
# user system elapsed
# 19.27 1.27 20.56
#Test 5 (big prime factor)
x <- as.bigz("12345678987654321321") # = 3 * 19 * 216590859432531953
system.time(x2 <- all_divisors(x^2))
#user system elapsed
#25.65 0.00 25.67

Major Update
Below is my latest R factorization algorithm. It is way faster and pays homage to the rle function.
Algorithm 3 (Updated)
library(gmp)
MyFactors <- function(MyN) {
myRle <- function (x1) {
n1 <- length(x1)
y1 <- x1[-1L] != x1[-n1]
i <- c(which(y1), n1)
list(lengths = diff(c(0L, i)), values = x1[i], uni = sum(y1)+1L)
}
if (MyN==1L) return(MyN)
else {
pfacs <- myRle(factorize(MyN))
unip <- pfacs$values
pv <- pfacs$lengths
n <- pfacs$uni
myf <- unip[1L]^(0L:pv[1L])
if (n > 1L) {
for (j in 2L:n) {
myf <- c(myf, do.call(c,lapply(unip[j]^(1L:pv[j]), function(x) x*myf)))
}
}
}
myf[order(asNumeric(myf))] ## 'order' is faster than 'sort.list'
}
Below are the new benchmarks (As Dirk Eddelbuettel says here, "Can't argue with empirics."):
Case 1 (large prime factors)
set.seed(100)
myList <- lapply(1:10^3, function(x) sample(10^6, 10^5))
benchmark(SortList=lapply(myList, function(x) sort.list(x)),
OrderFun=lapply(myList, function(x) order(x)),
replications=3,
columns = c("test", "replications", "elapsed", "relative"))
test replications elapsed relative
2 OrderFun 3 59.41 1.000
1 SortList 3 61.52 1.036
## The times are limited by "gmp::factorize" and since it relies on
## pseudo-random numbers, the times can vary (i.e. one pseudo random
## number may lead to a factorization faster than others). With this
## in mind, any differences less than a half of second
## (or so) should be viewed as the same.
x <- pow.bigz(2,256)+1
system.time(z1 <- MyFactors(x))
user system elapsed
14.94 0.00 14.94
system.time(z2 <- all_divisors(x)) ## system.time(factorize(x))
user system elapsed ## user system elapsed
14.94 0.00 14.96 ## 14.94 0.00 14.94
all(z1==z2)
[1] TRUE
x <- as.bigz("12345678987654321321")
system.time(x1 <- MyFactors(x^2))
user system elapsed
20.66 0.02 20.71
system.time(x2 <- all_divisors(x^2)) ## system.time(factorize(x^2))
user system elapsed ## user system elapsed
20.69 0.00 20.69 ## 20.67 0.00 20.67
all(x1==x2)
[1] TRUE
Case 2 (smaller numbers)
set.seed(199)
samp <- sample(10^9, 10^5)
benchmark(JosephDivs=sapply(samp, MyFactors),
DontasDivs=sapply(samp, all_divisors),
OldDontas=sapply(samp, Oldall_divisors),
replications=10,
columns = c("test", "replications", "elapsed", "relative"),
order = "relative")
test replications elapsed relative
1 JosephDivs 10 470.31 1.000
2 DontasDivs 10 567.10 1.206 ## with vapply(..., USE.NAMES = FALSE)
3 OldDontas 10 626.19 1.331 ## with sapply
Case 3 (for complete thoroughness)
set.seed(97)
samp <- sample(10^6, 10^4)
benchmark(JosephDivs=sapply(samp, MyFactors),
DontasDivs=sapply(samp, all_divisors),
CottonDivs=sapply(samp, get_all_factors),
ChaseDivs=sapply(samp, FUN),
replications=5,
columns = c("test", "replications", "elapsed", "relative"),
order = "relative")
test replications elapsed relative
1 JosephDivs 5 22.68 1.000
2 DontasDivs 5 27.66 1.220
3 CottonDivs 5 126.66 5.585
4 ChaseDivs 5 554.25 24.438
Original Post
The algorithm by #RichieCotton is a very nice R implementation. The brute force method will only get you so far and fails with large numbers. I have provided three algorithms that will meet different needs. The first one (is the original algorithm I posted in Jan 15 and has been updated slightly), is a stand-alone factorization algorithm which offers a combinatorial approach that is efficient, accurate, and can be easily translated into other languages. The second algorithm is more of a sieve that is very fast and extremely useful when you need the factorization of thousands of numbers quickly. The third is a short (posted above), yet powerful stand-alone algorithm that is superior for any number less than 2^70 (I scrapped almost everything from my original code). I drew inspiration from Richie Cotton's use of the plyr::count function (it inspired me to write my own rle function that has a very similar return as plyr::count), George Dontas's clean way of handling the trivial case (i.e. if (n==1) return(1)), and the solution provided by #Zelazny7 to a question I had regarding bigz vectors.
Algorithm 1 (original)
library(gmp)
factor2 <- function(MyN) {
if (MyN == 1) return(1L)
else {
max_p_div <- factorize(MyN)
prime_vec <- max_p_div <- max_p_div[sort.list(asNumeric(max_p_div))]
my_factors <- powers <- as.bigz(vector())
uni_p <- unique(prime_vec); maxp <- max(prime_vec)
for (i in 1:length(uni_p)) {
temp_size <- length(which(prime_vec == uni_p[i]))
powers <- c(powers, pow.bigz(uni_p[i], 1:temp_size))
}
my_factors <- c(as.bigz(1L), my_factors, powers)
temp_facs <- powers; r <- 2L
temp_facs2 <- max_p_div2 <- as.bigz(vector())
while (r <= length(uni_p)) {
for (i in 1:length(temp_facs)) {
a <- which(prime_vec > max_p_div[i])
temp <- mul.bigz(temp_facs[i], powers[a])
temp_facs2 <- c(temp_facs2, temp)
max_p_div2 <- c(max_p_div2, prime_vec[a])
}
my_sort <- sort.list(asNumeric(max_p_div2))
temp_facs <- temp_facs2[my_sort]
max_p_div <- max_p_div2[my_sort]
my_factors <- c(my_factors, temp_facs)
temp_facs2 <- max_p_div2 <- as.bigz(vector()); r <- r+1L
}
}
my_factors[sort.list(asNumeric(my_factors))]
}
Algorithm 2 (sieve)
EfficientFactorList <- function(n) {
MyFactsList <- lapply(1:n, function(x) 1)
for (j in 2:n) {
for (r in seq.int(j, n, j)) {MyFactsList[[r]] <- c(MyFactsList[[r]], j)}
}; MyFactsList}
It gives the factorization of every number between 1 and 100,000 in less than 2 seconds. To give you an idea of the efficiency of this algorithm, the time to factor 1 - 100,000 using the brute force method takes close to 3 minutes.
system.time(t1 <- EfficientFactorList(10^5))
user system elapsed
1.04 0.00 1.05
system.time(t2 <- sapply(1:10^5, MyFactors))
user system elapsed
39.21 0.00 39.23
system.time(t3 <- sapply(1:10^5, all_divisors))
user system elapsed
49.03 0.02 49.05
TheTest <- sapply(1:10^5, function(x) all(t2[[x]]==t3[[x]]) && all(asNumeric(t2[[x]])==t1[[x]]) && all(asNumeric(t3[[x]])==t1[[x]]))
all(TheTest)
[1] TRUE
Final Thoughts
#Dontas’s original comment about factoring large numbers got me thinking, what about really really large numbers… like numbers greater than 2^200. You will see that whichever algorithm you choose on this page, they will all take a very long time because most of them rely on gmp::factorize which uses the Pollard-Rho algorithm. From this question, this algorithm is only reasonable for numbers less than 2^70. I am currently working on my own factorize algorithm which will implement the Quadratic Sieve, which should take all of these algorithms to the next level.

A lot has changed in the R language since this question was originally asked. In version 0.6-3 of the numbers package, the function divisors was included that is very useful for getting all of the factors of a number. It will meet the needs of most users, however if you are looking for raw speed or you are working with larger numbers, you will need an alternative method. I have authored two new packages (partially inspired by this question, I might add) that contain highly optimized functions aimed at problems just like this. The first one is RcppAlgos and the other is RcppBigIntAlgos (formerly called bigIntegerAlgos).
RcppAlgos
RcppAlgos contains two functions for obtaining divisors of numbers less than 2^53 - 1 : divisorsRcpp (a vectorized function for quickly obtaining the complete factorization of many numbers) & divisorsSieve (quickly generates the complete factorization over a range). First up, we factor many random numbers using divisorsRcpp:
library(gmp) ## for all_divisors by #GeorgeDontas
library(RcppAlgos)
library(numbers)
options(scipen = 999)
set.seed(42)
testSamp <- sample(10^10, 10)
## vectorized so you can pass the entire vector as an argument
testRcpp <- divisorsRcpp(testSamp)
testDontas <- lapply(testSamp, all_divisors)
identical(lapply(testDontas, as.numeric), testRcpp)
[1] TRUE
And now, factor many numbers over a range using divisorsSieve:
system.time(testSieve <- divisorsSieve(10^13, 10^13 + 10^5))
user system elapsed
0.242 0.006 0.247
system.time(testDontasSieve <- lapply((10^13):(10^13 + 10^5), all_divisors))
user system elapsed
47.880 0.132 47.922
identical(lapply(testDontasSieve, asNumeric), testSieve)
[1] TRUE
Both divisorsRcpp and divisorsSieve are nice functions that are flexible and efficient, however they are limited to 2^53 - 1.
RcppBigIntAlgos
The RcppBigIntAlgos package (formerly called bigIntegerAlgos prior to version 0.2.0) links directly to the C library gmp and features divisorsBig which is designed for very large numbers.
library(RcppBigIntAlgos)
## testSamp is defined above... N.B. divisorsBig is not quite as
## efficient as divisorsRcpp. This is so because divisorsRcpp
## can take advantage of more efficient data types.
testBig <- divisorsBig(testSamp)
identical(testDontas, testBig)
[1] TRUE
And here are the benchmark as defined in my original post (N.B. MyFactors is replaced by divisorsRcpp and divisorsBig).
## Case 2
library(rbenchmark)
set.seed(199)
samp <- sample(10^9, 10^5)
benchmark(RcppAlgos=divisorsRcpp(samp),
RcppBigIntAlgos=divisorsBig(samp),
DontasDivs=lapply(samp, all_divisors),
replications=10,
columns = c("test", "replications", "elapsed", "relative"),
order = "relative")
test replications elapsed relative
1 RcppAlgos 10 5.236 1.000
2 RcppBigIntAlgos 10 12.846 2.453
3 DontasDivs 10 383.742 73.289
## Case 3
set.seed(97)
samp <- sample(10^6, 10^4)
benchmark(RcppAlgos=divisorsRcpp(samp),
RcppBigIntAlgos=divisorsBig(samp),
numbers=lapply(samp, divisors), ## From the numbers package
DontasDivs=lapply(samp, all_divisors),
CottonDivs=lapply(samp, get_all_factors),
ChaseDivs=lapply(samp, FUN),
replications=5,
columns = c("test", "replications", "elapsed", "relative"),
order = "relative")
test replications elapsed relative
1 RcppAlgos 5 0.083 1.000
2 RcppBigIntAlgos 5 0.265 3.193
3 numbers 5 12.913 155.578
4 DontasDivs 5 15.813 190.518
5 CottonDivs 5 60.745 731.867
6 ChaseDivs 5 299.520 3608.675
The next benchmarks demonstrate the true power of the underlying algorithm in the divisorsBig function. The number being factored is a power of 10, so the prime factoring step can almost be completely ignored (e.g. system.time(factorize(pow.bigz(10,30))) registers 0 on my machine). Thus, the difference in timing is due solely to how quickly the prime factors can be combined to produce all factors.
library(microbenchmark)
powTen <- pow.bigz(10, 30)
microbenchmark(divisorsBig(powTen), all_divisors(powTen), unit = "relative")
Unit: relative
expr min lq mean median uq max neval
divisorsBig(powTen) 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 100
all_divisors(powTen) 21.49849 21.27973 21.13085 20.63345 21.18834 20.38772 100
## Negative numbers show an even greater increase in efficiency
negPowTen <- powTen * -1
microbenchmark(divisorsBig(negPowTen), all_divisors(negPowTen), unit = "relative")
Unit: relative
expr min lq mean median uq max neval
divisorsBig(negPowTen) 1.00000 1.0000 1.0000 1.00000 1.00000 1.00000 100
all_divisors(negPowTen) 28.75275 28.1864 27.9335 27.57434 27.91376 30.16962 100
Very Large Numbers
With divisorsBig, obtaining the complete factorization with very large inputs is no problem. The algorithm dynamically adjusts based off of the input and applies different algorithms in different situations. We can also take advantage of multithreading if Lenstra's Elliptic Curve method or the Quadratic Sieve is utilized.
Here are some examples using n5 and n9 defined in this answer.
n5 <- as.bigz("94968915845307373740134800567566911")
system.time(print(divisorsBig(n5)))
Big Integer ('bigz') object of length 4:
[1] 1 216366620575959221 438925910071081891
[4] 94968915845307373740134800567566911
user system elapsed
0.162 0.003 0.164
n9 <- prod(nextprime(urand.bigz(2, 82, 42)))
system.time(print(divisorsBig(n9, nThreads=4)))
Big Integer ('bigz') object of length 4:
[1] 1
[2] 2128750292720207278230259
[3] 4721136619794898059404993
[4] 10050120961360479179164300841596861740399588283187
user system elapsed
1.776 0.011 0.757
Here is an example provided by #Dontas with one large prime and one smaller prime:
x <- pow.bigz(2, 256) + 1
divisorsBig(x, showStats=TRUE, nThreads=8)
Summary Statistics for Factoring:
115792089237316195423570985008687907853269984665640564039457584007913129639937
| Pollard Rho Time |
|--------------------|
| 479ms |
| Lenstra ECM Time | Number of Curves |
|--------------------|--------------------|
| 1s 870ms | 2584 |
| Total Time |
|--------------------|
| 2s 402ms |
Big Integer ('bigz') object of length 4:
[1] 1
[2] 1238926361552897
[3] 93461639715357977769163558199606896584051237541638188580280321
[4] 115792089237316195423570985008687907853269984665640564039457584007913129639937
Compare this to finding the prime factorization using gmp::factorize:
system.time(factorize(x))
user system elapsed
9.199 0.036 9.248
Lastly, here is an example with a large semiprime (N.B. since we know it's a semiprime, we skip the extended Pollard's rho algorithm as well as Lentra's elliptic curve method).
## https://members.loria.fr/PZimmermann/records/rsa.html
rsa79 <- as.bigz("7293469445285646172092483905177589838606665884410340391954917800303813280275279")
divisorsBig(rsa79, nThreads=8, showStats=TRUE, skipPolRho=T, skipECM=T)
Summary Statistics for Factoring:
7293469445285646172092483905177589838606665884410340391954917800303813280275279
| MPQS Time | Complete | Polynomials | Smooths | Partials |
|--------------------|----------|-------------|------------|------------|
| 2m 49s 174ms | 100 | 91221 | 5651 | 7096 |
| Mat Algebra Time | Mat Dimension |
|--------------------|--------------------|
| 14s 863ms | 12625 x 12747 |
| Total Time |
|--------------------|
| 3m 4s 754ms |
Big Integer ('bigz') object of length 4:
[1] 1
[2] 848184382919488993608481009313734808977
[3] 8598919753958678882400042972133646037727
[4] 7293469445285646172092483905177589838606665884410340391954917800303813280275279

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