Ulam numbers
With the following initial conditions:
u1 = 1 u2 = 2
we define the nth Ulam number un as the number that is greater than un-1 and
uniquely written as a sum of two different Ulam numbers ui (<n) and uj (<n).
- Task
Write a function to generate the n-th Ulam number.
- References
FreeBASIC
<lang freebasic>redim as uinteger ulam(1 to 2) ulam(1) = 1 : ulam(2) = 2
function get_ulam( n as uinteger, ulam() as uinteger ) as uinteger
dim as uinteger nu = ubound(ulam), c, r, s, t, i, usr if n <= nu then return ulam(n) 'if we have already calculated this one, just return it 'otherwise, calculate it and all intermediate terms redim preserve ulam(1 to n) for t = nu+1 to n i = ulam(t-1) while true i += 1 : c = 0 for r = 1 to t-2 for s = r+1 to t-1 usr = ulam(s)+ulam(r) if usr > i then exit for 'efficiency if usr = i then if c = 1 then continue while c += 1 end if next s next r if c = 0 then continue while 'I'm not 100% sure this is even possible... ulam(t) = i exit while wend next t return ulam(n)
end function
for i as uinteger = 1 to 4
print 10^i, get_ulam(10^i, ulam())
next i </lang>
- Output:
10 18 100 690 1000 12294 10000 132788
Go
Version 1
<lang go>package main
import "fmt"
func ulam(n int) int {
ulams := []int{1, 2} set := map[int]bool{1: true, 2: true} i := 3 for { count := 0 for j := 0; j < len(ulams); j++ { _, ok := set[i-ulams[j]] if ok && ulams[j] != (i-ulams[j]) { count++ if count > 2 { break } } } if count == 2 { ulams = append(ulams, i) set[i] = true if len(ulams) == n { break } } i++ } return ulams[n-1]
}
func main() {
for n := 10; n <= 10000; n *= 10 { fmt.Println("The", n, "\bth Ulam number is", ulam(n)) }
}</lang>
- Output:
The 10th Ulam number is 18 The 100th Ulam number is 690 The 1000th Ulam number is 12294 The 10000th Ulam number is 132788
Version 2
The above version is reasonably efficient and runs in about 2.9 seconds on my machine (Intel Core i7-8565U). The following version, which builds up a sieve as it goes along, is (astonishingly) about 40 times faster!
Although not shown here, the 100,000th Ulam number (1,351,223) is computed in about 13.5 seconds. <lang go>package main
import (
"fmt" "time"
)
func ulam(n int) int {
ulams := []int{1, 2} sieve := []int{1, 1} u := 2 for len(ulams) < n { s := u + ulams[len(ulams)-2] t := s - len(sieve) for i := 0; i < t; i++ { sieve = append(sieve, 0) } for i := 1; i <= len(ulams)-1; i++ { v := u + ulams[i-1] - 1 sieve[v]++ } index := -1 for i, e := range sieve[u:] { if e == 1 { index = u + i break } } u = index + 1 ulams = append(ulams, u) } return ulams[n-1]
}
func commatize(n int) string {
s := fmt.Sprintf("%d", n) if n < 0 { s = s[1:] } le := len(s) for i := le - 3; i >= 1; i -= 3 { s = s[0:i] + "," + s[i:] } if n >= 0 { return s } return "-" + s
}
func main() {
start := time.Now() for n := 1; n <= 10000; n *= 10 { s := "th" if n == 1 { s = "st" } fmt.Println("The", commatize(n), "\b"+s+" Ulam number is", commatize(ulam(n))) } fmt.Println("\nTook", time.Since(start))
}</lang>
- Output:
The 1st Ulam number is 1 The 10th Ulam number is 18 The 100th Ulam number is 690 The 1,000th Ulam number is 12,294 The 10,000th Ulam number is 132,788 Took 74.45373ms
Version 3
This version is even quicker than Version 2 and reduces the time needed to calculate the 10,000th and 100,000th Ulam numbers to about 40 milliseconds and 3.25 seconds respectively.
As mentioned in the Wren version 3 example, you need to know how much memory to allocate in advance. <lang go>package main
import (
"fmt" "time"
)
func ulam(n int) int {
if n <= 2 { return n } const MAX = 1_352_000 list := make([]int, MAX+1) list[0], list[1] = 1, 2 sums := make([]byte, 2*MAX+1) sums[3] = 1 size := 2 var query int for { query = list[size-1] + 1 for { if sums[query] == 1 { for i := 0; i < size; i++ { sum := query + list[i] t := sums[sum] + 1 if t <= 2 { sums[sum] = t } } list[size] = query size++ break } query++ } if size >= n { break } } return query
}
func commatize(n int) string {
s := fmt.Sprintf("%d", n) if n < 0 { s = s[1:] } le := len(s) for i := le - 3; i >= 1; i -= 3 { s = s[0:i] + "," + s[i:] } if n >= 0 { return s } return "-" + s
}
func main() {
start := time.Now() for n := 10; n <= 100000; n *= 10 { fmt.Println("The", commatize(n), "\bth Ulam number is", commatize(ulam(n))) } fmt.Println("\nTook", time.Since(start))
}</lang>
- Output:
The 10th Ulam number is 18 The 100th Ulam number is 690 The 1,000th Ulam number is 12,294 The 10,000th Ulam number is 132,788 The 100,000th Ulam number is 1,351,223 Took 3.226255944s
Haskell
Lazy List
<lang haskell> import Data.List
ulam
:: Integral i => Int -> i
ulam 1 = 1 ulam 2 = 2 ulam n
| n > 2 = ulams !! (n-1)
ulams
:: Integral n => [n]
ulams = 1:2:(nexts [2,1]) nexts us = u: (nexts (u:us))
where n = length us [u] = head . filter isSingleton . group . sort $ [v | i <- [0 .. n-2], j <- [i+1 .. n-1] , let s = us !! i , let t = us !! j , let v = s+t , v > head us ]
isSingleton :: [a] -> Bool isSingleton as
| length as == 1 = True | otherwise = False</lang>
Julia
<lang julia>function nthUlam(n)
ulams = [1, 2] memoized = Set([1, 2]) i = 3 while true count = 0 for j in 1:length(ulams) if i - ulams[j] in memoized && ulams[j] != i - ulams[j] (count += 1) > 2 && break end end if count == 2 push!(ulams, i) push!(memoized, i) length(ulams) == n && break end i += 1 end return ulams[n]
end
nthUlam(5)
for n in [10, 100, 1000, 10000]
@time println("The ", n, "th Ulam number is: ", nthUlam(n))
end
</lang>
- Output:
The 10th Ulam number is: 18 0.000657 seconds (27 allocations: 1.422 KiB) The 100th Ulam number is: 690 0.000959 seconds (39 allocations: 7.094 KiB) The 1000th Ulam number is: 12294 0.027564 seconds (52 allocations: 72.188 KiB) The 10000th Ulam number is: 132788 3.076024 seconds (63 allocations: 473.125 KiB)
Phix
<lang Phix>function ulam(integer n)
sequence ulams = {1, 2}, sieve = {1, 1} integer u := 2 while length(ulams)<n do integer s = u+ulams[$-1] sieve &= repeat(0,s-length(sieve)) for i=1 to length(ulams)-1 do s = u+ulams[i] t = sieve[s]+1 if t<=2 then sieve[s] = t end if end for u = find(1,sieve,u+1) ulams &= u end while return ulams[n]
end function
atom t0 = time() for p=0 to 4 do
integer n = power(10,p) printf(1,"The %,d%s Ulam number is %,d\n",{n,ord(n),ulam(n)})
end for ?elapsed(time()-t0)</lang>
- Output:
The 1st Ulam number is 1 The 10th Ulam number is 18 The 100th Ulam number is 690 The 1,000th Ulam number is 12,294 The 10,000th Ulam number is 132,788 "2.5s"
For comparison, Julia took 4.5s (9.3s on repl.it), Go took 4.9s, Wren (on tio) 27s,
Ring timed out (>60s) on tio before getting the 1,000th number,
REXX (also on tio) got to the 1,000th number in 12.3s but timed out before getting the 10,000th,
Raku (on repl.it) 9mins 50s,
FreeBASIC 17mins 44s, and I cancelled XPL0 (on EXPL32) after 53 minutes.
The Haskell entry does not compile for me on either tio or repl.it
The above algorithm can also yield "The 100,000th Ulam number is 1,351,223" in 1 minute and 40s, for me. (I fully expect translations of this better algorithm to run even faster, btw)
Raku
<lang perl6>my @ulams = 1, 2, &next-ulam … *;
sub next-ulam {
state $i = 1; state @sums = 0,1,1; my $last = @ulams[$i]; (^$i).map: { @sums[@ulams[$_] + $last]++ }; ++$i; quietly ($last ^.. *).first: { @sums[$_] == 1 };
}
for 1 .. 4 {
say "The {10**$_}th Ulam number is: ", @ulams[10**$_ - 1]
}</lang>
- Output:
The 10th Ulam number is: 18 The 100th Ulam number is: 690 The 1000th Ulam number is: 12294 The 10000th Ulam number is: 132788
REXX
This REXX version has several speed improvements. <lang rexx>/*REXX program finds and displays the Nth Ulam number (or any number of various numbers)*/ parse arg $ /*obtain optional argument from the CL.*/ if $= | $="," then $= 10 100 1000 10000 /*Not specified? Then use the defaults.*/
do k=1 for words($) x= Ulam( word($, k) ) /*define the first two Ulam numbers. */ say 'the ' commas(#)th(#) ' Ulam number is: ' commas(x) end /*k*/
exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg _; do jc=length(_)-3 to 1 by -3; _=insert(',', _, jc); end; return _ th: parse arg th; return word('th st nd rd', 1 + (th//10)*(th//100%10\==1)*(th//10<4)) /*──────────────────────────────────────────────────────────────────────────────────────*/ Ulam: parse arg n; @.1= 1; @.2=2; #= 2 /*1st two terms; #: sequence length. */
!.= 0; !.1= 1; !.2=1 /*semaphore for each term in sequence. */ z= 3 /*value of next possible term in seq. */ do until #==n cnt= 0 do j=1 for #; _= z - @.j /*_: short circuit value.*/ if !._ then if @.j\==_ then do; cnt= cnt + 1 if cnt>2 then leave end end /*j*/
if cnt==2 then do; #= # + 1 /*bump the number of terms*/ @.#= z; !.z= 1 /*add I to sequence; bool.*/ end z= z + 1 /*bump the next poss. term*/ end /*until*/ return @.#</lang>
- output when using the default input of: 10 100 1000 10000
the 10th Ulam number is: 18 the 100th Ulam number is: 690 the 1,000th Ulam number is: 12,294 the 10,000th Ulam number is: 132,788
- output (courtesy of Paul Kislanko's PC) when using the input of: 100000
the 100,000th Ulam number is: 1,351,223
Ring
<lang ring> load "stdlib.ring"
limit = 12500 Ulam = [] add(Ulam,1) add(Ulam,2)
for n = 3 to limit
flag = 0 count = 0 len = len(Ulam) for x = 1 to len-1 for y = x+1 to len if Ulam[x] + Ulam[y] = n flag = 1 count = count + 1 ok next next if flag = 1 and count = 1 add(Ulam,n) ln = len(Ulam) if ln = 10 see "The 10th Ulam number is: " + n + nl ok if ln = 100 see "The 100th Ulam number is: " + n + nl ok if ln = 1000 see "The 1000th Ulam number is: " + n + nl ok if ln = 10000 see "The 10000th Ulam number is: " + n + nl ok ok
next </lang> Output:
The 10th Ulam number is: 18 The 100th Ulam number is: 690 The 1000th Ulam number is: 12294 The 10000th Ulam number is: 132788
Wren
Version 1
<lang ecmascript>import "/set" for Set
var ulam = Fn.new() { |n|
var ulams = [1, 2] var set = Set.new(ulams) var i = 3 while (true) { var count = 0 for (j in 0...ulams.count) { if (set.contains(i - ulams[j]) && ulams[j] != (i - ulams[j])) { count = count + 1 if (count > 2) break } } if (count == 2) { ulams.add(i) set.add(i) if (ulams.count == n) break } i = i + 1 } return ulams[-1]
}
var n = 1 while (true) {
n = n * 10 System.print("The %(n)th Ulam number is %(ulam.call(n))") if (n == 10000) break
}</lang>
- Output:
The 10th Ulam number is 18 The 100th Ulam number is 690 The 1000th Ulam number is 12294 The 10000th Ulam number is 132788
Version 2
The above version is reasonably efficient and runs in about 21.6 seconds on my machine (Intel Core i7-8565U). The following version, which builds up a sieve as it goes along, is more than 3 times faster. <lang ecmascript>import "/seq" for Lst import "/fmt" for Fmt
var ulam = Fn.new { |n|
var ulams = [1, 2] var sieve = [1, 1] var u = 2 while (ulams.count < n) { var s = u + ulams[-2] sieve = sieve + ([0] * (s - sieve.count)) for (i in 1..ulams.count - 1) { var v = u + ulams[i-1] - 1 sieve[v] = sieve[v] + 1 } u = Lst.indexOf(sieve, 1, u) + 1 ulams.add(u) } return ulams[n-1]
}
var start = System.clock for (p in 0..4) {
var n = 10.pow(p) Fmt.print("The $,r Ulam number is $,d", n, ulam.call(n))
} System.print("\nTook %(System.clock - start) seconds.")</lang>
- Output:
The 1st Ulam number is 1 The 10th Ulam number is 18 The 100th Ulam number is 690 The 1,000th Ulam number is 12,294 The 10,000th Ulam number is 132,788 Took 6.366709 seconds.
Version 3
This version is even quicker than Version 2 and reduces the time needed to calculate the 10,000th Ulam number to about 3.65 seconds. It also makes the 100,000th Ulam number a viable proposition for the Wren interpreter coming in at about 6 minutes 50 seconds.
The only downside with this version is that you need to know how much memory to allocate in advance. <lang ecmascript>import "/fmt" for Fmt
var ulam = Fn.new { |n|
if (n <= 2) return n var max = 1352000 var list = List.filled(max+1, 0) list[0] = 1 list[1] = 2 var sums = List.filled(max*2+1, 0) sums[3] = 1 var size = 2 var query while (true) { query = list[size-1] + 1 while (true) { if (sums[query] == 1) { for (i in 0..size-1) { var sum = query + list[i] var t = sums[sum] + 1 if (t <= 2) sums[sum] = t } list[size] = query size = size + 1 break } query = query + 1 } if (size >= n) break } return query
}
var start = System.clock var n = 10 while (true) {
Fmt.print("The $,r Ulam number is $,d", n, ulam.call(n)) n = n * 10 if (n > 100000) break
} System.print("\nTook %(System.clock - start) seconds.")</lang>
- Output:
The 10th Ulam number is 18 The 100th Ulam number is 690 The 1,000th Ulam number is 12,294 The 10,000th Ulam number is 132,788 The 100,000th Ulam number is 1,351,223 Took 409.990502 seconds.
XPL0
Seeing "set" in the Go version and "sieve" in Phix, etc. lit a little light bulb. This version exploits those ideas and finds the 100,000th Ulam number in 24.7 seconds on a Pi4.
<lang XPL0>func Ulam(N); \Return Nth Ulam number int N; def Max = 1_352_000; \enough for 100_000th Ulam number int List(Max); \array of Ulam numbers char Sums(Max*2); \array: 0, 1, or more ways to sum Ulams int I, Size, Query, Sum, T; [if N <= 2 then return N; for I:= 0 to Max*2 do Sums(I):= 0; List(0):= 1; List(1):= 2; Sums(3):= 1; \only one way to sum Ulams: 1+2 = 3 Size:= 2; \start after first 2 Ulams repeat Query:= List(Size-1)+1; \possible next Ulam no.
loop [if Sums(Query) = 1 then \sums 1 way so it's next [for I:= 0 to Size-1 do \update Sums array with [Sum:= Query + List(I); \ all combos of sums T:= Sums(Sum)+1; \ but limit max count if T <= 2 then Sums(Sum):= T; ]; List(Size):= Query; \add Query to List Size:= Size+1; quit; ]; Query:= Query+1; \possible next Ulam no. ];
until Size >= N; return Query; ];
int N; [N:= 10; repeat Text(0, "The "); IntOut(0, N);
Text(0, "th Ulam number is "); IntOut(0, Ulam(N)); CrLf(0); N:= N*10;
until N > 100_000; ]</lang>
- Output:
The 10th Ulam number is 18 The 100th Ulam number is 690 The 1000th Ulam number is 12294 The 10000th Ulam number is 132788 The 100000th Ulam number is 1351223