### February 1, 2013

Before you can write the program, you have to figure out the rule that produces the sequence. The solution is the R sequence given by:

R1 = 1, Rn = Rn−1 + Sn−1

where Sn is the smallest positive integer not present in R0..n or S0..n−1.

At this point in his book Hofstadter is discussing the figure and ground of two sequences, similar to the foreground and background of an image. He characterizes the two sequences given above as a figure-figure sequence, with no ground, because the two sequences are complements of each other.

The program to calculate the R sequence is on the next page.

Pages: 1 2 3

### 20 Responses to “Hofstadter’s Sequence”

1. […] today’s Programming Praxis exercise, our goal is to write a program that generates the Hofstadter sequence. […]

```import Data.List

hofstadter = unfoldr (\(r:s:ss) -> Just (r, r+s:delete (r+s) ss)) [1..]
```
3. […] Question is from here. […]

4. Hamid said

Java solution here

5. Here is a java solution that only stores the Hofstader’s sequence instead of both sequences.

Vector<Integer> hof = new Vector<Integer>();

final int MAX = 20;
int nextS;
int hofCheckIndex;

nextS = 2;
hofCheckIndex = 1;

for(int i = 1; i < MAX; i++) fillHof();

System.out.println("Hof index: " + MAX + " Hof number: " + hof.lastElement());
}

public void fillHof(){
if(++nextS == hof.elementAt(hofCheckIndex)){
nextS++;
hofCheckIndex++;
}
}

6. Mike said

Python generator

```import itertools as it

hs = set()
h = 0
for s in it.filterfalse(hs.__contains__, it.count(1)):
h += s
yield h
```
7. dwjref said

C# solution

```public List Hofstadters(int n)
{
List results = new List(n);
int counter = 0;
while (results.Count < n)
if (results.Contains(++counter) == false)
results.Add(results.Count == 0 ? counter : results[results.Count - 1] + counter);

return results;
}
```
8. Richard Wagenknecht said

C# 2, not as elegant.

List sequenceNumbers = new List();
int i = 1;

while (i < 100)
{
if (sequenceNumbers.IndexOf(i + sequenceNumbers.LastOrDefault()) == -1)
{
++i;
}

while (sequenceNumbers.IndexOf(i) != -1)
{
++i;
}
}

9. Here is Python code with some optimization – we store available numbers (S sequence members) as tuples (min, max) and pops first available. This significantly reduces number of items stored in list.

```class Availables(object):
def __init__(self):
self.vals = []
self.max = 0

def append(self, low_value, high_value):
if self.max >= low_value:
raise Exception("Too low value")

if low_value > high_value:
return

self.vals.append((low_value, high_value))
self.max = high_value

def pop(self):
if len(self.vals) == 0:
return None

low, high = self.vals
if low == high:
del self.vals
else:
self.vals = low + 1, high

return low

def __len__(self):
return len(self.vals)

n = 1
avail = Availables()
while True:
yield n
if len(avail) == 0:
incr = n + 1
start = n + 2
else:
incr = avail.pop()
start = n + 1
avail.append(start, n + incr - 1)
n += incr

from itertools import islice
list(islice(hofstadter(), 100000, 100005)) # Get numbers from 100000-th to 100004-th

# Result is: [5028615179L, 5028715611L, 5028816044L, 5028916478L, 5029016913L], length of available number tuples list at 100004-th number is 99574
```
10. wwh said

Here’s my racket solution.

(define (hof length (current 1) (last 0) (s empty))
(cond
((> current length) “done”)
((equal? current 1) (printf “1~%”) (hof length (+ current 1)))
((equal? current 2) (printf “3~%”) (hof length (+ current 1)))
((equal? current 3) (printf “7~%”) (hof length (+ current 1) 7 (list 5 6)))
(else
(let ((this-hof (+ last (car s))))
(printf “~a~%” (+ last (car s)))
(hof
length
(+ current 1)
(+ last (car s))
(append (cdr s) (build-list (- this-hof last 1)
(lambda (x) (+ x last 1)))))))))

Even less pretty but faster C# version. Idea is to enumerate S elements and skip previously found R elements. Got lazy to optimize further. On my i7 finds 1M numbers in about 35ms.

```static IEnumerable<int> Hofstadter(int n) {
var init = new[] { 1, 3, 7 }; // first few elements are pre-set
var Rs = new Queue<int>(); // queue to store found elements of R
Rs.Enqueue(7); // pushing first value
var counter = 0;
var s = 5; // first unused element of S
var curR = 7;

while (true) {
if (counter < 3) { yield return init[counter++]; continue; }
var nextR = Rs.Dequeue();
do {
Rs.Enqueue(s + curR);
curR = s++ + curR;
if (counter++ >= n)
yield break;
yield return curR;
} while (s != nextR);
s++;
}
}
```
12. Jonathan said

Really interesting problem. I had a crack for a few hours (eheh…) and came up with a solution in Common Lisp on my blog.

The solution works for n = 5, but interestingly, it also works for n = 1 (where it finds one solution; 1), and n = 0 (where it finds no solutions).

13. Keith said

I hacked this out in Perl, and after the fact found it very similar to beardedone85’s Java solution
my \$n = shift;
my @R = (1); #Given
my \$S = 2; #Given
my \$r_index = 1;

for(my \$i = 1; \$i < \$n; \$i++){
#Step 1, first calculate the next R
push(@R, \$R[\$i-1] + \$S);
#Step 2, calculate next S
if(\$S == \$R[\$r_index] – 1){
\$S = \$R[\$r_index] + 1;
\$r_index++;
}
else{\$S++;}
}

return \@R;
}

14. Scott said

import java.util.Vector;

public static void main(String[] args){
Vector S = new Vector();
int N = 1000000, R = 2, n = 0;
int lastSkiped = 1;

while(N>n++){
System.out.println(S.lastElement()+” “+R);
if(R == S.get(lastSkiped)){
R++;
lastSkiped++;
}
}
System.out.println(S.lastElement()+” “+R);
}
}

Hofstadter mentions many sequences in his book. This particular one is also known as one of the “Hofstadter Figure-Figure sequences”.

16. Alcriss said

int sum = 0, shadow = 1, count = 2, ng = 1;

for (int i = 1; i <= count – ng; i++)
{
Console.Write(" " + sum + ", ");
if (i == count)
{
count = count + 1;
i = 0;
}
else if (sum == 150)
{
goto End;
}
ng = 0;
}
End:

17. Patrick Herrmann said

ffs = 1 : zipWith (+) ffs ffs’
where ffs’ = 2 : zipWith (+) ffs’ ffs”
ffs” = [2..] >>= twoOnes
twoOnes n = 2 : replicate n 1

18. Daniel said

An easy Python solution. Finds first n numbers in sequence by distributing every number in either fg or bg starting from i.

fg=; bg=; c=0; i=3; n=10

while(len(fg) c and fg[c]+bg[c] == i:
fg.append(i)
c += 1
else:
bg.append(i)
i += 1

The weird len(bg) > c comparison can be ommited if we give some more information to background (bg) and foreground (fg) since we can easily see that background is growing strictly faster:

fg=[1, 3]; bg=[2, 4]; c=1; i=5; n=10
while(len(fg) < n):
if fg[c]+bg[c] == i:
fg.append(i)
c += 1
else:
bg.append(i)
i += 1

19. Daniel said

Wow sorry for the double post but WP really mangled the code. I hope it’s better now.

An easy Python solution. Finds first n numbers in sequence by distributing every number in either fg or bg starting from i.

```fg=; bg=; c=0; i=3; n=10
while(len(fg) < n):
if len(bg) > c and fg[c]+bg[c] == i:
fg.append(i)
c += 1
else:
bg.append(i)
i += 1
```

The weird len(bg) > c comparison can be ommited if we give some more information to background (bg) and foreground (fg) since we can easily see that background is growing strictly faster:

```fg=[1, 3]; bg=[2, 4]; c=1; i=5; n=10
while(len(fg) < n):
if fg[c]+bg[c] == i:
fg.append(i)
c += 1
else:
bg.append(i)
i += 1
```