Stern-Brocot sequence: Difference between revisions
ReeceGoding (talk | contribs) m →While loop: Improved syntax. |
Stern-Brocot sequence in various BASIC dialents |
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All GCDs are 1. |
All GCDs are 1. |
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</pre> |
</pre> |
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==={{header|BASIC256}}=== |
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{{trans|FreeBASIC}} |
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<lang basic256>arraybase 1 |
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max = 2000 |
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global stern |
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dim stern(max+2) |
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subroutine SternBrocot() |
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stern[1] = 1 |
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stern[2] = 1 |
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i = 2 : n = 2 : ub = stern[?] |
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while i < ub |
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i += 1 |
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stern[i] = stern[n] + stern[n-1] |
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i += 1 |
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stern[i] = stern[n] |
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n += 1 |
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end while |
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end subroutine |
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function gcd(x, y) |
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while y |
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t = y |
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y = x mod y |
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x = t |
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end while |
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gcd = x |
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end function |
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call SternBrocot() |
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print "The first 15 are: "; |
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for i = 1 to 15 |
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print stern[i]; " "; |
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next i |
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print : print |
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print "Index First nr." |
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d = 1 |
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for i = 1 to max |
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if stern[i] = d then |
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print i; chr(9); stern[i] |
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d += 1 |
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if d = 11 then d = 100 |
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if d = 101 then exit for |
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i = 0 |
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end if |
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next i |
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print : print |
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d = 0 |
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for i = 1 to 1000 |
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if gcd(stern[i], stern[i+1]) <> 1 then |
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d = gcd(stern[i], stern[i+1]) |
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exit for |
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end if |
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next i |
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if d = 0 then |
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print "GCD of two consecutive members of the series up to the 1000th member is 1" |
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else |
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print "The GCD for index "; i; " and "; i+1; " = "; d |
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end if</lang> |
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{{out}} |
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<pre>Igual que la entrada de FreeBASIC.</pre> |
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==={{header|QBasic}}=== |
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{{trans|FreeBASIC}} |
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{{works with|QBasic|1.1}} |
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<lang QBasic>CONST max = 2000 |
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DIM SHARED stern(max + 2) |
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FUNCTION gcd (x, y) |
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WHILE y |
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t = y |
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y = x MOD y |
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x = t |
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WEND |
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gcd = x |
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END FUNCTION |
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SUB SternBrocot |
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stern(1) = 1 |
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stern(2) = 1 |
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i = 2: n = 2: ub = UBOUND(stern) |
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DO WHILE i < ub |
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i = i + 1 |
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stern(i) = stern(n) + stern(n - 1) |
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i = i + 1 |
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stern(i) = stern(n) |
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n = n + 1 |
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LOOP |
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END SUB |
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SternBrocot |
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PRINT "The first 15 are: "; |
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FOR i = 1 TO 15 |
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PRINT stern(i); " "; |
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NEXT i |
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PRINT : PRINT |
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PRINT " Index First nr." |
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d = 1 |
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FOR i = 1 TO max |
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IF stern(i) = d THEN |
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PRINT USING " ######"; i; stern(i) |
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d = d + 1 |
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IF d = 11 THEN d = 100 |
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IF d = 101 THEN EXIT FOR |
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i = 0 |
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END IF |
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NEXT i |
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PRINT : PRINT |
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d = 0 |
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FOR i = 1 TO 1000 |
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IF gcd(stern(i), stern(i + 1)) <> 1 THEN |
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d = gcd(stern(i), stern(i + 1)) |
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EXIT FOR |
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END IF |
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NEXT i |
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IF d <> 0 THEN |
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PRINT "GCD of two consecutive members of the series up to the 1000th member is 1" |
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ELSE |
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PRINT "The GCD for index "; i; " and "; i + 1; " = "; d |
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END IF</lang> |
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{{out}} |
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<pre>Igual que la entrada de FreeBASIC.</pre> |
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==={{header|Yabasic}}=== |
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{{trans|FreeBASIC}} |
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<lang yabasic>limite = 2000 |
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dim stern(limite+2) |
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sub SternBrocot() |
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stern(1) = 1 |
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stern(2) = 1 |
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i = 2 : n = 2 : ub = arraysize(stern(),1) |
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while i < ub |
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i = i + 1 |
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stern(i) = stern(n) + stern(n -1) |
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i = i + 1 |
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stern(i) = stern(n) |
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n = n + 1 |
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wend |
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end sub |
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sub gcd(p, q) |
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if q = 0 return p |
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return gcd(q, mod(p, q)) |
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end sub |
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SternBrocot() |
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print "The first 15 are: "; |
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for i = 1 to 15 |
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print stern(i), " "; |
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next i |
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print "\n\n Index First nr." |
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d = 1 |
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for i = 1 to limite |
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if stern(i) = d then |
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print i using "######", stern(i) using "######" |
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d = d + 1 |
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if d = 11 d = 100 |
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if d = 101 break |
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i = 0 |
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end if |
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next i |
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print : print |
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d = 0 |
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for i = 1 to 1000 |
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if gcd(stern(i), stern(i+1)) <> 1 then |
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d = gcd(stern(i), stern(i+1)) |
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break |
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end if |
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next i |
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if d = 0 then |
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print "GCD of two consecutive members of the series up to the 1000th member is 1" |
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else |
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print "The GCD for index ", i, " and ", i+1, " = ", d |
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end if</lang> |
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{{out}} |
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<pre>Igual que la entrada de FreeBASIC.</pre> |
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=={{header|BCPL}}== |
=={{header|BCPL}}== |
Revision as of 00:40, 8 March 2022
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
For this task, the Stern-Brocot sequence is to be generated by an algorithm similar to that employed in generating the Fibonacci sequence.
- The first and second members of the sequence are both 1:
- 1, 1
- Start by considering the second member of the sequence
- Sum the considered member of the sequence and its precedent, (1 + 1) = 2, and append it to the end of the sequence:
- 1, 1, 2
- Append the considered member of the sequence to the end of the sequence:
- 1, 1, 2, 1
- Consider the next member of the series, (the third member i.e. 2)
- GOTO 3
- ─── Expanding another loop we get: ───
- Sum the considered member of the sequence and its precedent, (2 + 1) = 3, and append it to the end of the sequence:
- 1, 1, 2, 1, 3
- Append the considered member of the sequence to the end of the sequence:
- 1, 1, 2, 1, 3, 2
- Consider the next member of the series, (the fourth member i.e. 1)
- The task is to
- Create a function/method/subroutine/procedure/... to generate the Stern-Brocot sequence of integers using the method outlined above.
- Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4)
- Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence.
- Show the (1-based) index of where the number 100 first appears in the sequence.
- Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.
Show your output on this page.
- Related tasks
- Ref
- Infinite Fractions - Numberphile (Video).
- Trees, Teeth, and Time: The mathematics of clock making.
- A002487 The On-Line Encyclopedia of Integer Sequences.
11l
<lang 11l>F stern_brocot(predicate = series -> series.len < 20)
V sb = [1, 1] V i = 0 L predicate(sb) sb [+]= [sum(sb[i .< i + 2]), sb[i + 1]] i++ R sb
V n_first = 15 print(("The first #. values:\n ".format(n_first))‘ ’stern_brocot(series -> series.len < :n_first)[0 .< n_first]) print()
V n_max = 10 L(n_occur) Array(1 .. n_max) [+] [100]
print((‘1-based index of the first occurrence of #3 in the series:’.format(n_occur))‘ ’(stern_brocot(series -> @n_occur !C series).index(n_occur) + 1))
print()
V n_gcd = 1000 V s = stern_brocot(series -> series.len < :n_gcd)[0 .< n_gcd] assert(all(zip(s, s[1..]).map((prev, this) -> gcd(prev, this) == 1)), ‘A fraction from adjacent terms is reducible’)</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179
360 Assembly
<lang 360asm>* Stern-Brocot sequence - 12/03/2019 STERNBR CSECT
USING STERNBR,R13 base register B 72(R15) skip savearea DC 17F'0' savearea SAVE (14,12) save previous context ST R13,4(R15) link backward ST R15,8(R13) link forward LR R13,R15 set addressability LA R4,SB+2 k=2; @sb(k) LA R2,SB+2 i=1; @sb(k-i) LA R3,SB+0 j=2; @sb(k-j) LA R1,NN/2 loop counter
LOOP LA R4,2(R4) @sb(k)++
LH R0,0(R2) sb(k-i) AH R0,0(R3) sb(k-i)+sb(k-j) STH R0,0(R4) sb(k)=sb(k-i)+sb(k-j) LA R3,2(R3) @sb(k-j)++ LA R4,2(R4) @sb(k)++ LH R0,0(R3) sb(k-j) STH R0,0(R4) sb(k)=sb(k-j) LA R2,2(R2) @sb(k-i)++ BCT R1,LOOP end loop LA R9,15 n=15 MVC PG(5),=CL80'FIRST' XDECO R9,XDEC edit n MVC PG+5(3),XDEC+9 output n XPRNT PG,L'PG print buffer LA R10,PG @pg LA R6,1 i=1 DO WHILE=(CR,R6,LE,R9) do i=1 to n LR R1,R6 i SLA R1,1 ~ LH R2,SB-2(R1) sb(i) XDECO R2,XDEC edit sb(i) MVC 0(4,R10),XDEC+8 output sb(i) LA R10,4(R10) @pg+=4 LA R6,1(R6) i++ ENDDO , enddo i XPRNT PG,L'PG print buffer LA R7,1 j=1 DO WHILE=(C,R7,LE,=A(11)) do j=1 to 11 IF C,R7,EQ,=F'11' THEN if j=11 then LA R7,100 j=100 ENDIF , endif LA R6,1 i=1 DO WHILE=(C,R6,LE,=A(NN)) do i=1 to nn LR R1,R6 i SLA R1,1 ~ LH R2,SB-2(R1) sb(i) CR R2,R7 if sb(i)=j BE EXITI then leave i LA R6,1(R6) i++ ENDDO , enddo i
EXITI MVC PG,=CL80'FIRST INSTANCE OF'
XDECO R7,XDEC edit j MVC PG+17(4),XDEC+8 output j MVC PG+21(7),=C' IS AT ' XDECO R6,XDEC edit i MVC PG+28(4),XDEC+8 output i XPRNT PG,L'PG print buffer LA R7,1(R7) j++ ENDDO , enddo j L R13,4(0,R13) restore previous savearea pointer RETURN (14,12),RC=0 restore registers from calling sav LTORG
NN EQU 2400 nn PG DC CL80' ' buffer XDEC DS CL12 temp for xdeco SB DC (NN)H'1' sb(nn)
REGEQU END STERNBR</lang>
- Output:
FIRST 15 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 FIRST INSTANCE OF 1 IS AT 1 FIRST INSTANCE OF 2 IS AT 3 FIRST INSTANCE OF 3 IS AT 5 FIRST INSTANCE OF 4 IS AT 9 FIRST INSTANCE OF 5 IS AT 11 FIRST INSTANCE OF 6 IS AT 33 FIRST INSTANCE OF 7 IS AT 19 FIRST INSTANCE OF 8 IS AT 21 FIRST INSTANCE OF 9 IS AT 35 FIRST INSTANCE OF 10 IS AT 39 FIRST INSTANCE OF 100 IS AT 1179
The nice part is the coding of the sequense: <lang c> k=2; i=1; j=2;
while(k<nn); k++; sb[k]=sb[k-i]+sb[k-j]; k++; sb[k]=sb[k-j]; i++; j++; } </lang>
Only five registers are used. No Horner's rule to access sequence items. <lang 360asm> LA R4,SB+2 k=2; @sb(k)
LA R2,SB+2 i=1; @sb(k-i) LA R3,SB+0 j=2; @sb(k-j) LA R1,NN/2 k=nn/2 'loop counter
LOOP LA R4,2(R4) @sb(k)++
LH R0,0(R2) sb(k-i) AH R0,0(R3) sb(k-i)+sb(k-j) STH R0,0(R4) sb(k)=sb(k-i)+sb(k-j) LA R3,2(R3) @sb(k-j)++ LA R4,2(R4) @sb(k)++ LH R0,0(R3) sb(k-j) STH R0,0(R4) sb(k)=sb(k-j) LA R2,2(R2) @sb(k-i)++ BCT R1,LOOP k--; if k>0 then goto loop </lang>
8080 Assembly
<lang 8080asm>puts: equ 9 ; CP/M syscall to print a string org 100h ;;; Generate the first 1200 elements of the Stern-Brocot sequence lxi b,600 ; 2 elements generated per loop lxi h,seq mov e,m ; Initialization inx h push h ; Save considered member pointer inx h ; Insertion pointer genseq: xthl ; Load considered member pointer mov d,e ; D = predecessor mov e,m ; E = considered member inx h ; Point at next considered member xthl ; Load insertion pointer mov a,d ; A = sum of both members add e mov m,a ; Append the sum inx h mov m,e ; Append the considered member inx h dcx b ; Decrement counter mov a,b ; Zero? ora c jnz genseq ; If not, generate next members of sequence pop h ; Remove pointer from stack ;;; Print first 15 members of sequence lxi d,seq mvi b,15 ; 15 members mvi h,0 p15: ldax d ; Get current member mov l,a call prhl ; Print member inx d ; Increment pointer dcr b ; Decrement counter jnz p15 ; If not zero, print another one lxi d,nl mvi c,puts call 5 ;;; Print indices of first occurrence of 1..10 lxi b,010Ah ; B = number, C = counter call fnext ;;; Print index of first occurrence of 100 lxi b,6401h call fnext ;;; Check if the GCD of first 1000 consecutive elements is 0 xra a ; Zero out 1001th element as end marker sta seq+1000 lxi h,seq ; Start of array mov e,m inx h gcdchk: mov d,e ; (D,E) = next pair mov e,m inx h mov a,e mov b,d ana a ; Reached the end? jz done call gcd ; If not, check GCD dcr a ; Check that it is 1 jz gcdchk ; If so, check next pair push h ; GCD not 1 - save pointer lxi d,gcdn ; Print message mvi c,puts call 5 pop h ; Calculate offset in array lxi d,-seq dad d jmp prhl ; Print offset of pair whose GCD is not 1 done: lxi d,gcdy ; Print OK message mvi c,puts jmp 5 ;;; GCD(A,B) gcd: cmp b rz ; If A=B, result = A jc b_le_a ; B>A? sub b ; If A>B, subtract B jmp gcd ; and loop b_le_a: mov c,a mov a,b sub c mov b,a mov a,c jmp gcd ;;; Print indices of occurrences of C numbers ;;; starting at B fnext: lxi d,seq fsrch: ldax d ; Get current member cmp b ; Is it the number we are looking for? inx d ; Increment number jnz fsrch ; If no match, check next number lxi h,-seq ; Match - subtract start of array dad d call prhl ; Print index inr b ; Look for next number dcr c ; If we need more numbers jnz fnext push d ; Save sequence pointer lxi d,nl ; Print newline mvi c,puts call 5 pop d ; Restore sequence pointer ret ;;; Print HL as ASCII number. prhl: push h ; Save all registers push d push b lxi b,pnum ; Store pointer to num string on stack push b lxi b,-10 ; Divisor prdgt: lxi d,-1 prdgtl: inx d ; Divide by 10 through trial subtraction dad b jc prdgtl mvi a,'0'+10 add l ; L = remainder - 10 pop h ; Get pointer from stack dcx h ; Store digit mov m,a push h ; Put pointer back on stack xchg ; Put quotient in HL mov a,h ; Check if zero ora l jnz prdgt ; If not, next digit pop d ; Get pointer and put in DE mvi c,9 ; CP/M print string call 5 pop b ; Restore registers pop d pop h ret db '*****' ; Placeholder for number pnum: db ' $' nl: db 13,10,'$' gcdn: db 'GCD not 1 at: $' gcdy: db 'GCD of all pairs of consecutive members is 1.$' seq: db 1,1 ; Sequence stored here</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 3 5 9 11 33 19 21 35 39 1179 GCD of all pairs of consecutive members is 1.
8086 Assembly
<lang asm>puts: equ 9 cpu 8086 bits 16 org 100h section .text ;;; Generate the first 1200 elemets of the Stern-Brocot sequence mov cx,600 ; 2 elements generated per loop mov si,seq mov di,seq+2 lodsb mov ah,al ; AH = predecessor genseq: lodsb ; AL = considered member add ah,al ; AH = sum xchg ah,al ; Swap them (AL = sum, AH = member) stosw ; Write sum and current considered member loop genseq ; Loop 600 times ;;; Print first 15 members of sequence mov si,seq mov cx,15 p15: lodsb ; Get member cbw call prax ; Print member loop p15 call prnl ;;; Print first occurrences of [1..10] mov al,1 mov cx,10 call find call prnl ;;; Print first occurrence of 100 mov al,100 mov cx,1 call find call prnl ;;; Check pairs of GCDs mov cx,1000 ; 1000 times mov si,seq lodsb gcdchk: mov ah,al ; AH = last member, AL = current member lodsb mov dx,ax ; Calculate GCD call gcd dec dl ; See if it is 1 jnz .fail ; If not, failure loop gcdchk ; Otherwise, check next pair mov dx,gcdy ; GCD of all pairs is 0 jmp pstr .fail: mov dx,gcdn ; GCD of current pair is not 0 call pstr mov ax,si sub ax,seq+1 jmp prax ;;; DL = gcd(DL,DH) gcd: cmp dl,dh jl .lt ; DL < DH jg .gt ; DL > DH ret .lt: sub dh,dl ; DL < DH jmp gcd .gt: sub dl,dh ; DL > DH jmp gcd ;;; Print indices of CX consecutive numbers starting ;;; at AL. find: mov di,seq push cx ; Keep loop counter mov cx,-1 repne scasb ; Find AL starting at [DI] pop cx ; Restore loop counter xchg si,ax ; Keep AL in SI mov ax,di ; Calculate index in sequence sub ax,seq call prax ; Output xchg si,ax ; Restore AL inc ax ; Increment loop find ; Keep going CX times ret ;;; Print newline prnl: mov dx,nl jmp pstr ;;; Print number in AX ;;; Destroys AX, BX, DX, BP prax: mov bp,10 ; Divisor mov bx,numbuf .loop: xor dx,dx ; DX = 0 div bp ; Divide DX:AX by 10; DX = remainder dec bx ; Move string pointer back add dl,'0' ; Make ASCII digit mov [bx],dl ; Write digit test ax,ax ; Any digits left? jnz .loop mov dx,bx pstr: mov ah,puts ; Print number string int 21h ret section .data gcdn: db 'GCD not 1 at: $' gcdy: db 'GCD of all pairs of consecutive members is 1.$' db '*****' numbuf: db ' $' nl: db 13,10,'$' seq: db 1,1</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 3 5 9 11 33 19 21 35 39 1179 GCD of all pairs of consecutive members is 1.
Action!
<lang Action!>PROC Generate(BYTE ARRAY seq INT POINTER count INT minCount,maxVal)
INT i
seq(0)=1 seq(1)=1 count^=2 i=1 WHILE count^<minCount OR seq(count^-1)#maxVal AND seq(count^-2)#maxVal DO seq(count^)=seq(i-1)+seq(i) seq(count^+1)=seq(i) count^==+2 i==+1 OD
RETURN
PROC PrintSeq(BYTE ARRAY seq INT count)
INT i
PrintF("First %I items:%E",count) FOR i=0 TO count-1 DO PrintB(seq(i)) Put(32) OD PutE() PutE()
RETURN
PROC PrintLoc(BYTE ARRAY seq INT seqCount
BYTE ARRAY loc INT locCount) INT i,j BYTE value
FOR i=0 TO locCount-1 DO j=0 value=loc(i) WHILE seq(j)#value DO j==+1 OD PrintF("%B appears at position %I%E",value,j+1) OD PutE()
RETURN
BYTE FUNC Gcd(BYTE a,b)
BYTE tmp
IF a1 THEN PrintF("GCD between %I and %I item is greater than 1",i+1,i+2) RETURN FI OD Print("GCD between all two consecutive items of the sequence is equal 1")
RETURN
PROC Main()
BYTE ARRAY seq(2000),loc=[1 2 3 4 5 6 7 8 9 10 100] INT count
Generate(seq,@count,1000,100) PrintSeq(seq,15) PrintLoc(seq,count,loc,11) PrintGcd(seq,1000)
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
First 15 items: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 appears at position 1 2 appears at position 3 3 appears at position 5 4 appears at position 9 5 appears at position 11 6 appears at position 33 7 appears at position 19 8 appears at position 21 9 appears at position 35 10 appears at position 39 100 appears at position 1179 GCD between all two consecutive items of the sequence is equal 1
Ada
<lang Ada>with Ada.Text_IO, Ada.Containers.Vectors;
procedure Sequence is
package Vectors is new Ada.Containers.Vectors(Index_Type => Positive, Element_Type => Positive); use type Vectors.Vector; type Sequence is record List: Vectors.Vector; Index: Positive; -- This implements some form of "lazy evaluation": -- + List holds the elements computed, so far, it is extended -- if the user tries to "Get" an element not yet computed; -- + Index is the location of the next element under consideration end record; function Initialize return Sequence is (List => (Vectors.Empty_Vector & 1 & 1), Index => 2); function Get(Seq: in out Sequence; Location: Positive) return Positive is -- returns the Location'th element of the sequence -- extends Seq.List (and then increases Seq.Index) if neccessary That: Positive := Seq.List.Element(Seq.Index); This: Positive := That + Seq.List.Element(Seq.Index-1); begin while Seq.List.Last_Index < Location loop
Seq.List := Seq.List & This & That; Seq.Index := Seq.Index + 1;
end loop; return Seq.List.Element(Location); end Get; S: Sequence := Initialize; J: Positive; use Ada.Text_IO;
begin
-- show first fifteen members Put("First 15:"); for I in 1 .. 15 loop Put(Integer'Image(Get(S, I))); end loop; New_Line; -- show the index where 1, 2, 3, ... first appear in the sequence for I in 1 .. 10 loop J := 1; while Get(S, J) /= I loop
J := J + 1;
end loop; Put("First" & Integer'Image(I) & " at" & Integer'Image(J) & "; "); if I mod 4 = 0 then
New_Line; -- otherwise, the output gets a bit too ugly
end if; end loop; -- show the index where 100 first appears in the sequence J := 1; while Get(S, J) /= 100 loop J := J + 1; end loop; Put_Line("First 100 at" & Integer'Image(J) & "."); -- check GCDs declare function GCD (A, B : Integer) return Integer is
M : Integer := A; N : Integer := B; T : Integer;
begin
while N /= 0 loop T := M; M := N; N := T mod N; end loop; return M;
end GCD; A, B: Positive; begin for I in 1 .. 999 loop
A := Get(S, I); B := Get(S, I+1); if GCD(A, B) /= 1 then raise Constraint_Error; end if;
end loop; Put_Line("Correct: The first 999 consecutive pairs are relative prime!"); exception when Constraint_Error => Put_Line("Some GCD > 1; this is wrong!!!") ; end;
end Sequence;</lang>
- Output:
First 15: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1; First 2 at 3; First 3 at 5; First 4 at 9; First 5 at 11; First 6 at 33; First 7 at 19; First 8 at 21; First 9 at 35; First 10 at 39; First 100 at 1179. Correct: The first 999 consecutive pairs are relative prime!
ALGOL 68
<lang algol68>BEGIN # find members of the Stern-Brocot sequence: starting from 1, 1 the #
# subsequent members are the previous two members summed followed by # # the previous # # iterative Greatest Common Divisor routine, returns the gcd of m and n # PROC gcd = ( INT m, n )INT: BEGIN INT a := ABS m, b := ABS n; WHILE b /= 0 DO INT new a = b; b := a MOD b; a := new a OD; a END # gcd # ; # returns an array of the Stern-Brocot sequence up to n # OP STERNBROCOT = ( INT n )[]INT: BEGIN [ 1 : IF ODD n THEN n + 1 ELSE n FI ]INT result; IF n > 0 THEN result[ 1 ] := result[ 2 ] := 1; INT next pos := 2; FOR i FROM 3 WHILE next pos < n DO INT p1 = result[ i - 1 ]; result[ next pos +:= 1 ] := p1 + result[ i - 2 ]; result[ next pos +:= 1 ] := p1 OD FI; result[ 1 : n ] END # STERNPROCOT # ; FLEX[ 1 : 0 ]INT sb := STERNBROCOT 1000; # start with 1000 elements # # if that isn't enough, more will be added later # # show the first 15 elements of the sequence # print( ( "The first 15 elements of the Stern-Brocot sequence are:", newline ) ); FOR i TO 15 DO print( ( whole( sb[ i ], -3 ) ) ) OD; print( ( newline, newline ) ); # find where the numbers 1-10 first appear # INT found 10 := 0; [ 10 ]INT pos 10; FOR i TO UPB pos 10 DO pos 10[ i ] := 0 OD; FOR i TO UPB sb WHILE found 10 < 10 DO INT sb i = sb[ i ]; IF sb i <= UPB pos 10 THEN IF pos 10[ sb i ] = 0 THEN # first occurance of this number # pos 10[ sb i ] := i; found 10 +:= 1 FI FI OD; print( ( "The first positions of 1..", whole( UPB pos 10, 0 ), " in the sequence are:", newline ) ); FOR i TO UPB pos 10 DO print( ( whole( i, -2 ), ":", whole( pos 10[ i ], 0 ), " " ) ) OD; print( ( newline, newline ) ); # find where the number 100 first appears # BOOL found 100 := FALSE; FOR i WHILE NOT found 100 DO IF i > UPB sb THEN # need more elements # sb := STERNBROCOT ( UPB sb * 2 ) FI; IF sb[ i ] = 100 THEN print( ( "100 first appears at position ", whole( i, 0 ), newline, newline ) ); found 100 := TRUE FI OD; # check that the pairs of elements up to 1000 are coprime # BOOL all coprime := TRUE; FOR i FROM 2 TO 1000 WHILE all coprime DO all coprime := gcd( sb[ i ], sb[ i - 1 ] ) = 1 OD; print( ( "Element pairs up to 1000 are ", IF all coprime THEN "" ELSE "NOT " FI, "coprime", newline ) )
END</lang>
- Output:
The first 15 elements of the Stern-Brocot sequence are: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 The first positions of 1..10 in the sequence are: 1:1 2:3 3:5 4:9 5:11 6:33 7:19 8:21 9:35 10:39 100 first appears at position 1179 Element pairs up to 1000 are coprime
ALGOL-M
<lang algolm>begin integer array S[1:1200]; integer i,ok;
integer function gcd(a,b); integer a,b; gcd :=
if a>b then gcd(a-b,b) else if a<b then gcd(a,b-a) else a;
integer function first(n); integer n; begin
integer i; i := 1; while S[i]<>n do i := i + 1; first := i;
end;
S[1] := S[2] := 1; for i := 2 step 1 until 600 do begin
S[i*2-1] := S[i] + S[i-1]; S[i*2] := S[i];
end;
write("First 15 numbers:"); for i := 1 step 1 until 15 do begin
if i-i/5*5=1 then write(S[i]) else writeon(S[i]);
end;
write(""); write("First occurrence:"); for i := 1 step 1 until 10 do write(i, " at", first(i)); write(100, " at", first(100));
ok := 1; for i := 1 step 1 until 999 do begin
if gcd(S[i], S[i+1]) <> 1 then begin write("gcd",S[i],",",S[i+1],"<> 1"); ok := 0; end;
end; if ok = 1 then write("The GCD of each pair of consecutive members is 1."); end</lang>
- Output:
First 15 numbers: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First occurrence: 1 at 1 2 at 3 3 at 5 4 at 9 5 at 11 6 at 33 7 at 19 8 at 21 9 at 35 10 at 39 100 at 1179 The GCD of each pair of consecutive members is 1.
APL
<lang APL>task←{
stern←{⍵{ ⍺←0 ⋄ ⍺⍺≤⍴⍵:⍺⍺↑⍵ (⍺+1)∇⍵,(+/,2⊃⊣)2↑⍺↓⍵ }1 1} seq←stern 1200 ⍝ Cache 1200 elements ⎕←'First 15 elements:',15↑seq ⎕←'Locations of 1..10:',seq⍳⍳10 ⎕←'Location of 100:',seq⍳100 ⎕←'All GCDs 1:','no' 'yes'[1+1∧.=2∨/1000↑seq]
}</lang>
- Output:
First 15 elements: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 Locations of 1..10: 1 3 5 9 11 33 19 21 35 39 Location of 100: 1179 All GCDs 1: yes
AppleScript
<lang applescript>use AppleScript version "2.4" use framework "Foundation" use scripting additions
STERN-BROCOT SEQUENCE -----------------
-- sternBrocot :: Generator [Int] on sternBrocot()
script go on |λ|(xs) set x to snd(xs) tail(xs) & {fst(xs) + x, x} end |λ| end script fmapGen(my head, iterate(go, {1, 1}))
end sternBrocot
TEST -------------------------
on run
set sbs to take(1200, sternBrocot()) set ixSB to zip(sbs, enumFrom(1)) script low on |λ|(x) 12 ≠ fst(x) end |λ| end script script sameFst on |λ|(a, b) fst(a) = fst(b) end |λ| end script script asList on |λ|(x) {fst(x), snd(x)} end |λ| end script script below100 on |λ|(x) 100 ≠ fst(x) end |λ| end script script fullyReduced on |λ|(ab) 1 = gcd(|1| of ab, |2| of ab) end |λ| end script unlines(map(showJSON, ¬ {take(15, sbs), ¬ take(10, map(asList, ¬ nubBy(sameFst, ¬ sortBy(comparing(fst), ¬ takeWhile(low, ixSB))))), ¬ asList's |λ|(fst(take(1, dropWhile(below100, ixSB)))), ¬ all(fullyReduced, take(1000, zip(sbs, tail(sbs))))}))
end run
--> [1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] --> [[1,32],[2,24],[3,40],[4,36],[5,44],[6,33],[7,38],[8,42],[9,35],[10,39]] --> [100,1179] --> true
GENERIC ------------------------
-- Absolute value. -- abs :: Num -> Num on abs(x)
if 0 > x then -x else x end if
end abs
-- Applied to a predicate and a list, `all` determines if all elements
-- of the list satisfy the predicate.
-- all :: (a -> Bool) -> [a] -> Bool
on all(p, xs)
tell mReturn(p) set lng to length of xs repeat with i from 1 to lng if not |λ|(item i of xs, i, xs) then return false end repeat true end tell
end all
-- comparing :: (a -> b) -> (a -> a -> Ordering)
on comparing(f)
script on |λ|(a, b) tell mReturn(f) set fa to |λ|(a) set fb to |λ|(b) if fa < fb then -1 else if fa > fb then 1 else 0 end if end tell end |λ| end script
end comparing
-- drop :: Int -> [a] -> [a]
-- drop :: Int -> String -> String
on drop(n, xs)
set c to class of xs if c is not script then if c is not string then if n < length of xs then items (1 + n) thru -1 of xs else {} end if else if n < length of xs then text (1 + n) thru -1 of xs else "" end if end if else take(n, xs) -- consumed return xs end if
end drop
-- dropWhile :: (a -> Bool) -> [a] -> [a]
-- dropWhile :: (Char -> Bool) -> String -> String
on dropWhile(p, xs)
set lng to length of xs set i to 1 tell mReturn(p) repeat while i ≤ lng and |λ|(item i of xs) set i to i + 1 end repeat end tell drop(i - 1, xs)
end dropWhile
-- enumFrom :: a -> [a]
on enumFrom(x)
script property v : missing value property blnNum : class of x is not text on |λ|() if missing value is not v then if blnNum then set v to 1 + v else set v to succ(v) end if else set v to x end if return v end |λ| end script
end enumFrom
-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f) set lst to {} set lng to length of xs repeat with i from 1 to lng set v to item i of xs if |λ|(v, i, xs) then set end of lst to v end repeat return lst end tell
end filter
-- fmapGen <$> :: (a -> b) -> Gen [a] -> Gen [b]
on fmapGen(f, gen)
script property g : gen property mf : mReturn(f)'s |λ| on |λ|() set v to g's |λ|() if v is missing value then v else mf(v) end if end |λ| end script
end fmapGen
-- fst :: (a, b) -> a
on fst(tpl)
if class of tpl is record then |1| of tpl else item 1 of tpl end if
end fst
-- gcd :: Int -> Int -> Int
on gcd(a, b)
set x to abs(a) set y to abs(b) repeat until y = 0 if x > y then set x to x - y else set y to y - x end if end repeat return x
end gcd
-- head :: [a] -> a
on head(xs)
if xs = {} then missing value else item 1 of xs end if
end head
-- iterate :: (a -> a) -> a -> Gen [a]
on iterate(f, x)
script property v : missing value property g : mReturn(f)'s |λ| on |λ|() if missing value is v then set v to x else set v to g(v) end if return v end |λ| end script
end iterate
-- length :: [a] -> Int
on |length|(xs)
set c to class of xs if list is c or string is c then length of xs else (2 ^ 29 - 1) -- (maxInt - simple proxy for non-finite) end if
end |length|
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell
end map
-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then y else x end if
end min
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn
-- nubBy :: (a -> a -> Bool) -> [a] -> [a]
on nubBy(f, xs)
set g to mReturn(f)'s |λ| script notEq property fEq : g on |λ|(a) script on |λ|(b) not fEq(a, b) end |λ| end script end |λ| end script script go on |λ|(xs) if (length of xs) > 1 then set x to item 1 of xs {x} & go's |λ|(filter(notEq's |λ|(x), items 2 thru -1 of xs)) else xs end if end |λ| end script go's |λ|(xs)
end nubBy
-- partition :: predicate -> List -> (Matches, nonMatches)
-- partition :: (a -> Bool) -> [a] -> ([a], [a])
on partition(f, xs)
tell mReturn(f) set ys to {} set zs to {} repeat with x in xs set v to contents of x if |λ|(v) then set end of ys to v else set end of zs to v end if end repeat end tell Tuple(ys, zs)
end partition
-- showJSON :: a -> String
on showJSON(x)
set c to class of x if (c is list) or (c is record) then set ca to current application set {json, e} to ca's NSJSONSerialization's dataWithJSONObject:x options:0 |error|:(reference) if json is missing value then e's localizedDescription() as text else (ca's NSString's alloc()'s initWithData:json encoding:(ca's NSUTF8StringEncoding)) as text end if else if c is date then "\"" & ((x - (time to GMT)) as «class isot» as string) & ".000Z" & "\"" else if c is text then "\"" & x & "\"" else if (c is integer or c is real) then x as text else if c is class then "null" else try x as text on error ("«" & c as text) & "»" end try end if
end showJSON
-- snd :: (a, b) -> b
on snd(tpl)
if class of tpl is record then |2| of tpl else item 2 of tpl end if
end snd
-- Enough for small scale sorts.
-- Use instead sortOn :: Ord b => (a -> b) -> [a] -> [a]
-- which is equivalent to the more flexible sortBy(comparing(f), xs)
-- and uses a much faster ObjC NSArray sort method
-- sortBy :: (a -> a -> Ordering) -> [a] -> [a]
on sortBy(f, xs)
if length of xs > 1 then set h to item 1 of xs set f to mReturn(f) script on |λ|(x) f's |λ|(x, h) ≤ 0 end |λ| end script set lessMore to partition(result, rest of xs) sortBy(f, |1| of lessMore) & {h} & ¬ sortBy(f, |2| of lessMore) else xs end if
end sortBy
-- tail :: [a] -> [a]
on tail(xs)
set blnText to text is class of xs if blnText then set unit to "" else set unit to {} end if set lng to length of xs if 1 > lng then missing value else if 2 > lng then unit else if blnText then text 2 thru -1 of xs else rest of xs end if end if
end tail
-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
set c to class of xs if list is c then if 0 < n then items 1 thru min(n, length of xs) of xs else {} end if else if string is c then if 0 < n then text 1 thru min(n, length of xs) of xs else "" end if else if script is c then set ys to {} repeat with i from 1 to n set v to xs's |λ|() if missing value is v then return ys else set end of ys to v end if end repeat return ys else missing value end if
end take
-- takeWhile :: (a -> Bool) -> [a] -> [a]
-- takeWhile :: (Char -> Bool) -> String -> String
on takeWhile(p, xs)
if script is class of xs then takeWhileGen(p, xs) else tell mReturn(p) repeat with i from 1 to length of xs if not |λ|(item i of xs) then ¬ return take(i - 1, xs) end repeat end tell return xs end if
end takeWhile
-- takeWhileGen :: (a -> Bool) -> Gen [a] -> [a]
on takeWhileGen(p, xs)
set ys to {} set v to |λ|() of xs tell mReturn(p) repeat while (|λ|(v)) set end of ys to v set v to xs's |λ|() end repeat end tell return ys
end takeWhileGen
-- Tuple (,) :: a -> b -> (a, b)
on Tuple(a, b)
{type:"Tuple", |1|:a, |2|:b, length:2}
end Tuple
-- unlines :: [String] -> String
on unlines(xs)
set {dlm, my text item delimiters} to ¬ {my text item delimiters, linefeed} set str to xs as text set my text item delimiters to dlm str
end unlines
-- zip :: [a] -> [b] -> [(a, b)]
on zip(xs, ys)
zipWith(Tuple, xs, ys)
end zip
-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
on zipWith(f, xs, ys)
set lng to min(|length|(xs), |length|(ys)) if 1 > lng then return {} set xs_ to take(lng, xs) -- Allow for non-finite set ys_ to take(lng, ys) -- generators like cycle etc set lst to {} tell mReturn(f) repeat with i from 1 to lng set end of lst to |λ|(item i of xs_, item i of ys_) end repeat return lst end tell
end zipWith</lang>
- Output:
[1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] [[1,32],[2,24],[3,40],[4,36],[5,44],[6,33],[7,38],[8,42],[9,35],[10,39]] [100,1179] true
AutoHotkey
<lang AutoHotkey>Found := FindOneToX(100), FoundList := "" Loop, 10
FoundList .= "First " A_Index " found at " Found[A_Index] "`n"
MsgBox, 64, Stern-Brocot Sequence
, % "First 15: " FirstX(15) "`n" . FoundList . "First 100 found at " Found[100] "`n" . "GCDs of all two consecutive members are " (GCDsUpToXAreOne(1000) ? "" : "not ") "one."
return
class SternBrocot {
__New() { this[1] := 1 this[2] := 1 this.Consider := 2 } InsertPair() { n := this.Consider this.Push(this[n] + this[n - 1], this[n]) this.Consider++ }
}
- Show the first fifteen members of the sequence. (This should be
- 1, 1, 2, 1, 3, 2, 3, 1, 4, 3,
- 5, 2, 5, 3, 4)
FirstX(x) {
SB := new SternBrocot() while SB.MaxIndex() < x SB.InsertPair() Loop, % x Out .= SB[A_Index] ", " return RTrim(Out, " ,")
}
- Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence.
- Show the (1-based) index of where the number 100 first appears in the sequence.
FindOneToX(x) {
SB := new SternBrocot(), xRequired := x, Found := [] while xRequired > 0 ; While the count of numbers yet to be found is > 0. { Loop, 2 ; Consider the second last member and then the last member. { n := SB[i := SB.MaxIndex() - 2 + A_Index] ; If number (n) has not been found yet, and it is less than the maximum number to ; find (x), record the index (i) and decrement the count of numbers yet to be found. if (Found[n] = "" && n <= x) Found[n] := i, xRequired-- } SB.InsertPair() ; Insert the two members that will be checked next. } return Found
}
- Check that the greatest common divisor of all the two consecutive members of the series up to
- the 1000th member, is always one.
GCDsUpToXAreOne(x) {
SB := new SternBrocot() while SB.MaxIndex() < x SB.InsertPair() Loop, % x - 1 if GCD(SB[A_Index], SB[A_Index + 1]) > 1 return 0 return 1
}
GCD(a, b) {
while b b := Mod(a | 0x0, a := b) return a
}</lang>
- Output:
First 15: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 First 1 found at 1 First 2 found at 3 First 3 found at 5 First 4 found at 9 First 5 found at 11 First 6 found at 33 First 7 found at 19 First 8 found at 21 First 9 found at 35 First 10 found at 39 First 100 found at 1179 GCDs of all two consecutive members are one.
BASIC
<lang basic>10 DEFINT A,B,I,J,S: DIM S(1200) 20 S(1)=1: S(2)=1 30 FOR I=2 TO 600 40 S(I*2-1)=S(I)+S(I-1) 50 S(I*2)=S(I) 60 NEXT I 70 PRINT "First 15 elements: "; 80 FOR I=1 TO 15: PRINT USING"# ";S(I);: NEXT I 85 PRINT 90 FOR I=1 TO 10 100 FOR J=1 TO 1200: IF S(J)<>I THEN NEXT J 110 PRINT "First";I;"at";J 120 NEXT I 130 FOR J=1 TO 1200: IF S(J)<>100 THEN NEXT J 140 PRINT "First 100 at";J 150 FOR I=2 TO 1000 160 A=S(I): B=S(I-1) 170 J=A: A=B: B=J MOD A: IF B THEN 170 180 IF A<>1 THEN PRINT "GCD <> 1 at ";I: STOP 190 NEXT I 200 PRINT "All GCDs are 1." 210 END</lang>
- Output:
First 15 elements: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179 All GCDs are 1.
BASIC256
<lang basic256>arraybase 1 max = 2000 global stern dim stern(max+2)
subroutine SternBrocot() stern[1] = 1 stern[2] = 1
i = 2 : n = 2 : ub = stern[?]
while i < ub i += 1 stern[i] = stern[n] + stern[n-1] i += 1 stern[i] = stern[n] n += 1 end while end subroutine
function gcd(x, y) while y t = y y = x mod y x = t end while gcd = x end function
call SternBrocot()
print "The first 15 are: "; for i = 1 to 15 print stern[i]; " "; next i
print : print print "Index First nr." d = 1 for i = 1 to max if stern[i] = d then print i; chr(9); stern[i] d += 1 if d = 11 then d = 100 if d = 101 then exit for i = 0 end if next i
print : print d = 0 for i = 1 to 1000 if gcd(stern[i], stern[i+1]) <> 1 then d = gcd(stern[i], stern[i+1]) exit for end if next i
if d = 0 then print "GCD of two consecutive members of the series up to the 1000th member is 1" else print "The GCD for index "; i; " and "; i+1; " = "; d end if</lang>
- Output:
Igual que la entrada de FreeBASIC.
QBasic
<lang QBasic>CONST max = 2000 DIM SHARED stern(max + 2)
FUNCTION gcd (x, y)
WHILE y t = y y = x MOD y x = t WEND gcd = x
END FUNCTION
SUB SternBrocot
stern(1) = 1 stern(2) = 1 i = 2: n = 2: ub = UBOUND(stern) DO WHILE i < ub i = i + 1 stern(i) = stern(n) + stern(n - 1) i = i + 1 stern(i) = stern(n) n = n + 1 LOOP
END SUB
SternBrocot
PRINT "The first 15 are: "; FOR i = 1 TO 15
PRINT stern(i); " ";
NEXT i
PRINT : PRINT PRINT " Index First nr." d = 1 FOR i = 1 TO max
IF stern(i) = d THEN PRINT USING " ######"; i; stern(i) d = d + 1 IF d = 11 THEN d = 100 IF d = 101 THEN EXIT FOR i = 0 END IF
NEXT i
PRINT : PRINT d = 0 FOR i = 1 TO 1000
IF gcd(stern(i), stern(i + 1)) <> 1 THEN d = gcd(stern(i), stern(i + 1)) EXIT FOR END IF
NEXT i
IF d <> 0 THEN
PRINT "GCD of two consecutive members of the series up to the 1000th member is 1"
ELSE
PRINT "The GCD for index "; i; " and "; i + 1; " = "; d
END IF</lang>
- Output:
Igual que la entrada de FreeBASIC.
Yabasic
<lang yabasic>limite = 2000 dim stern(limite+2)
sub SternBrocot()
stern(1) = 1 stern(2) = 1 i = 2 : n = 2 : ub = arraysize(stern(),1) while i < ub i = i + 1 stern(i) = stern(n) + stern(n -1) i = i + 1 stern(i) = stern(n) n = n + 1 wend
end sub
sub gcd(p, q)
if q = 0 return p return gcd(q, mod(p, q))
end sub
SternBrocot()
print "The first 15 are: "; for i = 1 to 15
print stern(i), " ";
next i
print "\n\n Index First nr." d = 1 for i = 1 to limite
if stern(i) = d then print i using "######", stern(i) using "######" d = d + 1 if d = 11 d = 100 if d = 101 break i = 0 end if
next i
print : print d = 0 for i = 1 to 1000
if gcd(stern(i), stern(i+1)) <> 1 then d = gcd(stern(i), stern(i+1)) break end if
next i
if d = 0 then
print "GCD of two consecutive members of the series up to the 1000th member is 1"
else
print "The GCD for index ", i, " and ", i+1, " = ", d
end if</lang>
- Output:
Igual que la entrada de FreeBASIC.
BCPL
<lang bcpl>get "libhdr"
manifest $( AMOUNT = 1200 $)
let gcd(a,b) =
a>b -> gcd(a-b, b), a gcd(a, b-a), a
let mkstern(s, n) be $( s!1 := 1
s!2 := 1 for i=2 to n/2 do $( s!(i*2-1) := s!i + s!(i-1) s!(i*2) := s!i $)
$)
let find(v, n, max) = valof
for i=1 to max if v!i=n then resultis i
let findwrite(v, n, max) be
writef("%I3 at %I4*N", n, find(v, n, max))
let start() be $( let stern = vec AMOUNT
mkstern(stern, AMOUNT) writes("First 15 numbers: ") for i=1 to 15 do writef("%N ", stern!i) writes("*N*NFirst occurrence:*N") for i=1 to 10 do findwrite(stern, i, AMOUNT) findwrite(stern, 100, AMOUNT) if valof $( for i=2 to AMOUNT unless gcd(stern!i, stern!(i-1)) = 1 resultis false resultis true $) then writes("*NThe GCD of each pair of consecutive members is 1.*N")
$)</lang>
- Output:
First 15 numbers: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First occurrence: 1 at 1 2 at 3 3 at 5 4 at 9 5 at 11 6 at 33 7 at 19 8 at 21 9 at 35 10 at 39 100 at 1179 The GCD of each pair of consecutive members is 1.
C
Recursive function. <lang c>#include <stdio.h>
typedef unsigned int uint;
/* the sequence, 0-th member is 0 */ uint f(uint n) { return n < 2 ? n : (n&1) ? f(n/2) + f(n/2 + 1) : f(n/2); }
uint gcd(uint a, uint b) { return a ? a < b ? gcd(b%a, a) : gcd(a%b, b) : b; }
void find(uint from, uint to) { do { uint n; for (n = 1; f(n) != from ; n++); printf("%3u at Stern #%u.\n", from, n); } while (++from <= to); }
int main(void) { uint n; for (n = 1; n < 16; n++) printf("%u ", f(n)); puts("are the first fifteen.");
find(1, 10); find(100, 0);
for (n = 1; n < 1000 && gcd(f(n), f(n+1)) == 1; n++); printf(n == 1000 ? "All GCDs are 1.\n" : "GCD of #%d and #%d is not 1", n, n+1);
return 0; }</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 are the first fifteen. 1 at Stern #1. 2 at Stern #3. 3 at Stern #5. 4 at Stern #9. 5 at Stern #11. 6 at Stern #33. 7 at Stern #19. 8 at Stern #21. 9 at Stern #35. 10 at Stern #39. 100 at Stern #1179. All GCDs are 1.
The above f()
can be replaced by the following, which is much faster but probably less obvious as to how it arrives from the recurrence relations.
<lang c>uint f(uint n)
{
uint a = 1, b = 0;
while (n) {
if (n&1) b += a;
else a += b;
n >>= 1;
}
return b;
}</lang>
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq;
static class Program {
static List<int> l = new List<int>() { 1, 1 };
static int gcd(int a, int b) { return a > 0 ? a < b ? gcd(b % a, a) : gcd(a % b, b) : b; }
static void Main(string[] args) { int max = 1000; int take = 15; int i = 1; int[] selection = new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100 }; do { l.AddRange(new List<int>() { l[i] + l[i - 1], l[i] }); i += 1; } while (l.Count < max || l[l.Count - 2] != selection.Last()); Console.Write("The first {0} items In the Stern-Brocot sequence: ", take); Console.WriteLine("{0}\n", string.Join(", ", l.Take(take))); Console.WriteLine("The locations of where the selected numbers (1-to-10, & 100) first appear:"); foreach (int ii in selection) { int j = l.FindIndex(x => x == ii) + 1; Console.WriteLine("{0,3}: {1:n0}", ii, j); } Console.WriteLine(); bool good = true; for (i = 1; i <= max; i++) { if (gcd(l[i], l[i - 1]) != 1) { good = false; break; } } Console.WriteLine("The greatest common divisor of all the two consecutive items of the" + " series up to the {0}th item is {1}always one.", max, good ? "" : "not "); }
}</lang>
- Output:
The first 15 items In the Stern-Brocot sequence: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 The locations of where the selected numbers (1-to-10, & 100) first appear: 1: 1 2: 3 3: 5 4: 9 5: 11 6: 33 7: 19 8: 21 9: 35 10: 39 100: 1,179 The greatest common divisor of all the two consecutive items of the series up to the 1000th item is always one.
C++
<lang cpp>
- include <iostream>
- include <iomanip>
- include <algorithm>
- include <vector>
unsigned gcd( unsigned i, unsigned j ) {
return i ? i < j ? gcd( j % i, i ) : gcd( i % j, j ) : j;
} void createSequence( std::vector<unsigned>& seq, int c ) {
if( 1500 == seq.size() ) return; unsigned t = seq.at( c ) + seq.at( c + 1 ); seq.push_back( t ); seq.push_back( seq.at( c + 1 ) ); createSequence( seq, c + 1 );
} int main( int argc, char* argv[] ) {
std::vector<unsigned> seq( 2, 1 ); createSequence( seq, 0 );
std::cout << "First fifteen members of the sequence:\n "; for( unsigned x = 0; x < 15; x++ ) { std::cout << seq[x] << " "; }
std::cout << "\n\n"; for( unsigned x = 1; x < 11; x++ ) { std::vector<unsigned>::iterator i = std::find( seq.begin(), seq.end(), x ); if( i != seq.end() ) { std::cout << std::setw( 3 ) << x << " is at pos. #" << 1 + distance( seq.begin(), i ) << "\n"; } }
std::cout << "\n"; std::vector<unsigned>::iterator i = std::find( seq.begin(), seq.end(), 100 ); if( i != seq.end() ) { std::cout << 100 << " is at pos. #" << 1 + distance( seq.begin(), i ) << "\n"; }
std::cout << "\n"; unsigned g; bool f = false; for( int x = 0, y = 1; x < 1000; x++, y++ ) { g = gcd( seq[x], seq[y] );
if( g != 1 ) f = true;
std::cout << std::setw( 4 ) << x + 1 << ": GCD (" << seq[x] << ", " << seq[y] << ") = " << g << ( g != 1 ? " <-- ERROR\n" : "\n" ); } std::cout << "\n" << ( f ? "THERE WERE ERRORS --- NOT ALL GCDs ARE '1'!" : "CORRECT: ALL GCDs ARE '1'!" ) << "\n\n"; return 0;
} </lang>
- Output:
First fifteen members of the sequence: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 is at pos. #1 2 is at pos. #3 3 is at pos. #5 4 is at pos. #9 5 is at pos. #11 6 is at pos. #33 7 is at pos. #19 8 is at pos. #21 9 is at pos. #35 10 is at pos. #39 100 is at pos. #1179 1: GCD (1, 1) = 1 2: GCD (1, 2) = 1 3: GCD (2, 1) = 1 4: GCD (1, 3) = 1 5: GCD (3, 2) = 1 6: GCD (2, 3) = 1 7: GCD (3, 1) = 1 8: GCD (1, 4) = 1 [...] 993: GCD (26, 21) = 1 994: GCD (21, 37) = 1 995: GCD (37, 16) = 1 996: GCD (16, 43) = 1 997: GCD (43, 27) = 1 998: GCD (27, 38) = 1 999: GCD (38, 11) = 1 1000: GCD (11, 39) = 1 CORRECT: ALL GCDs ARE '1'!
Clojure
<lang clojure>;; each step adds two items (defn sb-step [v]
(let [i (quot (count v) 2)] (conj v (+ (v (dec i)) (v i)) (v i))))
- A lazy, infinite sequence -- `take` what you want.
(def all-sbs (sequence (map peek) (iterate sb-step [1 1])))
- zero-based
(defn first-appearance [n]
(first (keep-indexed (fn [i x] (when (= x n) i)) all-sbs)))
- inlined abs; rem is slightly faster than mod, and the same result for positive values
(defn gcd [a b]
(loop [a (if (neg? a) (- a) a) b (if (neg? b) (- b) b)] (if (zero? b) a (recur b (rem a b)))))
(defn check-pairwise-gcd [cnt]
(let [sbs (take (inc cnt) all-sbs)] (every? #(= 1 %) (map gcd sbs (rest sbs)))))
- one-based index required by problem statement
(defn report-sb []
(println "First 15 Stern-Brocot members:" (take 15 all-sbs)) (println "First appearance of N at 1-based index:") (doseq [n [1 2 3 4 5 6 7 8 9 10 100]] (println " first" n "at" (inc (first-appearance n)))) (println "Check pairwise GCDs = 1 ..." (check-pairwise-gcd 1000)) true)
(report-sb)</lang>
- Output:
First 15 Stern-Brocot members: (1 1 2 1 3 2 3 1 4 3 5 2 5 3 4) First appearance of N at 1-based index: first 1 at 1 first 2 at 3 first 3 at 5 first 4 at 9 first 5 at 11 first 6 at 33 first 7 at 19 first 8 at 21 first 9 at 35 first 10 at 39 first 100 at 1179 Check pairwise GCDs = 1 ... true true
Clojure: Using Lazy Sequences
<lang clojure>(ns test-p.core) (defn gcd
"(gcd a b) computes the greatest common divisor of a and b." [a b] (if (zero? b) a (recur b (mod a b))))
(defn stern-brocat-next [p]
" p is the block of the sequence we are using to compute the next block This routine computes the next block " (into [] (concat (rest p) [(+ (first p) (second p))] [(second p)])))
(defn seq-stern-brocat
([] (seq-stern-brocat [1 1])) ([p] (lazy-seq (cons (first p) (seq-stern-brocat (stern-brocat-next p))))))
- First 15 elements
(println (take 15 (seq-stern-brocat)))
- Where numbers 1 to 10 first appear
(doseq [n (concat (range 1 11) [100])]
(println "The first appearnce of" n "is at index" (some (fn i k (when (= k n) (inc i))) (map-indexed vector (seq-stern-brocat)))))
- Check that gcd between 1st 1000 consecutive elements equals 1
- Create cosecutive pairs of 1st 1000 elements
(def one-thousand-pairs (take 1000 (partition 2 1 (seq-stern-brocat))))
- Check every pair has a gcd = 1
(println (every? (fn ith ith-plus-1 (= (gcd ith ith-plus-1) 1))
one-thousand-pairs))
</lang>
- Output:
(1 1 2 1 3 2 3 1 4 3 5 2 5 3 4) The first appearnce of 1 is at index 1 The first appearnce of 2 is at index 3 The first appearnce of 3 is at index 5 The first appearnce of 4 is at index 9 The first appearnce of 5 is at index 11 The first appearnce of 6 is at index 33 The first appearnce of 7 is at index 19 The first appearnce of 8 is at index 21 The first appearnce of 9 is at index 35 The first appearnce of 10 is at index 39 The first appearnce of 100 is at index 1179 true
CLU
<lang clu>stern = proc (n: int) returns (array[int])
s: array[int] := array[int]$fill(1, n, 1) for i: int in int$from_to(2, n/2) do s[i*2-1] := s[i] + s[i-1] s[i*2] := s[i] end return (s)
end stern
gcd = proc (a,b: int) returns (int)
while b ~= 0 do a, b := b, a//b end return (a)
end gcd
find = proc [T: type] (a: array[T], val: T) returns (int) signals (not_found)
where T has equal: proctype (T,T) returns (bool) for i: int in array[T]$indexes(a) do if a[i] = val then return (i) end end signal not_found
end find
start_up = proc ()
po: stream := stream$primary_output() s: array[int] := stern(1200) stream$puts(po, "First 15 numbers:") for i: int in int$from_to(1, 15) do stream$puts(po, " " || int$unparse(s[i])) end stream$putl(po, "") for i: int in int$from_to(1, 10) do stream$putl(po, "First " || int$unparse(i) || " at " || int$unparse(find[int](s, i))) end stream$putl(po, "First 100 at " || int$unparse(find[int](s, 100)))
begin for i: int in int$from_to(2, array[int]$high(s)) do if gcd(s[i-1], s[i]) ~= 1 then exit gcd_not_one(i) end end stream$putl(po, "The GCD of every pair of adjacent elements is 1.") end except when gcd_not_one(i: int): stream$putl(po, "The GCD of the pair at " || int$unparse(i) || " is not 1.") end
end start_up</lang>
- Output:
First 15 numbers: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179 The GCD of every pair of adjacent elements is 1.
Common Lisp
<lang lisp>(defun stern-brocot (numbers)
(declare ((or null (vector integer)) numbers)) (cond ((null numbers) (setf numbers (make-array 2 :element-type 'integer :adjustable t :fill-pointer t :initial-element 1))) ((zerop (length numbers)) (vector-push-extend 1 numbers) (vector-push-extend 1 numbers)) (t (assert (evenp (length numbers))) (let* ((considered-index (/ (length numbers) 2)) (considered (aref numbers considered-index)) (precedent (aref numbers (1- considered-index)))) (vector-push-extend (+ considered precedent) numbers) (vector-push-extend considered numbers)))) numbers)
(defun first-15 ()
(loop for input = nil then seq for seq = (stern-brocot input) while (< (length seq) 15) finally (format t "First 15: ~{~A~^ ~}~%" (coerce (subseq seq 0 15) 'list))))
(defun first-1-to-10 ()
(loop with seq = (stern-brocot nil) for i from 1 to 10 for index = (loop with start = 0 for pos = (position i seq :start start) until pos do (setf start (length seq) seq (stern-brocot seq)) finally (return (1+ pos))) do (format t "First ~D at ~D~%" i index)))
(defun first-100 ()
(loop for input = nil then seq for start = (length input) for seq = (stern-brocot input) for pos = (position 100 seq :start start) until pos finally (format t "First 100 at ~D~%" (1+ pos))))
(defun check-gcd ()
(loop for input = nil then seq for seq = (stern-brocot input) while (< (length seq) 1000) finally (if (loop for i from 0 below 999 always (= 1 (gcd (aref seq i) (aref seq (1+ i))))) (write-line "Correct. The GCDs of all the two consecutive numbers are 1.") (write-line "Wrong."))))
(defun main ()
(first-15) (first-1-to-10) (first-100) (check-gcd))</lang>
- Output:
First 15: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179 Correct. The GCDs of all the two consecutive numbers are 1.
Cowgol
<lang cowgol>include "cowgol.coh";
- Redefining these is enough to change the type and length everywhere,
- but arrays are 0-based so you need one extra element.
typedef Stern is uint8; # 8-bit math is enough for the numbers we need var stern: Stern[1201]; # Array containing Stern-Brocot sequence
- Fill up the Stern-Brocot array
sub GenStern() is
stern[1] := 1; stern[2] := 1;
var i: @indexof stern := 1; var last: @indexof stern := @sizeof stern / 2; while i <= last loop stern[i*2-1] := stern[i] + stern[i-1]; stern[i*2] := stern[i]; i := i + 1; end loop;
end sub;
- Find the first location of a given number
sub FindFirst(n: Stern): (i: @indexof stern) is
i := 1; while i < @sizeof stern and stern[i] != n loop i := i + 1; end loop;
end sub;
GenStern(); # Generate sequence
- Print the first 15 numbers
var i: @indexof stern := 1; while i <= 15 loop
print_i32(stern[i] as uint32); print_char(' '); i := i + 1;
end loop; print_nl();
- Print the first occurrence of 1..10
var j: Stern := 1; while j <= 10 loop
print_i32(FindFirst(j) as uint32); print_char(' '); j := j + 1;
end loop; print_nl();
- Print the first occurrence of 100
print_i32(FindFirst(100) as uint32); print_nl();
- Check that all GCDs of consecutive pairs are 1
sub gcd(a: Stern, b: Stern): (r: Stern) is
while a != b loop if a > b then a := a - b; else b := b - a; end if; end loop; r := a;
end sub;
i := 1; while i < @sizeof stern / 2 loop
if gcd(stern[i], stern[i+1]) != 1 then print("GCD not 1 at: "); print_i32(i as uint32); print_nl(); ExitWithError(); end if; i := i + 1;
end loop;
print("All GCDs are 1.\n");</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 3 5 9 11 33 19 21 35 39 1179 All GCDs are 1.
D
<lang d>import std.stdio, std.numeric, std.range, std.algorithm;
/// Generates members of the stern-brocot series, in order, /// returning them when the predicate becomes false. uint[] sternBrocot(bool delegate(in uint[]) pure nothrow @safe @nogc pred=seq => seq.length < 20) pure nothrow @safe {
typeof(return) sb = [1, 1]; size_t i = 0; while (pred(sb)) { sb ~= [sb[i .. i + 2].sum, sb[i + 1]]; i++; } return sb;
}
void main() {
enum nFirst = 15; writefln("The first %d values:\n%s\n", nFirst, sternBrocot(seq => seq.length < nFirst).take(nFirst));
foreach (immutable nOccur; iota(1, 10 + 1).chain(100.only)) writefln("1-based index of the first occurrence of %3d in the series: %d", nOccur, sternBrocot(seq => nOccur != seq[$ - 2]).length - 1);
enum nGcd = 1_000; auto s = sternBrocot(seq => seq.length < nGcd).take(nGcd); assert(zip(s, s.dropOne).all!(ss => ss[].gcd == 1), "A fraction from adjacent terms is reducible.");
}</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179
This uses a queue from the Queue/usage Task: <lang d>import std.stdio, std.algorithm, std.range, std.numeric, queue_usage2;
struct SternBrocot {
private auto sb = GrowableCircularQueue!uint(1, 1); enum empty = false; @property uint front() pure nothrow @safe @nogc { return sb.front; } uint popFront() pure nothrow @safe { sb.push(sb.front + sb[1]); sb.push(sb[1]); return sb.pop; }
}
void main() {
SternBrocot().drop(50_000_000).front.writeln;
}</lang>
- Output:
7004
Direct Version:
<lang d>void main() {
import std.stdio, std.numeric, std.range, std.algorithm, std.bigint, std.conv;
/// Stern-Brocot sequence, 0-th member is 0. T sternBrocot(T)(T n) pure nothrow /*safe*/ { T a = 1, b = 0; while (n) { if (n & 1) b += a; else a += b; n >>= 1; } return b; } alias sb = sternBrocot!uint;
enum nFirst = 15; writefln("The first %d values:\n%s\n", nFirst, iota(1, nFirst + 1).map!sb);
foreach (immutable nOccur; iota(1, 10 + 1).chain(100.only)) writefln("1-based index of the first occurrence of %3d in the series: %d", nOccur, sequence!q{n}.until!(n => sb(n) == nOccur).walkLength);
auto s = iota(1, 1_001).map!sb; assert(s.zip(s.dropOne).all!(ss => ss[].gcd == 1), "A fraction from adjacent terms is reducible.");
sternBrocot(10.BigInt ^^ 20_000).text.length.writeln;
}</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179 7984
EchoLisp
Function
<lang lisp>
- stern (2n ) = stern (n)
- stern(2n+1) = stern(n) + stern(n+1)
(define (stern n) (cond (( < n 3) 1) ((even? n) (stern (/ n 2))) (else (let ((m (/ (1- n) 2))) (+ (stern m) (stern (1+ m))))))) (remember 'stern) </lang>
- Output:
<lang lisp>
- generate the sequence and check GCD
(for ((n 10000))
(unless (= (gcd (stern n) (stern (1+ n))) 1) (error "BAD GCD" n))) → #t
- first items
(define sterns (cache 'stern)) (subvector sterns 1 16)
→ #( 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4)
- first occurences index
(for ((i (in-range 1 11))) (write (vector-index i sterns))) → 0 3 5 9 11 33 19 21 35 39
- 100
(writeln (vector-index 100 sterns)) → 1179
(stern 900000) → 446 (stern 900001) → 2479 </lang>
Stream
From A002487, if we group the elements by two, we get (uniquely) all the rationals. Another way to generate the rationals, hence the stern sequence, is to iterate the function f(x) = floor(x) + 1 - fract(x).
<lang lisp>
- grouping
(for ((i (in-range 2 40 2))) (write (/ (stern i)(stern (1+ i))))) → 1/2 1/3 2/3 1/4 3/5 2/5 3/4 1/5 4/7 3/8 5/7 2/7 5/8 3/7 4/5 1/6 5/9 4/11 7/10
- computing f(1), f(f(1)), etc.
(define (f x)
(let [(a (/ (- (floor x) -1 (fract x))))] (if (> a 1) (f a) (cons a a))))
(define T (make-stream f 1)) (for((i 19)) (write (stream-iterate T)))
→ 1/2 1/3 2/3 1/4 3/5 2/5 3/4 1/5 4/7 3/8 5/7 2/7 5/8 3/7 4/5 1/6 5/9 4/11 7/10 </lang>
Elixir
<lang elixir>defmodule SternBrocot do
def sequence do Stream.unfold({0,{1,1}}, fn {i,acc} -> a = elem(acc, i) b = elem(acc, i+1) {a, {i+1, Tuple.append(acc, a+b) |> Tuple.append(b)}} end) end def task do IO.write "First fifteen members of the sequence:\n " IO.inspect Enum.take(sequence, 15) Enum.each(Enum.concat(1..10, [100]), fn n -> i = Enum.find_index(sequence, &(&1==n)) + 1 IO.puts "#{n} first appears at #{i}" end) Enum.take(sequence, 1000) |> Enum.chunk(2,1) |> Enum.all?(fn [a,b] -> gcd(a,b) == 1 end) |> if(do: "All GCD's are 1", else: "Whoops, not all GCD's are 1!") |> IO.puts end defp gcd(a,0), do: abs(a) defp gcd(a,b), do: gcd(b, rem(a,b))
end
SternBrocot.task</lang>
- Output:
First fifteen members of the sequence: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1 first appears at 1 2 first appears at 3 3 first appears at 5 4 first appears at 9 5 first appears at 11 6 first appears at 33 7 first appears at 19 8 first appears at 21 9 first appears at 35 10 first appears at 39 100 first appears at 1179 All GCD's are 1
F#
The function
<lang fsharp> // Generate Stern-Brocot Sequence. Nigel Galloway: October 11th., 2018 let sb=Seq.unfold(fun (n::g::t)->Some(n,[g]@t@[n+g;g]))[1;1] </lang>
The Task
Uses Greatest_common_divisor#F.23 <lang fsharp> sb |> Seq.take 15 |> Seq.iter(printf "%d ");printfn "" [1..10] |> List.map(fun n->(n,(sb|>Seq.findIndex(fun g->g=n))+1)) |> List.iter(printf "%A ");printfn "" printfn "%d" ((sb|>Seq.findIndex(fun g->g=100))+1) printfn "There are %d consecutive members, of the first thousand members, with GCD <> 1" (sb |> Seq.take 1000 |>Seq.pairwise |> Seq.filter(fun(n,g)->gcd n g <> 1) |> Seq.length) </lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 (1, 1) (2, 3) (3, 5) (4, 9) (5, 11) (6, 33) (7, 19) (8, 21) (9, 35) (10, 39) 1179 There are 0 consecutive members, of the first thousand members, with GCD <> 1
Factor
Using the alternative function given in the C example for computing the Stern-Brocot sequence. <lang factor>USING: formatting io kernel lists lists.lazy locals math math.ranges prettyprint sequences ; IN: rosetta-code.stern-brocot
- fn ( n -- m )
[ 1 0 ] dip [ dup zero? ] [ dup 1 bitand zero? [ dupd [ + ] 2dip ] [ [ dup ] [ + ] [ ] tri* ] if -1 shift ] until drop nip ;
- search ( n -- m )
1 0 lfrom [ fn n = ] lfilter ltake list>array first ;
- first15 ( -- )
15 [1,b] [ fn pprint bl ] each "are the first fifteen." print ;
- first-appearances ( -- )
10 [1,b] 100 suffix [ dup search "First %3u at Stern #%u.\n" printf ] each ;
- gcd-test ( -- )
1,000 [1,b] [ dup 1 + [ fn ] bi@ gcd nip 1 = not ] filter empty? "" " not" ? "All GCDs are%s 1.\n" printf ;
- main ( -- ) first15 first-appearances gcd-test ;
MAIN: main</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 are the first fifteen. First 1 at Stern #1. First 2 at Stern #3. First 3 at Stern #5. First 4 at Stern #9. First 5 at Stern #11. First 6 at Stern #33. First 7 at Stern #19. First 8 at Stern #21. First 9 at Stern #35. First 10 at Stern #39. First 100 at Stern #1179. All GCDs are 1.
Forth
<lang forth>: stern ( n -- x : return N'th item of Stern-Brocot sequence)
dup 2 >= if 2 /mod swap if dup 1+ recurse swap recurse + else recurse then then
- first ( n -- x : return X such that stern X = n )
1 begin over over stern <> while 1+ repeat swap drop
- gcd ( a b -- a gcd b )
begin swap over mod dup 0= until drop
- task
( Print first 15 numbers ) ." First 15: " 1 begin dup stern . 1+ dup 15 > until drop cr ( Print first occurrence of 1..10 ) 1 begin ." First " dup . ." at " dup first . 1+ cr dup 10 > until drop ( Print first occurrence of 100 ) ." First 100 at " 100 first . cr ( Check that the GCD of each adjacent pair up to 1000 is 1 ) -1 2 begin dup stern over 1- stern gcd 1 = rot and swap 1+ dup 1000 > until swap if ." All GCDs are 1." drop else ." GCD <> 1 at: " . then cr
task bye</lang>
- Output:
First 15: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179 All GCDs are 1.
Fortran
Fortran IV
<lang fortran>* STERN-BROCOT SEQUENCE - FORTRAN IV
DIMENSION ISB(2400) NN=2400 ISB(1)=1 ISB(2)=1 I=1 J=2 K=2 1 IF(K.GE.NN) GOTO 2 K=K+1 ISB(K)=ISB(K-I)+ISB(K-J) K=K+1 ISB(K)=ISB(K-J) I=I+1 J=J+1 GOTO 1 2 N=15 WRITE(*,101) N 101 FORMAT(1X,'FIRST',I4) WRITE(*,102) (ISB(I),I=1,15) 102 FORMAT(15I4) DO 5 J=1,11 JJ=J IF(J.EQ.11) JJ=100 DO 3 I=1,K IF(ISB(I).EQ.JJ) GOTO 4 3 CONTINUE 4 WRITE(*,103) JJ,I 103 FORMAT(1X,'FIRST',I4,' AT ',I4) 5 CONTINUE END </lang>
- Output:
FIRST 15 FIRST 15 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 FIRST 1 AT 1 FIRST 2 AT 3 FIRST 3 AT 5 FIRST 4 AT 9 FIRST 5 AT 11 FIRST 6 AT 33 FIRST 7 AT 19 FIRST 8 AT 21 FIRST 9 AT 35 FIRST 10 AT 39 FIRST 100 AT 1179
Fortran 90
<lang fortran> ! Stern-Brocot sequence - Fortran 90
parameter (nn=2400) dimension isb(nn) isb(1)=1; isb(2)=1 i=1; j=2; k=2 do while(k.lt.nn) k=k+1; isb(k)=isb(k-i)+isb(k-j) k=k+1; isb(k)=isb(k-j) i=i+1; j=j+1 end do n=15 write(*,"(1x,'First',i4)") n write(*,"(15i4)") (isb(i),i=1,15) do j=1,11 jj=j if(j==11) jj=100 do i=1,k if(isb(i)==jj) exit end do write(*,"(1x,'First',i4,' at ',i4)") jj,i end do end </lang>
- Output:
First 15 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179
FreeBASIC
<lang freebasic>' version 02-03-2019 ' compile with: fbc -s console
- Define max 2000
Dim Shared As UInteger stern(max +2)
Sub stern_brocot
stern(1) = 1 stern(2) = 1
Dim As UInteger i = 2 , n = 2, ub = UBound(stern)
Do While i < ub i += 1 stern(i) = stern(n) + stern(n -1) i += 1 stern(i) = stern(n) n += 1 Loop
End Sub
Function gcd(x As UInteger, y As UInteger) As UInteger
Dim As UInteger t
While y t = y y = x Mod y x = t Wend
Return x
End Function
' ------=< MAIN >=------
Dim As UInteger i
stern_brocot
Print "The first 15 are: " ; For i = 1 To 15
Print stern(i); " ";
Next
Print : Print Print " Index First nr." Dim As UInteger d = 1 For i = 1 To max
If stern(i) = d Then Print Using " ######"; i; stern(i) d += 1 If d = 11 Then d = 100 If d = 101 Then Exit For i = 0 End If
Next
Print : Print d = 0 For i = 1 To 1000
If gcd(stern(i), stern(i +1)) <> 1 Then d = gcd(stern(i), stern(i +1)) Exit For End If
Next
If d = 0 Then
Print "GCD of two consecutive members of the series up to the 1000th member is 1"
Else
Print "The GCD for index "; i; " and "; i +1; " = "; d
End If
' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
The first 15 are: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 Index First nr. 1 1 3 2 5 3 9 4 11 5 33 6 19 7 21 8 35 9 39 10 1179 100 GCD of two consecutive members of the series up to the 1000th member is 1
Go
<lang go>package main
import (
"fmt"
"sternbrocot"
)
func main() {
// Task 1, using the conventional sort of generator that generates // terms endlessly. g := sb.Generator()
// Task 2, demonstrating the generator. fmt.Println("First 15:") for i := 1; i <= 15; i++ { fmt.Printf("%2d: %d\n", i, g()) }
// Task 2 again, showing a simpler technique that might or might not be // considered to "generate" terms. s := sb.New() fmt.Println("First 15:", s.FirstN(15))
// Tasks 3 and 4. for _, x := range []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100} { fmt.Printf("%3d at 1-based index %d\n", x, 1+s.Find(x)) }
// Task 5. fmt.Println("1-based indexes: gcd") for n, f := range s.FirstN(1000)[:999] { g := gcd(f, (*s)[n+1]) fmt.Printf("%d,%d: gcd(%d, %d) = %d\n", n+1, n+2, f, (*s)[n+1], g) if g != 1 { panic("oh no!") return } }
}
// gcd copied from greatest common divisor task func gcd(x, y int) int {
for y != 0 { x, y = y, x%y } return x
}</lang> <lang go>// SB implements the Stern-Brocot sequence. // // Generator() satisfies RC Task 1. For remaining tasks, Generator could be // used but FirstN(), and Find() are simpler methods for specific stopping // criteria. FirstN and Find might also be considered to satisfy Task 1, // in which case Generator would not really be needed. Anyway, there it is. package sb
// Seq represents an even number of terms of a Stern-Brocot sequence. // // Terms are stored in a slice. Terms start with 1. // (Specifically, the zeroth term, 0, given in OEIS A002487 is not represented.) // Term 1 (== 1) is stored at slice index 0. // // Methods on Seq rely on Seq always containing an even number of terms. type Seq []int
// New returns a Seq with the two base terms. func New() *Seq {
return &Seq{1, 1} // Step 1 of the RC task.
}
// TwoMore appends two more terms to p. // It's the body of the loop in the RC algorithm. // Generate(), FirstN(), and Find() wrap this body in different ways. func (p *Seq) TwoMore() {
s := *p n := len(s) / 2 // Steps 2 and 5 of the RC task. c := s[n] *p = append(s, c+s[n-1], c) // Steps 3 and 4 of the RC task.
}
// Generator returns a generator function that returns successive terms // (until overflow.) func Generator() func() int {
n := 0 p := New() return func() int { if len(*p) == n { p.TwoMore() } t := (*p)[n] n++ return t }
}
// FirstN lazily extends p as needed so that it has at least n terms. // FirstN then returns a list of the first n terms. func (p *Seq) FirstN(n int) []int {
for len(*p) < n { p.TwoMore() } return []int((*p)[:n])
}
// Find lazily extends p as needed until it contains the value x // Find then returns the slice index of x in p. func (p *Seq) Find(x int) int {
for n, f := range *p { if f == x { return n } } for { p.TwoMore() switch x { case (*p)[len(*p)-2]: return len(*p) - 2 case (*p)[len(*p)-1]: return len(*p) - 1 } }
}</lang>
- Output:
First 15: 1: 1 2: 1 3: 2 4: 1 5: 3 6: 2 7: 3 8: 1 9: 4 10: 3 11: 5 12: 2 13: 5 14: 3 15: 4 First 15: [1 1 2 1 3 2 3 1 4 3 5 2 5 3 4] 1 at 1-based index 1 2 at 1-based index 3 3 at 1-based index 5 4 at 1-based index 9 5 at 1-based index 11 6 at 1-based index 33 7 at 1-based index 19 8 at 1-based index 21 9 at 1-based index 35 10 at 1-based index 39 100 at 1-based index 1179 1-based indexes: gcd 1,2: gcd(1, 1) = 1 2,3: gcd(1, 2) = 1 3,4: gcd(2, 1) = 1 4,5: gcd(1, 3) = 1 ... 998,999: gcd(27, 38) = 1 999,1000: gcd(38, 11) = 1
Haskell
<lang haskell>import Data.List (elemIndex)
sb :: [Int] sb = 1 : 1 : f (tail sb) sb
where f (a : aa) (b : bb) = a + b : a : f aa bb
main :: IO () main = do
print $ take 15 sb print [ (i, 1 + (\(Just i) -> i) (elemIndex i sb)) | i <- [1 .. 10] <> [100] ] print $ all (\(a, b) -> 1 == gcd a b) $ take 1000 $ zip sb (tail sb)</lang>
- Output:
[1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] [(1,1),(2,3),(3,5),(4,9),(5,11),(6,33),(7,19),(8,21),(9,35),(10,39),(100,1179)] True
Or, expressed in terms of iterate:
<lang haskell>import Data.Function (on) import Data.List (nubBy, sortOn) import Data.Ord (comparing)
STERN-BROCOT SEQUENCE -----------------
sternBrocot :: [Int] sternBrocot = head <$> iterate go [1, 1]
where go (a : b : xs) = (b : xs) <> [a + b, b]
TEST -------------------------
main :: IO () main = do
print $ take 15 sternBrocot print $ take 10 $ nubBy (on (==) fst) $ sortOn fst $ takeWhile ((110 >=) . fst) $ zip sternBrocot [1 ..] print $ take 1 $ dropWhile ((100 /=) . fst) $ zip sternBrocot [1 ..] print $ (all ((1 ==) . uncurry gcd) . (zip <*> tail)) $ take 1000 sternBrocot</lang>
- Output:
[1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] [(1,1),(2,3),(3,5),(4,9),(5,11),(6,33),(7,19),(8,21),(9,35),(10,39)] [(100,1179)] True
J
We have two different kinds of list specifications here (length of the sequence, and the presence of certain values in the sequence). Also the underlying list generation mechanism is somewhat arbitrary. So let's generate the sequence iteratively and provide a truth valued function of the intermediate sequences to determine when we have generated one which is adequately long:
<lang J>sternbrocot=:1 :0
ind=. 0 seq=. 1 1 while. -. u seq do. ind=. ind+1 seq=. seq, +/\. seq {~ _1 0 +ind end.
)</lang>
(Grammatical aside: this is an adverb which generates a noun without taking any x/y arguments. So usage is: u sternbrocot
. J does have precedence rules, just not very many of them. Users of other languages can get a rough idea of the grammatical terms like this: adverb is approximately like a macro, verb approximately like a function and noun is approximately like a number. Also x and y are J's names for left and right noun arguments, and u and v are J's names for left and right verb arguments. An adverb has a left verb argument. There are some other important constraints but that's probably more than enough detail for this task.)
First fifteen members of the sequence:
<lang J> 15{.(15<:#) sternbrocot 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4</lang>
One based indices of where numbers 1-10 first appear in the sequence:
<lang J> 1+(10 e. ]) sternbrocot i.1+i.10 1 3 5 9 11 33 19 21 35 39</lang>
One based index of where the number 100 first appears:
<lang J> 1+(100 e. ]) sternbrocot i. 100 1179</lang>
List of distinct greatest common divisors of adjacent number pairs from a sternbrocot sequence which includes the first 1000 elements:
<lang J> ~.2 +./\ (1000<:#) sternbrocot 1</lang>
Java
This example generates the first 1200 members of the sequence since that is enough to cover all of the tests in the description. It borrows the gcd
method from BigInteger
rather than using its own.
<lang java5>import java.math.BigInteger;
import java.util.LinkedList;
public class SternBrocot { static LinkedList<Integer> sequence = new LinkedList<Integer>()Template:Add(1); add(1);;
private static void genSeq(int n){ for(int conIdx = 1; sequence.size() < n; conIdx++){ int consider = sequence.get(conIdx); int pre = sequence.get(conIdx - 1); sequence.add(consider + pre); sequence.add(consider); }
}
public static void main(String[] args){ genSeq(1200); System.out.println("The first 15 elements are: " + sequence.subList(0, 15)); for(int i = 1; i <= 10; i++){ System.out.println("First occurrence of " + i + " is at " + (sequence.indexOf(i) + 1)); }
System.out.println("First occurrence of 100 is at " + (sequence.indexOf(100) + 1));
boolean failure = false; for(int i = 0; i < 999; i++){ failure |= !BigInteger.valueOf(sequence.get(i)).gcd(BigInteger.valueOf(sequence.get(i + 1))).equals(BigInteger.ONE); } System.out.println("All GCDs are" + (failure ? " not" : "") + " 1"); } }</lang>
- Output:
The first 15 elements are: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] First occurrence of 1 is at 1 First occurrence of 2 is at 3 First occurrence of 3 is at 5 First occurrence of 4 is at 9 First occurrence of 5 is at 11 First occurrence of 6 is at 33 First occurrence of 7 is at 19 First occurrence of 8 is at 21 First occurrence of 9 is at 35 First occurrence of 10 is at 39 First occurrence of 100 is at 1179 All GCDs are 1
Stern-Brocot Tree
<lang java>import java.awt.*; import javax.swing.*;
public class SternBrocot extends JPanel {
public SternBrocot() { setPreferredSize(new Dimension(800, 500)); setFont(new Font("Arial", Font.PLAIN, 18)); setBackground(Color.white); }
private void drawTree(int n1, int d1, int n2, int d2, int x, int y, int gap, int lvl, Graphics2D g) {
if (lvl == 0) return;
// mediant int numer = n1 + n2; int denom = d1 + d2;
if (lvl > 1) { g.drawLine(x + 5, y + 4, x - gap + 5, y + 124); g.drawLine(x + 5, y + 4, x + gap + 5, y + 124); }
g.setColor(getBackground()); g.fillRect(x - 10, y - 15, 35, 40);
g.setColor(getForeground()); g.drawString(String.valueOf(numer), x, y); g.drawString("_", x, y + 2); g.drawString(String.valueOf(denom), x, y + 22);
drawTree(n1, d1, numer, denom, x - gap, y + 120, gap / 2, lvl - 1, g); drawTree(numer, denom, n2, d2, x + gap, y + 120, gap / 2, lvl - 1, g); }
@Override public void paintComponent(Graphics gg) { super.paintComponent(gg); Graphics2D g = (Graphics2D) gg; g.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON);
int w = getWidth();
drawTree(0, 1, 1, 0, w / 2, 50, w / 4, 4, g); }
public static void main(String[] args) { SwingUtilities.invokeLater(() -> { JFrame f = new JFrame(); f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); f.setTitle("Stern-Brocot Tree"); f.setResizable(false); f.add(new SternBrocot(), BorderLayout.CENTER); f.pack(); f.setLocationRelativeTo(null); f.setVisible(true); }); }
JavaScript
<lang JavaScript>(() => {
'use strict';
const main = () => {
// sternBrocot :: Generator [Int] const sternBrocot = () => { const go = xs => { const x = snd(xs); return tail(append(xs, [fst(xs) + x, x])); }; return fmapGen(head, iterate(go, [1, 1])); };
// TESTS ------------------------------------------ const sbs = take(1200, sternBrocot()), ixSB = zip(sbs, enumFrom(1));
return unlines(map( JSON.stringify, [ take(15, sbs), take(10, map(listFromTuple, nubBy( on(eq, fst), sortBy( comparing(fst), takeWhile(x => 12 !== fst(x), ixSB) ) ) ) ), listFromTuple( take(1, dropWhile(x => 100 !== fst(x), ixSB))[0] ), all(tpl => 1 === gcd(fst(tpl), snd(tpl)), take(1000, zip(sbs, tail(sbs))) ) ] )); };
// GENERIC ABSTRACTIONS -------------------------------
// Just :: a -> Maybe a const Just = x => ({ type: 'Maybe', Nothing: false, Just: x });
// Nothing :: Maybe a const Nothing = () => ({ type: 'Maybe', Nothing: true, });
// Tuple (,) :: a -> b -> (a, b) const Tuple = (a, b) => ({ type: 'Tuple', '0': a, '1': b, length: 2 });
// | Absolute value.
// abs :: Num -> Num const abs = Math.abs;
// Determines whether all elements of the structure // satisfy the predicate.
// all :: (a -> Bool) -> [a] -> Bool const all = (p, xs) => xs.every(p);
// append (++) :: [a] -> [a] -> [a] // append (++) :: String -> String -> String const append = (xs, ys) => xs.concat(ys);
// chr :: Int -> Char const chr = String.fromCodePoint;
// comparing :: (a -> b) -> (a -> a -> Ordering) const comparing = f => (x, y) => { const a = f(x), b = f(y); return a < b ? -1 : (a > b ? 1 : 0); };
// dropWhile :: (a -> Bool) -> [a] -> [a] // dropWhile :: (Char -> Bool) -> String -> String const dropWhile = (p, xs) => { const lng = xs.length; return 0 < lng ? xs.slice( until( i => i === lng || !p(xs[i]), i => 1 + i, 0 ) ) : []; };
// enumFrom :: a -> [a] function* enumFrom(x) { let v = x; while (true) { yield v; v = succ(v); } }
// eq (==) :: Eq a => a -> a -> Bool const eq = (a, b) => { const t = typeof a; return t !== typeof b ? ( false ) : 'object' !== t ? ( 'function' !== t ? ( a === b ) : a.toString() === b.toString() ) : (() => { const aks = Object.keys(a); return aks.length !== Object.keys(b).length ? ( false ) : aks.every(k => eq(a[k], b[k])); })(); };
// fmapGen <$> :: (a -> b) -> Gen [a] -> Gen [b] function* fmapGen(f, gen) { const g = gen; let v = take(1, g)[0]; while (0 < v.length) { yield(f(v)) v = take(1, g)[0] } }
// fst :: (a, b) -> a const fst = tpl => tpl[0];
// gcd :: Int -> Int -> Int const gcd = (x, y) => { const _gcd = (a, b) => (0 === b ? a : _gcd(b, a % b)), abs = Math.abs; return _gcd(abs(x), abs(y)); };
// head :: [a] -> a const head = xs => xs.length ? xs[0] : undefined;
// isChar :: a -> Bool const isChar = x => ('string' === typeof x) && (1 === x.length);
// iterate :: (a -> a) -> a -> Gen [a] function* iterate(f, x) { let v = x; while (true) { yield(v); v = f(v); } }
// Returns Infinity over objects without finite length // this enables zip and zipWith to choose the shorter // argument when one is non-finite, like cycle, repeat etc
// length :: [a] -> Int const length = xs => xs.length || Infinity;
// listFromTuple :: (a, a ...) -> [a] const listFromTuple = tpl => Array.from(tpl);
// map :: (a -> b) -> [a] -> [b] const map = (f, xs) => xs.map(f);
// nubBy :: (a -> a -> Bool) -> [a] -> [a] const nubBy = (p, xs) => { const go = xs => 0 < xs.length ? (() => { const x = xs[0]; return [x].concat( go(xs.slice(1) .filter(y => !p(x, y)) ) ) })() : []; return go(xs); };
// e.g. sortBy(on(compare,length), xs)
// on :: (b -> b -> c) -> (a -> b) -> a -> a -> c const on = (f, g) => (a, b) => f(g(a), g(b));
// ord :: Char -> Int const ord = c => c.codePointAt(0);
// snd :: (a, b) -> b const snd = tpl => tpl[1];
// sortBy :: (a -> a -> Ordering) -> [a] -> [a] const sortBy = (f, xs) => xs.slice() .sort(f);
// succ :: Enum a => a -> a const succ = x => isChar(x) ? ( chr(1 + ord(x)) ) : isNaN(x) ? ( undefined ) : 1 + x;
// tail :: [a] -> [a] const tail = xs => 0 < xs.length ? xs.slice(1) : [];
// take :: Int -> [a] -> [a] // take :: Int -> String -> String const take = (n, xs) => xs.constructor.constructor.name !== 'GeneratorFunction' ? ( xs.slice(0, n) ) : [].concat.apply([], Array.from({ length: n }, () => { const x = xs.next(); return x.done ? [] : [x.value]; }));
// takeWhile :: (a -> Bool) -> [a] -> [a] // takeWhile :: (Char -> Bool) -> String -> String const takeWhile = (p, xs) => xs.constructor.constructor.name !== 'GeneratorFunction' ? (() => { const lng = xs.length; return 0 < lng ? xs.slice( 0, until( i => lng === i || !p(xs[i]), i => 1 + i, 0 ) ) : []; })() : takeWhileGen(p, xs);
// takeWhileGen :: (a -> Bool) -> Gen [a] -> [a] const takeWhileGen = (p, xs) => { const ys = []; let nxt = xs.next(), v = nxt.value; while (!nxt.done && p(v)) { ys.push(v); nxt = xs.next(); v = nxt.value } return ys; };
// uncons :: [a] -> Maybe (a, [a]) const uncons = xs => { const lng = length(xs); return (0 < lng) ? ( lng < Infinity ? ( Just(Tuple(xs[0], xs.slice(1))) // Finite list ) : (() => { const nxt = take(1, xs); return 0 < nxt.length ? ( Just(Tuple(nxt[0], xs)) ) : Nothing(); })() // Lazy generator ) : Nothing(); };
// unlines :: [String] -> String const unlines = xs => xs.join('\n');
// until :: (a -> Bool) -> (a -> a) -> a -> a const until = (p, f, x) => { let v = x; while (!p(v)) v = f(v); return v; };
// Use of `take` and `length` here allows for zipping with non-finite // lists - i.e. generators like cycle, repeat, iterate.
// zip :: [a] -> [b] -> [(a, b)] const zip = (xs, ys) => { const lng = Math.min(length(xs), length(ys)); return Infinity !== lng ? (() => { const bs = take(lng, ys); return take(lng, xs).map((x, i) => Tuple(x, bs[i])); })() : zipGen(xs, ys); };
// zipGen :: Gen [a] -> Gen [b] -> Gen [(a, b)] const zipGen = (ga, gb) => { function* go(ma, mb) { let a = ma, b = mb; while (!a.Nothing && !b.Nothing) { let ta = a.Just, tb = b.Just yield(Tuple(fst(ta), fst(tb))); a = uncons(snd(ta)); b = uncons(snd(tb)); } } return go(uncons(ga), uncons(gb)); };
// MAIN --- return main();
})();</lang>
- Output:
[1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] [[1,1],[2,3],[3,5],[4,9],[5,11],[6,33],[7,19],[8,21],[9,35],[10,39]] [100,1179] true
jq
In jq 1.4, there is no equivalent of "yield" for unbounded streams, and so the following uses "until".
Foundations: <lang jq>def until(cond; update):
def _until: if cond then . else (update | _until) end; try _until catch if .== "break" then empty else . end ;
def gcd(a; b):
# subfunction expects [a,b] as input # i.e. a ~ .[0] and b ~ .[1] def rgcd: if .[1] == 0 then .[0] else [.[1], .[0] % .[1]] | rgcd end; [a,b] | rgcd ;</lang>
The A002487 integer sequence:
The following definition is in strict accordance with https://oeis.org/A002487: i.e. a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1). The n-th element of the Rosetta Code sequence (counting from 1) is thus a[n], which accords with the fact that jq arrays have an index origin of 0. <lang jq># If n is non-negative, then A002487(n)
- generates an array with at least n elements of
- the A002487 sequence;
- if n is negative, elements are added until (-n)
- is found.
def A002487(n):
[0,1] | until( length as $l | if n >= 0 then $l >= n else .[$l-1] == -n
end;
length as $l | ($l / 2) as $n | .[$l] = .[$n] | if (.[$l-2] == -n) then . else .[$l + 1] = .[$n] + .[$n+1]
end ) ;</lang> The tasks: <lang jq># Generate a stream of n integers beginning with 1,1... def stern_brocot(n): A002487(n+1) | .[1:n+1][];
- Return the index (counting from 1) of n in the
- sequence starting with 1,1,...
def stern_brocot_index(n):
A002487(-n) | length -1 ;
def index_task:
(range(1;11), 100) as $i | "index of \($i) is \(stern_brocot_index($i))" ;
def gcd_task:
A002487(1000) | . as $A | reduce range(0; length-1) as $i ( []; gcd( $A[$i]; $A[$i+1] ) as $gcd | if $gcd == 1 then . else . + [$gcd] end) | if length == 0 then "GCDs are all 1" else "GCDs include \(.)" end ;
"First 15 integers of the Stern-Brocot sequence",
"as defined in the task description are:",
stern_brocot(15),
"",
"Using an index origin of 1:",
index_task,
"",
gcd_task
</lang>
- Output:
<lang sh>$ jq -r -n -f stern_brocot.jq First 15 integers of the Stern-Brocot sequence as defined in the task description are: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4
Using an index origin of 1: index of 1 is 1 index of 2 is 3 index of 3 is 5 index of 4 is 9 index of 5 is 11 index of 6 is 33 index of 7 is 19 index of 8 is 21 index of 9 is 35 index of 10 is 39 index of 100 is 1179
GCDs are all 1</lang>
Julia
<lang julia>using Printf
function sternbrocot(f::Function=(x) -> length(x) ≥ 20)::Vector{Int}
rst = Int[1, 1] i = 2 while !f(rst) append!(rst, Int[rst[i] + rst[i-1], rst[i]]) i += 1 end return rst
end
println("First 15 elements of Stern-Brocot series:\n", sternbrocot(x -> length(x) ≥ 15)[1:15], "\n")
for i in (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100)
occurr = findfirst(x -> x == i, sternbrocot(x -> i ∈ x)) @printf("Index of first occurrence of %3i in the series: %4i\n", i, occurr)
end
print("\nAssertion: the greatest common divisor of all the two\nconsecutive members of the series up to the 1000th member, is always one: ") sb = sternbrocot(x -> length(x) > 1000) if all(gcd(prev, this) == 1 for (prev, this) in zip(sb[1:1000], sb[2:1000]))
println("Confirmed.")
else
println("Rejected.")
end</lang>
- Output:
First 15 elements of Stern-Brocot series: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] Index of first occurrence of 1 in the series: 1 Index of first occurrence of 2 in the series: 3 Index of first occurrence of 3 in the series: 5 Index of first occurrence of 4 in the series: 9 Index of first occurrence of 5 in the series: 11 Index of first occurrence of 6 in the series: 33 Index of first occurrence of 7 in the series: 19 Index of first occurrence of 8 in the series: 21 Index of first occurrence of 9 in the series: 35 Index of first occurrence of 10 in the series: 39 Index of first occurrence of 100 in the series: 1179 Assertion: the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one: Confirmed.
Kotlin
<lang scala>// version 1.1.0
val sbs = mutableListOf(1, 1)
fun sternBrocot(n: Int, fromStart: Boolean = true) {
if (n < 4 || (n % 2 != 0)) throw IllegalArgumentException("n must be >= 4 and even") var consider = if (fromStart) 1 else n / 2 - 1 while (true) { val sum = sbs[consider] + sbs[consider - 1] sbs.add(sum) sbs.add(sbs[consider]) if (sbs.size == n) break consider++ }
}
fun gcd(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)
fun main(args: Array<String>) {
var n = 16 // needs to be even to ensure 'considered' number is added println("First 15 members of the Stern-Brocot sequence") sternBrocot(n) println(sbs.take(15))
val firstFind = IntArray(11) // all zero by default firstFind[0] = -1 // needs to be non-zero for subsequent test for ((i, v) in sbs.withIndex()) if (v <= 10 && firstFind[v] == 0) firstFind[v] = i + 1 loop@ while (true) { n += 2 sternBrocot(n, false) val vv = sbs.takeLast(2) var m = n - 1 for (v in vv) { if (v <= 10 && firstFind[v] == 0) firstFind[v] = m if (firstFind.all { it != 0 }) break@loop m++ } } println("\nThe numbers 1 to 10 first appear at the following indices:") for (i in 1..10) println("${"%2d".format(i)} -> ${firstFind[i]}")
print("\n100 first appears at index ") while (true) { n += 2 sternBrocot(n, false) val vv = sbs.takeLast(2) if (vv[0] == 100) { println(n - 1); break } if (vv[1] == 100) { println(n); break } }
print("\nThe GCDs of each pair of the series up to the 1000th member are ") for (p in 0..998 step 2) { if (gcd(sbs[p], sbs[p + 1]) != 1) { println("not all one") return } } println("all one")
}</lang>
- Output:
First 15 members of the Stern-Brocot sequence [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] The numbers 1 to 10 first appear at the following indices: 1 -> 1 2 -> 3 3 -> 5 4 -> 9 5 -> 11 6 -> 33 7 -> 19 8 -> 21 9 -> 35 10 -> 39 100 first appears at index 1179 The GCDs of each pair of the series up to the 1000th member are all one
Lua
<lang Lua>-- Task 1 function sternBrocot (n)
local sbList, pos, c = {1, 1}, 2 repeat c = sbList[pos] table.insert(sbList, c + sbList[pos - 1]) table.insert(sbList, c) pos = pos + 1 until #sbList >= n return sbList
end
-- Return index in table 't' of first value matching 'v' function findFirst (t, v)
for key, value in pairs(t) do if v then if value == v then return key end else if value ~= 0 then return key end end end return nil
end
-- Return greatest common divisor of 'x' and 'y' function gcd (x, y)
if y == 0 then return math.abs(x) else return gcd(y, x % y) end
end
-- Check GCD of adjacent values in 't' up to 1000 is always 1 function task5 (t)
for pos = 1, 1000 do if gcd(t[pos], t[pos + 1]) ~= 1 then return "FAIL" end end return "PASS"
end
-- Main procedure local sb = sternBrocot(10000) io.write("Task 2: ") for n = 1, 15 do io.write(sb[n] .. " ") end print("\n\nTask 3:") for i = 1, 10 do print("\t" .. i, findFirst(sb, i)) end print("\nTask 4: " .. findFirst(sb, 100)) print("\nTask 5: " .. task5(sb))</lang>
- Output:
Task 2: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 Task 3: 1 1 2 3 3 5 4 9 5 11 6 33 7 19 8 21 9 35 10 39 Task 4: 1179 Task 5: PASS
MAD
<lang MAD> NORMAL MODE IS INTEGER
VECTOR VALUES FRST15 = $20HFIRST 15 NUMBERS ARE*$ VECTOR VALUES FRSTAT = $6HFIRST ,I3,S1,11HAPPEARS AT ,I4*$ VECTOR VALUES NUMBER = $I4*$ DIMENSION STERN(1200) STERN(1) = 1 STERN(2) = 1 R GENERATE FIRST 1200 MEMBERS OF THE STERN-BROCOT SEQUENCE THROUGH GENSEQ, FOR I = 1, 1, I .GE. 600 STERN(I*2-1) = STERN(I) + STERN(I-1)
GENSEQ STERN(I*2) = STERN(I)
R PRINT FIRST 15 VALUES OF STERN-BROCOT SEQUENCE PRINT FORMAT FRST15 THROUGH P15, FOR I = 1, 1, I .G. 15
P15 PRINT ON LINE FORMAT NUMBER, STERN(I)
R PRINT FIRST OCCURRENCE OF 1..10 THROUGH FRST10, FOR I = 1, 1, I .G. 10
FRST10 PRINT FORMAT FRSTAT, I, FIRST.(I)
PRINT FORMAT FRSTAT, 100, FIRST.(100) R SEARCH FOR FIRST OCCURRENCE OF N IN SEQUENCE INTERNAL FUNCTION(N) ENTRY TO FIRST. THROUGH SCAN, FOR K = 1, 1, I .G. 1200
SCAN WHENEVER N .E. STERN(K), FUNCTION RETURN K
END OF FUNCTION END OF PROGRAM</lang>
- Output:
FIRST 15 NUMBERS ARE 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 FIRST 1 APPEARS AT 1 FIRST 2 APPEARS AT 3 FIRST 3 APPEARS AT 5 FIRST 4 APPEARS AT 9 FIRST 5 APPEARS AT 11 FIRST 6 APPEARS AT 33 FIRST 7 APPEARS AT 19 FIRST 8 APPEARS AT 21 FIRST 9 APPEARS AT 35 FIRST 10 APPEARS AT 39 FIRST 100 APPEARS AT 1179
Mathematica / Wolfram Language
<lang Mathematica>sb = {1, 1}; Do[
sb = sb~Join~{Total@sbi - 1 ;; i, sbi} , {i, 2, 1000} ]
Take[sb, 15] Flatten[FirstPosition[sb, #] & /@ Range[10]] First@FirstPosition[sb, 100] AllTrue[Partition[Take[sb, 1000], 2, 1], Apply[GCD] /* EqualTo[1]]</lang>
- Output:
{1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4} {1, 3, 5, 9, 11, 33, 19, 21, 35, 39} 1179 True
Modula-2
<lang modula2>MODULE SternBrocot; FROM InOut IMPORT WriteString, WriteCard, WriteLn;
CONST Amount = 1200;
VAR stern: ARRAY [1..Amount] OF CARDINAL;
i: CARDINAL;
PROCEDURE GCD(a,b: CARDINAL): CARDINAL;
VAR c: CARDINAL;
BEGIN
WHILE b # 0 DO c := a MOD b; a := b; b := c; END; RETURN a;
END GCD;
PROCEDURE Generate;
VAR i: CARDINAL;
BEGIN
stern[1] := 1; stern[2] := 1; FOR i := 2 TO Amount DIV 2 DO stern[i*2 - 1] := stern[i] + stern[i-1]; stern[i*2] := stern[i]; END;
END Generate;
PROCEDURE FindFirst(n: CARDINAL): CARDINAL;
VAR i: CARDINAL;
BEGIN
FOR i := 1 TO Amount DO IF stern[i] = n THEN RETURN i; END; END;
END FindFirst;
PROCEDURE ShowFirst(n: CARDINAL); BEGIN
WriteString("First"); WriteCard(n,4); WriteString(" at "); WriteCard(FindFirst(n), 4); WriteLn;
END ShowFirst;
BEGIN
Generate; WriteString("First 15 numbers:"); FOR i := 1 TO 15 DO WriteCard(stern[i], 2); END; WriteLn; FOR i := 1 TO 10 DO ShowFirst(i); END; ShowFirst(100); WriteLn; FOR i := 2 TO Amount DO IF GCD(stern[i-1], stern[i]) # 1 THEN WriteString("GCD of adjacent elements not 1 at: "); WriteCard(i-1, 4); WriteLn; HALT; END; END; WriteString("The GCD of every pair of adjacent elements is 1."); WriteLn;
END SternBrocot.</lang>
- Output:
First 15 numbers: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179 The GCD of every pair of adjacent elements is 1.
Nim
We use an iterator and store the values in a sequence. To reduce memory usage, we could replace the sequence with a deque and pop the previous value at each round. But it’s no worth to complete the task.
<lang Nim>import math, strformat, strutils
iterator sternBrocot(): (int, int) =
## Yield index and value of the terms of the sequence. var sequence: seq[int] sequence.add 1 sequence.add 1 var index = 1 yield (1, 1) yield (2, 1) while true: sequence.add sequence[index] + sequence[index - 1] yield (sequence.len, sequence[^1]) sequence.add sequence[index] yield (sequence.len, sequence[^1]) inc index
- Fiind first 15 terms.
var res: seq[int] for i, n in sternBrocot():
res.add n if i == 15: break
echo "First 15 terms: ", res.join(" ") echo()
- Find indexes for 1..10.
var indexes: array[1..10, int] var toFind = 10 for i, n in sternBrocot():
if n in 1..10 and indexes[n] == 0: indexes[n] = i dec toFind if toFind == 0: break
for n in 1..10:
echo &"Index of first occurrence of number {n:3}: {indexes[n]:4}"
- Find index for 100.
var index: int for i, n in sternBrocot():
if n == 100: index = i break
echo &"Index of first occurrence of number 100: {index:4}" echo()
- Check GCD.
var prev = 1 index = 1 for i, n in sternBrocot():
if gcd(prev, n) != 1: break prev = n inc index if index > 1000: break
if index <= 1000:
echo "Found two successive terms at index: ", index
else:
echo "All consecutive terms up to the 1000th member have a GCD equal to one."</lang>
- Output:
First 15 terms: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 Index of first occurrence of number 1: 1 Index of first occurrence of number 2: 3 Index of first occurrence of number 3: 5 Index of first occurrence of number 4: 9 Index of first occurrence of number 5: 11 Index of first occurrence of number 6: 33 Index of first occurrence of number 7: 19 Index of first occurrence of number 8: 21 Index of first occurrence of number 9: 35 Index of first occurrence of number 10: 39 Index of first occurrence of number 100: 1179 All consecutive terms up to the 1000th member have a GCD equal to one.
Oforth
<lang Oforth>: stern(n) | l i |
ListBuffer new dup add(1) dup add(1) dup ->l n 1- 2 / loop: i [ l at(i) l at(i 1+) tuck + l add l add ] n 2 mod ifFalse: [ dup removeLast drop ] dup freeze ;
stern(10000) Constant new: Sterns</lang>
- Output:
>Sterns left(15) . [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] ok >10 seq apply(#[ dup . Sterns indexOf . printcr ]) 1 1 2 3 3 5 4 9 5 11 6 33 7 19 8 21 9 35 10 39 ok >Sterns indexOf(100) . 1179 ok >999 seq map(#[ dup Sterns at swap 1 + Sterns at gcd ]) conform(#[ 1 == ]) . 1 ok >
PARI/GP
<lang parigp> \\ Stern-Brocot sequence \\ 5/27/16 aev SternBrocot(n)={ my(L=List([1,1]),k=2); if(n<3,return(L)); for(i=2,n, listput(L,L[i]+L[i-1]); if(k++>=n, break); listput(L,L[i]);if(k++>=n, break)); return(Vec(L)); } \\ Find the first item in any list starting with sind index (return 0 or index). \\ 9/11/2015 aev findinlist(list, item, sind=1)={ my(idx=0, ln=#list); if(ln==0 || sind<1 || sind>ln, return(0)); for(i=sind, ln, if(list[i]==item, idx=i; break;)); return(idx); } { \\ Required tests: my(v,j); v=SternBrocot(15); print1("The first 15: "); print(v); v=SternBrocot(1200); print1("The first i@n: "); \\print(v); for(i=1,10, if(j=findinlist(v,i), print1(i,"@",j,", "))); if(j=findinlist(v,100), print(100,"@",j)); v=SternBrocot(10000); print1("All GCDs=1?: "); j=1; for(i=2,10000, j*=gcd(v[i-1],v[i])); if(j==1, print("Yes"), print("No")); } </lang>
- Output:
The first 15: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] The first i@n: 1@1, 2@3, 3@5, 4@9, 5@11, 6@33, 7@19, 8@21, 9@35, 10@39, 100@1179 All GCDs=1?: Yes
Pascal
<lang pascal>program StrnBrCt; {$IFDEF FPC}
{$MODE DELPHI}
{$ENDIF} const
MaxCnt = 10835282;{ seq[i] < 65536 = high(Word) }
//MaxCnt = 500*1000*1000;{ 2Gbyte -> real 0.85 s user 0.31 } type
tSeqdata = word;//cardinal LongWord pSeqdata = pWord;//pcardinal pLongWord tseq = array of tSeqdata;
function SternBrocotCreate(size:NativeInt):tseq; var
pSeq,pIns : pSeqdata; PosIns : NativeInt; sum : tSeqdata;
Begin
setlength(result,Size+1); dec(Size); //== High(result) pIns := @result[size];// set at end PosIns := -size+2; // negative index campare to 0 pSeq := @result[0];
sum := 1; pSeq[0]:= sum;pSeq[1]:= sum; repeat pIns[PosIns+1] := sum;//append copy of considered inc(sum,pSeq[0]); pIns[PosIns ] := sum; inc(pSeq); inc(PosIns,2);sum := pSeq[1];//aka considered until PosIns>= 0; setlength(result,length(result)-1);
end;
function FindIndex(const s:tSeq;value:tSeqdata):NativeInt; Begin
result := 0; while result <= High(s) do Begin if s[result] = value then EXIT(result+1); inc(result); end;
end;
function gcd_iterative(u, v: NativeInt): NativeInt; //http://rosettacode.org/wiki/Greatest_common_divisor#Pascal_.2F_Delphi_.2F_Free_Pascal var
t: NativeInt;
begin
while v <> 0 do begin t := u;u := v;v := t mod v; end; gcd_iterative := abs(u);
end;
var
seq : tSeq; i : nativeInt;
Begin
seq:= SternBrocotCreate(MaxCnt);
// Show the first fifteen members of the sequence.
For i := 0 to 13 do write(seq[i],',');writeln(seq[14]);
//Show the (1-based) index of where the numbers 1-to-10 first appears in the
For i := 1 to 10 do write(i,' @ ',FindIndex(seq,i),','); writeln(#8#32);
//Show the (1-based) index of where the number 100 first appears in the sequence.
writeln(100,' @ ',FindIndex(seq,100));
//Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.
i := 999; if i > High(seq) then i := High(seq); Repeat IF gcd_iterative(seq[i],seq[i+1]) <>1 then Begin writeln(' failure at ',i+1,' ',seq[i],' ',seq[i+1]); BREAK; end; dec(i); until i <0; IF i< 0 then writeln('GCD-test is O.K.'); setlength(seq,0);
end.</lang>
- Output:
1,1,2,1,3,2,3,1,4,3,5,2,5,3,41 @ 1,2 @ 3,3 @ 5,4 @ 9,5 @ 11,6 @ 33,7 @ 38,8 @ 42,9 @ 47,10 @ 57 100 @ 1179
GCD-test is O.K.
Perl
<lang perl>use strict; use warnings;
sub stern_brocot {
my @list = (1, 1); sub {
push @list, $list[0] + $list[1], $list[1]; shift @list;
}
}
{
my $generator = stern_brocot; print join ' ', map &$generator, 1 .. 15; print "\n";
}
for (1 .. 10, 100) {
my $index = 1; my $generator = stern_brocot; $index++ until $generator->() == $_; print "first occurrence of $_ is at index $index\n";
}
{
sub gcd {
my ($u, $v) = @_; $v ? gcd($v, $u % $v) : abs($u);
} my $generator = stern_brocot; my ($a, $b) = ($generator->(), $generator->()); for (1 .. 1000) {
die "unexpected GCD for $a and $b" unless gcd($a, $b) == 1; ($a, $b) = ($b, $generator->());
}
}</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 first occurrence of 1 is at index 1 first occurrence of 2 is at index 3 first occurrence of 3 is at index 5 first occurrence of 4 is at index 9 first occurrence of 5 is at index 11 first occurrence of 6 is at index 33 first occurrence of 7 is at index 19 first occurrence of 8 is at index 21 first occurrence of 9 is at index 35 first occurrence of 10 is at index 39 first occurrence of 100 is at index 1179
A slightly different method:
<lang perl>use ntheory qw/gcd vecsum vecfirst/;
sub stern_diatomic {
my ($p,$q,$i) = (0,1,shift); while ($i) { if ($i & 1) { $p += $q; } else { $q += $p; } $i >>= 1; } $p;
}
my @s = map { stern_diatomic($_) } 1..15; print "First fifteen: [@s]\n"; @s = map { my $n=$_; vecfirst { stern_diatomic($_) == $n } 1..10000 } 1..10; print "Index of 1-10 first occurrence: [@s]\n"; print "Index of 100 first occurrence: ", (vecfirst { stern_diatomic($_) == 100 } 1..10000), "\n"; print "The first 999 consecutive pairs are ",
(vecsum( map { gcd(stern_diatomic($_),stern_diatomic($_+1)) } 1..999 ) == 999) ? "all coprime.\n" : "NOT all coprime!\n";</lang>
- Output:
First fifteen: [1 1 2 1 3 2 3 1 4 3 5 2 5 3 4] Index of 1-10 first occurrence: [1 3 5 9 11 33 19 21 35 39] Index of 100 first occurrence: 1179 The first 999 consecutive pairs are all coprime.
Phix
sequence sb = {1,1} integer c = 2 function stern_brocot(integer n) while length(sb)<n do sb &= sb[c]+sb[c-1] & sb[c] c += 1 end while return sb[1..n] end function sequence s = stern_brocot(15) puts(1,"first 15:") ?s integer n = 16, k sequence idx = tagset(10) for i=1 to length(idx) do while 1 do k = find(idx[i],s) if k!=0 then exit end if n *= 2 s = stern_brocot(n) end while idx[i] = k end for puts(1,"indexes of 1..10:") ?idx puts(1,"index of 100:") while 1 do k = find(100,s) if k!=0 then exit end if n *= 2 s = stern_brocot(n) end while ?k s = stern_brocot(1000) integer maxgcd = 1 for i=1 to 999 do maxgcd = max(gcd(s[i],s[i+1]),maxgcd) end for printf(1,"max gcd:%d\n",{maxgcd})
- Output:
first 15:{1,1,2,1,3,2,3,1,4,3,5,2,5,3,4} indexes of 1..10:{1,3,5,9,11,33,19,21,35,39} index of 100:1179 max gcd:1
PicoLisp
Using the gcd function defined at Greatest_common_divisor#PicoLisp: <lang PicoLisp>(de nmbr (N)
(let (A 1 B 0) (while (gt0 N) (if (bit? 1 N) (inc 'B A) (inc 'A B) ) (setq N (>> 1 N)) ) B ) )
(let Lst (mapcar nmbr (range 1 2000))
(println 'First-15: (head 15 Lst)) (for N 10 (println 'First N 'found 'at: (index N Lst)) ) (println 'First 100 'found 'at: (index 100 Lst)) (for (L Lst (cdr L) (cddr L)) (test 1 (gcd (car L) (cadr L))) ) (prinl "All consecutive pairs are relative prime!") )</lang>
- Output:
First-15: (1 1 2 1 3 2 3 1 4 3 5 2 5 3 4) First 1 found at: 1 First 2 found at: 3 First 3 found at: 5 First 4 found at: 9 First 5 found at: 11 First 6 found at: 33 First 7 found at: 19 First 8 found at: 21 First 9 found at: 35 First 10 found at: 39 First 100 found at: 1179 All consecutive pairs are relative prime!
PL/I
<lang pli>sternBrocot: procedure options(main);
%replace MAX by 1200; declare S(1:MAX) fixed; /* find the first occurrence of N in S */ findFirst: procedure(n) returns(fixed); declare (n, i) fixed; do i=1 to MAX; if S(i)=n then return(i); end; end findFirst; /* find the greatest common divisor of A and B */ gcd: procedure(a, b) returns(fixed) recursive; declare (a, b) fixed; if b = 0 then return(a); return(gcd(b, mod(a, b))); end gcd; /* calculate S(i) up to MAX */ declare i fixed; S(1) = 1; S(2) = 1; do i=2 to MAX/2; S(i*2-1) = S(i) + S(i-1); S(i*2) = S(i); end; /* print first 15 elements */ put skip list('First 15 elements: '); do i=1 to 15; put edit(S(i)) (F(2)); end; /* find first occurrences of 1..10 and 100 */ do i=1 to 10; put skip list('First',i,'at',findFirst(i)); end; put skip list('First ',100,'at',findFirst(100)); /* check GCDs of adjacent pairs up to 1000th element */ do i=2 to 1000; if gcd(S(i-1),S(i)) ^= 1 then do; put skip list('GCD of adjacent pair not 1 at i=',i); stop; end; end; put skip list('All GCDs of adjacent pairs are 1.');
end sternBrocot;</lang>
- Output:
First 15 elements: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179 All GCDs of adjacent pairs are 1.
PL/M
<lang plm>100H: /* FIND LOCATION OF FIRST ELEMENT IN ARRAY */ FIND$FIRST: PROCEDURE (ARR, EL) ADDRESS;
DECLARE (ARR, N) ADDRESS, (EL, A BASED ARR) BYTE; N = 0;
LOOP:
IF A(N) = EL THEN RETURN N; ELSE N = N + 1; GO TO LOOP;
END FIND$FIRST;
/* CP/M CALL */ BDOS: PROCEDURE (FN, ARG);
DECLARE FN BYTE, ARG ADDRESS; GO TO 5;
END BDOS;
PRINT: PROCEDURE (STRING);
DECLARE STRING ADDRESS; CALL BDOS(9, STRING);
END PRINT;
/* PRINT NUMBER */ PRINT$NUMBER: PROCEDURE (N);
DECLARE S (6) BYTE INITIAL ('.....$'); DECLARE (N, P) ADDRESS, C BASED P BYTE; P = .S(5);
DIGIT:
P = P - 1; C = N MOD 10 + '0'; IF (N := N / 10) > 0 THEN GO TO DIGIT; CALL PRINT(P);
END PRINT$NUMBER;
/* GENERATE FIRST 1200 ELEMENTS OF STERN-BROCOT SEQUENCE */ DECLARE S (1201) BYTE, I ADDRESS; S(1) = 1; S(2) = 1; DO I = 2 TO 600;
S(I*2-1) = S(I) + S(I-1); S(I*2) = S(I);
END;
/* PRINT FIRST 15 ELEMENTS */ CALL PRINT(.'FIRST 15 ELEMENTS: $'); DO I = 1 TO 15;
CALL PRINT$NUMBER(S(I)); CALL PRINT(.' $');
END; CALL PRINT(.(13,10,'$'));
/* PRINT FIRST OCCURRENCE OF N */ PRINT$FIRST: PROCEDURE (N);
DECLARE N BYTE; CALL PRINT(.'FIRST $'); CALL PRINT$NUMBER(N); CALL PRINT(.' AT $'); CALL PRINT$NUMBER(FIND$FIRST(.S, N)); CALL PRINT(.(13,10,'$'));
END PRINT$FIRST;
DO I = 1 TO 10;
CALL PRINT$FIRST(I);
END; CALL PRINT$FIRST(100);
/* CHECK GCDS */ GCD: PROCEDURE (A, B) BYTE;
DECLARE (A, B, C) BYTE;
LOOP:
C = A; A = B; B = C MOD A; IF B <> 0 THEN GO TO LOOP; RETURN A;
END GCD;
DO I = 2 TO 1000;
IF GCD(S(I-1),S(I)) <> 1 THEN DO; CALL PRINT(.'GCD NOT 1 AT: $'); CALL PRINT$NUMBER(I); CALL BDOS(0,0); END;
END;
CALL PRINT(.'ALL GCDS ARE 1$'); CALL BDOS(0,0); EOF</lang>
- Output:
FIRST 15 ELEMENTS: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 FIRST 1 AT 1 FIRST 2 AT 3 FIRST 3 AT 5 FIRST 4 AT 9 FIRST 5 AT 11 FIRST 6 AT 33 FIRST 7 AT 19 FIRST 8 AT 21 FIRST 9 AT 35 FIRST 10 AT 39 FIRST 100 AT 1179 ALL GCDS ARE 1
PowerShell
<lang PowerShell>
- An iterative approach
function iter_sb($count = 2000) {
# Taken from RosettaCode GCD challenge function Get-GCD ($x, $y) { if ($y -eq 0) { $x } else { Get-GCD $y ($x%$y) } }
$answer = @(1,1) $index = 1 while ($answer.Length -le $count) { $answer += $answer[$index] + $answer[$index - 1] $answer += $answer[$index] $index++ } 0..14 | foreach {$answer[$_]}
1..10 | foreach {'Index of {0}: {1}' -f $_, ($answer.IndexOf($_) + 1)}
'Index of 100: {0}' -f ($answer.IndexOf(100) + 1)
[bool] $gcd = $true 1..999 | foreach {$gcd = $gcd -and ((Get-GCD $answer[$_] $answer[$_ - 1]) -eq 1)} 'GCD = 1 for first 1000 members: {0}' -f $gcd
} </lang>
- Output:
PS C:\> iter_sb 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 Index of 1: 1 Index of 2: 3 Index of 3: 5 Index of 4: 9 Index of 5: 11 Index of 6: 33 Index of 7: 19 Index of 8: 21 Index of 9: 35 Index of 10: 39 Index of 100: 1179 GCD = 1 for first 1000 members: True
PureBasic
<lang PureBasic>EnableExplicit Define.i i
If OpenConsole("")
PrintN("Stern-Brocot_sequence")
Else
End 1
EndIf
Procedure.i f(n.i)
If n<2 ProcedureReturn n ElseIf n&1 ProcedureReturn f(n/2)+f(n/2+1) Else ProcedureReturn f(n/2) EndIf
EndProcedure
Procedure.i gcd(a.i,b.i)
If b : ProcedureReturn gcd(b,a%b) : EndIf ProcedureReturn a
EndProcedure
Procedure.i ind(m.i)
Define.i i=1 While f(i)<>m : i+1 : Wend ProcedureReturn i
EndProcedure
Print("First 15 elements: ") For i=1 To 15
Print(Str(f(i))+Space(3))
Next PrintN(~"\n")
For i=1 To 10
PrintN(RSet(Str(i),3)+" is at pos. #"+Str(ind(i)))
Next PrintN("100 is at pos. #"+Str(ind(100))) PrintN("")
i=1 While i<1000 And gcd(f(i),f(i+1))=1 : i+1 : Wend If i=1000
PrintN("All GCDs are 1.")
Else
PrintN("GCD of "+Str(i)+" and "+Str(i+1)+" is not 1")
EndIf
Input()</lang>
- Output:
Stern-Brocot_sequence First 15 elements: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 is at pos. #1 2 is at pos. #3 3 is at pos. #5 4 is at pos. #9 5 is at pos. #11 6 is at pos. #33 7 is at pos. #19 8 is at pos. #21 9 is at pos. #35 10 is at pos. #39 100 is at pos. #1179 All GCDs are 1.
Python
Python: procedural
<lang python>def stern_brocot(predicate=lambda series: len(series) < 20):
"""\ Generates members of the stern-brocot series, in order, returning them when the predicate becomes false
>>> print('The first 10 values:', stern_brocot(lambda series: len(series) < 10)[:10]) The first 10 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3] >>> """
sb, i = [1, 1], 0 while predicate(sb): sb += [sum(sb[i:i + 2]), sb[i + 1]] i += 1 return sb
if __name__ == '__main__':
from fractions import gcd
n_first = 15 print('The first %i values:\n ' % n_first, stern_brocot(lambda series: len(series) < n_first)[:n_first]) print() n_max = 10 for n_occur in list(range(1, n_max + 1)) + [100]: print('1-based index of the first occurrence of %3i in the series:' % n_occur, stern_brocot(lambda series: n_occur not in series).index(n_occur) + 1) # The following would be much faster. Note that new values always occur at odd indices # len(stern_brocot(lambda series: n_occur != series[-2])) - 1)
print() n_gcd = 1000 s = stern_brocot(lambda series: len(series) < n_gcd)[:n_gcd] assert all(gcd(prev, this) == 1 for prev, this in zip(s, s[1:])), 'A fraction from adjacent terms is reducible'</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179
Python: More functional
An iterator is used to produce successive members of the sequence. (its sb variable stores less compared to the procedural version above by popping the last element every time around the while loop.
In checking the gcd's, two iterators are tee'd off from the one stream with the second advanced by one value with its call to next().
See the talk page for how a deque was selected over the use of a straightforward list' <lang python>>>> from itertools import takewhile, tee, islice >>> from collections import deque >>> from fractions import gcd >>> >>> def stern_brocot():
sb = deque([1, 1]) while True: sb += [sb[0] + sb[1], sb[1]] yield sb.popleft()
>>> [s for _, s in zip(range(15), stern_brocot())]
[1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4]
>>> [1 + sum(1 for i in takewhile(lambda x: x != occur, stern_brocot()))
for occur in (list(range(1, 11)) + [100])]
[1, 3, 5, 9, 11, 33, 19, 21, 35, 39, 1179] >>> prev, this = tee(stern_brocot(), 2) >>> next(this) 1 >>> all(gcd(p, t) == 1 for p, t in islice(zip(prev, this), 1000)) True >>> </lang>
Python: Composing pure (curried) functions
Composing and testing a Stern-Brocot function by composition of generic and reusable functional abstractions (curried for more flexible nesting and rearrangement).
<lang python>Stern-Brocot sequence
import math import operator from itertools import count, dropwhile, islice, takewhile
- sternBrocot :: Generator [Int]
def sternBrocot():
Non-finite list of the Stern-Brocot sequence of integers. def go(xs): [a, b] = xs[:2] return (a, xs[1:] + [a + b, b]) return unfoldr(go)([1, 1])
- ------------------------ TESTS -------------------------
- main :: IO ()
def main():
Various tests
[eq, ne, gcd] = map( curry, [operator.eq, operator.ne, math.gcd] )
sbs = take(1200)(sternBrocot()) ixSB = zip(sbs, enumFrom(1))
print(unlines(map(str, [
# First 15 members of the sequence. take(15)(sbs),
# Indices of where the numbers [1..10] first appear. take(10)( nubBy(on(eq)(fst))( sorted( takewhile( compose(ne(12))(fst), ixSB ), key=fst ) ) ),
# Index of where the number 100 first appears. take(1)(dropwhile(compose(ne(100))(fst), ixSB)),
# Is the gcd of any two consecutive members, # up to the 1000th member, always one ? every(compose(eq(1)(gcd)))( take(1000)(zip(sbs, tail(sbs))) ) ])))
- ----------------- GENERIC ABSTRACTIONS -----------------
- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
Right to left function composition. return lambda f: lambda x: g(f(x))
- curry :: ((a, b) -> c) -> a -> b -> c
def curry(f):
A curried function derived from an uncurried function. return lambda a: lambda b: f(a, b)
- enumFrom :: Enum a => a -> [a]
def enumFrom(x):
A non-finite stream of enumerable values, starting from the given value. return count(x) if isinstance(x, int) else ( map(chr, count(ord(x))) )
- every :: (a -> Bool) -> [a] -> Bool
def every(p):
True if p(x) holds for every x in xs return lambda xs: all(map(p, xs))
- fst :: (a, b) -> a
def fst(tpl):
First member of a pair. return tpl[0]
- head :: [a] -> a
def head(xs):
The first element of a non-empty list. return xs[0]
- nubBy :: (a -> a -> Bool) -> [a] -> [a]
def nubBy(p):
A sublist of xs from which all duplicates, (as defined by the equality predicate p) are excluded. def go(xs): if not xs: return [] x = xs[0] return [x] + go( list(filter( lambda y: not p(x)(y), xs[1:] )) ) return go
- on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
def on(f):
A function returning the value of applying the binary f to g(a) g(b) return lambda g: lambda a: lambda b: f(g(a))(g(b))
- tail :: [a] -> [a]
- tail :: Gen [a] -> [a]
def tail(xs):
The elements following the head of a (non-empty) list or generator stream. if isinstance(xs, list): return xs[1:] else: islice(xs, 1) # First item dropped. return xs
- take :: Int -> [a] -> [a]
- take :: Int -> String -> String
def take(n):
The prefix of xs of length n, or xs itself if n > length xs. return lambda xs: ( xs[0:n] if isinstance(xs, list) else list(islice(xs, n)) )
- unfoldr :: (b -> Maybe (a, b)) -> b -> Gen [a]
def unfoldr(f):
A lazy (generator) list unfolded from a seed value by repeated application of f until no residue remains. Dual to fold/reduce. f returns either None or just (value, residue). For a strict output list, wrap the result with list() def go(x): valueResidue = f(x) while valueResidue: yield valueResidue[0] valueResidue = f(valueResidue[1]) return go
- unlines :: [String] -> String
def unlines(xs):
A single string derived by the intercalation of a list of strings with the newline character. return '\n'.join(xs)
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
[1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] [(1, 1), (2, 3), (3, 5), (4, 9), (5, 11), (6, 33), (7, 19), (8, 21), (9, 35), (10, 39)] [(100, 1179)] True
R
For loop
<lang rsplus>
- Stern-Brocot sequence
- 12/19/16 aev
SternBrocot <- function(n){
V <- 1; k <- n/2; for (i in 1:k) { V[2*i] = V[i]; V[2*i+1] = V[i] + V[i+1];} return(V);
}
- Required tests:
require(pracma); { cat(" *** The first 15:",SternBrocot(15),"\n"); cat(" *** The first i@n:","\n"); V=SternBrocot(40); for (i in 1:10) {j=match(i,V); cat(i,"@",j,",")} V=SternBrocot(1200); i=100; j=match(i,V); cat(i,"@",j,"\n"); V=SternBrocot(1000); j=1; for (i in 2:1000) {j=j*gcd(V[i-1],V[i])} if(j==1) {cat(" *** All GCDs=1!\n")} else {cat(" *** All GCDs!=1??\n")} } </lang>
- Output:
> require(pracma) Loading required package: pracma *** The first 15: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 *** The first i@n: 1 @ 1 ,2 @ 3 ,3 @ 5 ,4 @ 9 ,5 @ 11 ,6 @ 33 ,7 @ 19 ,8 @ 21 ,9 @ 35 ,10 @ 39 ,100 @ 1179 *** All GCDs=1! >
While loop
Tasks like this are R's bread and butter. The previous solution uses smart mathematical tricks to generate the desired sequence with a for loop and uses a similarly clever for loop for the task involving gcds. However, R is smart enough to let us avoid this work by writing some much more idiomatic code.
As with the previous solution, we have used a library for our gcd function. In this case, we have used gmp. <lang rsplus>genNStern <- function(n) {
sternNums <- c(1, 1) i <- 2 while((endIndex <- length(sternNums)) < n) { #To show off R's vectorization, the following line is deliberately terse. #It assigns sternNums[i-1]+sternNums[i] to sternNums[endIndex+1] #and it assigns sternNums[i], the "considered" number, to sternNums[endIndex+2], now the end of the sequence. #Note that we do not have to initialize a big sternNums array to do this. #True to the algorithm, the new entries are appended to the end of the old sequence. sternNums[endIndex + c(1, 2)] <- c(sum(sternNums[c(i - 1, i)]), sternNums[i]) i <- i + 1 } sternNums[seq_len(n)]
}
- N=5000 was picked arbitrarily. The code runs very fast regardless of this number being much more than we need.
firstFiveThousandTerms <- genNStern(5000) match(1:10, firstFiveThousandTerms) match(100, firstFiveThousandTerms) all(sapply(1:999, function(i) gmp::gcd(firstFiveThousandTerms[i], firstFiveThousandTerms[i + 1])) == 1)</lang>
- Output:
> firstFiveThousandTerms <- genNStern(5000) > match(1:10, firstFiveThousandTerms) [1] 1 3 5 9 11 33 19 21 35 39 > match(100, firstFiveThousandTerms) [1] 1179 > all(sapply(1:999, function(i) gmp::gcd(firstFiveThousandTerms[i], firstFiveThousandTerms[i + 1])) == 1) [1] TRUE
Racket
<lang racket>#lang racket
- OEIS Definition
- A002487
- Stern's diatomic series
- (or Stern-Brocot sequence)
- a(0) = 0, a(1) = 1;
- for n > 0
- a(2*n) = a(n),
- a(2*n+1) = a(n) + a(n+1).
(define A002487
(let ((memo (make-hash '((0 . 0) (1 . 1))))) (lambda (n) (hash-ref! memo n (lambda () (define n/2 (quotient n 2)) (+ (A002487 n/2) (if (even? n) 0 (A002487 (add1 n/2)))))))))
(define Stern-Brocot A002487)
(displayln "Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4)") (for/list ((i (in-range 1 (add1 15)))) (Stern-Brocot i))
(displayln "Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence.") (for ((n (in-range 1 (add1 10))))
(for/first ((i (in-naturals 1)) #:when (= n (Stern-Brocot i))) (printf "~a first found at a(~a)~%" n i)))
(displayln "Show the (1-based) index of where the number 100 first appears in the sequence.") (for/first ((i (in-naturals 1)) #:when (= 100 (Stern-Brocot i))) i)
(displayln "Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.") (unless
(for/first ((i (in-range 1 1000)) #:unless (= 1 (gcd (Stern-Brocot i) (Stern-Brocot (add1 i))))) #t) (display "\tdidn't find gcd > (or otherwise ≠) 1"))</lang>
- Output:
Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4) (1 1 2 1 3 2 3 1 4 3 5 2 5 3 4) Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence. 1 first found at a(1) 2 first found at a(3) 3 first found at a(5) 4 first found at a(9) 5 first found at a(11) 6 first found at a(33) 7 first found at a(19) 8 first found at a(21) 9 first found at a(35) 10 first found at a(39) Show the (1-based) index of where the number 100 first appears in the sequence. 1179 Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one. didn't find gcd > (or otherwise ≠) 1
Raku
(formerly Perl 6)
<lang perl6>constant @Stern-Brocot = 1, 1, { |(@_[$_ - 1] + @_[$_], @_[$_]) given ++$ } ... *; put @Stern-Brocot[^15]; for flat 1..10, 100 -> $ix {
say "First occurrence of {$ix.fmt('%3d')} is at index: {(1+@Stern-Brocot.first($ix, :k)).fmt('%4d')}";
} say so 1 == all map ^1000: { [gcd] @Stern-Brocot[$_, $_ + 1] }</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First occurrence of 1 is at index: 1 First occurrence of 2 is at index: 3 First occurrence of 3 is at index: 5 First occurrence of 4 is at index: 9 First occurrence of 5 is at index: 11 First occurrence of 6 is at index: 33 First occurrence of 7 is at index: 19 First occurrence of 8 is at index: 21 First occurrence of 9 is at index: 35 First occurrence of 10 is at index: 39 First occurrence of 100 is at index: 1179 True
REXX
<lang rexx>/*REXX program generates & displays a Stern─Brocot sequence; finds 1─based indices; GCDs*/ parse arg N idx fix chk . /*get optional arguments from the C.L. */ if N== | N=="," then N= 15 /*Not specified? Then use the default.*/ if idx== | idx=="," then idx= 10 /* " " " " " " */ if fix== | fix=="," then fix= 100 /* " " " " " " */ if chk== | chk=="," then chk= 1000 /* " " " " " " */
if N>0 then say center('the first' N "numbers in the Stern─Brocot sequence", 70, '═') a= Stern_Brocot(N) /*invoke function to generate sequence.*/ say a /*display the sequence to the terminal.*/ say say center('the 1─based index for the first' idx "integers", 70, '═') a= Stern_Brocot(-idx) /*invoke function to generate sequence.*/ w= length(idx); do i=1 for idx
say 'for ' right(i, w)", the index is: " wordpos(i, a) end /*i*/
say say center('the 1─based index for' fix, 70, "═") a= Stern_Brocot(-fix) /*invoke function to generate sequence.*/ say 'for ' fix", the index is: " wordpos(fix, a) say if chk<2 then exit 0 say center('checking if all two consecutive members have a GCD=1', 70, '═') a= Stern_Brocot(chk) /*invoke function to generate sequence.*/
do c=1 for chk-1; if gcd(subword(a, c, 2))==1 then iterate say 'GCD check failed at index' c; exit 13 end /*c*/
say say '───── All ' chk " two consecutive members have a GCD of unity." exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ gcd: procedure; $=; do i=1 for arg(); $= $ arg(i) /*get arg list. */
end /*i*/ parse var $ x z .; if x=0 then x= z /*is zero case? */ x=abs(x) /*use absolute x*/ do j=2 to words($); y=abs( word($, j) ) if y=0 then iterate /*ignore zeros. */ do until y==0; parse value x//y y with y x /* ◄──heavy work*/ end /*until*/ end /*j*/ return x /*return the GCD*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ Stern_Brocot: parse arg h 1 f; $= 1 1; if h<0 then h= 1e9
else f= 0 f= abs(f) do k=2 until words($)>=h | wordpos(f, $)\==0 _= word($, k); $= $ (_ + word($, k-1) ) _ end /*k*/ if f==0 then return subword($, 1, h) return $</lang>
- output when using the default inputs:
══════════the first 15 numbers in the Stern─Brocot sequence═══════════ 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 ═════════════the 1-based index for the first 10 integers══════════════ for 1, the index is: 1 for 2, the index is: 3 for 3, the index is: 5 for 4, the index is: 9 for 5, the index is: 11 for 6, the index is: 33 for 7, the index is: 19 for 8, the index is: 21 for 9, the index is: 35 for 10, the index is: 39 ══════════════════════the 1-based index for 100═══════════════════════ for 100, the index is: 1179 ═════════checking if all two consecutive members have a GCD=1═════════ ───── All 1000 two consecutive members have a GCD of unity.
Ring
<lang ring>
- Project : Stern-Brocot sequence
limit = 1200 item = list(limit+1) item[1] = 1 item[2] = 1 nr = 2 gcd = 1 gcdall = 1 for num = 3 to limit
item[num] = item[nr] + item[nr-1] item[num+1] = item[nr] nr = nr + 1 num = num + 1
next showarray(item,15)
for x = 1 to 100
if x < 11 or x = 100 totalitem(x) ok
next
for n = 1 to len(item) - 1
if gcd(item[n],item[n+1]) != 1 gcdall = gcd ok
next
if gcdall = 1
see "Correct: The first 999 consecutive pairs are relative prime!" + nl
ok
func totalitem(p)
pos = find(item, p) see string(x) + " at Stern #" + pos + "." + nl
func showarray(vect,ln)
svect = "" for n = 1 to ln svect = svect + vect[n] + ", " next svect = left(svect, len(svect) - 2) see svect see nl
func gcd(gcd,b)
while b c = gcd gcd = b b = c % b end return gcd
</lang> Output:
1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 1 at Stern #1. 2 at Stern #3. 3 at Stern #5. 4 at Stern #9. 5 at Stern #11. 6 at Stern #33. 7 at Stern #19. 8 at Stern #21. 9 at Stern #35. 10 at Stern #39. 100 at Stern #1179. Correct: The first 999 consecutive pairs are relative prime!
Ruby
<lang ruby>def sb
return enum_for :sb unless block_given? a=[1,1] 0.step do |i| yield a[i] a << a[i]+a[i+1] << a[i+1] end
end
puts "First 15: #{sb.first(15)}"
[*1..10,100].each do |n|
puts "#{n} first appears at #{sb.find_index(n)+1}."
end
if sb.take(1000).each_cons(2).all? { |a,b| a.gcd(b) == 1 }
puts "All GCD's are 1"
else
puts "Whoops, not all GCD's are 1!"
end</lang>
- Output:
First 15: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1 first appears at 1. 2 first appears at 3. 3 first appears at 5. 4 first appears at 9. 5 first appears at 11. 6 first appears at 33. 7 first appears at 19. 8 first appears at 21. 9 first appears at 35. 10 first appears at 39. 100 first appears at 1179. All GCD's are 1
Scala
<lang scala>lazy val sbSeq: Stream[BigInt] = {
BigInt("1") #:: BigInt("1") #:: (sbSeq zip sbSeq.tail zip sbSeq.tail). flatMap{ case ((a,b),c) => List(a+b,c) }
}
// Show the results { println( s"First 15 members: ${(for( n <- 0 until 15 ) yield sbSeq(n)) mkString( "," )}" ) println for( n <- 1 to 10; pos = sbSeq.indexOf(n) + 1 ) println( s"Position of first $n is at $pos" ) println println( s"Position of first 100 is at ${sbSeq.indexOf(100) + 1}" ) println println( s"Greatest Common Divisor for first 1000 members is 1: " +
(sbSeq zip sbSeq.tail).take(1000).forall{ case (a,b) => a.gcd(b) == 1 } )
} </lang>
- Output:
First 15 members: 1,1,2,1,3,2,3,1,4,3,5,2,5,3,4 Position of first 1 is at 1 Position of first 2 is at 3 Position of first 3 is at 5 Position of first 4 is at 9 Position of first 5 is at 11 Position of first 6 is at 33 Position of first 7 is at 19 Position of first 8 is at 21 Position of first 9 is at 35 Position of first 10 is at 39 Position of first 100 is at 1179 Greatest Common Divisor for first 1000 members is 1: true
Sidef
<lang ruby># Declare a function to generate the Stern-Brocot sequence func stern_brocot {
var list = [1, 1] { list.append(list[0]+list[1], list[1]) list.shift }
}
- Show the first fifteen members of the sequence.
say 15.of(stern_brocot()).join(' ')
- Show the (1-based) index of where the numbers 1-to-10 first appears
- in the sequence, and where the number 100 first appears in the sequence.
for i (1..10, 100) {
var index = 1 var generator = stern_brocot() while (generator() != i) { ++index } say "First occurrence of #{i} is at index #{index}"
}
- Check that the greatest common divisor of all the two consecutive
- members of the series up to the 1000th member, is always one.
var generator = stern_brocot() var (a, b) = (generator(), generator()) {
assert_eq(gcd(a, b), 1) a = b b = generator()
} * 1000
say "All GCD's are 1"</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First occurrence of 1 is at index 1 First occurrence of 2 is at index 3 First occurrence of 3 is at index 5 First occurrence of 4 is at index 9 First occurrence of 5 is at index 11 First occurrence of 6 is at index 33 First occurrence of 7 is at index 19 First occurrence of 8 is at index 21 First occurrence of 9 is at index 35 First occurrence of 10 is at index 39 First occurrence of 100 is at index 1179 All GCD's are 1
Snobol
<lang snobol>* GCD function
DEFINE('GCD(A,B)') :(GCD_END)
GCD GCD = A
EQ(B,0) :S(RETURN) A = B B = REMDR(GCD,B) :(GCD)
GCD_END
- Find first occurrence of element in array
DEFINE('IDX(ARR,ELM)') :(IDX_END)
IDX IDX = 1 ITEST EQ(ARR<IDX>,ELM) :S(RETURN)
IDX = IDX + 1 :(ITEST)
IDX_END
- Declare array
SEQ = ARRAY(1200,1)
- Fill array with Stern-Brocot sequence
IX = 1
FILL IX = IX + 1
SEQ<IX * 2 - 1> = SEQ<IX> + SEQ<IX - 1> SEQ<IX * 2> = SEQ<IX> :S(FILL)
- Print first 15 elements
DONE IX = 1
S = "First 15 elements:"
P15 S = S " " SEQ<IX>
IX = IX + 1 LT(IX,15) :S(P15) OUTPUT = S
- Print first occurrence of 1..10 and 100
N = 1
FIRSTN OUTPUT = "First " N " at " IDX(SEQ,N)
N = N + 1 LT(N,10) :S(FIRSTN) OUTPUT = "First 100 at " IDX(SEQ,100)
- Test GCD between 1000 consecutive members
IX = 2
GCDTEST EQ(GCD(SEQ<IX - 1>,SEQ<IX>),1) :F(GCDFAIL)
IX = IX + 1 LT(IX,1000) :S(GCDTEST) OUTPUT = "All GCDs are 1." :(END)
GCDFAIL OUTPUT = "GCD is not 1 at " IX "."
END</lang>
- Output:
First 15 elements: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179 All GCDs are 1.
Swift
<lang swift>struct SternBrocot: Sequence, IteratorProtocol {
private var seq = [1, 1]
mutating func next() -> Int? { seq += [seq[0] + seq[1], seq[1]]
return seq.removeFirst() }
}
func gcd<T: BinaryInteger>(_ a: T, _ b: T) -> T {
guard a != 0 else { return b }
return a < b ? gcd(b % a, a) : gcd(a % b, b)
}
print("First 15: \(Array(SternBrocot().prefix(15)))")
var found = Set<Int>()
for (i, val) in SternBrocot().enumerated() {
switch val { case 1...10 where !found.contains(val), 100 where !found.contains(val): print("First \(val) at \(i + 1)") found.insert(val) case _: continue }
if found.count == 11 { break }
}
let firstThousand = SternBrocot().prefix(1000) let gcdIsOne = zip(firstThousand, firstThousand.dropFirst()).allSatisfy({ gcd($0.0, $0.1) == 1 })
print("GCDs of all two consecutive members are \(gcdIsOne ? "" : "not")one")</lang>
- Output:
First 15: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 7 at 19 First 8 at 21 First 6 at 33 First 9 at 35 First 10 at 39 First 100 at 1179 GCDs of all two consecutive members are one
Tcl
<lang tcl>
- !/usr/bin/env tclsh
package require generator ;# from tcllib
namespace eval stern-brocot {
proc generate Template:Count 100 { set seq {1 1} set n 0 while {[llength $seq] < $count} { lassign [lrange $seq $n $n+1] a b lappend seq [expr {$a + $b}] $b incr n } return $seq }
proc genr {} { yield [info coroutine] set seq {1 1} while {1} { set seq [lassign $seq a] set b [lindex $seq 0] set c [expr {$a + $b}] lappend seq $c $b yield $a } }
proc Step {a b args} { set c [expr {$a + $b}] list $a [list $b {*}$args $c $b] }
generator define gen {} { set cmd [list 1 1] while {1} { lassign [Step {*}$cmd] a cmd generator yield $a } }
namespace export {[a-z]*} namespace ensemble create
}
interp alias {} sb {} stern-brocot
- a simple adaptation of gcd from http://wiki.tcl.tk/2891
proc coprime {a args} {
set gcd $a foreach arg $args { while {$arg != 0} { set t $arg set arg [expr {$gcd % $arg}] set gcd $t if {$gcd == 1} {return true} } } return false
}
proc main {} {
puts "#1. First 15 members of the Stern-Brocot sequence:" puts \t[generator to list [generator take 16 [sb gen]]]
puts "#2. First occurrences of 1 through 10:" set first {} set got 0 set i 0 generator foreach x [sb gen] { incr i if {$x>10} continue if {[dict exists $first $x]} continue dict set first $x $i if {[incr got] >= 10} break } foreach {a b} [lsort -integer -stride 2 $first] { puts "\tFirst $a at $b" }
puts "#3. First occurrence of 100:" set i 0 generator foreach x [sb gen] { incr i if {$x eq 100} break } puts "\tFirst $x at $i"
puts "#4. Check first 1k elements for common divisors:" set prev [expr {2*3*5*7*11*13*17*19+1}] ;# a handy prime set i 0 generator foreach x [sb gen] { if {[incr i] >= 1000} break if {![coprime $x $prev]} { error "Element $i, $x is not coprime with $prev!" } set prev $x } puts "\tFirst $i elements are all pairwise coprime"
}
main </lang>
- Output:
#1. First 15 members of the Stern-Brocot sequence: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 #2. First occurrences of 1 through 10: First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 #3. First occurrence of 100: First 100 at 1179 #4. Check first 1k elements for common divisors: First 1000 elements are all pairwise coprime
VBScript
<lang VBScript>sb = Array(1,1) i = 1 'considered j = 2 'precedent n = 0 'loop counter Do ReDim Preserve sb(UBound(sb) + 1) sb(UBound(sb)) = sb(UBound(sb) - i) + sb(UBound(sb) - j) ReDim Preserve sb(UBound(sb) + 1) sb(UBound(sb)) = sb(UBound(sb) - j) i = i + 1 j = j + 1 n = n + 1 Loop Until n = 2000
WScript.Echo "First 15: " & DisplayElements(15)
For k = 1 To 10 WScript.Echo "The first instance of " & k & " is in #" & ShowFirstInstance(k) & "." Next
WScript.Echo "The first instance of " & 100 & " is in #" & ShowFirstInstance(100) & "."
Function DisplayElements(n) For i = 0 To n - 1 If i < n - 1 Then DisplayElements = DisplayElements & sb(i) & ", " Else DisplayElements = DisplayElements & sb(i) End If Next End Function
Function ShowFirstInstance(n) For i = 0 To UBound(sb) If sb(i) = n Then ShowFirstInstance = i + 1 Exit For End If Next End Function</lang>
- Output:
First 15: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 The first instance of 1 is in #1. The first instance of 2 is in #3. The first instance of 3 is in #5. The first instance of 4 is in #9. The first instance of 5 is in #11. The first instance of 6 is in #33. The first instance of 7 is in #19. The first instance of 8 is in #21. The first instance of 9 is in #35. The first instance of 10 is in #39. The first instance of 100 is in #1179.
Visual Basic .NET
<lang vbnet>Imports System Imports System.Collections.Generic Imports System.Linq
Module Module1
Dim l As List(Of Integer) = {1, 1}.ToList()
Function gcd(ByVal a As Integer, ByVal b As Integer) As Integer Return If(a > 0, If(a < b, gcd(b Mod a, a), gcd(a Mod b, b)), b) End Function
Sub Main(ByVal args As String()) Dim max As Integer = 1000, take As Integer = 15, i As Integer = 1, selection As Integer() = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100} Do : l.AddRange({l(i) + l(i - 1), l(i)}.ToList) : i += 1 Loop While l.Count < max OrElse l(l.Count - 2) <> selection.Last() Console.Write("The first {0} items In the Stern-Brocot sequence: ", take) Console.WriteLine("{0}" & vbLf, String.Join(", ", l.Take(take))) Console.WriteLine("The locations of where the selected numbers (1-to-10, & 100) first appear:") For Each ii As Integer In selection Dim j As Integer = l.FindIndex(Function(x) x = ii) + 1 Console.WriteLine("{0,3}: {1:n0}", ii, j) Next : Console.WriteLine() : Dim good As Boolean = True : For i = 1 To max If gcd(l(i), l(i - 1)) <> 1 Then good = False : Exit For Next Console.WriteLine("The greatest common divisor of all the two consecutive items of the" & " series up to the {0}th item is {1}always one.", max, If(good, "", "not ")) End Sub
End Module</lang>
- Output:
The first 15 items In the Stern-Brocot sequence: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 The locations of where the selected numbers (1-to-10, & 100) first appear: 1: 1 2: 3 3: 5 4: 9 5: 11 6: 33 7: 19 8: 21 9: 35 10: 39 100: 1,179 The greatest common divisor of all the two consecutive items of the series up to the 1000th item is always one.
Wren
<lang ecmascript>import "/math" for Int import "/fmt" for Fmt
var sbs = [1, 1]
var sternBrocot = Fn.new { |n, fromStart|
if (n < 4 || (n % 2 != 0)) Fiber.abort("n must be >= 4 and even.") var consider = fromStart ? 1 : (n/2).floor - 1 while (true) { var sum = sbs[consider] + sbs[consider - 1] sbs.add(sum) sbs.add(sbs[consider]) if (sbs.count == n) return consider = consider + 1 }
}
var n = 16 // needs to be even to ensure 'considered' number is added System.print("First 15 members of the Stern-Brocot sequence:") sternBrocot.call(n, true) System.print(sbs.take(15).toList)
var firstFind = List.filled(11, 0) firstFind[0] = -1 // needs to be non-zero for subsequent test var i = 0 for (v in sbs) {
if (v <= 10 && firstFind[v] == 0) firstFind[v] = i + 1 i = i + 1
} while (true) {
n = n + 2 sternBrocot.call(n, false) var vv = sbs[-2..-1] var m = n - 1 var outer = false for (v in vv) { if (v <= 10 && firstFind[v] == 0) firstFind[v] = m if (firstFind.all { |e| e != 0 }) { outer = true break } m = m + 1 } if (outer) break
} System.print("\nThe numbers 1 to 10 first appear at the following indices:") for (i in 1..10) Fmt.print("$2d -> $d", i, firstFind[i])
System.write("\n100 first appears at index ") while (true) {
n = n + 2 sternBrocot.call(n, false) var vv = sbs[-2..-1] if (vv[0] == 100) { System.print("%(n - 1).") break } if (vv[1] == 100) { System.print("%(n).") break }
}
System.write("\nThe GCDs of each pair of the series up to the 1,000th member are ") var p = 0 while (p < 1000) {
if (Int.gcd(sbs[p], sbs[p + 1]) != 1) { System.print("not all one.") return } p = p + 2
} System.print("all one.")</lang>
- Output:
First 15 members of the Stern-Brocot sequence: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] The numbers 1 to 10 first appear at the following indices: 1 -> 1 2 -> 3 3 -> 5 4 -> 9 5 -> 11 6 -> 33 7 -> 19 8 -> 21 9 -> 35 10 -> 39 100 first appears at index 1179. The GCDs of each pair of the series up to the 1,000th member are all one.
zkl
<lang zkl>fcn SB // Stern-Brocot sequence factory --> Walker
{ Walker(fcn(sb,n){ a,b:=sb; sb.append(a+b,b); sb.del(0); a }.fp(L(1,1))) }
SB().walk(15).println();
[1..10].zipWith('wrap(n){ [1..].zip(SB())
.filter(1,fcn(n,sb){ n==sb[1] }.fp(n)) }) .walk().println();
[1..].zip(SB()).filter1(fcn(sb){ 100==sb[1] }).println();
sb:=SB(); do(500){ if(sb.next().gcd(sb.next())!=1) println("Oops") }</lang>
- Output:
L(1,1,2,1,3,2,3,1,4,3,5,2,5,3,4) L(L(L(1,1)),L(L(3,2)),L(L(5,3)),L(L(9,4)),L(L(11,5)),L(L(33,6)),L(L(19,7)),L(L(21,8)),L(L(35,9)),L(L(39,10))) L(1179,100)
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