Add tasks for all the new languages
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Tina Müller
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PROGRAM CATALAN
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!$DOUBLE
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DIM CATALAN[50]
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FUNCTION ODD(X)
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ODD=FRC(X/2)<>0
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END FUNCTION
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PROCEDURE GETCATALAN(L)
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LOCAL J,K,W
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LOCAL DIM PASTRI[100]
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L=L*2
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PASTRI[0]=1
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J=0
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WHILE J<L DO
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J+=1
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K=INT((J+1)/2)
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PASTRI[K]=PASTRI[K-1]
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FOR W=K TO 1 STEP -1 DO
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PASTRI[W]+=PASTRI[W-1]
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END FOR
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IF NOT(ODD(J)) THEN
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K=INT(J/2)
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CATALAN[K]=PASTRI[K]-PASTRI[K-1]
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END IF
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END WHILE
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END PROCEDURE
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BEGIN
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LL=15
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GETCATALAN(LL)
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FOR I=1 TO LL DO
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WRITE("### ####################";I;CATALAN[I])
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END FOR
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END PROGRAM
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(define dim 100)
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(define-syntax-rule (Tidx i j) (+ i (* dim j)))
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;; generates Catalan's triangle
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;; T (i , j) = T(i-1,j) + T (i, j-1)
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(define (T n)
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(define i (modulo n dim))
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(define j (quotient n dim))
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(cond
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((zero? i) 1) ;; left column = 1
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((= i j) (T (Tidx (1- i) j))) ;; diagonal value = left value
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(else (+ (T (Tidx (1- i) j)) (T (Tidx i (1- j)))))))
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(remember 'T #(1))
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;; take elements on diagonal = Catalan numbers
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(for ((i (in-range 0 16))) (write (T (Tidx i i))))
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→ 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
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' version 15-09-2015
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' compile with: fbc -s console
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#Define size 31 ' (N * 2 + 1)
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Sub pascal_triangle(rows As Integer, Pas_tri() As ULongInt)
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Dim As Integer x, y
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For x = 1 To rows
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Pas_tri(1,x) = 1
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Pas_tri(x,1) = 1
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Next
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For x = 2 To rows
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For y = 2 To rows + 1 - x
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Pas_tri(x, y) = pas_tri(x - 1 , y) + pas_tri(x, y - 1)
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Next
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Next
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End Sub
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' ------=< MAIN >=------
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Dim As Integer count, row
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Dim As ULongInt triangle(1 To size, 1 To size)
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pascal_triangle(size, triangle())
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' 1 1 1 1 1 1
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' 1 2 3 4 5 6
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' 1 3 6 10 15 21
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' 1 4 10 20 35 56
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' 1 5 15 35 70 126
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' 1 6 21 56 126 252
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' The Pascal triangle is rotated 45 deg.
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' to find the Catalan number we need to follow the diagonal
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' for top left to bottom right
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' take the number on diagonal and subtract the number in de cell
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' one up and one to right
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' 1 (2 - 1), 2 (6 - 4), 5 (20 - 15) ...
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Print "The first 15 Catalan numbers are" : print
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count = 1 : row = 2
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Do
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Print Using "###: #########"; count; triangle(row, row) - triangle(row +1, row -1)
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row = row + 1
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count = count + 1
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Loop Until count > 15
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' empty keyboard buffer
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While InKey <> "" : Wend
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Print : Print "hit any key to end program"
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Sleep
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End
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const n = 15
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var t = newSeq[int](n + 2)
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t[1] = 1
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for i in 1..n:
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for j in countdown(i, 1): t[j] += t[j-1]
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t[i+1] = t[i]
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for j in countdown(i+1, 1): t[j] += t[j-1]
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stdout.write t[i+1] - t[i], " "
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: pascal(n) [ 1 ] #[ dup 0 + 0 rot + zipWith(#+) ] times(n) ;
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: catalan(n) pascal(n 2 * ) at(n 1+) n 1+ / ;
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constant N = 15 -- accurate to 30, nan/inf for anything over 514 (bigatom version is below).
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sequence catalan = {}, -- (>=1 only)
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p = repeat(1,N+1)
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atom p1
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for i=1 to N do
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p1 = p[1]*2
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catalan = append(catalan,p1-p[2])
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for j=1 to N-i+1 do
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p1 += p[j+1]
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p[j] = p1
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end for
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-- ?p[1..N-i+1]
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end for
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?catalan
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-- FreeBASIC said:
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--' 1 1 1 1 1 1
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--' 1 2 3 4 5 6
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--' 1 3 6 10 15 21
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--' 1 4 10 20 35 56
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--' 1 5 15 35 70 126
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--' 1 6 21 56 126 252
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--' The Pascal triangle is rotated 45 deg.
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--' to find the Catalan number we need to follow the diagonal
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--' for top left to bottom right
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--' take the number on diagonal and subtract the number in de cell
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--' one up and one to right
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--' 1 (2 - 1), 2 (6 - 4), 5 (20 - 15) ...
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--
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-- The first thing that struck me was it is twice as big as it needs to be,
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-- something like this would do...
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-- 1 1 1 1 1 1
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-- 2 3 4 5 6
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-- 6 10 15 21
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-- 20 35 56
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-- 70 126
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-- 252
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-- It is more obvious from the upper square that the diagonal on that, which is
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-- that same as column 1 on this, is twice the previous, which on the second
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-- diagram is in column 2. Further, once we have calculated the value for column
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-- one above, we can use it immediately to calculate the next catalan number and
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-- do not need to store it. Lastly we can overwrite row 1 with row 2 etc in situ,
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-- and the following shows what we need for subsequent rounds:
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-- 1 1 1 1 1
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-- 3 4 5 6
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-- 10 15 21
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-- 35 56
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-- 126 (unused)
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include builtins\bigatom.e
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function catalanB(integer n) -- very very fast!
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sequence catalan = {},
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p = repeat(1,n+1)
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bigatom p1
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if n=0 then return 1 end if
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for i=1 to n do
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p1 = ba_multiply(p[1],2)
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catalan = append(catalan,ba_sub(p1,p[2]))
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for j=1 to n-i+1 do
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p1 = ba_add(p1,p[j+1])
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p[j] = p1
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end for
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end for
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return catalan[n]
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end function
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atom t0 = time()
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string sc100 = ba_sprint(catalanB(100))
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printf(1,"%d: %s (%3.2fs)\n",{100,sc100,time()-t0})
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atom t0 = time()
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string sc250 = ba_sprint(catalanB(250))
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printf(1,"%d: %s (%3.2fs)\n",{250,sc250,time()-t0})
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n=15
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cat = list(n+2)
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cat[1]=1
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for i=1 to n
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for j=i+1 to 2 step -1
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cat[j]=cat[j]+cat[j-1]
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next
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cat[i+1]=cat[i]
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for j=i+2 to 2 step -1
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cat[j]=cat[j]+cat[j-1]
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next
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see "" + (cat[i+1]-cat[i]) + " "
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next
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func catalan(num) {
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var t = [0, 1];
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range(1, num).map { |i|
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range(i, 1, -1).each {|j| t[j] += t[j-1]};
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t[i+1] = t[i];
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range(i+1, 1, -1).each {|j| t[j] += t[j-1]};
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t[i+1] - t[i];
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}
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}
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say catalan(15).join(' ');
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def binomial(n; k):
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if k > n / 2 then binomial(n; n-k)
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else reduce range(1; k+1) as $i (1; . * (n - $i + 1) / $i)
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end;
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# Direct (naive) computation using two numbers in Pascal's triangle:
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def catalan_by_pascal: . as $n | binomial(2*$n; $n) - binomial(2*$n; $n-1);
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$ jq -n -c -f Catalan_numbers_Pascal.jq
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[0,0]
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[1,1]
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[2,2]
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[3,5]
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[4,14]
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[5,42]
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[6,132]
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[7,429]
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[8,1430]
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[9,4862]
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[10,16796]
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[11,58786]
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[12,208012]
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[13,742900]
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[14,2674440]
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[15,9694845]
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[30,3814986502092304]
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[31,14544636039226880]
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[510,5.491717746183512e+302]
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[511,null]
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