Floyd's triangle: Difference between revisions

<lang julia>#floyd(n) creates an n-row floyd's triangle counting from 1 to (n/2+.5)*n

function floyd(n)

x = 1

dig(x,line,n) = (while line < n; x+=line; line+= 1 end; return ndigits(x)+1)

for line = 1:n, i = 1:line; print(lpad(x,dig(x,line,n)," ")); x+=1; i==line && print("\n") end

end</lang>

<pre>julia> floyd(5)

1

2 3

4 5 6

7 8 9 10

11 12 13 14 15

julia> floyd(14)

1

Here is another solution that makes use of the fact that the number in the (i,j)th position in the array is equal to the sum of j and the binomial coefficient (j,2). This

number should be padded according to the number of digits

in the coefficient (n,j).

<lang julia>floyd(n) =

pprint([join([lpad(j+binomial(i,2), (j==1?0:1)+ndigits(j+binomial(n,2)), " ")

for j=1:i])

for i=1:n])

pprint(matrix) = for i = 1:size(matrix,1) println(join(matrix[i,:])) end

</lang>

{{Out}}

<pre>julia> floyd(5)

1

2 3

4 5 6

7 8 9 10

11 12 13 14 15</pre>