# Filling graphics in Mathematica

• March 11th 2009, 09:49 AM
bkarpuz
Filling graphics in Mathematica
Dear friends hi again,

I am using the following code:
Code:

`Manipulate[Plot[{p-Floor[k]^Floor[k]/(Floor[k]+1)^(Floor[k]+1)(Floor[k]-1),1,0},{p,0,2},PlotRange->{{0,2},{0,2}}],{k,1,10}]`
The code above gives the following graphic:
My first question is how can I fill the graphic in the following form:
And the second question is how can I plot the following graphic?
Or simply, how can I put the graph of $x=1$ into the first graph?

Thanks.
• March 12th 2009, 06:45 AM
bkarpuz
I could do what I wanted with the following codes, I am giving them below.
Code:

```kmax = 10; Manipulate[  Show[Graphics[{White, Polygon[{{2, 0}, {0, 0}, {0, 1}, {2, 1}}]}],   Graphics[{Cyan,     Polygon[{{2, 0}, {k^k/(k + 1)^(k + 1) (k - l),       0}, {1 + k^k/(1 + k)^(k + 1) (k - l), 1}, {2, 1}}]}]], {k, 1,   kmax, 1}, {l, 0, k - 1, 1}] Manipulate[  Show[Graphics[{White, Polygon[{{2, 0}, {0, 0}, {0, 1}, {2, 1}}]}],   Graphics[{Magenta, Line[{{1, 1}, {2, 1}}]}]], {k, 1, kmax, 1}, {l,   0, k - 1, 1}] Manipulate[  Show[Graphics[{White, Polygon[{{2, 0}, {0, 0}, {0, 1}, {2, 1}}]}],   Graphics[Table[If[s == k - l, Line[{{1, 1}, {2, 1}}], If[s == 0,       Line[{{1, 0}, {1, 1}}],       Line[{{1, 1},         If[(k - l)/(k - l - s) > 2, {2, (2 s - k + l)/s}, {(k - l)/(           k - l - s), 0}]}]]], {s, 0, k - l}]]], {k, 1, kmax, 1}, {l,   0, k - 1, 1}] Manipulate[  Show[Graphics[{White, Polygon[{{2, 0}, {0, 0}, {0, 1}, {2, 1}}]}],   Graphics[{Magenta, Polygon[{{2, 0}, {1, 0}, {1, 1}, {2, 1}}]}]], {k,   1, kmax, 1}, {l, 0, k - 1, 1}]```
And all in one version
Code:

```kmax = 10; Manipulate[  Show[If[c == 1,   Show[Graphics[{White, Polygon[{{2, 0}, {0, 0}, {0, 1}, {2, 1}}]}],     Graphics[{Cyan,       Polygon[{{2, 0}, {k^k/(k + 1)^(k + 1) (k - l),         0}, {1 + k^k/(1 + k)^(k + 1) (k - l), 1}, {2, 1}}]}]],   If[c == 2,     Show[Graphics[{White, Polygon[{{2, 0}, {0, 0}, {0, 1}, {2, 1}}]}],       Graphics[{Cyan,       Polygon[{{2, 0}, {k^k/(k + 1)^(k + 1) (k - l),           0}, {1 + k^k/(1 + k)^(k + 1) (k - l), 1}, {2, 1}}]}],     Graphics[{Magenta, Line[{{1, 1}, {2, 1}}]}]],     If[c == 3,     Show[Graphics[{White,         Polygon[{{2, 0}, {0, 0}, {0, 1}, {2, 1}}]}],       Graphics[{Cyan,         Polygon[{{2, 0}, {k^k/(k + 1)^(k + 1) (k - l),           0}, {1 + k^k/(1 + k)^(k + 1) (k - l), 1}, {2, 1}}]}],       Graphics[       Table[If[s == k - l, Line[{{1, 1}, {2, 1}}], If[s == 0,           Line[{{1, 0}, {1, 1}}],           Line[{{1, 1},             If[(k - l)/(k - l - s) > 2, {2, (2 s - k + l)/s}, {(               k - l)/(k - l - s), 0}]}]]], {s, 0, k - l}]]],     Show[Graphics[{White,         Polygon[{{2, 0}, {0, 0}, {0, 1}, {2, 1}}]}],       Graphics[{Cyan,         Polygon[{{2, 0}, {k^k/(k + 1)^(k + 1) (k - l),           0}, {1 + k^k/(1 + k)^(k + 1) (k - l), 1}, {2, 1}}]}],       Graphics[{Magenta,         Polygon[{{2, 0}, {1, 0}, {1, 1}, {2, 1}}]}]]]]]], {k, 1, kmax,   1}, {l, 0, k - 1, 1}, {c, 1, 4, 1}]```
Is there a way to simplify the latter code?

Thanks.