Generate a modulo rosace

Question

The cool stuff

The following rosace can help calculate numbers modulo 7.

In order to do that, you must start at 0 and turn clockwise a number of steps given by the first digit. Then, for each successive digit, follow the arrow and then turn clockwise the number of steps given by that digit.

Here is how you proceed for number 294:

1. You start at circle 0.
2. You turn clockwise the number of steps given by the first digit (which is a 2, you end up at 2).
3. You follow the arrow there (you end up at 6).
4. You turn clockwise the number of steps given by the second digit (which is a 9, you end up at 1).
5. You follow the arrow there (you end up at 3).
6. You turn clockwise the number of steps given by the third number (which is 4, you end up at 0).
7. 294 mod 7 = 0 (meaning 294 is multiple of 7).

(Video explanation if you still didn't get it)

The goal

Figure out how that works (I know but I won't tell you).

Create a program that takes a number n in parameter and that generates a rosace for mod n.

The rosace can be displayed by any means you want (ASCII, generate PNG, generate SVG, ...) as long as it could be used by a 8 years old child (so no list of rules, I want a picture)!

You can use strait lines, even if it's a bit less clear than what I made for the example, but you must show clearly the numbers that point to themselves with some kind of tail-biting arrow.

Test cases

(I only give the links between the numbers, feel free to edit my question once your program sucessfully generates them)

mod 2:

0 -> 0
1 -> 0

mod 3:

0 -> 0
1 -> 1
2 -> 2

mod 4:

0 -> 0
1 -> 2
2 -> 0
3 -> 2

mod 5:

0 -> 0
1 -> 0
2 -> 0
3 -> 0
4 -> 0

mod 6:

0 -> 0
1 -> 4
2 -> 2
3 -> 0
4 -> 4
5 -> 2

mod 7:

0 -> 0
1 -> 3
2 -> 6
3 -> 2
4 -> 5
5 -> 1
6 -> 4

mod 8:

0 -> 0
1 -> 2
2 -> 4
3 -> 6
4 -> 0
5 -> 2
6 -> 4
7 -> 6

mod 9:

0 -> 0
1 -> 1
2 -> 2
3 -> 3
4 -> 4
5 -> 5
6 -> 6
7 -> 7
8 -> 8

mod 10:

0 -> 0
1 -> 0
2 -> 0
3 -> 0
4 -> 0
5 -> 0
6 -> 0
7 -> 0
8 -> 0
9 -> 0

Rules

This is code-golf, shortest code in bytes wins.

As usual, loopholes and cheats are forbidden.

Answer 1

Mathematica, 192 bytes

This type of challenge (nontrivial mathematical computation together with high-level graphics output) is what Mathematica is made for!

(d=#;r_~t~n_:=r{Sin[#],Cos[#]}&[6.3n/d];m=Mod[10#,d]&;Graphics@{Array[#~Text~t[9,#]&,d,0],Array[Arrow@{t[9,#+1/7],t[9,#+6/7]}&,d],Array[Arrow@BezierCurve@{t[8,#+1/9],{0,0},t[8,m@#-1/9]}&,d]})&

Expanded and explained:

[1] (
[2] d = #; r_~t~n_ := r {Sin[#], Cos[#]} &[6.3 n/d]; m = Mod[10 #, d] &;
[3] Graphics@{
[4]   Array[#~Text~t[9, #] &, d, 0],
[5]   Array[Arrow@{t[9, # + 1/7], t[9, # + 6/7]} &, d],
[6]   Array[Arrow@BezierCurve@{t[8, # + 1/9], {0, 0}, t[8, m@# - 1/9]} &, d]
[7]          }
[8] ) &

Lines 1 and 8 delimit an unnamed function of one argument. Lines 3 and 7 delimit several commands that output graphics.

Lines 2 stores the input as d; defines a binary function t giving the coordinates of a point n/d of the way around the circle of radius r, clockwise from the top (in the spirit of this site, I saved a byte by rounding 2π to 6.3!); and defines a unary function m calculating the destination of the arrow starting at its argument.

Line 4 renders the numbers 0 to d–1 equally spaced around the circle of radius 9 (the exact radius unimportant, chosen to maximize aesthetics subject to fitting in one byte).

Line 5 renders the clockwise straight arrows around the circumference of the circle. The 1/7 and 6/7 leave enough space to read the numbers.

Line 6 renders the curved arrows from each number to (10 times the number modulo d). BezierCurve automatically draws a Bézier curve using the given control points. Fortunately, using the origin as a single interior control point produces reasonable output.

Sample output (note that the cases 9, 10, and 11 are trivial in different ways):

d = 7

d = 8

d = 9

d = 10

d = 11

d = 12

d = 13

d = 37

This last input was chosen because 37 divides 10^3–1, and so the interior arrows (not counting the obligatory self-arrow from 0 to 0) form lots of triangular cycles.

Answer 2

Python 2, 294 bytes

n=input()
r,R="",range(n)
for i in R:r+=["   ","__ "][10*i%n==i]
r+="\n"
for i in R:r+=["   ","|/ "][10*i%n==i]
print r
print">>".join(map(str,R))
for i in R:
    o,r=10*i%n,""
    for j in R:
        r+="|"
        if i<=j<o:r+=">>"
        elif i>j>=o:r+="<<"
        else:r+="  "
    print r

Prints the diagram in this format:

__                   
|/                   
0>>1>>2>>3>>4>>5>>6
|  |  |  |  |  |  |  
|  |>>|>>|  |  |  |  
|  |  |>>|>>|>>|>>|  
|  |  |<<|  |  |  |  
|  |  |  |  |>>|  |  
|  |<<|<<|<<|<<|  |  
|  |  |  |  |<<|<<|  

I don't know if this format is okay, therefore I'll leave this answer as invalid for the time being. Yay, it's valid!

Try it on repl.it!

Answer 3

PHP+SVG, 500 Bytes

little arrow for connections between same values

<svg viewBox=0,0,500,500><def><marker id=t viewBox=0,0,9,9 refX=1 refY=5 orient=auto><path d=M0,0L10,5L0,10z /></marker></def><circle cx=250 cy=250 r=150 fill=#ccc /><?for($i=0;$i<$n=$_GET[n];$i++){$w=deg2rad((($i*10%$n)-$i)*360/$n);echo"<g transform=rotate(".$i*360/$n.",250,250)><path d=M250,110L".(250+(sin($w)*135)).",".(250-cos($w)*145)." fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>$i</text></g>";}?></svg>

to see the arrows to the same values I use this color value rgba(255,0,0,0.3) . it is a possibility to short it down.

expanded

    <svg viewBox=0,0,500,500>
<def>
<marker id=t viewBox=0,0,10,10 refX=1 refY=5 markerWidth=5 markerHeight=5 orient=auto>
<path d=M0,0L10,5L0,10z />
</marker>
</def>
<circle cx=250 cy=250 r=150 fill=#ccc stroke=#a00 />
<?php for($i=0;$i<$n=$_GET[n];$i++){
$w=deg2rad((($i*10%$n)-$i)*360/$n);
echo"<g transform=rotate(".$i*360/$n.",250,250)>
<path d=M250,110L".(250+(sin($w)*135)).",".(250-cos($w)*145)." fill=none stroke=#0f0 marker-end=url(#t) />
<circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) />
<text x=250 y=105 text-anchor=middle>$i</text>
</g>";
}?>
</svg>

output for n=45

<svg viewBox=0,0,500,500><def><marker id=t viewBox=0,0,10,10 refX=1 refY=5 markerWidth=5 markerHeight=5 orient=auto><path d=M0,0L10,5L0,10z /></marker></def><circle cx=250 cy=250 r=150 fill=#ccc stroke=#a00 /><g transform=rotate(0,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>0</text></g><g transform=rotate(8,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>1</text></g><g transform=rotate(16,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>2</text></g><g transform=rotate(24,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>3</text></g><g transform=rotate(32,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>4</text></g><g transform=rotate(40,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>5</text></g><g transform=rotate(48,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>6</text></g><g transform=rotate(56,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>7</text></g><g transform=rotate(64,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>8</text></g><g transform=rotate(72,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>9</text></g><g transform=rotate(80,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>10</text></g><g transform=rotate(88,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>11</text></g><g transform=rotate(96,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>12</text></g><g transform=rotate(104,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>13</text></g><g transform=rotate(112,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>14</text></g><g transform=rotate(120,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>15</text></g><g transform=rotate(128,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>16</text></g><g transform=rotate(136,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>17</text></g><g transform=rotate(144,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>18</text></g><g transform=rotate(152,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>19</text></g><g transform=rotate(160,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>20</text></g><g transform=rotate(168,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>21</text></g><g transform=rotate(176,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>22</text></g><g transform=rotate(184,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>23</text></g><g transform=rotate(192,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>24</text></g><g transform=rotate(200,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>25</text></g><g transform=rotate(208,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>26</text></g><g transform=rotate(216,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>27</text></g><g transform=rotate(224,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>28</text></g><g transform=rotate(232,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>29</text></g><g transform=rotate(240,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>30</text></g><g transform=rotate(248,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>31</text></g><g transform=rotate(256,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>32</text></g><g transform=rotate(264,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>33</text></g><g transform=rotate(272,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>34</text></g><g transform=rotate(280,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>35</text></g><g transform=rotate(288,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>36</text></g><g transform=rotate(296,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>37</text></g><g transform=rotate(304,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>38</text></g><g transform=rotate(312,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>39</text></g><g transform=rotate(320,250,250)><path d=M250,110L250,105 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>40</text></g><g transform=rotate(328,250,250)><path d=M250,110L378.39262969985,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>41</text></g><g transform=rotate(336,250,250)><path d=M250,110L329.35100905948,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>42</text></g><g transform=rotate(344,250,250)><path d=M250,110L170.64899094052,367.30746418437 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>43</text></g><g transform=rotate(352,250,250)><path d=M250,110L121.60737030015,205.19253581563 fill=none stroke=#0f0 marker-end=url(#t) /><circle cx=250 cy=100 r=10 fill=rgba(255,0,0,0.3) /><text x=250 y=105 text-anchor=middle>44</text></g></svg>

320 Bytes working with rect

<svg viewBox=0,0,<?=$w=30+10*$n=$_GET[n]?>,<?=20*$n?>><?for($i=0;$i<$n;$i++)echo"<rect x=0 y=".($y=$i*20)." fill=none width=$w height=20 stroke=grey /><text x=5 y=".($y+17).">$i</text><path d=M".($x=$i*10+25).",".($y+17)."L$x,".($m=($i*10%$n)*20+5)." fill=none stroke=blue /><circle cx=$x cy=$m r=2 fill=#0f0 />"?></svg>

expanded

<svg viewBox=0,0,<?=$w=30+10*$n=$_GET[n]?>,<?=20*$n?>>
<?for($i=0;$i<$n;$i++)
echo"<rect x=0 y=".($y=$i*20)." fill=none width=$w height=20 stroke=grey />
<text x=5 y=".($y+17).">$i</text>
<path d=M".($x=$i*10+25).",".($y+17)."L$x,".($m=($i*10%$n)*20+5)." fill=none stroke=blue />
<circle cx=$x cy=$m r=2 fill=#0f0 />"?>
</svg>

output for n=72

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Answer 4

Python 2, 540 464 431 bytes

Some golfing like using shorter variable names, variable substitution, list comprehension and changing everything to white (except the text). Biggest save was changing precomputed positions to dynamically (see L).

from cv2 import*
from numpy import*
D=1024
G=zeros((D,D,3),uint8)
n=7
w=(1,1,1)
R=range(n)
c=circle
I=int
L=lambda i,r=400:(I(sin(2*pi*i/n)*r+D/2),I(D/2-cos(2*pi*i/n)*r))
for i in R:
 p,q=L(i),L((i*10)%n)
 line(G,p,L((i+1)%n),w,5)
 line(G,p,q,w,5)
 f=lambda z:I(.15*p[z]+.85*q[z])
 c(G,[(f(0),f(1)),L(i,450)][p==q],25,w,-1)
for i in R:
 c(G,L(i),50,w,-1)
 putText(G,`i`,(L(i)[0]-30,L(i)[1]+30),0,2,(0,0,0),2)
imwrite('m.png',G*255)

L calculates the positions of the circles by distance to origin for the big ones containing the numbers and the outer small ones that indicate self pointing.

First loop draws the connections: 1st line is the circle around and 2nd line is in the interior, a small circle is added to show direction or self pointing.

Second loop puts big circle and number.

It is obviously not as nice as the Mathematica answers, but everything made from scratch.

Answer 5

Mathematica, 124 121 bytes

Graph[Join@@Thread[{k->Mod[k+1,#],k->Mod[10k,#]}~Table~{k,0,#-1}],VertexLabels->"Name",GraphLayout->"CircularEmbedding"]&

Creates a figure as a graph with directed edges. The graph output now follows the same pattern except counter-clockwise. I prefer Greg Martin's solution a lot more since the output is much more aesthetically pleasing.

A less visually pleasant graph can be generated for 82 bytes using

Graph[Join@@({k->Mod[k+1,#],k->Mod[10k,#]}~Table~{k,0,#-1}),VertexLabels->"Name"]&

For d = 8,

Answer 6

Batch, 394 bytes

@echo off
set/an=%1-1
set s=
for /l %%j in (0,1,%n%)do call:h %%j
echo %s%
for /l %%i in (0,1,%n%)do call:r %%i %1
exit/b
:h
set "s=%s%^>%1^>"
exit/b
:r
set/ar=%1*10%%%2
set s=
for /l %%j in (0,1,%n%)do call:l %%j %1
echo %s%
exit/b
:l
if %1==%r% set "s=%s%^>%1^<"&exit/b
if %2 geq %1 if %1 geq %r% set "s=%s%^<%1%^<"&exit/b
if %2 leq %1 if %1 leq %r% goto h
set "s=%s% %1 "

Escaping in Batch is ugly at the best of times. Example output for 7:

>0>>1>>2>>3>>4>>5>>6>
>0< 1  2  3  4  5  6
 0 >1>>2>>3< 4  5  6
 0  1 >2>>3>>4>>5>>6<
 0  1 >2<<3< 4  5  6
 0  1  2  3 >4>>5< 6
 0 >1<<2<<3<<4<<5< 6
 0  1  2  3 >4<<5<<6<

Answer 7

Python 2 + graphviz, 147 bytes

from graphviz import*
d=Digraph()
k=input()
p=d.edge
for j in[d.node(`i`,`i`)or i for i in range(k)]:
 p(`j`,`-~j%k`);p(`j`,`10*j%k`)
d.render("o")

Doesn't always draw a circle, outputs a pdf file called o

Answer 8

Haskell, 350 bytes

Inspired by @Loovjo, I also use ASCII art. This works for numbers under 100000 (or something like that).

g k="\t"++(foldl (++) ""$map (\y->show y++"|") [0..k])
n k=length$g k
r z x=(show x)++":\t"++(replicate f ' ')++(replicate d s)++"\n" where y=mod (x*10) z;(a,b)=(min x y,max x y);s=(case compare y x of LT->'<';EQ->'.';GT->'>');f=n (a-1)-1;d=n b-f-2
w z=(g (z-1))++"\n"++(foldl1 (++) (map (r z) [0..(z-1)]))
main=do{n<-getLine;putStrLn (w$read n);}

Basically you point from x to (x*10)%n.

You can try this here. But since codepad doesn't support input, change q to the value of n you want, and resubmit. (Note that forking doesn't work, so copy and resubmit from main page). The code submitted there is a bit different, because the above version takes input from console.

I hope the output is intuitive. Compression suggestions are welcome (particularly if that beats 294 ;)).