# Enigmatic Code

Programming Enigma Puzzles

## Enigma 1728: Problem tackling

From New Scientist #2895, 15th December 2012 [link]

Joe was in his school’s rugby team. While practising tackling, Joe remembers standing on the goal line 25 feet from his friend Ken, who was by the corner flag. When the master blew his whistle, Ken ran along the touchline and Joe had to try to tackle him. Ken runs at half a mile an hour faster than Joe, who could only manage 12 miles an hour. So Joe never caught up with Ken.

What was the nearest Joe came to Ken?

It is pleasing that Enigma 1728 should be published on 12.12.12, as 1728 = 12³.

[enigma1728]

### 14 responses to “Enigma 1728: Problem tackling”

1. Jim Randell 12 December 2012 at 6:05 pm

Assuming I understand they layout of rugby pitches correctly. This is a solution that uses the functions I recently added to the enigma.py library to find numerical approximations of functions. (It is implemented using a Golden Section Search minimiser). It runs in 45ms.

from math import sin, cos, tan, radians, sqrt
from enigma import find_min, printf

# convert mph to fps
def fps(s):
return (s * 22.0) / 15.0

# Joe runs at 12 mph
vj = fps(12.0)

# Ken runs at 12.5 mph
vk = fps(12.5)

# min distance for an angle of theta
def min_d(theta):

# distance at time t if J runs at an angle of theta
def d(t):
dk = 25.0 / tan(a) - vk * t
dj = 25.0 / sin(a) - vj * t
return sqrt(dk ** 2 + dj ** 2 - 2 * dk * dj * cos(a))

r = find_min(d, 0.0, 120.0)
return r.fv

r = find_min(min_d, 0.0, 90.0)
printf("min distance = {r.fv} at theta = {r.v}")


Solution: Assuming Joe manages the minimum achievable distance, the closest he comes to Ken is 7 ft.

• Jim Randell 12 December 2012 at 11:17 pm

Here’s a symbolic solution using SymPy.

from sympy import symbols, solve, Rational, sqrt
from enigma import printf

# convert mph to fps
def fps(s):
return s * Rational(22, 15)

# Joe runs at 12 mph
vj = fps(12)

# Ken runs at 12.5 mph
vk = fps(Rational(25, 2))

# distance at time t if J runs at a y velocity of vjy
def d(t, vjy):
ky = vk * t
vjx = sqrt(vj ** 2 - vjy ** 2)
(jx, jy) = (25 - vjx * t, vjy * t)
return sqrt(jx ** 2 + (ky - jy) ** 2)

(t, v) = symbols('t v')
# differentiate the distance
eq1 = d(t, v).diff(t)
# and solve to find when the distance is at a minimum
for r1 in solve(eq1, t):
# so the min distance (for a y velocity of v) is...
eq2 = d(r1, v).simplify()

# so now we need the minimum of that, so we differentiate it...
for r2 in solve(eq2.diff(v), v):
# and find the value for the minimum distance
m = eq2.subs(v, r2)
# and the x,y components of J's velocity
x = sqrt(vj ** 2 - r2 ** 2)
printf("min distance = {m} at vjx/vjy = {q}", q=x/r2)

• Jim Randell 16 January 2013 at 10:46 am By using the observation that the minimum distance occurs when J is travelling directly towards K, you can deduce the equation for the minimum distance as:

d = 25 / sin(θ) – 24 / tan(θ)

or more generally:

d = g(1/sin(θ) – r/tan(θ))

where r is the ratio vj / vk and g is the initial distance between Joe and Ken.

By differentiation we find this is minimised in the case where cos(θ) = r.

And the solution follows.

2. ahmet sarqcoglu 14 December 2012 at 6:18 pm

hi, can we use any programming language here?

• Jim Randell 14 December 2012 at 7:53 pm

If you’d like to share a programmatic solution, then feel free. In whatever language you choose. (See the “Notes” page for some tips on posting source code in WordPress).

At the moment I tend to post my solutions in Python as I find it lends itself to concise readable code (but then I initially started writing solutions in the somewhat more esoteric Perl), and is freely available on many platforms (and even pre-installed on some), but it’s always interesting to see other languages too, and assuming it’s reasonably self-explanatory I should be able to understand the algorithm.

3. ahmet saracoğlu 15 December 2012 at 12:39 pm
 private double ConvertTo(double velocity)
{
return velocity * 5280 / 3600;
}

private void button1_Click(object sender, EventArgs e)
{
double radian = Math.PI / 180, angle = 0, distanceX = 0;
double YKen = 0, YJoe = 0, feet = 0, Vken = 12.5, VJoe = 12, VJx = 0, VJy = 0, t = 0,
difference = 0, min = 25, ang = 0, tmin = 0, distant = 0;
string fortext = "";

TextWriter tw = new StreamWriter("enigma.txt");

// close the stream

Vken=ConvertTo(Vken);
VJoe = ConvertTo(VJoe);

for (angle = 0; angle < 90; angle++)
{
VJx = VJoe * Math.Cos(angle * radian);
VJy = VJoe * Math.Sin(angle * radian);

fortext = fortext +"ANGLE:"+angle.ToString() + "\r\n";

for (distanceX = 0; distanceX <= 25; distanceX+=1)
{
t = distanceX / VJx;
YJoe = VJy * t;
YKen = Vken * t;
difference = Math.Sqrt(Math.Pow((YKen-YJoe),2)+Math.Pow((25-distanceX),2));

fortext = fortext + " TIME PASSED:" + t.ToString();
fortext = fortext + " DISTANCE:" + difference.ToString() + "\r\n";

if (difference < min)
{
distant = distanceX;
min = difference;
ang = angle;
tmin = t;
}
}
}
button1.Text = min.ToString()+ "  "+ang.ToString()+" "+tmin.ToString();

tw.WriteLine(fortext);
tw.Close();

}

4. ahmet saracoğlu 15 December 2012 at 5:19 pm

This is the first python program whose belong to me, I have learned today a little

def ConvertTo(velocity):
return velocity*5280/3600

def FindMinDistance(Vken,VJoe):
import math
minumum=25
Vk=ConvertTo(Vken) #miles/hour is converted to feet/seconds with that function
Vj=ConvertTo(VJoe)
for angle in range(0,89):
Vjx=VJoe*math.cos(angle*rdn)# Joe has velocity at x axis, and this is calculated
Vjy=VJoe*math.sin(angle*rdn)
for distanceX in range (0,25):
t=distanceX/Vjx #Joe is running distanceX feet and time elapsed.
Yjoe=Vjy*t
Yken=Vken*t #Ken's distance to his starting point for 't' time
difference = math.sqrt(math.pow((Yken-Yjoe),2)+math.pow((25-distanceX),2))#Distance between them
if (difference < minumum):  minumum = difference
ang = angle# Joe's velocity vector angle
tmin = t
print minumum

• Jim Randell 18 December 2012 at 11:29 pm

I think that’s a good start. Python is fairly easy to pick up (especially if you already know some other programming language), and has relatively few gotchas (although division using the / operator is one of them).

5. arthurvause 17 December 2012 at 11:53 am

Converting the speeds to feet per second, this diagram derives the distance between Joe and Ken.

Then let Wolfram Alpha work out the minimum distance.

• arthurvause 17 December 2012 at 4:46 pm

I have published an analytic solution to the problem on my blog.

However, I can’t work out why the form of the solution should be that the sine of the angle that Joe runs at is in proportion to the speeds of Joe and Ken.

• arthurvause 18 December 2012 at 3:11 pm

The analytic solution now has a derivation of the general formula for the minimum distance, and the derivation of the sine of the angle being the proportion of the speeds

• Jim Randell 18 December 2012 at 5:48 pm

Very neat… I didn’t know you could get maths laid out so nicely in Blogger.

• Jim Randell 29 January 2013 at 9:03 am

And I’ve just discovered you can use LaTeX markup directly in WordPress $i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right>$

Did it work? It did for me.

(Although without the ability to preview/edit comments you might have to be careful when using this feature).

6. Hugh Casement 13 June 2016 at 10:08 am

In roughly 4.675 seconds Ken runs 600/7 feet and Joe 24/25 as far.
I think I’m right in saying that if their starting positions (and roles) were reversed, the diagram would be the same triangle. In about 4.87 seconds Ken would run the extra 7, making 625/7 ft = 25/24 times as much as Joe’s 600/7.
Of course it all assumes that rugby players are more intelligent than the dog that always runs directly towards its master: this old dog can’t work out the equation of the pursuit curve.

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