MaplePrimes - Newest Questions and Posts

Finding the Shortest Path Between Two Points

callumneily — Thu, 20 Aug 2026 19:57:07 Z

Defining Our Goal

Suppose you place two points on the x-y plane and ask a seemingly simple question: What is the shortest path connecting them?

The "obvious" answer is a straight line (the green curve), but how can we show that this is truly the optimal path? Imagine two fixed points (x₁,y₁) and (x₂,y₂) connected by some curve y(x).

To find the length of this curve, we zoom in on an infinitesimally small segment. If the curve changes by small amounts dx horizontally and dy vertically, then the Pythagorean theorem gives the tiny arclength ds:

Here, y' = dy/dx is the derivative/slope of y(x) with respect to x. The total length S of the curve, often referred to as the "action", is found by integrating these arclengths:

Notice that, unlike ordinary functions that take numbers as inputs, S[y] takes an entire curve y(x) as input and outputs a single number, S. Such a function is called a functional.

Our goal is to find the curve y(x) that minimizes the length S[y]. Instead of asking how the function (or "functional") S[y] changes for some change in x, we must ask how the total path length changes if we slightly deform the curve y(x) itself. This is the central idea behind a branch of calculus called "Calculus of Variations".

Digging Deeper

Let's predict that the shortest path between two curves is y(x). Without knowing that y(x) is a straight line (we haven't proved this yet!), our first prediction is probably incorrect. So, let's say the true ideal path is a nearby path y(x) + ε*η(x), where η(x) is the shape of the function that corrects our original prediction, and ε scales this correction function (i.e. controls how large the deformation is). To ensure that the endpoints remain the same, we require η(x₁) = η(x₂) = 0.

Substituting this new path into the expression for S[y] above, we get the path length S as a function of ε:

Notice that if ε = 0, we recover the original path y(x). If the original path does minimize S, then S has a minimum at ε = 0. That is,

So, to find the minimizing curve y(x), we find a curve y(x) such that dS/dε = 0 at ε = 0. Bringing the derivative inside of the integral for S[y] and using the chain rule, this means:

We can integrate this by parts using the substitutions:

Since we originally defined η(x₁) = η(x₂) = 0, the first term on the right disappears. We're left with:

Now, this must be true for any η(x) we choose, meaning:

Squaring both sides and rearranging to solve for y', we get that:

Hence, y(x) has a constant slope and is therefore a straight line. Note that there are multiple ways to prove that the shortest path between two points is a straight line (such as using the Triangle Inequality or algebraic arguments under Euclidean geometry), but this method of using least action has proven to have much deeper physical significance to more sophisticated questions, as discussed below.

Important Distinction

As many of you have probably noticed, this argument only holds in flat Euclidean space (the space most of us are used to, like the xyz plane). In curved spaces, the analogue of a straight line is called a geodesic.

The general idea is still the same, but the formula for distance changes depending on the geometry of the space.

For example, consider the surface of the Earth. If you want the shortest route from Toronto to London while remaining on Earth's surface, you don't follow what looks like a straight line on a flat map. You follow part of a "great circle".

On a sphere of radius R, for example, the infinitesimal distance is given using spherical coordinates:

So instead of minimizing the flat-space length

we minimize:

Applying the same calculus-of-variations machinery gives the geodesics of the sphere, which turn out to be great circles.

This becomes especially interesting in general relativity. Spacetime itself is curved, and free-falling objects follow geodesics through that curved spacetime. In that sense, Earth's orbit around the Sun can be thought of not as Earth being forced away from a straight path, but as Earth following the natural geodesic of curved spacetime.

Deeper Physical Significance

What we proved above is a remarkable fact, as the idea of minimizing the "action" functional is a fundamental characteristic of systems in nature. Instead of minimizing distance, physical systems obey the same mathematics to minimize quantities involving energy and time. For example, a hanging rope will form a shape called a catenary (a type of hyperbolic cosine function) to minimize its gravitational potential energy, not a parabola (which approximates the motion of projectiles) as one might think.

Similarly, light bends when moving between materials like air and water to minimize travel time.

So, never doubt the power of seemingly "simple" or intuitive mathematical results, as this elementary question of minimizing distance between two points uses the same mathematics that governs planetary motion, quantum fields, and spacetime itself!

AI Assistant with Anthropic Claude Max and other providers

OptimusMaplePrime — Sun, 16 Aug 2026 18:14:35 Z

Can the AI Assistant in the Maple 2026 GUI be configured to use Anthropic's Claude Max plans, local AIs (e.g., via LM Studio), or other providers?

Interupting Maple not working

C_R — Sun, 16 Aug 2026 17:18:37 Z

I am only occasionally using Maple versions with the new ribon user interface and noticed about 2 weeks ago that I cannot interrupt for loops under these interfaces. For example this one

for i to 1000 do
    i^i;
end do

I re-run the code today (after installing windows updates) and could interrupt before the screen was filled with output but not after executing the code a second time (without restart).

Is that reproducible on other installations?

Are there other commands that cannot be interrupted?
If that is known, are there workarounds?

Update:
I have restarted Maple and have two worksheets open with the same code. I can repeatedly interrupt in one worksheet but not in other

What type of matrix normalisation is this

Ronan — Fri, 14 Aug 2026 05:22:23 Z

Could anyone tell me what type of matrix normalisation this is and it it built into Maple? I found this in a paper and Google AI said this is the process of normailsation being used.

restart

>	with(LinearAlgebra):

>	J:=Matrix([[ 1 , 0 , 0 ], [ 0 , 1 , 0 ], [ 0 , 0 , -1 ]]);

(1)

Matrix N has already been scaled, so the rows are of equal magnitude x^2 + y^2 - z^2= k

>	N:=Matrix(3, 3, [[4, -3, 2], [-1/2sqrt(70), -3/10sqrt(70), 1/5sqrt(70)], [1/11sqrt(77), 6/11sqrt(77), 2/11sqrt(77)]])

(2)

>	for i to 3 do add(N[i,j]^2,j=1..2)-N[i,3]^2; end do

(3)

Normalisation process to produce C

>	A:=N.J.N^%T

(4)

>	dA:=DiagonalMatrix(1/~(sqrt~(abs(Diagonal((A))))));

(5)

>	C:=(dA.A.dA);

(6)

evalf(C)

(7)

Download 2026-08-14_Q_What_Type_of_Matrix_Normalisation.mw

Maple transactions new issue?

Steven_Huang — Thu, 13 Aug 2026 18:25:08 Z

I notice that Maple transactions journal does not publish any new issue ever since March this year. Based on the previous recording it is supposed to publish issue each season. When is it going to promote a new issue? My research is supposed to publish there and I believe I have made some interesting progress and it is of interest to the community.

Poisoned Wine Bottle

Thu, 13 Aug 2026 16:59:30 Z

You are hosting a big party tonight and prepared 1000 bottles of wine. A spiteful neighbor sneaks in and poisons exactly one bottle. The poison is colorless, tasteless, and takes about an hour to take effect and kill. Your party also starts in an hour, and you don’t want to throw all the wine away.

You have 10 mice and time for only one round of tasting. In that round, a mouse can taste from any number of bottles. How can you identify that single poisoned bottle among 1000?

Pause here and try to solve the problem yourself!

An obvious solution might be to group the wine into 10 batches of 100 bottles and have each mouse drink from one batch. But if a mouse dies, you have only narrowed it down to 100 bottles, and there is no time for a second round. Whatever test you run, you only get one shot.

Ten mice. A thousand bottles. The problem sounds impossible – until you realize each mouse isn’t just a taster.

The exact idea that powers computers is also what solves our wine puzzle.

A 300-Year-Old Idea

Some background information before we solve the problem.

In 1703, Gottfried Leibniz published a paper describing how every number can be written using only two symbols: 0 and 1. To him, the concept felt almost divine – an entire universe created out of nothingness and unity. But for over two centuries, binary remained an idea trapped on paper.

Everything changed when a young engineer named Claude Shannon realized that 0 and 1 map perfectly onto the physical states of an electrical switch: off and on. That single insight laid the groundwork for digital circuits, eventually powering every smartphone, laptop, and text message on the planet.

Yet binary is more than just how machines store information. It is a way of extracting information.

0 and 1 not only reflect the underlying logic of switches, but they also correspond to every yes/no question you ask: “did this mouse die, or not?”. And 10 binary digits can cover 2¹⁰ = 1024 different possibilities.

1024 is more than 1000.

The Solution

Step 1: Relabel the bottles in binary

Write each bottle’s number, 1 through 1000, as a 10-digit binary number, padding the front with 0s. For example, bottle 17 is 10001 in binary, so its label becomes 0000010001. Every bottle now carries a unique 10-digit barcode of 0s and 1s.

Step 2: Assign each mouse a digit

Line up the mice and assign each one a digit: mouse #1 owns the leftmost digit, mouse #10 the rightmost. Then, run the tasting by the simple rule of a mouse drinks from a bottle if and only if its digit in that bottle’s label is a 1.

For bottle 17, only mouse #6 and mouse #10 take a sip. No two bottles are sampled by the same combination of mice; each bottle’s binary label is its unique drinking pattern.

Step 3: Read the answer off the casualties

Wait an hour. Then, write a 1 in every position with a dead mouse and a 0 otherwise. The string you end up with is the binary label of the poisoned bottle.

One More Sip

Before you pop the corks, one last question: was the binary system essential here, or could we have used a different method?

Here’s a food for thought: suppose your neighbor used a cheaper poison that kicks in 30 minutes instead of an hour, now there is time for a second round of tasting.

Now each mouse has three possible outcomes instead of two: dies in round 1, dies in round 2, or survives. Binary is the wrong language here; you want base 3.

The scheme goes as follows. Label the bottles in base 3, give each mouse a digit position, and follow one rule: if digit 1, drink in round 1; if digit 2, drink in round 2; if digit 0, sit out. Each mouse’s fate is its digit on the poisoned bottle’s label.

The ternary system’s power grows exponentially. Ten mice can now handle 3¹⁰ = 59049 bottles – our 1000 can be covered by seven mice (3⁷ = 2187). Generalized, with r rounds, the whole construction runs in base r+1.

That is the lesson hidden in the wine cellar. The binary system is more than the foundation of our telecom network, it is also a way of thinking – a reminder that any question, no matter how large, can be answered by a patient sequence of yes and no. You walked into an impossible evening with ten mice and walked out with 999 bottles of perfectly good wine.

Enjoy the party.