Angles correspond to their opposite sides. Interior angle is defined as the angle formed between two adjacent sides of a triangle. If, $\alpha$, $\beta$ and $\gamma$ are three interior angles in a triangle, then $\alpha+\beta+\gamma = 180^\circ$ This basic geometrical property of a triangle is often used as a formula in geometry in some special cases. For ∆ABC, Angle sum property of triangle declares that. The interior, or inside, angles of a triangle always add up to 180 degrees. What is the sum of all angles in a triangle? What is the sum of interior angles of a Heptagon? The relation between angular defect and the triangle's area was first proven by Johann Heinrich Lambert.[4]. If the exterior angle of a triangle measures 130 , then what is the sum of the non-adjacent interior angles of that triangle? In addition to your geometry skills, you’ll be able to polish your algebra skills as you set up and solve equations on some of the tougher triangles. Thus, the sum of the interior angles of a triangle is 180°. Specifically, the sum of the angles is. Corresponding and Alternate Angles are also congruent angles. For a spherical triangle, the sum of the angles is greater than 180° and can be up to 540°. pg. 360 degrees. In the differential geometry of surfaces, the question of a triangle's angular defect is understood as a special case of the Gauss-Bonnet theorem where the curvature of a closed curve is not a function, but a measure with the support in exactly three points – vertices of a triangle. Your first step should be to set up an equation where the sum of the angles adds up to 180̊. Since X and, $$\angle J$$ are remote interior angles in relation to the 120° angle, you can use the formula. Learn to apply the angle sum property and the exterior angle theorem, solve for 'x' to determine the indicated interior and exterior angles. Tags: Question 7 . What is the sum of interior angles of a Nonagon? This means that the largest angle is opposite the longest side, while the smallest angle is opposite the shortest side. That should be || to base of the triangle (Side B… Sum of Interior Angles … List of trigonometric identities § Angle sum and difference identities, https://en.wikipedia.org/w/index.php?title=Sum_of_angles_of_a_triangle&oldid=997636359, Short description is different from Wikidata, Articles to be expanded from November 2013, Creative Commons Attribution-ShareAlike License, This page was last edited on 1 January 2021, at 14:37. The diagram below shows the interior and exterior angles of a triangle.. Sum of the Angle Measures in a Triangle is 180 ° - Justify. Ultimately, the answer was proven to be positive: in other spaces (geometries) this sum can be greater or lesser, but it then must depend on the triangle. The area of a … This set includes a 1 foldable interactive graphic organizer (INB) and 2 posters (8.5 X 11). Proof 1 uses the fact that the alternate interior angles formed by a transversal with two parallel lines are congruent. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"And the inside angle at each vertex of a polygon is always equal to (180° - deflection angle), because (new direction - old direction) + (reverse direction - new direction) = (reverse direction - old direction), which is always 180°.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-why-is-the-sum-of-the-interior-angles-of-a-triangle-equal-to-180'\u003eread more\u003c/a\u003e"},"name":"🗯Why is the sum of the interior angles of a triangle equal to 180? Sum Of The Interior Angles Of A Triangle 2 - Displaying top 8 worksheets found for this concept.. You can do this. Step-by-step explanation: Sum of interior angles of a polygon=180. Exterior angle is defined as the angle formed between a side of triangle and an adjacent side extending outward. Maybe it's a piece you'd been looking for on and off for a while. In a triangle, the exterior angle is always equal to the sum of the interior opposite angle. Angle sum of a triangle, which appears to be more common and is more concise; Triangle postulate, which is the technical name of this topic, and is how Wolfram MathWorld refers to it; Sum of angles of a triangle, the current name. [1] In the presence of the other axioms of Euclidean geometry, the following statements are equivalent:[2], The sum of the angles of a hyperbolic triangle is less than 180°. It is also possible to calculate the measure of each angle if the polygon is regular by dividing the sum by the number of sides. Proof 2 uses the exterior angle theorem. Step 4 : What do you notice about how the angles fit together around a point ? It follows that a 180-degree rotation is a half-circle. Interior Angles of a Triangle Rule. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"180°Triangle/Sum of interior angles\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-sum-of-the-interior-angle-of-a-triangle'\u003eread more\u003c/a\u003e"},"name":"🗯What is the sum of the interior angle of a triangle? Angles between adjacent sides of a triangle are referred to as interior angles in Euclidean and other geometries. Example 8 : In a right triangle, apart from the right angle, the other two angles are x + 1 and 2x + 5. find the angles of the triangle. This one's y. Theorem 1: Angle sum property of triangle states that the sum of interior angles of a triangle is 180°. Therefore, straight angle ABD measures 180 degrees. Quick Answer: What Is The Significance Of The Baseball In The Movie Knives Out? And the inside angle at each vertex of a polygon is always equal to (180° – deflection angle), because (new direction – old direction) + (reverse direction – new direction) = (reverse direction – old direction), which is always 180°. All the black and white shapes in the figure on the left are hyperbolic triangles. You can solve for Y. $$120° = 45° + x \\ 120° - 45° = x \\ 75° = x. The influence of this problem on mathematics was particularly strong during the 19th century. Grab each of the points, and move them to increase or decrease their respective angles. Properties of Interior Angles The sum of the three interior angles in a triangle is always "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"Answer and Explanation: Because of the fact that the sum of the three interior angles of a triangle must be 180 degrees, a triangle could not have two right angles.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-can-a-triangle-have-two-right-angles'\u003eread more\u003c/a\u003e"},"name":"🗯Can a triangle have two right angles? Q.2.$$ Now, since the sum of all interior angles of a triangle is 180°. Angle sum property of a triangle Theorem 1: The angle sum property of a triangle states that the sum of interior angles of a triangle is 180°. Therefore, straight angle ABD measures 180 degrees. How do you prove that the sum of all angles in a triangle is 180? When a ray of sunlight. So: angles A are the same ; angles B are the same ; And you can easily see that A + C + B does a complete rotation from one side of the straight line to the other, or 180° Thanks, Laurdecl talk 11:47, 9 … These free geometry worksheets will introduce you to the Triangle Sum Theorem, as you find the measurements of the interior angles of a triangle. Defined in the differentially-geometrical sense. But there exist other angles outside the triangle which we call exterior angles.. We know that in a triangle, the sum of all three interior angles is always equal to 180 degrees. If, $\alpha$, $\beta$ and $\gamma$ are three interior angles in a triangle, then $\alpha+\beta+\gamma = 180^\circ$ This basic geometrical property of a triangle is often used as a formula in geometry in some special cases. Proof 3 uses the idea of transformation specifically rotation. In this triangle ∠ x, ∠y and ∠z are all interior angles. Sum of the Interior Angles of a Triangle. Interior Angles and Polygons: The measures of the interior angles of any triangle must add up to {eq}180^\circ {/eq}. This is also called angle sum property of a triangle . The sum of all of the interior angles can be found using the formula S = (n – 2)*180. For an ideal triangle, a generalization of hyperbolic triangles, this sum is equal to zero. The sum of the measures of the interior angles of a triangle equals 180º. Question 5: Do interior angles add up to 180º? It is not to be confused with, "Angle sum theorem" redirects here. The sum of the interior angles is always 180° implies, ∠ x + ∠y + ∠z = 180°. By interior angle sum property of triangles, ∠A + ∠B + ∠C = 180 0 ⇒ ∠A + 60 0 + 45 0 = 180 0 ⇒ ∠A + 105 0 = 180 0 ⇒ ∠A = 180 -105 0 ⇒ ∠A = 75 0. And again, try it for the square: The sum of the three angles of a triangle equals 180°. Any two triangles will be similar if their corresponding angles tend to be congruent and length of their sides will be proportional. The Triangle Sum Theorem states that the interior angles of a triangle always add up to 180̊. The angles in any given triangle have two key properties: 1. Answer: No, and here’s an example to illustrate. Finding the Number of Sides of a Polygon. Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. Label the angles A, B, and C. Try it first with our equilateral triangle: (n - 2) × 180 °(3 - 2) × 180 °Sum of interior angles = 180 ° Sum of angles of a square. Suppose a triangle had angles of 120, 45 and 15. Exterior Angle Property of a Triangle Theorem. Answer and Explanation: Because of the fact that the sum of the three interior angles of a triangle must be 180 degrees, a triangle could not have two right angles. It was conjectured for a long time that the parallel postulate (also called Euclid's fifth postulate) followed from the first four axioms of … {"@context":"https://schema.org","@type":"FAQPage","mainEntity":[{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"The sum of three sides of a triangle is known as perimeter.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-sum-of-all-3-sides-of-a-triangle'\u003eread more\u003c/a\u003e"},"name":"🗯What is the sum of all 3 sides of a triangle? Is the sum of any two angles of a triangle always greater than the third angle? Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. “SSS” is when we know three sides of the triangle, and want to find the missing angles….To solve an SSS triangle:use The Law of Cosines first to calculate one of the angles.then use The Law of Cosines again to find another angle.and finally use angles of a triangle add to 180° to find the last angle. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"\"SSS\" is when we know three sides of the triangle, and want to find the missing angles.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-how-do-you-find-an-angle-of-a-triangle-with-3-sides'\u003eread more\u003c/a\u003e"},"name":"🗯How do you find an angle of a triangle with 3 sides? Quick Answer: What Was The Number One Song For 2019? In hyperbolic geometry, the interior angle sum of a triangle is less than $180^{\circ}$, whilst in elliptic geometry, the interior angle sum is more than $180^{\circ}$. 90 degrees. An exterior angle of a triangle is equal to the sum of the opposite interior angles. )? You can solve for Y. In a triangle, the angle opposite to longer side is larger and vice-versa. The rotation from A to D forms a straight line and measures 180 degrees. Thus, the sum of the interior angles of a triangle is 180°.. Do all Triangle angles equal 180? From the above diagram, we can say that the triangle has three interior angles. I've drawn an arbitrary triangle right over here. To Prove :- ∠4 = ∠1 + ∠2 Proof:- From • Since rotations are rigid transformations, angle measure is preserved and m∠ABC = m∠A'B'C' and m∠B'A'C' = m∠B''A''C''. Here are three proofs for the sum of angles of triangles. Displaying top 8 worksheets found for - Sum Of Interior Angles In A Triangle. Since X and, $$\angle J$$ are remote interior angles in relation to the 120° angle, you can use the formula. New questions in Math. a triangle have 3 sides n=3 So if the measure of this angle is a, the measure of this angle over here is b, and the measure of this angle is c, we know that a plus b plus c is equal to 180 degrees. Sum of all the interior angles of the triangle … 'There has to be a light blue sky piece somewhere here...' When we're working with triangles, sometimes we have missing puzzle pieces. Q. How tall is Jacquees? Some of the worksheets for this concept are Sum of the interior angles of a triangle 2 directions, 4 angles in a triangle, 4 the exterior angle theorem, Triangle, Sum of the interior angles of a triangle, Triangle, Relationship between exterior and remote interior angles, Name geometry polygons n. A. Every triangle has six exterior angles (two at each vertex are equal in measure). "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"An interior angle is located within the boundary of a polygon.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-formula-for-interior-angles'\u003eread more\u003c/a\u003e"},"name":"🗯What is the formula for interior angles? How are we supposed … An exterior angle of a triangle is equal to the sum of the opposite interior angles. 90. Hence , the sum of all interior angles of a triangle is 180° regardless of the type of triangle . In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). Proof: Sum of all the angles of a triangle is equal to … How do you find an angle of a triangle with 3 sides? The sum of two sides of a triangle is greater than the third side. BLINDING LIGHTS. Triangle angle sum theorem: Which states that, the sum of all the three interior angles of a triangle is equal to 180 degrees. What is the sum of the measures of the two interior angles of a right triangle, not including the right angle (∠ sum of rt. So, the sum of three exterior angles added to the sum of three interior angles always gives three straight angles. Each interior angle of a regular octagon is = 135 °. Exit Ticket (5 minutes) Lesson Summary The sum of the measures of the remote interior angles of a triangle is equal to the measure of the related exterior angle. 80. 180° Step 6 : Describe th… An exterior angle is supplementary to its adjacent triangle interior angle. Thus, the sum of the interior angles of a triangle is 180°.. 100. Sum of interior angles in a triangle. Author: Robert Paterson. Note that spherical geometry does not satisfy several of Euclid's axioms (including the parallel postulate.). answer choices . Exterior Angle Theorem – Explanation & Examples. Can A Doctor Change Your Medication Without Telling You? 4: Interactive fol actress K CallanIn. 60 degrees. The sum of the angles can be arbitrarily small (but positive). We already know that the sum of the interior angles of a triangle add up to 180 degrees. B. Proof. An exterior angle is supplementary to its adjacent triangle interior angle. As you can see from the picture below, if you add up all of the angles in a triangle the sum must equal 180 ∘ . In the given figure, the side BC of ∆ABC is extended. The two sides of the triangle that are by the right angle are called the legs... and the side opposite of the right angle is called the hypotenuse. Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"To prove the above property of triangles, draw a line \\overleftrightarrow {PQ} parallel to the side BC of the given triangle.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-how-do-you-prove-that-the-sum-of-all-angles-in-a-triangle-is-180'\u003eread more\u003c/a\u003e"},"name":"🗯How do you prove that the sum of all angles in a triangle is 180? C. step 2: Tear off each “ corner ” of the interior angles of the three interior always! Answer choices angles, taken one at each vertex, always sum up to 180 degrees its triangle. Measure of 90°,  angle sum theorem '' redirects here all of the angles of a triangle n... Of hyperbolic triangles for right triangles, n=3, so sum-of-interior-angles = 3 * –! Small ( but positive ) 8 worksheets found for - sum of all interior angles ' B C! Up an equation where the sum of the interior angles, QR is produced to S.... 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Worksheet 2 - this angle Worksheet features 12 different triangles smallest angle is defined as the set points! ∠Y and ∠z are all interior angles of a triangle is a half-circle ∠ x, ∠y and ∠z all., 45 and 15 boundary of sum of interior angles of a triangle triangle and an adjacent side extending outward three points also... We already know that the sum of any triangle is planar, meaning it is equal... Where the sum of interior angles of a triangle are congruent for an ideal triangle, third. About how the angles a, B, and C. step 2: Tear off each “ corner of! Specifically rotation blue triangle, the measure of 90° enclosed by the triangle is a 3-sided figure three! Triangle measures 130, then what is the sum of the interior angles of the …... 1 foldable interactive graphic organizer ( INB ) and 2 posters ( x!, or inside, angles of that triangle makes money, what is x are in Movie... 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Suppose you’re on a game show, and you’re given the choice of three doors,

Behind one door is a car; behind the two others, goats. You pick a door, say No. 1, and the host of the show opens another door, say No. 3, which has a goat.

He then says to you, “Do you want to pick door No. 2?”.

Question: What would you do?

Is it to your advantage to switch your choice from door 1 to door 2? Is it to your advantage, in the long run, for a large number of game-tries, to switch to the other door?

Now whatever your answer is, I want you to check/prove your answer by a Monte Carlo simulation of this problem. Make a plot of your simulation for $\rm{ngames}=100000$ repeat of this game, that shows, in the long run, on average, what the probability of winning this game is if you switch your choice, and what is the probability of winning, if you do not switch to the other door.

As you see in the figure, although you may initially win by not switching your choice, in the long run, on average, you will lose, if you don’t switch your choice.

Here is an example MATLAB implementation, monteGame.m,

%rng('shuffle');
rng(1235);
nExperiments=100000; % the number of times the experiment is simulated
TotalWinsWithoutSwitch = 0;
TotalWinsWithSwitch = 0;
AllDoors = [1,2,3];
averageNumberOfWinsWithoutSwitch = zeros(nExperiments,1);
averageNumberOfWinsWithSwitch = zeros(nExperiments,1);

for iExperiment = 1:nExperiments

if mod(iExperiment,10000)==0; disp(['simulating experiment number ', num2str(iExperiment)]); end

ContainsCar = zeros(3,1); % first create the three doors, assuming none contains the car, hence all are zero valued.
doorWithCar = randi([1,3],1,1); % Now put the car behind one of the three doors randomly.
myChoice = randi([1,3],1,1); % Now make a choice of door randomly.

excludedDoor = AllDoors(AllDoors~=doorWithCar & AllDoors~=myChoice); % find which door should be opened by the host
if length(excludedDoor)>1
excludedDoor = excludedDoor(randi([1,2])); % pick one of the empty doors randomly
end

% find the chance of winning by no door switching
if myChoice==doorWithCar
TotalWinsWithoutSwitch = TotalWinsWithoutSwitch + 1;
end

% find the chance of winning by door switching
myChoice = AllDoors(AllDoors~=myChoice & AllDoors~=excludedDoor); % update my choice to the alternative door after host's exclusion of one of the doors.
if myChoice==doorWithCar
TotalWinsWithSwitch = TotalWinsWithSwitch + 1;
end

averageNumberOfWinsWithoutSwitch(iExperiment) = TotalWinsWithoutSwitch / iExperiment;
averageNumberOfWinsWithSwitch(iExperiment) = TotalWinsWithSwitch / iExperiment;

end

% Now draw and export figures

figure('visible','on'); hold on; box on;
plot( averageNumberOfWinsWithoutSwitch ...
, 'color', 'blue' ...
, 'linewidth', 1.5 ...
);
plot( averageNumberOfWinsWithSwitch ...
, 'color', 'red' ...
, 'linewidth', 1.5 ...
);
axis([0,nExperiments,0,1]);
line1 = line( [1 nExperiments], [0.66666666 0.66666666] ...
, 'LineStyle', '--' ...
, 'color','green' ...
, 'LineWidth',1.5 ...
);
line2 = line( [1 nExperiments], [0.33333333 0.33333333] ...
, 'LineStyle','--' ...
, 'color','green' ...
, 'LineWidth',1.5 ...
);
xlabel('Experiment Repeat Number');
ylabel('Average Probability of Winning');

legend( {'No Door Switch','With Door Switch','theoretical prediction','theoretical prediction'} ...
, 'Location', 'southeast' ...
);
uistack(line1,'bottom');
uistack(line2,'bottom');

set(gca,'xscale','log');

set( gca ...
, 'XMinorTick','on','YMinorTick','on' ...
, 'FontSize', 13 ...
, 'LineWidth', 1.25 ...
);

saveas(gcf,'MontyGameResult.png');

%close(gcf);

Here is an example Python implementation, monteGame.py,

#!/usr/bin/env python

import numpy as np
import matplotlib.pyplot as plt

ngames = 100000

chance_of_success_by_switching = np.zeros(ngames,dtype=np.double)
chance_of_success_by_not_switching = np.zeros(ngames,dtype=np.double)

options = [1,2,3]

for i in range(ngames):
door_host_will_open = options.copy()
door_with_car = np.random.randint(1,4)
door_host_will_open.remove(door_with_car)
door_i_choose = np.random.randint(1,4)
if (door_i_choose!=door_with_car): door_host_will_open.remove(door_i_choose)

door_to_switch = options.copy()
door_to_switch.remove(door_i_choose)
door_to_switch.remove(door_host_will_open[0])

if i==0:
if (door_to_switch[0]==door_with_car):
chance_of_success_by_switching[i] = 1.0
else:
chance_of_success_by_not_switching[i] = 1.0
else:
if (door_to_switch[0]==door_with_car):
chance_of_success_by_switching[i] = (chance_of_success_by_switching[i-1]*np.double(i) + 1.0)/np.double(i+1)
chance_of_success_by_not_switching[i] = chance_of_success_by_not_switching[i-1]*np.double(i)/np.double(i+1)
else:
chance_of_success_by_not_switching[i] = (chance_of_success_by_not_switching[i-1]*np.double(i) + 1.0)/np.double(i+1)
chance_of_success_by_switching[i] = chance_of_success_by_switching[i-1]*np.double(i)/np.double(i+1)

trials = np.linspace(1,ngames,ngames)
plt.semilogx(trials, chance_of_success_by_switching, 'r-')
plt.hold('on')
plt.semilogx(trials, chance_of_success_by_not_switching, 'b-')
plt.legend(['switch choice' , 'not switch choice'])
plt.xlabel('Game Number')
plt.ylabel('Average probability of winnig')
plt.title('{} Monte Hall Games: On average, you win if your switch!'.format(ngames))
plt.savefig('MontyGameResult.png')
plt.show()


# Can f''(x_0) = 0 if f'(x_0) =/= 0?

1. Dec 6, 2015

### FaroukYasser

I was wandering, can $$\ \frac { d^{ 2 }y }{ dx^{ 2 } } | _{ x={ x }_{ 0 } }=\quad 0$$ if $$\frac { dy }{ dx }| _{ x={ x }_{ 0 } }\neq \quad 0$$ and if so, what does this translate to geometrically?

2. Dec 6, 2015

### HallsofIvy

Staff Emeritus
Yes, of course. That is the same as saying that a function can have the value "0" at a point where its derivative is not 0. And it is very easy to give an example of that- suppose y= x- 1. That is a straight line such that y(1)= 1-1= 0 but its derivative is the constant 1 which is never 0. To relate that to your question about first and second derivatives, take an "anti-derivative". An anti-derivative of x- 1 is y= (1/2)x^2- x+ 1. Now, y'= x- 1 which is 0 at x= 1 while the second derivative is 1.

An nth derivative tells how fast the n-1 derivative is changing and is NOT related to the actual value of that n-1 derivative.

3. Dec 6, 2015

### Samy_A

$y=x-x_0$

4. Dec 6, 2015

### FaroukYasser

Thanks. But I was hoping for a non linear function. Or in other words, I am wandering what this means geometrically. I know that if a point is an inflection point then its second derivative is 0 but the converse doesn't necessarily hold. In other words, what does f''(x_0) = 0 tell us about that point other than it being a likely inflection point.

Thanks :)

5. Dec 6, 2015

### FactChecker

This means that the slope of the curve is not changing at that point
This means the slope is not zero at the point. So @Samy_A 's example of a sloped straight line satisfies both conditions at every point.

6. Dec 6, 2015

### FactChecker

Nothing. That is exactly what it means. sin(x) has an inflection point at every x= n * pi. Those are all inflection points with non-zero slopes.

7. Dec 6, 2015

### WWGD

Actually, take a degree-2 or higher polynomial other than $x^n$, i.e. $a_nx^n+...+a_1x+a_0 ; a_j$ not all 0 $n>j\geq 0$, I would say the probability of having both f'(x) and f''(x)=0 is 0 under "reasonable" choices of pdf..

8. Dec 7, 2015

### FaroukYasser

I see. Thanks all!

9. Dec 22, 2015

### LCKurtz

The second derivative geometrically identifies concavity. f'' > 0 or f''<0 on an interval identifies concave up or down, respectively. In a case where f'' is continuous, the only way for concavity to switch is for f'' to pass through zero. So f'' = 0 at a point might be a case where f'' is changing sign, indicating an inflection point (change of concavity). Of course, f'' might not change sign nearby so it might not be an inflection point. I wouldn't say such a point is a "likely" inflection point but a "possible" inflection point.


Upper and Lower Bounds of $\emptyset$

From some reading, I've noticed that $\sup(\emptyset)=\min(S)$, but $\inf(\emptyset)=\max(S)$, given that $\min(S)$ and $\max(S)$ exist, where $S$ is the universe in which one is working. Is there some inherent reasoning/proof as to why this is? It seems strange to me that an upper bound of a set would be smaller than a lower bound of the same set.

-

The supremum is defined to be the least upper bound. An upper bound for a set $K$ is defined to be an element $b$ such that for all $k\in K$, $k\leq b$. If $K$ is empty, then the condition "for all $k\in K$, $k\leq b$" is true by vacuity (you have an implication of the form "if $k$ is an element of the empty set, then $\lt$something happens$\gt$", and the antecedent is always false so the implication is always true).

Therefore, every element of $S$ is an upper bound for $\emptyset$; since $\sup(\emptyset)$ is the least upper bound, that means that $\sup(\emptyset)=\min($upper bounds of $\emptyset) = \min(S)$.

Similarly, the infimum is the greatest lower bound, and every element of $S$ is a lower bound for $\emptyset$, so $\inf(\emptyset) = \max($lower bounds of $\emptyset) = \max(S)$.

Yes, it is somewhat counterintuitive that you have $\sup(\emptyset)\leq\inf(\emptyset)$, but in fact the empty set is the only one for which you can have $\sup(\emptyset)\lt \inf(\emptyset)$. The empty set often causes counterintuitive results.

-
Interesting, thanks for that explanation. –  yunone Sep 1 '10 at 5:20

A quick counterpoint to Arturo's answer: yes, the fact that $\operatorname{sup}(\emptyset) = -\infty < \infty = \operatorname{inf}(\emptyset)$ seems strange at first. But it is designed, among other things, to preserve an intuitive and useful property of infima and suprema:

Let $S$ and $T$ be subsets of $\mathbb{R}$ with $S \subset T$. Then

$\operatorname{inf}(T) \leq \operatorname{inf}(S)$

and

$\operatorname{sup}(T) \geq \operatorname{sup}(S)$.

Indeed, by adding elements to a set, its infimum can only get smaller and its supremum can only get larger. Now if you take $S = \emptyset$ and let $T$ range through even all one-element subsets of $\mathbb{R}$, you see that in order for these inequalities to hold we must define $\operatorname{inf}(\emptyset)$ and $\operatorname{sup}(\emptyset)$ as we did.

-

A very natural explanation of the correct definition for the values of min and max on empty sets arises from their dual universal definitions - analogous to the universal GCD, LCM definitions that I presented in a post here.

First some notation. $\;$ Write $\ \ \rm x \le S \;\iff\; x \le s,\;\: \forall\: s \in S,\;$ and dually for $\rm\; x \ge S$

DEFINITION of $\:$ min $\quad$ $\quad\rm x \le S \;\iff\; x \le min\ S$

DEFINITION of max $\:\quad$ $\quad\rm x \ge S \;\iff\; x \ge max\ S$

For min, when $\;\rm S = \emptyset\;$ is empty, the first clause $\;\rm x \le S\;$ in the min definition is vacuously true, hence the definition reduces to $\;\rm x \le \min \emptyset\;$ for all $\;\rm x\;$. Hence $\;\rm \min \emptyset = \infty\;$. Dually $\;\rm \max\emptyset = -\infty\;$.

As I remarked in said GCD, LCM post, such universal definitions often facilitate slick proofs. For some nontrivial examples of min, max flavor consider the slick proofs of the integrality of various products of binomial coefficients by employing the floor function, e.g. see Joe Roberts: Elementary Number Theory. Instead I close with an analogous slick GCD, LCM proof from my mentioned post (see it for further details).

Generally, in any domain, we have the following dual universal definitions of LCM and GCD:

DEFINITION of LCM $\quad$ If $\quad\rm a,b\ |\ c \;\iff\; [a,b]\ |\ c \quad$ then $\quad\rm [a,b] \;\;$ is an LCM of $\;\rm a,b$

DEFINITION of GCD $\quad$ If $\quad\rm c\ |\ a,b \;\iff\; c\ |\ (a,b) \quad$ then $\quad\rm (a,b) \;$ is an GCD of $\;\;\rm a,b$

Note: that $\;\rm a,b\ |\ [a,b] \;$ follows by putting $\;\rm c = [a,b] \;$ in the definition. Dually $\;\rm (a,b)\ |\ a,b \;$

Such $\iff$ definitions provide slick unified proofs of both arrow directions, e.g. the fundamental

THEOREM $\rm\quad (a,b)\ =\ ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.

Proof: $\rm\quad\quad d\ |\ a,b \;\iff\; a,b\ |\ ab/d \;\iff\; [a,b]\ |\ ab/d \;\iff\; d\ |\ ab/[a,b] \quad\;\;$ QED

The conciseness of this proof arises by exploiting to the hilt the $\iff$ definition of LCM and GCD. Compare to less concise / general / illuminating proofs in many number theory textbooks.

-
@Bill: I think where you say "min" and "max" in the first two paragraphs you mean "inf" and "sup"; the empty set has no minimum and no maximum. –  Arturo Magidin Feb 1 '11 at 16:33
@Arturo: above min,max are synonyms for inf,sup, cf. their universal definitions. –  Bill Dubuque Feb 1 '11 at 20:30
@Bill: That's nonstandard use of min and max, as far as I am aware; I don't see what I'm supposed to compare, though... –  Arturo Magidin Feb 1 '11 at 20:41
@Arturo: You're supposed to refer to the universal definitions of min, max that I am using above. –  Bill Dubuque Feb 1 '11 at 21:59


# What does this 'L' and upside down 'L' symbol mean?

I need help finding out what the following symbols are called and what they do. I searched up math symbols but couldn't find them anywhere near there.

$$\lceil{-3.14}\rceil=$$

$$\lfloor{-3.14}\rfloor=$$

• Try Detexify. – lhf Nov 3 '15 at 10:22

They are ceil and floor values, that is they are the closest integers one below and one above respectively , see link.

• Ohhh! Thanks so much – Xirol Nov 3 '15 at 9:56

The notation $\lfloor x \rfloor$ (known as ‘the floor function’) denotes the largest integer less than or equal to $x \in \mathbb{R}$. Examples include $\lfloor7\rfloor$ = $7$, $\lfloor2.5\rfloor$ = $2$, $\lfloor\pi\rfloor$= $3$ and $\lfloor−2.5\rfloor$ = $-3$.

The notation $\lceil x \rceil$ (known as ‘the ceiling function’) denotes the smallest integer greater than or equal to $x \in \mathbb{R}$. So using the same examples as before $\lceil7\rceil$ = $7$, $\lceil2.5\rceil$ = $3$, $\lceil\pi\rceil$= $4$ and $\lceil−2.5\rceil$ = $-2$.

$$\lceil{-3.14}\rceil=-3$$

$$\lfloor{-3.14}\rfloor=-4$$

• For completeness, there is also the Nearest Integer function: $\lfloor x \rceil$. – YoTengoUnLCD Dec 22 '15 at 22:04
• @YoTengoUnLCD Oh wow, I never heard of that one before; thanks for pointing it out :) Perhaps you should make this as an answer, and give some examples for it (maybe the same numerical figures as I used for comparison). It's still vaguely related to OP's question. Besides, I would like to see this function in action! Thank you. – BLAZE Dec 22 '15 at 22:09
• Done! Thanks for the suggestion :-). – YoTengoUnLCD Dec 22 '15 at 22:20

Related to the two mentioned functions, there is the perhaps less common

$\text{Nearest integer function } \lfloor x\rceil$.

$$\lfloor 3.2\rceil=3\\ \lfloor 3.6\rceil=4\\ \lfloor -1.2\rceil=-1$$

The function is ambiguous at numbers of the form $r+\frac 1 2, r\in \bf Z$, and so some kind of note should be made when using this function, clarifying what's to be done in those cases.

Note that we can set

$$\lfloor x\rceil_1=\left \lfloor x+\frac 1 2\right\rfloor\\ \\ \lfloor x\rceil_2=\left \lceil x-\frac 1 2\right\rceil\\$$

The first one rounds half integers up, and the second rounds them down.


# Condition Number of Rectangular Matrices

The 2-norm condition number can be easily extended to rectangular matrices. I'm wondering if the inequality for the product of matrices still holds in that case, i.e.,

$\operatorname{cond}(AB) \leq \operatorname{cond}(A)\operatorname{cond}(B)$

• Yes, this is still true for rectangular matrices both in the spectral norm and in other norms. It's a good exercise so I won't ruin the challenge for you. A hint would be start by considering the definition of $\mathrm{cond}(A)$ that you're using. – Brian Borchers Mar 25 '16 at 0:04

I guess I figured out the answer to my question.

Suppose the SVD of $A = U \Sigma V^\ast$ (where $V^\ast$ is the conjugate transpose of the matrix $V$). Noting the fact that the unitary transformations $U$ and $V$ preserve the 2-norm, $\|Ax\|_{2}$ for any unit vector $x$ can be written as

\begin{align*} \frac{\|Ax\|_{2}}{\|x\|_{2}} = \|Ax\|_{2} &= \|U \Sigma V^\ast x\|_{2} \\ &= \| \Sigma V^\ast x\|_{2} \\ &= \| V^\ast \Sigma x\|_{2} \\ &= \|\Sigma x\|_{2} \end{align*}

Hence $\|Bx\|_{2} \leq \sigma_\max(B)$ and $\|Bx\|_{2} \geq \sigma_\min(B)$.

For $y = Bx$,

\begin{equation*} \|ABx\|_{2} = \|Ay\|_{2} \leq \sigma_\text{max}(A) \|y\|_{2} \leq \sigma_\max(A) \sigma_\max(B) \end{equation*}

Similarly,

\begin{equation*} \|ABx\|_{2} = \|Ay\|_{2} \geq \sigma_\min(A) \|y\|_{2} \geq \sigma_\min(A) \sigma_\min(B) \end{equation*}

Since the above statements are true for all $x$, they are true both for the minimum and maximum. Thus, \begin{equation*} \sigma_\max(AB) = \max \|ABx\|_{2} \leq \sigma_\max(A) \sigma_\max(B) \end{equation*}

and \begin{equation*} \sigma_\min(AB) = \min \|ABx\|_{2} \geq \sigma_\min(A) \sigma_\min(B) \end{equation*}

On dividing the above two equations, we obtain \begin{equation*} \operatorname{cond}_{2}(AB) \leq \operatorname{cond}_{2}(A) \operatorname{cond}_{2}(B) \end{equation*}

• You can accept your own answer. – nicoguaro Apr 24 '16 at 3:44


# Place $8$ rooks on a $10\times 10$ board.

The Problem:-

"In chess, a rook attacks any piece in the same row or column as the rook, provided no other piece is between them. In how many ways can $8$ rooks be placed on a $[8\times8]$ chessboard so that no two attack each other? What about $8$ rooks on a $10\times10$ board?"

I believe I have an answer for the first part of the question. When placing the first rook, there are 8 places on any particular column (or row) to place the rook, leaving just 7 places on a different column (or row) for the next rook, and so on, providing 8! possible ways to place the rooks in such a way that they cannot attack each other ($P(8,8) = 8!/(8-8)! = 8!$).
However, I am not sure I fully understand how this would work for a board where there are more rows and columns than pieces (such as on a 10x10 board). Does it become $P(10,8) = 10!/(10-8)! = 10!/2$ ? If so, why? If not, how should I approach this problem?

This problem was found in "Introduction to Combinatorics and Graph Theory" by David Guichard.

• Only one rook can go on each row and on each column. If there are two more rows and columns there are two rows and two columns with no rooks on them. just declare at the begininning that there will be 2 rows which won't be used. There are ${10 \choose 2}$ ways of doing that and ${10 \choose 2}$ ways of choosing the columns. So there are ${10\choose 2}^2 *D$ ways of doing this if $D$ is the number of ways to do it on an 8x8 board. – fleablood Nov 7 '17 at 18:45

So for the first question you need to place a rook in each column. The rook in the first column can go in $8$ places, the rook in the second column then has $7$ possible rows etc.

In the second part, you need first to choose the $8$ columns out of the $10$ available, and then place a rook in each of the columns. If once again you move from left to right, counting the number of possibilities, you should be fine.

Rather than talking about "any particular" column or row, you should make a habit of defining the scheme more closely (the first, second rather than "any" etc). This will help to avoid a great deal of confusion and can also highlight any double-counting.

• For clarification, to determine the number of possible combinations for placing 8 rooks on a 10x10 board, we could calculate 10 choose 8 to determine how many 8x8 boards we could have, and then use that to determine how many total combinations there are, by multiplying the number from 10 choose 8 (45) with the number of possible permutations on the board (8!)? – oath Nov 7 '17 at 18:54
• @oath With $\binom {10}{8}$ columns you effectively reduce to a $10\times 8$ board - but then you have to find a position in each column (there are ten positions to start with, and each choice reduces the possibilities for the next choice by one) – Mark Bennet Nov 7 '17 at 18:56
• Ooooh, I see. So with this line of thought, would $10\choose 8$ multiplied by permutations $P(10,8)$ give the correct answer ($10\choose 8$ gives 45, and the permutation $P(10,8)$ gives $10!/2!$) (the answer provided by @Bram28)? – oath Nov 7 '17 at 19:15

For the second part, if we first focus on the columns the rooks are in: since there cannot be two rooks in the same column, the rooks must appear in $8$ different columns. And you can choose $8$ columns out of $10$ in

$$10 \choose 8$$

ways. Once, we've chosen the $8$ columns, let's place the rooks in those columns. Well, you can place the first rook in $10$ different rows, the next in $9$, etc. until the last one, which can be place in $3$ different rows, so we can place those $8$ rooks in:

$$\frac{10!}{2!} = \frac{10!}{2}$$

ways. This leads to a grand total of:

$${10 \choose 8} \cdot \frac{10!}{2}$$

ways of placing $8$ rooks on a $10 \times 10$ board.

• As a variation on this, we can (independently) choose eight columns and eight rows to be populated, each in $$10 \choose 8$$ different ways. Having done that, the chosen rows and columns form a board equivalent to the 8 x 8 case, which the OP has already answered (8!). Overall, then, for 10 x 10 case, there are $${10 \choose 8} ^ 2 \cdot 8!$$ different ways to place the 8 rooks. – John Bollinger Nov 7 '17 at 22:20
• @JohnBollinger Ooh, nice, thanks! – Bram28 Nov 7 '17 at 22:25

This is a different approach than what's been offered so far, but that might help some people understand the problem differently.

I would solve this by first assuming order mattered when placing the rooks (so placing on $A1$ then $B2$ is different than $B2$ then $A1$), then adjusting my answer to compensate.

The number of squares you can place the first rook on is $10^2$. Wherever you place it, you effectively turn your board into a $9 \times 9$ board with no rooks on it.

Now that we've lost a row and a column from our first rook, the number of squares you can place the second rook on is $9^2$.

Carrying on for each rook, we get:

$$10^2 \cdot 9^2 \cdot ... \cdot 3^2$$

Ways to place our rooks (don't forget we only have $8$ rooks, so we don't go all the way to $1^2$). Remember though, we assumed order matters. In reality, it does not. However, there's only $8!$ different ways to order our rooks, so we can divide our initial answer by $8!$, leaving:

$$(10^2 \cdot 9^2 \cdot ... \cdot 3^2) / 8!$$

Ways to place our rooks.

For the first case of placing 8 rooks on an 8x8 board your result of 8! is correct.

For the second case of placing 8 rooks on a 10x10 board I find the following approach to be the simplest way to do the calculations:

There are 10*9/2 = 45 ways to choose which two rows you will not be using. Likewise there are 45 ways to choose which two columns you will not be using.

Once you have chosen which rows and columns not to use there are 8! ways to place the rooks in the remaining squares.

That gives a total of 8!*45² = 81648000 possibilities.


# Combinatorial Proof for Binomial Identity: $\sum_{k = 0}^n \binom{k}{p} = \binom{n+1}{p+1}$ [duplicate]

I am studying combinatorics and I came across the identity $$\sum\limits_{k=0}^n \binom kp =\binom {n+1}{p+1}.$$ I have read the algebraic proof and it does not appeal to me. Is there an elegant counting trick we can use to arrive at this identity?

## marked as duplicate by Jack D'Aurizio, user230715, Macavity, vonbrand, 6005Aug 25 '15 at 20:55

• In case $p=0$? More than one question mark reads as an insulting tone. @WillJagy – Thomas Andrews Aug 25 '15 at 2:37
• @WillJagy I do realize that $\binom kp = 0$ if $k <p$ $(k \in \mathbb{Z})$. I guess it is more elegant to write it that way. And everything is well defined so there is no issues. – Miz Aug 25 '15 at 2:40
• @GeorgeSimpson: I disagree with this duplication mark. This question is explicitly asking for a combinatorial proof contrary to the referred question. So, two of three answers of the referred question are to exlude as proper answer for this question. The combinatorial answers here which are quite ok for this question do hardly answer the referred binomial identity. – Markus Scheuer Aug 26 '15 at 8:21
• @MarkusScheuer Looking among linked and related questions it seems that there are several other questions about combinatorial proof of the same identity. For example: 321022, 497413 1332282. If needed, we can discuss this further in chat. – Martin Sleziak Aug 29 '15 at 7:59

What do you get when you take a sequence of ones and zeros of length $n+1$ that has $k+1$ ones and remove the rightmost $1$ and everything to its right?

Example: $100100\rightarrow 100$

You get a sequence that has $k$ ones, the length varies between $k$ and $n$ depending on the actual sequence.

In fact this function is a bijection between the sequences of length $n+1$ and $k+1$ ones and the sequences of length $n$ or less and $k$ ones.

This establishes $\binom{n+1}{k+1}=\sum\limits_{j=k}^{n}\binom{j}{k}$

I'd prefer to omit the zero terms and prove $$\sum_{k = p}^n \binom{k}{p} = \binom{n+1}{p+1}.$$

Suppose we have $n+1$ people standing in a line so that we can refer to Person 1, Person 2, ..., Person $n+1$.

The righthand side counts the number of committees of size $p+1$ that can be formed out of these $n+1$ people.

The lefthand side counts the same thing, but it does so in cases.

The first case is that Person $p+1$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $p$ people in line, which can be done in $\binom{p}{p}$ ways.

The second case is that Person $p+2$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $p+1$ people in line, which can be done in $\binom{p+1}{p}$ ways.

Continuing in this way, the last case would be that Person $n+1$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $n$ people in line, which can be done in $\binom{n}{p}$ ways.

Summing over these cases yields the desired identity.

Here is another combinatorial approach using lattice paths.

We can interpret $\binom{n+1}{p+1}$ as number of lattice paths of length $n+1$ containing $p+1$ horizontal $(1,0)$-steps and $n-p$ vertical $(0,1)$-steps.

This is valid because there are $\binom{n+1}{p+1}$ choices to select $p+1$ steps in horizontal direction leaving the remaining $n-p$ steps for the vertical direction.

Let's consider all paths from $(0,0)$ to $(p+1,n-p)$. We know there are $\binom{n+1}{p+1}$ different paths.

On the other hand each of these paths has to cross the vertical line going through $(p,0)$. The crossing points are $(p,0), (p,1), \ldots, (p,n-p)$. This way we partition the set of lattice paths into sets containing $\binom{k}{p}$ elements with $p\leq k \leq n$ establishing the identity

The RHS counts the ways to select $p+1$ places in a line of $n+1$ places.
The LHS counts the ways to select $p$ places in a line of $k$ places, summed over all values of $k: 0\leq k\leq n$.   (More strictly, $k: p\leq k\leq n$, since there are no ways to do this for $k<p$).
Interpreting $k+1$ as the position of the last, or $(p+1)^{st}$, place, we notice that these both count ways to perform the exact same task.   Ergo they are equivalent.
$$\sum_{k=p}^n \dbinom{k}{p} = \binom{n+1}{p+1}$$


# Group Combinations Problem

#### mathceleb

If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

Any thoughts?

#### undefined

MHF Hall of Honor
If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

Any thoughts?
I haven't worked this out rigorously, but I think the probability is $$\displaystyle \displaystyle \frac{1}{\binom{149}{2}}$$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $$\displaystyle \displaystyle \binom{149}{2}$$ ways to choose two people out of the remaining 149...

#### Soroban

MHF Hall of Honor
Hello, mathceleb!

We have a group of 150 people, broken into groups of 3, and only 3 are blue.
What is the probability that all 3 blue people are put in the same group?

The 150 people are divided into 50 groups of 3 people each.

There are: .$$\displaystyle \dfrac{150!}{(3!)^{50}}$$ possible groupings.

To have the 3 blue people in one group, there is: .$$\displaystyle {3\choose3} = 1$$ way.

The other 147 people are divided into 49 groups of 3: .$$\displaystyle \dfrac{147!}{(3!)^{49}}$$ ways.
. . Hence, there are: .$$\displaystyle \dfrac{147!}{(3!)^{49}}$$ ways to have the 3 blue people in one group.

The probability that all 3 blue people are in one group is:

. . $$\displaystyle \dfrac{\dfrac{147!}{(3!)^{49}}} {\dfrac{150!}{(3!)^{50}}} \;=\;\dfrac{147!}{(3!)^{49}}\cdot\dfrac{(3!)^{50}}{150!} \;=\;\dfrac{6}{150\cdot149\cdot148} \;=\;\dfrac{1}{551,\!300}$$

#### Plato

MHF Helper
the probability is $$\displaystyle \displaystyle \frac{1}{\binom{149}{2}}$$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $$\displaystyle \displaystyle \binom{149}{2}$$ ways to choose two people out of the remaining 149...
This correct.

The 150 people are divided into 50 groups of 3 people each.
There are: .$$\displaystyle \dfrac{150!}{(3!)^{50}}$$ possible groupings.

Those are ordered partions.

There are $$\displaystyle \dfrac{150!}{(3!)^{50}(50!)}$$ unordered partitions.

There are $$\displaystyle \dfrac{147!}{(3!)^{49}(49!)}$$ of those where the blues are together.

Note that $$\displaystyle \dfrac{\dfrac{147!}{(3!)^{49}(49!)}}{ \dfrac{150!}{(3!)^{50}(50!)}}=\dfrac{1}{\binom{149}{2}}$$

#### undefined

MHF Hall of Honor
Hmm since Soroban's answer differs from mine I decided to do a little more research.

There are: .$$\displaystyle \dfrac{150!}{(3!)^{50}}$$ possible groupings.
I believe this should be $$\displaystyle \dfrac{\left(\dfrac{150!}{(3!)^{50}}\right)}{50!}$$ because the groups of 3 can be permuted.

...Hence, there are: .$$\displaystyle \dfrac{147!}{(3!)^{49}}$$ ways to have the 3 blue people in one group.
Likewise I believe this should be $$\displaystyle \dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)}{49!}$$.

Leading to:

$$\displaystyle \dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)\cdot50!}{49!\cdot\left(\dfrac{150!}{(3!)^{50}}\right)} = 50\left(\dfrac{1}{551,\!300}\right) = \dfrac{1}{11026} = \displaystyle \frac{1}{\binom{149}{2}}$$

See here (a PDF file) for some more info; under subheading "Partitioning".

Edit: Plato answered a lot quicker than I did, didn't refresh the page.

Similar threads


## [Math][Python] Project Euler #99 and logarithmic representation

Comparing two numbers written in index form like $$2^{11}$$ and $$3^7$$ is not difficult, as any calculator would confirm that $$2^{11} = 2048 < 3^7 = 2187$$. However, confirming that $$632382^{518061} > 519432^{525806}$$ would be much more difficult, as both numbers contain over three million digits. Using base_exp.txt (right click and 'Save Link/Target As...'), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.

This is a problem where logarithmic representation of number can be used. We compute power not for numbers, but for their logarithmic representation. And they consumes way less RAM. Also, this is much faster. In fact, John Napier invented logarithms to ease multiplication operation.

Since we only need info, which number is bigger, logarithmic representations are enough: they obey this law as well.

#!/usr/bin/env python3
import math

f=open(filename,"r")
f.close()

return [item.rstrip() for item in ar]

tmp=123**45
print ("generic algo:", tmp)
# log with other base would also work!
# but here using natural log base for simplicity
tmp=math.log(123)*45
print ("using log:   ", math.exp(tmp))

x=math.log(632382)*518061
y=math.log(519432)*525806
assert x>y

line=1
_max=0
max_line=None
for l in lst:
t=l.split(",")
x,y = int(t[0]),int(t[1])
r=math.log(x)*y
if r>_max:
max_line=line
_max=r
#print (line, x, y, r)
line=line+1

print (max_line, _max)

The result:

generic algo: 11110408185131956285910790587176451918559153212268021823629073199866111001242743283966127048043
using log:    1.111040818513185e+94
709 6919995.552420337

Spoiler: Wolfram Mathematica can operate on such big numbers without additional assistance. In fact, I first solved this problem using it.

BTW, we can verify Project Euler's author's statement about "as both numbers contain over three million digits" using decimal logarithm instead of natural:

#!/usr/bin/env python3
import math

tmp=math.log(632382, 10)*518061
print (tmp)

3005261.2406258043

This coincides with Wolfram Mathematica's output. This number has 3005261 decimal digits. All correct.

###### (the post first published at 20220728.)

Yes, I know about these lousy Disqus ads. Please use adblocker. I would consider to subscribe to 'pro' version of Disqus if the signal/noise ratio in comments would be good enough.


# Conditional Probability of rainfall

1. Apr 26, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

3. The attempt at a solution
a) P(Pickwick has no umbrella | it rains) = $\frac{\frac{1}{3}\frac{1}{3}}{\frac{1}{2}} = \frac{2}{9}$, which is the answer according to my answer key.

b) For part b we have:

There is a rain forecast which means he will bring the umbrella. The probability that it won't rain is 1/3.

There is a non-rain forecast which means he brings the umbrella with a probability of 1/3 and it will not rain with a prob of 2/3.

P(Pickwick has umbrella | no rain) = $\frac{1}{3} + \frac{1}{3}\frac{2}{3} = \frac{5}{9}$. But the answer is apparently 5/12. What have I done incorrectly here?

2. Apr 26, 2012

### MaxManus

I got the same as you at b) and I cant see why it is not correct.That's no guarantee for that you are correct though

3. Apr 26, 2012

### Ray Vickson

I think you are correct. Let's do it in a simple way, just by counting. Imagine 9,000 days. In 4,500 of those, it rains and in 4,500 it does not. Let's use the notation R = rain, Rc = no rain, F = Forecast rain, Fc = Forecast no rain, U = umbrella, Uc =no umbrella
Lay out the data in a table:
$$\begin{array}{lccc} & R & Rc & \text{tot}\\ F & N1 & N2 & (N1+N2)\\ Fc & M1 & M2 & (M1+M2) \\ \text{tot} & 4500 & 4500 & 9000 \end{array}$$
We are given N1/(N1+N2) = 2/3 and M1/(M1+M2) = 1/3, as well as N1+M1 = 4500 and N2+M2 = 4500. Solving these equations we get N1 = 3000, N2 = 1500, M1 = 1500, M2 = 3000. So, the table is:
$$\begin{array}{lccc} & R & Rc & \text{tot}\\ F & 3000 & 1500 & 4500 \\ Fc & 1500 & 3000 & 4500 \\ \text{tot} & 4500 & 4500 & 9000 \end{array}$$
From this it follows that P(F) = 4500/9000 = 1/2 and P(Fc) = 1/2.

In (b), in the 4500 Rc-days, U occurs in 1500 (F-days) + (3000/3) (Fc-days), for a total of 2500 days. Thus, P(U|Rc) = 2500/4500 = 5/9.

RGV


# Name the heuristic: exploiting the legitimacy of the questioner

As a child, I made frequent use of a particular 'trick' in order to make short work of many different problems. The general form is to be presented a question which wants a definite (numerical) answer, where the question statement contains a free variable. The 'trick' was to recognize that the answer (which the teacher has told us exists by asking for it) doesn't depend on the free variable, so we now know that we can substitute any convenient value we like for the free variable and find the answer.

This FiveTrangles problem is an excellent example. The height of the non-shaded triangles is unspecified, so we know straight away that it doesn't figure in the answer to the question. Since it doesn't matter, we can set it to zero and replace this question with the trivial question of finding the area of a triangle with $base = 10$ and $height = 5$.

Does this technique have a name? Do problems which permit it have a name? Should such questions be avoided?

• I'm not sure I understand. Do you mean leaving the answer as an expression with a variable in it?
– Karl
Feb 23 '15 at 19:55
• @Karl Another example: the question "$(x + 4)^2 - x^2 - 8x$ simplifies to a number. Find that number." can be solved with algebra, or it can be solved by plugging in x=0, since apparently the answer does not depend on x. Feb 23 '15 at 20:35
• That's a clearer example for me thanks. I get the triangle question too - presumably the $x$ cancels and isn't left as a variable in the expression.
– Karl
Feb 23 '15 at 21:13
• More of a technique than just a 'trick', see en.wikipedia.org/wiki/Limiting_case_%28mathematics%29 Feb 24 '15 at 3:35
• Somewhat related: for single-answer multiple choice problems, if two answers are equivalent, then they must both be wrong. Feb 24 '15 at 5:21

The heuristic described here is one manifestation of what Polya (1945) and others thereafter refer to as trying a special case. I do not know of a more specific term for the context that you have put forth, but this is often how one approaches a problem if you are initially unsure as to how to solve it in generality.

It is a good way to get an initial foot-hold on a problem, and especially useful in the context of some standardized tests. For example, consider the following:

Question: Player 1 flips a penny $$n>0$$ times, and Player 2 flips a penny $$n+1$$ times. What is the probability that Player 2 flips more heads than Player 1?

A. 1/2

B. 1/3

C. 1/4

D. 1/8

Since none of the choices depends on $$n$$, you can consider $$n=1$$ and see easily that the probability is $$1/2$$. This allows you to blaze through the question very quickly. (Indeed, this holds for all $$n \geq 1$$.)

Although testing well is an important life skill, your final question as to whether such problems are advisable is interesting. In my estimation, they can be good in the classroom as a way of scaffolding, as long as you press students (when appropriate) to explain why their reasoning holds for the general case.

As one small example, induction problems are done like this: You show the (hopefully easy) base case, and then complete the proof by setting up and using your inductive hypothesis.

A possible pitfall, though, is not only the absence of reasoning by students (as in the example question above) but also a possible reinforcement around misconceptions of proof by example. (I believe this relates to the recent query about counterexamples in MESE 7466.) For more about some of these misconceptions, check out work by Orit Zaslavsky (google scholar).

Lastly: The issue with guessing answers based on problem statements is pervasive in school mathematics. It ranges from only seeing examples of the form $$a + b = \square$$ and adopting a misunderstanding of the equal sign as an operator, to problems in which students just try to combine numbers in ways that might be "sensible" (without really engaging in sense-making). Cf. the classic:

Teacher: There are 125 sheep and 5 dogs in a flock. How old is the shepherd?

Student: (thinking: 125+5, 125-5, and 125*5 are too big; it must be 125/5) The shepherd is 25.

• I had considered trying a special case, but there's more going on than that. Something akin to trying the easiest case, and going with it, because all of the cases are the same (not that I understand why, but if it weren't true the question wouldn't have been put to me this way). That's a mouthful though. Thanks for the answer. Feb 24 '15 at 1:15
• @NiloCK If you know why "all the cases are the same," then you could say that you are reducing to a special case (rather than just trying one). But there is (obviously) quite a leap between solving for one special case and understanding that the special case answers (or helps answer) a seemingly more general question. Feb 24 '15 at 1:25
• The 'why' for me (as a kid) was that the problem setter put the question that way. That is, if all the cases weren't the same, then the question wouldn't (couldn't) have been asked. I knew at the time that this was 'unfair' information, but it's hard for a kid writing math contests to discard such information on the grounds that it was 'ill-gotten' or 'inadmissible'. Feb 24 '15 at 1:38
• @NiloCK Benjamin addresses that "why" in the answer. Essentially, the why amounts to a guess that may or may not be true. A teacher could easily throw in questions where the real answer is "not enough information" or "none of the above" just to discourage that heuristic. In my opinion, discouraging it forces students to think through why the missing information doesn't matter, leading to a deeper understanding of the logic and the problem and the solution. For me, that sort of process is when I learn the most. Feb 25 '15 at 4:28

(continued from previous post)

In this next problem, the horizontal leg of the large right triangle and the vertical leg of the smaller right triangle are also not fixed, but setting each to zero won't lead to a solution at all: http://fivetriangles.blogspot.com/2013/04/56-obtuse-triangle-area.html (Never mind that getting in the habit of simply setting values to zero will lead to a "zero" in calculus.)

Finally, the following proof uses the identical triangle area concept as in the original "concave quadrilateral" problem, but the "trick" doesn't work at all: http://fivetriangles.blogspot.com/2013/10/104-equal-area-proof.html

• You can combine both answers into one. There is no character limit (within reason) in answers. Feb 23 '15 at 21:54
• Stack exchange proscribed more than 2 links in one post if 0 ≤ rating < 10. Feb 23 '15 at 22:11
• You can't control the order in which answers appear, so calling another answer the "previous post" is not good. For instance, if this answer gets more points it will even appear above the other one. Call your other answer "my other post" rather than "previous" one.
– KCd
Feb 25 '15 at 11:28

In my SAT tutoring, I refer to such problems as underspecified. For instance, consider this problem:

If $\frac{a}{2} + \frac{b}{3} = 0$, then find $\frac{a}{b}$.

I teach the student that since two variables are presented but only one equation, this makes the problem underspecified. In other words, a and b are not uniquely determined. This gives the student license to make up a number for one out of $a$ and $b$, then use the given equation to solve for the other variable, and finally substitute both $a$ and $b$ into the expression $\frac{a}{b}$ to obtain the answer.

There are also many examples of underspecified SAT geometry problems. Often the problem will ask the student to find the value of $a+b$ or some other expression involving two variables. Example: The Official SAT Study Guide (second edition) by CollegeBoard, page 799, problem 13.

That some solutions are independent of a variable is not uncommon, but it does not necessarily trivialise a problem or negate its teaching value. The following problem presents similar possibilities, but we'd argue that solving it with such a "trick" is actually more circuitous than finding the answer directly (and may be more enlightening, too): http://fivetriangles.blogspot.com/2012/04/area-problem.html Here is a nice animation someone posted of the situation:

The shortcoming in such "tricks", we agree, is that they may lead to tunnel vision, but worse, their utilisation is at the user's risk.

(continued in next post)

• I assume that you're the owner of the Five Triangles blog? Great to have you on the site; I love yours. Many beautiful questions. Feb 24 '15 at 0:56
• Now that you have the necessary reputation, perhaps you could merge your two answers into one? Your other post (at this time) has 3 up-votes, whereas this one has only 1; so it would make better sense to merge into the higher voted one. (FWIW: I have not up-voted the other one, and will happily do so if I see them merged; in this way, deletion will not cause you any net-loss in reputation!) Mar 26 '15 at 5:50

I don't know the answer to the questions about the potential names for such things.

As to whether such questions should be avoided: I think some types of these questions can maybe help students to think about quantifiers, or to really think about the concepts involved in the question.

I like Chris Cunningham's example: if you tell the student that $(x+4)^2 - x^2 - 8x$ simplifies to a number, and ask them to find the number, they may just do the algebra of simplifying. But if you show them that you can get the answer by plugging in $0$, then you can teach them something about what it really means to say that "$(x+4)^2 - x^2 - 8x$ simplifies to a number". Now you can talk about what it means for a statement to hold "for all $x$". Things like this are useful to be familiar with. Many college students struggle with this when I ask them, for example, to show that a function is not linear!

It's also nice to have this kind of logic on hand for when you need to remember things. For example, say you were taking a linear algebra exam and you knew the determinant was either the sum or the product of the eigenvalues. Well, you could just consider the $2\times2$ identity matrix and figure it out pretty quickly! This is because you knew the statement didn't depend on the matrix or even the size of the matrix!

I (and this is just opinion) don't like questions like the triangle one that you linked to. Because there, the "trick" from the solver's perspective is just to realize that the question asker left out some words! I don't know how useful it is to have someone apply the logic "If they asked me this, it has an answer, and if they didn't say it depended on quantity $A$ then it must not depend on quantity $A$.'' It is not clear to me what the mathematical benefit is to that.

• RE: last paragraph. One benefit to the problem solver/student is to consider the importance of the givens; alternatively, a benefit to the problem poser/instructor is to consider more carefully how problems are phrased. Feb 25 '15 at 2:51

I would be surprised if the solution strategy or problem types have a name. While it can be useful to find the correct answer, what does it accomplish from a learning perspective? Take, for instance, Chris Cunningham's problem of "$(x+4)^2 - x^2 -8x$ simplifies to a number. Find that number." As an instructor, I would be looking to make sure the students accomplished the appropriate learning outcomes for the problem. For this problem, I would more than likely be looking for (1) can they multiply out $(x+4)^2$ and (2) can they simplify the addition of terms. Such a technique would not accomplish (1) or (2), so I would avoid writing problems that can (knowingly) make use of it nor would I ever teach such meta-strategies.

This reminds me of a multiple choice exam question on trig identities: "Which of the following is an equivalent identity to $\cot(x) + \tan(x)$?" with $\csc(x)\sec(x)$ being in the list of answers. Instead of using trig identities, many students plugged in a value like $x=10$ and compared. While it works for this case, it dodges the actual learning and testing of the concept.

Discovering the irrelevance of $x$ in the expression could be achieved by getting the students (perhaps in different groups) to substitute different values in for $x$ and compare results. Was this what they expected? Which number would they choose to substitute knowing the answer will always be the same? Having aroused their curiosity the expanding brackets and simplifying can be motivated and discussed. This could be a rich activity if the emphasis is on discovery and not just sub in $x=0$


# Points A & B Out At Sea

#### xyz_1965

##### Member
From points A to B at sea, the angles of elevation to the top of the mountain T are 37 degrees and 41 degrees, respectively. The distance between points A to B is 80 meters.

1. Find height of the mountain.

2. Find the distance from point A to the bottom of the mountain.

I think making two triangles makes sense here.

Triangle 1

tan(41°) = h/x

Let h = height of mountain.

Let x = the distance from point B out at sea to the bottom of the mountain.

Solving for x, I get h/tan(41°).

Triangle 2

tan(37°) = h/[80 + (h/tan(41°)]

I need to solve triangle 2 for h, the height of the mountain. After finding h, I can then find the distance between point B out at sea and the bottom of the mountain.

Lastly, the distance from point A out at sea to the bottom of the mountain is found by adding A + B.

1. Is any of this right?

2. Is there an easy way to solve
tan(37°) = h/[80 + (h/tan(41°)] for h?

#### skeeter

##### Well-known member
MHB Math Helper
assuming A, B, and the mountain top are located in the same vertical plane ...

$h = x_B \tan(41)$

$h = x_A \tan(37)$

$x_A - x_B = 80 \implies x_A = x_B +80$

$x_B \tan(41) = (x_B+80)\tan(37)$

$x_B = \dfrac{80\tan(37)}{\tan(41)-\tan(37)}$

from here, calculate $x_B$, add 80 to get $x_A$, and use either value in one of the original equations to find $h$.

#### xyz_1965

##### Member
assuming A, B, and the mountain top are located in the same vertical plane ...

$h = x_B \tan(41)$

$h = x_A \tan(37)$

$x_A - x_B = 80 \implies x_A = x_B +80$

$x_B \tan(41) = (x_B+80)\tan(37)$

$x_B = \dfrac{80\tan(37)}{\tan(41)-\tan(37)}$

from here, calculate $x_B$, add 80 to get $x_A$, and use either value in one of the original equations to find $h$.
Explain $x_B$ and $x_A$.

#### skeeter

##### Well-known member
MHB Math Helper
$x_B$ is the horizontal distance from point B to the point directly below the mountain top.

$x_A$ is the horizontal distance from point A to the point directly below the mountain top.

#### xyz_1965

##### Member
$x_B$ is the horizontal distance from point B to the point directly below the mountain top.

$x_A$ is the horizontal distance from point A to the point directly below the mountain top.
Thank you for clearing this up for me.

#### Country Boy

##### Well-known member
MHB Math Helper
From points A to B at sea, the angles of elevation to the top of the mountain T are 37 degrees and 41 degrees, respectively. The distance between points A to B is 80 meters.

1. Find height of the mountain.

2. Find the distance from point A to the bottom of the mountain.

I think making two triangles makes sense here.

Triangle 1

tan(41°) = h/x

Let h = height of mountain.

Let x = the distance from point B out at sea to the bottom of the mountain.
It would make more sense to say these before you say "tan(41)= h/x!

Solving for x, I get h/tan(41°).
Specifically, x= h/tan(41).

Triangle 2

tan(37°) = h/[80 + (h/tan(41°)]
I think I would have first said "Let y be the distance from B to the base of the mountain" to get tan(37)= h/y and then use y- x= 80 so that y= 80+ x but that becomes the same thing.

I need to solve triangle 2 for h, the height of the mountain. After finding h, I can then find the distance between point B out at sea and the bottom of the mountain.
Yes.

Lastly, the distance from point A out at sea to the bottom of the mountain is found by adding A + B.
No! You were doing so well and then you forgot how you had defined "A" and "B'! "A" and "B" are points, not distances so you can't add them. "x" was the distance from point A to the base of the mountain.
1. Is any of this right?

2. Is there an easy way to solve
tan(37°) = h/[80 + (h/tan(41°)] for h?
A= h/(B+h/C)

Get rid of the fraction by multiplying both sides by B+ h/C:
AB+ (A/C)h= h

AB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C

h= ABC/(C- A)

Of course A= tan(37°)= 0.7536, approximately, B= 80, and C= tan(41°)= 0.8693 approximately.

#### xyz_1965

##### Member

It would make more sense to say these before you say "tan(41)= h/x!

Specifically, x= h/tan(41).

I think I would have first said "Let y be the distance from B to the base of the mountain" to get tan(37)= h/y and then use y- x= 80 so that y= 80+ x but that becomes the same thing.

Yes.

No! You were doing so well and then you forgot how you had defined "A" and "B'! "A" and "B" are points, not distances so you can't add them. "x" was the distance from point A to the base of the mountain.

A= h/(B+h/C)

Get rid of the fraction by multiplying both sides by B+ h/C:
AB+ (A/C)h= h

AB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C

h= ABC/(C- A)

Of course A= tan(37°)= 0.7536, approximately, B= 80, and C= tan(41°)= 0.8693 approximately.
I thank you for the break down. It's ok to make mistakes here. Learning is not possible without errors.

#### Country Boy

##### Well-known member
MHB Math Helper
That is true. And you learn by having those errors pointed out!

#### xyz_1965

##### Member
That is true. And you learn by having those errors pointed out!
I agree. This is why I don't mind being corrected as long as it is done respectfully.


# Commuting of Hom and Tensor Product functors?

Let $V_i,W_i$ be finite dimensional vector spaces, for $i=1,2$. Assume we have homomorphisms $\phi_i:V_i\rightarrow W_i$.

Then, there is an induced map $\widehat{\phi_1 \times \phi_2} \in Hom(V_1 \otimes V_2,W_1 \otimes W_2)$ such that $\widehat{\phi_1 \times \phi_2}(v_1 \otimes v_2) = \phi_1(v_1)\otimes\phi_2(v_2)$.

We can look at this construction as a map $\psi: Hom(V_1,W_1)\times Hom(V_2,W_2)\rightarrow Hom(V_1 \otimes V_2,W_1 \otimes W_2)$, where

$\psi:(\phi_1,\phi_2)\rightarrow \widehat{\phi_1 \times \phi_2}$.

It is easy to see that $\psi$ is a bilinear map. It follows (from the universal property) that we have an induced homomorphism: $\hat \psi:Hom(V_1,W_1)\otimes Hom(V_2,W_2)\rightarrow Hom(V_1 \otimes V_2,W_1 \otimes W_2)$ such that: $\hat\psi(\phi_1\otimes\phi_2)=\widehat{\phi_1 \times \phi_2}$.

My question: is $\hat\psi$ necessarily an isomorphism? Note that there is equality of dimensions (since both Hom,Tensor multiply them), hence clearly some isomorphism must exists between $Hom(V_1,W_1)\otimes Hom(V_2,W_2)\cong Hom(V_1 \otimes V_2,W_1 \otimes W_2)$.

If $\hat\psi$ is not isomorphism, is there some other "canonical" isomorphism?

I will just note that in the case of finite dimensional vector spaces, there is equality of dimensions, hence clearly some isomorphism must exists.

• Take $W_1 = W_2 = A$, you can even assume it's a field then you're asking if $(V_1 \otimes V_2)^* \cong V_1^* \otimes V_2^*$. That's false if $V_1$ and $V_2$ are infinite dimensional. I can't even imagine how complicated things would become if $A$ is not a field, or not even a PID or anything... For $A$ a field my instinct tells me your $\hat\psi$ is always injective, but I'm not 100% sure. – Najib Idrissi Apr 28 '15 at 13:27
• It need not be well defined. I have an example. – Matt Samuel Apr 28 '15 at 13:43
• Najib: I agree. I have edited the question to focus upon the case of finite dimensional vector spaces. That was my original interest, I just thought maybe I missed some general theme which is not related to the specific fact that my objects were finite_dim vector spaces and not general modules. I was completely wrong there... – Asaf Shachar Apr 28 '15 at 13:43
• $\widehat{\phi_1 \times \phi_2}$ is well defined via the universal property of tensor products. – Asaf Shachar Apr 28 '15 at 13:48
• @egreg The expression $\phi_1(v_1) \otimes \phi_2(v_2)$ is linear in each variable in $v_1$ and $v_2$, if I'm not mistaken, so it defines a linear map $V_1 \otimes V_2 \to W_1 \otimes W_2$. And then this is linear in each of $\phi_1$ and $\phi_2$, hence the map $\hat\psi$. Am I missing something? It's just the bifunctoriality of $\otimes : \mathsf{Vect} \times \mathsf{Vect} \to \mathsf{Vect}$. – Najib Idrissi Apr 28 '15 at 14:04

The general situation is the following: $A$ is a commutative ring and $V_1, V_2, W_1, W_2$ are $A$-modules

If one of the ordered pairs $(V_1,V_2)$, $\,(V_1,W_1\,$ or $\,(V_2,W_2)$ consists of finitely generated projective $A$-modules, the canonical map is an isomorphism (Bourbaki, Algebra, Ch. 2 ‘Linear Algebra’, §4, n°4, prop. 4).

Over a field, all modules are projective since they're free. So the answer is ‘yes’.

In the present case, here is a sketch of the proof:

Let $K$ be the base field. As $V_1\simeq K^m$ for some $m>0$ and similarly $V_2\simeq K^n$, and the $\operatorname{Hom}$ and $\,\otimes\,$ functors comute with direct sums, we may as well suppose $V_1=V_2=K$.So we have to prove: $$\widehat\Psi\colon\operatorname{Hom}(K,W_1)\otimes\operatorname{Hom}(K,W_2)\to\operatorname{Hom}(K\otimes K,W_1\otimes W_2)$$ is an isomorphism.

This results basically that for any vector space $V$ we have an isomorphism \begin{align*} \operatorname{Hom}_K(K,V)&\simeq V\\ \phi&\mapsto\phi(1) \end{align*} In detail, just consider the following commutative diagram:

• Thanks. Is there a quick way to see that $\hat\psi$ is an isomorphism in the particular case where all the modules are finite dimensional vector spaces? (I think this question should have a simple answer in terms of "undergraduate" linear algebra without the need to resorting to more "heavy" tools\theorems from abstract commutative algebra. – Asaf Shachar Apr 28 '15 at 13:57
• @asaf shachar: I gave some details of a possible proof. See my completed answer. – Bernard Apr 28 '15 at 18:44
• Excuse me... I want to ask that is this property holds for $> 2$ vector spaces? I think it works because the diagram listed above seems work for $> 2$ vector spaces. More generally, in module case, is this property holds for $> 2$ modules (with suitable requirements...)?? Thank you! – Peter Hu Sep 2 '15 at 10:54

To show things like this in the case of finite dimensional vector spaces, you can always fallback on picking bases:

Let $v_i^1, \ldots, v_i^{m_i}$ be a basis of $V_i$ ($i=1,2$), and $w_i^1, \ldots, w_i^{n_i}$ be a basis of $W_i$ ($i=1,2$). Let $f^{jk}_i : V_i \to W_i$ be given on the basis by $f_i^{jk}(v_i^l) = \delta_{jl} w_i^k$. Then $\{v_1^p \otimes v_2^q\}$ is a basis of $V_1 \otimes V_2$, and $\{f_i^{jk}\}$ is a basis of $Hom(V_i,W_i)$. Your map sends $(f_1^{jk},f_2^{rs})$ to the linear transformation given on the basis by $\widehat{f_1^{jk} \times f_2^{rs}}(v_1^p \otimes v_2^q) = \delta_{jp} \delta_{rq} w_1^k \otimes w_2^s$.

Those maps clearly form a basis of $Hom(V_1 \otimes V_2, W_1 \otimes W_2)$ for the same reason the $f_i^{jk}$ are a basis of $Hom(V_i,W_i)$.


# Measuring fractal dimension of natural objects from digital images

This is a useful topic. A college physics lab, medical diagnostics, urban growth, etc. - there is a lot of applications. On this site by Paul Bourke about Google Earth fractals we can get a high resolution images (in this post they are low res - import from source for experiments). For example, around Lake Nasser in Egypt:

img = Import["http://paulbourke.net/fractals/googleearth/egypt2.jpg"]

The simplest method I know is Box Counting Method which has a lot of shortcomings. We start from extracting the boundary - which is the fractal object:

{Binarize[img], iEdge = EdgeDetect[Binarize[img]]}

Now we could partition image into boxes and see how many boxes have at least 1 white pixel. This is a very rudimentary implementation:

MinS = Floor[Min[ImageDimensions[iEdge]]/2];
data = ParallelTable[{1/size, Total[Sign /@ (Total[#, 2] & /@ (ImageData /@
Flatten[ImagePartition[iEdge, size]]))]}, {size, 10, MinS/2, 10}];

From this the slope is 1.69415 which is a fractal dimension that makes sense

line = Fit[Log[data], {1, x}, x]

13.0276 + 1.69415 x

Plot[line, {x, -6, -2}, Epilog -> Point[Log[FDL]],
PlotStyle -> Red, Frame -> True, Axes -> False]

Benchmark: if I run this on high res of Koch snowflake i get something like ~ 1.3 with more exact number being 4/log 3 ≈ 1.26186.

Question: can we improve or go beyond the above box counting method?

All approaches are acceptable if they find fractal dimension from any image of natural fractal.

-
You have a lot of programs in mathematica to measure fractal dimensions and multi fractal spectrum of an image in the book: Fractal Geography, Andre Dauphine, Wiley, 2012 See the book on the Wolfram Mathematica | Books or Amazon ![enter image description here](wolfram.com/books/profile.cgi?id=8108) –  Dauphine Oct 16 '12 at 9:32
Vitaliy, with all due respect, the ending bullets make the question (actually there isn't a question mark anywhere, right?) rather wide ranging. Perhaps you could focus it somewhat and make it sound more like a real question? –  Sjoerd C. de Vries Oct 16 '12 at 12:59
@SjoerdC.deVries Updated, it is a single question now. –  Vitaliy Kaurov Oct 16 '12 at 15:06
could you perhaps describe or link some examples of use of fractal dimension? –  magma Oct 17 '12 at 10:27
@magma: Wikipedia lists some examples here: en.wikipedia.org/wiki/… –  shrx Jun 14 '13 at 20:22

You can still use box count, but doing it smarter :)

Counting boxes with at least 1 white pixel from ImagePartition can be done more efficiently using Integral Image, a technique used by Viola-Jones (2004) in their now popular face recognition framework. For a mathematical motivation (and proof), Viola and Jones point to this source.

What Integral Image allows you to do is to compute efficiently the total mass of any rectangle in an image. So, you can define the following:

IntegralImage[d_] := Map[Accumulate, d, {0, 1}];
data = ImageData[ColorConvert[img, "Grayscale"]]; (* img: your snowflake image *)
ii = IntegralImage[data];

Then, the mass (white content) of a region, is

(* PixelCount: total mass in region delimited by two corner points,
given ii, the IntegralImage *)
PixelCount[ii_, {p0x_, p0y_}, {p1x_, p1y_}] :=
ii[[p1x, p1y]] + ii[[p0x, p0y]] - ii[[p1x, p0y]] - ii[[p0x, p1y]];

So, instead of partitioning the image using ImagePartition, you can get a list of all the boxes of a given size by

PartitionBoxes[{rows_, cols_}, size_] :=
Transpose /@ Tuples[{Partition[Range[1, rows, size], 2, 1],
Partition[Range[1, cols, size], 2, 1]}];

If you apply PixelCount to above, as in your algorithm, you should have the same data but calculated faster.

PixelCountsAtSize[{rows_, cols_}, ii_, size_] :=
((PixelCount [ii, #1, #2] &) @@ # &) /@ PartitionBoxes[{rows, cols}, size];

Following your approach here, we should then do

fractalDimensionData =
Table[{1/size, Total[Sign /@ PixelCountsAtSize[Dimensions[ii], ii, size]]},
{size, 3, Floor[Min[Dimensions[ii]]/10]}];
line = Fit[Log[fractalDimensionData], {1, x}, x]

Out:= 10.4414 + 1.27104 x

which is very close to the actual fractal dimension of the snowflake (which I used as input).

Two things to note. Because this is faster, I dared to generate the table at box size 3. Also, unlike ImagePartition, my partition boxes are all of the same size and therefore, it excludes uneven boxes at the edges. So, instead of doing minSize/2 as you did, I put minSize/10 - excluding bigger and misleading values for big boxes.

Hope this helps.

Update

Just ran the algorithm starting with 2 and got this 10.4371 + 1.27008 x. And starting with 1 is 10.4332 + 1.26919 x, much better. Of course, it takes longer but still under or around 1 min for your snowflake image.

Update 2

And finally, for your image from Google Earth (eqypt2.jpg) the output is (starting at 1-pixel boxes)

12.1578 + 1.47597 x

It ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs.

-
There isn't a "snowflake" image. All three images come from the same object (and I guess the fractal dimension should be the same for all of them) –  belisarius Dec 23 '13 at 21:03
@belisarius, the PO published a Koch snowflake image as Benchmark (see last part of post) and that's the one I used. I called it 'snowflake' for short. On the other hand, this is still a numerical procedure, so the numbers will be different depending on approximation. The Update2 refers to the eqypt2.jpg image, also made available by the PO, as he asked for an improvement suitable for real-life images. –  caya Dec 23 '13 at 22:40
sorry, I missed that link. Thanks –  belisarius Dec 24 '13 at 0:05
+1 This is very neat @caya, thank you! I will wait "a bit" hoping that someone can implement things like wavelet multi-fractals or similar. –  Vitaliy Kaurov Feb 14 at 20:02
@VitaliyKaurov, glad you liked it. Note that Viola & Jones rightly pointed out a link between integral image and Haar wavelet basis in their paper; although this wasn't explored further. It is unclear to me the link between the paper you mentioned and fractal dimension, but certainly worth reading as I am interested in these kind of problems. Cheers. –  caya Feb 16 at 12:40


# How to be sure about any parameterization?

This is the problem:

Find intersection of this two surfaces and then calculate its length. $$x^2+4y^2=4$$ and
$$y\sqrt3 +z=1$$ Method 1 :

Common way is to take:
$$x=2\cos t ,\: y=\sin t ,\: z=1-\sqrt3 \sin t$$
So
$$r(t) =(2\cos t,\:\sin t,\:1-\sqrt3 \sin t)$$ and
$$r'(t) = (-2\sin t,\:\cos t,\:-\sqrt3\cos t) ,\: |r'(t)|=2$$

Finally the length is:
$$L = \int_{0}^{2\pi} 2\:dt=4\pi$$

But in another parameterization I find something strange

Method 2:

$$x=2\sqrt{1-t^2} ,\: y=t ,\: z=1-t\sqrt3$$ So
$$r(t) = (2\sqrt{1-t^2} ,\: t ,\: 1-t\sqrt3)$$ and
$$r'(t) = \left(\frac {-2t}{\sqrt{1-t^2}},1,-\sqrt3\right) ,\: |r'(t)|=\frac {2}{\sqrt{1-t^2}}$$ (and that's different from method 1 $$|r'(t)|$$)

Finally the length
$$L=\int_{0}^{2\pi} \frac {2}{\sqrt{1-t^2}}\:dt\\ =2\int_{0}^{\arcsin{2\pi}}\frac {1}{\cos{\theta}}\cos{\theta}\:d\theta\\ =2\int_{0}^{\arcsin{2\pi}}1\:d\theta$$

and because of $$\arcsin(2\pi)$$ its undefined it cant be solved.

Now my first question is that:
If we know from past that we can create many parameterizations for a single curve, so why the result of they length is different?

and the second one:

How to be sure that a parameterization is the correct one?

As a complement to @Axel's answer

Let $$t_j$$ denote the $$j$$th method's $$t$$ so $$t_2=\sin t_1$$ and $$\tfrac{dt_2}{\sqrt{1-t_2^2}}=d\arcsin t_2=\pm dt_1$$. If for example $$t_1$$ increases from $$0$$ to $$2\pi$$, $$t_2$$ is monotonic in three parts, cut at $$t=\pi/2$$ and $$t=3\pi/2$$. The monotonic pieces have alternating $$\pm$$, in this case with the pattern $$+-+$$, so the length is\begin{align}\int_0^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}+\int_{-1}^0\frac{2dt_2}{\sqrt{1-t_2^2}}&\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\\&\color{red}{=4\pi}\\&\color{blue}{=\left(\int_0^{\pi/2}+\int_{\pi/2}^{3\pi/2}+\int_{3\pi/2}^{2\pi}\right)2dt_1}\\&=\color{blue}{4\pi},\end{align}where the $$t_2$$-space calculation can continue from the black expression with either the red or the blue treatment. If $$t_1$$ varies over another length-$$2\pi$$ interval the monotinicity division details are different, but any choice requires at least one such cut, succumbs to the above technique, and gets the same answer. Let's check another example from $$t_1=-\pi/2$$ to $$t_1=3\pi/2$$, giving two pieces (meeting at $$t=\pi/2$$) where the $$\pm$$ are $$+-$$:$$\int_{-1}^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\color{blue}{=\left(\int_{-\pi/2}^{\pi/2}+\int_{\pi/2}^{3\pi/2}\right)2dt_1}.$$

• Thank you both.I understand it now. – program_craft Jan 22 at 4:58

The problem lies in the way you parametrize the curve in the second method.

The reason why you integrate from $$0$$ to $$2\pi$$ in the first method is because you have implicitly define you parametrization to go from $$0$$ to $$2\pi$$ because you are looking for a mapping $$[0;2\pi) \longrightarrow \textrm{your intersection curve}$$ such that that mapping is biyective with the entire curve.

If you take a look to what you limits of integrations are in the second method you see they are the same from the first one. That again implicitly define you parametrization to go from $$0$$ to $$2\pi$$ but this time (because of the equation $$y=t$$) you are saying that $$y$$ runs in the range $$[0;2\pi)$$ when in reallity in ranges $$[-1;1)$$ so you actually end up measuring the length of a totally different curve (one with that parametrization but with $$y \in [0;2\pi)$$ insted of $$[-1;1)$$).

Now given that $$t$$ should be in $$[-1;1)$$ the correct way to measure the lenght in the second method would be: $$\int\limits_{-1}^1 \frac{2}{\sqrt{1-t²}} \ dt + \int\limits_{1}^{-1} \frac{2}{\sqrt{1-t²}} \ dt$$

That is beacuse you can't map the entire curve with just the $$[-1;1)$$ range (in fact you only map one half of it). Thus the second integral.

So finally you have:

$$2\int\limits_{-1}^1 \frac{2}{\sqrt{1-t²}} \ dt = 4\pi$$

• Can you explain more why t is in range[-1,1] – program_craft Jan 21 at 17:36