text:
## How do you get motivated / inspired every day?

I often need blood pumping or motivation to make me think at the beginning of the day to start. I listen to some podcast / music or watch videos like this:

How do you ensure you maintain a growth mindset during the day?

Do you have any special tricks or tips that you are willing to share?

## dnd 5e – How can I create a character inspired by WoW's Illidan Stormrage?

I want to make a D&D 5e character inspired by Illidan Stormrage, a character from World of Warcraft.

I love Illidan's appearance, but it seems impossible to copy. His skills are amazing. I have a little idea about how to build it, but my construction never fits well, so I need help to build the character.

For those who do not know:

1. He is a double-handled night elf, with demon horns and wings.

I have 2 ideas on how to recreate this. One way is to use a sorcerer who can launch Alter Self at will to create horns. The other method is to make a Draconic Bloodline sorcerer, who gets wings (draconic but similar). Both methods are mutually exclusive.

2. It has a high attack speed, deals a lot of damage and does not use armor.

My idea for attacks / damages is to use a class with additional attack or barbaric reprisals. To deal with the lack of armor, there are things like wizard's armor, Draconic Resistance, or Disarmed Defense.

3. Abilities:

• His best ability is metamorphosis; He can become a demon.
For this, my idea is to launch true polymorph on myself.

• Fire Aura (DH in WoW, as well as certain talents in HotS and a main basic skill in WC3, a DoT that treats aura for fire damage)
My idea: fire shield

• Evasion
My idea: rogue and monk classes have evasion, but I think that just having a high AC is enough, like 18-20; wizard's armor It seems enough to achieve this.

If anyone knows how to create a character like Illidan, especially how it looks, I would like to play with a drow with wings and horns. Please help me build this character.

## A Quori is not destroyed when its host is

When we look at the statistics blocks for Hashalaq, Kalaraq and Tsucora Quori, which are quori that currently do not have a body, we see an ability called "Possession", which does exactly that, and includes the text

Possession lasts until the body falls to 0 hit points, the quori ends as a bonus action, or the quori is expelled for an effect such as dispelling the spell from evil and good. When the possession ends, the quori reappears in an unoccupied space less than 5 feet from the body.

(emphasis mine)

As this text states that when a possessed body falls to 0 hp, the quori reappears outside it, we can infer that this is also true for Inspired. Also, if you assume that they act exactly like these, then the Quori would not even necessarily be sent back to the darkness of dreams, simply expelled from their body.

## nt. number theory: a problem inspired by the definition of tau numbers and a divisibility relationship related to the powers of two

It is (I suppose this easy fact is well known) obvious that an integer $$n> 1$$ it's a power of two $$n = 2α$$, where $$alpha geq 1$$ it's whole, if only yes $$n$$ satisfies the divisibility ratio $$( text {rad} (n) cdot varphi (n)) mid n tag {1}$$
where $$text {rad} (n) = prod_ {p mid n} p$$ denotes the product of the division of distinct prime numbers $$n$$ Y $$varphi (n)$$ It is Euler's totient function.

(A draft of the test is that for integers $$n> 1$$ one can write $$text {rad} (n) varphi (n) = n prod_ {p mid n} (p-1)$$ so our whole $$n$$ satisfied $$(1)$$ have $$omega (n) = 1$$ different prime factor in its factorization, this arithmetic function $$omega (x)$$ count the number of different prime factors, and one has $$p-1 = 1$$ implies that our whole has the form $$n = 2α$$, where $$alpha geq 1$$ it's whole)

Observations A) One knows that the following series is convergent $$sum_ {n = 1} ^ infty frac {1} {2 ^ n}$$, therefore, the sum of the reciprocals of the solutions of $$(1)$$ it's convergent (really $$n = 1$$ It will also be a solution and therefore the previous series has value $$2$$ when we consider that $$1$$ it is also a power of two $$1 = 2 0$$ and solution of $$(1)$$) B) We note that Euler's totient function is a counting function, because it counts positive integers $$k$$ until $$n$$ that satisfy $$(n, k) = 1$$.

This week I met the beautiful article (1) in which problems related to the one known as tau numbers (See the introductory section of (1), this magazine is an open-access electronic magazine, in the first paragraph the author remembers the definition, or see the linked Wikipedia). From the definition of tau numbers, (see if you want Wikipedia Refeasible Number) I was inspired to define the following variation of tau numbers, or if you prefer a variation of the divisibility ratio $$(1)$$ as $$tau (n) = sigma_0 (n) = sum_ {1 leq d mid n} d$$ It is also a counting function.

Definition. For integers $$n> 1$$, I say that $$n$$ is a rad-refactorable number yes and only if it satisfies $$( text {rad} (n) cdot tau (n)) mid n, tag {2}$$
where $$tau (n)$$ denotes the function of number of divisors.

The first terms of this sequence are the following where I have considered adding like this $$n = 1$$ as a solution $$(2)$$

$$a_1 = 1, a_2 = 8, a_3 = 9, a_4 = 72, a_5 = 128, a_6 = 384, a_7 = 625, ldots$$

Therefore, we have redefined solutions as those positive integers $$n geq 1$$ satisfactory $$(2)$$ in order to include $$a_1 = 1$$ as a legitimate retractable radial number.

Question. Can you prove or disprove that? $$sum _ { substack { text {positive integers} a_m \ text {which are rad-refactorible}}} frac {1} {a_m}$$
is it convergent? Thank you.

I have calculated more terms in the sequence, I think it is not easy to link the terms $$frac {1} {a_m}$$ using a geometric series, and I don't know if there are infinite refactorizable radial numbers.

## References

(1) Joshua Zelinsky, Tau numbers: a partial proof of a conjecture and other results, Journal of Integer Sequences, vol. 5 (2002), Article 02.2.8.

## Universal property: why should Microtek Greenburg, a residential paradise inspired by nature, invest?

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On top of that, it offers yoga and aerobics, swimming pool, clubhouse, basketball court, meditation court, health club and gym. The collection of rainwater is undoubtedly one of the most outstanding features of the Microtek Greenburg Ready to move project. An experienced real estate agent like Orion could provide the required assistance related to providing all the information related to this project.

## Find solutions and get a first statement for these diofanthine equations inspired by certain figurative numbers

Yesterday I was thinking about speculative relationships between certain figurative numbers, please see if you need the tables and references of the article
MathWorld Encyclopedia Figurative number.

I was wondering about speculative relationships of the type: first I take prism base the hexadecimal number and $$z$$ as height, we represent the hexadecimal number of this base with the integer $$x$$. Secondly we do the same, we multiply $$z$$ by a centered pentagonal number using $$and$$ in his representation to build a second prism. We finally tested the comparison

$$z cdot left ( text {hexadecimal number} right) + z cdot left ( text {centered pentagonal number} right) = text {Octahedral number of Haüy}$$
where the variable for the representation of our octahedral number of Haüy in RHS is precisely the height $$z$$. Such an equation for positive integers $$x, y, z geq 1$$ It is equivalent to
$$18x ^ 2z + 18xz + 15y ^ 2z + 15yz-8z ^ 3 + 12z ^ 2-4z + 6 = 0 tag {1}.$$

Another equation that I have considered was

$$z cdot left ( text {hexadecimal number} right) -z cdot left ( text {centered pentagonal number} right) = text {Octahedral number of Haüy}$$
that is equivalent to

$$18 x ^ 2 z + 18 x z – 15 y ^ 2 z – 15 and z – 8 z ^ 3 + 12 z ^ 2 – 16 z + 6 = 0 tag {2}.$$

Question. Can you find solutions for $$(1)$$ or $$(2)$$? Please see the computational facts below. What can be a first step / professional statement that can be deduced in the study of these equations? $$(1)$$ or $$(2)$$? I ask about how to obtain a first professional proposition (which is an interesting first statement) for some of these equations, when the study of these equations is evoked. Thank you.

Fact. If we divide by $$2$$ Genuine equations, if I'm not mistaken, it's easy to show that we just need to verify the equations for odd integers $$z geq 1$$ and integers $$and geq 1$$ such the corresponding centered pentagonal number is an even integer, since the other cases are absurd. I cannot deduce a similar statement for the congregation $$equiv text {mod} 3$$ (when we divide the equation by $$3$$) For each equation $$(1)$$ Y $$(2)$$ The Wolfram Alpha online calculator gives me partial leads and the polynomial discriminant, but I don't know if it is useful to solve the problem in my question.

Computational facts

1) I can't find solutions $$(x, y, z)$$ of the equation $$(1)$$ for integers $$x, y, z geq 1$$ when our integers travel the segments $$1 leq x leq 300$$ Y $$1 leq and leq 300$$ Y $$1 leq z leq 300$$.

2) Similarly for the second equation and the same integer segments, our difference of figurative numbers $$(2)$$, I can only find the solutions $$(x, y, z) = (1,1,1)$$ Y $$(x, y, z) = (31,34,1)$$

## dnd 5e – After the game tests: Planescape Inspired Lady of Pain Patron, is it balanced?

A couple of weeks ago I tried a subclass inspired by Planescape for the sorcerer.
Since then, I have been able to incorporate the comments offered by the community, as well as play the tests with the player who asked me to make this character possible for her. For a bit of context, our campaign has become a kind of adventure for planar rangers. We have an evocation assistant, a forester from Horizon Walker and now the sorcerer from LoP. We started the campaign at 5th level and since then we have progressed to 9th. The initial goal of the subclass was to provide another front-line option other than Hexblade that is more defensive when the Hexblade is offensive.

For this we have simplified the central function, The shadow of serenity to offer the same CA bonus while exchanging the ability to make creatures vulnerable to damage by a few Hellish Rebuke castings.

I still worry about the actual wording of certain features to maintain consistency with the published materials, as well as the overall balance of the Subclass at the highest levels.

You made a silent pact with the Lady of Pain and her cage. As a shadow of His serenity, you have been accused of maintaining the balance between the planes and preserving the privacy of the Lady. As part of this agreement, you have the following restrictions and benefits:

• You should never disclose the nature of your covenant, even under penalty of death.

• You must protect Dabu and the city of Sigil, and never cause harm.

• Alcohol and other intoxicants do not have a diminished effect on the pain you feel. On the contrary, their tolerance to pain has increased twice that of most mortal creatures.

• You can understand the strange visual language of Dabu, but you can not communicate using it.

Expanded spell list

1st: Shield, inflict Wounds.

2nd: Cloud of daggers, silence.

4th: stone skin, death room

5th: binding planar, Geas

Features

Starting at the first level, you can like an action project a sinister shadow that hides your enemy form within a 10-foot radius centered on you for 8 hours, which gives you several benefits:

• You gain a bonus to your CA equal to half of your Charisma modifier,

• In reaction to being hit by a creature within your aura, you can
launches a version of the Hellish Rebuke spell. The level of the spell is
equivalent to the amount of Serenity & # 39; s Shadow charges that
spend. This spell can not exceed the 5th level (6d10) and you must have
The ability to cast the spell to the corresponding corresponding.
level. The damage by this reaction is considered a magical touch.
and use your spell to save DC.

• You can spend 1 charge of Serenity & # 39; s Shadow to take the Hide or
Undock the action as a bonus action.

Dexterity (stealth controls).

The Shadow of Serenity has a number of charges equal to your wizard level +1. Once the Duration of the Shadow ends, you end it soon as an action, or if you use all the Serenity Shadow charges, you must finish a long rest to use the Shadow again.

At level 10, the aura of the shadow grants CA bonus to the allies within the range equal to half of your Charisma modifier rounded down (minimum of +1). The radius of the Serenity Shadow expands to 30 feet.

At level 14, the shadow aura gives you twice the Dexterity skill bonus (stealth controls) and allows allies on your radio to take the Unlink or Hide action as a bonus action.

Breaking off

On the sixth level, your body has adapted to recover from frightful injuries. You have an advantage in saving death shots. Also, if you have hit points equal to or less than half of your maximum hit points, you can use an additional action to throw a Hit Die to recover hit points, the number of hit points retrieved is equal to the Hit die roll. given more your Charism. modifier

Planar goalkeeper

At level 14, you can cast the plane change spell once without using a spell slot or material components as long as you address Sigil. You can use this function to travel to another place if you own the cost of the material component and the spell slot does it normally. In reaction to the plane change spell cast by another creature, you can counteract that spell immediately by using a level 7 spell slot. Once you use this function, you must finish a long rest before using it again.

Lady of Pain Patron Eldritch Invocations

The radius of the Shadow feature of your Serenity expands 30 feet. Also, while the shadow is active, you can cast the Dark spell by spending 3 charges from the Shadow of Serenity.

Berk & # 39; s Bain

Prerequisites: Level 5, Lady of Pain Patron

You are becoming more cruel to your enemies, since you now have a critical hit with an attack roll of 19 or 20, while the Serenity Shadow feature is active, in addition, if you are not, you acquire the ability of Intimidation .

Tormentor

Prerequisites: Level 10, Patron of the Lady of Pain, Covenant of the Leaf

You have learned to foster a fear of abnormal and persistent pain in your enemies.
Once per turn, you can inflict additional psychic damage of 5d6 on a creature that you hit with an attack if you have an advantage on the attack roll. The attack must use your covenant weapon.

You do not need an advantage in the attack roll if another enemy of the target is within 5 feet of it, that enemy is not incapacitated and you have no disadvantage in the attack roll.

I thought you were dead for sure!

Prerequisites: Level 12, Lady of Sorrow Patron

Injuries are a daily occurrence in your line of work, allowing you to obtain more information about anatomy and medicine. You gain skill in medicine ability, in addition, when you use the Separation function, you can now add twice the Charisma Modifier and you can reattach a cut limb by holding a cut body part (other than your head) in the place where it fell. Once you use this function to reposition a limb, you can not use it again until you have finished a long rest.

Prerequisites: Level 18, Lady of Pain Patron

You gain mastery of the specific dimensions of the pocket in which the Lady of Pain fills Berks who dislike it. You can cast the maze spell once without using a spell slot. The labyrinth seems to be a circular series of platforms, roads and portals. The labyrinth can also be cast on a group of willing creatures, just like your Charisma modifier. The objects and inanimate objects that remain in the labyrinth will remain there. If the maze is used in this way, you must complete a DC 20 Intelligence test to finish the spell before. Once this Invocation has been used for any of the functions, you must take a long break to regain access to the maze.

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## A graph theory question inspired by a question about Morse functions in \$ mathbb {S} ^ 2 \$

This question arose from my attempt to solve the problem posed in this fascinating question about the gradient of a soft function in $$mathbb {S} ^ 2$$.

Fix a finite set, $$P$$, of an even number of points $$p_1, ldots, p_N in mathbb {S} ^ 2$$. Leave $$C (p_i) = {q in mathbb {S} ^ 2: q cdot p_i = 0 }$$ Be set of large circles with axis of $$p_i$$.

We make the following assumptions

• $$C (p_i) neq C (p_j)$$ for $$p neq j$$
• $$C (p_i) cap C (p_j) cap C (p_k) = emptyset$$ for different $$p_i, p_j, p_k$$.

Now let $$V = {q in mathbb {S} ^ 2: q in C (p_i) cap C (p_j), mbox {some} p_i neq p_j }$$. This is the set of {vertices}.
Leave
$$Gamma = bigcup_ {p in P} C (p)$$
and let $$E$$ be the set of components of $$Gamma backslash V$$. This is the set of edges. Clearly, $$(V, E)$$ It is a flat graph. Finally, let's go $$F$$ be the set of the components of $$mathbb {S} ^ 2 backslash Gamma$$. This is the set of faces. We can think of $$(F, E)$$ as the dual chart for $$(V, E)$$.

Keep in mind that the antipodal map $$A: mathbb {S} ^ 2 to mathbb {S} ^ 2$$ Induces an involution in the sets. $$V$$, $$E$$ Y $$F$$ which we will also denote by $$A$$.

Now suppose we have a map $$sigma: F to {0, ldots, N }$$ with the property that

• $$sigma (f) + sigma (A (f)) = N$$
• Yes $$f_1$$ Y $$f_2$$ they are adjacent faces, then $$| sigma (f_1) – sigma (f_2) | = 1$$ (that is to say., $$sigma$$ Jump through one through an edge.

Look at the construction ensuring that each vertex. $$v$$ It limits four edges and thus one has four faces. $$f_1$$, $$f_2$$, $$f_3$$ Y $$f_4$$ meet in $$v$$, then, until the re-labeling $$sigma (f_1) = sigma (f_2)$$, $$sigma (f_3) = sigma (f_2) -1$$ Y $$sigma (f_4) = sigma (f_2) + 1 = sigma (f_3) + 2$$.

Now let $$F_e = {f in F: sigma (f) mbox {even} }$$ and y $$F_o = {f in F: sigma (f) mbox {odd} }.$$
Leave
$$F_o & # 39; = {f in F_o: sigma (f) geq 3 mbox {y} sigma (A (f)) geq 3 }.$$
As $$N$$ it's even, we have $$A (F_e) = F_e$$ Y $$A (F_o) = F_o$$.

by $$F_o & # 39; & # 39;$$ some unspecified subset of $$F_o & # 39;$$, we are interested in the following problem:

Issue: Determine if $$F_o & # 39; & # 39; cup F_e$$ contains a connected path (in the dual graph $$(F, E)$$) connecting antipodal faces.

When $$N = 2$$ or $$N = 4$$, $$F_o & # 39;$$ it is empty and it is easy to see any path between the antipodal points in $$F_e$$ must go through $$F_o$$ – That is, there is no route as requested by the problem. I have no idea what can happen when $$N = 6$$. My suspicion is that for $$F_o & # 39; & # 39;$$ big enough one can have such a way.


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# error in estimating the mean Princeville, Illinois

The sample standard deviation s = 10.23 is greater than the true population standard deviation Ïƒ = 9.27 years. In each of these scenarios, a sample of observations is drawn from a large population. Scenario 2. Student approximation when Ïƒ value is unknown Further information: Student's t-distribution Â§Confidence intervals In many practical applications, the true value of Ïƒ is unknown.

You can see an example of this generalization from some of the numbers generated in that last example: (2) As the confidence level(1âˆ’Î±)100% increases, thenecessary sample size increases. It will be shown that the standard deviation of all possible sample means of size n=16 is equal to the population standard deviation, Ïƒ, divided by the square root of the Footer bottom Explorable.com - Copyright Â© 2008-2016. As an example of the use of the relative standard error, consider two surveys of household income that both result in a sample mean of \$50,000.

In the absence of the complete database, we may wish to estimate by taking a random sample of, say, n=25 students and computing the sample average (call this ). So you've got another 10,000 trials. With n = 2 the underestimate is about 25%, but for n = 6 the underestimate is only 5%. Thank you to...

The reason for this is that the limits for the confidence interval are now found by subtracting and adding the maximum error of the estimate from/to the sample mean. However, different samples drawn from that same population would in general have different values of the sample mean, so there is a distribution of sampled means (with its own mean and Follow us! Then the WMU population average GPA is estimated as 3.05 with an estimated standard error of .

But let's say we eventually-- all of our samples we get a lot of averages that are there that stacks up, that stacks up there, and eventually will approach something that Because these 16 runners are a sample from the population of 9,732 runners, 37.25 is the sample mean, and 10.23 is the sample standard deviation, s. Note that it is not the same as the probability distribution of the population. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

The distribution of these 20,000 sample means indicate how far the mean of a sample may be from the true population mean. Search More Info . Standard error of the mean This section will focus on the standard error of the mean. If you know the variance you can figure out the standard deviation.

The researcher's goal is to estimate Î¼ so that the error is no larger than 3 mm Hg.(By the way,Îµis typically called themaximum error of the estimate.)That is, her goal is What's your standard deviation going to be? The sample proportion of 52% is an estimate of the true proportion who will vote for candidate A in the actual election. For the purpose of hypothesis testing or estimating confidence intervals, the standard error is primarily of use when the sampling distribution is normally distributed, or approximately normally distributed.

But it's going to be more normal. The maximum error of the estimate is given by the formula for E shown. The ages in that sample were 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55. The mean of all possible sample means is equal to the population mean.

The data set is ageAtMar, also from the R package openintro from the textbook by Dietz et al.[4] For the purpose of this example, the 5,534 women are the entire population The standard error estimated using the sample standard deviation is 2.56. So if I know the standard deviation-- so this is my standard deviation of just my original probability density function, this is the mean of my original probability density function. The mean of our sampling distribution of the sample mean is going to be 5.

Please bear in mind the difference between , which is the estimate of the error in a single measurement, i.e. A t*-value is one that comes from a t-distribution with n - 1 degrees of freedom. Population Standard Deviation Known If the population standard deviation, sigma is known, then the mean has a normal (Z) distribution. In an example above, n=16 runners were selected at random from the 9,732 runners.

But anyway, hopefully this makes everything clear and then you now also understand how to get to the standard error of the mean.Sampling distribution of the sample mean 2Sampling distribution example A medical research team tests a new drug to lower cholesterol. Next, consider all possible samples of 16 runners from the population of 9,732 runners. ISBN 0-7167-1254-7 , p 53 ^ Barde, M. (2012). "What to use to express the variability of data: Standard deviation or standard error of mean?".

Then the mean here is also going to be 5. And if it confuses you let me know. Next: Exercises Up: Sampling Distribution of the Previous: Drawing from a Nonnormal 2003-09-08 menuMinitabÂ®Â 17Â SupportWhat is the standard error of the mean?Learn more about Minitab 17Â  The standard error of the mean LoginSign UpPrivacy Policy Stats: Estimating the Mean You are estimating the population mean, mu, not the sample mean, x bar.

Secondly, the standard error of the mean can refer to an estimate of that standard deviation, computed from the sample of data being analyzed at the time. With it we need only make measurements, then estimate the population mean from Eq.(11) and the population standard deviation from Eq.(12). The margin of error and the confidence interval are based on a quantitative measure of uncertainty: the standard error. It is useful to compare the standard error of the mean for the age of the runners versus the age at first marriage, as in the graph.

Blackwell Publishing. 81 (1): 75â€“81. For the runners, the population mean age is 33.87, and the population standard deviation is 9.27. If Ïƒ is not known, the standard error is estimated using the formula s x ¯   = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample As a result, we need to use a distribution that takes into account that spread of possible Ïƒ's.

How to cite this article: Siddharth Kalla (Sep 21, 2009). Scenario 1. JSTOR2340569. (Equation 1) ^ James R. We take 10 samples from this random variable, average them, plot them again.

A natural way to describe the variation of these sample means around the true population mean is the standard deviation of the distribution of the sample means. Notice the formula is the same as for a population mean when the population standard deviation is known. Now to show that this is the variance of our sampling distribution of our sample mean we'll write it right here. When n is equal to-- let me do this in another color-- when n was equal to 16, just doing the experiment, doing a bunch of trials and averaging and doing

format: plain


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# Aptitude  HCF and LCM  Practice Q&A

1. 504 can be expressed as a product of primes as

 2 × 2 × 3 × 3 × 7 × 7 2 × 3 × 3 × 3 × 7 × 7 2 × 3 × 3 × 3 × 3 × 7 2 × 2 × 2 × 3 × 3 × 7

2. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

 4 5 6 8

3. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

 28 32 40 64

4. What is the HCF of 1/3,2/3, and 1/4 ?

 2/3 1/3 1/4 1/12

5. The H.C.F of two 9/10,12/25,18/35 and 21/40 is:

 $\dfrac{3}{5}$ $\dfrac{252}{5}$ $\dfrac{3}{1400}$ $\dfrac{63}{700}$

6. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

 $\dfrac{55}{601}$ $\dfrac{601}{55}$ $\dfrac{11}{120}$ $\dfrac{120}{11}$

7. Which of the following has the most number of divisors?

 99 101 176 182

8. The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

 1677 1683 2523 3363

9. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

 3 13 23 33

10. What is the least number which when divided by 8, 12, 15 and 20 leaves in each case a remainder of 5 ?

 125 117 132 112

*Click on the QNo to display a Question.

 Total Ans

 Problems on Ages Prev Next Races and games
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text:
# MONTE.COUNT function

If condition is missing return the global count. Otherwise increment the cell value when the condition is true.


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# Image reconstruction in diffuse optical imaging

27pages on
this wiki

## Linear image reconstructionEdit

The forward problem, discussed above, involves calculating simulated data y (which may be CW, time-domain or frequency-domain data), given a forward operator F and knowledge of the sources q and internal optical properties x (which may include µa and µ's). The forward problem may therefore be formulated as:

$y=F(x)$. (3)

To reconstruct an image, it is necessary to solve the inverse problem, which is to calculate the internal optical properties x, given data y and sources q (Arridge 1999):

$x=F^{-1}(y)$. (4)

This is a non-linear problem but it can be linearised if the actual optical properties x are close to an initial estimate x0 and the measured data y is close to the simulated measurements y0. This is typically the case in difference imaging where measurements are taken before and after a small change in the optical properties. Then we can expand (3) about x0 in a Taylor series:

$y=y_0 + F'(x_0)(x-x_0)+F''(x_0(x-x_0)^2+...$. (5)

where and are the first and second-order Fréchet derivatives of F, respectively. The Fréchet derivative is a linear integral operator mapping functions in the image space to functions in the data space. In some cases, such as in section 3.3.1, the kernel of the integral operator is known in terms of the Green's functions. This approach was taken by Boas et al. (1994).

More generally, the forward problem can be solved by a numeric technique by representing the Fréchet derivatives and by matrices J (the Jacobian) and H (the Hessian), respectively. Equation 5 can then be linearised by neglecting higher order terms and considering changes in the optical properties and data to give the linear problem (6):

$\Delta y = \mathbf{J} \Delta x$. (6)

Linearising the change in intensity in this way gives rise to the Born approximation; linearising the change in log intensity instead is the Rytov approximation and may lead to improved images by reducing the dynamic range of y. Either way, image reconstruction consists of the problem of inverting the matrix J, or some form of normalised J. This may be large, underdetermined and ill-posed but standard matrix inversion methods can be used. These techniques differ in the way in which the matrix inversion is regularised to suppress the effect of measurement noise and modelling errors. Perhaps the most common techniques are truncated singular value decomposition, Tikhonov regularisation, and the algebraic reconstruction technique (ART) (Gaudette et al. 2000). The Moore-Penrose inverse J-1=JT(J JT)-1 offers a more efficient inversion if J is underdetermined and leads naturally to a Tikhonov-type formulation (7) where I is the identity matrix and λ is a regularisation parameter. J itself is calculated from the forward model, most efficiently using the adjoint method of Arridge and Schweiger (1995). The linear inverse problem can then be expressed as:

$\Delta x = \mathbf{J}^T (\mathbf{J J}^T + \lambda \mathbf{I})^{-1} \Delta y$. (7)

## Non-linear image reconstructionEdit

If the inverse problem cannot be assumed to be that of reconstructing the difference between two similar states, but is instead a reconstruction from a single acquisition to obtain absolute values of the optical properties, the linear approximation cannot be justified and the full non-linear problem must be solved. This is commonly the case in breast imaging (Dehghani et al. 2003), and in static imaging of the brain (Bluestone et al. 2001, Hintz et al. 2001, Hebden et al. 2002). To solve the non-linear problem, we define an objective function ψ, which represents the difference between the measured data y and data which is simulated using the forward model F(x). If $\hat x$ is the distribution of optical parameters which minimises ψ, then it is also the model which best fits the data, and is therefore taken to be the image. However, because the problem is ill-posed, it must be regularised, and can be expressed as:

$\psi = \left \Vert y - f(x)\right \Vert ^2 + \alpha \left \Vert \Pi \right \Vert ^2$, (8)

where α is a regularisation parameter, and Π represents prior information. This may be very simple, for example Π = I, in which case (8) becomes an iterative solution of (7). However, it is increasingly common for Π to include anatomical and other information, see section 3.4.3. $\left \Vert . \right \Vert ^2$ represents the L2-norm and gives the least squares solution. Equation 8 can be extended to include the covariance of the data CP and the covariance of the image CQ as follows:

$\psi = (y-f(x))^T C_P (y-f(x)) + \alpha (\Pi ^T C_Q \Pi)$. (9)

This approach gives a weighted least squares solution which reduces the influence of noise and cross-talk both in the data and in the image. Measuring CP and CQ may not be straightforward. Equation 9 is a non-linear minimisation problem which is generally solved either by a Newton method such as the Levenberg-Marqardt algorithm, or by a gradient method such as conjugate gradients (Arridge 1999).

## UniquenessEdit

No general uniqueness results exist for optical imaging, because of the range of different unknowns (µa, µ's, refractive index, coupling coefficients, anisotropy, geometry, etc.) and the range of different measurables (time-domain or frequency-domain measurements or CW measurements of intensity alone). Some useful results do exist, however, to illustrate situations which may not yield unique solutions (Isakov 1993).

The first important uniqueness result directly relevant to the diffusion equation was derived by Arridge and Lionheart (1998) and developed further by Arridge (1999). They demonstrated that measurements of intensity alone cannot separate the effects of absorption and scatter. Furthermore, if the refractive index is also unknown, even frequency-domain or time-domain measurements are not sufficient for uniqueness, although Matcher (1999) showed that uniqueness may be restored if the underlying model is taken to be the P1 approximation. The specific case of optical mammography using a slab geometry was analysed by Romanov and He (2000), who showed that if full time-domain data are available then reflection measurements (where the sources and detectors are on the same side of the slab) can uniquely determine a piecewise continuous distribution of either µa or µ's if the other parameter is given, but transmission measurements can distinguish µa and µ's if both are unknown. In an interesting application of uniqueness theory, Corlu et al. (2003) used a uniqueness argument to determine which wavelengths gave the optimal separation between different chromophores in spectroscopic imaging, and to reconstruct directly for chromophore concentration.

Arridge and Lionheart (1998) illustrated their uniqueness result with an example of two distributions of µa and µ's which gave indistinguishable intensity data on the boundary, and Hoenders (1997) described a whole class of solutions which could not be distinguished from one another. More specifically, Schweiger and Arridge (1999a) plotted the objective function of the inverse problem against µa and µ's for a range of different datatypes extracted from the TPSF and showed that, where certain combinations of datatypes gave a well-behaved minimum, measurements of intensity alone gave an objective function with an extended minimum at approximately µa µ's = constant, suggesting that µa and µ's could not be separated.

However, as Hoenders (1997) points out, the existence of a solution to an inverse problem depends on more than just the uniqueness. It also depends on other issues including the quality of the data and the statistical behaviour of the problem. In particular, even if a problem is non-unique, it may be possible to distinguish between two identical data sets if appropriate prior information is available, whether it is used implicitly in the form of regularisation or normalisation of the Jacobian (Pei et al. 2001, Xu et al. 2002), or explicitly. This may enable the theoretical results outlined above to be reconciled with experimental work by Schmitz et al. (2002) and Pei et al. (2001), and by Jiang et al. (2002) and Xu et al. (2002) who have reconstructed images showing some separation between µa and µ's from intensity data alone. It remains to be seen how non-uniqueness affects reconstruction of clinical images, where the potential for a reconstruction to converge to an incorrect solution may be unacceptable.

## Use of prior informationEdit

Advances in modelling and reconstruction techniques cannot overcome the fact that diffuse optical imaging is a non-unique, ill-posed, underdetermined problem and that this puts a limit on the image quality, particularly the spatial resolution, which can be achieved. One way to improve the quality of the image reconstruction is to make maximum use of prior information, which can be obtained from anatomical imaging techniques or by considering the physics and the physiology of the problem. This reduces the effect of the ill-posedness by improving the accuracy of the model, and makes the problem better determined by allowing the small number of measurements which are obtained to be used in a more effective way. For example, Ntziachristos et al. (2000) used an MR image of a breast to provide the location of a tumour a priori. Using this approach, the spatial resolution of the optical image effectively becomes that of the MR image. The reduction in quantitative accuracy due to the partial volume effect seen in optical images reconstructed without prior information does not occur, and truly quantitative images of the haemodynamic properties of the tumour can be obtained.

The most straightforward way to include prior information into the image reconstruction is to use an anatomically realistic forward model (section 3.3.3). This increases the accuracy of the forward model so that it can represent the measurements more precisely, thereby improving the image reconstruction. Further improvements can be made by including information about the covariance of the data CP, and of the image CQ, as in (9). The diagonal of CP is simply the variance of each measurement, and gives an indication of the reliability of that measurement. Off-diagonal elements are the covariance between each pair of measurements and give an indication of the interdependence between measurements. Introducing CP into (9) transforms the fit between the data and the model into a basis where all measurements are independent. CQ contains information about the predicted smoothness of the image.

Prior knowledge of the anatomy can also be incorporated into the inverse problem. In the example of a breast tumour given above, the change in optical properties is assumed to come from either a region of interest defined from anatomy (Ntziachristos et al. 2002) or be heavily biased towards that region (Brooksby et al. 2003, Li et al. 2003). The degradation in image quality due to uncertainties in the anatomical prior, and the best way to minimise this degradation, are as yet uncertain (Schweiger and Arridge 1999b). Including prior anatomical information has advantages for both linear and non-linear reconstruction (Boas et al. 2004a)

Another approach is to express (8) as a statistical inverse problem in a Bayesian framework. Bayes theorem can be written as p(x|y)p(y|x)p(x) where p(x|y) is the posterior probability of obtaining the image x given the data y, p(y|x) is the likelihood of obtaining y given x (which is obtained by solving the forward problem), and p(x) is the prior density, which in these terms is simply an estimate, made before carrying out the experiment, of the most likely image. If we assume that the distribution of errors is Gaussian, then Bayes theorem takes the form

$p(x|y) \propto exp(-\left \Vert y - f(x)) \right \Vert ^2) exp(-\left \Vert \Pi \right \Vert^2)$. (10)

where the likelihood $p(y|x) \propto exp(- \left \Vert y - f(x) \right \Vert ^2)$ and the prior $p(y|x) \propto exp(- \left \Vert \Pi \right \Vert ^2)$. The value of x which maximises (10) is the maximum a posteriori (MAP) estimate. Maximising (10) is equivalent to minimising its negative logarithm, which is (8), but can offer further approaches which deal with prior information and covariance estimates in a direct and intuitive manner (Kwee 1999, Kohlemainen 2001, Mosegaard and Sambridge 2002, Evans and Stark 2002, Oh et al. 2002).

Recently, efforts have been made to incorporate Monte Carlo techniques into the inverse problem. Standard Monte Carlo techniques are too slow to be used for practical image reconstruction, although faster methods are available. One example is the Metropolis-Hastings algorithm which is an example of a Markov Chain Monte Carlo (MCMC) technique, in which individual photon steps are drawn from a biased random walk which is chosen to minimise the number of discarded photons (Mosegaard and Sambridge 2002). These techniques have been used as the basis of image reconstruction algorithms in EIT (Kaipio et al. 2000, West et al. 2004), partly because the Monte Carlo simulations form a probability density which can be interpreted as the Bayesian maximum a posteriori estimate, leading very naturally to a statistical expression of the inverse problem (Evans and Stark 2002).

The approaches discussed above assume that the prior information, whether in terms of the data covariance or the image statistics, is governed by Gaussian statistics. This is not necessarily the case: a Poisson model for the data may be more appropriate if photon counting errors are significant, and assuming that the pixel values are distributed as Gaussian random variables forces the image to be homogeneously and isotropically smooth. The latter issue has been identified as a significant weakness and has been addressed in two ways. First, Kaipio et al. (1999) introduced an inhomogeneous, anisotropic image covariance term which allowed, for example, a change in optical properties in the brain to be correlated more closely with neighbouring pixels in the brain than with those in the skull. A second approach is to minimise the total variation, defined as the integral of the absolute gradient of the image, rather than the L2-norm (Borsic 2002). The L2-norm forces the image to be smooth, and is optimal if the optical properties are distributed as a Gaussian random field. If the distribution of optical properties is known to be piecewise constant, the L2-norm will smooth the image, whereas minimising total variation will reduce oscillations in the image but still allow boundaries to be sharp. Other non-Gaussian priors may be considered.

## Methods that exploit symmetryEdit

Some improvements to speed and robustness of the inverse problem can be obtained for systems that possess symmetry. Examples are rotational symmetry in a cylinder or sphere, or translational symmetry in a slab. It is easiest to develop this idea for the analytical version of the linear inversion kernels based on Green's functions. Green's functions are the kernels of the inverse solver for a PDE, and when the domain possesses symmetry, the PDE may often be solved by a separation of variables technique, which implies that the inversion kernels can be expressed as the product of functions in the separated variables. In the work of Markel and Schotland (2001), the slab geometry gives rise to a product of transverse plane waves and one-dimensional integral equations in the perpendicular direction. It is this decoupling of transverse propagating waves, and perpendicular scalar waves that allows faster inversion of a set of one dimensional integral equations, rather than one coupled 3D integral equation. The solution of these linear equations builds up a solution in the basis of the transverse component, which can then be transformed into image space by a fast transform method. Similar methods apply in the cylindrical and spherical geometries where the transverse components are Fourier and spherical harmonic waves respectively. Furthermore the case where only part of the data is sampled can be included by convolution with a masking function. Finally Markel et al. (2003) considered extension to the nonlinear case by applying an iterative method, linearised around the symmetric case.

As well as the analytical case, a discrete implementation such as FEM can also be developed. This approach was developed by Metherall et al. (1996) for EIT, and applied to optical tomography by Hampel and Freyer (1998).

## Solving for coupling coefficientsEdit

The measured data in optical tomography depends not only on the properties of the object under examination, but also on the characteristics of the source and detector fibres, instabilities in the source power and detector efficiency, and on the efficiency with which light is coupled into and out of the medium. A well-designed calibration procedure can minimise the effects of the optical fibres and the instrumentation (Hillman et al. 2000, McBride et al. 2001) but characterising the coupling between the fibres and the tissue can be more difficult. One approach which has been successfully adopted, particularly with regard to optical topography, is difference imaging, in which images are reconstructed using the difference between measurements recorded at two states in such a way that coupling effects cancel (Section 4.1). Alternatively, images can be reconstructed using temporal data only (Hebden et al. 2002), although this may reduce the ability to separate the absorption and scattering properties. These approaches are not always appropriate, however, and so techniques have been developed to correct for coupling effects during image reconstruction.

Schmitz et al. (2000) and Boas et al. (2001) introduced a method for solving for the unknown coupling coefficients as part of the image reconstruction problem. Each source and detector was associated with a complex coupling term. A measurement Mij between source i and detector j was modelled as Mij = SiDjmij where Si and Dj are the source and detector scaling factors and mij is the ideal measurement, depending only on the internal optical properties of the object. The reconstruction problem is then posed in terms of minimising the difference between the measured data and SiDjmij, where Si and Dj are solved for in the inverse problem. Images could be reconstructed successfully if up to 80% uncertainty in amplitude was added to simulated data (Boas et al. 2001). The technique has since been extended to allow small errors in optode position to be corrected (Culver et al. 2003, Stott et al. 2003) and has been used for dynamic breast imaging (Intes et al. 2003). Vilhunen et al. (2004) have implemented a related method which can be used when the sources and detectors have rotational symmetry. Oh et al. (2002) used a Bayesian formulation to reconstruct 3D images non-linearly by treating the unknown source-detector coupling coefficients as unwanted nuisance parameters.

## Use of dynamic informationEdit

Image reconstruction is generally applied to the spatial domain but additional information can be obtained by examining the temporal evolution of the signal. Physiological signals from the heart rate, ventilation and blood vessels are time-dependent. If the image acquisition time is short compared to the physiological fluctuations, an image time series can be obtained, which can then be processed to give dynamic images showing parameters such as the covariance or the amplitude and phase at a particular frequency of physiological interest (Barbour et al. 2001, Graber et al. 2002).

A different situation occurs when the image acquisition rate is not short compared to the physiological changes. In this case, the optical properties change during the acquisition of each image and so a straightforward static image reconstruction would not be valid. The Kalman filter technique has been used to model the image space as a state whose properties evolve with time in a known manner, the details of which are supplied as prior information. It may be expressed as:

$x_{t+1} = \mathbf{K} x_t + n_t$, (11)

where xt is the image x at time t, K is the state transition matrix and nt is additive noise at time t. In the simplest case, K is the identity matrix, in which case the update is a random walk. This approach helps to condition the inverse problem by using the current state, and knowledge of how it evolves, to constrain the possible solutions for the next state. It can be used to image events which occur on a shorter timescale than the image acquisition by calculating an updated image using the Kalman filter when only a limited subset of data has been acquired. This is ideally suited for systems which acquire data from sources sequentially but the detectors simultaneously, in which case an updated state can be estimated when each source is activated. This approach has been successfully applied to both simulated data (Kohlemainen et al. 2003) and real data (Prince et al. 2003) in optical tomography. Eppstein et al. (2001, 2002) have developed an image reconstruction technique for fluorescence tomography which uses the Kalman filter to update each iteration of a static non-linear reconstruction procedure.

## Recovery of object shapeEdit

Another approach to optimising the use of data in optical tomography is not to reconstruct for a large number of image pixels, which leads to a highly underdetermined problem, but instead to reconstruct for the boundaries and properties of internal regions which can be assumed to have piecewise constant optical properties.

One method to solve for the boundaries of internal regions is to formulate a minimisation problem where the image parameters are the Fourier coefficients of the smooth region boundaries. This was first solved assuming the optical properties of the regions are known a priori (Kohlemainen et al. 1999), and later extended to solve simultaneously for both the region boundaries and the optical properties (Kohlemainen et al. 2000a). A further refinement was to use an initial static image reconstruction to identify the number of regions and their approximate locations before finally solving for the boundaries and optical properties of the regions (Kohlemainen et al. 2000b). In 3D, it may be more appropriate to assume that regions of interest are ellipsoidal and to solve for the parameters of the ellipse rather than Fourier components (Kilmer et al. 2003).

A promising technique for the inverse shape estimation problem is provided by representing internal regions using a level set function. This is a function which is 1 inside the regions and 0 elsewhere. This method assumes that the optical properties of the background and regions of interest are known, but does not place constraints on the number of regions or their shapes. Dorn (2002) implemented this method, thresholding an initial spatial reconstruction to provide a starting estimate for the internal objects which are then described using a level set whose boundaries evolve as the reconstruction progresses.

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text:
Argumentation, series and convergence

Consider $\sum_{n=1}^{\infty} \frac{1}{n^{3}}$. We know that it converges.

Given $k\in\left(0,\infty\right)$. Is it then "okay" to say that there exists a $j\in\mathbb{N}$ such that $\sum_{n=j}^{\infty} \frac{1}{n^{3}}<k$? Or does it need some extra argumentation?

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I agree since it is an elementary fact that a series converges only if its terms tend to $0$. –  robjohn Sep 26 '12 at 18:02
The fact $\sum\limits_{n=1}^{\infty} \frac{1}{n^{3}}$ that converges means that converges the sequence of partial sums $S_N=\sum\limits _{n=1}^{N} \frac{1}{n^{3}}$


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# Tag Info

0

Yes, a function $f$ is said to be negligible if for every polynomial function $p(n)$ there exits some constant $N$ such that $f(n) < \frac{1}{p(n)}$ for all $n > N$. If $\frac{1}{n!} < \frac{1}{p(n)}$ then $n! > p(n)$, for all polynomials $p(n)$ and suitable $N$ such that $n>N$. Thus, you'd only have to prove the second segment of the ...

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Hint: you can notice that $n! > 2^n$ (except for very small $n$).

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To be negligible is not a property of fixed numbers, it is a property of functions of some parameter. It does not make sense to talk about a probability being negligible unless you give a parameter relative to which the probability is negligible. In cryptography we will often consider probability being negligible in the security parameter of the given ...

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First of all, keep in mind that all meaningful probabilities must be between $0$ and $1$. In particular, this means that, if $X$ and $Y$ are probabilities, $0 \le XY \le \min(X, Y)$. In cryptography, a probability is considered "negligible" if it is very small. What actually counts as "negligible" depends on context, but typically, we're talking about ...

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# Failure of Newton's corpuscular theory and success of photon theory of light

Corpuscular theory of light States that:

Light is made up of small discrete particles called "corpuscles" (little particles) which travel in a straight line with a finite velocity.

According to Einstein,

Light is composed of small particle are known as photons. And energy of a photon is given by $$E=h\nu$$.

Corpuscular theory suggests particle nature of light. And existence of photons also suggests the particle nature of light. I see no difference between them. But why existence of photon is accepted but not corpuscles?


text:
CarTalk's Homophones Puzzler: A Programmatic Solution

The puzzler:

This was sent in by a fellow named Dan O'Leary. He came upon a common one-syllable, five-letter word recently that has the following unique property. When you remove the first letter, the remaining letters form a homophone of the original word, that is a word that sounds exactly the same. Replace the first letter, that is, put it back and remove the second letter and the result is yet another homophone of the original word. And the question is, what's the word?

Now I'm going to give you an example that doesn't work. Let's look at the five-letter word, 'wrack.' W-R-A-C-K, you know like to 'wrack with pain.' If I remove the first letter, I am left with a four-letter word, 'R-A-C-K.' As in, 'Holy cow, did you see the rack on that buck! It must have been a nine-pointer!' It's a perfect homophone. If you put the 'w' back, and remove the 'r,' instead, you're left with the word, 'wack,' which is a real word, it's just not a homophone of the other two words.

I've used CMU Pronouncing Dictionary and modified it (manually) to keep alphabetical strings only.

My code:

def read_dict(file_path):
"""Reads 'CMU Pronouncing Dictionary' text file.

returns: dict mapping from a word to a string describing
its pronounciation

file_path: str
"""
fin = open(file_path)
pron_dict = {}

for line in fin:
index = line.find(' ')  # e.g. "AA  EY2 EY1"
word = line[:index]
pron = line[index+2:]
pron_dict[word] = pron

return pron_dict

def are_homophones(pron_dict, word1, word2, word3):
"""Returns whether 3 words are homophones.

pron_dict: dict
"""

for word in [word1, word2, word3]:
if not word in pron_dict:
return False

return pron_dict[word1] == pron_dict[word2] == pron_dict[word3]

def find_homophones_words(pron_dict):
"""Returns a list of words, where:

* word
* word with the first item omitted
* word with the second item omitted

are all homophones and existing in pron_dict.

pron_dict: dict
"""

homophones = []

for word in pron_dict:
first_word = word[1:]  # word with the first letter omitted
second_word = word[0] + word[2:] # word with the second letter omitted
if are_homophones(pron_dict, word, first_word, second_word):
homophones.append(word)

return homophones

homophones = find_homophones_words(pron_dict)

for word in homophones:
print([word, word[1:], word[0] + word[2:]])

Notes:

• I haven't, strictly, followed the puzzler. The code finds words satisfying the main requirements, except that the word is "one-syllable and five-letters".

• I'm a hobbyist and a beginner.

• I don't only want to optimize the code, but also my skills and style. Therefore any notes, advice and suggestions aren't only welcomed, but also encouraged!

• So, did you find the word this way? :) – Graipher Aug 22 '16 at 20:41
• @Graipher Oh! You are right, it's originally, an exercise from "Think Python". The author wanted us to list words satisfying these requirements. I think I should search the list to find only one-syllable ,five-letters words. :) – Mahmood Muhammad Nageeb Aug 22 '16 at 20:46

There's an efficiency improvement you could make to are_homophones: As is, you check that all three words are in pron_dict before checking that they're the same. However, it's better practice to catch errors using try blocks rather than trying to prevent them from happening (see this Stack Overflow post). A cleaner/faster way way to write this would be

def are_homophones(pron_dict, word1, word2, word3):
try:
return pron_dict[word1] == pron_dict[word2] == pron_dict[word3]
except KeyError:
return False

It's also good practice to use with blocks to open files so that Python automatically handles closing the file. With this change, read_dict becomes

def read_dict(file_path):
pron_dict = {}

with open(file_path) as fin:
for line in fin:
index = line.find(' ')  # e.g. "AA  EY2 EY1"
word = line[:index]
pron = line[index+2:]
pron_dict[word] = pron

return pron_dict

• That's good, I'll implement it. Thank you! :) – Mahmood Muhammad Nageeb Aug 22 '16 at 21:29

Some of your names are on the edge. They are descriptive enough that there is no problem with this short script, but it could be difficult in a longer script. For example, fin is what a fish has. Yes, I understand that it is file-in, but even that isn't very descriptive. I would try to find a name that describes what the file is for.

In read_dict(), you have an overly-complicated way to find the word and pronunciation. There is a format for the file. It has a word, two spaces, then its pronunciation. The easy way to do it is to use the built-in partition() method:

word, _, pron = line.partition("  ")

The _ is there because .partition() includes the separator in what it returns, but we are ignoring it.

When opening files, it is a good idea to use a with block. That takes care of closing the file if something bad happens.

find_puzzler_words() doesn't need to return a list. All you need is something that will give a for loop the right values. The more memory-efficient solution is to use a generator function. That means, instead of appending each word to res, yield it:

def find_puzzler_words(d):
...
# res = []  can be deleted now

for word in d:
...
if ...:
if ...:
yield word  # Instead of res.append(word)

It is shorter and more memory-efficient.

Note: I have attempted to put this in beginner language, but I'm afraid I probably didn't do a very good job. If you don't understand it now, I am open to answering any questions you may have.

Your program leaves out some things such as multiple pronunciations of the same word. A slight difficulty is that the duplicate words are something like "UPDATE(1)", where "(1)" isn't really part of the word. You can use a regular expression for this:

[^($$\d+$$)]*

Which means:

[^...]*  -> Anything except ... any number of times.  The ... is:
(...) -> A group which is:
$$-> A literal ( \d+ -> A number one or more times$$ -> A literal )

That way, the first word is matched as all normal characters except the group of characters such as (1), (2), (10), etc. You can then compile that regex and test each word with it. Now we have "UPDATE" and "UPDATE(1)" going under the same dictionary entry. Since we are now accounting for multiple pronunciations, we need a container for those pronunciations. I suggest a set. To make the dictionary create the sets automatically as they are needed, use collections.defaultdict:

from collections import default
import re

"""Reads 'CMU Pronouncing Dictionary' text file.

returns: dict mapping from a word to a set of strings describing
its pronunciations

path: str
"""

regex = re.compile(r"[^($$\d+$$)]*")
result = defaultdict(set)

with open(path) as f_in:
for line in f_in:
t = line.split()
word = regex.match(t[0].lower()).group()
pron = ' '.join(t[1:])

return result

Now that we have multiple pronunciations, we need to account for that in are_homophones(). That's actually very simple. Just change:

if d[word] == d[f_o_word] == d[s_o_word]:

to:

if d[word] & d[f_o_word] & d[s_o_word]:

You see, & (when used between sets) creates a new set of where the two sets intersect. If an intersection of all three sets still has something in it, we append word to res.

• I've refactored the code, after deleting the old question. And I don't use any single-character variable names anymore. Can you re-read it, if it doesn't bother you. I appreciate your review. – Mahmood Muhammad Nageeb Aug 22 '16 at 22:25
• I did re-read your question, but I didn't catch that. I have taken that out, and I also added a little bit more. – zondo Aug 22 '16 at 22:47