GingerAle
Oct 14, 2018

#2**+1 **

Solution:

The Traveling Salesman Problem (TSP) is **not** NP-complete. It is not solvable, nor is its solution verifiable in **polynomial time**. The time complexity for brute force analysis is **O(n!).**

\({}\)

For this question, a close approximation for brute force analysis time is \( \dfrac {(11-1)!} { (7-1)!}* (8.5E-3) = 42.85 \text { seconds.} \)

For comparison, if there were (21) cities then the computation time would exceed (910,761) **years**.

Here is an animated graphic giving a visual representation for the analysis of a seven-city TSP.

https://en.wikipedia.org/wiki/Travelling_salesman_problem#Computing_a_solution

The Wiki article also gives a comprehensive overview for the Traveling Salesman Problem.

GA

GingerAleFeb 8, 2020

#9**+1 **

Solution for #2:

\({}\)

Solve this using the derangement formula:

\(!n = \left [\dfrac{n!}{e} \right] \qquad | \qquad \text { where [ ] is the nearest integer, and (e) is Euler’s Number ~(2.71828...).} \\\)

*In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element appears in its original position. In other words, a derangement is a permutation that has no fixed points. *Source: https://en.wikipedia.org/wiki/Derangement

\(!6 = \left [\dfrac{6!}{e} \right] = 265 \; \text {sets where no color occupies its original position.}\\ \)

GA

GingerAleFeb 5, 2020

#3**0 **

~~The two solutions presented above are wrong.~~

**Apologies: I’m so use to seeing wrong answers, and the large drop in the x value (weight) are common for these types of questions;**

**I failed to check and verify my own work. **

**These types of questions require careful reading to understand what it is asking for. **

Such questions are common in science and statistics, both in academics and in the real-world.

**The mathematical solution seems paradoxical, but it’s not; it’s the language of the question that gives this illusion.** Banana Paradox

\( {\text { Deleted Equation }}\)

\(\text { Here’s the correct equation. }\\ x=8+\dfrac{90}{100}x\\ x=80.0 \;Lbs\\ \text { The weight of the grapefruit is $80.0$ Lbs } \)

This now agrees with Badada’s and EP’s solutions above, and Dragan’s solution below.

Related question, with an expanded solution method.

https://web2.0calc.com/questions/help-asap-thanks_2#r5

GA

GingerAleJan 21, 2020

#17**+1 **

You should know, Mr. BB, that using “brovos” for a woman is often considered mockery. The correct form is “brave” and “brava” is the singular form.

In any case, Melody's *brave* should be directed here, https://web2.0calc.com/questions/six-6-sided-dice-are-rolled-what-is-the-probability, where Melody and I *monkey around* with this exact question.

GA

GingerAleJan 2, 2020

#4**+1 **

**Happy New Year, JB. My biggest fan. **

I originally wrote that post with three scenes. I deleted the funniest scene because the Shanghai’d Post Production Supervisor or Shanghai’d Technical Consultant would have factored it to invisibility. ** **

I’ll send you a copy of the deleted scene.

-------------

This post is one in a set that I’ve periodically monitored for view counts.

https://web2.0calc.com/questions/express-big-integers-into-smallest-possible-values

Creation date and time: 2019-01-10 @10:14:51UTC

In its lifetime of 346 days, the post collected **388 views** with 13 views in the 15 days prior to its hiding by the Administrator on 12/21/19 @ ~ 14:12:52 UTC.

Posted question is class category (1) –informal, simple question.

Subjective Question classes:

*Category (0) –Mr. BB Interest Rate questions or arcane questions optimally answered with Monty Carlo style or tabulated, logical resolution computer code.

*Category (1) – Informal, simple non-homework, or curiosity based math question.

*Category (2) – Formal, simple homework.

*Category (3) – Formal, intermediate homework.

*Category (4) – Formal, advanced homework.

*Category (5) – Formal, AoPS (or related) homework.

*Category (6) – Formal, basic and intermediate physics, engineering, or chemistry.

*Category (7) – Formal, advanced physics, engineering, or chemistry.

*Category (8) – Formal, advanced collegiate mathematics or specialized math. (These are very rare on this forum.)

*Category (9) – Social or non-math, Off Topic posts (includes posts with random characters).

**View count statistics for all visible questions (posts) dated 1/10/2019: **

**Mean: 205.88**

**Median: 175**

**Mode(s): 317, 138, 141, 182, 191, 328, 152**

Minimum: 105

Maximum: 461

Range: 356

Count: 50 (Number of visible posted threads)

At least (10) posts were hidden by one or more moderators (including the auto-moderator). Several of the hidden posts were by a (now banned) member named Synth.

Comparative analysis:

The view counts for the post are 85.2% higher than average for all visible questions posted on 01/10/19. This post is about 366% higher than the sample mean for all (Cat 1) non-homework, informally asked questions. This post has 3.8% of all views for all visible posts made on 1/10/2019.

Other interesting stats, contrast the reasons for this.

GA

GingerAleDec 31, 2019

#2**+1 **

**Solution:**

One method for visualizing a solution is to view the cards in a 4 X 4 matrix, where the columns represent the players and the rows represent the cards in the players' hands. **Valid conditions for success occur when the players have single “A” cards located in any one of the four positions of their respective hands.**

\(\left[ {\begin{array}{cccc} A & A & A & A \\ X & X & X & X \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {First arrangement of “A} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{cccc} A & A & A & X \\ X & X & X & A \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{cccc} A & A & X & A \\ X & X & A & X \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {Fifth arrangement} \\ \left[ {\begin{array}{cccc} X & X & X & X \\ X & X & X & X \\ X & X & X & X \\ A & A & A & A \\ \end{array} } \right] \text {Last arrangement} \)

**From this, it’s easy to see there are (4^4) = 256 arrangements of success. **

Dividing the number of successes by the total number of combinations

**(4 ^{4}) / (nCr(16,4))**, gives

**-------------------------------------------------**

Here are two other solution methods for this question:

https://web2.0calc.com/questions/probability-help_11#r5

Here, Melody presents the total arrangements __without__ the *target* cards then adds the arrangements of the *target* cards.

https://web2.0calc.com/questions/probability-help_11#r13

Here, Rom presents the arrangements of each hand __without__ the *target* cards then adds the arrangements of the *target* cards to each hand in sequence.

GA

GingerAleDec 7, 2019