tag:blogger.com,1999:blog-57303912023-10-02T10:54:22.566-07:00Blaine's Puzzle BlogWeekly discussion on the NPR puzzler, brain teasers, math problems and more.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-5730391.post-85944618239771743042012-07-28T13:05:00.000-07:002012-07-28T13:07:01.836-07:00GeekDad Puzzle of the Week: Contiguous Consonants<a href="http://2.bp.blogspot.com/-Wwx7oWBuP4Q/UA69McmXdwI/AAAAAAAAKqg/5F8xjN7zvG4/s1600/ContiguousConsonants.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="152" src="http://2.bp.blogspot.com/-Wwx7oWBuP4Q/UA69McmXdwI/AAAAAAAAKqg/5F8xjN7zvG4/s200/ContiguousConsonants.png" width="160" style="border: 8px solid rgb(17, 174, 195);" alt="breastroke, matchstick, corkscrew, postscript"/></a><a href="http://www.wired.com/geekdad/2012/07/puzzle-contiguous-consonants/">GeekDad Puzzle of the Week: Contiguous Consonants</a>: <br />
<blockquote><b>Q: </b><i>[GeekDad had]</i> a lot of time to work out several phrases that incorporated words with multiple adjacent non-vowels or "contiguous consonants." For purposes of this puzzle, please consider the letter "y" strictly as a consonant. In parentheses, after each phrase is the number of words, and each word's count of contiguous consonants.<br />
<ul><li>Ice-free, super tall buildings in Scranton (3w/6c)</li>
<li>Encoding long words in a fixed orbit (3w/6c)</li>
<li>Crazy fish-studier’s two wheeled transport (3w/5c)</li>
<li>Melodic equivalents to the "Queen of Diamonds" (Condon)(3w/6c)</li>
<li>Sufficiently valuable magic during the America’s Cup (3w/5c)</li>
<li>Where playing Beethoven on your iPhone was invented (3w/5c)</li>
<li>Artificial disk-flip game (2w/5c)</li>
<li>Rotational energy "battery," 10<sup>-10</sup> meters across (2w/5c)</li>
</ul></blockquote>For those that have struggled with the math problems, this might be more up your alley. The hardest one, in my opinion, is the 4th one; I'm not completely happy with my answer. Which ones do you find tricky? Remember don't give anything away since this is a contest with a prize. Feel free to read the <a href="http://www.wired.com/geekdad/2012/07/puzzle-contiguous-consonants/">full puzzle details on the GeekDad site</a> and submit your answers by Friday for a chance at the $50 prize.<br/><br/><b>Edit: </b>The deadline has passed and I've posted our answers in the comments. I'm still waiting to see the intended answers, especially for #4.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com4tag:blogger.com,1999:blog-5730391.post-82010752979273162702012-07-21T08:21:00.000-07:002012-07-21T08:21:54.307-07:00GeekDad Puzzle of the Week: Dog Siblings<a href="http://3.bp.blogspot.com/-mrMtOxMy9Co/UAaqp-gbvdI/AAAAAAAAKp4/rj6B2G7Gp70/s1600/BlackLabs.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img alt="Black Labs, mrpattersonsir@flickr" border="0" height="128" src="http://3.bp.blogspot.com/-mrMtOxMy9Co/UAaqp-gbvdI/AAAAAAAAKp4/rj6B2G7Gp70/s320/BlackLabs.png" style="border: 8px solid rgb(17, 174, 195);" width="160" /></a><a href="http://www.wired.com/geekdad/2012/07/geekdad-puzzle-of-the-week-dog-siblings/">GeekDad Puzzle of the Week: Dog Siblings</a>: <br />
<blockquote>
<b>Q: </b>A guy with a black lab said that his dog, Selkie, has five brothers and sisters in town. "But I’ve never run into one of them," he said. "I wonder what are the chances of that?"<br />
<br />
Imagine that each of the six dogs goes out somewhere an average of once every three days. And imagine that between trails and parks and fields there are 200 places a dog can go, all (let's say...) with equal probability.<br />
<br />
If it's been exactly two years — 730 days — since Selkie's owner picked her up from the litter, what are the chances that during this time Selkie would NOT see a doggie sibling?<br />
For extra credit, what are the chances over the same time that any sibling will meet any other sibling?</blockquote>
I've got my answers which I will reveal after the deadline. In the meantime, feel free to solve it and submit your answer to GeekDad.<br />
<br />
<strong>Edit:</strong> The deadline was Friday, so here is how I went about solving the puzzle.<br />
<br />
The key to this puzzle is figure out the chances of dogs <strong><em>not</em></strong> meeting on one day. From there it is easy to figure out the chance of them <strong><em>not </em></strong>meeting for 730 days. And then if necessary, you can figure out the probability of the opposite case (meeting) by subtracting from 100%.<br />
<br />
<h4>
Part 1 - Selkie doesn't meet a doggie sibling</h4>
In order for a dog to be at a specific location, they must be out (with 1/3 probability) and at that specific spot (1/200 probability). That means there is a 1/600 chance of a specific dog being out at a specific location. Thinking of the negative probability, that means there is a 599/600 chance that a dog is *not* at a specific location.<br />
<br />
Selkie will *not* meet another dog on a specific day if,<br />
1) Selkie is at home (2/3)<br />
2) Selkie is out at any location (1/3) and dog 1 is not there (599/600) and dog 2 is not there (599/600) and dog 3 is not there...<br />
<br />
In other words, the chance that Selkie doesn't meet any other dog on a specific day is:<br />
2/3 + 1/3 x (599/600)^5 ≈ 99.7231466062%<br />
<br />
And the chance that Selkie doesn't meet any dogs for 2 years (730 days) is:<br />
[ 2/3 + 1/3 x (599/600)^5 ]^730 ≈ 13.2148023616%<br />
<br />
<blockquote><b>A: </b>(Part 1) The chance that Selkie does NOT see a doggie sibling is around 13.2148%</blockquote>
<h4>
<br />Part 2 - Chances that any sibling meets another sibling</h4>
This is the much tougher question. First let's make a table of probabilities of having 0 dogs out, 1 dog out, 2 dogs out, etc.<br />
<br />
The chance that 'n' dogs are out on any one day is C(6,n) x (1/3)^n x (2/3)^(6-n).<br />
And given 'n' dogs are out, the chance that they do NOT meet is 200/200 x 199/200 x 198/200 ... (using the number of dogs that are out.)<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="http://4.bp.blogspot.com/-OX8ZkMgUfss/UArHEMsjG6I/AAAAAAAAKqE/QcBuhY3jYuI/s1600/DoggieChart.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="117" src="http://4.bp.blogspot.com/-OX8ZkMgUfss/UArHEMsjG6I/AAAAAAAAKqE/QcBuhY3jYuI/s400/DoggieChart.png" width="400" /></a></div>
<br />
Thus, the chance that no dogs meet on a single day is around 99.1717389885%<br />
The chance that no dogs meet for 730 days is (99.1717389885%)^730 ≈ 0.2307745729%<br />
Subtracting from 100%, you get the probability that at least two siblings will meet (100% - 0.230774573%) = 99.769225427%<br />
<br />
<blockquote><b>A: </b>(Part 2) The chance that any of the siblings meet during those 2 years is around 99.7692%</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com1tag:blogger.com,1999:blog-5730391.post-79851307582422409922012-07-18T04:51:00.001-07:002012-07-18T05:37:08.581-07:00GeekDad Puzzle of the Week: Math Trolls<a href="http://1.bp.blogspot.com/-neKn6nw7Ego/UAahahQGsRI/AAAAAAAAKpo/VAlgAtWF2Kc/s1600/TrollBridge.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="160" src="http://1.bp.blogspot.com/-neKn6nw7Ego/UAahahQGsRI/AAAAAAAAKpo/VAlgAtWF2Kc/s320/TrollBridge.png" width="128" alt="Troll Bridge Sign"/></a><a href="http://www.wired.com/geekdad/2012/07/geekdad-puzzle-of-the-week-solution-math-trolls/">GeekDad Puzzle of the Week: Math Trolls</a>: <br />
<blockquote><b>Q: </b>Nora is taking a trip to visit her Grandmother in northernmost New York State this week, to bring her some freshly picked berries. On the way there, she has to cross a total of 30 bridges, and under each of the these bridges lives a troll. Each troll is aware of their bridge number, and either demands or gives berries based upon the rarest or most applicable description of their bridge. They demand or give berries according to the following schedule:<br />
<ul><li>Trolls under odd numbered bridges demand half of your berries. </li>
<li>Trolls under even numbered bridges demand 20 berries. </li>
<li>Trolls under prime numbered bridges give you half again the number of berries you are carrying. </li>
<li>Trolls under perfect square numbered bridges demand a quarter of your berries. </li>
<li>Trolls under perfect cube numbered bridges give you the number of berries you are carrying, doubling your number of berries. </li>
</ul>If trolls round up in their demands (i.e., if you have 57 berries at the foot of a bridge best described as odd numbered, you will cross it with 28 berries), what is the minimum number of berries Nora must start with so that she ends up with 1,000 berries when she arrives at her Grandmother’s house?</blockquote>The solution has already been posted on the GeekDad website. If you want to solve it yourself, read no further. A detailed breakdown of my solution is given in the comments.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com4tag:blogger.com,1999:blog-5730391.post-9729275556528646272012-07-04T09:59:00.000-07:002012-07-08T07:39:54.230-07:00GeekDad Puzzle of the Week: Poaching Berries<a href="http://3.bp.blogspot.com/-nXfHof6byJQ/T_R09rWO28I/AAAAAAAAKow/t4xPgVxuvOg/s1600/RedRaspberries.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img alt="Red Raspberries, photofarmer@Flickr" border="0" height="107" src="http://3.bp.blogspot.com/-nXfHof6byJQ/T_R09rWO28I/AAAAAAAAKow/t4xPgVxuvOg/s320/RedRaspberries.jpg" style="border: 8px solid rgb(17, 174, 195);" width="160" /></a><a href="http://www.wired.com/geekdad/2012/07/geekdad-puzzle-week-poaching-berries/">GeekDad Puzzle of the Week: Poaching Berries</a>: <br />
<blockquote>
<b>Q: </b>Leif and Kestrel are willing to give GeekDad a 20% cut of berries they poach for turning a blind eye. Imagine that Leif picks five berries per poach and Kestrel picks three berries per poach and that they attempt to poach once every day, with the exception of any day just after they’ve been caught. Now imagine that each time they poach berries, they have a 15% chance of getting caught. How many berries can GeekDad expect to eat each each week, averaged over time?</blockquote>
I'll post my thoughts on the answer after the answer is revealed (generally early next week).<br />
<br />
<b>Edit: </b>The <a href="http://www.wired.com/geekdad/2012/07/geekdad-puzzle-of-the-week-answer-poaching-berries/">following answer</a> was what I submitted to end up winning the puzzle:<br />
<br />
Let's assume a few things:<br />
1) Leif and Kestrel start in a state where they haven't ever been caught.<br />
2) Each time they are caught, the berries that day don't count. Also, they must skip the next day until trying again.<br />
<br />
There are essentially 3 states that each child could be in. Since the probabilities for each child are the same we can group them together, though in reality the puzzle allows for one child to be caught and the other not. The results work out the same, so let's just simplify it to a pair of children trying to poach 8 berries a day.<br />
<br />
The children could be:<br />
1) Caught the day before and therefore have no chance of getting berries that day (must sit out).<br />
2) Caught today (15%, if not caught the day before)<br />
3) Not caught today (85%, if not caught the day before)<br />
<br />
Each day, the chance of being caught the day before is just carried forward. So on day 2, the chance is 15% they are sitting out. That leaves 85% chance they are attempting poaching. Of that 85%, there's a 15% chance they are caught (0.85 x 0.15 = 12.75%) and an 85% chance they poach successfully (0.85 x 0.85 = 72.25%).<br />
<br />
If you repeat this, the next day (Day 3) they have a 12.75% chance of sitting out and a 87.25% chance of attempting poaching. Day 4, the chance of sitting out is 0.15 x 87.25% = 13.0875% and not sitting out is 86.9125%. Day 5, the chance of sitting out is 0.15 x 86.9125% = 13.036875% and not sitting out is 86.963125%. You can continue this progression and you will see that the chance they are sitting out approaches a value of about 13.043478%. The chance they are caught that day is equivalently about 13.043478%. That leaves a 73.913043% chance they are able to poach 8 berries with an expected return of 5.913043478 berries a day.<br />
<br />
That equates to approximately 41.39130435 berries a week. With a 20% "commission" after awhile you will be eating approximately 8.27826087 berries each week.<br />
<br />
<strong>Note:</strong> For a more accurate answer (rather than just a decimal approximation) we can solve this algebraically as follows:<br />
<br />
Let p be the chance that you ARE sitting out.
<br />
Let q be the chance you are NOT sitting out.
<br />
<br />
Together these are mutually exclusive and therefore add up to 100% (or mathematically we say 1)
<br />
p + q = 1
<br />
p = 1 - q
<br />
<br />
The chance you are NOT sitting out, but CAUGHT is 0.15q
<br />
The chance you are NOT sitting out, and SUCCESSFULLY POACHED is 0.85q
<br />
<br />
We know that eventually the two values p and 0.15q end up being the same, so equate them
<br />
p = 0.15q
<br />
<br />
Substitute in 1-q:
<br />
1 - q = 0.15q
<br />
<br />
Rearrange:
<br />
1 = 1.15q
<br />
q = 1/1.15
<br />
<br />
The chance we are caught that day is 0.15q<br />
0.15q = 0.15(1/1.15) = 0.15/1.15 = 15/115 = 3/23
<br />
And the chance we poach some berries is 0.85q
<br />
0.85q = 0.85(1/1.15) = 0.85/1.15 = 85/115 = 17/23
<br />
<br />
For the sake of completeness that means you have:
<br />
3/23 = chance child is sitting out
<br />
3/23 = chance child was caught today
<br />
17/23 = chance child was able to poach successfully
<br />
<br />
Now multiply this last number by 8 berries attempted times 7 days and then times 1/5 (20% commission) to get the expected number of berries poached each week.
<br />
Berries per week = 17/23 x 8 x 7 x 1/5
<br />
<br />
If you reduce that to a fraction you end up with:
<br />
952/115 = 8 32/115 berries each week.<br />
<br />
<blockquote><b>A: </b>8 32/115 berries each week.<br/>
8.2<u>7826086956521739130434</u>... (underlined portion repeats indefinitely)</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com7tag:blogger.com,1999:blog-5730391.post-16696092587308840102012-06-16T14:57:00.002-07:002012-06-18T06:03:58.417-07:00GeekDad Puzzle of the Week: When Are the Odds Even?<a href="http://4.bp.blogspot.com/-FCyzq0gVmBg/T9z-QwDTq1I/AAAAAAAAKnw/BEHzwC8_6sY/s1600/Marbles.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img alt="Black and white marbles" border="0" height="120" src="http://4.bp.blogspot.com/-FCyzq0gVmBg/T9z-QwDTq1I/AAAAAAAAKnw/BEHzwC8_6sY/s320/Marbles.jpg" title="" width="120" style="border: 8px solid rgb(17,174,195);"/></a><a href="http://www.wired.com/geekdad/2012/06/geekdad-puzzle-of-the-week-when-are-the-odds-even/">GeekDad Puzzle of the Week: When Are the Odds Even?</a>:<br/><blockquote><b>Q: </b>If we have a bag containing equal numbers of black and white marbles, and we pull out <b>one</b> marble, the odds of it being black are even. If we have a bag containing 120 marbles, 85 of which are black, the odds of us pulling out <b>two</b> marbles and them both being black is also even — (85/120)x(84/119) = 0.5 or 50%.<br />
<br />
If the largest bag we have can hold 1,000,000 marbles, for how many sets of marbles (i.e., the 120 marbles described above are one set) can we pull <b>two</b> marbles and have a 50% chance of them being the same designated color? Are there any sets of marbles for which we can pull <b>three</b> marbles and have a 50% chance of them being the same designated color? If so, how many?</blockquote>After the solution is revealed, I'll post the details of my answer.<br/><br/><b>Edit: </b><a href="http://www.wired.com/geekdad/2012/06/geekdad-puzzle-of-the-week-solution-when-are-the-odds-even/">GeekDad Puzzle Solution:</a><br/>In the first case you are essentially looking for integer solutions to:<br/>
a(a-1) = 2b(b-1)<br/><br/>There are EIGHT sets under 1 million that will result in even odds when 2 balls are drawn.<br/><br/>
4 marbles (3 black) --> 4 x 3 = 2(3 x 2)<br/>
21 marbles (15 black) --> 21 x 20 = 2(15 x 14)<br/>
120 marbles (85 black) --> 120 x 119 = 2(85 x 84)<br/>
697 marbles (493 black) --> 697 x 696 = 2(493 x 492)<br/>
4,060 marbles (2,871 black) --> 4,060 x 4,059 = 2(2,871 x 2,870)<br/>
23,661 marbles (16,731 black) --> 23,661 x 23,660 = 2(16,731 x 16,730)<br/>
137,904 marbles (97,513 black) --> 137,904 x 137,903 = 2(97,513 x 97,512)<br/>
803,761 marbles (568,345 black) --> 803,761 x 803,760 = 2(568,345 x 568,344)<br/><br/>
Interestingly, the next number in each sequence can be computed as follows:<br/>
a(n) = 6a(n-1) - a(n-2) - 2<br/><br/>
So for example, the next numbers in the sequence would be:<br/>
Total balls: 6 x 803,761 - 137,904 - 2 = 4,684,660 marbles<br/>
Black balls: 6 x 568,345 - 97,513 - 2 = 3,312,555 black<br/><br/>
Integer sequences: <a href="http://oeis.org/A011900">A011900</a> and <a href="http://oeis.org/A046090">A046090</a><br/><br/>
In the second case you are looking for integer solutions to:</br>
a(a-1)(a-2) = 2b(b-1)(b-2)<br/><br/>
There is only ONE set under 1 million that will result in even odds when 3 balls are drawn.</br><br/>
6 marbles (5 black) --> 6 x 5 x 4 = 2(5 x 4 x 3)<br/><br/>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com5tag:blogger.com,1999:blog-5730391.post-2279265264096044802012-06-03T23:11:00.001-07:002018-01-23T01:54:55.521-08:00GeekDad Puzzle of the Week: Waffle Cuts<a href="http://www.wired.com/geekdad/2012/06/geekdad-puzzle-of-the-week-solution-waffle-cuts/">GeekDad Puzzle of the Week: Waffle Cuts</a>: <a href="https://drive.google.com/file/d/1fgzZryjCBRPpHYVayoE5Au3AzN2KaqjO/view?usp=sharing" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="98" width="98" src="http://3.bp.blogspot.com/-IF8wudlJO2Q/T8oZKaZFYYI/AAAAAAAAKls/WNIch4YVyUQ/s320/Waffle.png" alt="Waffle, no cuts" /></a><blockquote><b>Q: </b>If we only cut along the ridges of a circular waffle, and if each cut traverses the waffle in a straight line from edge to edge, how many different ways can the waffle be cut?<br/><i><b>Note: </b>rotations, horizontal flips, and vertical flips of a set of cuts should only be counted once.</i></blockquote>
Given that there are 6 places to cut vertically and 6 places to cut horizontally, that's a total of 12 cut lines. If you allow for any combination of these 12 lines to be cut or not, you have a total of 2^12 = 4096 ways to divide the waffle. But of course, the puzzle asks for the number of <b><i>unique</i></b> ways to cut the waffle, not including any mirrored or rotationally symmetric sets of cuts.<br/><br/>After the official answer to the puzzle is posted, I'll post my solution here.<br/><br/><b>Edit: </b>The solution is posted, but just the number without any detail. Also, I disagree with their counting of the "no cuts at all" solution as one of the ways to "cut" the waffle. In any case, a full detailing of my solution along with an enumeration of all <b>665 (or 666)</b> ways to uniquely cut the waffle can be found in <a href="https://drive.google.com/file/d/1fgzZryjCBRPpHYVayoE5Au3AzN2KaqjO/view?usp=sharing">Blaine's Solution to the GeekDad Waffle Puzzle</a>.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com2tag:blogger.com,1999:blog-5730391.post-35659360413084035432012-05-26T21:48:00.000-07:002012-06-02T06:55:43.325-07:00GeekDad Puzzle of the Week Answer: That Darn Achilles<a href="http://www.wired.com/geekdad/2012/05/gpotw-answer-that-darn-achilles/">That Darn Achilles</a>: <blockquote><b>Q: </b>You, Paris, have the luxury of launching an arrow at faraway Achilles either from the ramparts above the Hesperian Gate at a height of exactly 8 meters. Or you can stand atop Priam’s palace. This gains you another 7 meters of launch height, but it costs you 15 meters of horizontal distance. If the arrow leaves your bow at a somewhat modest 70 meters per second, are you best taking your pot-shot at far-off Achilles from the ramparts or the palace? Which perch offers the farthest reach?</blockquote><a href="http://3.bp.blogspot.com/-MHgjb5Jfl3Y/T8MA-Pfkg2I/AAAAAAAAKlc/c4mjuQVeNOw/s1600/AchillesArrowTrajectories.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="58" src="http://3.bp.blogspot.com/-MHgjb5Jfl3Y/T8MA-Pfkg2I/AAAAAAAAKlc/c4mjuQVeNOw/s320/AchillesArrowTrajectories.png" width="160" /></a><br />
I had to look up the formulas for determining the maximum range of a projectile when fired on uneven ground. The following page was invaluable.<br />
<a href="http://en.wikipedia.org/wiki/Range_of_a_projectile">Wikipedia: Range of a Projectile</a><br />
<br />
Because several things weren't stated, I'm going to assume that we can use acceleration of gravity (g) at sea level, we can assume no wind resistance and also assume that the ground is level between the target and the firing point (except for the elevation change and horizontal offset provided by the ramparts (0,8) and the palace (-15, 15).<br />
<br />
While the ideal case (firing on even ground) results in an optimal angle of 45°, when you are firing from a height, then you want to angle down slightly to maximize distance. I won't bore you with going through the details on that page, but basically there's an equation for the horizontal and vertical positions at time t, given an initial angle (theta) and velocity v. You can then set the final height to be 0 and solve for t. Using that you can get an equation for distance given an angle and by taking the derivative and setting it to zero, you can get a formula for the optimal angle to get the longest distance.<br />
<br />
Rampart (x0,y0) = (0,8)<br />
Palace (x0,y0) = (-15,15)<br />
Velocity (v) = 70 m/s<br />
Gravity (g) = 9.80665 m/s^2<br />
<br />
Optimal angle (θ) = cos-1 [ √(2*g*y0 + v^2) / (2*g*y0 + 2v^2) ]<br />
<br />
Rampart angle (θ) = 0.777519 Radians or 44.54853°<br />
Palace angle (θ) = 0.7708234 Radians or 44.164927°<br />
Distance (d) = [ v*cos θ [v*sin θ + √((v*sin θ)^2 + 2*g*y0) ] /g + x0<br />
<br />
Rampart distance = 507.5979m<br />
Palace distance = 499.44231m<br />
<br />
<blockquote><b>A: </b>When firing from the ramparts (8m), the optimal angle is around 44.55° and will net you a distance of 507.6 meters. <br />
<br />
When firing from the palace (15m), the optimal angle is around 44.16° but because of the -15m offset you only reach 499.4 meters.</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com0tag:blogger.com,1999:blog-5730391.post-30444567485932666532012-05-13T21:31:00.000-07:002012-06-02T06:55:43.322-07:00GeekDad Puzzle of the Week Answer: Coffee Conundrum II<a href="http://www.wired.com/geekdad/2012/05/puzzle-answer-coffee-conundrum-ii/">Coffee Conundrum II</a>: <blockquote><b>Q: </b>Each cup of coffee gives me a jolt and then the jolt decays across time according to the following equation (t in minutes): Jitters=10-[(t-10)^2]/10<br />
So at minute 10 after consuming a cup of coffee (which, for the purposes of this puzzle happens instantly), I reach a maximum of 10 jitters. At a combined 20 jitters, I go catatonic. With what frequency can I instantly consume coffee without the combined jitters passing this important tipping point?</blockquote><a href="http://1.bp.blogspot.com/-jnrs85mlI4Y/T8L-o9oFR3I/AAAAAAAAKlU/PceccxElFZk/s1600/Coffee%2BJitters.png" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="85" width="160" src="http://1.bp.blogspot.com/-jnrs85mlI4Y/T8L-o9oFR3I/AAAAAAAAKlU/PceccxElFZk/s320/Coffee%2BJitters.png" /></a>The diagram shows the exact tipping point situation. Focus on the red curve which is the second cup of coffee. Right at the middle of the parabola, the second cup is at its maximum (10 jitters). At this point, the effects of the first cup are diminishing and the effects of the third cup are increasing. Notice, because of symmetry, they have the exact same value when they cross. The tipping point will be when cup 1 is contributing 5 jitters, cup 2 is at its maximum of 10 jitters and cup 3 is also contributing 5 jitters (5+10+5 = 20). (Note you can extend this diagram out to 4 cups, 5 cups, etc. but you'll never have more than 3 cups contributing to the total jitters at once. If you did, you'd definitely be over the tipping point of 20.)<br />
<br />
As stated in the puzzle, the height of one parabolic curve is given by the formula:<br />
Jitters = 10 - [ ( t - 10 ) ^ 2 ] / 10<br />
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Solving this for a Jitters value of 5 we have:<br />
5 = 10 - [ ( t - 10 ) ^ 2] / 10<br />
-5 = -[ ( t - 10 ) ^ 2] / 10<br />
5 = [ ( t - 10 ) ^ 2] / 10<br />
50 = ( t - 10 )^2<br />
t - 10 = √50<br />
t = 10 ± √50<br />
t = 10 ± 5√2<br />
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That means that the coffee Jitters are at a height of 5 either 5√2 seconds before the peak or 5√2 seconds after the peak. This also happens to be the minimum frequency between cups (5√2 seconds or approximately 7.071068 seconds apart) to avoid jitters.<br />
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If you can handle exactly 20 jitters without going catatonic, then you could handle drinking cups of coffee every 5√2 seconds (≈7.071068 seconds)<br />
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If you must stay under 20 jitters, then you would have to pick a frequency just over every 5√2 seconds.<blockquote><b>A: </b>Frequency of 5√2 seconds (or approximately every 7.0710678118654752440084436210484903928483593768847403 seconds)</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com0tag:blogger.com,1999:blog-5730391.post-54763412231259273192012-03-11T22:02:00.000-07:002012-06-02T06:55:43.327-07:00GeekDad Puzzle of the Week Answer: Bigger than Googol!<a href="http://www.wired.com/geekdad/2012/03/geekdad-puzzle-of-the-week-solution-bigger-then-googol/">Bigger than Googol!</a>: <blockquote><b>Q: </b>What is the first Fibonacci number bigger than a googol? </blockquote>This was relatively easy in Haskell:<br />
<br />
let fib = 0 : 1 : zipWith (+) fib (tail fib)<br />
let answer = [ (n, fib!!n) | n <- [1..1000], (fib!!(n-1) <= 10^100) && (fib!!n > 10^100) ]<br />
answer<br />
(481,14913169640232740127827512057302148063648650711209401966150219926546779697987984279570098768737999681)<br />
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<blockquote><b>A: </b>F(481) is the first Fibonacci number greater than googol. The numerical value is given above, but just for fun here's that number in words:<br />
fourteen duotrigintillion, nine hundred thirteen untrigintillion, one hundred sixty-nine trigintillion, six hundred forty novemvigintillion, two hundred thirty-two octovigintillion, seven hundred forty septemvigintillion, one hundred twenty-seven sesvigintillion, eight hundred twenty-seven quinquavigintillion, five hundred twelve quattuorvigintillion, fifty-seven tresvigintillion, three hundred two duovigintillion, one hundred forty-eight unvigintillion, sixty-three vigintillion, six hundred forty-eight novemdecillion, six hundred fifty octodecillion, seven hundred eleven septendecillion, two hundred nine sexdecillion, four hundred one quindecillion, nine hundred sixty-six quattuordecillion, one hundred fifty tredecillion, two hundred nineteen duodecillion, nine hundred twenty-six undecillion, five hundred forty-six decillion, seven hundred seventy-nine nonillion, six hundred ninety-seven octillion, nine hundred eighty-seven septillion, nine hundred eighty-four sextillion, two hundred seventy-nine quintillion, five hundred seventy quadrillion, ninety-eight trillion, seven hundred sixty-eight billion, seven hundred thirty-seven million, nine hundred ninety-nine thousand, six hundred eighty-one</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com0