 Brain Teasers Forum
 → Viewing Profile: Reputation: plasmid
Welcome to BrainDen.com  Brain Teasers Forum
Welcome to BrainDen.com  Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :) 
Community Stats
 Group VIP
 Active Posts 1484
 Profile Views 5877
 Member Title Senior Lolcat
 Age Age Unknown
 Birthday Birthday Unknown

Gender
Male
User Tools
#341631 Otto never loved Yolanda
Posted by plasmid on 18 February 2015  05:04 PM
#341630 Universal Truth Machine
Posted by plasmid on 18 February 2015  04:55 PM
The UTM would have to declare it as false. It certainly couldn't call it true. And because calling it false would not pose a paradox that I can see (because it's an AND statement it would be ok to call it false if the lefthand side is false), I don't think the UTM would be forced to call it unanswerable.
However there's no reason a third party couldn't call the statement true.
But this is sort of dangerously skirting on the issue of whether a selfreferential statement can be considered to be true or false simply on the grounds that such an answer would be logically consistent. By such logic, the statement "This statement is true" might then be called either true or false as you see fit. A third party could call it true, but I guess they wouldn't be forced to do so.
 1
#341504 Segments on a line
Posted by plasmid on 06 February 2015  07:45 PM
 1
#341480 The ngon eats out
Posted by plasmid on 31 January 2015  04:35 PM
Draw a logic grid of people at the restaurant along the horizontal axis in the order in which they're sitting around the ngon, and whose dish they received on the vertical axis. In the figure below, if Joe had Sam's dish placed in front of him, you would fill in the green square. If you were to rotate the table, that could be represented by shifting the entire grid down (with any cells that move off the bottom of the grid wrapping back around to the top).
Since no one received their own dish, you can X out all the cells along the diagonal  they wouldn't be allowed in an initial distribution of dishes.
Suppose Joe did receive Sam's dish. Then consider all the cells along that diagonal through the green cell running parallel to the main diagonal (the cells representing Joe getting Sam's dish, Sue getting Al's dish, Amy getting Joe's dish, etc.) Rotating the table to give Joe his dish would be shifting the entire logic grid down by two cells, so if any other cells along that diagonal had been filled, that rotation would make two dishes fall on the main diagonal so two people would get their correct dish. So if person X gets dish Y, you can't use any other cells along the diagonal through (X, Y) parallel to the main diagonal.
There are n diagonals parallel to the main diagonal (wrapping around the edges of the grid), and we crossed out the main diagonal at the outset because no one started out with their own dish, so there are (n minus 1) diagonals on which to place n dishes, which of course can't be done.
 2
#341208 Empty a peg and marry a Princess (difficult)
Posted by plasmid on 02 December 2014  08:34 AM
1) Can you clear out a peg?
2) How would you do it?
A little surprisingly, I can answer the second question but not the first.
Working backwards like this, you can color all of the final solution states red (the states where one of the pegs has zero rings, which is the perimeter of the triangle), and call red "color #0". From each red dot, color every dot that can reach a red dot in one move (and is not already colored red) orange, or "color #1". From each orange dot, find every uncolored dot that could lead to an orange dot in one move and color it yellow, or "color #2". Continuing to do so will either fill the triangle or eventually reach a point where no more uncolored cells can be reached. In this example, I've colored the R=15 trinagle with colors according to the spectral sequence. For any given starting state, if it is colored, then you can reach a solution state on the edge of the triangle within the number of steps corresponding to the color of that state  simply look at all the states that could follow the current state and move to one whose color number is smaller than the current color number until you reach color #0 on the perimeter. This not only ensures you reach a solution, but that you do so in the minimum number of steps.
 1
#340959 Upside Down Cake
Posted by plasmid on 06 November 2014  03:51 AM
After a cut is made at any arbitrary point X, the cutter will eventually make it around the cake and cut out a piece containing X and flip that point into a new position Y such that the next time around the cake the cutter will land on Y. Y is irrational, but it's equal to a rational number (an integer in fact) times the irrational slice size, so it will be reached in a finite number of cuts. It's possible to have two irrational numbers that are integer multiples of each other, such as sqrt(2) and sqrt(8), so that 2 cuts of size sqrt(2) would reach the point sqrt(8).
 1
#340958 Marked Hat
Posted by plasmid on 06 November 2014  03:28 AM
 1
#340259 Freebirds
Posted by plasmid on 20 September 2014  03:40 PM
 1
#339452 optimal game strat
Posted by plasmid on 12 August 2014  04:40 AM
I tried modifying the code so it generates a matrix of steps to get from each starting point to each goal, just like the matrix I posted earlier, and runs from 1 to 30 so I could directly compare the matrices they produce. They differ at some spots, one of which is the path going from 11 to 3. For some reason it took four steps and went 11/11=1, 11+11=22, 22/11=2, 2+1=3. Obviously a shorter path would be 11/11=1, 1+1=2, 1+2=3.
Apparently there are two equally short paths from 11 to 2: {11/11=1, 1+1=2} and {11+11=22, 22/11=2}, and the algorithm ended up storing the second path as the most efficient path to get to 2. However, since it didn't store the path to 2 that creates 1 as an intermediate rather than 22, it wasn't able to discover that you could reach 3 from a path going to 2 through 1 that had a useful intermediate already generated.
 1
#339390 optimal game strat
Posted by plasmid on 06 August 2014  07:21 AM
1+1=2
2*2=4
2+1=3
4+3=7
With the algorithm, after you get one path with
1, 2, 4
and another path with
1, 2, 3
it looks like the first of those sets wouldn't allow you to add 4 in the next round because there is already a different path to reach 4, and the second of those sets wouldn't allow you to add 3 in the next round because there is already a different path to reach 3. (This is from the next if (exists($m>{$result})); line.) I'm wondering if it would have a way of discovering that path to 7?
For the most part our algorithms seem to agree, but when I set the max size for both algorithms to 30, mine said you can go from 25 to 13 in 4 steps (from the matrix in my earlier post) while yours said that 25 to 13 took 6 steps (when I commented out the
if ($start{$s}{$STEPS} == $start{$start_best}{$STEPS}) {and subsequent close bracket surrounding the print statement). A path with four steps is:
25/25 = 1
25+1 = 26
1+1 = 2
26/2 = 13
 1
#339107 I'm not a rebellious missionary
Posted by plasmid on 16 July 2014  06:14 AM
Clad in cloaks of four or three
Straight and narrow I'll not be
Thus my fate is cast
Sawn asunder as you're fain
In coffin uninterred lie slain
'neath ice with bloodstained ball and chain
Gouged until the last
 0
#339028 Sacrifice Mice
Posted by plasmid on 13 July 2014  06:54 PM
Whoever starts off with the mice sends a mouse clockwise N times, where N is the day of his execution, and each of the other prisoners will let the mouse continue on its way so all prisoners will see a mouse going clockwise N times. The starting prisoner then sends a mouse counterclockwise, with everyone passing the mouse along, to let everyone know they've reached the end of the number.
To get the mice to the next prisoner, the prisoner who started off with the mice sends one mouse clockwise and one mouse counterclockwise. The next prisoner (going clockwise) will see a mouse going clockwise after the counterclockwise mouse signaled the end of a number, and that will be his cue to keep the mice and become the next prisoner to signal his execution day. The other prisoners will see the mouse going counterclockwise (which will be the second time in a row they see one going counterclockwise), and will know that it's not their turn to signal yet and will let the counterclockwise mouse continue along until it reaches the next person to signal their date.
Then just repeat that process for all the prisoners. It should work as long as the distances between cells aren't vastly different  if they were, then on the last step where the mice are released at the same time going in opposite directions, it might cause the counterclockwise mouse to arrive at the next cell going clockwise before the clockwise mouse.
 1
#338758 Max. # of convex/concave hexagons possible using 6 points
Posted by plasmid on 27 June 2014  05:17 AM
1. ABCDFE
2. ABCEDF
3. ABCEFD
4. ABCFED
5. ABCFDE
6. ABDECF
7. ABDEFC
8. ABDFEC
9. ABECFD
10. ABEDFC
11. ABEFCD
12. ABEFDC
13. ABFCDE
14. ABFEDC
15. ACBDEF
16. ACBEDF
17. ACBEFD
18. ACBFDE
19. ACBFED
ACDEFB (is equivalent to ABFEDC)
20. ACDFBE
ACDFEB (is equivalent to ABEFDC)
21. ACEBDF
ACEFDB (is equivalent to ABDFEC)
22. ACFBED
23. ACFEBD
ACFEDB (is equivalent to ABDEFC)
24. ADBCEF
25. ADBECF
ADBEFC (is equivalent to ACFEBD)
26. ADCBFE
27. ADCFBE
ADCFEB (is equivalent to ABEFCD)
28. ADEBCF
ADEBFC (is equivalent to ACFBED)
ADEFBC (is equivalent to ACBFED)
ADEFCB (is equivalent to ABCFED)
29. ADFCBE
And there aren't any more novel solutions with that orientation of points.
I haven't come up with a way to tell if that's the maximum possible; so far I can only guarantee that it's no greater than 48 if I understand the rules correctly.
 1
#338697 First 2014 puzzle. Dissection of a triangle
Posted by plasmid on 22 June 2014  10:09 PM
Now consider what happens if you move point P up or down the line down the center of the triangle the line between A and F and between C and D. There will still be symmetry about that line, so the areas of A and F are equal, the areas of B and E are equal, and the areas of C and D are equal if point P is anywhere along that line, so (A+C+E) still equals (B+D+F). By symmetry, if you were to start with P in the center and were to move it along any of the other two lines going through the middle (the line between B and C and between F and E, or the line between A and B and between D and E) then the sum of areas remains equal.
So placing P at any point on the dashed lines in the figure will lead to subdivisions with area (A+C+E) = (B+D+F). Also notice that any point in the triangle can be reached by picking a point #1 on one of the dotted lines, a point #2 on another dotted line, and moving partway from point #1 to point #2. Next we'll consider what happens if you were to place P somewhere between two points on different dotted lines. To make things easier later on, define points #1 and #2 by drawing a line through the desired ultimate location of point P and parallel to the nearest edge of the equilateral triangle. For example, if you wanted to place P somewhere within the area of B in the figure above, you could place point #1 on the line between F and A and point #2 on the line between B and C, so #1 P and #2 form a line parallel to the side of the large equilateral triangle touching A and B.
I'll refer to the subdivisions of the triangle as A1, B1, C1, etc. when P is at point #1 and A2, B2, etc. when P is at point #2. And I'll refer to the base of a triangle (the side that's in common with the large equilateral triangle) as bA1, bB1, bA2, bB2, etc. and the height of a triangle (the line perpendicular to the base and going toward P) as hA1, hC1, hE1, hA2, hC2, hE2. Notice that hA1 would be the same as hB1, hC1=hD1, hA2=hB2, etc.
When P is at point #1 the area of A1 is bA1*hA1 / 2, and when P is at point #2 the area of A2 is bA2*hA2 / 2. If P has moved some fraction f of the way from point #1 to point #2, then the area of Af would be
((1f)bA1 + fbA2) * ((1f)hA1 + fhA2) / 2
=
(1f)^2 (bA1*hA1) / 2
+ (ff^2) (bA1*hA2) / 2
+ (ff^2) (bA2*hA1) / 2
+ f^2 (bA2*hA2) / 2
Notice that that has terms bA1*hA1 and bA2*hA2 which are interpretable and we know based on the equality when P is at point #1 or #2 that those terms by themselves will lead to equal sums of alternating triangles, but it also has terms bA1*hA2 and bA2*hA1 which will require some sorting out. I'll define the term deltaA as (bA1*hA2)+(bA2*hA1) so the equation for the area becomes
(1f)^2 (bA1*hA1) / 2
+ f^2 (bA2*hA2) / 2
+ (ff^2) deltaA / 2
In order to generate a useful equation with those mixed terms, you can start off by writing an equation saying that the sum of areas (A1+C1+E1) equals (B1+D1+F1) and (A2+C2+E2) equals (B2+D2+F2).
bA1*hA1 + bC1*hC1 + bE1*hE1 = bB1*hA1 + bD1*hC1 + bF1*hE1
bA2*hA2 + bC2*hC2 + bE2*hE2 = bB2*hA2 + bD2*hC2 + bF2*hE2
Combine terms that are multiplied by a common height
(bA1bB1)hA1 + (bC1bD1)hC1 + (bE1bF1)hE1 = 0
(bA2bB2)hA2 + (bC2bD2)hC2 + (bE2bF2)hE2 = 0
For simplicity in the upcoming section, I'll rename those terms in the equations
GH + IJ + KL = 0
MN + OP + QR = 0
Note that
(G+M) (H+N) + (I+O) (J+P) + (K+Q) (L+R)
= (GH + GN + MH + MN) + (IJ + IP + OJ + OP) + (KL + KR + QL + QR)
= (GH + IJ + KL) + (MN + OP + QR) + (GN + MM + IP + OJ + KR + QL)
= 0 + 0 + (GN + MM + IP + OJ + KR + QL)
= (bA1bB1)hA2 + (bA2bB2)hA1 + (bC1bD1)hC2 + (bC2bD2)hC1 + (bE1bF1)hE2 + (bE2bF2)hE1
= bA1*hA2 + bA2*hA1  bB1*hA2  bB2*hA1 + ....
= deltaA  deltaB + deltaC  deltaD + deltaE  deltaF
The term at the beginning of that last block (G+M) (H+N) + (I+O) (J+P) + (K+Q) (L+R) is precisely the lengths you would get if you were to consider an equilateral triangle that's twice as large as the original triangle and with the point P moved halfway from #1 to #2. Remember that points #1 and #2 are positioned such that they form a line parallel to one edge of the large equilateral triangle and are on the two dotted lines that lie farthest apart. That means that the triangle described by (G+M) (H+N) + (I+O) (J+P) + (K+Q) (L+R) has its point P lying halfway between #1 and #2. That means it lies on the dotted line between those two. That means that, since it lies on a dotted line, the sum of its alternating subtriangles must be equal. That means that
(G+M) (H+N) + (I+O) (J+P) + (K+Q) (L+R) = 0
And that means that
deltaA  deltaB + deltaC  deltaD + deltaE  deltaF = 0
And going back to the equation of the area of a subtriangle after moving point P some fraction of the way between points #1 and #2, that means that the sum of areas for all intermediate points are equal. QED.
Edit: added the figure back
 1
#338691 First 2014 puzzle. Dissection of a triangle
Posted by plasmid on 22 June 2014  04:55 PM
Drat. I just looked over that again and saw that I divided instead of multiplying when manipulating that equation after introducing the deltaA term, so it won't work.
 1
 Brain Teasers Forum
 → Viewing Profile: Reputation: plasmid
 Board Guidelines