GP, GQ, GR, GS: The Math Factor Catches Up (For Now)
A bit lazy, but we’re pretty far behind. Herewith, are
GP: Switcheroo!
GQ: Durned Ants
GR: VIth Anniversary Special
GS: I Met a Man
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Margaret Myatt said,
March 9, 2010 at 1:30 pm
I download your podcasts and I love getting the puzzles and trying them out, but I never seem to get the answers. Please have answers within the same podcast. I’m getting frustrated not hearing anwsers. How about the 1000 medium togas, the man who was n years old and born in the year n squared. 44 years old?
and, I definitely want to know the pattern to 0 1 8 10 19 90 and it can’t have another number.
Help! Thanks, love your podcast! margaret mmyatt01757@yahoo.com
Mr H said,
April 2, 2010 at 2:14 am
Margaret
You have to remember that the podcasts are re-broadcasted radio segments where the whole point is that you have a week (or more!) to work out the answer before the next programme.
Mr H
neil said,
April 16, 2010 at 5:51 am
Hi
I have another number to add to the sequence of 0 1 8 10 19 90 it’s 10^24, which is a Yotta (http://en.wikipedia.org/wiki/Yotta)
And I have to say I love the podcast, it’s awesome.
neil
Neil Smith said,
April 20, 2010 at 8:54 am
In the GR episode the sequence 0 1 8 10 19 and 90 can be extended in two ways.
Firstly with 10^24 which is yotta and then the extension after that is 10^-18 atto.
The other is 10^-24 which is yocto.
Both of these extensions then lead back to one, making the sequence recur infinately with 0 1 8 10 19 90 10^24 10^-18 1 8 10 19 90 10^24 10^-18 1 …
Don McBrien said,
July 11, 2011 at 5:25 pm
I liked the Switcheroo puzzle. However, on the program you announced that it takes about 400 visits to the switch to obtain an answer. My simulation says an average 170 days! Here’s a sample output:
Prisoner 2 is the accountant
Prisoner Claimed Accountant
Day Selected Switch Tallied
1 10 Y
2 6
3 7
4 3
5 10
6 8
7 2 Y
8 9 Y
9 3
10 6
11 5
12 6
13 5
14 8
15 11
16 8
17 1
18 9
19 10
20 7
21 0
22 2 Y
23 5 Y
24 1
25 3
26 9
27 1
28 10
29 3
30 1
31 5
32 5
33 7
34 0
35 8
36 10
37 0
38 3
39 7
40 4
41 9
42 5
43 10
44 6
45 1
46 1
47 2 Y
48 7 Y
49 11
50 5
51 2 Y
52 3 Y
53 7
54 11
55 8
56 10
57 8
58 9
59 1
60 11
61 2 Y
62 10
63 5
64 2
65 10
66 5
67 0 Y
68 11
69 0
70 11
71 7
72 10
73 4
74 9
75 8
76 10
77 10
78 11
79 5
80 9
81 8
82 11
83 5
84 3
85 10
86 1
87 5
88 10
89 3
90 10
91 1
92 5
93 8
94 10
95 11
96 7
97 4
98 11
99 10
100 4
101 3
102 5
103 6
104 7
105 6
106 3
107 9
108 4
109 6
110 2 Y
111 6 Y
112 2 Y
113 1 Y
114 11
115 9
116 3
117 4
118 2 Y
119 5
120 7
121 1
122 7
123 1
124 9
125 5
126 4 Y
127 8
128 1
129 4
130 6
131 10
132 11
133 3
134 4
135 6
136 9
137 11
138 4
139 6
140 5
141 10
142 4
143 11
144 4
145 3
146 8
147 11
148 7
149 11
150 5
151 7
152 4
153 0
154 8
155 5
156 5
157 4
158 2 Y
159 11 Y
160 0
161 0
162 9
163 11
164 8
165 5
166 6
167 9
168 5
169 2 Y
170 3
171 2
172 4
173 7
174 2
175 8 Y
176 2 Y
Stephen Morris said,
July 12, 2011 at 1:44 pm
Don,
I’ve just tried to calculate the expected duration, I make it 168 551/2310, so about 170 as you had it.
If you want to try this yourself don’t read on just yet.
An event which occurs with probability p will occur on average after 1/p occasions. For example if you repeatedly throw a die then the chance of a six is 1/6. You would expect to wait 1/(1/6) = 6 throws. This makes sense, if you kept throwing the die then one in six throws would be a six so you expect the average wait for the next six to be six throws.
In this puzzle there is one ‘counter’ and 11 ‘signallers’. We are looking for the following sequence:
first new signaller (probability 11/12 so expected wait is 12/11 visits)
counter (probability 1/12 so expected wait is 12)
second new signaller (probability 10/12 so expected wait is 12/10)
counter (probability 1/12 so expected wait is 12)
…..
eleventh new signaller (probability 1/12 so expected wait is 12/1)
counter (probability 1/12 so expected wait is 12)
The total expected wait is 12( 1 + 1/2 + 1/3 … 1/11) + 11×12 which I make 168 551/2310. (I did the sum by hand so apologies if it’s not quite right)
Cheers,
Steve
Stephen Morris said,
July 12, 2011 at 2:11 pm
I don’t think we ever gave the answer to ‘Durned Ants’. I think it was from a Peter Winkler book, which has a helpful diagram. The problem is that you have an invasion of ants in your house. You want to sleep soundly without being disturbed by them. You know that they will drown instantly if they attempt to cross water. How can you guarantee a good night’s sleep? First of all you could build a moat around the bed. But then they could climb up the walls, along the ceiling and then drop down. You could add a roof over your bed, suspended from the ceiling. But then they could crawl down the supports, under the roof and drop down on you. You could support the roof by posts outside the moat. But then they could crawl up the posts, crawl underneath the roof and drop down on you. You could build a gutter around the roof, but then they could drop onto the outer lip of the gutter, crawl underneath it and drop down on you. So what do you do? Answer in the spoiler tag. [spoiler]You do build a gutter, but it is inside the roof. The outer lip is connected to the roof. An ant could crawl underneath the gutter to reach inner lip but then it would be stuck. [/spoiler]
Shawn said,
March 22, 2012 at 3:20 pm
For the switch puzzle… [spoiler] don’t you have to have to have each signaler flip the right switch twice and have the counter count to 22 (twice the number of signalers)? The counter wouldn’t be able to know whether the switch started out up or down, since he’s not necessarily going to be the first person brought into the room. If it started off up, the counter could count to 11 (the number of signalers), but since the first time was just to establish the switch in the down position, it could be that one of the prisoners had not yet been to the room. However, if the counter decides that he’s just going to wait until he flips the switch 12 times (one more time than the number of signalers), he would be waiting forever if the switch did in fact start down, because there are only 11 signalers. There aren’t enough signalers to switch it up 12 times. If each signaler is instructed to flip up the right switch twice, then the right switch will be flipped up 22 times. If the switch started off down, then the counter will have counted to 22 eventually and can declare that everyone has been in the room. If the switch started up, then when the counter counts to 22, the switch will have been switched up 21 times. However, we know that 10 people could only switch 20 times, so the counter can be certain that the last person must have been to the room at least once, regardless of whether the switch started up or down. So the counter can then safely declare that everyone had been in the room at least once. [/spoiler]
Shawn said,
May 13, 2012 at 12:14 pm
Although I guess… [spoiler] …if the everybody sees the switches before they begin the “game”, maybe the counter can be somebody who remembered the initial position of one of the switches and that can be the switch used for signaling. Then they can decide that the other position of that switch will be used to indicate the first opportunity a prisoner has had to flip the switch, and they can avoid having to have almost everybody flip the switch twice.[/spoiler]