harey

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On 10/31/2015 at 3:39 AM, jasen said:
If I'm lucky I will win in a few step, but if not, it will take an infinite time to win.
A good question would be "How many steps on average until I win?"

There is a strip of N (N>13) squares. In the middle is the SILVER DOLLAR.
Two players alternatively place a penny on an empty square.Then, at each turn, a player must:
 move one coin one or more squares to the left, observing:
 the coin cannot move out of the strip
 the coin cannot jump on another coin
 the coin cannot jump over another coin
 or 
 pocket the leftmost coinBest strategy?

@Plasmid I translated your notation into mine, as far as we are gone, we have same results (excepted some doublets).
SpoilerRound 1, Person N
...(new) L M N
L = 2M 2 1 3
M = 2L 1 2 3
L = 3M/2 3 2 5 (a)
L = 3M 3 1 4 (b)
M = 3L/2 2 3 5
M = 3L 1 3 4
M = L/3 3 1 4 (b)
M = 2L/3 3 2 5 (a)
M = 4L/3 3 4 7
M = 5L/3 3 5 8
I thought about interpolation, too, but I did not go this way estimating there is not enough data.

I fear there is no solution.
SpoilerRound 1, A:
A knows his number if he sees (1,2). As the difference (1) is already used, he must have the sum.
1Aa: (3,1,2)
1Ab: (3,2,1)(And multiples.)
In all other cases, A must pass.
Round 1, B:
The same way:
1Ba: (1,3,2)
1Bb: (2,3,1)B reads A's thoughts, useful in following cases:
from 1Aa: (3,!2,1) > 1Bc: (3,4,1)
from 1Ab: (3,!1,2) > 1Bd: (3,5,2)In all other cases, B must pass.
Round 1, C:
As previously:
1Ca: (1,2,3)
1Cb: (2,1,3)Let's assume C sees (3,4,?).
If the solution is (3,4,1), B would have announced it. So he has 7.In general:
from 1Aa: (3,1,!2) > 1Cc: (3,1,4)
from 1Ab: (3,2,!1) > 1Cd: (3,2,5)
from 1Ba: (1,3,!2) > 1Ce: (1,3,4)
from 1Bb: (2,3,!1) > 1Cf: (2,3,5)
from 1Bc: (3,4,!1) > 1Cg: (3,4,7)
from 1Bd: (3,5,!2) > 1Ch: (3,5,8)In all other cases, C must pass.
Round 2, A:
from 1Ba: (!1,3,2) > 2Aa: (5,3,2)
from 1Bb: (!2,3,1) > 2Ab: (4,3,1)
from 1Bc: (!3,4,1) > 2Ac: (5,4,1)
from 1Bd: (!3,5,2) > 2Ad: (7,5,2)
from 1Ca: (!1,2,3) > 2Ae: (5,2,3)
from 1Cb: (!2,1,3) > 2Af: (4,1,3)
from 1Cc: (!3,1,4) > 2Ag: (5,1,4)
from 1Cd: (!3,2,5) > 2Ah: (7,2,5)
from 1Ce: (!1,3,4) > 2Ai: (7,3,4)
from 1Cf: (!2,3,5) > 2Aj: (8,3,5)
from 1Cg: (!3,4,7) > 2Ak: (11,4,7)
from 1Ch: (!3,5,8) > 2Al: (13,5,8)Here, I give up. Python:
Spoilerdef A(new,old):
for l in Z:
if l[0]==old: Z.append([new,old,l[3]+l[4],l[3],l[4],'',l[2],l[3],l[4]])
#for l in Z: print(*l)def B(new,old):
for l in Z:
if l[0]==old: Z.append([new,old,l[2],l[2]+l[4],l[4],'',l[2],l[3],l[4]])def C(new,old):
for l in Z:
if l[0]==old: Z.append([new,old,l[2],l[3],l[2]+l[3],'',l[2],l[3],l[4]])
##
Z=[ ['A1',' ',3,1,2], ['B1',' ',1,3,2], ['C1',' ',1,2,3],
['A1',' ',3,2,1], ['B1',' ',2,3,1], ['C1',' ',2,1,3]]
B('B1','A1')
C('C1','A1')
C('C1','B1')
A('A2','B1')
A('A2','C1')
B('B2','A2')
B('B2','C1')
C('C2','A2')
C('C2','B2')for l in Z:
print(l[0],end=' ')
print('{:3d}'.format(l[2]),end=' ')
print('{:3d}'.format(l[3]),end=' ')
print('{:3d}'.format(l[4]),end=' ')
if len(l)>5:
print(' from',l[1],end=' ')
print('{:3d}'.format(l[6]),end=' ')
print('{:3d}'.format(l[7]),end=' ')
print('{:3d}'.format(l[8]),end='')
print()bingo=[]
for l in Z:
if l[0]=='C2' and not l[4] in bingo: bingo.append(l[4])
bingo.sort()
print('\nPossible bingos for C2:',*bingo)
Result:
A1 3 1 2
B1 1 3 2
C1 1 2 3
A1 3 2 1
B1 2 3 1
C1 2 1 3
B1 3 5 2 from A1 3 1 2
B1 3 4 1 from A1 3 2 1
C1 3 1 4 from A1 3 1 2
C1 3 2 5 from A1 3 2 1
C1 1 3 4 from B1 1 3 2
C1 2 3 5 from B1 2 3 1
C1 3 5 8 from B1 3 5 2
C1 3 4 7 from B1 3 4 1
A2 5 3 2 from B1 1 3 2
A2 4 3 1 from B1 2 3 1
A2 7 5 2 from B1 3 5 2
A2 5 4 1 from B1 3 4 1
A2 5 2 3 from C1 1 2 3
A2 4 1 3 from C1 2 1 3
A2 5 1 4 from C1 3 1 4
A2 7 2 5 from C1 3 2 5
A2 7 3 4 from C1 1 3 4
A2 8 3 5 from C1 2 3 5
A2 13 5 8 from C1 3 5 8
A2 11 4 7 from C1 3 4 7
B2 5 7 2 from A2 5 3 2
B2 4 5 1 from A2 4 3 1
B2 7 9 2 from A2 7 5 2
B2 5 6 1 from A2 5 4 1
B2 5 8 3 from A2 5 2 3
B2 4 7 3 from A2 4 1 3
B2 5 9 4 from A2 5 1 4
B2 7 12 5 from A2 7 2 5
B2 7 11 4 from A2 7 3 4
B2 8 13 5 from A2 8 3 5
B2 13 21 8 from A2 13 5 8
B2 11 18 7 from A2 11 4 7
B2 1 4 3 from C1 1 2 3
B2 2 5 3 from C1 2 1 3
B2 3 7 4 from C1 3 1 4
B2 3 8 5 from C1 3 2 5
B2 1 5 4 from C1 1 3 4
B2 2 7 5 from C1 2 3 5
B2 3 11 8 from C1 3 5 8
B2 3 10 7 from C1 3 4 7
C2 5 3 8 from A2 5 3 2
C2 4 3 7 from A2 4 3 1
C2 7 5 12 from A2 7 5 2
C2 5 4 9 from A2 5 4 1
C2 5 2 7 from A2 5 2 3
C2 4 1 5 from A2 4 1 3
C2 5 1 6 from A2 5 1 4
C2 7 2 9 from A2 7 2 5
C2 7 3 10 from A2 7 3 4
C2 8 3 11 from A2 8 3 5
C2 13 5 18 from A2 13 5 8
C2 11 4 15 from A2 11 4 7
C2 5 7 12 from B2 5 7 2
C2 4 5 9 from B2 4 5 1
C2 7 9 16 from B2 7 9 2
C2 5 6 11 from B2 5 6 1
C2 5 8 13 from B2 5 8 3
C2 4 7 11 from B2 4 7 3
C2 5 9 14 from B2 5 9 4
C2 7 12 19 from B2 7 12 5
C2 7 11 18 from B2 7 11 4
C2 8 13 21 from B2 8 13 5
C2 13 21 34 from B2 13 21 8
C2 11 18 29 from B2 11 18 7
C2 1 4 5 from B2 1 4 3
C2 2 5 7 from B2 2 5 3
C2 3 7 10 from B2 3 7 4
C2 3 8 11 from B2 3 8 5
C2 1 5 6 from B2 1 5 4
C2 2 7 9 from B2 2 7 5
C2 3 11 14 from B2 3 11 8
C2 3 10 13 from B2 3 10 7Possible bingos for C2: 5 6 7 8 9 10 11 12 13 14 15 16 18 19 21 29 34
None of these numbers divides 148.
Notice that it is always the prisoner with the highest number who can know, so there is no solution in the form [a*148,b*148,148].
Please comment, especially if you found it correct

Same as rocdocmac
SpoilerBefore reaching 20 (n) stairs, you can either:
 go up 1 stair and apply the solution for 19 (n1) stairs  or 
 go up 2 stairs and apply the solution for 18 (n2) stairs.This gives the formula: f(n)=f(n1)+f(n2). (Looks like Fibonacci...)
For 1 stair, there is 1 possibility.
For 2 stairs, there is 2 possibilities. => FibonacciCombinatorics:
SpoilerMaximum number of double stair steps is given by the integer division of number of stairs by 2.
Complete sigle stair steps.
Take all permutations with repetitions:Spoilerfrom math import factorial as MF
FIBO_1,FIBO_2=1,1
# staircase will take the values 1,2,3,...20 (yes, 20)
for staircase in range(1,21):
FIBO_1,FIBO_2=FIBO_2,FIBO_1+FIBO_2
combi=0
doubles=staircase//2# double will take values 0,1,2...doubles (see staircase)
for double in range(0,doubles+1):
single=staircasedouble*2
combi=combi+MF(single+double)//MF(double)//MF(single)
# remove # in the next line for intermediate results
# print(staircase,single,double,combi)print('stairs =',staircase,'combinations =',combi,'Fibonacci =',FIBO_1)
Output:
Spoilerstairs = 1 combinations = 1 Fibonacci = 1
stairs = 2 combinations = 2 Fibonacci = 2
stairs = 3 combinations = 3 Fibonacci = 3
stairs = 4 combinations = 5 Fibonacci = 5
stairs = 5 combinations = 8 Fibonacci = 8
stairs = 6 combinations = 13 Fibonacci = 13
stairs = 7 combinations = 21 Fibonacci = 21
stairs = 8 combinations = 34 Fibonacci = 34
stairs = 9 combinations = 55 Fibonacci = 55
stairs = 10 combinations = 89 Fibonacci = 89
stairs = 11 combinations = 144 Fibonacci = 144
stairs = 12 combinations = 233 Fibonacci = 233
stairs = 13 combinations = 377 Fibonacci = 377
stairs = 14 combinations = 610 Fibonacci = 610
stairs = 15 combinations = 987 Fibonacci = 987
stairs = 16 combinations = 1597 Fibonacci = 1597
stairs = 17 combinations = 2584 Fibonacci = 2584
stairs = 18 combinations = 4181 Fibonacci = 4181
stairs = 19 combinations = 6765 Fibonacci = 6765
stairs = 20 combinations = 10946 Fibonacci = 10946 
16 hours ago, plasmid said:
Comment on your answer
04 [2] B: "C is more than 18 years old."
05 [A] E: "C is less than 18 years old." => (b1) E=!BThat's not necessarily true... you could have E=B if C is exactly 18 years old, which is what I had in my answer. And the overall solution runs into a problem: A (liar) is 19 years old, but B (truthful) said C (truthful) is >18. (I'm assuming the OP doesn't want you to invoke finer granularity than integer ages and say that C's birthday is earlier than A's so they can both be 19 but with different truthiness.)
17 hours ago, plasmid said:Comment on your answer
04 [2] B: "C is more than 18 years old."
05 [A] E: "C is less than 18 years old." => (b1) E=!BThat's not necessarily true... you could have E=B if C is exactly 18 years old, which is what I had in my answer. And the overall solution runs into a problem: A (liar) is 19 years old, but B (truthful) said C (truthful) is >18. (I'm assuming the OP doesn't want you to invoke finer granularity than integer ages and say that C's birthday is earlier than A's so they can both be 19 but with different truthiness.)
Thanks for finding the problem. I was so sure that if X contradicts Y, they must be in different categories  it did not occur to me that can be both liars.

Do not worry, on my first attempt, I did not manage it to write it clearly enough that I myself could read and understand it. When you asked your question, I checked my solution written about a year ago and wondered whether it would not be easier to start from the beginning. I have rewritten it and found a kind of notation:
SpoilerMy idea was to separate them in two groups of same of same confidence and then assume one group are liars and the other tell true. It worked only partly, but it worked:
01 [5] E: "A is more than 21 years old."
02 [8] C: "A is 19 years old." => (a1) E=!C04 [2] B: "C is more than 18 years old."
05 [A] E: "C is less than 18 years old." => (b1) E=!B
06
07 Assume (c1) E=true:
08 (d1) B=!E=liar
09 (e1) C=!E=liar11 Reverse what B states:
12 [2] B: "C<=18" => C=true (The younger is true,
13 [7] B: "E>20" => E=liar the older is false.)
14 Contradicts both assumption and e1, so:16 (c2) E=false
17 (d2) B=!E=true
18 (e2) C=!E=true20 Rewriting as true statements:
21 [ 2] B: "C>18"
22 [ 3] C: "D<22"
23 [!5] E: "A<=21"
24 [ 7] B: "E<20"
25 [ 8] C: "A=19"
26 [!A] E: "C>=18"28 (L1): [7] and (c2): L<20
30 [1] A: "B>20" translates B=liar, contradicts (L1) => A=liar
31 [9] D: "B=20" translates B=liar, contradicts (L1) => D=liar33 [L1] [8]: L<20 and L>=19 > L=19.
35 A=liar, 19 [8]
36 B=true, B<=L
37 C=true, C=L [2], [A]
38 D=liar, L<D<22 [3]
39 E=liar, L=19 [8]We both excluded the possibility E being true, but considered different implications.
You will certainly get the best answer. but maybe you are wanting to rewrite it using my notation (and possibly improve it). I thought about a chart, but too much work.
P.S. Can someone change in the title ONE in ONCE?
Thanks in advance.

17 hours ago, plasmid said:
I think I've got a solution, but not everyone ends up being restricted to one exact age.
Does that mean I messed up somewhere?
Bonus question: Answer your question.
Why don't you post your solution?

On an island, every statement is true if the islander is aged less than L and false if he is at least L years old. Find their ages.
[1] A: "B is more than 20 years old."
[2] B: "C is more than 18 years old."
[3] C: "D is less than 22 years old."
[4] D: "E is not 17 years old."
[5] E: "A is more than 21 years old."
[6] A: "D is more than 16 years old."
[7] B: "E is less than 20 years old."
[8] C: "A is 19 years old."
[9] D: "B is 20 years old."
[A] E: "C is less than 18 years old." 
There be sixtyandfour flowersdeluce (in a grid 8x8), and the riddle is to show how I may remove six of these so that there may yet be an even number of the flowers in every row and every column.
I am not able to remove 6 of them: interactive version
What am I missing?
Solution. that does not help me.
I got it now, to late to delete.

I did some research...
At first, the drawing Nick made is somewhat confusing. What he calls "forward force" should be "thrust" and what he calls "thrust" should be "drag". Just google "drag thrust weight lift".
If thrust=drag and lift=weight, the plane will fly at constant speed at constant height (plenty of pages).
Now, the question is how much thrust we need to generate lift=weight (or a little better).
On the end of the article https://en.wikipedia.org/wiki/Lifttodrag_ratio, there is a table: latest aircrafts have a coefficient over 19.

I fear there are multiple solutions.
Babysnoot is correct, just the only reason to proclaim Jester truthteller is that there is no further contradiction in the system.
For Bear=truthteller, Drummer=liar, others=mix (it does not matter whether what they say is true or false), there is no contradiction in the system, neither.
Does someone agree?

A table with English words by the set of all combinations (with repetition) of 7 letters. However, even with some tricks like sorting the letters, I do not think my computer would give me the answer overnight.
Remains the MonteCarlo method. However, I am pretty sure it is not the expected answer.
A hint?

3 hours ago, harey said:
Assume that passengers 298 ask (if necessary) the forgetful passenger to leave his seat and THIS passenger then choses a seat at random.
When the last passenger boards, 98 are correctly seated. The forgetful passenger occupies either his own seat or that of the 100th passenger.
So p=50%.
Read "passenger 299" instead of "298".

Spoiler
Assume that passengers 298 ask (if necessary) the forgetful passenger to leave his seat and THIS passenger then choses a seat at random.
When the last passenger boards, 98 are correctly seated. The forgetful passenger occupies either his own seat or that of the 100th passenger.
So p=50%.

You forget the lift by air depression on the wings.
When the space shuttle (which is basically a plane) takes off vertically, as long as the thrust is less than 10 [N/kg], it does not move. When the thrust exceeds this value, it moves up. [If you then cut the engines, it slows down, stops and falls like a stone.] That's what your equations describe.
When a plane flies horizontally at a constant speed, the gravitation is compensated by depression on wings  if you cut the engines, the plane glides. You still can keep it at constant horizontal speed, it slowly loses potential energy, but does not vertically fall like a stone. This implies that on constant speed/height, you need to furnish less energy than by a vertical takeoff and therefore less thrust.
Another approach: The energy of a cruising plane is constant, so the furnished energy must be equal to the energy lost. As the lost energy is partly transformed to lift on the wings...

Spoiler
Monk M1 starts at 7:00 am at the bottom going up.
Monk M2 starts at 7:00 am at the top going down.
If there is just one path, they must meet.

And what about this:
Spoiler1: 100 pay 10 with 100
2: 50 20 20 pay 9 with 20
3: 50 20 10 1 pay 8 with 10
4: 50 20 1 1 1 pay 9 with 20
5: 50 10 1 1 1 1 pay 8 with 10
6: 50 1 1 1 1 1 1 pay 10 with 50
7: 20 20 1 1 1 1 1 1 pay 9 with 20
8: 20 10 1 1 1 1 1 1 1 pay 8 with 10
9: 20 1 1 1 1 1 1 1 1 1 pay 18 with 20
1 1 1 1 1 1 1 1 1 1 1 
Thinking it over and over again, I always finish with a problem I cannot solve.
The holes are numbered 1, 2, 3....
Three hunters check:
1 2 3
3 4 5
5 6 7
....On day n, they will have checked up to the hole 2 * n +1
The groundhog starts in the hole k and moves to the right.
On day n, he will be in the hole k+nQuestion 1: Will the hunters catch the groundhog (and if so, when)?
2 * n +1 grows faster than k + 1, I already have the answer. Nevertheless:
2 * n + 1 = k + n
n ⁼ k + 1
No matter how high the number of the starting hole the groundhog choses, it will be caught.Question 2: Is there a starting hole so that the groundhog is not discovered on day n?
k + n > 2 * n + 1
k > n + 1
No matter how long the hunters hunt, it always is possible that the groundhog started in a hole leading him outside the checked area. 
On 6/6/2020 at 12:13 PM, EventHorizon said:
Never mind how big the area you chose, I always can say "Bad luck, the groundhog is somewhere about 17 holes outside the defined area."
I agree that the size of the cleared area will > inf. Just the size is expressed as a number. No matter how big it is, there always is a bigger number.

I think I can do better:
Spoiler01: 100 pay 10 with 100
02: 50 20 20 pay 5 with 50
03: 20 20 20 20 5 pay 3 with 5
04: 20 20 20 20 1 1 pay 5 with 20
05: 20 20 20 10 5 1 1 pay 3 with 5
06: 20 20 20 10 1 1 1 1 pay 8 with 10
07: 20 20 20 1 1 1 1 1 1 pay 9 with 20
08: 20 20 10 1 1 1 1 1 1 1 pay 8 with 10
09: 20 20 1 1 1 1 1 1 1 1 1 pay 18 with 20
10: 20 1 1 1 1 1 1 1 1 1 1 1 pay 18 with 20But not sure it is the max.

Please delete.

Still confused.
SpoilerSay the numbers are 37 111 148.
On the end of the first round C thinks:
 I can have 11137=74 or 111+37=148.
 If the guy with 111 sees 37 and 74, he announces 37+74=111 (as he knows he cannot have 7437=37).
 As he did not announce 111, I only can have 148.
So why the second round?

Spoiler
Label the square ABCD clockwise starting at the upper left corner.
Put Y next to the angle of 70 (I will use both X and Y as angles and as points).
BAY=1809070=20
DAX=902045=25Rotate ABY over AY.
Rotate ADX over AX.ABY and ADX will exactly cover AXY  proof is left to the reader as exercise.
> X=1804570=65
 1
Gold chests
in New Logic/Math Puzzles
Posted
I get 13.
Take one coin of each chest.
a) You get two Gold, one Silver:
Silver is identified, remains to distinguish all_Gold and Gold_Silver.
Take 10 more coins from the first G:
 if they are all gold, you identified all_Gold, the remaining is Gold_Silver
 if you get a silver coin, you identified Gold_Silver, the remaining is all_Gold
b) You get one Gold, two Silver:
Same proceeding, permute Gold/Silver.