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A Child and a Staircase

15 replies to this topic

#11 Rockmond

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Posted 21 June 2012 - 05:21 PM

I have a clarification question. If there are 10 steps and a landing at the bottom and the top, the child would have to take 11 steps if taken one at a time, right?
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#12 jim

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Posted 22 June 2012 - 01:38 AM

442.
Since we can solve this with either Fibonacii numbers or combinations, we can conclude that
F10 = C(10,0) + C(9.1)+ ...+c(5.5).

More generally Fn = C(n,0) + C(n-1.1) + C(n-2,2) + ... as long as the first number is no les than the second.

NOTE If you interpet the question to mean 10 steps plus the upper landing the answer would be F11, a point one poster raised. In addition if we distinquish between first strride right foot and first stride left we would double our answer.
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#13 bonanova

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Posted 22 June 2012 - 04:56 AM

Congrats to witzar! I just counted them.
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Vidi vici veni.

#14 ujjagrawal

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Posted 22 June 2012 - 05:36 AM

Solved. Congrats to all of you who got it correct.

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#15 wilbloodworth

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Posted 06 January 2013 - 05:27 PM

I think 89 is correct.  Here's a simple C# implementation that you can test out if you want to use LinqPad:

void Main()
{
countWays(10).Dump();
}

int countWays(int steps)
{
if (steps < 0)
return 0;
if (steps == 0)
return 1;

if (_map.ContainsKey(steps))
return _map[steps];

_map[steps] = countWays(steps - 1) + countWays(steps - 2);

return _map[steps];
}

private readonly Dictionary<int, int> _map = new Dictionary<int, int>();

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#16 wolfgang

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Posted 06 January 2013 - 06:31 PM

Spoiler for I think...they should be

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