This is a classic puzzle which can look simple.
Given 25 horses, find the best 3 horses with minimum number of races. Each race can have only 5 horses. You don't have a timer.
Now the challange is how we can do it in 7 races.
Solution
We will have 5 races with all 25 horses
Let the results be
a1,a2,a3,a4,a5
b1,b2,b3,b4,b5
c1,c2,c3,c4,c5
d1,d2,d3,d4,d5
e1,e2,e3,e4,e5
Where a1 faster than a2 , a2 faster than a3 etc and
We need to consider only the following set of horses
a1,a2,a3,
b1,b2,b3,
c1,c2,c3,
d1,d2,d3,
e1,e2,e3,
Race 6
We race a1,b1,c1,d1 abd e1
Let speed(a1)>speed(b1)>speed(c1)>speed(d1)>speed(e1)
We get a1 as the fastest horse
We can ignore d1,d2,d3,e1,e2 and e3
a2,a3,
b1,b2,b3,
c1,c2,c3,
Race 7
Race a2,a3,b1,b2 and c1
The first and second will be second and third of the whole set
Subscribe to:
Post Comments (Atom)
6 comments:
"The second and third will be second and third of the whole set"
shouldn't it be "The FIRST and SECOND will be second and third of the whole set" ?
Thanks Shamsi,
I have corrected it.
RACE#1 - 1 2 3 4 5
RACE#2 - R1 6 7 8 9
RACE#3 - R2 R2 10 11 12
RACE#7 - R3 R3 R6 R6 15
RACE#6 - R5 R5 14 15 javascript:void(0)16
RACE#5 - R4 17 18 19 20
RACE#4 - 21 22 23 24 25
The R#(Race#) represent the winner of the previous race.
in race 1 - 3 we get the 2 best horse
in race 3 - 6 we also get the 2 best horse
so in race 7, 2 best horse in round 1-3 + 2 best horse in 4-6 + the one that did not yet compete
Why didn't you consider b3 too as the b3 < c1 relationship is not established yet??
@Akshay,
Why didn't you consider b3 too as the b3 < c1 relationship is not established yet??
Because b3 is slower than a1, b1 and b2 so it can never be in the top 3.
Post a Comment