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Suppose you want to move from $0$ to $100$ on the number line. In each step, you either move right by a unit distance or you take a shortcut. A shortcut is simply a pre-specified pair of integers $i,\:j \:\text{with}\: i <j$. Given a shortcut $(i,j)$, if you are at position $i$ on the number line, you may directly move to $j$. Suppose $T(k)$ denotes the smallest number of steps needed to move from $k$ to $100$. Suppose further that there is at most $1$ shortcut involving any number, and in particular, from $9$ there is a shortcut to $15$. Let $y$ and $z$ be such that $T(9) = 1 + \min(T(y),T(z))$. Then the value of the product $yz$ is _____.

asked in Algorithms by Veteran (101k points)
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2 Answers

+19 votes
Best answer
$T(k)$ is the smallest number of steps needed to move from $k$ to $100$.

Now, it is given that $y$ and $z$ are two numbers such that,

$T(9) = 1 + min (T(y), T(z))$ , i.e.,

$T(9) = 1 + min ($Steps from $y$ to $100$, Steps from $z$ to $100)$, where $y$ and $z$ are two possible values that can be reached from $9$.

One number that can be reached from $9$ is $10$, which is the number obtained if we simply move one position right on the number line. Another number is $15$, the shortcut path from $9$, as given in the question. So, we have two paths from $9$, one is $10$ and the other is $15$.

Therefore, the value of $y$ and $z$ is $10$ and $15$ (either variable may take either of the values).

Thus, $yz = 150$.
answered by Loyal (8.4k points)
edited by
0
nice
0
what a nice approach @Regina phalange
+31 votes
$T(9) =$ Distance from $9$ to $100$
$T(9)=1+ \min(T(y),T(z))=1+$min(Distance from $y$ to $100$ , Distance from $z$ to $100$)

There are only two such values where we can reach from $9$ , one is simple step to right on number line , i.e $10$ and another is $15$ (given shortcut)

Hence , $y=10$ , $z=15$
$yz=10 \times 15 = 150$
answered by Active (3.6k points)
edited by
0
Can you please explain this question once again , I am not getting it precisely .
0
I am not getting .. Please explain anyone
0

it is solved like this

T(1)=1+min(T(y),T(z))=1+1=2

T(2)=1+min(T(y),T(z))=1+2=3

T(3)=1+min(T(y),T(z))=1+3=4

.

.

.

T(9)=10

Now, shortcut for T(9)=15

So, yz=10*15=150

but how could it be say , it goes from 0 to 100?

@ Srinath

+3
Solution of the equation

T(9)=1+min(T(y),T(z))

Now, 9 to 15 there is a shortcut.

Now, if I take solution like that

T(9)=1+min(T(10),T(15))

    =1+T(15)

that means , T(9) to T(15) there exists 1 step, i.e. why we are adding 1.


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