{
"translations" :{
"en": {
"sc1": "To find the cost of binary search in the worst case, we can model the running time as a recurrence and then find the closed-form solution. Each recursive call to binarySearch cuts the size of the array approximately in half, so we can model the worst-case cost as follows, assuming for simplicity that $n$ is a power of two.
$T(n) = T(n/2) + 1, T(1) = 1$",
"sc2.1": "If we expand the recurrence, we will find that we can do so only $\\log n$ times before we reach the base case, and each expansion adds one to the cost. For a problem of size $n$, we have $1$ unit of work plus the amount of work required for one subproblem of size $n/2$",
"sc2.2": "
$T(n) = 1 + T(n/2)$",
"sc3.1": "For a problem of size $n/2$, we have $1$ unit of work plus the amount of work required for one subproblem of size $n/4$",
"sc3.2": "
$T(n) = 1 + (1 + T(n/4))$",
"sc4.1": "For a problem of size $n/4$, we have $1$ unit of work plus the amount of work required for one subproblem of size $n/8$",
"sc4.2": "
$T(n) = 1 + (1 + (1 + T(n/8)))$",
"sc5.1": "For a problem of size $n/8$, we have $1$ unit of work plus the amount of work required for one subproblem of size $n/16$",
"sc5.2": "
$T(n) = 1 + (1 + (1 + (1 + T(n/16)))$",
"sc6.1": "This pattern will continue till we reach a subproblem of size $1$",
"sc6.2": "
$T(n) = 1 + (1 + (1 + (1 + (1 + (...))))$",
"sc7": "Thus, the closed form solution of $T(n) = T(n/2) + 1$ can be modeled by the summation $\\displaystyle\\sum_{i=0}^{\\log{n}}1$",
"sc8": "Finally, we have the closed form solution of $T(n) = T(n/2) + 1,\\ T(1) = 1$ evaluates to $\\log{n} + 1$ or $\\Theta(\\log{n})$"
}
}
}