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A divide-and-conquer algorithm handles a small input directly; otherwise it divides the input into smaller instances, recursively solves them, and combines their answers. With $a$ equally sized subproblems of size approximately $n/b$ and nonrecursive work $f(n)$, its recurrence relation has the form $T(n)=aT(n/b)+f(n)$, with constant-size base cases, $a\ge 1$ and $b>1$.

Write $p={\log }_{b}a$. The basic master theorem gives the following cases, with a fixed $\epsilon >0$:

For the common extended middle case, $f(n)=\Theta (n^p{\log }^{k}n)$ with fixed $k\ge 0$ gives $T(n)=\Theta (n^p{\log }^{k+1}n)$. A recursion tree explains the three cases: work is dominated by its leaves, spread across comparable levels, or dominated by its top levels. The regularity condition in the last case is part of the theorem, not an optional simplification.

There are three unnumbered cases inside this leaf.

For the nine-child recurrence, a recursion tree has $9^j$ subproblems of size about $n/3^j$. There are $\Theta (n^2)$ constant-size leaves, and the total unit work over earlier levels is a geometric sum of the same order. Hence $T(n)=\Theta (n^2)$. This is also the first case of the master theorem, since ${\log }_{3}9=2$ and the additive work is constant.

For the decrementing recurrence, summing from the supplied base at three givesso $T(n)=\Theta (n^2)$ here as well. Values at the smaller base indices remain one as specified.

For the recurrence that squares its recursive result, set $S(n)={\log }_{2}T(n)$. Then $S(n)=1+2S(\lfloor n/2\rfloor )$ above the base range. LetAfter $d$ expansions, $S(n)=1+2+\cdots +2^{d-1}=2^d-1$, giving the exact answerFor powers of two $n\ge 4$ this is $2^{n/2-1}$. For arbitrary positive integer $n$, the exact values form plateaux: $T(4)=\cdots =T(7)=2$, while $T(8)=\cdots =T(15)=8$. In general $T(n)=2^{\Theta (n)}$. It is not $\Theta (2^{n/2})$ uniformly over integer $n$: near the top of a plateau the exponent is closer to $n/4$. Keeping the integer halving is essential; the ordinary linear master theorem does not apply to squaring $T$.

For the first recurrence relation, the master theorem compares $n^2$ with $n^{{\log }_{2}6}$. Since $2<{\log }_{2}6$, the recursive leaves dominate and the answer is $\Theta (n^{{\log }_{2}6})$. More explicitly, the work at depth $j$ is $n^2(6/4{)}^j$, a growing geometric series that stops at depth ${\log }_{2}n-O(1)$, where its sum has that order.

For the second recurrence relation, each step subtracts two and adds one until reaching a base case. For integral $n>4$ there are $\lceil (n-4)/2\rceil$ steps, so $T(n)=1+\lceil (n-4)/2\rceil$ and the answer is $\Theta (n)$.

The third set of printed requirements is inconsistent. The specified base values and its recurrence relation give $T(n)=1$ identically, as strong induction immediately verifies: $4\cdot 1-4\cdot 1+1=1$. In particular $T(8)=1$, while the extra condition demands $T(8)>2T(4)=2$. Thus no function satisfies all the printed conditions. Dropping that extra inequality gives $\Theta (1)$.

The likely intended growth calculation can nevertheless be made precise, without treating it as a solution of the contradictory data. On dyadic arguments put $U_j=T(2^j)$. Its equation has the general formbecause $U_j-1$ has characteristic equation $(r-2{)}^2=0$. The printed values $U_1=U_2=1$ force $A=B=0$. With suitably different initial values, the condition $U_j>2U_{j-1}$ for arbitrarily large $j$ instead requires $B{2}^j-1>0$, hence $B>0$ and $U_j=\Theta (j{2}^j)$, or $T(n)=\Theta (n\log n)$. That last order describes corrected initial data, not the paper as printed. All three bullets belong to this single marked part.

The three answers, in source order, are:

The change of variable in the third case handles the additive shift in the recursive argument exactly; it is not an unjustified deletion of the +1. These Big theta notation conclusions use the usual admissible input sizes, or bounded integer rounding of subproblem sizes.

$\Theta (n^{{\log }_{2}7})$, approximately $\Theta (n^{2.8074})$, arithmetic operations. Strassen's algorithm replaces an $n\times n$ matrix multiplication by seven products of half-size blocks and a constant number of block additions and subtractions. ConsequentlyThe Master theorem applies with $a=7$, $b=2$, since $2<{\log }_{2}7$. Padding an arbitrary matrix dimension to the next power of two changes the dimension by less than a factor of two and preserves this asymptotic bound. The cost model counts scalar arithmetic as constant time; it is not a bit-complexity claim for arbitrarily large integer entries.

A recursion tree records recursive calls as nodes and their direct recursive subcalls as children. Its node labels can record the work done outside those subcalls. Summing the work over levels or charging it to nodes proves a total running time bound. For $T(n)=aT(n/b)+f(n)$, level $j$ has approximately $a^j$ instances of size $n/b^j$, so its nonrecursive work is $a^{j}f(n/b^j)$. Summing those quantities explains the cases of the master theorem.

The matrix multiplication algorithm of Volker Strassen using seven half-size block products and quadratically many scalar additions per recursion level. Its recurrence is $T(n)=7T(n/2)+\Theta (n^2)$, giving $\Theta (n^{{\log }_{2}7})$ arithmetic operations by the Master theorem.^[ref]