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In 2.42， every round minus 1. so it is (n!).
In 2.43， every round is n. so it is (n^n).
for board-size=8, (8^8)/(8!)≈416T
Not solved yet but share my thinking process:
Q(k): number of solutions of board size n * k
T(k): number of steps required to calculate Q(k)
original approach: T(k) = T(k-1) + n * Q(k-1)
T(0) = 1
louis's approach: T(k) = n * (T(k-1) + Q(k-1)) = n * T(k-1) + n * Q(k-1)
T(0) = 1
So louis's approach requires signifcantly more steps. How many more? It's tempted to say a factor of N^N due to tree recursion but it may not be the actual case.
Found this analysis for reference: https://wernerdegroot.wordpress.com/2015/08/01/sicp-exercise-2-43/
So, I arrived at the same result as @woofy and in the end I got these results:
Time complexity of Ex2.42 queens is n^4 Time complexity of Louis Reasoner's queens is n^n
So, roughly for Louis's approach time taken will be n^(n-4)T for n>4 and equal to T for n<=4. These results are confirmed by actually calculating execution time of both procedures and comparing them.
Given n is input size and k is k as defined in recursive iterations of
(queen-cols (- k 1))
original solution:
T(k) = T(k-1)*n, T(0) = 1
T(n) = n*T(n-1) + n*T(n-2) + ... + 1
T(n) = n!n
Louis's solution:
τ(k) = τ(k-1)^n, τ(0)=1
τ(k) = τ(k-1)^n + τ(n-2)^n + ... + 1
τ(k) = n^n
Comparing these solutions:
τ/T, τ=n^n, T=n!n
τ/T = n^n/n!n
τ/T = n^(n-1)/n! ≈ n^n
So, Louis's solution is n^(n-1)/n! times less efficient than the original which is exponentially worse on the order of n^n
Find the link given by @woofy helpful: https://wernerdegroot.wordpress.com/2015/08/01/sicp-exercise-2-43/
Exchanging the order of the mapping in the flatmap results in queen-cols being re-evaluated for every item in (enumerate-interval 1 board-size). Therefore the whole work has to be duplicated board-size times at every recursion level. Since there are always board-size recursions this means that the whole work will be duplicated bord-size^board-size times.
Therefore if the function would take T time to run for board-size=8 with correct ordering, with the interchanged ordering it will take (8^8)T to run.