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We are given an array with n values.
Example: [1,4,5,6,6]
For each index i of the array a ,we construct a new element of array b such that,
b[i]= [a[i]/1] + [a[i+1]/2] + [a[i+2]/3] + ⋯ + [a[n]/(n−i+1)] where [.] denotes the greatest integer function.
We are given an integer k as well.
We have to find the minimum i such that b[i] ≤ k.
I know the brute-force O(n^2) algorithm (to create the array - 'b'), can anybody suggest a better time complexity and way solve it?
For example, for the input [1,2,3],k=3, the output is 1(minimum-index).
Here, a[1]=1; a[2]=2; a[3]=3;
Now, b[1] = [a[1]/1] + [a[2]/2] + [a[3]/3] = [1/1] + [2/2] + [3/3] = 3;
b[2] = [a[2]/1] + [a[3]/2] = [2/1] + [3/2] = 3;
b[3] = [a[3]/1] = [3/1] = 3 (obvious)
Now, we have to find the index i such that b[i]<=k , k='3' , also b[1]<=3, henceforth, 1 is our answer! :-)
Constraints : - Time limits: -(2-seconds) , 1 <= a[i] <= 10^5, 1 <=
n <= 10^5, 1 <= k <= 10^9
Here's an O(n √A)-time algorithm to compute the b array where n is the number of elements in the a array and A is the maximum element of the a array.
This algorithm computes the difference sequence of the b array (∆b = b[0], b[1] - b[0], b[2] - b[1], ..., b[n-1] - b[n-2]) and derives b itself as the cumulative sums. Since the differences are linear, we can start with ∆b = 0, 0, ..., 0, loop over each element a[i], and add the difference sequence for [a[i]], [a[i]/2], [a[i]/3], ... at the appropriate spot. The key is that this difference sequence is sparse (less than 2√a[i] elements). For example, for a[i] = 36,
>>> [36//j for j in range(1,37)]
[36, 18, 12, 9, 7, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
>>> list(map(operator.sub,_,[0]+_[:-1]))
[36, -18, -6, -3, -2, -1, -1, -1, 0, -1, 0, 0, -1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
We can derive the difference sequence from a subroutine that, given a positive integer r, returns all maximal pairs of positive integers (p, q) such that pq ≤ r.
See complete Python code below.
def maximal_pairs(r):
p = 1
q = r
while p < q:
yield (p, q)
p += 1
q = r // p
while q > 0:
p = r // q
yield (p, q)
q -= 1
def compute_b_fast(a):
n = len(a)
delta_b = [0] * n
for i, ai in enumerate(a):
previous_j = i
for p, q in maximal_pairs(ai):
delta_b[previous_j] += q
j = i + p
if j >= n:
break
delta_b[j] -= q
previous_j = j
for i in range(1, n):
delta_b[i] += delta_b[i - 1]
return delta_b
def compute_b_slow(a):
n = len(a)
b = [0] * n
for i, ai in enumerate(a):
for j in range(n - i):
b[i + j] += ai // (j + 1)
return b
for n in range(1, 100):
print(list(maximal_pairs(n)))
lst = [1, 34, 3, 2, 9, 21, 3, 2, 2, 1]
print(compute_b_fast(lst))
print(compute_b_slow(lst))
This probably cannot reach the efficiency of David Eisenstat's answer but since I spent quite a long time figuring out an implementation, I thought I'd leave it up anyway. As it is, it seems about O(n^2).
The elements of b[i] may be out of order, but sections of them are not:
[a[1]/1] + [a[2]/2] + [a[3]/3]
|------ s2_1 -----|
|-s1_1-|
[a[2]/1] + [a[3]/2]
|------ s2_2 -----|
|-s1_2-|
[a[3]/1]
|-s1_3-|
s2_1 < s2_2
s1_1 < s1_2 < s1_3
Binary search for k on s1. Any result with an s1_i greater than k will rule out a section of ordered rows (rows are b_is).
Binary search for k on s2 on the remaining rows. Any result with an s2_i greater than k will rule out a section of ordered rows (rows are b_is).
This wouldn't help much since in the worst case, we'd have O(n^2 * log n) complexity, greater than O(n^2).
But we can also search horizontally. If we know that b_i ≤ k, then it will rule out both all rows with greater or equal length and the need to search smaller s(m)s, not because smaller s(m)s cannot produce a sum >= k, but because they will necessarily produce one with a higher i and we are looking for the minimum i.
JavaScript code:
var sum_width_iterations = 0
var total_width_summed = 0
var sum_width_cache = {}
function sum_width(A, i, width){
let key = `${i},${width}`
if (sum_width_cache.hasOwnProperty(key))
return sum_width_cache[key]
sum_width_iterations++
total_width_summed += width
let result = 0
for (let j=A.length-width; j<A.length; j++)
result += ~~(A[j] / (j + 1 - i))
return sum_width_cache[key] = result
}
function get_b(A){
let result = []
A.map(function(a, i){
result.push(sum_width(A, i, A.length - i))
})
return result
}
function find_s_greater_than_k(A, width, low, high, k){
let mid = low + ((high - low) >> 1)
let s = sum_width(A, mid, width)
while (low <= high){
mid = low + ((high - low) >> 1)
s = sum_width(A, mid, width)
if (s > k)
high = mid - 1
else
low = mid + 1
}
return [mid, s]
}
function f(A, k, l, r){
let n = A.length
if (l > r){
console.log(`l > r: l, r: ${l}, ${r}`)
return [n + 1, Infinity]
}
let width = n - l
console.log(`\n(call) width, l, r: ${width}, ${l}, ${r}`)
let mid = l + ((r - l) >> 1)
let mid_width = n - mid
console.log(`mid: ${mid}`)
console.log('mid_width: ' + mid_width)
let highest_i = n - mid_width
let [i, s] = find_s_greater_than_k(A, mid_width, 0, highest_i, k)
console.log(`hi_i, s,i,k: ${highest_i}, ${s}, ${i}, ${k}`)
if (mid_width == width)
return [i, s]
// either way we need to look left
// and down
console.log(`calling left`)
let [li, ls] = f(A, k, l, mid - 1)
// if i is the highest, width is
// the width of b_i
console.log(`got left: li, ls, i, high_i: ${li}, ${ls}, ${i}, ${highest_i}`)
if (i == highest_i){
console.log(`i == highest_i, s <= k: ${s <= k}`)
// b_i is small enough
if (s <= k){
if (ls <= k)
return [li, ls]
else
return [i, s]
// b_i is larger than k
} else {
console.log(`b_i > k`)
let [ri, rs] = f(A, k, mid + 1, r)
console.log(`ri, rs: ${ri}, ${rs}`)
if (ls <= k)
return [li, ls]
else if (rs <= k)
return [ri, rs]
else
return [i, s]
}
// i < highest_i
} else {
console.log(`i < highest_i: high_i, i, s, li, ls, mid, mid_width, width, l, r: ${highest_i}, ${i}, ${s}, ${li}, ${ls}, ${mid}, ${mid_width}, ${width}, ${l}, ${r}`)
// get the full sum for this b
let b_i = sum_width(A, i, n - i)
console.log(`b_i: ${b_i}`)
// suffix sum is less than k
// so we cannot rule out either side
if (s < k){
console.log(`s < k`)
let ll = l
let lr = mid - 1
let [lli, lls] = f(A, k, ll, lr)
console.log(`ll, lr, lli, lls: ${ll}, ${lr}, ${lli}, ${lls}`)
// b_i is a match so we don't
// need to look to the right
if (b_i <= k){
console.log(`b_i <= k: i, b_i: ${i}, ${b_i}`)
if (lls <= k)
return [lli, lls]
else
return [i, b_i]
// b_i > k
} else {
console.log(`b_i > k: i, b_i: ${i}, ${b_i}`)
let rl = mid + 1
let rr = r
let [rri, rrs] = f(A, k, rl, rr)
console.log(`rl, rr, rri, rrs: ${rl}, ${rr}, ${rri}, ${rrs}`)
// return the best of right
// and left sections
if (lls <= k)
return [lli, lls]
else if (rrs <= k)
return [rri, rrs]
else
return [i, b_i]
}
// suffix sum is greater than or
// equal to k so we can rule out
// this and all higher rows (`b`s)
// that share this suffix
} else {
console.log(`s >= k`)
let ll = l
// the suffix rules out b_i
// and above
let lr = i - 1
let [lli, lls] = f(A, k, ll, lr)
console.log(`ll, lr, lli, lls: ${ll}, ${lr}, ${lli}, ${lls}`)
let rl = highest_i + 1
let rr = r
let [rri, rrs] = f(A, k, rl, rr)
console.log(`rl, rr, rri, rrs: ${rl}, ${rr}, ${rri}, ${rrs}`)
// return the best of right
// and left sections
if (lls <= k)
return [lli, lls]
else if (rrs <= k)
return [rri, rrs]
else
return [i, b_i]
}
}
}
let lst = [1, 2, 3, 1]
// b [3, 3, 3, 1]
lst = [ 1, 34, 3, 2, 9, 21, 3, 2, 2, 1]
// b [23, 41, 12, 13, 20, 22, 4, 3, 2, 1]
console.log(
JSON.stringify(f(lst, 20, 0, lst.length)))
console.log(`sum_width_iterations: ${sum_width_iterations}`)
console.log(`total_width_summed: ${total_width_summed}`)
Why should calculating b[i] lead to O(n²)? If i = 1, it takes n steps. If i = n, it takes one step to calculate b[i]...
You could improve your calculation when you abort the sum on the condition Sum > k.
Let a in N^n
Let k in N
for (i1 := 1; i1 <= n; i1++)
b := 0
for (i2 :=i1; i2 <= n; i2++) // This loop is the calculation of b[i]
b := b + ceil(a[i2]/(i2 + 1))
if (b > k)
break
if (i2 == n)
return i1
I am studying the flow of control using recursion in Merge Sort.
The particular algorithm I used is:
MergeS(ar, p, r){
1. if p<r{
2. k = floor[(p+r)/2]
3. MergeS(ar, p , k)
4. MergeS(ar, k+1, r) //in the schematic diagram I have written this as mergeS(,,)
5. Merge(ar, p ,k, r)
6. }
7.}
Merge(ar, p, k, r){
8. n1 = k-p+1
9. n2 = r-p
10. let L[1...n1+1] and R[1....n2+1] be new arrays
11. for i=1 to n1
12. L[i] = ar[p+i-1]
13. for j=1 to n2
14. R[j] = ar[k+j]
15. L[n1+1] = Infinity
16. R[n2+1] = Infinity
17. i = 1
18. j = 1
19. for t = p to r
20. if L[i] <= R[j]
21. A[t] = L[i]
22. i = i+1
23. else
24. A[t] = R[j]
25. j = j+1
30. }
MergeS(ar, k+1, r) is exactly same as merge(ar, k+1, r). I used the lowercase m in the latter only for better visual clarity in the schematic diagram.
For understanding purpose, I took an array example-- 43, 32, 56, 12, 4.
The flow of recursion calls is schematically shown below till the first call to Merge is made and the code inside that block is executed.
---1:MergeS(ar, 0, 5)-----
if(0<5):true ; set k=(0+5)/2
---Pause 1:MergeS(ar, 0, 5)-----Call-->2:MergeS(ar, 0, 2)---------
if(0<2):true ; set k=(0+2)/2
---Pause 2:MergeS(ar, 0, 2)-----Call-->3:MergeS(ar, 0, 1)---------
if(0<1):true ; set k=(0+1)/2
---Pause 3:MergeS(ar, 0, 1)-----Call-->4:MergeS(ar, 0, 0)---------
if(0<0):false
---Resume 3:MergeS(ar, 0, 1)-----Call-->5:mergeS(ar, 0+1, 5)---------
if(1<5):true ; set k=(1+5)/2
---Pause 5:mergeS(ar, 0+1, 5)-----Call-->6:MergeS(ar, 1, 3)---------
if(1<3):true ; set k=(1+3)/2
---Pause 6:MergeS(ar, 1, 3)-----Call-->7:MergeS(ar, 1, 2)---------
if(1<2):true ; set k=(1+2)/2
---Pause 7:MergeS(ar, 1, 2)-----Call-->8:MergeS(ar, 1, 1)---------
if(1<1):false
---Resume 7:MergeS(ar, 1, 2)------Call-->9:mergeS(ar, 1+1, 5)---------
if(2<5):true ; set k=(2+5)/2
---Pause 9:mergeS(ar, 2, 5)--------Call-->10:Merge(ar, 2, 3)---------
if(2<3):true ; set k=(2+3)/2
---Pause 10:Merge(ar, 2, 3)-------Call-->11:Merge(ar, 2, 2)----------
if(2<2):false
---Resume 10:MergeS(ar, 2, 3)-----Call-->12:mergeS(ar, 2+1, 5)----------
if(3<5):true ; set k=(3+5)/2
---Pause 12:mergeS(ar, 2+1, 5)------Call-->13:MergeS(ar, 5, 5 )---------
if(5<5):false
---Resume 12:mergeS(ar, 2+1, 5)-------Call-->Merge(ar, 3, 4, 5)----------
n1 = 4-3+1=2
n2 = 5-4=1
for(i = 0 to i=1):
iteration1:
L[0] = ar[3+0-1]= ar[2] //value 56 is assigned to ar[2]
iteration2:
L[1] = ar[3+1-1] = ar[3] //value 12 is assigned to ar[3]
for(j=0 to j=0):
iteration1:
R[0] = ar[4+0] = ar[4] //value 4 is assigned to ar[4]
x = 0;
y = 0;
for(t=0 to t=1):
iteration1:
/* if(L[0]<= R[0]) //56<=4:false */
else
ar[0] = R[0] //4 is assigned to ar[0]
y = y+1
iteration2:
if(L[0]<=R[1]): //56<=infinity
ar[1] = L[0] // 56 is assigned to ar[1]
After the preceding step, I got a value 4 in ar[0] and a value 56 in ar[2]. My guess is that this is wrong.
I need help understanding where I went wrong and and an explanation for that.
The indices in the algorithm in the question need to be modified in the code for simpler use. I am not saying that the algorithm is wrong, it's absolutely correct, only needs modification.
The tweaked code of Merge:
private static void Merge(int[] ar, int p, int k, int r) {
int n1 = k-p+1;
int n2 = r-k;
int[] L = new int[n1+1];
int[] R = new int[n2+1];
for(int i=0; i<n1; i++){
L[i] = ar[p+i]; // p+i and not p+i-1
}
for(int i=0; i<n2; i++){
R[i] = ar[k+i+1]; //k+1+i and not k+i
}
// Use of Tnteger.MAX_VALUE
L[n1]=Integer.MAX_VALUE;
R[n2]=Integer.MAX_VALUE;
int i = 0;
int j = 0;
for(int t=p; t<=r; t++){ //carefully set initialization and t<=r and not simply t<r
if(L[i]<r[j]){
ar[t] = L[i];
i++;
}
else{
ar[t] = R[j];
j++;
}
}
}
}
The correct flow of recursion calls is shown below:(the changes from the above flow of control diagram are italicised)
---1:MergeS(ar, 0, 5)-----
if(0<5):true ; set k=(0+5)/2
---Pause 1:MergeS(ar, 0, 5)-----Call-->2:MergeS(ar, 0, 2)---------
if(0<2):true ; set k=(0+2)/2
---Pause 2:MergeS(ar, 0, 2)-----Call-->3:MergeS(ar, 0, 1)---------
if(0<1):true ; set k=(0+1)/2
---Pause 3:MergeS(ar, 0, 1)-----Call-->4:MergeS(ar, 0, 0)---------
if(0<0):false
---Break 4:MergeS(ar, 0, 0)-----Call-->5:mergeS(ar, 0+1, 1)---------
if(1<_1_):_false_ ;
---Break 5:mergeS(ar, 0+1, 1)-----Call-->Merge(ar,0,0,1)
After the execution of Merge function with parameters ar, 0, 0, 1, the elements at first two indices of the array 43 32 56 12 4 are swapped.
Maintaining a stack of function calls came in handy.
I'm learning about dynamic programming via the 0-1 knapsack problem.
I'm getting some weird Nulls out from the function part1. Like 3Null, 5Null etc. Why is this?
The code is an implementation of:
http://www.youtube.com/watch?v=EH6h7WA7sDw
I use a matrix to store all the values and keeps, dont know how efficient this is since it is a list of lists(indexing O(1)?).
This is my code:
(* 0-1 Knapsack problem
item = {value, weight}
Constraint is maxweight. Objective is to max value.
Input on the form:
Matrix[{value,weight},
{value,weight},
...
]
*)
lookup[x_, y_, m_] := m[[x, y]];
part1[items_, maxweight_] := {
nbrofitems = Dimensions[items][[1]];
keep = values = Table[0, {j, 0, nbrofitems}, {i, 1, maxweight}];
For[j = 2, j <= nbrofitems + 1, j++,
itemweight = items[[j - 1, 2]];
itemvalue = items[[j - 1, 1]];
For[i = 1, i <= maxweight, i++,
{
x = lookup[j - 1, i, values];
diff = i - itemweight;
If[diff > 0, y = lookup[j - 1, diff, values], y = 0];
If[itemweight <= i ,
{If[x < itemvalue + y,
{values[[j, i]] = itemvalue + y; keep[[j, i]] = 1;},
{values[[j, i]] = x; keep[[j, i]] = 0;}]
},
y(*y eller x?*)]
}
]
]
{values, keep}
}
solvek[keep_, items_, maxweight_] :=
{
(*w=remaining weight in knapsack*)
(*i=current item*)
w = maxweight;
knapsack = {};
nbrofitems = Dimensions[items][[1]];
For[i = nbrofitems, i > 0, i--,
If[keep[[i, w]] == 1, {Append[knapsack, i]; w -= items[[i, 2]];
i -= 1;}, i - 1]];
knapsack
}
Clear[keep, v, a, b, c]
maxweight = 5;
nbrofitems = 3;
a = {5, 3};
b = {3, 2};
c = {4, 1};
items = {a, b, c};
MatrixForm[items]
Print["Results:"]
results = part1[items, 5];
keep = results[[1]];
Print["keep:"];
Print[keep];
Print["------"];
results2 = solvek[keep, items, 5];
MatrixForm[results2]
(*MatrixForm[results[[1]]]
MatrixForm[results[[2]]]*)
{{{0,0,0,0,0},{0,0,5 Null,5 Null,5 Null},{0,3 Null,5 Null,5 Null,8 Null},{4 Null,4 Null,7 Null,9 Null,9 Null}},{{0,0,0,0,0},{0,0,Null,Null,Null},{0,Null,0,0,Null},{Null,Null,Null,Null,Null}}}
While your code gives errors here, the Null problem occurs because For[] returns Null. So add a ; at the end of the outermost For statement in part1 (ie, just before {values,keep}.
As I said though, the code snippet gives errors when I run it.
In case my answer isn't clear, here is how the problem occurs:
(
Do[i, {i, 1, 10}]
3
)
(*3 Null*)
while
(
Do[i, {i, 1, 10}];
3
)
(*3*)
The Null error has been reported by acl. There are more errors though.
Your keep matrix actually contains two matrices. You need to call solvek with the second one: solvek[keep[[2]], items, 5]
Various errors in solvek:
i -= 1 and i - 1 are more than superfluous (the latter one is a coding error anyway). The i-- in the beginning of the For is sufficient. As it is now you're decreasing i twice per iteration.
Append must be AppendTo
keep[[i, w]] == 1 must be keep[[i + 1, w]] == 1 as the keep matrix has one more row than there are items.
Not wrong but superfluous: nbrofitems = Dimensions[items][[1]]; nbrofitems is already globally defined
The code of your second part could look like:
solvek[keep_, items_, maxweight_] :=
Module[{w = maxweight, knapsack = {}, nbrofitems = Dimensions[items][[1]]},
For[i = nbrofitems, i > 0, i--,
If[keep[[i + 1, w]] == 1, AppendTo[knapsack, i]; w -= items[[i, 2]]]
];
knapsack
]
I have aproblem:
Thread::tdlen: Objects of unequal length in {Null} {} cannot be combined. >>
It seems to occur in the while test which makes no sense at all since I am onlu comparing numbers...?
The program is a program to solve the 0-1 knapsack dynamic programming problem though I use loops, not recursion.
I have put some printouts and i can only think that the problem is in the while loop and it doesnt make sense.
(* 0-1 Knapsack problem
item = {value, weight}
Constraint is maxweight. Objective is to max value.
Input on the form:
Matrix[{value,weight},
{value,weight},
...
]
*)
lookup[x_, y_, m_] := m[[x, y]];
generateTable[items_, maxweight_] := {
nbrofitems = Dimensions[items][[1]];
keep = values = Table[0, {j, 0, nbrofitems}, {i, 1, maxweight}];
For[j = 2, j <= nbrofitems + 1, j++,
itemweight = items[[j - 1, 2]];
itemvalue = items[[j - 1, 1]];
For[i = 1, i <= maxweight, i++,
{
x = lookup[j - 1, i, values];
diff = i - itemweight;
If[diff > 0, y = lookup[j - 1, diff, values], y = 0];
If[itemweight <= i ,
{If[x < itemvalue + y,
{values[[j, i]] = itemvalue + y; keep[[j, i]] = 1;},
{values[[j, i]] = x; keep[[j, i]] = 0;}]
},
y(*y eller x?*)]
}
]
];
{values, keep}
}
pickItems[keep_, items_, maxweight_] :=
{
(*w=remaining weight in knapsack*)
(*i=current item*)
w = maxweight;
knapsack = {};
nbrofitems = Dimensions[items][[1]];
i = nbrofitems + 1;
x = 0;
While[i > 0 && x < 10,
{
Print["lopp round starting"];
x++;
Print["i"];
Print[i];
Print["w"];
Print[w];
Print["keep[i,w]"];
Print[keep[[i, w]]];
If[keep[[i, w]] == 1,
{Append[knapsack, i];
Print["tjolahej"];
w -= items[[i - 1, 2]];
i -= 1;
Print["tjolahopp"];
},
i -= 1;
];
Print[i];
Print["loop round done"];
}
knapsack;
]
}
Clear[keep, v, a, b, c]
maxweight = 5;
nbrofitems = 3;
a = {5, 3};
b = {3, 2};
c = {4, 1};
items = {a, b, c};
MatrixForm[items]
results = generateTable[items, 5];
keep = results[[1]][[2]];
Print["keep:"];
MatrixForm[keep]
Print["------"];
results2 = pickItems[keep, items, 5];
MatrixForm[results2]
This is not really an answer to the specific question being asked, but some hints on general situations when this error occurs. The short answer is that this is a sign of passing lists of unequal lengths to some Listable function, user-defined or built-in.
Many of Mathematica's built-in functions are Listable(have Listable attribute). This basically means that, given lists in place of some or all arguments, Mathematica automatically threads the function over them. What really happens is that Thread is called internally (or, at least, so it appears). This can be illustrated by
In[15]:=
ClearAll[f];
SetAttributes[f,Listable];
f[{1,2},{3,4,5}]
During evaluation of In[15]:= Thread::tdlen: Objects of unequal length in
f[{1,2},{3,4,5}] cannot be combined. >>
Out[17]= f[{1,2},{3,4,5}]
You can get the same behavior by using Thread explicitly:
In[19]:=
ClearAll[ff];
Thread[ff[{1,2},{3,4,5}]]
During evaluation of In[19]:= Thread::tdlen: Objects of unequal length in
ff[{1,2},{3,4,5}] cannot be combined. >>
Out[20]= ff[{1,2},{3,4,5}]
In case of Listable functions, this is a bit more hidden though. Some typical examples would include things like {1, 2} + {3, 4, 5} or {1, 2}^{3, 4, 5} etc. I discussed this issue in a bit more detail here.
Try this version:
pickItems[keep_, items_, maxweight_] := Module[{},
{(*w=remaining weight in knapsack*)(*i=current item*)w = maxweight;
knapsack = {};
nbrofitems = Dimensions[items][[1]];
i = nbrofitems + 1;
x = 0;
While[i > 0 && x < 10,
{
Print["lopp round starting"];
x++;
Print["i"];
Print[i];
Print["w"];
Print[w];
Print["keep[i,w]"];
Print[keep[[i, w]]];
If[keep[[i, w]] == 1,
{
Append[knapsack, i];
Print["tjolahej"];
w -= items[[i - 1, 2]];
i -= 1;
Print["tjolahopp"];
},
i -= 1;
];
Print[i];
Print["loop round done"]
};
knapsack
]
}
]
no errors now, but I do not know what it does really :)
Can you do something like Python's yield statement in Mathematica, in order to create generators? See e.g. here for the concept.
Update
Here's an example of what I mean, to iterate over all permutations, using only O(n) space: (algorithm as in Sedgewick's Algorithms book):
gen[f_, n_] := Module[{id = -1, val = Table[Null, {n}], visit},
visit[k_] := Module[{t},
id++; If[k != 0, val[[k]] = id];
If[id == n, f[val]];
Do[If[val[[t]] == Null, visit[t]], {t, 1, n}];
id--; val[[k]] = Null;];
visit[0];
]
Then call it it like:
gen[Print,3], printing all 6 permutations of length 3.
As I have previously stated, using Compile will given faster code. Using an algorithm from fxtbook, the following code generates a next partition in lexicographic ordering:
PermutationIterator[f_, n_Integer?Positive, nextFunc_] :=
Module[{this = Range[n]},
While[this =!= {-1}, f[this]; this = nextFunc[n, this]];]
The following code assumes we run version 8:
ClearAll[cfNextPartition];
cfNextPartition[target : "MVM" | "C"] :=
cfNextPartition[target] =
Compile[{{n, _Integer}, {this, _Integer, 1}},
Module[{i = n, j = n, ni, next = this, r, s},
While[Part[next, --i] > Part[next, i + 1],
If[i == 1, i = 0; Break[]]];
If[i == 0, {-1}, ni = Part[next, i];
While[ni > Part[next, j], --j];
next[[i]] = Part[next, j]; next[[j]] = ni;
r = n; s = i + 1;
While[r > s, ni = Part[next, r]; next[[r]] = Part[next, s];
next[[s]] = ni; --r; ++s];
next
]], RuntimeOptions -> "Speed", CompilationTarget -> target
];
Then
In[75]:= Reap[PermutationIterator[Sow, 4, cfNextPartition["C"]]][[2,
1]] === Permutations[Range[4]]
Out[75]= True
This is clearly better in performance than the original gen function.
In[83]:= gen[dummy, 9] // Timing
Out[83]= {26.067, Null}
In[84]:= PermutationIterator[dummy, 9, cfNextPartition["C"]] // Timing
Out[84]= {1.03, Null}
Using Mathematica's virtual machine is not much slower:
In[85]:= PermutationIterator[dummy, 9,
cfNextPartition["MVM"]] // Timing
Out[85]= {1.154, Null}
Of course this is nowhere near C code implementation, yet provides a substantial speed-up over pure top-level code.
You probably mean the question to be more general but the example of iterating over permutations as given on the page you link to happens to be built in to Mathematica:
Scan[Print, Permutations[{1, 2, 3}]]
The Print there can be replaced with any function.