Adding and subtracting from elements in a list

Question

I'm trying to solve an aparently simple problem.

Given a list of length n, {a1,b1,c1,d1...} I want to perform two simple operations, and form a new list from each one: {a1+1,b1-1,c1,d1...}, {a1,b1+1,c1-1,d1...}, {a1,b1,c1+1,d1-1...} and so on, and {a1-1,b1+1,c1,d1...} and so on.

The list elements must not be greater or less than certain values after the addition or substraction. In {a1+1,b1-1,c1,d1...}, {a1,b1+1,c1-1,d1...}, {a1,b1,c1+1,d1-1...} each element must be constrained: a1<max and b1>0, b1<max and c1>0, etc.

Using this lists I finally use the original list and the n new lists to get a list of the form: {Join[list,new1],Join[list,new2],...}

To solve this problem I wrote a code that although works, takes a very long time to calculate for several sets of list {{a1,b1,c1,d1...}, {a2,b2,c2,d2...}...}. Here I post an example using only a list with two sublists, with four elements each one: {{1, 2, 0, 2}, {2, 2, 1, 1}}

lst1 = {{1, 2, 0, 2}, {2, 2, 1, 1}};
dim = Partition[Range[Length@First@lst1], 2, 1];
list = li[]; (*to save the "composed" lists*)

(lst = #;
lst2 = Partition[lst, 2, 1];

lst31 = MapThread[If[#1 < 2 && #2 > 0, {#1 + 1, #2 - 1}, {#1, #2}] &,Transpose@lst2];
lst41 = Union@MapThread[ReplacePart[lst, {#1[[1]]-> #2[[1]], #1[[2]]-> #2[[2]]}]&, {dim,lst31}];

lst32 = MapThread[If[#1 > 0 && #2 < 2, {#1 - 1, #2 + 1}, {#1, #2}]&,Transpose@lst2];
lst42 = Union@MapThread[ReplacePart[lst, {#1[[1]]-> #2[[1]], #1[[2]]-> #2[[2]]}]&, {dim, lst32}];

lst61 = Join[lst, #] & /@ lst41;
lst62 = Join[lst, #] & /@ lst42;
lst6 = Union@Join[lst61, lst62];
list = li[list, lst6];
) & /@ lst1;

list = List @@ Flatten@list

Any advice to improve the performance of the code would be greatly appreciated!

Edit

Based on comments, I tried to explain better the second part of my problem giving a numerical example.

Edit 2

I have added a more precise description about the constraints of the elements of the lists.

Answer 1

I don't understand the second part of your question so I shall address the first.
Here is a method that should be reasonably fast and versatile.

lst = CharacterRange["a", "j"];
n = Length @ lst;
offsets = NestList[RotateRight, PadRight[{1, -1}, n], n - 2];
lst2 = lst + # & /@ offsets;

lst2 // Column

{1+a,-1+b,c,d,e,f,g,h,i,j}
{a,1+b,-1+c,d,e,f,g,h,i,j}
{a,b,1+c,-1+d,e,f,g,h,i,j}
{a,b,c,1+d,-1+e,f,g,h,i,j}
{a,b,c,d,1+e,-1+f,g,h,i,j}
{a,b,c,d,e,1+f,-1+g,h,i,j}
{a,b,c,d,e,f,1+g,-1+h,i,j}
{a,b,c,d,e,f,g,1+h,-1+i,j}
{a,b,c,d,e,f,g,h,1+i,-1+j}

---

Here is a faster method based on techniques described here.

lst = CharacterRange["a", "j"];
n = Length @ lst;
offsets = Sum[DiagonalMatrix[SparseArray[{}, n - 1, 1 - 2 i], i, {n - 1, n}], {i, 0, 1}];
lst2 = ConstantArray[lst, n - 1] + offsets;

Answer 2

For completeness (since it's also mentioned in the link provided by Mr. Wizard), here is an approach using Band:

lst = CharacterRange["a", "j"];
Outer[Times, Most[1 & /@ lst], lst] + 
 SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> -1}, {# - 1, #} &@
   Length[lst]]

$\left( \begin{array}{cccccccccc} \text{a}+1 & \text{b}-1 & \text{c} & \text{d} & \text{e} & \text{f} & \text{g} & \text{h} & \text{i} & \text{j} \\ \text{a} & \text{b}+1 & \text{c}-1 & \text{d} & \text{e} & \text{f} & \text{g} & \text{h} & \text{i} & \text{j} \\ \text{a} & \text{b} & \text{c}+1 & \text{d}-1 & \text{e} & \text{f} & \text{g} & \text{h} & \text{i} & \text{j} \\ \text{a} & \text{b} & \text{c} & \text{d}+1 & \text{e}-1 & \text{f} & \text{g} & \text{h} & \text{i} & \text{j} \\ \text{a} & \text{b} & \text{c} & \text{d} & \text{e}+1 & \text{f}-1 & \text{g} & \text{h} & \text{i} & \text{j} \\ \text{a} & \text{b} & \text{c} & \text{d} & \text{e} & \text{f}+1 & \text{g}-1 & \text{h} & \text{i} & \text{j} \\ \text{a} & \text{b} & \text{c} & \text{d} & \text{e} & \text{f} & \text{g}+1 & \text{h}-1 & \text{i} & \text{j} \\ \text{a} & \text{b} & \text{c} & \text{d} & \text{e} & \text{f} & \text{g} & \text{h}+1 & \text{i}-1 & \text{j} \\ \text{a} & \text{b} & \text{c} & \text{d} & \text{e} & \text{f} & \text{g} & \text{h} & \text{i}+1 & \text{j}-1 \\ \end{array} \right)$