Is a stack machine with a forward read iterator Turing complete?

It is well known that a machine with a single stack as only unlimited storage is not Turing complete, if it can only read from the top of the stack. I want a machine which is (slightly) more powerful than a stack machine, but still not Turing complete. (I wonder whether there exists a non-Turing complete machine, which can deterministically simulate any non-deterministic pushdown automata with an only polynomial slow-down.) The most benign (straightforward) extension that came to my mind was a (single) forward read iterator.

Let me elaborate the implementation details, to make it clear what I mean by a forward read iterator. A singly linked list can be used for implementing a stack. Let the list be implemented by a pointer pTop, which is either zero, or points to an SList node. An SList node consists of a payload field value and a pointer field pNext, where pNext is either zero, or points to an SList node. Let the forward read iterator be implemented by a pointer pRead, which is either zero, or points to an SListnode. The pointers pTop and pRead cannot be accessed directly, but can only be used via the following methods:

• Push(val) creates a new SList node n with n.value = val and n.pNext = pTop, and sets pTop = &n.

• Pop() aborts if pTop == 0 or pRead == pTop. Otherwise it reads val = pTop->value and pTopNext = pTop->pNext, frees the SList node pointed to by pTop, sets pTop = pTopNext and returns val.

• ReadBegin() sets pRead = pTop.

• ReadNext() aborts if pRead == 0. Otherwise it reads val = pRead->value, sets pRead = pRead->pNext and returns val.

• ReadFinished() returns true if pRead == 0, and false otherwise.

Your model is Turing-complete, unfortunately.

You can simulate a queue in your data structure using the following algorithm. It introduced 3 new stack symbols: $d, x, y$.

Enqueue(val) is just Push(val).

For Dequeue():

• ReadBegin().

• Count the number of anything else - number of $d$ in the whole stack (which should be always non-negative). Push $y$ or pop $x$ for every $d$, and push $x$ or pop $y$ for anything else. Always prefer pop to push. Finally there won't be any $y$ in the stack and the result will be the number of $x$ on the top of the stack.

• ReadBegin().

• While pTop is a $x$:

• Repeat ReadNext() until it returned something other than $x$ and $d$.

• Pop().

• Push a $d$.

• The last result of ReadNext() is returned as the result of Dequeue.

The proof is straightforward. Check the revision history for a more complicated version firstly reducing it to a two-way version.

StackExchange Computer Science Q#41246, answer score: 6