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If (ij) is the exit then stop -maze successfully solved. 6. If maze [i,j + 1] is free, then add 1 to j, store East (as the move direction) and repeat from step 4. 7. if maze [i + 1,j) is free, then add 1 to i, store South and repeat from step 4. 8. If maze[i,j- 1] is free, then subtract 1 from j, store West and repeat from step 4. 9. If maze [i- 1,j] is free, then subtract 1 from i, store North and repeat from step 4. 10. A dead-end square has been reached. If at the start then no path is possible - take a helicopter out; otherwise backtrack by moving in the reverse direction to the last move, changing (i,j) appropriately and repeat from step 4.

Hence '(x*y)*x' would not work. It is true that the last unit in the routine is the name d, but one can only access its value since d is declared locally and is out of range when we leave the procedure. To obtain a ref type variable as the result needs a more advanced feature as explained in the final chapter. 11 Write a procedure with ref parameters to calculate the inner product ~ a[i] *b[i] of two vectors a, b. A procedure may have proc type variables as parameters. An earlier problem gives an example: To sum a series involving a variable until the terms are smaller than a prescribed epsilon, it is sufficient to know epsilon, the first term, the value of the variable, and a procedure that, given n, supplies a multiplying factor to obtain the (n + 1)th term from the nth.

Instead of repeatedly testing whether a move strays out of bounds, we plant an extra hedge right around the maze. This means that more space will be required to represent the maze, but time is saved when running the program as the test statements and their repetition are removed. The trade-off between space and time is frequently encountered in program construction. To plant this hed,~e, if the maze had bounds 1 :m, 1:n then we must declare an enlarged maze with bounds O:m + 1, O:n + 1. Rows 0 and m + 1, and columns 0 and n + 1 have to be filled with obstacles.