## OpenJML Examples: Verify This 2019 2A

This is Challenge 2A from the 2019 VerifyThis competition.

The full text of the challenge is here.

The solution is here.

In brief, given an array of integers, for each position of the array find the nearest position with a smaller index that has a smaller value.

``````
// This is Challenge 2A from VerifyThis 2019

public class Challenge2A2019 {

//@ public normal_behavior
//@ assigns \nothing;
//@ ensures \result.length == input.length;
//@ ensures \forall int i; 0 <= i < \result.length; \result[i] < i; // left-neighbor is to the left
//@ ensures \forall int i; 0 <= i < \result.length; \result[i] != -1 ==> input[\result[i]] < input[i]; // LN has smaller value
//@ ensures \forall int i; 0 <= i < \result.length; \forall int j; \result[i] < j < i; input[j] >= input[i]; // LN is closest smaller value
public int[] leftNeighbors(int[] input) {

int[] left = new int[input.length];
int[] stack = new int[input.length];
int height = 0;

//@ loop_invariant 0 <= x <= input.length;
//@ loop_invariant 0 <= height <= x;
//@ loop_invariant \forall int i; 0 <= i < x; left[i] < i;  // so far, all left-neighbors are to the left
//@ loop_invariant \forall int i; 0 <= i < x; left[i] != -1 ==> input[left[i]] < input[i]; // so far, all LNs have smaller values
//@ loop_invariant \forall int i; 0 <= i < x; \forall int j; left[i] < j < i; input[j] >= input[i]; // so far, all LNs are closest smaller values
//@ loop_invariant \forall int i; 0 <= i < height; 0 <= stack[i] < x; // all stack values are legitimate positions
//@ loop_invariant x > 0 ==> height > 0;
//@ loop_invariant height > 0 ==> stack[height-1] == x-1; // x is always one more than top of stack (after the first iteration)
//@ loop_invariant \forall int i; 1 <= i < height; \forall int j; stack[i-1] < j < stack[i]; input[j] >= input[stack[i]]; // items missing from stack are larger than something to their right
//@ loop_invariant height > 0 ==> \forall int j; 0 <= j < stack; input[j] >= input[stack]; // items missing from the stack in the first segment are larger than stack
//@ loop_decreases input.length - x;
for (int x=0; x<input.length; x++) {
//@ loop_invariant 0 <= height <= x;
//@ loop_invariant \forall int i; 0 <= i < x; left[i] < i;   // so far, all left-neighbors are to the left
//@ loop_invariant \forall int i; 0 <= i < height; 0 <= stack[i] < x;  // so far, all LNs have smaller values
//@ loop_invariant \forall int i; 1 <= i < height; \forall int j; stack[i-1] < j < stack[i]; input[j] >= input[stack[i]];  // so far, all LNs are closest smaller values
//@ loop_invariant height > 0 ==> \forall int j; stack[height-1] < j < x; input[j] >= input[x]; // Everything between top of stack and x is larger than va
//@ loop_invariant height == 0 ==> \forall int j; 0 <= j < x; input[j] >= input[x]; // If height is 0, everything to the left is smaller than value at current position
//@ loop_decreases height;
while (height > 0 && input[stack[height-1]] >= input[x]) {
height--;
}

if (height == 0) {
left[x] = -1;
} else {
left[x] = stack[height-1];
}
stack[height++] = x;
}
return left;
}
}

``````