There is a line in the code that is basically 7 - X where X is the index. So The Descending order of my nodes was. 5 3 6 4 2 1. So what I had to ...
Since we control the order of the indexes with our stdin , we need to put the nodes in descending order to beat this level, based on their value, not index.
7-X, where X is each of the six node values, come in descending order.
The user must enter both an index * into the table and the sum accumulated by
I managed to see where i was wrong, on the line 8048e46: b9 07 00 00 00 mov $0x7,%ecx can be written as f(x)=7-x where x is an index.
void phase_3(char *input) { #if defined(PROBLEM) int index, val, x = 0;
Phase 6 is harder then the previous phases.
phase_[1-6] take exactly one parameter, the pointer(start address) of the input. string, let's call it ...
Phase 4 - The snippet Phase_4 While glancing through the dump we observe few
Their is a line in the code that is basically 7 - X where X is the index. So The Descending order of my nodes was.