From 175d1af3bf850fd0816a730215e028045d82e037 Mon Sep 17 00:00:00 2001
From: Malfurious <m@lfurio.us>
Date: Sun, 25 Feb 2024 12:53:04 -0500
Subject: Writeup BraekerCTF 2024 / e

Signed-off-by: Malfurious <m@lfurio.us>
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 docs/writeups/2024/BraekerCTF/misc/e.txt | 103 +++++++++++++++++++++++++++++++
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+"Grrrrr".  This robot just growls.  The other bots tell you that it is angry
+because it can't count very high.  Can you teach it how?
+
+
+
+Overview
+--------
+The challenge provides a C++ source file and a netcat port.  The program code
+appears to be a short series of prompts, which passing all of them will print
+the flag.  At each stage, our input is taken as a single-precision floating
+point number which must pass various rounding error and precision checks.
+
+
+
+Level 1
+-------
+bool flow_start() {
+    // Get user input
+    float a = get_user_input("Number that is equal to two: ");
+
+    // Can't be two
+    if (a <= 2)
+            return false;
+
+    // Check if equal to 2
+    return (unsigned short)a == 2;
+}
+
+I saw some solutions take advantage of the fact that large numbers would
+"overflow" when truncated via the (unsigned short) cast, giving a valid input
+like 65538 (0x10002).
+
+My solution leveraged the floating point truncation (aka: loss of decimal
+places): 2.0000002384
+
+
+
+Level 2
+-------
+bool round_2() {
+    float total = 0;
+
+    // Sum these numbers to 0.9
+    for (int i = 0; i < 9; i++)
+            total += 0.1;
+
+    // Add user input
+    total += get_user_input("Number to add to 0.9 to make 1: ");
+
+    // Check if equal to one
+    return total == 1.0;
+}
+
+During the for-loop, precision errors accumulate and the total will overshoot
+0.9.  Less than 0.1 must be given: 0.09999990
+
+
+
+Level 3
+-------
+bool level_3() {
+    float total = 0;
+
+    unsigned int *seed;
+    vector<float> n_arr;
+
+    // Random seed
+    seed = (unsigned int *)getauxval(AT_RANDOM);
+    srand(*seed);
+
+    // Add user input
+    add_user_input(&n_arr, "Number to add to array to equal zero: ");
+
+    // Add many random integers
+    for (int i = 0; i < 1024 * (8 + rand() % 1024); i++)
+            n_arr.push_back((rand() % 1024) + 1);
+
+    // Add user input
+    add_user_input(&n_arr, "Number to add to array to equal zero: ");
+
+    // Get sum
+    for (int i = 0; i < n_arr.size(); i++)
+            total += n_arr[i];
+
+    // Check if equal to zero
+    return total == 0;
+}
+
+Many random numbers between [1, 1024] are summed up in this function, and we are
+asked for two more, for all of them to sum to zero.  Since the range of random
+numbers is known (<=1024), we can provide relatively large numbers to squeeze
+out the randomness.
+
+Given the fixed-precision, yet floating decimal point of floats, adding a large
+value to a small value can potentially reduce the smaller value to zero as its
+exponent and mantissa are adjusted to match the other.
+
+10000000000000000
+-10000000000000000
+
+
+
+brck{Th3_3pS1l0n_w0rkS_In_M15t3riOuS_W4yS}
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