RSA剪枝攻击技术详解<\/h1>

1. 基础剪枝攻击<\/h2>

1.1 基本概念<\/h3>
剪枝攻击是一种针对RSA密码系统的攻击方法，当攻击者获得部分关于私钥的信息时，可以通过逐步排除不可能的解空间来恢复完整的私钥。<\/p>

1.2 简单剪枝案例<\/h3>

给定条件：<\/p>

p*q = n<\/li>

p^q = xor_result<\/li> <\/ul>

对于每一位p^q的结果，可能的组合有4种：<\/p>

1^0<\/li>
0^1<\/li>
1^1<\/li>

0^0<\/li> <\/ol>

1.3 搜索策略<\/h3>

从低位开始搜索<\/strong>：从最低有效位(LSB)向最高有效位(MSB)逐步构建p和q的值<\/li>

剪枝条件<\/strong>：

将p和q剩余位全部填充为1时，必须满足p*q > n<\/li>
将p和q剩余位全部填充为0时，必须满足p*q < n<\/li>
p和q的低k位乘积的低k位必须等于n的低k位<\/li> <\/ul> <\/li> <\/ol>
1.4 示例脚本框架<\/h3>
def<\/span> simple_prune<\/span>(n, xor_result): <\/span><\/span> # 初始化p和q的可能值<\/span> <\/span><\/span> p_candidates =<\/span> [0<\/span>] <\/span><\/span> q_candidates =<\/span> [0<\/span>] <\/span><\/span> <\/span><\/span> for<\/span> bit_pos in<\/span> range(xor_result.<\/span>bit_length()): <\/span><\/span> new_p =<\/span> [] <\/span><\/span> new_q =<\/span> [] <\/span><\/span> current_bit =<\/span> (xor_result >><\/span> bit_pos) &<\/span> 1<\/span> <\/span><\/span> <\/span><\/span> for<\/span> p, q in<\/span> zip(p_candidates, q_candidates): <\/span><\/span> # 尝试所有可能的组合<\/span> <\/span><\/span> for<\/span> p_bit in<\/span> [0<\/span>, 1<\/span>]: <\/span><\/span> for<\/span> q_bit in<\/span> [0<\/span>, 1<\/span>]: <\/span><\/span> if<\/span> (p_bit ^<\/span> q_bit) ==<\/span> current_bit: <\/span><\/span> temp_p =<\/span> p |<\/span> (p_bit <<<\/span> bit_pos) <\/span><\/span> temp_q =<\/span> q |<\/span> (q_bit <<<\/span> bit_pos) <\/span><\/span> <\/span><\/span> # 剪枝条件检查<\/span> <\/span><\/span> if<\/span> check_prune_conditions(temp_p, temp_q, n, bit_pos): <\/span><\/span> new_p.<\/span>append(temp_p) <\/span><\/span> new_q.<\/span>append(temp_q) <\/span><\/span> <\/span><\/span> p_candidates, q_candidates =<\/span> new_p, new_q <\/span><\/span> <\/span><\/span> return<\/span> p_candidates, q_candidates <\/span><\/span> <\/span><\/span>def<\/span> check_prune_conditions<\/span>(p, q, n, bit_pos): <\/span><\/span> # 填充剩余位为0检查<\/span> <\/span><\/span> p_min =<\/span> p <\/span><\/span> q_min =<\/span> q <\/span><\/span> if<\/span> p_min *<\/span> q_min ><\/span> n: <\/span><\/span> return<\/span> False<\/span> <\/span><\/span> <\/span><\/span> # 填充剩余位为1检查<\/span> <\/span><\/span> p_max =<\/span> p |<\/span> ((1<\/span> <<<\/span> (n.<\/span>bit_length() -<\/span> bit_pos)) -<\/span> 1<\/span>) <\/span><\/span> q_max =<\/span> q |<\/span> ((1<\/span> <<<\/span> (n.<\/span>bit_length() -<\/span> bit_pos)) -<\/span> 1<\/span>) <\/span><\/span> if<\/span> p_max *<\/span> q_max <<\/span> n: <\/span><\/span> return<\/span> False<\/span> <\/span><\/span> <\/span><\/span> # 检查低k位乘积<\/span> <\/span><\/span> mask =<\/span> (1<\/span> <<<\/span> (bit_pos +<\/span> 1<\/span>)) -<\/span> 1<\/span> <\/span><\/span> if<\/span> (p *<\/span> q) &<\/span> mask !=<\/span> n &<\/span> mask: <\/span><\/span> return<\/span> False<\/span> <\/span><\/span> <\/span><\/span> return<\/span> True<\/span> <\/span><\/span><\/code><\/pre>2. 首尾剪枝攻击<\/h2> 2.1 基本概念<\/h3> 当给定p^q_rev（p与q的反方向二进制异或值）和n=p*q时，可以从两端向中间同时搜索。<\/p> 2.2 搜索策略<\/h3> 从两端向中间搜索<\/strong>：同时考虑最高位和最低位<\/li> 当前位分析<\/strong>：如果当前最高位为"1"： p该位为1，q对应低位为0<\/li> p该位为0，q对应低位为1<\/li> <\/ul> <\/li> 如果当前最高位为"0"： p该位为0，q对应低位为0<\/li> p该位为1，q对应低位为1<\/li> <\/ul> <\/li> <\/ul> <\/li> <\/ol> 2.3 剪枝条件<\/h3> 将p、q未搜索到的位全填0，乘积应小于n<\/li> 将p、q未搜索到的位全填1，乘积应大于n<\/li> p、q低k位乘积的低k位应与n的低k位相同<\/li> <\/ol> 2.4 示例脚本框架<\/h3> def<\/span> bidirectional_prune<\/span>(n, xor_rev_result): <\/span><\/span> bit_length =<\/span> n.<\/span>bit_length() <\/span><\/span> # 初始化候选对，包含高低位<\/span> <\/span><\/span> candidates =<\/span> [{'high_p'<\/span>: 0<\/span>, 'high_q'<\/span>: 0<\/span>, 'low_p'<\/span>: 0<\/span>, 'low_q'<\/span>: 0<\/span>, 'mask'<\/span>: 0<\/span>}] <\/span><\/span> <\/span><\/span> for<\/span> step in<\/span> range((bit_length +<\/span> 1<\/span>) \/\/<\/span> 2<\/span>): <\/span><\/span> new_candidates =<\/span> [] <\/span><\/span> current_high_bit =<\/span> (xor_rev_result >><\/span> (bit_length -<\/span> 1<\/span> -<\/span> step)) &<\/span> 1<\/span> <\/span><\/span> current_low_bit =<\/span> (xor_rev_result >><\/span> step) &<\/span> 1<\/span> <\/span><\/span> <\/span><\/span> for<\/span> cand in<\/span> candidates: <\/span><\/span> # 处理高位<\/span> <\/span><\/span> high_options =<\/span> [] <\/span><\/span> if<\/span> current_high_bit ==<\/span> 1<\/span>: <\/span><\/span> high_options.<\/span>extend([(1<\/span>, 0<\/span>), (0<\/span>, 1<\/span>)]) <\/span><\/span> else<\/span>: <\/span><\/span> high_options.<\/span>extend([(0<\/span>, 0<\/span>), (1<\/span>, 1<\/span>)]) <\/span><\/span> <\/span><\/span> # 处理低位<\/span> <\/span><\/span> low_options =<\/span> [] <\/span><\/span> if<\/span> current_low_bit ==<\/span> 1<\/span>: <\/span><\/span> low_options.<\/span>extend([(1<\/span>, 0<\/span>), (0<\/span>, 1<\/span>)]) <\/span><\/span> else<\/span>: <\/span><\/span> low_options.<\/span>extend([(0<\/span>, 0<\/span>), (1<\/span>, 1<\/span>)]) <\/span><\/span> <\/span><\/span> # 组合所有可能性<\/span> <\/span><\/span> for<\/span> (hp, hq) in<\/span> high_options: <\/span><\/span> for<\/span> (lp, lq) in<\/span> low_options: <\/span><\/span> new_high_p =<\/span> cand['high_p'<\/span>] |<\/span> (hp <<<\/span> (bit_length -<\/span> 1<\/span> -<\/span> step)) <\/span><\/span> new_high_q =<\/span> cand['high_q'<\/span>] |<\/span> (hq <<<\/span> step) <\/span><\/span> new_low_p =<\/span> cand['low_p'<\/span>] |<\/span> (lp <<<\/span> step) <\/span><\/span> new_low_q =<\/span> cand['low_q'<\/span>] |<\/span> (lq <<<\/span> (bit_length -<\/span> 1<\/span> -<\/span> step)) <\/span><\/span> new_mask =<\/span> cand['mask'<\/span>] |<\/span> (1<\/span> <<<\/span> step) |<\/span> (1<\/span> <<<\/span> (bit_length -<\/span> 1<\/span> -<\/span> step)) <\/span><\/span> <\/span><\/span> # 构建完整的p和q候选<\/span> <\/span><\/span> temp_p =<\/span> new_high_p |<\/span> new_low_p <\/span><\/span> temp_q =<\/span> new_high_q |<\/span> new_low_q <\/span><\/span> <\/span><\/span> # 剪枝检查<\/span> <\/span><\/span> if<\/span> check_bidirectional_conditions(temp_p, temp_q, n, new_mask, bit_length): <\/span><\/span> new_candidates.<\/span>append({ <\/span><\/span> 'high_p'<\/span>: new_high_p, <\/span><\/span> 'high_q'<\/span>: new_high_q, <\/span><\/span> 'low_p'<\/span>: new_low_p, <\/span><\/span> 'low_q'<\/span>: new_low_q, <\/span><\/span> 'mask'<\/span>: new_mask <\/span><\/span> }) <\/span><\/span> <\/span><\/span> candidates =<\/span> new_candidates <\/span><\/span> <\/span><\/span> return<\/span> candidates <\/span><\/span> <\/span><\/span>def<\/span> check_bidirectional_conditions<\/span>(p, q, n, mask, bit_length): <\/span><\/span> # 计算未确定的位数<\/span> <\/span><\/span> unknown_mask =<\/span> ((1<\/span> <<<\/span> bit_length) -<\/span> 1<\/span>) ^<\/span> mask <\/span><\/span> <\/span><\/span> # 填充0检查<\/span> <\/span><\/span> p_min =<\/span> p <\/span><\/span> q_min =<\/span> q <\/span><\/span> if<\/span> p_min *<\/span> q_min ><\/span> n: <\/span><\/span> return<\/span> False<\/span> <\/span><\/span> <\/span><\/span> # 填充1检查<\/span> <\/span><\/span> p_max =<\/span> p |<\/span> unknown_mask <\/span><\/span> q_max =<\/span> q |<\/span> unknown_mask <\/span><\/span> if<\/span> p_max *<\/span> q_max <<\/span> n: <\/span><\/span> return<\/span> False<\/span> <\/span><\/span> <\/span><\/span> # 检查已知位乘积<\/span> <\/span><\/span> known_mask =<\/span> mask |<\/span> ((1<\/span> <<<\/span> (bit_length \/\/<\/span> 2<\/span> +<\/span> 1<\/span>)) -<\/span> 1<\/span>) <\/span><\/span> if<\/span> (p *<\/span> q) &<\/span> known_mask !=<\/span> n &<\/span> known_mask: <\/span><\/span> return<\/span> False<\/span> <\/span><\/span> <\/span><\/span> return<\/span> True<\/span> <\/span><\/span><\/code><\/pre>3. p^(q>>kbits)攻击<\/h2> 3.1 思路一：高位向低位搜索<\/h3> 条件：<\/h4> n = p*q<\/li> leak = p^(q>>kbits)<\/li> <\/ul> 搜索策略：<\/h4> p的高kbits位可以从leak的高kbits位直接获得<\/li> 从第kbits位后开始搜索：如果leak的当前位为1： p为1，q为0<\/li> p为0，q为1<\/li> <\/ul> <\/li> 如果leak的当前位为0： p为1，q为1<\/li> p为0，q为0<\/li> <\/ul> <\/li> <\/ul> <\/li> <\/ol> 剪枝条件：<\/h4> 将p和q剩余位全填1，必须满足p*q > n<\/li> 将p和q剩余位全填0，必须满足p*q < n<\/li> 结束条件：n mod p == 0<\/li> <\/ol> 示例脚本框架：<\/h4> def<\/span> high_to_low_prune<\/span>(n, leak, kbits): <\/span><\/span> # 提取p的高kbits位<\/span> <\/span><\/span> p_high =<\/span> leak >><\/span> (leak.<\/span>bit_length() -<\/span> kbits) <\/span><\/span> p_high <<=<\/span> (n.<\/span>bit_length() -<\/span> kbits) <\/span><\/span> <\/span><\/span> candidates =<\/span> [p_high] <\/span><\/span> <\/span><\/span> for<\/span> bit_pos in<\/span> range(n.<\/span>bit_length() -<\/span> kbits -<\/span> 1<\/span>, -<\/span>1<\/span>, -<\/span>1<\/span>): <\/span><\/span> new_candidates =<\/span> [] <\/span><\/span> current_bit =<\/span> (leak >><\/span> bit_pos) &<\/span> 1<\/span> <\/span><\/span> <\/span><\/span> for<\/span> p in<\/span> candidates: <\/span><\/span> # 尝试两种可能性<\/span> <\/span><\/span> for<\/span> p_bit in<\/span> [0<\/span>, 1<\/span>]: <\/span><\/span> temp_p =<\/span> p |<\/span> (p_bit <<<\/span> bit_pos) <\/span><\/span> <\/span><\/span> # 推导q的对应位<\/span> <\/span><\/span> q_bit =<\/span> (p_bit ^<\/span> current_bit) <\/span><\/span> <\/span><\/span> # 剪枝检查<\/span> <\/span><\/span> if<\/span> check_high_low_conditions(temp_p, n, bit_pos): <\/span><\/span> new_candidates.<\/span>append(temp_p) <\/span><\/span> <\/span><\/span> candidates =<\/span> new_candidates <\/span><\/span> <\/span><\/span> # 检查有效解<\/span> <\/span><\/span> for<\/span> p in<\/span> candidates: <\/span><\/span> if<\/span> n %<\/span> p ==<\/span> 0<\/span>: <\/span><\/span> return<\/span> p, n \/\/<\/span> p <\/span><\/span> <\/span><\/span> return<\/span> None<\/span>, None<\/span> <\/span><\/span> <\/span><\/span>def<\/span> check_high_low_conditions<\/span>(p, n, bit_pos): <\/span><\/span> # 填充剩余位为0<\/span> <\/span><\/span> p_min =<\/span> p <\/span><\/span> q_min =<\/span> 0<\/span> <\/span><\/span> if<\/span> p_min *<\/span> q_min ><\/span> n: <\/span><\/span> return<\/span> False<\/span> <\/span><\/span> <\/span><\/span> # 填充剩余位为1<\/span> <\/span><\/span> p_max =<\/span> p |<\/span> ((1<\/span> <<<\/span> bit_pos) -<\/span> 1<\/span>) <\/span><\/span> q_max =<\/span> ((1<\/span> <<<\/span> (n.<\/span>bit_length() -<\/span> p.<\/span>bit_length())) -<\/span> 1<\/span>) <\/span><\/span> if<\/span> p_max *<\/span> q_max <<\/span> n: <\/span><\/span> return<\/span> False<\/span> <\/span><\/span> <\/span><\/span> return<\/span> True<\/span> <\/span><\/span><\/code><\/pre>3.2 思路二：迭代恢复法<\/h3> 算法步骤：<\/h4> 从leak的高kbits位得到p的高kbits位<\/li> 用n除以p的高kbits位得到q的高位估计<\/li> 利用q的高位和leak恢复p的下kbits位<\/li> 重复这个过程直到恢复完整的p和q<\/li> <\/ol> 示例脚本框架：<\/h4> def<\/span> iterative_recovery<\/span>(n, leak, kbits): <\/span><\/span> p =<\/span> leak >><\/span> (leak.<\/span>bit_length() -<\/span> kbits) <\/span><\/span> p <<=<\/span> (n.<\/span>bit_length() -<\/span> kbits) <\/span><\/span> <\/span><\/span> recovered_bits =<\/span> kbits <\/span><\/span> while<\/span> recovered_bits <<\/span> n.<\/span>bit_length(): <\/span><\/span> # 估计q的高位<\/span> <\/span><\/span> q_high =<\/span> n \/\/<\/span> (p >><\/span> (p.<\/span>bit_length() -<\/span> recovered_bits)) <\/span><\/span> <\/span><\/span> # 从leak恢复p的下kbits位<\/span> <\/span><\/span> next_p_bits =<\/span> (leak >><\/span> (recovered_bits -<\/span> kbits)) ^<\/span> (q_high >><\/span> (q_high.<\/span>bit_length() -<\/span> kbits)) <\/span><\/span> next_p_bits &=<\/span> (1<\/span> <<<\/span> kbits) -<\/span> 1<\/span> <\/span><\/span> <\/span><\/span> # 更新p<\/span> <\/span><\/span> p |=<\/span> (next_p_bits <<<\/span> (n.<\/span>bit_length() -<\/span> recovered_bits -<\/span> kbits)) <\/span><\/span> recovered_bits +=<\/span> kbits <\/span><\/span> <\/span><\/span> if<\/span> n %<\/span> p ==<\/span> 0<\/span>: <\/span><\/span> return<\/span> p, n \/\/<\/span> p <\/span><\/span> return<\/span> None<\/span>, None<\/span> <\/span><\/span><\/code><\/pre>4. nextprime剪枝攻击<\/h2> 4.1 特殊条件<\/h3> p和q的关系：q = nextprime(reverse(p))<\/li> 已知n = p*q<\/li> 可能已知q的低位信息<\/li> <\/ol> 4.2 攻击步骤<\/h3> 第一步：爆破低6位<\/h4> p*q的低6位必须等于n的低6位<\/li> 爆破p的低6位，计算对应的q的高6位<\/li> <\/ol> 第二步：爆破高6位<\/h4> 由于q = nextprime(reverse(p))，且nextprime偏移通常<2000<\/li> 从q的低6位可以推断p的高6位的逆序<\/li> 爆破依据： p高6位后填0，q高6位后填0，乘积 < n<\/li> p高6位后填9，q高6位后填9，乘积 > n<\/li> <\/ul> <\/li> <\/ol> 第三步：中间位搜索<\/h4> 从两端向中间搜索十进制位<\/li> 每次搜索10×10种可能（0-9）<\/li> 剪枝条件：未搜索位填0，乘积 < n<\/li> 未搜索位填9，乘积 > n<\/li> 低k位乘积的低k位 = n的低k位<\/li> <\/ul> <\/li> <\/ol> 4.3 示例脚本框架<\/h3> def<\/span> nextprime_prune<\/span>(n, q_low_bits, p_low_bits=<\/span>None<\/span>): <\/span><\/span> # 第一步：爆破p的低6位<\/span> <\/span><\/span> if<\/span> p_low_bits is<\/span> None<\/span>: <\/span><\/span> for<\/span> p_low in<\/span> range(10<\/span>**<\/span>6<\/span>): <\/span><\/span> # 计算q的高6位<\/span> <\/span><\/span> q_high =<\/span> n \/\/<\/span> p_low <\/span><\/span> q_high =<\/span> int(str(q_high)[:6<\/span>]) if<\/span> q_high ><\/span> 0<\/span> else<\/span> 0<\/span> <\/span><\/span> <\/span><\/span> # 检查低6位乘积<\/span> <\/span><\/span> if<\/span> (p_low *<\/span> q_high) %<\/span> 10<\/span>**<\/span>6<\/span> ==<\/span> n %<\/span> 10<\/span>**<\/span>6<\/span>: <\/span><\/span> break<\/span> <\/span><\/span> <\/span><\/span> # 第二步：爆破p的高6位<\/span> <\/span><\/span> for<\/span> offset in<\/span> range(-<\/span>2000<\/span>, 2000<\/span>): <\/span><\/span> q_low =<\/span> q_low_bits +<\/span> offset <\/span><\/span> if<\/span> q_low <<\/span> 0<\/span>: <\/span><\/span> continue<\/span> <\/span><\/span> <\/span><\/span> p_high_rev =<\/span> q_low <\/span><\/span> p_high =<\/span> int(str(p_high_rev)[::-<\/span>1<\/span>]) <\/span><\/span> <\/span><\/span> # 剪枝检查<\/span> <\/span><\/span> p_test_min =<\/span> p_high *<\/span> 10<\/span>**<\/span>(p_low.<\/span>bit_length()) <\/span><\/span> q_test_min =<\/span> q_high *<\/span> 10<\/span>**<\/span>(q_low.<\/span>bit_length()) <\/span><\/span> if<\/span> p_test_min *<\/span> q_test_min ><\/span> n: <\/span><\/span> continue<\/span> <\/span><\/span> <\/span><\/span> p_test_max =<\/span> (p_high +<\/span> 1<\/span>) *<\/span> 10<\/span>**<\/span>(p_low.<\/span>bit_length()) -<\/span> 1<\/span> <\/span><\/span> q_test_max =<\/span> (q_high +<\/span> 1<\/span>) *<\/span> 10<\/span>**<\/span>(q_low.<\/span>bit_length()) -<\/span> 1<\/span> <\/span><\/span> if<\/span> p_test_max *<\/span> q_test_max <<\/span> n: <\/span><\/span> continue<\/span> <\/span><\/span> <\/span><\/span> # 找到可能的p高6位<\/span> <\/span><\/span> break<\/span> <\/span><\/span> <\/span><\/span> # 第三步：中间位搜索<\/span> <\/span><\/span> candidates =<\/span> [{'high'<\/span>: p_high, 'low'<\/span>: p_low}] <\/span><\/span> total_digits =<\/span> len(str(n)) \/\/<\/span> 2<\/span> <\/span><\/span> known_high_digits =<\/span> len(str(p_high)) <\/span><\/span> known_low_digits =<\/span> len(str(p_low)) <\/span><\/span> <\/span><\/span> for<\/span> pos in<\/span> range(known_high_digits, total_digits -<\/span> known_low_digits): <\/span><\/span> new_candidates =<\/span> [] <\/span><\/span> <\/span><\/span> for<\/span> cand in<\/span> candidates: <\/span><\/span> for<\/span> d_high in<\/span> range(10<\/span>): # p的下一个高位数字<\/span> <\/span><\/span> for<\/span> d_low in<\/span> range(10<\/span>): # p的下一个低位数字<\/span> <\/span><\/span> new_p_high =<\/span> cand['high'<\/span>] *<\/span> 10<\/span> +<\/span> d_high <\/span><\/span> new_p_low =<\/span> cand['low'<\/span>] +<\/span> d_low *<\/span> (10<\/span> **<\/span> (pos +<\/span> known_low_digits)) <\/span><\/span> <\/span><\/span> # 计算对应的q<\/span> <\/span><\/span> p_rev =<\/span> int(str(new_p_high).<\/span>zfill(pos +<\/span> known_high_digits +<\/span> 1<\/span>)[::-<\/span>1<\/span>] +<\/span> <\/span><\/span> str(new_p_low)[known_low_digits:]) <\/span><\/span> q_cand =<\/span> nextprime(p_rev) <\/span><\/span> <\/span><\/span> # 剪枝检查<\/span> <\/span><\/span> p_min =<\/span> new_p_high *<\/span> (10<\/span> **<\/span> (total_digits -<\/span> pos -<\/span> 1<\/span>)) +<\/span> new_p_low <\/span><\/span> q_min =<\/span> q_high *<\/span> (10<\/span> **<\/span> (total_digits -<\/span> pos -<\/span> 1<\/span>)) +<\/span> q_low <\/span><\/span> if<\/span> p_min *<\/span> q_min ><\/span> n: <\/span><\/span> continue<\/span> <\/span><\/span> <\/span><\/span> p_max =<\/span> (new_p_high +<\/span> 1<\/span>) *<\/span> (10<\/span> **<\/span> (total_digits -<\/span> pos -<\/span> 1<\/span>)) -<\/span> 1<\/span> +<\/span> new_p_low <\/span><\/span> q_max =<\/span> (q_high +<\/span> 1<\/span>) *<\/span> (10<\/span> **<\/span> (total_digits -<\/span> pos -<\/span> 1<\/span>)) -<\/span> 1<\/span> +<\/span> q_low <\/span><\/span> if<\/span> p_max *<\/span> q_max <<\/span> n: <\/span><\/span> continue<\/span> <\/span><\/span> <\/span><\/span> # 检查已知位乘积<\/span> <\/span><\/span> known_mask =<\/span> 10<\/span> **<\/span> (pos +<\/span> known_low_digits +<\/span> 1<\/span>) <\/span><\/span> if<\/span> (p_min *<\/span> q_min) %<\/span> known_mask !=<\/span> n %<\/span> known_mask: <\/span><\/span> continue<\/span> <\/span><\/span> <\/span><\/span> new_candidates.<\/span>append({'high'<\/span>: new_p_high, 'low'<\/span>: new_p_low}) <\/span><\/span> <\/span><\/span> candidates =<\/span> new_candidates <\/span><\/span> <\/span><\/span> # 验证候选<\/span> <\/span><\/span> for<\/span> cand in<\/span> candidates: <\/span><\/span> p =<\/span> cand['high'<\/span>] *<\/span> (10<\/span> **<\/span> (total_digits -<\/span> known_high_digits)) +<\/span> cand['low'<\/span>] <\/span><\/span> if<\/span> n %<\/span> p ==<\/span> 0<\/span>: <\/span><\/span> return<\/span> p, n \/\/<\/span> p <\/span><\/span> <\/span><\/span> return<\/span> None<\/span>, None<\/span> <\/span><\/span><\/code><\/pre>5. 实际应用案例<\/h2> 5.1 帕鲁杯"江枫渔火对愁眠"<\/h3> 问题特点<\/strong>：<\/p> 混淆了按位或(|)和按位异或(^)操作<\/li> 需要正确识别leak1^leak2=p^q<\/li> <\/ul> 解决方案<\/strong>：<\/p> 确认实际是异或操作<\/li> 应用常规剪枝方法<\/li> <\/ol> 5.2 DASCTF 2023 ezRSA<\/h3> 问题特点<\/strong>：<\/p> 求出的n结尾是2且长度不足<\/li> 实际n应该是计算结果加上N<\/li> <\/ul> 解决方案<\/strong>：<\/p> 识别n可能被模减过<\/li> 恢复完整n = 计算结果 + N<\/li> 使用Franklin-Reiter攻击解flag<\/li> <\/ol> 6. 总结与最佳实践<\/h2> 正确识别操作<\/strong>：确认给定的信息是异或(^)还是或(|)操作<\/li> 搜索方向选择<\/strong>：从低位向高位：适合简单剪枝<\/li> 从两端向中间：适合首尾剪枝<\/li> 从高位向低位：适合p^(q>>kbits)情况<\/li> <\/ul> <\/li> 剪枝条件<\/strong>：始终检查填充0和填充1的边界条件<\/li> 验证已知位的乘积匹配<\/li> <\/ul> <\/li> 特殊关系处理<\/strong>： nextprime关系需要十进制处理<\/li> 逆序关系需要调整搜索策略<\/li> <\/ul> <\/li> 调试技巧<\/strong>：记录搜索过程中的候选数量<\/li> 设置合理的递归深度限制<\/li> 对中间结果进行验证<\/li> <\/ul> <\/li> <\/ol> 通过掌握这些剪枝技术，可以有效地解决多种RSA变体问题，特别是在CTF竞赛中遇到的部分密钥泄露场景。<\/p>

RSA剪枝攻击技术详解<\/h1>

1. 基础剪枝攻击<\/h2>

1.1 基本概念<\/h3> 剪枝攻击是一种针对RSA密码系统的攻击方法，当攻击者获得部分关于私钥的信息时，可以通过逐步排除不可能的解空间来恢复完整的私钥。<\/p>

2. 首尾剪枝攻击<\/h2>

2.1 基本概念<\/h3> 当给定p^q_rev（p与q的反方向二进制异或值）和n=p*q时，可以从两端向中间同时搜索。<\/p>

3. p^(q>>kbits)攻击<\/h2>

3.1 思路一：高位向低位搜索<\/h3>

3.2 思路二：迭代恢复法<\/h3>

4. nextprime剪枝攻击<\/h2>

4.2 攻击步骤<\/h3>

5. 实际应用案例<\/h2>

1.1 基本概念<\/h3>
剪枝攻击是一种针对RSA密码系统的攻击方法，当攻击者获得部分关于私钥的信息时，可以通过逐步排除不可能的解空间来恢复完整的私钥。<\/p>

2.1 基本概念<\/h3>
当给定p^q_rev（p与q的反方向二进制异或值）和n=p*q时，可以从两端向中间同时搜索。<\/p>