encryption - 校验和的可视化表示算法(例如 SHA)

标签 encryption cryptography sha ascii-art

我想生成类似 VisualHostKey 的东西对于 SHA 校验和。但它应该适用于任何十六进制校验和。

生成的工件可以是 ASCII 艺术、2D 调色板,或者只是 PNG 中的一些随机垃圾。我个人喜欢 VisualHostKey 方法,但我愿意接受建议。

这个想法是能够仅使用人眼快速识别两个校验和是相同的。当面对一堆金额时,请快速找到您要找的金额。

最佳答案

您可以使用实际的 OpenSSH VisualHostKey 代码,该代码位于 key_fingerprint_randomart() 中。 key.c 中的函数OpenSSH source code 中的文件.该算法相当简单,可以将任何字节数组作为输入。在 OpenSSH 中,输入是 key 的加密散列;你也可以这样做。

(如 OpenSSH 源代码中所定义,该函数还需要一个指向 key 结构本身的指针,但这仅用于在图片顶部打印 key 的类型和大小。)

事实上,由于代码是免费许可的,让我在这里只包含一个副本。这是从 OpenSSH 6.1 中提取的,$OpenBSD: key.c,v 1.99 2012/05/23 03:28:28 djm Exp $ :

/*
 * Copyright (c) 2000, 2001 Markus Friedl.  All rights reserved.
 * Copyright (c) 2008 Alexander von Gernler.  All rights reserved.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions
 * are met:
 * 1. Redistributions of source code must retain the above copyright
 *    notice, this list of conditions and the following disclaimer.
 * 2. Redistributions in binary form must reproduce the above copyright
 *    notice, this list of conditions and the following disclaimer in the
 *    documentation and/or other materials provided with the distribution.
 *
 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND ANY EXPRESS OR
 * IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES
 * OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED.
 * IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY DIRECT, INDIRECT,
 * INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
 * NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
 * DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
 * THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
 * (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF
 * THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
 */

/*
 * Draw an ASCII-Art representing the fingerprint so human brain can
 * profit from its built-in pattern recognition ability.
 * This technique is called "random art" and can be found in some
 * scientific publications like this original paper:
 *
 * "Hash Visualization: a New Technique to improve Real-World Security",
 * Perrig A. and Song D., 1999, International Workshop on Cryptographic
 * Techniques and E-Commerce (CrypTEC '99)
 * sparrow.ece.cmu.edu/~adrian/projects/validation/validation.pdf
 *
 * The subject came up in a talk by Dan Kaminsky, too.
 *
 * If you see the picture is different, the key is different.
 * If the picture looks the same, you still know nothing.
 *
 * The algorithm used here is a worm crawling over a discrete plane,
 * leaving a trace (augmenting the field) everywhere it goes.
 * Movement is taken from dgst_raw 2bit-wise.  Bumping into walls
 * makes the respective movement vector be ignored for this turn.
 * Graphs are not unambiguous, because circles in graphs can be
 * walked in either direction.
 */

/*
 * Field sizes for the random art.  Have to be odd, so the starting point
 * can be in the exact middle of the picture, and FLDBASE should be >=8 .
 * Else pictures would be too dense, and drawing the frame would
 * fail, too, because the key type would not fit in anymore.
 */
#define FLDBASE         8
#define FLDSIZE_Y       (FLDBASE + 1)
#define FLDSIZE_X       (FLDBASE * 2 + 1)
static char *
key_fingerprint_randomart(u_char *dgst_raw, u_int dgst_raw_len, const Key *k)
{
        /*
         * Chars to be used after each other every time the worm
         * intersects with itself.  Matter of taste.
         */
        char    *augmentation_string = " .o+=*BOX@%&#/^SE";
        char    *retval, *p;
        u_char   field[FLDSIZE_X][FLDSIZE_Y];
        u_int    i, b;
        int      x, y;
        size_t   len = strlen(augmentation_string) - 1;

        retval = xcalloc(1, (FLDSIZE_X + 3) * (FLDSIZE_Y + 2));

        /* initialize field */
        memset(field, 0, FLDSIZE_X * FLDSIZE_Y * sizeof(char));
        x = FLDSIZE_X / 2;
        y = FLDSIZE_Y / 2;

        /* process raw key */
        for (i = 0; i < dgst_raw_len; i++) {
                int input;
                /* each byte conveys four 2-bit move commands */
                input = dgst_raw[i];
                for (b = 0; b < 4; b++) {
                        /* evaluate 2 bit, rest is shifted later */
                        x += (input & 0x1) ? 1 : -1;
                        y += (input & 0x2) ? 1 : -1;

                        /* assure we are still in bounds */
                        x = MAX(x, 0);
                        y = MAX(y, 0);
                        x = MIN(x, FLDSIZE_X - 1);
                        y = MIN(y, FLDSIZE_Y - 1);

                        /* augment the field */
                        if (field[x][y] < len - 2)
                                field[x][y]++;
                        input = input >> 2;
                }
        }

        /* mark starting point and end point*/
        field[FLDSIZE_X / 2][FLDSIZE_Y / 2] = len - 1;
        field[x][y] = len;

        /* fill in retval */
        snprintf(retval, FLDSIZE_X, "+--[%4s %4u]", key_type(k), key_size(k));
        p = strchr(retval, '\0');

        /* output upper border */
        for (i = p - retval - 1; i < FLDSIZE_X; i++)
                *p++ = '-';
        *p++ = '+';
        *p++ = '\n';

        /* output content */
        for (y = 0; y < FLDSIZE_Y; y++) {
                *p++ = '|';
                for (x = 0; x < FLDSIZE_X; x++)
                        *p++ = augmentation_string[MIN(field[x][y], len)];
                *p++ = '|';
                *p++ = '\n';
        }

        /* output lower border */
        *p++ = '+';
        for (i = 0; i < FLDSIZE_X; i++)
                *p++ = '-';
        *p++ = '+';

        return retval;
}

除了 const Key *k 之外,它似乎对其余 OpenSSH 代码没有显着的依赖性。参数,仅在一行上用作 key_type() 的参数和 key_size()函数(或宏?)。非标类型 u_charu_int似乎只是 unsigned char 的别名和 unsigned int分别和 xcalloc()函数似乎只是标准 calloc() 的替代品或包装器.

关于encryption - 校验和的可视化表示算法(例如 SHA),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10940854/

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