题目
Introduction
Brainfuck is one of the most well-known esoteric programming languages. But it can be hard to understand any code longer that 5 characters. In this kata you have to solve that problem.
Description
In this kata you have to write a function which will do 3 tasks:
Optimize the given Brainfuck code.Check it for mistakes.Translate the given Brainfuck programming code into C programming code.
More formally about each of the tasks:
- Your function has to remove from the source code all useless command sequences such as: '+-', '<>', '[]'. Also it must erase all characters except +-<>,.[].
Example:
"++--+." -> "+."
"[][+++]" -> "[+++]"
"<>><" -> ""
If the source code contains unpaired braces, your function should return "Error!" string.
Your function must generate a string of the C programming code as follows:
- Sequences of the X commands + or - must be replaced by *p += X;\n or *p -= X;\n.
Example:
"++++++++++" -> "*p += 10;\n"
"------" -> "*p -= 6;\n"
- Sequences of the Y commands > or < must be replaced by p += Y;\n or p -= Y;\n.
Example:
">>>>>>>>>>" -> "p += 10;\n"
"<<<<<<" -> "p -= 6;\n"
- . command must be replaced by putchar(*p);\n.
Example:
".." -> "putchar(*p);\nputchar(*p);\n"
- , command must be replaced by *p = getchar();\n.
Example:
"," -> "*p = getchar();\n"
- [ command must be replaced by if (p) do {\n. ] command must be replaced by } while (p);\n.
Example:
"[>>]" ->
if (*p) do {\n
p += 2;\n
} while (*p);\n
- Each command in the code block must be shifted 2 spaces to the right accordingly to the previous code block.
Example:
"[>>[<<]]" ->
if (*p) do {\n
p += 2;\n
if (*p) do {\n
p -= 2;\n
} while (*p);\n
} while (*p);\n
Examples
Input:
+++++[>++++.<-]
Output:
*p += 5;
if (*p) do {
p += 1;
*p += 4;
putchar(*p);
p -= 1;
*p -= 1;
} while (*p);
Sample Tests
def testing(code, expected):
result = brainfuck_to_c(code)
test.assert_equals(result, expected)
test.describe("general tests")
test.it("basic")
testing("++++", "*p += 4;\n")
testing("----", "*p -= 4;\n")
testing(">>>>", "p += 4;\n");
testing("<<<<", "p -= 4;\n");
testing(".", "putchar(*p);\n");
testing(",", "*p = getchar();\n");
testing("[[[]]", "Error!");
testing("[][]", "");
testing("[.]", "if (*p) do {\n putchar(*p);\n} while (*p);\n");
testing("[]][", "Error!");
testing("++ ++", "*p += 4;\n");
testing("> <<", "p -= 1;\n");
testing("[[.]]", "if (*p) do {\n if (*p) do {\n putchar(*p);\n } while (*p);\n} while (*p);\n");
我的解法
最初版：
#!/usr/bin/python
def brainfuck_sum(l):
r = ''
if l['key'] == '+':
r = '*p += ' + str(l['value']) + ';\n'
if l['key'] == '-':
r = '*p -= ' + str(l['value']) + ';\n'
if l['key'] == '<':
r = 'p -= ' + str(l['value']) + ';\n'
if l['key'] == '>':
r = 'p += ' + str(l['value']) + ';\n'
return r
def brainfuck_add(i):
r = ''
if i == ',':
r = "*p = getchar();\n"
if i == '.':
r = "putchar(*p);\n"
if i == '[':
r = "if (*p) do {\n"
if i == ']':
r = "} while (*p);\n"
return r
def brainfuck_retract(i, p):
r = ''
p_count_l = p[0:i].count('[')
p_count_r = p[0:i+1].count(']')
p_count = p_count_l - p_count_r
for t in range(p_count):
r += ' '
return r
def brainfuck_to_c(source_code):
p = source_code
for i in p:
if i not in ['+', '-', '<', '>', '.', ',', '[', ']']:
p = p.replace(i, '')
while '+-' in p or '-+' in p or '<>' in p or '><' in p or '[]' in p:
p = p.replace('+-', '').replace('-+', '').replace('<>', '').replace('><', '').replace('[]', '')
if '[' in p or ']' in p:
if p.count('[') != p.count(']') or p.index('[') > p.index(']'):
return 'Error!'
r = ''
l = {'key': 'p', 'value': 0}
for i in range(len(p)):
r += brainfuck_retract(i, p)
r += brainfuck_add(p[i])
if p[i] == '<' or '>' or '+' or '-':
if p[i] == l['key']:
l['value'] += 1
else:
r += brainfuck_sum(l)
l['key'] = p[i]
l['value'] = 1
else:
r += brainfuck_sum(l)
l['key'] = ''
l['value'] = 0
if l['value'] != 0:
r += brainfuck_sum(l)
return r
部分优化版本：
解法一代码太过于冗长，做边界测试的时候直接内存过载，所以要优化代码。
简单看了一下，决定要把while 循环的部分拿掉，优化思路，看看代码执行情况。
#!/usr/bin/python
def brainfuck_sum(l):
r = ''
if l['key'] == '+-':
if l['value'] > 0:
r = '*p += ' + str(l['value']) + ';\n'
elif l['value'] < 0:
r = '*p -= ' + str(-l['value']) + ';\n'
if l['key'] == '<>':
if l['value'] < 0:
r = 'p -= ' + str(-l['value']) + ';\n'
elif l['value'] > 0:
r = 'p += ' + str(l['value']) + ';\n'
return r
def brainfuck_add(i):
r = ''
if i == ',':
r = "*p = getchar();\n"
if i == '.':
r = "putchar(*p);\n"
if i == '[':
r = "if (*p) do {\n"
if i == ']':
r = "} while (*p);\n"
return r
def brainfuck_retract(i, p):
r = ''
p_count_l = p[0:i].count('[')
p_count_r = p[0:i+1].count(']')
p_count = p_count_l - p_count_r
for t in range(p_count):
r += ' '
return r
def brainfuck_to_c(source_code):
p = source_code
print(p)
for i in p:
if i not in ['+', '-', '<', '>', '.', ',', '[', ']']:
p = p.replace(i, '')
if '[]' in p:
p = p.replace('[]', '')
if '[' in p or ']' in p:
if p.count('[') != p.count(']') or p.index('[') > p.index(']'):
return 'Error!'
r = ''
l = {'key': '', 'value': 0}
for i in range(len(p)):
if p[i] in '+-':
if l['key'] != '+-':
r += brainfuck_sum(l)
l['key'] = '+-'
l['value'] = 0
l['value'] += (1 if (p[i] == '+') else -1)
elif p[i] in '<>':
if l['key'] != '<>':
r += brainfuck_sum(l)
l['key'] = '<>'
l['value'] = 0
l['value'] += (-1 if (p[i] == '<') else 1)
elif p[i] in ['.', ',', '[', ']']:
r += brainfuck_retract(i, p)
r += brainfuck_add(p[i])
if l['value'] != 0:
r += brainfuck_sum(l)
print(r)
return r
目前最终版：
解法二解决了内存过载的问题，但在边界测试的时候，仍然会遇到Max Buffer Size Reached (1.5 MiB)，这个说明代码尽管优化了一部分，但还需要继续优化。但仔细思考了一下，还是把while循环的部分添加上，只是作为单列的代码（这一部分是我觉得题目是有争议的，我没想清楚，思考部分在最后）。最终形成了以下解法：
#!/usr/bin/python
def brainfuck_sum(l):
res = ''
if l['key'] == '+-':
if l['value'] > 0:
res = '*p += ' + str(l['value']) + ';\n'
elif l['value'] < 0:
res = '*p -= ' + str(-l['value']) + ';\n'
elif l['key'] == '<>':
if l['value'] < 0:
res = 'p -= ' + str(-l['value']) + ';\n'
elif l['value'] > 0:
res = 'p += ' + str(l['value']) + ';\n'
else:
pass
return res
def brainfuck_add(i):
r = ''
if i == ',':
r = "*p = getchar();\n"
elif i == '.':
r = "putchar(*p);\n"
elif i == '[':
r = "if (*p) do {\n"
elif i == ']':
r = "} while (*p);\n"
else:
pass
return r
def brainfuck_retract(i, p):
r = ''
p_count_l = p[0:i].count('[')
p_count_r = p[0:i+1].count(']')
p_count = p_count_l - p_count_r
for t in range(p_count):
r += ' '
return r
def brainfuck_replace(p):
while '+-' in p or '-+' in p or '<>' in p or '><' in p or '[]' in p:
p = p.replace('+-', '').replace('-+', '').replace('<>', '').replace('><', '').replace('[]', '')
return p
def brainfuck_to_c(source_code):
p = source_code
print(p)
for i in p:
if i not in ['+', '-', '<', '>', '.', ',', '[', ']']:
p = p.replace(i, '')
p = brainfuck_replace(p)
if '[' in p or ']' in p:
if p.count('[') != p.count(']') or p.index('[') > p.index(']'):
return 'Error!'
if p == '':
return ''
r = ''
l = {'key': '', 'value': 0}
for i in range(len(p)):
if p[i] in '+-':
if l['key'] != '+-':
r += brainfuck_sum(l)
l['key'] = '+-'
l['value'] = 0
l['value'] += (1 if (p[i] == '+') else -1)
elif p[i] in '<>':
if l['key'] != '<>':
r += brainfuck_sum(l)
l['key'] = '<>'
l['value'] = 0
l['value'] += (-1 if (p[i] == '<') else 1)
elif p[i] in ['.', ',', '[', ']']:
r += brainfuck_retract(i, p)
r += brainfuck_add(p[i])
if l['value'] != 0:
r += brainfuck_sum(l)
return r
思考：
我最终把while循环部分添加上了，主要是这里面我觉得题目是有争议的。
例如题目要求："Your function has to remove from the source code all useless command sequences such as: '+-', '<>', '[]'. Also it must erase all characters except +-<>,.[]."
Is this rule recursive?
For example:
"<[+-]>" -> "<[]>"
"<[+-]>" -> ""
Which one is right?
(I guess the second is right, just want to be sure.)