Basic_CSI_RE_Test.asm

; pertinent MASM disassembly excerpt
loc_417925:				; CODE XREF: sub_417900:loc_417A5Dj
		cmp	[ebp+var_18], 1
		jnz	loc_417A62
		push	offset aPleaseEnterAPa ; "Please Enter a password "
		call	sub_4114AB
		add	esp, 4
		push	offset off_428B40
		push	7
		lea	eax, [ebp+var_C]
		push	eax
		call	sub_41104B
		add	esp, 0Ch
		movsx	eax, byte ptr [ebp+var_C]
		add	eax, 79h
		cdq
		sub	eax, edx
		sar	eax, 1
		mov	byte ptr [ebp+var_C], al
		movsx	eax, byte ptr [ebp+var_C+1]
		movsx	ecx, byte ptr [ebp+var_C]
		add	eax, ecx
		shl	eax, 1
		cdq
		mov	ecx, 3
		idiv	ecx
		sub	eax, 1Eh
		mov	byte ptr [ebp+var_C+1],	al
		mov	byte ptr [ebp+var_C+2],	65h
		movsx	eax, byte ptr [ebp+var_C+3]
		movsx	ecx, byte ptr [ebp+var_C+1]
		lea	eax, [eax+ecx+1]
		cdq
		sub	eax, edx
		sar	eax, 1
		mov	byte ptr [ebp+var_C+3],	al
		movsx	eax, [ebp+var_8]
		movsx	ecx, byte ptr [ebp+var_C+3]
		add	eax, ecx
		movsx	edx, byte ptr [ebp+var_C]
		lea	eax, [eax+edx+2]
		cdq
		mov	ecx, 3
		idiv	ecx
		mov	[ebp+var_8], al
		movsx	eax, [ebp+var_7]
		movsx	ecx, [ebp+var_8]
		add	eax, ecx
		cdq
		sub	eax, edx
		sar	eax, 1
		mov	[ebp+var_7], al
		movsx	eax, byte ptr [ebp+var_C]
		cmp	eax, 76h
		jnz	short loc_417A0B
		movsx	eax, byte ptr [ebp+var_C+1]
		cmp	eax, 74h
		jnz	short loc_417A0B
		movsx	eax, byte ptr [ebp+var_C+2]
		cmp	eax, 65h
		jnz	short loc_417A0B
		movsx	eax, byte ptr [ebp+var_C+3]
		cmp	eax, 73h
		jnz	short loc_417A0B
		movsx	eax, [ebp+var_8]
		cmp	eax, 70h
		jnz	short loc_417A0B
		movsx	eax, [ebp+var_7]
		cmp	eax, 72h
		jnz	short loc_417A0B
		push	offset aGoodJob	; "good	job"
		call	sub_4114AB

impl.rb

# reimplementation in Ruby
s = ARGV[0].dup
s[0] = ((s[0]  + 0x79) >> 1)
s[1] = (((s[1] + s[0]) << 1)/3)-0x1e
s[2] = 0x65
s[3] =  (s[3] + s[1] + 1) >> 1
s[4] = ((s[4] + s[3] + s[0] + 2)/3)
s[5] = (s[4] + s[5]) >> 1
puts s

rev.rb

# generate the lexigraphically first solution (aprox.)
t = "vtespr"
s  = "a"*6
s[0]  = (t[0] << 1) - 0x79
s[1] = (((t[1] + 0x1e)*3)>>1)-t[0]
s[2] = 0x65
s[3] = ((t[3] << 1) - 1 - t[1])
s[4] = t[4]*3 - 2 - t[3] - t[0]
s[5] = (t[5] << 1) - t[4]
puts s
# n.b. not originally stated, but they wanted an English word as a solution